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Santos\\ \href{mailto:dsantos@ccp.edu}{dsantos@ccp.edu}} \date{\today \quad VERSION} %%%%%%%%%%%%%%%%%%%% \makeatletter \newcommand{\twocoltoc}{% \section*{\contentsname \@mkboth{% }{}}% \begin{multicols}{2} \@starttoc{toc}% \end{multicols}} \makeatother %%%%%%%%%%%% \makeindex \usepackage{polynom, maplestd2e} \setlength{\fboxrule}{2pt} \begin{document} \pagenumbering{roman} \pagestyle{fancy} % Front matter may follow here \begin{frontmatter} \maketitle \clearpage \begin{quote} Copyright \copyright{} 2007 David Anthony SANTOS. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled ``GNU Free Documentation License''. \end{quote} \clearpage \clearpage \twocoltoc{} \end{frontmatter} \clearpage \section*{Preface} \markboth{}{} \addcontentsline{toc}{chapter}{Preface} \markright{Preface} There are very few good Calculus books, written in English, available to the American reader. Only \cite{Har}, \cite{Kla}, \cite{Apo}, \cite{Olm}, and \cite{Spi} come to mind. The situation in Precalculus is even worse, perhaps because Precalculus is a peculiar American animal: it is a review course of all that which should have been learned in High School but was not. A distinctive American slang is thus called to describe the situation with available Precalculus textbooks: they stink! I have decided to write these notes with the purpose to, at least locally, for my own students, I could ameliorate this situation and provide a semi-rigorous introduction to precalculus. \bigskip I try to follow a more or less historical approach. My goal is to not only present a coherent view of Precalculus, but also to instill appreciation for some elementary results from Precalculus. Thus I do not consider a student (or for that matter, {\em an instructor}) to be educated in Precalculus if he cannot demonstrate that $\sqrt{2}$ is irrational;\footnote{Plato's dictum comes to mind: ``He does not deserve the appellative {\em man} who does not know that the diagonal of a square is inconmensurable with its side.} that the equation of a non-vertical line on the plane is of the form $y=mx+k$, and conversely; that lines $y=m_1x+k_1$ and $y=m_2x+k_2$ are perpendicular if and only if $m_1m_2=-1$; that the curve with equation $y=x^2$ is a parabola, etc. \bigskip I do not claim a 100\% rate of success, or that I stick to the same paradigms each semester,\footnote{I don't, in fact, I try to change emphases from year to year.} but a great number of students seem genuinely appreciative what I am trying to do. \bigskip I start with sets of real numbers, in particular, intervals. I try to make patent the distinction between rational and irrational numbers, and their decimal representations. Usually the students reaching this level have been told fairy tales about $\sqrt{2}$ and $\pi$ being irrational. I prove the irrationality of the former using Hipassus of Metapontum's proof.\footnote{I wonder how many of my colleagues know how to prove that $\pi$ is irrational? Transcendental? Same for $e$, $\log 2$, $\cos 1$, etc. How many tales are the students told for which the instructor does not know the proof?} \bigskip After sets on the line, I concentrate on distance on the line. Absolute values are a good place (in my opinion) to introduce sign diagrams, which are a technique that will be exploited in other instances, as for example, in solving rational and absolute-value inequalities. \bigskip The above programme is then raised to the plane. I derive the distance formula from the Pythagorean Theorem. It is crucial, in my opinion, to make the students understand that these formul\ae\ do not appear by fiat, but that are obtained from previous concepts. \bigskip Depending on my mood, I either move to the definition of functions, or I continue to various curves. Let us say for the sake of argument that I have chosen to continue with curves. \bigskip Once the distance formula is derived, it is trivial to talk about circles and semi-circles. The graph of $y=\sqrt{1-x^2}$ is obtained. This is the first instance of the translation {\em Geometry-to-Algebra} and {\em Algebra-to-Geometry} that the students see, that is, they are able to tell what the equation of a given circle looks like, and vice-versa, to produce a circle from an equation. \bigskip Now, using similar triangles and the distance formula once again, I move on to lines, proving that the canonical equation of a non-vertical line is of the form $y=mx+k$ and conversely. I also talk about parallel and normal lines, proving\footnote{The Pythagorean Theorem once again!} that two non-vertical lines are perpendicular if and only if the product of their slopes is $-1$. In particular, the graph of $y=x$, $y=-x$, and $y=|x|$ are obtained. \bigskip The next curve we study is the parabola. First, I give the locus definition of a parabola. We use a T-square and a string in order to illustrate the curve produced by the locus definition. It turns out to be a sort-of ``U''-shaped curve. Then, using the distance formula again, we prove that one special case of these parabolas has equation $y=x^2$. The graph of $x=y^2$ is obtained, and from this the graph of $y=\sqrt{x}$. \bigskip Generally, after all this I give my first exam. \bigskip We now start with functions. A {\em function} is defined by means of the following five characteristics: \begin{enumerate} \item a set of inputs, called the {\em domain} of the function; \item a set of all {\em possible} outputs, called the {\em target set} of the function; \item a name for a typical input (colloquially referred to as the {\em dummy variable}); \item a name for the function; \item an assignment rule or formula that assigns to every element of the domain a unique element of the target set. \end{enumerate} All these features are collapsed into the notation $$\fun{f}{x}{f(x)}{\dom{f}}{\target{f}}. $$ Defining functions in such a careful manner is necessary. Most American books focus only on the assignment rule (formula), but this makes a mess later on in abstract algebra, linear algebra, computer programming etc. For example, even though the following four functions have the same formula, they are all different: $$\fun{a}{x}{x^2}{\reals }{\reals }; \qquad \fun{b}{x}{x^2}{\co{0; +\infty}}{\reals };$$ $$ \fun{c}{x}{x^2}{\reals }{\co{0; +\infty}}; \qquad \fun{d}{x}{x^2}{\co{0; +\infty}}{\co{0; +\infty}}; $$for $a$ is neither injective nor surjective, $b$ is injective but not surjective, $c$ is surjective but not injective, and $d$ is a bijection. \bigskip I first focus on the domain of the function. We study which possible sets of real numbers can be allowed so that the output be a real number. \bigskip I then continue to graphs of functions and functions defined by graphs.\footnote{This last means, given a picture in $\reals ^2$ that passes the vertical line test, we derive its domain and image by looking at its shadow on the $x$ and $y$ axes.} At this point, of course, there are very functional curves of which the students know the graphs: only $x\mapsto x$, $x\mapsto |x|$, $x\mapsto x^2$, $x\mapsto \sqrt{x}$, $x\mapsto \sqrt{1-x^2}$, piecewise combinations of them, etc., but they certainly can graph a function with a finite (and extremely small domain). The repertoire is then extended by considering the following transformations of a function $f$: $x\mapsto -f(x)$, $x\mapsto f(-x)$, $x\mapsto Vf(Hx+h)+v$, $x\mapsto |f(x)|$, $x\mapsto f(|x|)$, $x\mapsto f(-|x|)$. These last two transformations lead a discussion about even and odd functions. The floor, ceiling, and the decimal part functions are also now introduced. \bigskip The focus now turns to the assignment rule of the function, and is here where the algebra of functions (sum, difference, product, quotient, composition) is presented. Students are taught the relationship between the various domains of the given functions and the domains of the new functions obtained by the operations. \bigskip Composition leads to iteration, and iteration leads to inverse functions. The student now becomes familiar with the concepts of injective, surjective, and bijective functions. The relationship between the graphs of a function and its inverse are explored. It is now time for the second exam. \bigskip The distance formula is now powerless to produce the graph of more complicated functions. The concepts of {\em monotonicity} and {\em convexity} of a function are now introduced. Power functions (with strictly positive integral exponents are now studied. The global and local behaviour of them is studied, obtaining a catalogue of curves $y = x^{n}$, $n\in\naturals$. \bigskip After studying power functions, we now study polynomials. The study is strictly limited to polynomials whose splitting field is $\reals$.\footnote{I used to make a brief incursion into some ancillary topics of the theory of equations, but this makes me digress too much from my plan of {\em Algebra-Geometry-Geometry-Algebra}, and nowadays I am avoiding it. I have heard colleagues argue for Ruffini's Theorem, solely to be used in one example of Calculus I, the factorisation of a cubic or quartic polynomial in optimisation problems, but it seems hardly worth the deviation for only such an example.} \bigskip We now study power functions whose exponent is a strictly negative integer. In particular, the graph of the curve $xy=1$ is deduced from the locus definition of the hyperbola. Studying the monotonicity and concavity of these functions, we obtain a catalogue of curves $y = x^{-n}$, $n\in\naturals$. \bigskip Rational functions are now introduced, but only those whose numerators and denominators are polynomials splitting in $\reals$. The problem of graphing them is reduced to examining the local at the zeroes and poles, and their global behaviour. \bigskip I now introduce formul\ae\ of the type $x\mapsto x^{1/n}$, $n\in \integers \setminus \{0\}$, whose graphs I derived by means of inverse functions of $x\mapsto x^{n}$, $n\in \integers$. This concludes the story of Precalculus I as I envision it, and it is time for the third exam, usually during the last week of classes. A comprehensive final exam is given during final-exam week. \bigskip These notes are in constant state of revision. I would greatly appreciate comments, additions, exercises, figures, etc., in order to help me enhance them. \bigskip \hfill David A. Santos \chapter*{To the Student} \markboth{}{} \addcontentsline{toc}{chapter}{To the Student} \markright{To the Student} These notes are provided for your benefit as an attempt to organise the salient points of the course. They are a {\em very terse} account of the main ideas of the course, and are to be used mostly to refer to central definitions and theorems. The number of examples is minimal. The {\em motivation} or informal ideas of looking at a certain topic, the ideas linking a topic with another, the worked-out examples, etc., are given in class. Hence these notes are not a substitute to lectures: {\bf you must always attend to lectures}. The order of the notes may not necessarily be the order followed in the class. \bigskip There is a certain algebraic fluency that is necessary for a course at this level. These algebraic prerequisites would be difficult to codify here, as they vary depending on class response and the topic lectured. If at any stage you stumble in Algebra, seek help! I am here to help you! \bigskip Tutoring can sometimes help, but bear in mind that whoever tutors you may not be familiar with my conventions. Again, I am here to help! On the same vein, other books may help, but the approach presented here is at times unorthodox and finding alternative sources might be difficult. \bigskip Here are more recommendations: \begin{itemize} \item Read a section before class discussion, in particular, read the definitions. \item Class provides the informal discussion, and you will profit from the comments of your classmates, as well as gain confidence by providing your insights and interpretations of a topic. {\bf Don't be absent! } \item I encourage you to form study groups and to discuss the assignments. Discuss among yourselves and help each other but don't be {\em parasites!} Plagiarising your classmates' answers will only lead you to disaster! \item Once the lecture of a particular topic has been given, take a fresh look at the notes of the lecture topic. \item Try to understand a single example well, rather than ill-digest multiple examples. \item Start working on the distributed homework ahead of time. \item {\bf Ask questions during the lecture.} There are two main types of questions that you are likely to ask. \begin{enumerate} \item {\em Questions of Correction: Is that a minus sign there?} If you think that, for example, I have missed out a minus sign or wrote $P$ where it should have been $Q$,\footnote{My doctoral adviser used to say ``I said $A$, I wrote $B$, I meant $C$ and it should have been $D$!} then by all means, ask. No one likes to carry an error till line XLV because the audience failed to point out an error on line I. Don't wait till the end of the class to point out an error. Do it when there is still time to correct it! \item {\em Questions of Understanding: I don't get it!} Admitting that you do not understand something is an act requiring utmost courage. But if you don't, it is likely that many others in the audience also don't. On the same vein, if you feel you can explain a point to an inquiring classmate, I will allow you time in the lecture to do so. The best way to ask a question is something like: ``How did you get from the second step to the third step?'' or ``What does it mean to complete the square?'' Asseverations like ``I don't understand'' do not help me answer your queries. If I consider that you are asking the same questions too many times, it may be that you need extra help, in which case we will settle what to do outside the lecture. \end{enumerate} \item Don't fall behind! The sequence of topics is closely interrelated, with one topic leading to another. \item You will need square-grid paper, a ruler (preferably a T-square), some needle thread, and a compass. \item The use of calculators is allowed, especially in the occasional lengthy calculations. However, when graphing, you will need to provide algebraic/analytic/geometric support of your arguments. The questions on assignments and exams will be posed in such a way that it will be of no advantage to have a graphing calculator. \item Presentation is critical. Clearly outline your ideas. When writing solutions, outline major steps and write in complete sentences. As a guide, you may try to emulate the style presented in the scant examples furnished in these notes. \end{itemize} \clearpage \section*{Notation}\markboth{}{} \begin{tabular}{ll} $\in$ & Belongs to. \\ $\not\in$ & Does not belong to. \\ $\forall$ & For all (Universal Quantifier). \\ $\exists$ & There exists (Existential Quantifier). \\ $\varnothing$ & Empty set. \\ $P \implies Q$ & $P$ implies $Q$. \\ $P \Leftrightarrow Q$ & $P$ if and only if $Q$. \\ $\naturals$ & The Natural Numbers $\{0, 1, 2, 3, \ldots \}$. \\ $\integers$ & The Integers $\{ \ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\}$. \\ $\rationals$ & The Rational Numbers. \\ $\reals$ & The Real Numbers. \\ ${\complex }$ & The Complex Numbers. \\ $A^n$ & The set of $n$-tuples $\{(a_1, a_2, \ldots, a_n)|a_k\in A\}$. \\ $]a; b[$ & The open finite interval $\{x \in {\reals}: a < x < b\}$. \\ $[a; b]$ & The closed interval $\{x \in {\reals}: a \leq x \leq b\}$. \\ $]a; b]$ & The semi-open interval $\{x \in {\reals}: a < x \leq b\}$. \\ $[a; b[$ & The semi-closed interval $\{x \in {\reals}: a \leq x < b\}$. \\ $]a; +\infty[$ & The infinite open interval $\{x \in {\reals}: x > a\}$. \\ $]-\infty; a]$ & The infinite closed interval $\{x \in {\reals}: x \leq a\}$. \\ $\sum _{k = 1} ^n a_k $ & The sum $ a_1 + a_2 + \cdots + a_{n - 1}+ a_n $. \end{tabular} \clearpage %%%%%%%%This one in case I decide to use again the fancy numbers on the side \renewcommand{\chaptermark}{\markboth{\chaptername\ \thechapter}} \renewcommand{\sectionmark}{\markright} \chapter{The Line} \Opensolutionfile{ans}[ansPreCalcI1] \pagenumbering{arabic} \setcounter{page}{1} This chapter introduces essential notation and terminology that will be used throughout these notes. The focus of this course will be the real numbers, of which we assume the reader has passing familiarity. We will review some of the properties of real numbers as a way of having a handy vocabulary that will be used for future reference. \section{Sets and Notation} \begin{df}We will mean \index{set} by a {\em set} a collection of well defined members or {\em elements}. A {\em subset} is a sub-collection of a set. We denote that $B$ is a subset of $A$ by the notation $B \subseteqq A$ or sometimes $B \subset A$.\footnote{There is no agreement relating the choice. Some use $\subset$ to denote strict containment, that is, $A\subseteqq B $ but $A\neq B$. In the case when we want to denote strict containment we will simply write $A\varsubsetneqq B$.}\end{df} Some sets of numbers will be referred to so often that they warrant special notation. Here are some of the most common ones. \begin{tabular}{ll} $\varnothing$ & Empty set. \\ $\naturals$ & The Natural Numbers $\{0, 1, 2, 3, \ldots \}$.\\ $\integers$ & The Integers $\{ \ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\}$.\ \\ $\rationals$ & The Rational Numbers. \\ $\reals$ & The Real Numbers. \\ $\complex $ & The Complex Numbers. \\ \end{tabular} \begin{rem} Observe that $\naturals \subseteq\integers \subseteq\rationals \subseteq\reals \subseteq \complex$. \end{rem} From time to time we will also use the following notation, borrowed from set theory and logic. \begin{tabular}{ll} $\in$ & Is in. Belongs to. Is an element of.\\ $\not\in$ & Is not in. Does not belong to. Is not an element of.\\ $\forall$ & For all (Universal Quantifier). \\ $\exists$ & There exists (Existential Quantifier). \\ $P \implies Q$ & $P$ implies $Q$. \\ $P \Leftrightarrow Q$ & $P$ if and only if $Q$. \\ \end{tabular} \begin{exa} $-1\in\integers$ but $\frac{1}{2}\not\in \integers$. \end{exa} \begin{df}Let $A$ be a set. If $a$ belongs to the set $A$, then we write $a\in A$, read ``$a$ is an element of $A$.'' If $a$ does not belong to the set $A$, we write $a\not\in A$, read ``$a$ is not an element of $A$.'' The set that has no elements, that is {\em empty set}, will be denoted by $\varnothing$. \index{set!empty} \end{df} There are various ways of alluding to a set. We may use a description, or we may list its elements individually. \begin{exa} The sets $$ A=\{x\in \integers: x^2 \leq 9\}, \qquad B=\{x\in\integers: |x|\leq 3\},\qquad C=\{-3,-2,-1,0,1,2,3\} $$are identical. The first set is the set of all integers whose square lies between $1$ and $9$ inclusive, which is precisely the second set, which again is the third set. \end{exa} \begin{exa} Consider the set $$A=\{2,9,16, \ldots , 716\}, $$where the elements are in arithmetic progression. How many elements does it have? Is $401\in A$? Is $514\in A$? What is the sum of the elements of $A$? \end{exa} \begin{solu} Observe that the elements have the form $$ 2 = 2+7\cdot 0, \quad 9 = 2+7\cdot 1, \quad 16 = 2+7\cdot 2, \quad \ldots , $$thus the general element term has the form $2+7n$. Now, $$2+7n=716 \implies n=102.$$ This means that there are $103$ elements, since we started with $n=0$. \bigskip If $2+7k=401$, then $k=57$, so $401\in A$. On the other hand, $2+7a=514 \implies a=\dfrac{512}{7}$, which is not integral, and hence $514\nin A$. \bigskip To find the sum of the arithmetic progression we will use a trick due to the great German mathematician K. F. Gau\ss\ who presumably discovered it when he was in first grade. To add the elements of $A$, put $$ S= 2 + 9 + 16 + \cdots + 716. $$Observe that the sum does not change if we sum it backwards, so $$S=716 + 709+702 + \cdots + 16 + 9 + 2. $$Adding both sums and grouping corresponding terms, $$\begin{array}{lll} 2S & = & (2+716) + (9+709) + (16+702) + \cdots + (702+16) + (709+9) + (716+2)\\ & = & 718 + 718 + 718 + \cdots + 718 + 718 + 718 \\ & = & 718\cdot 103, \end{array}$$since there are $103$ terms. We deduce that $$ S=\dfrac{718\cdot 103}{2}=36977. $$ \end{solu} \vspace{1cm} \begin{figure}[!hptb] \begin{minipage}{3cm}$$\psset{unit=1pc} \pscircle[fillstyle=hlines, fillcolor=brown](-1,0){2}\pscircle[fillstyle=hlines, fillcolor=brown](1,0){2} \uput[d](-1,-2){A}\uput[d](1,-2){B} $$\meinecaption{1}{$A\cup B$} \label{fig:a_union_b} \end{minipage} \hfill \begin{minipage}{3cm}$$ \psset{unit=1pc} \pscircle(-1,0){2}\pscircle(1,0){2} \uput[d](-1,-2){A}\uput[d](1,-2){B} \pscustom[fillstyle=solid,fillcolor=blue]{\psarc(1,0){2}{120}{240}\psarc(-1,0){2}{270}{60}} $$\meinecaption{1}{$A\cap B$} \label{fig:a_intersection_b} \end{minipage} \hfill \begin{minipage}{3cm}$$\psset{unit=1pc} \pscircle[fillstyle=solid,fillcolor=blue](-1,0){2}\pscircle(1,0){2} \uput[d](-1,-2){A}\uput[d](1,-2){B} \pscustom[fillstyle=solid,fillcolor=white]{\psarc(1,0){2}{120}{240}\psarc(-1,0){2}{270}{60}} $$\meinecaption{1}{$A\setminus B$} \label{fig:a_minus_b} \end{minipage} \hfill \end{figure} \vspace{1cm} We now define some operations with sets. \begin{df} The {\em union} of two sets $A$ and $B$, is the set $$A\cup B = \{x:(x\in A)\ \mathrm{or}\ (x\in B)\}.$$ This is read ``$A$ union $B$.'' See figure \ref{fig:a_union_b}. \bigskip The {\em intersection} of two sets $A$ and $B$, is $$A\cap B = \{x:(x\in A)\ \mathrm{and} \ (x\in B)\}.$$ This is read ``$A$ intersection $B$.'' See figure \ref{fig:a_intersection_b}. \bigskip The {\em difference} of two sets $A$ and $B$, is $$A\setminus B = \{x:(x\in A)\ \mathrm{and}\ (x\not\in B)\}.$$ This is read ``$A$ set minus $B$.'' See figure \ref{fig:a_minus_b}. \end{df} \renewcommand{\arraystretch}{2} \begin{table}[!hptb]\begin{center}\begin{tabular}{llc}{\bf Interval Notation} & {\bf Set Notation} & {\bf Graphical Representation} \\ $\cc{a; b}$ & $\{x\in \reals: a \leq x \leq b\}$\footnote{This is ``the set of all real numbers $x$ such that $a$ is less than or equal to $x$, and $x$ is less than or equal to $b$.''} & $ \psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=brown]{*-*}(-5,0)(5,0)\uput[d](-5,0){a}\uput[d](5,0){b} $ \\ $\oo{a; b}$ & $\{x\in \reals: a < x < b\}$ & $ \psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=brown]{o-o}(-5,0)(5,0)\uput[d](-5,0){a}\uput[d](5,0){b} $ \\ $\co{a; b}$ & $\{x\in \reals: a \leq x < b\}$ & $ \psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=brown]{*-o}(-5,0)(5,0)\uput[d](-5,0){a}\uput[d](5,0){b} $ \\ $\oc{a; b}$ & $\{x\in \reals: a < x \leq b\}$ & $ \psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=brown]{o-*}(-5,0)(5,0)\uput[d](-5,0){a}\uput[d](5,0){b} $ \\ $\oo{a; +\infty}$ & $\{x\in \reals: x>a\}$ & $ \psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=brown]{o->}(-5,0)(5,0)\uput[d](-5,0){a}\uput[d](5,0){+\infty} $ \\ $\co{a; +\infty}$ & $\{x\in \reals: x\geq a\}$ & $ \psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=brown]{*->}(-5,0)(5,0)\uput[d](-5,0){a}\uput[d](5,0){+\infty} $ \\ $\oo{-\infty; b}$ & $\{x\in \reals: x< b\}$ & $ \psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=brown]{<-o}(-5,0)(5,0)\uput[d](-5,0){-\infty}\uput[d](5,0){b} $ \\ $\oc{-\infty; b}$ & $\{x\in \reals: x\leq b\}$ & $ \psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=brown]{<-*}(-5,0)(5,0)\uput[d](-5,0){-\infty}\uput[d](5,0){b} $ \\ $\oo{-\infty; +\infty}$ & $\reals$ & $ \psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=brown]{<->}(-5,0)(5,0)\uput[d](-5,0){-\infty}\uput[d](5,0){+\infty} $ \\ \end{tabular}\meinecaption{.25}{Intervals.} \label{tab:intervals}\end{center}\end{table} \bigskip \begin{exa} Let $A = \{1, 2, 3, 4, 5, 6\}$, and $B = \{1, 3, 5, 7, 9\}$. Then $$ A\cup B = \{1, 2, 3, 4, 5, 6, 7, 9\},\quad A\cap B = \{1, 3, 5\}, \quad A\setminus B = \{2, 4, 6\},\quad B\setminus A = \{7, 9\}.$$\end{exa} \begin{exa} Consider the sets of arithmetic progressions $$ A=\{3,9,15, \ldots , 681\}, \qquad B=\{9, 14, 19, \ldots ,564\}. $$How many elements do they share, that is, how many elements does $A\cap B$ have? \end{exa} \begin{solu} The members of $A$ have common difference $6$ and the members of $B$ have common difference $5$. Since the least common multiple of $6$ and $5$ is $30$, and $9$ is the smallest element that $A$ and $B$ have in common, every element in $A\cap B$ has the form $9+30k$. We then need the largest $k\in\naturals$ satisfying the inequality $$ 9+30k\leq 564 \implies k \leq 18.5, $$and since $k$ is integral, the largest value it can achieve is $18$. Thus $A\cap B$ has $18+1=19$ elements, where we have added $1$ because we start with $k=0$. In fact, $$ A\cap B = \{9, 39, 69, \ldots , 549 \}. $$ \end{solu} . \begin{df} An {\em interval}\index{interval} $I$ is a subset of the real numbers with the following property: if $s\in I$ and $t\in I$, and if $s < x < t$, then $x\in I$. In other words, intervals are those subsets of real numbers with the property that every number between two elements is also contained in the set. Since there are infinitely many decimals between two different real numbers, intervals with distinct endpoints contain infinitely many members. Table \ref{tab:intervals} shews the various types of intervals. \end{df} Observe that we indicate that the endpoints are included by means of shading the dots at the endpoints and that the endpoints are excluded by not shading the dots at the endpoints. \footnote{It may seem like a silly analogy, but think that in $[a;b]$ the brackets are ``arms'' ``hugging'' $a$ and $b$, but in $]a;b[$ the ``arms'' are repulsed. ``Hugging'' is thus equivalent to including the endpoint, and ``repulsing'' is equivalent to excluding the endpoint.} \begin{exa} If $A = \cc{-10; 2}$, $B = \oo{-\infty; 1}$, then $$A \cap B = \co{-10; 1}, \quad A \cup B = \oc{-\infty; 2}, \quad A \setminus B = \cc{1; 2}, \quad B \setminus A = \oo{-\infty; -10}. $$ \end{exa} \begin{exa} Let $A = \cc{1 - \sqrt{3}; 1 + \sqrt{2}}$, $B =\co{\frac{\pi}{2}; \pi}$. By approximating the endpoints to three decimal places, we find $1 - \sqrt{3} \approx -0.732,$ $1 + \sqrt{2} \approx 2.414,$ $\frac{\pi}{2} \approx 1.571,$ $\pi \approx 3.142$. Thus $$A \cap B = \cc{\frac{\pi}{2}; 1 + \sqrt{2}}, \quad A \cup B = \co{ 1 - \sqrt{3}; \pi}, \quad A \setminus B = \co{1 - \sqrt{3}; \frac{\pi}{2}}, \quad B \setminus A = \oo{ 1 + \sqrt{2};\pi}. $$ \end{exa} We conclude this section by defining some terms for future reference. \begin{df} Let $a\in \reals $. We say that the set $\N{a}\subseteqq \reals $ is a {\em neighbourhood} of $a$ if there exists an open interval $I$ centred at $a$ such that $I \subseteqq \N{a}$. In other words, $\N{a}$ is a neighbourhood of $a$ if there exists a $\delta > 0$ such that $\oo{a-\delta; a+\delta}\subseteqq \N{a}$. This last condition may be written in the form $$ \left\{x\in \reals : |x-a|<\delta\right\}\subseteqq \N{a}. $$ If $\N{a}$ is a neighbourhood of $a$, then we say that $\N{a}\setminus \{a\}$ is a {\em deleted neighbourhood of $a$}.\index{neighbourhood}\index{neighbourhood!deleted} \end{df} This means that $\N{a}$ is a neighbourhood of $a$ if $a$ has neighbours left and right. \begin{exa} The interval $\oo{0;1}$ is neighbourhood of all of its points. The interval $\cc{0;1}$, on the contrary, is a neighbourhood of all of its points, with the exception of its endpoints $0$ and $1$, since $0$ does not have left neighbours in the interval and $1$ does not have right neighbours on the interval. \end{exa} \vspace{1cm} \begin{figure}[!hptb] \begin{minipage}{5cm} \centering \psaxes[arrows={->},linewidth=2pt,yAxis=false,labels=none, arrowscale=2.1,labelsep=-20pt](0,0)(-2.5,0)(2.7,0) \pstGeonode[PointName=none](0,0){O}(1,0){A}(-1,0){A'} \uput{13pt}[d](O){$a$} \uput{10pt}[d](A'){$a-\delta$} \uput{10pt}[d](A){$a+\delta$} \rput(A'){\psline[linewidth=2pt,linecolor=magenta](-.1,.25)(0,.25)(0,-.25)(-.1,-.25)} \rput(A){\psline[linewidth=2pt,linecolor=magenta](.1,.25)(0,.25)(0,-.25)(.1,-.25)} \psdots[dotscale=1.5,linecolor=brown](O) \meinecaption{1}{ Neighbourhood of $a$.} \end{minipage} \hfill \begin{minipage}{5cm} \centering \psaxes[arrows={->},linewidth=2pt,yAxis=false,labels=none, arrowscale=2.1,labelsep=-20pt](0,0)(-2.5,0)(2.7,0) \pstGeonode[PointName=none](0,0){O}(1,0){A}(-1,0){A'} \uput{12pt}[d](O){$a$} \uput{10pt}[d](A'){$a-\delta$} \psdots[dotscale=1.5,linecolor=brown](O) \rput(A'){\psline[linewidth=2pt,linecolor=magenta](-.1,.25)(0,.25)(0,-.25)(-.1,-.25)} \rput(O){\psline[linewidth=2pt,linecolor=magenta](-.1,.25)(0,.25)(0,-.25)(-.1,-.25)} \meinecaption{1}{Sinistral neighbourhood of $a$.} \end{minipage} \hfill \begin{minipage}{5cm} \centering \psaxes[arrows={->},linewidth=2pt,yAxis=false,labels=none, arrowscale=2.1,labelsep=-20pt](0,0)(-2.5,0)(2.7,0) \pstGeonode[PointName=none](0,0){O}(1,0){A}(-1,0){A'} \uput{12pt}[d](O){$a$} \uput{10pt}[d](A){$a+\delta$} \psdots[dotscale=1.5,linecolor=brown](O) \rput(A){\psline[linewidth=2pt,linecolor=magenta](.1,.25)(0,.25)(0,-.25)(.1,-.25)} \rput(O){\psline[linewidth=2pt,linecolor=magenta](.1,.25)(0,.25)(0,-.25)(.1,-.25)} \meinecaption{1}{Dextral neighbourhood of $a$.} \end{minipage} \end{figure} We may now extend the definition of neighbourhood. \begin{df} Let $a\in \reals $. We say that the set $V\subseteqq \reals $ is a {\em dextral neighbourhood } or {\em right-hand neighbourhood} of $a$ if there exists a $\delta > 0$ such that $\co{a; a+\delta}\subseteqq V$. We say that the set $V'\subseteqq \reals $ is a {\em sinistral neighbourhood } or {\em left-hand neighbourhood} of $a$ if there exists a $\delta ' > 0$ such that $\oc{a-\delta '; a}\subseteqq V'$. \end{df} The following result will be used later. \begin{lem} Let $(a, b)\in\reals^2, a < b$. Then every number of the form $\lambda a + (1 - \lambda)b, \ \lambda \in [0; 1]$ belongs to the interval $[a; b]$. Conversely, if $x\in [a; b]$ then we can find a $\lambda \in [0; 1]$ such that $x = \lambda a + (1 - \lambda)b.$ \label{lem:betwixt}\end{lem} \begin{pf} Clearly $\lambda a + (1 - \lambda)b = b + \lambda(a - b)$ and since $a - b < 0,$ $$ b = b + 0(a - b) \geq b + \lambda(a - b) \geq b + 1(a - b) = a ,$$whence the first assertion follows. Assume now that $x\in [a; b]$. Solve the equation $x = \lambda a + (1 - \lambda)b$ for $\lambda$ obtaining $\lambda = \frac{x - b}{b - a}$. All what remains to prove is that $0 \leq \lambda \leq 1$, but this is evident, as $ 0 \leq x - b \leq b - a.$ This concludes the proof. \end{pf} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} List all the elements of the set $$\{x\in \integers: 1 \leq x^2 \leq 100, \quad x \ \mathrm{is\ divisible\ by\ } 3 \}. $$ \begin{answer} This is the set $\{-9,-6,-3,0,3,6,9\}$. \end{answer} \end{pro} \begin{pro} Determine the set $$\{x\in \naturals: x^2-x=6 \} $$explicitly. \begin{answer}We have $$x^2-x=6\implies x^2-x-6=0 \implies (x-3)(x+2)=0\implies x\in \{-2,3\}. $$ Since $-2\nin\naturals$, we deduce that $$\{x\in \naturals: x^2-x=6, \}=\{3\}. $$ \end{answer} \end{pro} \begin{pro}Determine all the fractions lying strictly between $2$ and $3$ that have denominator $6$, that is, determine the set $$\{x\in \naturals: 2 < \dfrac{x}{6}<3 \} $$ explicitly. \begin{answer}We have $$ 2 < \dfrac{x}{6}<3 \implies 12 < x < 18 \implies x\in \{13, 14, 15, 16, 17\}. $$ \end{answer} \end{pro} \begin{pro} Let $A = \{a, b, c, d, e, f\}$ and $B = \{a, e, i, o, u\}$. Find $A \cup B$, $A \cap B$, $A \setminus B$ and $B \setminus A.$ \begin{answer} $A\cup B = \{a, b, c, d, e, f, i, o, u\}$, $A \cap B = \{a, e\}$, $A \setminus B = \{b, c, d, f\}$, $B\setminus A = \{i, o, u \}$ \end{answer} \end{pro} \begin{pro} Describe the following sets explicitly by either providing a list of their elements or an interval. \begin{multicols}{2} \begin{enumerate} \item $\{x\in \reals: x^3 = 8\}$ \item $\{x\in \reals: |x|^3 = 8\}$ \item $\{x\in \reals: |x| = -8\}$ \item $\{x\in \reals: |x| < 4\}$ \item $\{x\in \integers: |x| < 4\}$ \item $\{x\in \reals: |x| < 1\}$ \item $\{x\in \integers: |x| < 1\}$ \item $\{x\in \integers: x^{2002} < 0\}$ \end{enumerate}\end{multicols} \begin{answer} (i) $\{2\}$, (ii) $\{-2, 2\}$, (iii) $\varnothing$, (iv) $]-4; 4[$, (v) $\{-3, -2, -1, 0, 1, 2, 3\}$, (vi) $]-1;1[$, (vii) $\{0\}$, (viii) $\varnothing$ \end{answer} \end{pro} \begin{pro} Describe explicitly the set $$\{x\in\integers: x < 0, 1000 < x^2 < 2003\}$$ by listing its elements. \begin{answer} $\{-32, -33, -34,$ $ -35, -36, $ $-37, -38, -39,$ $ -40, -41, -42, -43, -44\}$\end{answer} \end{pro} \begin{pro} The set $S$ is formed according to the following rules: \begin{enumerate} \item $2$ belongs to $S$; \item if $n$ is in $S$ then $n+5$ is also in $S$; \item if $n$ is in $S$ then $3n$ is also in $S$. \end{enumerate} Find the largest integer in the set $$ \{1,2,3,\ldots , 2008\} $$that does not belong to $S$. \begin{answer} Observe that applying $k$ times the second rule, $n+5k$ is in $S$. Similarly, $3^k\cdot 2$ is in $S$ by applying $k$ times the third rule. Since $2$ is in $S$, $2+5k$ is in $S$, that is, numbers that leave remainder $2$ upon division by $5$ are in $S$. This means that $$\{2, 7, 12, \cdots 2002, 2007\}\subseteq S.$$ Since $3\cdot 2=6$ is in $S$, then the numbers $6+5k=1+5(k+1)$ are in $S$, that is, numbers $6$ or higher that leave remainder $1$ upon division by $5$. Thus the numbers $$\{6,11,16, \cdots 2001, 2006\}\subseteq S. $$ Since $3\cdot 6=18$ is in $S$, then the numbers $18+5k=3+5(k+3)$ are in $S$, that is, numbers $18$ or higher that leave remainder $3$ upon division by $5$. Thus the numbers $$\{18,23,28, \cdots 2003, 2008\}\subseteq S. $$ Since $3\cdot 18=54$ is in $S$, then the numbers $54+5k=4+5(k+10)$ are in $S$, that is, numbers $54$ or higher that leave remainder $4$ upon division by $5$. Thus the numbers $$\{54,59,64, \cdots 2004\}\subseteq S. $$Now, we claim that there are no multiples of $5$ in $S$. For by combining the rules every number in $S$ has the form $3^a\cdot 2 + 5b$, with $a\geq 0$, $b\geq 0$ integers. Since $3^a\cdot 2$ is never a multiple of $5$, this establishes the claim. Hence the largest element of $$ \{1,2,3,\ldots , 2008\} $$ not in $S$ is $2005$. \end{answer} \end{pro} \begin{pro} Use the trick of Gau\ss\ to prove that $$ 1+2+3+\cdots + n=\dfrac{n(n+1)}{2}. $$ \end{pro} \begin{pro} Let $C = \oo{-5; 5}$, $D = \oo{-1; +\infty }$. Find $C\cap D$, $C \cup D$, $C \setminus D$, and $D \setminus C$. \begin{answer} $\oo{-1;5}$, $\oo{-5; +\infty }$, $\oc{-5;-1}$, $\co{5; +\infty}$ \end{answer} \end{pro} \begin{pro} Let $C = \oo{-5; 3}$, $D = \co{4; +\infty }$. Find $C\cap D$, $C \cup D$, $C \setminus D$, and $D \setminus C$. \begin{answer} $\varnothing$, $\oo{-5;3}\cup \co{4; +\infty }$, $\oo{-5;3}$, $\co{4; +\infty}$ \end{answer} \end{pro} \begin{pro} Let $C = \co{-1; -2 + \sqrt{3}}$, $D = \cc{-0.5; \sqrt{2}-1}$. Find $C\cap D$, $C \cup D$, $C \setminus D$, and $D \setminus C$. \begin{answer} $\co{-0.5;-2 + \sqrt{3}}$, $\cc{-1; \sqrt{2}-1}$, $\co{-1;-0.5}$, $\cc{-2 + \sqrt{3};\sqrt{2}-1}$ \end{answer} \end{pro} \begin{pro} Consider $101$ different points $x_1, x_2, \ldots, x_{101}$ belonging to the interval $[0; 1[$. Shew that there are at least two say $x_i$ and $x_j, i \neq j$, such that $$|x_i - x_j| \leq \frac{1}{100} $$ \end{pro} \begin{pro}[Dirichlet's Approximation Theorem] Shew that $\forall x\in\reals, \ \forall N \in \naturals, N > 1, \ \exists (h\in\naturals , k\in\naturals )$ with $0 < k \leq N$ such that $$\absval{x - \frac{h}{k}} < \frac{1}{Nk}.$$ \begin{answer} Hint: Consider the $N + 1$ numbers $tx - \lfloor tx \rfloor , t = 0, 1, 2, \ldots , N.$ \end{answer} \end{pro} \end{multicols} \section{Rational Numbers and Irrational Numbers} Let us start by considering the strictly positive natural numbers. Primitive societies needed to count objects, say, their cows or sheep. Though some societies, like the Yanomame indians in Brazil or members of the CCP English and Social Sciences Department\footnote{Among these, many are Philosophers, who, though unsuccessful in finding their Philosopher's Stone, have found renal calculi.} cannot count above $3$, the need for counting is indisputable. In fact, many of these societies were able to make the following abstraction: add to a pile one pebble (or stone) for every sheep, in other words, they were able to make one-to-one correspondences. In fact, the word {\em Calculus} comes from the Latin for ``stone.'' \bigskip Breaking an object into almost equal parts (that is, {\em fractioning} it) justifies the creation of the positive rational numbers. In fact, most ancient societies did very well with just the strictly positive rational numbers. The problems of counting and of counting broken pieces were solved completely with these numbers. \bigskip As societies became more and more sophisticated, the need for new numbers arose. For example, it is believed that the introduction of negative quantities arose as an accounting problem in Ancient India. Fair enough, write $+1$ if you have a rupee---or whatever unit that ancient accountant used---in your favour. Write $-1$ if you owe one rupee. Write $0$ if you are rupeeless. \bigskip Thus we have constructed $\naturals$, $\integers$ and $\rationals$. In $\rationals$ we have, so far, a very elegant system of numbers which allows us to perform four arithmetic operations (addition, subtraction, multiplication, and division)\footnote{ ``Reeling and Writhing, of course, to begin with, ''the Mock Turtle replied, ``and the different branches of Arithmetic—--Ambition, Distraction, Uglification, and Derision.'' }and that has the notion of ``order'', which we will discuss in a latter section. A formal definition of the rational numbers is the following. \begin{df} The set of rational numbers $\rationals$ is the set of quotients of integers where a denominator $0$ is not allowed. In other words: $$ \rationals = \left\{\dfrac{a}{b}: a\in \integers, b \in \integers, b \neq 0\right\}. $$ \end{df} Notice also that $\rationals$ has the wonderful property of {\em closure}, meaning that if we add, subtract, multiply or divide any two rational numbers (with the exclusion of division by $0$), we obtain as a result a rational number, that is, we stay within the same set. \bigskip Since $a=\dfrac{a}{1}$, every integer is also a rational number, in other words, $\integers \subseteq \rationals$. Notice that every finite decimal can be written as a fraction, for example, we can write the decimal $3.14$ as $$3.14=\dfrac{314}{100}=\dfrac{157}{50}. $$What about non-finite decimals? Can we write them as a fraction? The next example shews how to convert an infinitely repeating decimal to fraction from. \begin{exa} Write the infinitely repeating decimal $0.3\overline{45} = 0.345454545\ldots$ as the quotient of two natural numbers. \end{exa} \begin{solu} The trick is to obtain multiples of $x=0.345454545\ldots$ so that they have the same infinite tail, and then subtract these tails, cancelling them out.\footnote{That this cancellation is meaningful depends on the concept of {\em convergence}, of which we may talk more later.} So observe that $$10x = 3.45454545\ldots; 1000x= 345.454545\ldots \implies 1000x - 10x = 342 \implies x = \dfrac{342}{990} = \dfrac{19}{55}. $$ \end{solu} By mimicking the above examples, the following should be clear: decimals whose decimal expansions terminate or repeat are rational numbers. Since we are too cowardly to prove the next statement,\footnote{The curious reader may find a proof in many a good number theory book, for example \cite{HarWri}} we prefer to call it a \begin{fact} Every rational number has a terminating or a repeating decimal expansion. Conversely, a real number with a terminating or repeating decimal expansion must be a rational number. Moreover, a rational number has a terminating decimal expansion if and only if its denominator is of the form $2^m5^n$, where $m$ and $n$ are natural numbers. \end{fact} From the above fact we can tell, without actually carrying out the long division, that say, $\dfrac{1}{1024}=\dfrac{1}{2^{10}}$ has a terminating decimal expansion, but that, say, $\dfrac{1}{6}$ does not. \bigskip Is every real number a rational number? Enter the Pythagorean Society in the picture, whose founder, Pythagoras lived 582 to 500 BC. This loony sect of Greeks forbade their members to eat beans. But their lunacy went even farther. Rather than studying numbers to solve everyday ``real world problems''---as some misguided pedagogues insist---they tried to understand the very essence of numbers, to study numbers in the abstract. At the beginning it seems that they thought that the ``only numbers'' were rational numbers. But one of them, Hipassos of Metapontum, was able to prove that the length of hypotenuse of a right triangle whose legs\footnote{The appropriate word here is ``cathetus.''} had unit length could not be expressed as the ratio of two integers and hence, it was {\em irrational}. \begin{thm}\label{thm:hipassos}[Hipassos of Metapontum]$\sqrt{2}$ is irrational. \end{thm} \begin{pf} Assume there is $s\in \rationals$ such that $s^2=2$. We can find integers $m, n\neq 0$ such that $s=\dfrac{m}{n}$. The crucial part of the argument is that we can choose $m, n$ such that this fraction be in least terms, and hence, $m, n$ cannot be both even. Now, $n^2s^2=m^2$, that is $2n^2=m^2$. This means that $m^2$ is even. But then $m$ itself must be even, since the product of two odd numbers is odd. Thus $m=2a$ for some non-zero integer $a$ (since $m\neq 0$). This means that $2n^2 = (2a)^2=4a^2 \implies n^2=2a^2$. This means once again that $n$ is even. But then we have a contradiction, since $m$ and $n$ were not both even. \end{pf} \vspace{1cm} \begin{figure}[!hptb] \centering \psaxes[arrows={->},linewidth=2pt,yAxis=false,labelFontSize=\tiny, arrowscale=2.1,labelsep=-20pt](0,0)(-2.5,0)(2.7,0) \pstGeonode[PointName=none](0,0){O}(1,0){A}(0,1){B}(1.414214,0){C}(1,1){D} \pstArcOAB{O}{C}{D}\psline(O)(B)(D)(A)(O)(D)\uput[d](C){{\tiny $\sqrt{2}$}}\meinecaption{.5}{Theorem \ref{thm:hipassos}.} \end{figure} \begin{rem}The above theorem says that the set $\reals\setminus \rationals$ of irrational numbers is non-empty. This is one of the very first theorems ever proved. It befits you, dear reader, if you want to be called mathematically literate, to know its proof. \end{rem} Suppose that we knew that every strictly positive natural number has a {\em unique} factorisation into primes. Then if $n$ is not a perfect square we may deduce that, in general, $\sqrt{n}$ is irrational. For, if $\sqrt{n}$ were rational, there would exist two strictly positive natural numbers $a, b$ such that $\sqrt{n}=\dfrac{a}{b}$. This implies that $na^2=b^2$. The dextral side of this equality has an even number of prime factors, but the sinistral side does not, since $n$ is not a perfect square. This contradicts unique factorisation, and so $\sqrt{n}$ must be irrational. \begin{rem} From now on we will accept the result that $\sqrt{n}$ is irrational whenever $n$ is a positive non-square integer. \end{rem} The shock caused to the other Pythagoreans by Hipassos' result was so great (remember the Pythagoreans were a cult), that they drowned him. Fortunately, mathematicians have matured since then and the task of burning people at the stake or flying planes into skyscrapers has fallen into other hands. \begin{exa} Give examples, if at all possible, of the following. \begin{enumerate} \item the sum of two rational numbers giving an irrational number. \item the sum of two irrationals giving an irrational number. \item the sum of two irrationals giving a rational number. \item the product of a rational and an irrational giving an irrational number. \item the product of a rational and an irrational giving a rational number. \item the product of two irrationals giving an irrational number. \item the product of two irrationals giving a rational number. \end{enumerate} \end{exa} \begin{solu} \begin{enumerate} \item This is impossible. The rational numbers are closed under addition and multiplication. \item Take both numbers to be $\sqrt{2}$. Their sum is $2\sqrt{2}$ which is also irrational.\item Take one number to be $\sqrt{2}$ and the other $-\sqrt{2}$. Their sum is $0$, which is rational. \item take the rational number to be $1$ and the irrational to be $\sqrt{2}$. Their product is $1\cdot\sqrt{2} = \sqrt{2}$. \item Take the rational number to be $0$ and the irrational to be $\sqrt{2}$. Their product is $0\cdot\sqrt{2} = 0$. \item Take one irrational number to be $\sqrt{2}$ and the other to be $\sqrt{3}$. Their product is $\sqrt{2}\cdot \sqrt{3} = \sqrt{6}$. \item Take one irrational number to be $\sqrt{2}$ and the other to be $\dfrac{1}{\sqrt{2}}$. Their product is $\sqrt{2}\cdot \dfrac{1}{\sqrt{2}} = 1$. \end{enumerate} \end{solu} After the discovery that $\sqrt{2}$ was irrational, suspicion arose that there were other irrational numbers. In fact, Archimedes suspected that $\pi$ was irrational, a fact that wasn't proved till the XIX-th Century by Lambert. The ``irrationalities'' of $\sqrt{2}$ and $\pi$ are of two entirely ``different flavours,'' but we will need several more years of mathematical study\footnote{Or in the case of people in the English and the Social Sciences Departments, as many lifetimes as a cat.} to even comprehend the meaning of that assertion. \bigskip Irrational numbers, that is, the set $\reals \setminus \rationals$, are those then having infinite non-repeating decimal expansions. Of course, by simply ``looking'' at the decimal expansion of a number we can't tell whether it is irrational or rational without having more information. Your calculator probably gives about $9$ decimal places when you try to compute $\sqrt{2}$, say, it says $\sqrt{2}\approx 1.414213562$. What happens after the final $2$ is the interesting question. Do we have a pattern or do we not? \begin{exa} We expect a number like $$0.100100001000000001\ldots , $$ where there are $2, 4, 8, 16, \ldots$ zeroes between consecutive ones, to be irrational, since the gaps between successive $1$'s keep getting longer, and so the decimal does not repeat. For the same reason, the number $$ 0.123456789101112\ldots, $$which consists of enumerating all strictly positive natural numbers after the decimal point, is irrational. This number is known as the {\em Champernowne-Mahler} number. \end{exa} \begin{exa} Prove that $\sqrt[4]{2}$ is irrational. \end{exa} \begin{solu} If $\sqrt[4]{2}$ were rational, then there would be two non-zero natural numbers, $a, b$ such that $$\sqrt[4]{2}=\dfrac{a}{b}\implies \sqrt{2}=\dfrac{a^2}{b^2}. $$ Since $\dfrac{a}{b}$ is rational, $\dfrac{a^2}{b^2}=\dfrac{a}{b}\cdot \dfrac{a}{b}$ must also be rational. This says that $\sqrt{2}$ is rational, contradicting Theorem \ref{thm:hipassos}. \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Write the infinitely repeating decimal $0.\overline{123} = 0.123123123\ldots$ as the quotient of two positive integers. \begin{answer} If $x = 0.123123123\ldots$ then $1000x = 0.123123123\ldots $ giving $1000x-x = 123 $, since the tails cancel out. This results in $x = \dfrac{123}{999} = \dfrac{41}{333}$. \end{answer} \end{pro} \begin{pro} Prove that $\sqrt{8}$ is irrational. \begin{answer} If $\sqrt{8}=2\sqrt{2}$ were rational, then there would exist strictly positive natural numbers $a, b$ such that $2\sqrt{2}=\dfrac{a}{b}$, which entails that $\sqrt{2}=\dfrac{a}{2b}$, a rational number, a contradiction. \end{answer} \end{pro} \begin{pro} Assuming that $\sqrt{6}$ is irrational, prove that $\sqrt{2}+\sqrt{3}$ must be irrational. \begin{answer} If $\sqrt{2}+\sqrt{3}$ were rational, then there would exist strictly positive natural numbers $a, b$ such that $\sqrt{2}+\sqrt{3}=\dfrac{a}{b}$, which entails that $$(\sqrt{2}+\sqrt{3})^2=\dfrac{a^2}{b^2}\implies 2 + 2\sqrt{6}+3 = \dfrac{a^2}{b^2}\implies \sqrt{6}=\dfrac{a^2}{2b^2}-\dfrac{5}{2}.$$The dextral side of the last equality is a rational number, but the sinistral side is presumed irrational, a contradiction. \end{answer} \end{pro} \begin{pro} Suppose that you are given a finite string of integers, say, $12345$. Can you find an {\em irrational} number whose first five decimal digits after the decimal point are $12345$? \begin{answer} Yes! There multiple ways of doing this. An idea is to take the first decimal digits of $\sqrt{2}$, remove them, and supplant them with the given string. For example, for $12345$ we proceed as follows: $$ \sqrt{2}\approx 1.414213562\ldots \implies \dfrac{\sqrt{2}}{10^6}\approx 0.000001414213562\ldots. $$ Then the number $$\dfrac{\sqrt{2}}{10^6}+0.12345 $$is an irrational number whose first five digits after the decimal point are $12345$. \bigskip Another idea is to form the number $$ 0.1234501234500123450000123450000000012345\ldots $$where one puts $2^k$ zeroes between appearances of the string $12345$. \end{answer} \end{pro} \begin{pro} Find a rational number between the irrational numbers $\sqrt{2}$ and $\sqrt{3}$. \begin{answer} There are infinitely many answers. Since $\sqrt{2}<1.5$ and $1.7<\sqrt{3}$, we may take, say, $1.6$. Of course, using the mentioned inequalities we may take also $1.61$, $1.601$, $1.52$, etc. \end{answer} \end{pro} \begin{pro} Find an irrational number between the irrational numbers $\sqrt{2}$ and $\sqrt{3}$. \begin{answer}There are infinitely many answers. One may take the average, $\dfrac{\sqrt{2}+\sqrt{3}}{2}$. \end{answer} \end{pro} \begin{pro} Find an irrational number between the rational numbers $\dfrac{1}{10}$ and $\dfrac{1}{9}$. \begin{answer} Since $\dfrac{1}{10}=0.1$ and $0.111<\dfrac{1}{9}=$, we may take, say $0.110100100001000000001\ldots$, where there are $2^k, k=1,2,\ldots$ $0$'s between consecutive $1$'s. Another approach can be taking $\sqrt{2}-1.314$, since $\sqrt{2}-1.314<0.1003$. \end{answer} \end{pro} \end{multicols} \section{Operations with Real Numbers} The set of real numbers is furnished with two operations $+$ (addition) and $\cdot$ (multiplication) that satisfy the following axioms. \begin{axi}[Closure] $$x\in \reals\quad \mathrm{and}\quad y\in \reals \implies x+y\in \reals\quad \mathrm{and}\quad xy\in \reals .$$ \end{axi} This axiom tells us that if we add or multiply two real numbers, then we stay within the realm of real numbers. Notice that this is not true of division, for, say, $1\div 0$ is the division of two real numbers, but $1\div 0$ is not a real number. This is also not true of taking square roots, for, say, $-1$ is a real number but $\sqrt{-1}$ is not. \begin{axi}[Commutativity] $$x\in \reals\quad \mathrm{and}\quad y\in \reals \implies x+y=y+x\quad \mathrm{and}\quad xy=yx.$$ \end{axi} This axiom tells us that order is immaterial when we add or multiply two real numbers. Observe that this axiom does not hold for division, because, for example, $1\div 2 \neq 2\div 1$. \begin{axi}[Associativity] $$x\in \reals, y\in \reals\quad \mathrm{and}\quad z\in \reals \implies x+(y+z)=(x+y)+z\quad \mathrm{and}\quad (xy)z=x(yz).$$ \end{axi} This axiom tells us that in a string of successive additions or multiplications, it is immaterial where we put the parentheses. Observe that subtraction is not associative, since, for example, $(1-1)-1\neq 1-(1-1)$. \begin{axi}[Additive and Multiplicative Identity] There exist two unique elements, $0$ and $1$, with $0\neq 1$, such that $\forall x\in\reals$, $$ 0+x=x+0=x, \quad \mathrm{and} \quad 1\cdot x = x\cdot 1 =x. $$ \end{axi} \begin{axi}[Existence of Opposites and Inverses] For all $x\in \reals$ $\exists -x\in \reals$, called the {\em opposite} of $x$, such that $$ x+(-x)=(-x)+x=0. $$For all $y\in \reals\setminus \{0\}$ $\exists y^{-1}\in\reals\setminus \{0\} $, called the {\em multiplicative inverse} of $y$, such that $$y\cdot y^{-1}=y^{-1}\cdot y=1. $$ \end{axi} In the axiom above, notice that $0$ does not have a multiplicative inverse, that is, division by $0$ is not allowed. Why? Let us for a moment suppose that $0$ had a multiplicative inverse, say $0^{-1}$. We will obtain a contradiction as follows. First, if we multiply any real number by $0$ we get $0$, so, in particular, $0\cdot 0^{-1}=0$. Also, if we multiply a number by its multiplicative inverse we should get $1$, and hence, $0\cdot 0^{-1}=1$. This gives $$ 0 = 0\cdot 0^{-1}=1, $$in contradiction to the assumption that $0\neq 1$. \begin{axi}[Distributive Law] For all real numbers $x,y,z$, there holds the equality $$ x\cdot (y+z) =x\cdot y + x\cdot z.$$ \end{axi} \begin{rem} It is customary in Mathematics to express a product like $x\cdot y$ by juxtaposition, that is, by writing together the letters, as in $xy$, omitting the product symbol $\cdot$. From now on we will follow this custom. \end{rem} The above axioms allow us to obtain various algebraic identities, of which we will demonstrate a few. \begin{thm}[Difference of Squares Identity] For all real numbers $a, b$, there holds the identity $$ a^2-b^2=(a-b)(a+b). $$ \label{thm:diff_of_sqs} \end{thm} \begin{pf}Using the distributive law twice, $$(a-b)(a+b)=a(a+b)-b(a+b)=a^2+ab-ba-b^2=a^2+ab-ab-b^2=a^2-b^2. $$ \end{pf} Here is an application of the above identity. \begin{exa} Given that $2^{32}-1$ has exactly two divisors $a$ and $b$ satisfying the inequalities $$50< a 0$ be an integer. Then for all real numbers $x, y$ \begin{equation}\label{eq:diff_of_binoms} x^n-y^n = (x-y)(x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \cdots + x^2y^{n-3}+xy^{n-2}+y^{n-1}).\end{equation} For example, $$x^5-y^5=(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4), \qquad x^5+y^5=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4). $$ See problem \ref{pro:finite-geom-sum}. \begin{thm}[Perfect Squares Identity] For all real numbers $a, b$, there hold the identities $$ (a+b)^2=a^2+2ab+b^2\qquad \mathrm{and}\qquad (a-b)^2=a^2-2ab+b^2.$$ \label{thm:perfect_sqs} \end{thm} \begin{pf}Expanding using the distributive law twice, $$(a+b)^2= (a+b)(a+b)=a(a+b)+b(a+b)=a^2+ab+ba+b^2=a^2+2ab+b^2. $$ To obtain the second identity, replace $b$ by $-b$ in the just obtained identity: $$(a-b)^2=(a+(-b))^2=a^2+2a(-b)+(-b)^2=a^2-2ab+b^2. $$ \end{pf} \begin{exa}\label{exa:sum_prod_nums} The sum of two numbers is $7$ and their product is $3$. Find the sum of their squares and the sum of their cubes. \end{exa} \begin{solu} Let the two numbers be $a, b$. Then $a+b=7$ and $ab=3$. Then $$49 = (a+b)^2=a^2+2ab+b^2=a^2+b^2+6 \implies a^2+b^2=49-6=43. $$ Also, $$a^3+b^3=(a+b)(a^2+b^2-ab)=(7)(43-3)=280. $$ Thus the sum of their squares is $43$ and the sum of their cubes is $280$. \end{solu} \vspace*{2cm} \begin{figure}[h] \centering \psset{unit=1pc} \def\bigsquare{\uput[l](-1,0){$x$}\uput[u](0,1){$x$}\pspolygon[linewidth=2pt,linecolor=brown](-1,-1)(-1,1)(1,1)(1,-1)} \def\bigsquarea{\pspolygon[linewidth=2pt,linecolor=brown](-1,-1)(-1,1)(1,1)(1,-1)} \def\bigrectangle{ \uput[l](-1,.5){$a$}\pspolygon[linewidth=2pt,fillcolor=yellow,fillstyle=solid](-1,1)(1,1)(1,0)(-1,0)} \def\smallrectanglea{\pspolygon[linewidth=2pt,fillcolor=yellow,fillstyle=solid](-1,1.2)(1,1.2)(1,.7)(-1,.7)} \def\smallrectangleb{\pspolygon[linewidth=2pt,fillcolor=yellow,fillstyle=solid](-1,.3)(1,.3)(1,-.2)(-1,-.2)} \def\bigpicture{\pspolygon[linewidth=2pt,linecolor=brown](-1,-1)(-1,1)(1,1)(1,-1)\pspolygon[linewidth=2pt,fillcolor=yellow,fillstyle=solid](-1,-1)(1,-1)(1,-1.5)(-1,-1.5) \pspolygon[linewidth=2pt,fillcolor=yellow,fillstyle=solid](1,-1)(1,1)(1.5,1)(1.5,-1)\pspolygon[fillstyle=vlines](1.5,-1.5)(1.5,-2)(2,-2)(2,-1.5)} %\renewcommand{\arraystretch}{.2} \rput(-4,0){\begin{tabular}{c}\bigsquare\\ $+$ \\ \bigrectangle \end{tabular}} \rput(-1.2,0){$=$} \rput(1,0){\begin{tabular}{c}\bigsquarea\\ $+$ \\ \smallrectanglea\smallrectangleb \end{tabular}} \rput(3.5,0){$=$} \rput(5.5,0){\bigpicture} \meinecaption{1}{Completing the square: $x^2+ax=\left(x+\dfrac{a}{2}\right)^2-\left(\dfrac{a}{2}\right)^2$.}\label{fig:completing-square} \end{figure} The following method, called {\em Sophie Germain's trick}\footnote{Sophie Germain (1776--1831) was an important French mathematician of the French Revolution. She pretended to be a man in order to study Mathematics. At the time, women were not allowed to matriculate at the \'{E}cole Polytechnique, but she posed as a M. Leblanc in order to obtain lessons from Lagrange.} is useful to convert some expressions into differences of squares.\index{Sophie Germain}\index{Sophie Germain's trick} \begin{exa} We have $$ \begin{array}{lll}x^4+x^2+1 & = & x^4+2x^2+1 -x^2 \\ & = & (x^2+1)^2-x^2\\ & = & (x^2+1-x)(x^2+1+x). \end{array}$$ \end{exa} \begin{exa} We have $$ \begin{array}{lll}x^4+4 & = & x^4+4x^2 + 4-4x^2\\ & = & (x^2+2)^2-4x^2\\ & = & (x^2+2-2x)(x^2+2+2x). \end{array}$$ \end{exa} Sophie Germain's trick is often used in factoring quadratic trinomials, where it is often referred to as the technique of {\em completing the square}, which has the geometric interpretation given in figure \ref{fig:completing-square}. We will give some examples of factorisations that we may also obtain with the trial an error method commonly taught in elementary algebra.\index{completing the square} \begin{exa} We have $$ \begin{array}{lll} x^2 -8x-9 & = & x^2-8x +16-9-16 \\ & = & (x-4)^2-25\\ & = & (x-4)^2-5^2\\ & = & (x-4-5)(x-4+5)\\ & =& (x-9)(x+1). \end{array}$$ Here to complete the square, we looked at the coefficient of the linear term, which is $-8$, we divided by $2$, obtaining $-4$, and then squared, obtaining $16$. \end{exa} \begin{exa} We have $$ \begin{array}{lll} x^2+4x-117 & = & x^2+4x+4-117-4\\ & = & (x+2)^2-11^2\\ & = & (x+2-11)(x+2+11)\\ & = & (x-9)(x+13). \end{array}$$ Here to complete the square, we looked at the coefficient of the linear term, which is $4$, we divided by $2$, obtaining $2$, and then squared, obtaining $4$. \end{exa} \begin{exa} We have $$a^2+ab+b^2 = a^2+ab+\dfrac{b^2}{4}-\dfrac{b^2}{4}+b^2 =a^2+ab+\dfrac{b^2}{4}+\dfrac{3b^2}{4}=\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}. $$ Here to complete the square, we looked at the coefficient of the linear term (in $a$), which is $b$, we divided by $2$, obtaining $\dfrac{b}{2}$, and then squared, obtaining $\dfrac{b^2}{4}$. \end{exa} \begin{exa} Factor $2x^2+3x-8$ into linear factors by completing squares. \end{exa} \begin{solu} First, we force a $1$ as coefficient of the square term: $$2x^2+3x-8=2\left(x^2+\dfrac{3}{2}x-4\right) . $$ Then we look at the coefficient of the linear term, which is $\dfrac{3}{2}$. We divide it by $2$, obtaining $\dfrac{3}{4}$, and square it, obtaining $\dfrac{9}{16}$. Hence $$\begin{array}{lll}2x^2+3x-8 & = & 2\left(x^2+\dfrac{3}{2}x-4\right) \\ & = & 2\left(x^2+\dfrac{3}{2}x+ \dfrac{9}{16}-\dfrac{9}{16}-4\right) \\ & = & 2\left(\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{16}-4\right)\\ & = & 2\left(\left(x+\dfrac{3}{2}\right)^2-\dfrac{73}{16}\right)\\ & = & 2\left(x+\dfrac{3}{2}-\dfrac{\sqrt{73}}{4}\right)\left(x+\dfrac{3}{2}+\dfrac{\sqrt{73}}{4}\right).\\ \end{array} $$ \end{solu} \begin{thm}[Perfect Cubes Identity] For all real numbers $a, b$, there hold the identities $$ (a+b)^3=a^3+3a^2b+3ab^2+b^3\qquad \mathrm{and}\qquad (a-b)^3=a^3-3a^2b+3ab^2-b^3.$$ \label{thm:perfect_pubes} \end{thm} \begin{pf}Expanding, using Theorem \ref{thm:perfect_sqs}, $$\begin{array}{lll}(a+b)^3 & = & (a+b)(a+b)^2\\ & = & (a+b)(a^2+2ab+b^2)\\ & = & a(a^2+2ab+b^2)+b(a^2+2ab+b^2)\\ & = & a^3+2a^2b+ab^2+ba^2+2ab^2+b^3\\ & = & a^3+3a^2b+3ab^2+b^3.\\ \end{array} $$ The second identity is obtained by replacing $b$ with $-b$: $$ (a-b)^3=(a+(-b))^3=a^2+3a^2(-b)+3a(-b)^2+(-b)^3=a^3-3a^2b+3ab^2-b^3. $$ \end{pf} It is often convenient to rewrite the above identities as $$ (a+b)^3=a^3+b^3+3ab(a+b), \qquad (a-b)^3=a^3-b^3-3ab(a-b). $$ \begin{exa} Redo example \ref{exa:sum_prod_nums} using Theorem \ref{thm:perfect_pubes}. \end{exa} \begin{solu} Again, let the two numbers the two numbers $a, b$ satisfy $a+b=7$ and $ab=3$. Then $$ 343=7^3=(a+b)^3=a^3+b^3+3ab(a+b)=a^3+b^3+3(3)(7)\implies a^3+b^3=343-63=280, $$as before. \end{solu} The results of Theorems \ref{thm:perfect_sqs} and \ref{thm:perfect_pubes} generalise in various ways. In Appendix \ref{chap:bino_mio} we present the binomial theorem, which provides the general expansion of $(a+b)^n$ when $n$ is a positive integer. \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Expand and collect like terms: $$\left(\dfrac{2}{x}+\dfrac{x}{2}\right)^2-\left(\dfrac{2}{x}-\dfrac{x}{2}\right)^2.$$ \begin{answer} We have, $$\begin{array}{lll} \left(\dfrac{2}{x}+\dfrac{x}{2}\right)^2-\left(\dfrac{2}{x}-\dfrac{x}{2}\right)^2 & = & \left(\dfrac{4}{x^2}+2+\dfrac{x^2}{4}\right)-\left(\dfrac{4}{x^2}-2+\dfrac{x^2}{4}\right)\\ & = & 4. \end{array}$$ \end{answer} \end{pro} \begin{pro}\label{pro:simu_eq1} Find all the real solutions to the system of equations $$ x+y=1, \qquad xy=-2. $$ \begin{answer} We have $$1=(x+y)^2=x^2+y^2+2xy = x^2+y^2-4 \implies x^2+y^2=5. $$ Hence, $$(x-y)^2=x^2+y^2-2xy =5-2(-2)=9 \implies x-y = \pm 3.$$ In the first case, $$x+y=1, x-y = 3 \implies x=2, \quad y = -1. $$ In the second case, $$x+y=1, x-y = -3 \implies x=-1, \quad y = 2. $$ \end{answer} \end{pro} \begin{pro}\label{pro:simu_eq2} Find all the real solutions to the system of equations $$ x^3+y^3=7, \qquad x+y=1. $$ \begin{answer} We have $$7 = x^3+y^3=(x+y)(x^2-xy+y^2)=x^2-xy+y^2. $$ Also, $$ 1=(x+y)^2=x^2+y^2+2xy. $$ This gives $$ 6=(x^2-xy+y^2)-(x^2+2xy+y^2)=-3xy \implies -2=xy. $$ Hence we have the system $$x+y=1, \quad xy=-2, $$which was already solved in problem \ref{pro:simu_eq1}. \end{answer} \end{pro} \begin{pro}Compute $$ 1^2-2^2+3^2-4^2+\cdots +99^2-100^2. $$ \begin{answer}We have $$\begin{array}{lll} 1^2-2^2+3^2-4^2+\cdots +99^2-100^2 & = & (1-2)(1+2) + (3-4)(3+4) + \cdots + (99-100)(99+100)\\ & = & -(1+2+3+4+\cdots +99+100)\\ \end{array}$$ To compute the sum of the arithmetic progression $1+2+3+4+\cdots + 99+100$, use Gau\ss\ 's trick: if $S=1+2+3+4+\cdots + 99+100$, then $S=100+99+\cdots + 2+1$. Hence $$2S=(1+100)+(2+99)+ \cdots + (99+2)+(100+1)=101\cdot 100 \implies S=5050. $$ This means that $$ 1^2-2^2+3^2-4^2+\cdots +99^2-100^2=-5050. $$ \end{answer} \end{pro} \begin{pro} Let $n\in\naturals$. Find all prime numbers of the form $n^3-8$. \begin{answer}Since $n^3-8=(n-2)(n^2+2n+4)$, for it to be a prime one needs either $n-2=1 \implies n=3$ or $n^2+2n+4=1$, but this last equation does not have integral solutions. Hence $3^3-8=19$ is the only such prime. \end{answer} \end{pro} \begin{pro} Compute $1234567890^2-1234567889\cdot 1234567891$ {\em mentally}. \begin{answer} Put $x=1234567890$. Then $$1234567890^2-1234567889\cdot 1234567891=x^2-(x-1)(x+1)=x^2-(x^2-1)=1.$$ \end{answer} \end{pro} \begin{pro}The sum of two numbers is $3$ and their product is $9$. What is the sum of their reciprocals? \begin{answer}If the numbers are $x, y$ then $x+y=3$ and $xy=9$. This gives $$ \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}=\dfrac{3}{9}=\dfrac{1}{3}. $$ \end{answer} \end{pro} \begin{pro} Given that $$ 1,\ 000, \ 002,\ 000, \ 001$$ has a prime factor greater than $9000$, find it. \begin{answer} We have $$ \begin{array}{lll}1,\ 000, \ 002,\ 000, \ 001 & = & 10^{12} + 2\cdot 10^6 + 1\\ & = & (10^6 + 1)^2 \\ & = & ((10^2)^3 + 1)^2 \\ & = & (10^2 + 1)^2((10^2)^2 - 10^2 + 1)^2 \\ & = & 101^29901^2,\\ \end{array}$$whence the prime sought is $9901$. \end{answer} \end{pro} \begin{pro}\label{pro:sum_of_sqs} Let $a, b, c$ be arbitrary real numbers. Prove that $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca). $$ \end{pro} \begin{pro}\label{pro:sum_of_pubes} Let $a, b, c$ be arbitrary real numbers. Prove that $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca). $$ \begin{answer} One can expand the dextral side and obtain the sinistral side. $$x^3 + y^3 = (x + y)^3 - 3xy(x + y)$$twice: $$ \begin{array}{lll} a^3 + b^3 + c^3 - 3abc & = & (a + b)^3 + c^3 - 3ab(a + b) - 3abc \\ & = & (a + b + c)^3 - 3(a + b)c(a + b + c) - 3ab(a + b + c) \\ & = & (a + b + c)((a + b + c)^2 - 3ac - 3bc - 3ab) \\ & = & (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca). \end{array} $$ \end{answer} \end{pro} \begin{pro} The numbers $a, b, c$ satisfy $$ a+b+c=-6, \qquad ab+bc+ca=2, \qquad a^3+b^3+c^3=6. $$ Find $abc$. \begin{answer} From problem \ref{pro:sum_of_sqs}, $$ 36=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=a^2+b^2+c^2+4\implies a^2+b^2+c^2=32. $$ From problem \ref{pro:sum_of_pubes}, $$abc= \dfrac{a^3+b^3+c^3-(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}{3}=\dfrac{6-(-6)(32-2)}{3}=62. $$ \end{answer} \end{pro} \begin{pro}Compute $$\sqrt{(1 000 000)(1 000 001)(1 000 002)(1 000 003) + 1}$$ without a calculator. \begin{answer} Put $x = 1\ 000\ 000 = 10^6$. Then $$x(x + 1)(x + 2)(x + 3) = x(x + 3)(x + 1)(x + 2) = (x^2 + 3x)(x^2 + 3x + 2).$$ Again, put $y = x^2 + 3x.$ Then $$x(x + 1)(x + 2)(x + 3) + 1 = (x^2 + 3x)(x^2 + 3x + 2) + 1 = y(y + 2) + 1 = (y + 1)^2.$$ All this gives, $$\begin{array}{lll}\sqrt{x(x + 1)(x + 2)(x + 3) + 1} & = & y + 1\\ & = & x^2 + 3x + 1 \\ & = & 10^{12} + 3\cdot 10^6 + 1 \\ & = & 1\ 000\ 003\ 000 \ 001. \end{array} $$ \end{answer} \end{pro} \begin{pro} Find two positive integers $a, b$ such that $$\sqrt{5+2\sqrt{6}}=\sqrt{a}+\sqrt{b}.$$ \begin{answer} We have, $$\sqrt{5+2\sqrt{6}} = \sqrt{2+2\sqrt{6}+ 3} = \sqrt{(\sqrt{2}+\sqrt{3})^2}=\sqrt{2}+\sqrt{3} $$ \end{answer} \end{pro} \begin{pro} If $a, b, c, d,$ are real numbers such that $$a^2 + b^2 + c^2 + d^2 = ab + bc + cd + da,$$ prove that $a = b = c = d$. \begin{answer} Transposing, $$a^2 - ab + b^2 - bc + c^2 - dc + d^2 - da = 0,$$or $$\frac{a^2}{2} - ab + \frac{b^2}{2} + \frac{b^2}{2} - bc + \frac{c^2}{2} + \frac{c^2}{2} - dc + \frac{d^2}{2} + \frac{d^2}{2} - da + \frac{a^2}{2} = 0.$$ Factoring, $$\frac{1}{2}(a - b)^2 + \frac{1}{2}(b - c)^2 +\frac{1}{2}(c - d)^2 +\frac{1}{2}(d - a)^2 = 0.$$ As the sum of positive quantities is zero only when the quantities themselves are zero, we obtain $a = b, b = c, c = d, d = a,$ which proves the assertion. \end{answer} \end{pro} \begin{pro} Find all real solutions to the equation $$(x+y)^2=(x-1)(y+1).$$ \begin{answer} We have $$\begin{array}{lll} (x+y)^2=(x-1)(y+1) & \implies & (x-1+y+1)^2=(x-1)(y+1)\\ & \implies & (x-1)^2+2(x-1)(y+1)+(y+1)^2=(x-1)(y+1) \\ & \implies & (x-1)^2+(x-1)(y+1)+(y+1)^2=0\\ & \implies & \left(x-1 +\dfrac{y+1}{2}\right)^2+\dfrac{3(y+1)^2}{4}=0. \end{array}$$ This last is a sum of squares, which can only be zero if $$ x-1 +\dfrac{y+1}{2}=0, \qquad y+1=0 \implies x=1, y=-1. $$ Thus $(x,y)=(1,-1)$ is the only solution. \end{answer} \end{pro} \begin{pro} Let $a, b, c$ be real numbers with $a+b+c=0$. Prove that $$ \dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}=\dfrac{a^3}{bc}+ \dfrac{b^3}{ca}+ \dfrac{c^3}{ab}.$$ \begin{answer} Observe that $$\begin{array}{lll}\dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a} & = & \dfrac{a^2+b^2}{-c}+\dfrac{b^2+c^2}{-a}+\dfrac{c^2+a^2}{-b}\\ & = & -a^2\left(\dfrac{1}{b}+\dfrac{1}{c}\right) -b^2\left(\dfrac{1}{c}+\dfrac{1}{a}\right) -c^2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\\ & = &-a^2\left(\dfrac{c+b}{bc}\right) -b^2\left(\dfrac{a+c}{ca}\right) -c^2\left(\dfrac{b+a}{ab}\right)\\ & = & \dfrac{a^3}{bc}+ \dfrac{b^3}{ca}+ \dfrac{c^3}{ab},\\ \end{array}$$as was to be shewn. \end{answer} \end{pro} \begin{pro}\label{pro:finite-geom-sum} Prove that if $a\in\reals$, $a\neq 1$ and $n\in\naturals\setminus \{0\}$, then \begin{equation}\label{eq:geom_sum}1 + a + a^2 + \cdots a^{n-1} = \dfrac{1 - a^{n}}{1 -a}. \end{equation}Then deduce that if $n$ is a strictly positive integer, it follows $$x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \cdots + xy^{n-2} + y^{n-1}).$$ \begin{answer} Put $S = 1 + a + a^2 + \cdots + a^{n-1}.$ Then $aS = a + a^2 + \cdots + a^{n-1} + a^n.$ Thus $$S - aS = (1 + a + a^2 + \cdots +a^{n-1}) - (a + a^2 + \cdots + a^{n-1} + a^n) = 1 - a^n,$$ and from $(1-a)S = S-aS=1 - a^n $ we obtain the result. \bigskip By making the substitution $a = \frac{y}{x}$ in the preceding identity, we see that $$1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2 + \cdots + \left(\frac{y}{x}\right)^{n-1} = \dfrac{1- \left(\frac{y}{x}\right)^n}{1-\frac{y}{x}} $$ we obtain $$\left(1-\frac{y}{x}\right)\left(1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2 + \cdots + \left(\frac{y}{x}\right)^{n-1}\right) = 1- \left(\frac{y}{x}\right)^n, $$ or equivalently, $$\left(1-\frac{y}{x}\right)\left(1 + \frac{y}{x} + \frac{y^2}{x^2} + \cdots +\frac{y^{n-1}}{x^{n-1}}\right) = 1- \frac{y^n}{x^n}. $$ Multiplying by $x^n$ both sides, $$x\left(1-\frac{y}{x}\right) x^{n-1}\left(1 + \frac{y}{x} + \frac{y^2}{x^2} + \cdots +\frac{y^{n-1}}{x^{n-1}}\right) = x^n\left(1- \frac{y^n}{x^n}\right), $$ which is $$x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + \cdots + xy^{n-2} + y^{n-1}), $$ yielding the result. \end{answer} \end{pro} \begin{pro} Prove that the product of two sums of squares is a sum of squares. That is, let $a, b, c, d$ be integers. Prove that you can find integers $A, B$ such that $$ (a^2+b^2)(c^2+d^2)=A^2+B^2. $$ \begin{answer} Observe that $$ (ac+bd)^2+(ad-bc)^2=a^2c^2+b^2d^2+a^2d^2+b^2c^2 =(a^2+b^2)(c^2+d^2).$$ \end{answer} \end{pro} \begin{pro} Prove that if $a, b, c$ are real numbers, then $$ (a+b+c)^3-3(a+b)(b+c)(c+a)=a^3+b^3+c^3. $$ \end{pro} \begin{pro} If $a, b, c$ are real numbers, prove that $a^5+b^5+c^5$ equals $$ (a+b+c)^5-5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca). $$ \end{pro} \end{multicols} \section{Order on the Line} \begin{rem} \fcolorbox{red}{cyan}{ \begin{minipage}{.90\linewidth} \noindent\textcolor{red}{\textbf{Vocabulary Alert!}} We will call a number $x$ {\em positive} if $x\geq 0$ and {\em strictly positive} if $x>0$. Similarly, we will call a number $y$ {\em negative} if $y\leq 0$ and {\em strictly negative} if $y<0$. This usage differs from most Anglo-American books, who prefer such terms as {\em non-negative} and {\em non-positive}. \end{minipage}} \end{rem} The set of real numbers $\reals$ is also endowed with a relation $>$ which satisfies the following axioms. \begin{axi}[Trichotomy Law]\label{axi:trichotomy} For all real numbers $x, y$ exactly one of the following holds: $$ x> y, \quad x=y, \quad \mathrm{or}\quad y>x.$$ \end{axi} \begin{axi}[Transitivity of Order]\label{axi:transitivity} For all real numbers $x, y, z $, $$\mathrm{if}\quad x> y \quad \mathrm{and}\quad y>z \quad \mathrm{then}\quad x>z.$$ \end{axi} \begin{axi}[Preservation of Inequalities by Addition]\label{axi:preservation-of-ineqs-by-addition} For all real numbers $x, y, z $, $$\mathrm{if} \quad x> y \quad \mathrm{then}\quad x+z>y+z.$$ \end{axi} \begin{axi}[Preservation of Inequalities by Positive Factors]\label{axi:preservation-of-ineqs} For all real numbers $x, y, z $, $$\mathrm{if} \quad x> y \quad \mathrm{and}\quad z>0 \quad \mathrm{then}\quad xz>yz.$$ \end{axi} \begin{axi}[Inversion of Inequalities by Negative Factors]\label{axi:preservation-of-ineqs} For all real numbers $x, y, z $, $$\mathrm{if} \quad x> y \quad \mathrm{and}\quad z<0 \quad \mathrm{then}\quad xzx$. $x\leq y$ means that either $y>x$ or $y=x$, etc. \end{rem} The above axioms allow us to solve several inequality problems. \begin{exa}\label{exa:ineq-1} Solve the inequality $$ 2x-3 < -13. $$ \end{exa} \begin{solu} We have $$2x-3 < -13 \implies 2x < -13 + 3 \implies 2x < -10. $$The next step would be to divide both sides by $2$. Since $2>0$, the sense of the inequality is preserved, whence $$2x<-10 \implies x < \dfrac{-10}{2} \implies x<-5. $$ \end{solu} \begin{exa}\label{exa:ineq-2} Solve the inequality $$ -2x-3 \leq -13. $$ \end{exa} \begin{solu} We have $$-2x-3\leq -13 \implies -2x \leq -13 + 3 \implies -2x \leq -10. $$The next step would be to divide both sides by $-2$. Since $-2<0$, the sense of the inequality is inverted, and so $$-2x\leq -10 \implies x \geq \dfrac{-10}{-2} \implies x\geq -5. $$ \end{solu} The method above can be generalised for the case of a product of linear factors. To investigate the set on the line where the inequality \begin{equation} (a_1x+b_1)\cdots (a_nx+b_n)>0, \label{eq:diag-signos} \end{equation}holds, we examine each individual factor. By trichotomy, for every $k$, the real line will be split into the three distinct zones $$ \{x\in \reals : a_kx+b_k > 0\}\cup \{x\in \reals : a_kx+b_k = 0\}\cup\{x\in \reals : a_kx+b_k < 0\}. $$ We will call the real line with punctures at $x=-\dfrac{a_k}{b_k}$ and indicating where each factor changes sign the {\em sign diagram} corresponding to the inequality (\ref{eq:diag-signos}). \begin{exa} Consider the inequality $$ x^2+2x-35<0. $$ \begin{enumerate} \item Form a sign diagram for this inequality. \item Write the set $\{x\in\reals: x^2+2x-35<0\}$ as an interval or as a union of intervals. \item Write the set $\left\{x\in\reals: x^2+2x-35\geq 0\right\}$ as an interval or as a union of intervals. \item Write the set $\left\{x\in\reals: \dfrac{x+7}{x-5}\geq 0\right\}$ as an interval or as a union of intervals. \item Write the set $\left\{x\in\reals: \dfrac{x+7}{x-5}\leq -2\right\}$ as an interval or as a union of intervals. \end{enumerate} \end{exa} \begin{solu} \noindent \begin{enumerate} \item Observe that $x^2+2x-35=(x-5)(x+7)$, which vanishes when $x=-7$ or when $x=5$. In neighbourhoods of $x=-7$ and of $x=5$, we find: $$\begin{array}{|l|l|l|l|}\hline x\in & \oo{-\infty; -7} & \oo{-7;5} & \oo{5;+\infty} \\ \hline x+7 & - & + & + \\ \hline x-5 & - & - & + \\ \hline (x+7)(x-5) & + & - & + \\ \hline \end{array}$$ On the last row, the sign of the product $(x+7)(x-5)$ is determined by the sign of each of the factors $x+7$ and $x-5$. \item From the sign diagram above we see that $$\{x\in\reals: x^2+2x-35<0\} = \oo{-7;5}. $$ \item From the sign diagram above we see that $$\left\{x\in\reals: x^2+2x-35\geq 0\right\}= \oc{-\infty; -7}\cup \co{5;+\infty}. $$ Notice that we include both $x=-7$ and $x=5$ in the set, as $(x+7)(x-5)$ vanishes there. \item From the sign diagram above we see that $$\left\{x\in\reals: \dfrac{x+7}{x-5}\geq 0\right\}= \oc{-\infty; -7}\cup \oo{5;+\infty}. $$ Notice that we include $x=-7$ since $ \dfrac{x+7}{x-5}$ vanishes there, but we do not include $x=5$ since there the fraction $\dfrac{x+7}{x-5}$ would be undefined. \item We must add fractions: $$ \dfrac{x+7}{x-5}\leq -2 \iff \dfrac{x+7}{x-5}+2\leq 0 \iff \dfrac{x+7}{x-5}+\dfrac{2x-10}{x-5}\leq 0 \iff \dfrac{3x-3}{x-5}\leq 0. $$ We must now construct a sign diagram puncturing the line at $x=1$ and $x=5$: $$\begin{array}{|l|l|l|l|}\hline x\in & \oo{-\infty; 1} & \oo{1;5} & \oo{5;+\infty} \\ \hline 3x-3 & - & + & + \\ \hline x-5 & - & - & + \\ \hline \dfrac{3x-3}{x-5} & + & - & + \\ \hline \end{array}$$ We deduce that $$\left\{x\in\reals: \dfrac{x+7}{x-5}\leq -2\right\} = \co{1;5}. $$ Notice that we include $x=1$ since $\dfrac{3x-3}{x-5}$ vanishes there, but we exclude $x=5$ since there the fraction $\dfrac{3x-3}{x-5}$ is undefined. \end{enumerate} \end{solu} \begin{exa} Determine the following set explicitly: $\{x\in\reals : -x^2 + 2x -2\geq 0\}$. \end{exa} \begin{solu} The equation $-x^2 + 2x -2=0$ does not have rational roots. To find its roots we either use the quadratic formula, or we may complete squares. We will use the latter method: $$ -x^2 + 2x -2 = -(x^2-2x) -2 = -(x^2-2x+1)-2+1 = -(x-1)^2-1. $$ Therefore, $$-x^2 + 2x -2\geq 0 \iff -(x-1)^2-1\geq 0 \iff -((x-1)^2+1)\geq 0. $$ This last inequality is impossible for real numbers, as the expression $-((x-1)^2+1)$ is strictly negative. Hence, $$\{x\in\reals : -x^2 + 2x -2\geq 0\} =\varnothing. $$ {\em Aliter:} The discriminant of $-x^2+2x-2$ is $2^2-4(-1)(-2)=-4<0$, which means that the equation has complex roots. Hence the quadratic polynomial keeps the sign of its leading coefficient, $-1$, and so it is always negative. \end{solu} \begin{exa}\label{exa:desi-comp-cuad} Determine explicitly the set $\{x\in\reals : 32x^2-40x+9>0\}$. \end{exa} \begin{solu} The equation $32x^2-40x+9=0$ does not have rational roots. To find its roots we will complete squares: $$\begin{array}{lll} 32x^2-40x+9 & = & 32\left(x^2-\dfrac{5}{4}x+\dfrac{9}{32}\right)\\ & = & 32\left(x^2-\dfrac{5}{4}x + \dfrac{5^2}{8^2}+\dfrac{9}{32}- \dfrac{5^2}{8^2}\right) \\ & = & 32\left(\left(x-\dfrac{5}{8}\right)^2-\dfrac{7}{64}\right)\\ & = & 32\left(x-\dfrac{5}{8}-\dfrac{\sqrt{7}}{8}\right)\left(x-\dfrac{5}{8}+\dfrac{\sqrt{7}}{8}\right) . \end{array}$$ We may now form a sign diagram, puncturing the line at $x=\dfrac{5}{8}-\dfrac{\sqrt{7}}{8}$ and at $x=\dfrac{5}{8}+\dfrac{\sqrt{7}}{8}$: $$\begin{array}{|l|l|l|l|}\hline x\in & \oo{-\infty;\dfrac{5}{8}-\dfrac{\sqrt{7}}{8}} & \oo{\dfrac{5}{8}-\dfrac{\sqrt{7}}{8};\dfrac{5}{8}+\dfrac{\sqrt{7}}{8}} & \oo{\dfrac{5}{8}+\dfrac{\sqrt{7}}{8};+\infty} \\ \hline \left(x-\dfrac{5}{8}+\dfrac{\sqrt{7}}{8}\right) & - & + & + \\ \hline \left(x-\dfrac{5}{8}-\dfrac{\sqrt{7}}{8}\right)& - & - & + \\ \hline \left(x-\dfrac{5}{8}+\dfrac{\sqrt{7}}{8}\right)\left(x-\dfrac{5}{8}-\dfrac{\sqrt{7}}{8}\right) & + & - & + \\ \hline \end{array}$$ We deduce that $$\left\{x\in\reals: 32x^2-40x+9> 0\right\} = \oo{-\infty; \dfrac{5}{8}-\dfrac{\sqrt{7}}{8}} \cup \oo{\dfrac{5}{8}+\dfrac{\sqrt{7}}{8};+\infty}. $$ \end{solu} Care must be taken when transforming an inequality, as a given transformation may introduce spurious solutions. \begin{exa} Solve the inequality $$ 2\sqrt{1-x} - \sqrt{x+1} \geq \sqrt{x}. $$ \end{exa} \begin{solu} For the square roots to make sense, we must have $$ x\in \oc{-\infty;1} \cap \co{-1;+\infty}\cap \co{0;+\infty} \implies x\in \cc{0;1}.$$ Squaring both sides of the inequality, transposing, and then squaring again, $$ 4(1-x)-4\sqrt{1-x^2}+ x+1 > x \implies 5-4x>4\sqrt{1-x^2}\implies 25 -40x+16x^2>16-16x^2 \implies 32x^2-40x+9>0.$$ This last inequality has already been solved in example \ref{exa:desi-comp-cuad}. Thus we want the intersection $$ \left( \oo{-\infty; \dfrac{5}{8}-\dfrac{\sqrt{7}}{8}} \cup \oo{\dfrac{5}{8}+\dfrac{\sqrt{7}}{8};+\infty}\right)\cap \cc{0;1}= \co{0;\frac{5}{8}-\frac{\sqrt{7}}{8}}.$$ \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Consider the set $$ \{x\in\reals: x^2 - x -6 \leq 0\}. $$ \begin{enumerate} \item Draw a sign diagram for this set. \item Using the obtained sign diagram, write the set $$ \{x\in\reals: x^2 - x -6 \leq 0\} $$ as an interval or as a union of intervals. \item Using the obtained sign diagram, write the set $$ \left\{x\in\reals: \dfrac{x-3}{x+2} \geq 0\right\} $$ as an interval or as a union of intervals. \end{enumerate} \begin{answer} \bigskip\noindent \begin{enumerate} \item Observe that $x^2-x-6=(x-3)(x+2)$. Hence, in a neighbourhood of $x=-2$ and $x=3$, we have: $$\begin{array}{|l|l|l|l|}\hline x\in & \oo{-\infty}{-2} & \oo{-2;3} & \oo{3;+\infty} \\ \hline x+2 & - & + & + \\ \hline x-3 & - & - & + \\ \hline (x+2)(x-3) & + & - & + \\ \hline \end{array}$$ \item From the diagram we deduce that the desired set is $\cc{-2;3}$. \item From the diagram we deduce that the desired set is $\oo{-\infty ; -2}\cup \co{3 ; +\infty}$. \end{enumerate} \end{answer} \end{pro} \begin{pro} Write the set $$\left\{ x\in \reals : \dfrac{x^2+x-6}{x^2-x-6} \geq 0\right\}$$ as an interval or as a union of intervals. \begin{answer} $\{x\in\reals: x\in \oc{-\infty; -3}\cup \oc{-2;2}\cup \oo{3;+\infty}\}$. \end{answer} \end{pro} \begin{pro} Give an explicit description of the set $$\{x\in\reals : x^2 - x -4\geq 0\}.$$ \begin{answer} As the equation $x^2 - x -4=0$ does not have rational roots, we complete squares to find its roots: $$ x^2-x-4 = x^2-x+\dfrac{1}{4}-\dfrac{1}{4}-4= \left(x-\dfrac{1}{2}\right)^2-\dfrac{17}{4}=\left(x-\dfrac{1}{2}-\dfrac{\sqrt{17}}{2}\right)\left(x-\dfrac{1}{2}+\dfrac{\sqrt{17}}{2}\right). $$ Therefore $$x^2-x-4 \geq 0\iff \left(x-\dfrac{1}{2}-\dfrac{\sqrt{17}}{2}\right)\left(x-\dfrac{1}{2}+\dfrac{\sqrt{17}}{2}\right)\geq 0. $$ We may now form a sign diagram, puncturing the real line at $x=\dfrac{1}{2}-\dfrac{\sqrt{17}}{2}$ and at $x=\dfrac{1}{2}+\dfrac{\sqrt{17}}{2}$: $$\begin{array}{|l|l|l|l|}\hline x\in & \oo{-\infty; \dfrac{1}{2}-\dfrac{\sqrt{17}}{2}} & \oo{\dfrac{1}{2}-\dfrac{\sqrt{17}}{2};\dfrac{1}{2}+\dfrac{\sqrt{17}}{2}} & \oo{\dfrac{1}{2}+\dfrac{\sqrt{17}}{2};+\infty} \\ \hline \left(x-\dfrac{1}{2}-\dfrac{\sqrt{17}}{2}\right) & - & + & + \\ \hline \left(x-\dfrac{1}{2}+\dfrac{\sqrt{17}}{2}\right)& - & - & + \\ \hline \left(x-\dfrac{1}{2}-\dfrac{\sqrt{17}}{2}\right)\left(x-\dfrac{1}{2}+\dfrac{\sqrt{17}}{2}\right) & + & - & + \\ \hline \end{array}$$ We deduce that $$\left\{x\in\reals: x^2-x-4\geq 0\right\} = \oo{-\infty; \dfrac{1}{2}-\dfrac{\sqrt{17}}{2}} \cup \oo{\dfrac{1}{2}+\dfrac{\sqrt{17}}{2};+\infty}. $$ \end{answer} \end{pro} \begin{pro}Write the set $$\left\{x\in\reals: x^2-x-6\leq 0\right\} \ \cap \ \left\{x\in\reals: \dfrac{1-x}{x+3}\geq 1\right\} $$ in interval notation. \begin{answer} $\cc{-2;-1}$ \end{answer} \end{pro} \begin{pro} Solve the inequality $\sqrt{x^2-4x+3}\geq -x+2$. \begin{answer} $\co{3;+\infty}$ \end{answer} \end{pro} \begin{pro} Solve the inequality $\dfrac{1-\sqrt{1-4x^2}}{x}> \dfrac{1}{2}$. \begin{answer} $\oc{\dfrac{12}{25};\dfrac{1}{2}}$ \end{answer} \end{pro} \begin{pro} Solve the inequality $$\sqrt{2x+1}+\sqrt{2x-5}\geq \sqrt{5-2x}.$$ \begin{answer} $\left\{\dfrac{5}{2}\right\}$ \end{answer} \end{pro} \begin{pro} Find the least positive integer $n$ satisfying the inequality $$ \sqrt{n+1}-\sqrt{n}<\dfrac{1}{10}. $$ \begin{answer} From the identity $x^2-y^2=(x-y)(x+y)$ and using the fact that $\sqrt{n}<\sqrt{n+1}$, we obtain $$ n+1-n=1 \implies (\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})=1 \implies \sqrt{n+1}-\sqrt{n} = \dfrac{1}{\sqrt{n+1}+\sqrt{n}} >\dfrac{1}{2\sqrt{n+1}}. $$ Hence, $$\dfrac{1}{2\sqrt{n+1}} < \dfrac{1}{10} \implies 5 < \sqrt{n+1} \implies 2524. $$ Since $5.1^2 = 26.01>26$, we have $\sqrt{26}<5.1$. Hence, $$ \sqrt{26}-\sqrt{25}< 5.1-5 = \dfrac{1}{10}, $$and so $n=25$ fulfills the inequality. \end{answer} \end{pro} \begin{pro} Determine the values of the real parameter $t$ such that the set $$A_t = \left\{x\in\reals : (t-1)x^2+tx+\dfrac{t}{4}=0\right\} $$ \begin{enumerate} \item be empty; \item have exactly one element; \item have exactly two elements. \end{enumerate} \begin{answer}The quadratic equation will not have any real solutions as long as its determinant be strictly negative: $$ t^2-(t-1)(t)<0 \implies t\in\oo{-\infty;0}.$$ Hence $A_t=\varnothing \iff t\in\oo{-\infty;0}$. The set will have exactly one element either when $t-1=0$, which means that the equation reduces to a linear one, or if the quadratic equation has a repeated root, which occurs when its discriminant vanishes: $$ t^2-(t-1)(t)=0 \implies t=0.$$ Thus the set has exactly one element when $t=1$ and when $t=0$, and it is seen that $$A_1=\{x\in \reals x+\dfrac{1}{4}=0\} =\{-\dfrac{1}{4}\}, \qquad A_0=\{x\in \reals -x^2=0\} =\{0\}. $$ The quadratic equation will have exactly two real solutions when its discriminant is strictly positive: $$t-1 \neq 0, t^2-(t-1)(t)>0 \implies t\in\oo{0;1}\cup \oo{1;+\infty}.$$ In this case, $$A_t = \left\{x\in\reals : (t-1)x^2+tx+\dfrac{t}{4}=0\right\} =\left\{\dfrac{4\sqrt{t}-4t}{8t-8}, \dfrac{-4\sqrt{t}-4t}{8t-8}\right\}. $$ \end{answer} \end{pro} \begin{pro} List the elements of the set $$ \left\{x\in\integers : \min \left(x+2,4-\dfrac{x}{3}\right) \geq 1\right\}. $$ \begin{answer} $\{-1,0,1,2,3,4,5,6,7,8,9\}$ \end{answer} \end{pro} \begin{pro} Demonstrate that for all real numbers $x>0$ it is verified that $$ 2x^3-6x^2+\dfrac{11}{2}x+1>0. $$ \begin{answer} The inequality is obtained at once from $$ 2x^3-6x^2+\dfrac{11}{2}x+1> = x\left(\left(\sqrt{2}x-\dfrac{3}{\sqrt{2}}\right)^2+1\right) +1. $$ \end{answer} \end{pro} \begin{pro} Demonstrate that for all real numbers $x$ it is verified that $$ x^8-x^5+x^2-x+1>0. $$ \begin{answer} Either $x\in \oc{-\infty;0}$, or $x\in \oc{0;1}$, or $x\in \oo{1;+\infty}$. In the first case the inequality is obvious, since for $x\leq 0$ $$ x^8\geq 0, -x^5\geq 0, x^2\geq 0, -x\geq 0 \implies x^8-x^5+x^2-x+1>0. $$ In the second case we regroup $$ x^8-x^5+x^2-x+1 = x^8 +x^2(1-x^3) + (1-x)>0. $$ In the third case we have $$x^8-x^5+x^2-x+1 = x^5(x^3-1)+x(x-1)+1>0 $$ \end{answer} \end{pro} \begin{pro} The values of $a, b, c,$ and $d$ are $1, 2, 3$ and $4$ but not necessarily in that order. What is the largest possible value of $ab + bc + cd + da$? \begin{answer} $25$ \end{answer} \end{pro} \begin{pro} Prove that if $r \geq s \geq t$ then $$ r^2 - s^2 + t^2 \geq (r - s + t)^2. $$ \end{pro} \end{multicols} \section{Absolute Value}\label{sec:abs_val_eqns} We start with a definition. \begin{df} Let $x\in \reals$.\index{absolute value} The {\em absolute value of $x$}---denoted by $|x|$---is defined by \begin{center} \fcolorbox{blue}{white}{ $ |x| = \left\{ \begin{tabular}{ll} $-x$ & \rm{if} \ $x < 0$, \\ $x$ & \rm{if} \ $x \geq 0$. \end{tabular} \right. $} \end{center} \end{df}The absolute value of a real number is thus the distance of that real number to $0$, and hence $|x-y|$ is the distance between $x$ and $y$ on the real line. The absolute value of a quantity is either the quantity itself or its opposite. \begin{exa} Write without absolute value signs: \begin{enumerate} \item $|\sqrt{3} - 2|$, \item $|\sqrt{7} - \sqrt{5}|$, \item $||\sqrt{7} - \sqrt{5}| - |\sqrt{3} - 2||$ \end{enumerate} \end{exa}\begin{solu} We have \begin{enumerate} \item since $2 > 1.74 > \sqrt{3}$, we have $|\sqrt{3} - 2| = 2 - \sqrt{3}$. \item since $\sqrt{7} >\sqrt{5}$, we have $|\sqrt{7} - \sqrt{5}| = \sqrt{7} - \sqrt{5}$. \item by virtue of the above calculations, $$||\sqrt{7} - \sqrt{5}| - |\sqrt{3} - 2|| = |\sqrt{7} - \sqrt{5} - (2 - \sqrt{3})| = |\sqrt{7} + \sqrt{3} -\sqrt{5} - 2|.$$The question we must now answer is whether $\sqrt{7} + \sqrt{3} >\sqrt{5} + 2$. But $\sqrt{7} + \sqrt{3} > 4.38 > \sqrt{5} + 2$ and hence $$|\sqrt{7} + \sqrt{3} -\sqrt{5} - 2| = \sqrt{7} + \sqrt{3} -\sqrt{5} - 2. $$ \end{enumerate} \end{solu} \begin{exa} Let $x>10$. Write $|3-|5-x||$ without absolute values. \end{exa} \begin{solu} We know that $|5-x|= 5-x$ if $5-x\geq 0$ or that $|5-x|= -(5-x)$ if $5-x< 0$. As $x>10$, we have then $|5-x|=x-5$. Therefore $$|3-|5-x|| =|3-(x-5)|=|8-x|. $$In the same manner , either $|8-x|= 8-x$ if $8-x\geq 0$ or $|8-x|= -(8-x)$ if $8-x< 0$. As $x>10$, we have then $|8-x|=x-8$. We conclude that $x>10$, $$ |3-|5-x|| = x-8. $$ \end{solu} The method of sign diagrams from the preceding section is also useful when considering expressions involving absolute values. \begin{exa} Find all real solutions to $|x + 1| + |x + 2| - |x - 3| = 5$. \end{exa} \begin{solu} The vanishing points for the absolute value terms are $x = -1$, $x = -2$ and $x = 3$. Notice that these are the points where the absolute value terms change sign. We decompose $\reals$ into (overlapping) intervals with endpoints at the places where each of the expressions in absolute values vanish. Thus we have $$\reals = ]-\infty; -2] \ \cup \ [-2; -1] \ \cup [-1; 3] \ \cup [3; +\infty[.$$ We examine the sign diagram $$\begin{array}{|l|l|l|l|l|} x\in & ]-\infty; -2] & [-2;-1] & [-1; 3] & [3; +\infty[ \\ \hline |x + 2| = & -x-2 & x + 2 & x + 2 & x + 2 \\ \hline |x + 1| =& -x -1 & -x-1 & x + 1 & x + 1 \\ \hline |x - 3| =& -x + 3 & -x+3 & -x + 3 & x - 3 \\ \hline |x + 2| + |x + 1| - |x - 3| = & -x-6 & x-2 & 3x & x+6 \\ \hline \end{array} $$ Thus on $]-\infty; -2]$ we need $-x-6 = 5$ from where $x = -11$. On $[-2; -1]$ we need $x-2 = 5$ meaning that $x = 7$. Since $7\not\in [-2;-1]$, this solution is spurious. On $[-1;3]$ we need $3x = 5$, and so $x = \dfrac{5}{3}$. On $[3; +\infty[$ we need $x + 6= 5$, giving the spurious solution $x = -1$. Upon assembling all this, the solution set is $$\left\{ -11, \dfrac{5}{3}\right\}.$$ \end{solu} We will now demonstrate two useful theorems for dealing with inequalities involving absolute values. \begin{thm}\label{thm:|x|within_t} Let $t\geq 0$. Then $$|x| \leq t \iff -t \leq x \leq t. $$ \end{thm} \begin{pf}Either $\absval{x} =x$, or $\absval{x}=-x$. If $\absval{x} =x$, $$\absval{x}\leq t \iff x\leq t \iff -t \leq 0\leq x \leq t. $$ If $\absval{x} =-x$, $$\absval{x}\leq t \iff -x\leq t \iff -t \leq x \leq 0 \leq t. $$ \end{pf} \begin{exa} Solve the inequality $|2x-1|\leq 1$. \end{exa} \begin{solu} From theorem \ref{thm:|x|within_t}, $$|2x-1|\leq 1 \iff -1 \leq 2x-1 \leq 1 \iff 0 \leq 2x \leq 2 \iff 0 \leq x \leq 1 \iff x\in \cc{0;1}. $$ The solution set is the interval $\cc{0;1}$. \end{solu} \begin{thm}\label{thm:|x|without_t} Let $t\geq 0$. Then $$|x| \geq t \iff x \geq t \qquad \mathrm{or} \qquad x \leq -t. $$ \end{thm} \begin{pf}Either $\absval{x} =x$, or $\absval{x}=-x$. If $\absval{x} =x$, $$\absval{x}\geq t \iff x\geq t. $$ If $\absval{x} =-x$, $$\absval{x}\geq t \iff -x\geq t \iff x \leq -t. $$ \end{pf} \begin{exa} Solve the inequality $|3+2x|\geq 1$. \end{exa} \begin{solu} From theorem \ref{thm:|x|without_t} , $$|3+2x|\geq 1 \implies 3+2x \geq 1 \qquad \mathrm{or} \qquad 3+2x \leq -1 \implies x\geq -1 \qquad \mathrm{or} \qquad x \leq -2. $$ The solution set is the union of intervals $\oc{-\infty;-2}\cup\co{-1;+\infty}$. \end{solu} \begin{exa} Solve the inequality $|1-|1-x||\geq 1$. \end{exa} \begin{solu} We have $$ |1-|1-x||\geq 1\iff 1-|1-x|\geq 1 \quad \mathrm{or} \quad 1-|1-x|\leq -1. $$ Solving the first inequality, $$ 1-|1-x|\geq 1 \iff -|1-x|\geq 0 \implies x=1, $$since the quantity $-|1-x|$ is always negative. Solving the second inequality, $$ 1-|1-x|\leq -1 \iff -|1-x|\leq -2 \iff |1-x|\geq 2 \iff 1-x \geq 2 \quad \mathrm{or} \quad 1-x \leq -2 \implies x\in \co{3;+\infty}\cup \oc{-\infty;-1} $$and thus $$ \{x\in\reals : |1-|1-x||\geq 1\}= \oc{-\infty;-1} \cup \{1\}\cup \co{3;+\infty}. $$ \end{solu} We conclude this section with a classical inequality involving absolute values. \begin{thm}[Triangle Inequality] Let $a, b$ be real numbers. Then \begin{equation}\fcolorbox{blue}{white}{ $|a + b| \leq |a| + |b|$.}\end{equation} \label{tri_ineq} \end{thm} \begin{pf}Since clearly $-|a| \leq a \leq |a| $ and $-|b| \leq b \leq |b| $, from Theorem \ref{thm:|x|within_t}, by addition, $$-|a| \leq a \leq |a| $$to$$-|b| \leq b \leq |b| $$we obtain $$-(|a| + |b| ) \leq a + b \leq (|a| + |b|),$$whence the theorem follows. \end{pf} \begin{cor} Let $a, b$ be real numbers. Then \begin{equation} \fcolorbox{blue}{white}{$ ||a| - |b|| \leq |a - b|$}.\end{equation} \label{tri_ineq_2} \end{cor} \begin{pf} We have $$|a| = |a - b + b| \leq |a - b| + |b|,$$giving $$|a| - |b| \leq |a - b|.$$Similarly, $$|b| = |b - a + a| \leq |b - a| + |a| = |a - b| + |a|,$$gives $$|b| - |a| \leq |a - b|.$$The stated inequality follows from this. \end{pf} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Write without absolute values: $|\ \sqrt{3} - \sqrt{|2 - \sqrt{15}|} \ |$ \begin{answer} $ \sqrt{3}-\sqrt{\sqrt{15} - 2} $ \end{answer} \end{pro} \begin{pro} Write without absolute values if $x>2$: $|x - |1-2x||$. \begin{answer} For $x > \frac{1}{2}$, we have $ |1-2x| = 2x-1$. Thus $|x-|1-2x|| = |x-(2x-1)| = |-x+1|$. If $x>1$ then $|-x+1| = x-1$. In conclusion, for all $x>1$ (and {\em a fortiori $x>2$}, we have $|x-|1-2x||= x-1$. \end{answer} \end{pro} \begin{pro} If $x < -2$ prove that $|1 - |1 + x|| = -2 - x$. \begin{answer} If $x < -2$ then $1 + x < -1$ and hence $|1 + x| = -(1 + x) = -1 - x$. Thus $|1 - |1 + x|| = |1 - (-1 - x)| = |2 + x|$. But since $x < -2$, $x + 2 < 0$ and so $|2 + x| = -2 - x$. We conclude that $|1 - |1 + x|| = -2 - x.$ \end{answer} \end{pro} \begin{pro} Let $a, b$ be real numbers. Prove that $|ab|=|a||b|$. \end{pro} \begin{pro} Let $a\in\reals$. Prove that $\sqrt{a^2}=|a|$. \end{pro} \begin{pro} Let $a\in\reals$. Prove that $a^2=|a|^2=|a^2|$. \end{pro} \begin{pro} Solve the inequality $|1-2x|<3$. \begin{answer} Se tiene $$ |1-2x|<3 \iff -3 < 1-2x < 3 \iff -4 < -2x < 2 \iff -1 < x <2 \iff x\in \oo{-1;2}, $$ en donde ha of recordarse que el dividir a una desigualdad por una cantidad negativa, se invierte el sentido of la desigualdad. \end{answer} \end{pro} \begin{pro} How many real solutions are there to the equation $$|x^2-4x| = 3\ ?$$ \begin{answer} Four. Either $x^2-4x=-3$ or$x^2-4x=3$. Thus $x\in\{1,3,2-\sqrt{7}, 2+\sqrt{7}\}$. \end{answer} \end{pro} \begin{pro} Solve the following absolute value equations: \begin{enumerate} \item $|x-3| + |x+2| = 3 $, \item $ |x-3| + |x+2| = 5 $, \item $ |x-3| + |x+2| = 7 $. \end{enumerate} \begin{answer} We know that $$ |x-3|=\pm (x-3)\qquad \mathrm{and \quad that}\ \qquad |x+2|=\pm (x+2). $$ We puncture the real line at $x=3$ and at $x=-2$, that is, where the absolute value terms change sign. We have $$\begin{array}{|l|l|l|l|}\hline x\in & \oo{-\infty; -2} & \oo{-2;3} & \oo{3;+\infty} \\ \hline |x-3| =& 3-x & 3-x & x-3 \\ \hline |x+2|= & -x-2 & x+2 & x+2 \\ \hline |x-3| + |x+2|= & -2x+1 & 5 & 2x-1 \\ \hline \end{array}$$ Therefore , $$|x-3| + |x+2|= \left\{ \begin{array}{ll} -2x+1 & \mathrm{if}\ \ x \leq -2, \\ 5 & \mathrm{if}\ \ -2\leq x \leq 3, \\ 2x-1 & \mathrm{if}\ \ x \geq 3. \end{array} \right. $$ \begin{enumerate} \item To solve $|x-3| + |x+2| = 3 $, we need $$-2x+1 =3 \quad \mathrm{if}\ x \leq -2, \quad 5=3 \quad \mathrm{if}\ -2 \leq x \leq 3, \quad 2x-1=3 \quad \mathrm{if}\ x \geq 3. $$ The first equation gives $x=-1$. As $-1\nin\oc{-\infty ; -2}$, this is a spurious solution. The second equation is a contradiction. In the third equation, $2x-1=3 \implies x=2$, which is also spurious since $2\nin \co{3;+\infty}$. Therefore the equation $|x-3| + |x+2| = 3 $ does not have real solutions. \item To solve $|x-3| + |x+2| = 3 $, we need $$-2x+1 =5 \quad \mathrm{if}\ x \leq -2, \quad 5=5 \quad \mathrm{if}\ -2 \leq x \leq 3, \quad 2x-1=5 \quad \mathrm{if}\ x \geq 3. $$ The first equation gives $x=-2$. As $-2\in\oc{-\infty ; -2}$, which is a legitimate solution. The second equation is a tautology, which means that all the elements in the interval $\cc{-2;3}$ are solutions. In the third equation $2x-1=5 \implies x=3$, which is also a legitimate solution, since $3\in \co{3;+\infty}$. Therefore the equation $|x-3| + |x+2| = 5 $ has an infinite number of real solutions, all in the interval $\cc{-2;3}$.\item To solve $|x-3| + |x+2| = 7 $, we need $$-2x+1 =7 \quad \mathrm{if}\ x \leq -2, \quad 5=7 \quad \mathrm{if}\ -2 \leq x \leq 3, \quad 2x-1=7 \quad \mathrm{if}\ x \geq 3. $$ The first equation gives $x=-3$. As $-3\in\oc{-\infty ; -2}$, this is a legitimate solution. The second equation is a contradiction. In the third equation $2x-1=7 \implies x=4$, which is also a legitimate solution since $4\in \co{3;+\infty}$. Therefore the equation $|x-3| + |x+2| = 3 $ has exactly two real solutions: $$ \left\{x\in \reals : |x-3|+|x+2|=7\right\} = \{-3,4\}. $$ \end{enumerate} \end{answer} \end{pro} \begin{pro} Find all the real solutions of the equation $$x^2-2|x+1|-2=0.$$ \begin{answer} $x\in\{-2,1+\sqrt{5}\}$ \end{answer} \end{pro} \begin{pro} Find all the real solutions to $|5x - 2| = |2x + 1|$. \begin{answer} We have $$\begin{array}{lll}|5x - 2| = |2x + 1| & \iff & (5x - 2 = 2x + 1) \ \mathrm{or}\ (5x - 2 = -(2x + 1)) \\ & \iff & (x = 1) \ \mathrm{or}\ (x = \dfrac{1}{7}) \\ & \iff & x\in \left\{\dfrac{1}{7}, 1\right\} \end{array}$$ \end{answer} \end{pro} \begin{pro} Find all real solutions to $|x - 2| + |x - 3| = 1$. \begin{answer} The first term vanishes when $x = 2$ and the second term vanishes when $x = 3$. We decompose $\reals$ into (overlapping) intervals with endpoints at the places where each of the expressions in absolute values vanish. Thus we have $$\reals = \oc{-\infty; 2} \ \cup \ \cc{2; 3} \ \cup \co{3; +\infty}.$$ We examine the sign diagram $$\begin{array}{|l|l|l|l|} x\in & \oc{-\infty; 2} & \cc{2;3} & \co{3; +\infty} \\ \hline |x - 2| = & -x+2 & x - 2 & x - 2 \\ \hline |x - 3| =& -x + 3 & -x+3 & x - 3 \\ \hline |x - 2| + |x - 3| = & -2x + 5 & 1 & 2x - 5 \\ \hline \end{array} $$ Thus on $\oc{-\infty; 2}$ we need $-2x + 5 = 1$ from where $x = 2$. On $\cc{2; 3}$ we obtain the identity $1 = 1$. This means that all the numbers on this interval are solutions to this equation. On $\co{3; +\infty}$ we need $2x - 5 = 1$ from where $x = 3$. Upon assembling all this, the solution set is $\{ x: x\in\cc{2; 3}\}.$ \end{answer} \end{pro} \begin{pro} Find the set of solutions to the equation $$|x| + |x - 1| = 2.$$ \begin{answer} $\{-\frac{1}{2}, \frac{3}{2}\}$ \end{answer} \end{pro} \begin{pro} Find the solution set to the equation $$|x| + |x - 1| = 1.$$ \begin{answer} $\{x|x \in \cc{0; 1}\}$ \end{answer} \end{pro} \begin{pro} Find the solution set to the equation $$|2x| + |x - 1| - 3|x + 2| = 1.$$ \begin{answer} $\{-1\}$ \end{answer} \end{pro} \begin{pro} Find the solution set to the equation $$|2x| + |x - 1| - 3|x + 2| = -7.$$ \begin{answer} $\co{1;+\infty}$ \end{answer} \end{pro} \begin{pro} Find the solution set to the equation $$|2x| + |x - 1| - 3|x + 2| = 7.$$ \begin{answer} $\oc{-\infty; -2}$ \end{answer} \end{pro} \begin{pro} If $x < 0$ prove that $\left|x - \sqrt{(x - 1)^2}\right| = 1 - 2x.$ \end{pro} \begin{pro} Find the real solutions, if any, to $|x^2 - 3x| = 2$. \begin{answer} $\{\frac{3}{2}+\frac{\sqrt{17}}{2},\frac{3}{2}-\frac{\sqrt{17}}{2} , 1, 2\}$ \end{answer} \end{pro} \begin{pro} Find the real solutions, if any, to $x^2 -2|x| + 1 = 0$. \begin{answer} $\{-1,1\}$ \end{answer} \end{pro} \begin{pro} Find the real solutions, if any, to $x^2 -|x| - 6 = 0$. \begin{answer} $\{-3,-2, 2, 3\}$ \end{answer} \end{pro} \begin{pro} Find the real solutions, if any, to $x^2 = |5x - 6|$. \begin{answer} $\{-6, 1,2,3\}$ \end{answer} \end{pro} \begin{pro} Prove that if $x\leq -3$, then $|x+3| - |x-4|$ is constant. \begin{answer} We have $$|x+3| = \left\{ \begin{tabular}{ll} $-x-3$ & \rm{if} \ $x+3 < 0$, \\ $x+3$ & \rm{if} \ $x+3 \geq 0$. \end{tabular} \right. \qquad |x-4| = \left\{ \begin{tabular}{ll} $-x+4$ & \rm{if} \ $x-4 < 0$, \\ $x-4$ & \rm{if} \ $x-4 \geq 0$. \end{tabular} \right. $$This means that when $x<-3$ $$|x+3|- |x-4| = (-x-3) - (-x+4) = -7, $$a constant. Since at $x=-3$ we also obtain $-7$, the result holds true for the larger interval $x\leq -3$. \end{answer} \end{pro} \begin{pro}Solve the equation $$\left|\dfrac{2x}{x-1}\right| = |x+1|.$$ \begin{answer} There are four solutions: $\{-1-\sqrt{2}, -1+\sqrt{2}, 1-\sqrt{2}, -1+\sqrt{2}\}$. \end{answer} \end{pro} \begin{pro}Write the set $$\{x\in\reals: |x+1|-|x-2|=-3\}$$ in interval notation. \begin{answer} $\oc{-\infty;-1}$. \end{answer} \end{pro} \begin{pro}Let $x, y$ real numbers. Demonstrate that the maximum and the minimum of $x$ and $y$ are given by $$\max (x, y) = \dfrac{x+y+\absval{x-y}}{2}$$ and $$\min (x, y) = \dfrac{x+y-\absval{x-y}}{2}.$$ \begin{answer}Clearly, $\max (x, y) + \min (x, y) = x+y$. Now, either $\absval{x-y} = x-y$ and so $x\geq y$, which signifies that $\max (x, y)-\min (x, y) = x-y$, or $\absval{x-y} = -(x-y)=y-x$, which means that $y\geq x$ and thus $\max (x, y)-\min (x, y) = y-x$. At any rate, $\max(x,y)-\min (x, y)=\absval{x-y}$. Solving the system of equations $$\begin{array}{lll}\max (x, y) + \min (x, y) & = & x+y\\ \max(x,y)-\min (x, y) & = &\absval{x-y},\\ \end{array} $$for $\max(x,y)$ and $\min (x,y)$, we obtain the result. \end{answer} \end{pro} \begin{pro} Solve the inequality $|x-1||x+2|>4$. \begin{answer} $\{x\in\reals : |x-1||x+2|>4 \}=\oc{-\infty;-3}\cup\co{2;+\infty}$ \end{answer} \end{pro} \begin{pro} Solve the inequality $\dfrac{|2x^2-1|}{x^2-x-2} >\dfrac{1}{2}$. \begin{answer} $\oo{-4;-1}\cup \oo{2;5}$ \end{answer} \end{pro} \end{multicols} \section{Completeness Axiom} The alert reader may have noticed that the smaller set of rational numbers satisfies all the arithmetic axioms and order axioms of the real numbers given in the preceding sections. Why then, do we need the larger set $\reals$? In this section we will present an axiom that characterises the real numbers. \begin{df} A number $u$ is an {\em upper bound} for a set of numbers $A$ if for all $a\in A$ we have $a \leq u$. The smallest such upper bound is called the {\em supremum} of the set $A$. Similarly, a number $l$ is a {\em lower bound} for a set of numbers $B$ if for all $b\in B$ we have $l \leq b$. The largest such lower bound is called the {\em infimum} of the set $B$. \end{df} The real numbers have the following property, which further distinguishes them from the rational numbers. \begin{axi}[Completeness of $\reals$] Any set of real numbers which is bounded above has a supremum. Any set of real numbers which is bounded below has a infimum. \index{Completeness Axiom} \end{axi} \vspace{1cm} \begin{figure}[!hptb] $$\psset{unit=.5} \psaxes[yAxis=false, ](0, 0)(-7,0)(7, 0) \psline[linestyle=dotted]{->}(7,0)(11,0) \psline[linestyle=dotted]{->}(-7,0)(-11,0) \uput[dr](11,0){+\infty}\uput[dl](-11,0){-\infty} $$\meinecaption{1}{The Real Line.} \label{fig:the_real_line} \end{figure} \bigskip Observe that the rational numbers are not complete. For example, there is no largest rational number in the set $$\{x\in\rationals: x^2<2\}$$ since $\sqrt{2}$ is irrational and for any good rational approximation to $\sqrt{2}$ we can always find a better one. \bigskip Geometrically, each real number can be viewed as a point on a straight line. We make the convention that we orient the real line with $0$ as the origin, the positive numbers increasing towards the right from $0$ and the negative numbers decreasing towards the left of $0$, as in figure \ref{fig:the_real_line}. The Completeness Axiom says, essentially, that this line has no ``holes.'' \bigskip We append the object $+\infty$, which is larger than any real number, and the object $-\infty$, which is smaller than any real number. Letting $x\in\reals$, we make the following conventions. \begin{equation} (+\infty) + (+\infty) = +\infty \end{equation} \begin{equation} (-\infty) + (-\infty) = -\infty \end{equation} \begin{equation} x + (+\infty) = +\infty \end{equation} \begin{equation} x + (-\infty) = -\infty \end{equation} \begin{equation} x(+\infty) = +\infty\ \ \ \mathrm{if} \ x> 0 \end{equation} \begin{equation} x(+\infty) = -\infty\ \ \ \mathrm{if} \ x< 0 \end{equation} \begin{equation} x(-\infty) = -\infty\ \ \ \mathrm{if} \ x> 0 \end{equation} \begin{equation} x(-\infty) = +\infty\ \ \ \mathrm{if} \ x< 0 \end{equation} \begin{equation} \label{eq:1/big}\dfrac{x}{\pm \infty} = 0 \end{equation} Observe that we leave the following undefined: $$\dfrac{\pm \infty}{\pm \infty}, \ \ \ \ (+\infty) + (-\infty), \ \ \ 0(\pm \infty). $$ \chapter{The Plane} \section{Sets on the Plane} \begin{df} Let $A , B$, be subsets of real numbers. Their {\em Cartesian Product} $A\times B$ is defined and denoted by $$A\times B = \{(a, b): a\in A, b\in B\},$$ that is, the set of all ordered pairs whose elements belong to the given sets. \end{df} \begin{rem} In the particular case when $A=B$ we write $$A\times A = A^2.$$ \end{rem} \begin{exa} If $A = \{-1, -2\}$ and $B = \{-1, 2\}$ then $$A\times B = \{(-1, -1), (-1, 2), (-2, -1), (-2,2 ))\}, $$ $$B\times A = \{(-1,-1), (-1, -2), (2, -1), (2, -2)\}, $$ $$A^2 = \{(-1, -1), (-1, -2), (-2,-1), (-2, -2)\}, $$ $$B^2 = \{(-1, -1), (-1, 2), (2, -1), (2, 2)\}. $$ Notice that these sets are all different, even though some elements are shared. \end{exa} \begin{exa} $(-1, 2)\in \integers^2$ but $(-1, \sqrt{2})\not\in \integers^2$. \end{exa} \begin{exa} $(-1, \sqrt{2})\in \integers\times\reals$ but $(-1, \sqrt{2})\not\in \reals\times\integers$. \end{exa} \begin{df} $\reals^2 = \reals \times \reals$---the {\em real Cartesian Plane}---- is the set of all ordered pairs $(x, y)$ of real numbers. \end{df} We represent the elements of $\reals^2$ graphically as follows. Intersect perpendicularly two copies of the real number line. These two lines are the {\em axes}. Their point of intersection---which we label $O = (0, 0)$--- is the {\em origin}. To each point $P$ on the plane we associate an ordered pair $P = (x, y)$ of real numbers. Here $x$ is the {\em abscissa}\footnote{From the Latin {\em linea abscissa} or line cut-off.}, which measures the horizontal distance of our point to the origin, and $y$ is the {\em ordinate}, which measures the vertical distance of our point to the origin. The points $x$ and $y$ are the {\em coordinates} of $P$. This manner of dividing the plane and labelling its points is called the {\em Cartesian coordinate system.} The horizontal axis is called the $x$-{\em axis} and the vertical axis is called the $y$-{\em axis}. It is therefore sufficient to have two numbers $x$ and $y$ to completely characterise the position of a point $P = (x, y)$ on the plane $\reals^2$. \begin{df} Let $a\in \reals$ be a constant. The set $$ \{(x,y)\in\reals ^2: x=a\} $$is a vertical line. \end{df} \begin{df} Let $b\in \reals$ be a constant. The set $$ \{(x,y)\in\reals ^2: y=b\} $$ is a horizontal line. \end{df} Figures \ref{fig: line -vertical} and \ref{fig: line -horizontal} give examples of vertical and horizontal lines. \vspace{2cm} \begin{figure}[h] \begin{minipage}{4.5cm} $$ \psset{unit=.6pc} \psaxes[labels=none, linewidth=1.2pt]{<->}(0,0)(-7,-7)(7,7) \psline[linewidth=2pt,linecolor=brown]{<->}(3,-6.5)(3,6.5)$$ \meinecaption{1.5}{Line $x=3$.}\label{fig: line -vertical} \end{minipage} \hfill \begin{minipage}{4.5cm} $$ \psset{unit=.6pc} \psaxes[labels=none,linewidth=1.2pt]{<->}(0,0)(-7,-7)(7,7) \psline[linewidth=2pt,linecolor=brown]{<->}(-6.5,-1)(6.5, -1)$$ \meinecaption{1.5}{Line $y=-1$.}\label{fig: line -horizontal} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.8pc} \psaxes[, linewidth=1.5pt,labelFontSize=\tiny]{<->}(0,0)(-5.5,-5.5)(5.5,5.5) \pscustom[linestyle=dashed, fillstyle=solid, fillcolor=brown]{\psline[linestyle=dashed](1, 4)(1, 2)(3, 2)(3, 4)(1, 4) } $$ \meinecaption{1.5}{Example \ref{exa:square_2}. }\label{fig:square_2} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.8pc} \rput(0,-1){ \psaxes[, linewidth=1.5pt,labelFontSize=\tiny]{<->}(0,0)(-5.5,-5.5)(5.5,5.5)\pscustom[linestyle=solid, fillstyle=solid, fillcolor=brown]{\psline[linestyle=solid](1, -3)(3, -3)(3, 5)(2.75,5.4)(2.5,5)(2.25,5.4)(2,5)(1.75,5.4)(1.5,5)(1.25,5.4)(1, 5)(1, -3) }} $$ \meinecaption{1.5}{Example \ref{exa:strip_1}}. \label{fig:strip_1} \end{minipage} \end{figure} \begin{exa} Draw the Cartesian product of intervals $$\mathscr{R} = \oo{1;3}\times \oo{2;4}= \{(x, y)\in\reals^2: 1 < x < 3,\quad 2 < y < 4\}.$$ \label{exa:square_2}\end{exa}\begin{solu} The set is bounded on the left by the vertical line $x=1$ and bounded on the right by the vertical line $x=3$, excluding the lines themselves. The set is bounded above by the horizontal line $y=4$ and below by the horizontal line $y=2$, excluding the lines themselves. The set is thus a square minus its boundary, as in figure \ref{fig:square_2}. \end{solu} \begin{exa} Sketch the region $$\mathscr{R} = \{(x, y)\in\reals^2: 1 < x < 3,\quad 2 < y < 4\}.$$ \label{exa:square_2}\end{exa} \begin{solu} The region is a square, excluding its boundary. The graph is shewn in figure \ref{fig:square_2}, where we have dashed the boundary lines in order to represent their exclusion. \end{solu} \begin{exa}\label{exa:strip_1} The region $$\mathscr{R} = \cc{1;3}\ \times\ \co{-3;+\infty}$$ is the infinite half strip on the plane sketched in figure \ref{fig:strip_1}. The boundary lines are solid, to indicate their inclusion. The upper boundary line is toothed, to indicate that it continues to infinity. \end{exa} \begin{exa} A quadrilateral has vertices at $A = (5, -9)$,$B = (2, 3),$ $C = (0, 2),$ and $D = (-8, 4)$. Find the area, in square units, of quadrilateral $ABCD$. \label{polygon}\end{exa} \begin{solu} Enclose quadrilateral $ABCD$ in right $\triangle AED$, and draw lines parallel to the $y$-axis in order to form trapezoids $AEFB$, $FBCG$, and right $\triangle GCD$, as in figure \ref{polygon_1}. The area $[ABCD]$ of quadrilateral $ABCD$ is thus given by $$\renewcommand{\arraystretch}{1.5} \begin{array}{lll} [ABCD] & = & [AED] - [AEFB] - [FBCG] - [GCD] \\ & = & \frac{1}{2}(AE)(DE) - \frac{1}{2}(FE)(FB + AE) - \\ & & \qquad - \frac{1}{2}(GF)(GC + FB) - \frac{1}{2}(DG)(GC) \\ & = & \frac{1}{2}(13)(13) - \frac{1}{2}(3)(13 + 1) - \frac{1}{2}(2)(2 + 1) - \frac{1}{2}(8)(2) \\ & = & 84.5 - 21 - 3 - 8 \\ & = & 52.5. \end{array} $$ \end{solu} \vspace{2cm} \begin{figure}[!hptb] $$\psset{unit=.75pc} \pspolygon[linecolor=brown, linewidth=2pt](-8, 4)(0, 2)(2, 3)(5, -9) \psline[linecolor=brown, linewidth=2pt](5,-9)(5,4)(-8,4) \psline[linecolor=brown, linewidth=2pt](2,4)(2,3) \psline[linecolor=brown, linewidth=2pt](0,4)(0,2) \psaxes[labelFontSize=\tiny, linewidth=2pt](0,0)(-11,-11)(11,11) \psdots[dotstyle=*,dotscale=1](-8, 4)(0, 2)(2, 3)(5, -9)(-8,4)(5,4)(2,4)(0,4) \uput[90](-8,4){D}\uput[200](0,2){C}\uput[0](2, 3){B}\uput[0](5, - 9){A} \uput[0](5,4){E}\uput[90](2,4){F}\uput[60](0,4){G} $$ \meinecaption{3}{Example \ref{polygon}.}\label{polygon_1} \end{figure} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Sketch the following regions on the plane. \begin{enumerate} \item $R_1 = \{(x, y)\in\reals^2: x \leq 2\}$ \item $R_2 = \{(x, y)\in\reals^2: y \geq -3\}$ \item $R_3 = \{(x, y)\in\reals^2: |x| \leq 3, |y| \leq 4\}$ \item $R_4 = \{(x, y)\in\reals^2: |x| \leq 3, |y| \geq 4\}$ \item $R_5 = \{(x, y)\in\reals^2: x \leq 3, y \geq 4\}$ \item $R_6 = \{(x, y)\in\reals^2: x \leq 3, y \leq 4\}$ \end{enumerate} \end{pro} \begin{pro} Find the area of $\triangle ABC$ where $A = (-1, 0)$, $B = (0, 4)$ and $C = (1, -1)$. \begin{answer} $4.5$ square units. \end{answer} \end{pro} \begin{pro} Let $A = [-10; 5], B = \{5, 6, 11\}$ and $C = ]-\infty ; 6[$. Answer the following true or false. \begin{multicols}{2} \begin{enumerate} \item $5\in A$. \item $6 \in C$. \item $(0, 5, 3) \in A\times B\times C$. \item $(0, -5, 3) \in A\times B\times C$. \item $(0, 5, 3) \in C\times B\times C$. \item $A\times B\times C \subseteq C\times B\times C$. \item $A\times B\times C \subseteq C^3$. \end{enumerate} \end{multicols} \begin{answer} TFTFTTF \end{answer} \end{pro} \begin{pro} True or false: $(\reals\setminus \{0\})^{2} = \reals^2 \setminus \{(0,0)\}$. \begin{answer} False. $(\reals\setminus \{0\})^{2}$ consists of the plane minus the axes. $\reals^2 \setminus \{(0,0)\}$ consists of the plane minus the origin. \end{answer} \end{pro} \end{multicols} \section{Distance on the Real Plane} In this section we will deduce some important formul\ae\ from analytic geometry. Our main tool will be the Pythagorean Theorem from elementary geometry.\vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{4cm} \centering \psset{unit=1.5pc} \pstGeonode[PointName=none](-2,-2){A}(2,2){B}(2,-2){C} \pstLineAB[linecolor=brown,linewidth=2pt]{A}{B}\pstLineAB[linecolor=blue,linewidth=2pt,linestyle=dotted]{A}{C} \pstLineAB[linecolor=blue,linewidth=2pt,linestyle=dotted]{B}{C} \uput[ur](2,2){$B(x_2,y_2)$} \uput[dl](-2,-2){$A(x_1,y_1)$} \uput[dr](2,-2){$C(x_2,y_1)$} \uput[d](0,-2){$\absval{x_2-x_1}$} \uput[r](2,0){$\absval{y_2-y_1}$} \meinecaption{2}{Distance between two points.}\label{fig:distance_formula} \end{minipage} \hfill \begin{minipage}{4cm} \centering \psset{unit=1.5pc} \pstGeonode[PointName=none](-2,-2){A}(2,2){B}(2,-2){C} \pstLineAB[linecolor=brown,linewidth=2pt]{A}{B}\pstLineAB[linecolor=blue,linewidth=2pt,linestyle=dotted]{A}{C} \pstLineAB[linecolor=blue,linewidth=2pt,linestyle=dotted]{B}{C} \uput[ur](B){$B(x_2,y_2)$} \uput[dl](A){$A(x_1,y_1)$} \uput[dr](C){$C(x_2,y_1)$} \pstMiddleAB[PointName=none]{A}{B}{M_C} \pstMiddleAB[PosAngle=0]{B}{C}{M_A} \pstMiddleAB[PosAngle=-90]{C}{A}{M_B} \uput[ul](M_C){$(x,y)$} \psline(M_B)(M_C)(M_A) \meinecaption{2}{Midpoint of a line segment.}\label{fig: point -medio} \end{minipage} \hfill \begin{minipage}{4cm} \centering \psset{unit=1.5pc} \pstGeonode[PointName=none](-2,-2){A}(2,2){B}(2,-2){C} \pstLineAB[linecolor=brown,linewidth=2pt]{A}{B}\pstLineAB[linecolor=blue,linewidth=2pt,linestyle=dotted]{A}{C} \pstLineAB[linecolor=blue,linewidth=2pt,linestyle=dotted]{B}{C} \uput[ur](B){$B(x_2,y_2)$} \uput[dl](A){$A(x_1,y_1)$} \uput[dr](C){$C(x_2,y_1)$} \pstTranslation[DistCoef=.3,PosAngle=-30]{A}{B}{A}[P] \pcline[offset=12pt,arrows={|-|}](A)(P)\ncput*[nrot=:U]{$m$} \pcline[offset=12pt,arrows={|-|}](P)(B)\ncput*[nrot=:U]{$n$} \pstProjection[PosAngle=-90]{A}{C}{P}[Q]\pstProjection[PosAngle=0]{B}{C}{P}[R] \psline(Q)(P)(R) \meinecaption{2}{Division of a segment.}\label{fig:joachimstal} \end{minipage} \end{figure} \begin{thm}[Distance Between Two Points on the Plane] The distance between the points $A = (x_1, y_1), B = (x_2, y_2)$ in $\reals ^2$ is given by $$ \fcolorbox{blue}{white}{$AB= \distance{(x_1, y_1)}{(x_2, y_2)} := \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}.$} $$ \end{thm} \begin{pf} Consider two points on the plane, as in figure \ref{fig:distance_formula}. Constructing the segments ${CA}$ and $BC$ with $C=(x_2,y_1)$, we may find the length of the segment ${AB}$, that is, the distance from $A$ to $B$, by utilising the Pythagorean Theorem: $$ AB^2 = AC^2 + BC^2 \implies AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2.}$$ \end{pf} \begin{exa} The point $(x, 1)$ is at distance $\sqrt{11}$ from the point $(1, -x)$. Find all the possible values of $x$. \end{exa} \begin{solu} We have, $$\begin{array}{cccc} & \distance{(x, 1)}{(1, -x)} & = & \sqrt{11} \\ \Longleftrightarrow & \sqrt{(x - 1)^2 + (1 + x)^2} & = & \sqrt{11} \\ \Longleftrightarrow & (x - 1)^2 + (1 + x)^2 & = & 11 \\ \Longleftrightarrow & 2x^2 + 2 & = & 11. \\ \end{array} $$ Hence, $x = -\frac{3\sqrt{2}}{2}$ or $x = \frac{3\sqrt{2}}{2}$. \end{solu} \begin{exa} Find the point equidistant from $A = (-1, 3), B = (2, 4)$ and $C = (1, 1)$. \end{exa} \begin{solu} Let $(x, y)$ be the point sought. Then $$\distance{(x, y)}{(-1, 3)} = \distance{(x, y)}{(2, 4)} \implies (x + 1)^2 + (y - 3)^2 = (x - 2)^2 + (y - 4)^2,$$ $$\distance{(x, y)}{(-1, 3)} = \distance{(x, y)}{(1,1)} \implies (x + 1)^2 + (y - 3)^2 = (x - 1)^2 + (y - 1)^2.$$ Expanding, we obtain the following linear equations: $$2x + 1 - 6y + 9 = -4x + 4 - 8y + 16,$$ $$2x + 1 - 6y + 9 = -2x + 1 - 2y + 1,$$ or $$6x + 2y = 10,$$ $$4x - 4y = -8.$$ We easily find that $(x, y) = \left(\frac{3}{4}, \frac{11}{4}\right).$ \end{solu} \begin{exa}\label{exa:lattice-triangle} We say that a point $(x, y)\in \reals^2$ is a {\em lattice point} if $x\in \integers$ and $y\in \integers$. Demonstrate that no equilateral triangle on the plane may have its three vertices as lattice points.\end{exa} \begin{solu}Since a triangle may be translated with altering its angles, we may assume, without loss of generality, that we are considering $\triangle ABC$ with $A(0,0)$, $B(b,0)$, $C(m, n)$, with integers $b>0$, $m>0$ and $n>0$, as in figure \ref{fig:lattice-triangle}. If $\triangle ABC$ were equilateral , then $$ AB=BC=CA \implies b=\sqrt{(m-b)^2+n^2}=\sqrt{m^2+n^2}. $$ Squaring and expanding, $$ b^2=m^2-2bm+b^2+n^2=m^2+n^2. $$ From $BC=CA$ we deduce that $$-2bm+b^2 =0 \implies b(b-2m)\implies b=2m, $$ as we are assuming that $b>0$. Hence, $$ b^2 = m^2+n^2 =\dfrac{b^2}{4}+n^2 \implies n=\dfrac{\sqrt{3}}{2}b. $$ Since we are assuming that $b\neq 0$, $n$ cannot be an integer, since $\sqrt{3}$ is irrational. \end{solu} \vspace{2cm} \begin{figure}[!hptb] \centering\psset{unit=.7pc} \pstGeonode[PointName=none](0,0){A}(4,0){B}(4;60){C} \psaxes[linewidth=2pt,labels=none,ticks=none,arrows={->}](0,0)(-7,-7)(7,7) \pspolygon[linewidth=2pt,linecolor=brown](A)(B)(C)\psdots(A)(B)(C) \uput[u](C){\tiny$(m,n)$} \uput[d](B){\tiny$(b,0)$} \meinecaption{2}{Example \ref{exa:lattice-triangle}.}\label{fig:lattice-triangle} \end{figure} \begin{thm}[Midpoint of a Line Segment] The point $\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$ lies on the line joining $A(x_1,y_1)$ and $B(x_2,y_2)$, and it is equidistant from both points.\label{thm: point -medio} \end{thm} \begin{pf}First observe that it is easy to find the midpoint of a vertical or horizontal line segment. The interval $\cc{a;b}$ has length $b-a$. Hence, its midpoint is at $a+\dfrac{b-a}{2}=\dfrac{a+b}{2}$. \bigskip Let $(x,y)$ be the midpoint of the line segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$. With $C(x_2,y_1)$, form the triangle $\triangle ABC$, right-angled at $C$. From $(x,y)$, consider the projections of this point onto the line segments $AC$ and $BC$. Notice that these projections are parallel to the legs of the triangle and so these projections pass through the midpoints of the legs. Since $AC$ is a horizontal segment, its midpoint is at $M_B=(\frac{x_1+x_2}{2},y_1)$. As $BC$ is a horizontal segment, its midpoint is $M_A=(x_2, \frac{y_1+y_2}{2})$. The result is obtained on noting that $(x,y)$ must have the same abscissa as $M_B$ and the same ordinate as $M_A$. \end{pf} In general, we have the following result. \begin{thm}[Joachimstal's Formula] The point $P$ which divides the line segment $AB$, with $A(x_1,y_1)$ and $B(x_2,y_2)$, into two line segments in the ratio $m:n$ has coordinates $$\left(\dfrac{nx_1+mx_2}{m+n}, \dfrac{ny_1+my_2}{m+n}\right).$$ \end{thm} \begin{pf}The proof proceeds along the lines of Theorem \ref{thm: point -medio}. First we consider the interval $\cc{a;b}$. Suppose that $a 0 $ units per second and simultaneously, the car at point $B$ travels upwards at constant speed, at a rate of $b > 0$ units per second. How many units apart are these cars after $t > 0$ seconds? \begin{answer} $\sqrt{4x^2 + (a + b)^2t^2}$ \end{answer} \end{pro} \begin{pro}Point $C$ is at $\dfrac{3}{5}$ of the distance from $A(1,5)$ to $B(4,10)$ on the segment $AB$ (and closer to $B$ than to $A$). Find $C$. \begin{answer} $C$ is the point that divides $AB$ in the ratio $3:2$. By Joachimstal's formula, $$C=\left(\dfrac{2\cdot 1 + 3\cdot 4}{3+2}, \dfrac{2\cdot 5 + 3\cdot 10}{3+2}\right) = \left(\dfrac{14}{5},8\right). $$ \end{answer} \end{pro} \begin{pro} For which value of $x$ is the point $(x,1)$ at distance $2$ del from the point $(0,2)$? \begin{answer} We have $$ \sqrt{(x-0)^2+(1-2)^2}=2 \implies \sqrt{x^2+1}=2 \implies x^2+1=4 \implies x^2=3 \implies x=\pm \sqrt{3}. $$ \end{answer} \end{pro} \begin{pro} A bug starts at the point $(-1,-1)$ and wants to travel to the point $(2,1)$. In each quadrant, and on the axes, it moves with unit speed, except in quadrant II, where it moves with half the speed. Which route should the bug take in order to minimise its time? The answer is {\bf not} a straight line from $(-1,-1)$ to $(2,1)$! \begin{answer} The bug should travel along two line segments: first from $(-1,1)$ the origin, and then from the origin to $(2,1)$ avoiding quadrant II altogether. For, if $a>0, b>0$ then the line segment joining $(-b, 0)$ and $(a, 0)$ lies in quadrant II, it is $\sqrt{a^2+b^2}$ long, and the bug spends an amount of time equal to $\dfrac{\sqrt{a^2+b^2}}{2}$ on this line. But a path on the axes from $(-b, 0)$ to $(a, 0)$ is $a+b$ units long and the bug spends an amount of time equal to $a+b$ there. Thus as long as $$a+b \leq \dfrac{a^2+b^2}{2}$$the bug should avoid quadrant II completely. But by the Arithmetic-Mean-Geometric-Mean Inequality we have $$2ab \leq a^2 + b^2 \implies (a+b)^2 = a^2 + 2ab + b^2 \leq 2a^2+2b^2 \implies a+b \leq \sqrt{2}\sqrt{a^2+b^2}, $$which means that as long as the speed of the bug in quadrant II is $< \dfrac{1}{\sqrt{2}}$ then the bug will better avoid quadrant II. Since $\dfrac{1}{2}<\dfrac{1}{\sqrt{2}}$, this follows in our case. \end{answer} \end{pro} \begin{pro} Find the point equidistant from $(-1, 0)$, $(1, 0)$ and $(0, 1/2)$. \begin{answer} $(0, -3/4)$ \end{answer} \end{pro} \begin{pro} Find the coordinates of the point symmetric to $(a, b)$ with respect to the point $(b, a)$. \begin{answer} $\dis{\left(2b - a, 2a - b\right)}$ \end{answer} \end{pro} \begin{pro} Demonstrate that the diagonals of a rectangle are congruent. \begin{answer} Without loss of generality assume that the rectangle $ABCD$ has vertices at $A(0, 0)$, $B(u,0)$, $C(u, v)$ and $D(0,v)$. Its diagonals are $AC$ and $BD$, which results in $$AC =\sqrt{(u-0)^2+(v-0)^2} =\sqrt{u^2+v^2}, $$and $$BD =\sqrt{(u-0)^2+(0-v)^2} =\sqrt{u^2+v^2}, $$demonstrating their equality. \end{answer} \end{pro} \begin{pro}Prove that the diagonals of a parallelogram bisect each other.. \begin{answer}Without loss of generality, assume that the parallelogram $ABCD$ has vertices at $A(0, 0)$, $B(u,0)$, $C(u+w, v)$ and $D(w,v)$. The coordinates of the midpoint of the segment $AC$ are $\left(\dfrac{u+w}{2},\dfrac{v}{2}\right)$, which are the coordinates of the midpoint of $BD$, demonstrating the result. \end{answer} \end{pro} \begin{pro} A fly starts at the origin and goes $1$ unit up, $1/2$ unit right, $1/4$ unit down, $1/8$ unit left, $1/16$ unit up, etc., {\em ad infinitum.} In what coordinates does it end up? \begin{answer} Its $x$ coordinate is $$\frac{1}{2} - \frac{1}{8} + \frac{1}{32} - \cdots = \frac{\frac{1}{2}}{1 - \frac{-1}{4}} = \frac{2}{5}.$$ Its $y$ coordinate is $$1 - \frac{1}{4} + \frac{1}{16} - \cdots = \frac{1}{1 - \frac{-1}{4}} = \frac{4}{5}.$$Therefore, the fly ends up in $$\left(\frac{2}{5}, \frac{4}{5}\right).$$ Here we have used the fact the sum of an infinite geometric progression with common ratio $r$, with $|r|<1$ and first term $a$ is $$ a + ar + ar^2 + ar^3 + \cdots = \dfrac{a}{1-r}. $$ \end{answer} \end{pro} \begin{pro} Find the coordinates of the point which is a quarter of the way from $(a, b)$ to $(b, a)$. \begin{answer} $(\frac{3a + b}{4}, \frac{3b + a}{4})$ \end{answer} \end{pro} \begin{pro} Find the coordinates of the point symmetric to $(-a, b)$ with respect to: (i) the $x$-axis, (ii) the $y$-axis, (iii) the origin. \begin{answer} $(a, b)$ ; $(-a, -b)$ ; $(a, -b)$ \end{answer} \end{pro} \begin{pro}[Minkowski's Inequality]Prove that if $(a, b), (c, d) \in \reals^2$, then $$\sqrt{(a + c)^2 + (b + d)^2} \leq \sqrt{a^2 + b^2} + \sqrt{c^2 + d^2}.$$Equality occurs if and only if $ad = bc.$ \label{minkowski} \begin{answer} It is enough to prove this in the case when $a, b, c, d$ are all positive. To this end, put $O = (0,0)$, $L = (a, b)$ and $M = (a+c, b+d)$. By the triangle inequality $OM \leq OL + LM$, where equality occurs if and only if the points are collinear. But then $$\sqrt{(a+c)^2 + (b+d)^2} = OM \leq OL + LM = \sqrt{a^2+b^2}+ \sqrt{c^2+d^2},$$and equality occurs if and only if the points are collinear, that is $\dfrac{a}{b} = \dfrac{c}{d}$. \end{answer} \end{pro} \begin{pro} Prove the following generalisation of Minkowski's Inequality. If $(a_k, b_k) \in (\reals\setminus \{0\})^{2}, 1 \leq k \leq n$, then $$\sum _{k = 1} ^n \sqrt{a_k ^2 + b_k ^2} \geq \sqrt{\left(\sum _{k = 1} ^n a_k\right)^2 + \left(\sum _{k = 1} ^n b_k\right)^2}. $$Equality occurs if and only if $$\frac{a_1}{b_1} = \frac{a_2}{b_2} = \cdots = \frac{a_n}{b_n}.$$ \end{pro} \begin{pro}[AIME 1991] Let $P = \{a_1, a_2, \ldots, a_n\}$ be a collection of points with $$0 < a_1 < a_2 < \cdots < a_n < 17.$$ Consider $$S_n = \min _P \sum _{k = 1} ^n\sqrt{(2k - 1)^2 + a_k ^2},$$where the minimum runs over all such partitions $P$. Shew that exactly one of $S_2, S_3, \ldots , S_n, \ldots$ is an integer, and find which one it is. \begin{answer} Use the above generalisation of Minkowski's Inequality and the fact that $17^2 + 144^2= 145^2$. The desired value is $S_{12}$. \end{answer} \end{pro} \end{multicols} \section{Circles} The distance formula gives an algebraic way of describing points on the plane. \begin{thm}The equation of a circle with radius $R>0$ and centre $(x_0,y_0)$ is \begin{equation} (x - x_0)^2 + (y - y_0)^2 = R^2.\end{equation} This is called the {\em canonical equation} of the circle with centre $((x_0,y_0))$ and radius $R$.\label{eq:canonical_circle} \end{thm} \begin{pf} The point $(x, y)$ belongs to the circle with radius $R>0$ if and only if its distance from the centre of the circle is $R$. This requires $$ \begin{array}{llll} \iff & \distance{(x, y)}{(x_0, y_0)} & = & R \\ \iff & \sqrt{(x - x_0)^2 + (y - y_0)^2} & = & R \\ \iff & (x - x_0)^2 + (y - y_0)^2 & = & R^2\\ \end{array},$$obtaining the result. See figure \ref{fig:equation-circle}.\end{pf} \vspace*{2cm} \begin{figure}[!hptb] \begin{minipage}{4cm} \centering \psset{unit=1pc} \pstGeonode[PointName=none](0,0){O}(2,2){A} \pstCircleOA{O}{A} \pstLineAB[arrows={->},arrowsize=.5]{O}{A}\uput[d](0,0){$(x_0,y_0)$} \uput[dr](1,1){$R$}\meinecaption{3}{The circle.}\label{fig:equation-circle} \end{minipage} \hfill \begin{minipage}{4cm} \centering \psset{unit=1pc} \psaxes[linewidth=1.5pt,labelFontSize=\tiny]{<->}(0,0)(-6.5,-6.5)(6.5,6.5) \pscircle[linewidth=2pt, linecolor=brown](-1,2){3} \psdots[dotstyle=*,dotscale=1](-1,2)(2,2)(-4,2)(-1,-1)(-1,5) \meinecaption{3}{Example \ref{ex:circle_1}. } \label{fig:circle_1} \end{minipage} \hfill \begin{minipage}{4cm} \centering \psset{unit=1pc} \psaxes[arrows={->},linewidth=1.7pt,labelFontSize=\tiny](0,0)(-6.5,-6.5)(6.5,6.5) \pstGeonode[PointName=none](-1,3){T} \rput(T){\pscircle[linewidth=2pt,linecolor=brown](0,0){2}\psdots(2,0)(-2,0)(0,2)(0,-2)} \meinecaption{3}{ Example \ref{exa:circulo-completar-cuad}.}\label{fig:circulo-completar-cuad} \end{minipage} \end{figure} \begin{exa} The equation of the circle with centre $(-1, 2)$ and radius $3$ is $(x + 1)^2 + (y - 2)^2 = 9$. Observe that the points $(-1 \pm 3, 2)$ and $(-1, 2\pm 3)$ are on the circle. Thus $(-4, 2)$ is the left-most point on the circle, $(2,2)$ is the right-most, $(-1, -1)$ is the lower-most, and $(-1, 5)$ is the upper-most. The circle is shewn in figure \ref{fig:circle_1}.\label{ex:circle_1}\end{exa} \begin{exa}\label{exa:circulo-completar-cuad} Trace the circle of equation $$ x^2+2x+y^2-6y=-6.$$ \end{exa} \begin{solu} Completing squares, $$ x^2+2x+y^2-6y=-6 \implies x^2+2x{\bf +1}+y^2-6y{\bf +9}=-6{\bf +1+9} \implies (x+1)^2+(y-3)^2=4, $$from where we deduce that the centre of the circle is $(-1,3)$ and the radius is $2$. The point $(-1+2,3)=(1,3)$ lies on the circle, two units to the right of the centre. The point $(-1-2,3)=(-3,3)$ lies on the circle, two units to the left of the centre. The point $(-1,3+2)=(-1,5)$ lies on the circle, two unidades above the centre. The point $(-1,3-2)=(-1,1)$ lies on the circle, two unidades below the centre. See figure \ref{fig:circulo-completar-cuad}. \end{solu} \begin{exa}\label{exa:circle_given_diameter} A diameter of a circle has endpoints $(-2, -1)$ and $(2,3)$. Find the equation of this circle and graph it. \end{exa} \begin{solu} The centre of the circle lies on the midpoint of the diameter, thus the centre is $\dis{\left(\dfrac{-2+2}{2}, \dfrac{-1+3}{2}\right) = (0,1)}$. The equation of the circle is $$x^2 + (y - 1)^2 = R^2. $$ To find the radius, we observe that $(2,3)$ lies on the circle, thus $$ 2^2 + (3 - 1)^2 = R^2 \implies R = 2\sqrt{2}. $$ The equation of the circle is finally $$ x^2 + (y - 1)^2 = 8. $$Observe that the points $(0\pm 2\sqrt{2},1)$, $(0,1\pm 2\sqrt{2})$, that is, the points \mbox{$(2\sqrt{2},1)$}, \mbox{$(-2\sqrt{2},1)$}, \mbox{$(0, 1+2\sqrt{2})$}, \mbox{$(0, 1-2\sqrt{2})$}, \mbox{$(-2,-1)$}, and \mbox{$(2, 3)$} all lie on the circle. The graph appears in figure \ref{fig:circle_given_diameter}. \end{solu} \begin{exa}\label{exa:region-del-circulo} Draw the plane region $$\{(x,y)\in\reals ^2: x^2 + y^2 \leq 4, \quad \absval{x}\geq 1 \}.$$ \end{exa} \begin{solu}Observe that $|x|\geq 1$ $ \iff x\geq 1$ o $x \leq -1$. The region lies inside the circle with centre $(0,0)$ and radius $2$, to the right of the vertical line $x=1$ and to the left of the vertical line $x=-1$. See figure \ref{fig:region-del-circulo}. \end{solu} \begin{exa} Find the equation of the circle passing through $(1, 1)$, $(0, 1)$ and $(1, 2).$ \label{exa:circle_through_3_points}\end{exa} \begin{solu} Let $(h, k)$ be the centre of the circle. Since the centre is equidistant from $(1,1)$ and $(0,1)$, we have, $$(h - 1)^2 + (k - 1)^2 = h^2 + (k - 1)^2, \implies h^2-2h+1 =h^2 \implies h=\dfrac{1}{2}. $$ Since he centre is equidistant from $(1,1)$ and $(1,2)$, we have, $$(h - 1)^2 + (k - 1)^2 = (h - 1)^2 + (k - 2)^2 \implies k^2-2k+1=k^2-4k+4 \implies k=\dfrac{3}{2}.$$ The centre of the circle is thus $(h, k) = (\frac{1}{2}, \frac{3}{2}).$ The radius of the circle is the distance from its centre to any point on the circle, say, to $(0, 1)$: $$\sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{3}{2} - 1\right)^2} = \frac{\sqrt{2}}{2}.$$The equation sought is finally $$\left(x - \frac{1}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 = \frac{1}{2}.$$ See figure \ref{fig:circle_through_3_points}. \end{solu} \vspace{3cm} \begin{figure}[!hptb] \begin{minipage}{4cm} \centering \psset{unit=1pc} \psgrid[subgriddiv=0,griddots=5,gridlabels=0.5](-5,-5)(5,5) \pstGeonode[PointName=none](-2,-1){A}(2,3){B}(0,1){O} \pstCircleAB[PointName=none,linewidth=2pt,linecolor=brown]{A}{B} \meinecaption{3}{ Example \ref{fig:circle_given_diameter}.}\label{fig:circle_given_diameter} \end{minipage} \hfill \begin{minipage}{4cm} \centering \psset{unit=1pc} \psgrid[subgriddiv=0,griddots=5,gridlabels=0.5](-5,-5)(5,5) \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(2,0){A}(-1,-1){l}(-1,1){l'}(1,-1){r}(1,1){r'} \pstCircleOA[linewidth=2pt]{O}{A} \pstInterLC[PointName=none]{l}{l'}{O}{A}{M}{M'} \pstInterLC[PointName=none]{r}{r'}{O}{A}{N}{N'} \pscustom[fillstyle=solid,fillcolor=brown]{\pstArcOAB{O}{M'}{M}\pstLineAB{M}{M'}} \pscustom[fillstyle=solid,fillcolor=brown]{\pstArcOAB{O}{N}{N'}\pstLineAB{N'}{N}} \meinecaption{3}{ Example \ref{exa:region-del-circulo}.}\label{fig:region-del-circulo} \end{minipage} \hfill \begin{minipage}{4cm} \centering \psset{unit=1pc} \psgrid[subgriddiv=0,griddots=5,gridlabels=0.5](-5,-5)(5,5) \pstGeonode[PointName=none,PointSymbol=none](1,1){A}(0,1){B}(1,-2){C} \pstCircleABC[PointName=none,linewidth=2pt,linecolor=brown]{A}{B}{C}{O} \meinecaption{3}{ Example \ref{exa:circle_through_3_points}.}\label{fig:circle_through_3_points} \end{minipage} \end{figure} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Prove that the points $(4, 2)$ and $(-2, -6)$ lie on the circle with centre at $(1, -2)$ and radius $5$. Prove, moreover, that these two points are diametrically opposite. \end{pro} \begin{pro} A diameter $AB$ of a circle has endpoints $A = (1, 2)$ and $B = (3, 4)$. Find the equation of this circle. \begin{answer} $(x - 2)^2 + (y - 3)^2 = 2$ \end{answer} \end{pro} \begin{pro} Find the equation of the circle with centre at $(-1, 1)$ and passing through $(1, 2).$ \label{ex:circle_4}\begin{answer} We must find the radius of this circle. Since the radius is the distance from the centre of the circle to any point on the circle, we see that the required radius is $$\sqrt{(-1 - 1)^2 + (1 - 2)^2} = \sqrt{5} \approx 2.236.$$The equation sought is thus $$(x + 1)^2 + (y - 1)^2 = 5.$$ \end{answer} \end{pro} \begin{pro} Rewrite the following circle equations in canonical form and find their centres $C$ and their radius $R$. Draw the circles. Also, find at least four points belonging to each circle.\\ \begin{enumerate} \item $x^2 + y^2 - 2y = 35$, \item $x^2 + 4x + y^2 - 2y = 20$, \item $x^2 + 4x + y^2 - 2y = 5$, \item $2x^2 - 8x + 2y^2 = 16$, \item $4x^2 + 4x + \frac{15}{2} + 4y^2 - 12y = 0$ \item $3x^2 + 2x\sqrt{3} + 5 + 3y^2 - 6y\sqrt{3} = 0$ \end{enumerate} \begin{answer} (1) $x^2 + (y - 1)^2 = 36, C = (0, 1), R = 6$. (2) $(x + 2 )^2 + (y - 1)^2 = 25, C = (-2, 1), R = 5$, (3) $(x + 2)^2 + (y - 1 )^2 = 10, C = (-2, 1), R = \sqrt{10}$, (4) $(x - 2)^2 + y^2 = 12, C = (2, 0), R = 2\sqrt{3}$ (5) $(x + \frac{1}{2})^2 + (y - \frac{3}{2} )^2 = \frac{5}{8}, C = (-\frac{1}{2}, \frac{3}{2}), R = \sqrt{\frac{5}{8}} $, (6) $(x + \frac{1}{\sqrt{3}})^2 + (y - \frac{\sqrt{3}}{3})^2 = \frac{5}{3}, C = (-\frac{\sqrt{3}}{3}, \sqrt{3}), R = \sqrt{\frac{5}{3}} $ \end{answer} \end{pro} \begin{pro} Let $$R_1 = \{(x, y)\in\reals^2| x^2 + y^2 \leq 9\},$$ $$R_2 = \{(x, y)\in\reals^2| (x + 2)^2 + y^2 \leq 1\},$$ $$R_3 = \{(x, y)\in\reals^2| (x - 2)^2 + y^2 \leq 1\},$$ $$R_4 = \{(x, y)\in\reals^2| x^2 + (y + 1)^2 \leq 1\},$$ $$R_5 = \{(x, y)\in\reals^2| |x| \leq 3, |y| \leq 3\},$$ $$R_6 = \{(x, y)\in\reals^2| |x| \geq 2, |y| \geq 2\}.$$ Sketch the following regions. \begin{enumerate} \item $R_1\setminus (R_2 \cup R_3 \cup R_4)$. \item $R_5 \setminus R_1$ \item $R_1 \setminus R_6$ \item $R_2 \cup R_3 \cup R_6$ \end{enumerate} \end{pro} \begin{pro} Find the equation of the circle passing through $(-1, 2)$ and centre at $(1, 3)$. \begin{answer} $(x - 1)^2 + (y - 3)^2 = 5$ \end{answer} \end{pro} \begin{pro} Find the canonical equation of the circle passing through $(-1, 1)$, $(1, -2)$, and $(0, 2)$. \begin{answer} $(x - \frac{9}{10})^2 + (y - \frac{1}{10})^2 = \frac{221}{50} $ \end{answer} \end{pro} \begin{pro} Let $a,b,c$ be real numbers with $a^2 > 4b$. Construct a circle with diameter at the points $(1,0)$ and $(-a, b)$. Shew that the intersection of this circle with the $x$-axis are the roots of the equation $x^2 + ax + b = 0$. Why must we impose $a^2 > 4b$? \end{pro} \begin{pro}\label{pro:face} Draw $$(x^2+y^2-100)((x-4)^2+y^2-4)((x+4)^2+y^2-4)(x^2+(y+4)^2-4)=0. $$ \begin{answer}This is asking to draw the circles $x^2+y^2=100$, $(x+4)^2+y^2=4$, $(x-4)^2+y^2=4$, $x^2+(y+4)^2=4$, all in the same set of axes.The picture appears in figure \ref{fig:face}. \vspace*{2cm} \begin{figurehere} \centering \psset{unit=.4pc} \psaxes[linewidth=2pt,labels=none](0,0)(-10.5,-10.5)(10.5,10.5) \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(4,0){O1}(-4,0){O2}(0,-4){O3} \pscircle[linewidth=2pt,linecolor=brown](O){10} \pscircle[linewidth=2pt,linecolor=blue](O1){2} \pscircle[linewidth=2pt,linecolor=blue](O2){2} \pscircle[linewidth=2pt,linecolor=brown](O3){2} \meinecaption{2}{Problem \ref{pro:face}.}\label{fig:face} \end{figurehere} \end{answer} \end{pro} \end{multicols} \section{Semicircles} Given a circle of centre $(a, b)$ and radius $R>0$, its canonical equation is $$ (x-a)^2+(y-b)^2=R^2. $$ Solving for $y$ we gather $$(y-b)^2=R^2- (x-a)^2 \implies y=b\pm \sqrt{R^2- (x-a)^2 }. $$ If we took the $+$ sign on the square root, then the values of $y$ will lie above the line $y=b$, and hence $y =b+ \sqrt{R^2- (x-a)^2 } $ is the equation of the upper semicircle with centre at $(a, b)$ and radius $R>0$. Also, $y =b- \sqrt{R^2- (x-a)^2 } $ is the equation of the lower semicircle. \bigskip In a similar fashion, solving for $x$ we obtain, $$(x-a)^2=R^2- (y-b)^2 \implies x=a\pm \sqrt{R^2- (y-b)^2 }. $$ Taking the $+$ sign on the square root, the values of $x$ will lie to the right of the line $x=a$, and hence $x =a+ \sqrt{R^2- (y-b)^2 } $ is the equation of the right semicircle with centre at $(a, b)$ and radius $R>0$. Similarly, $x =a- \sqrt{R^2- (y-b)^2 } $ is the equation of the left semicircle. \vspace*{2cm} \begin{figure}[h] \begin{minipage}{5cm} \centering \psset{unit=1pc} \psaxes[arrows=->,linewidth=2pt,labelFontSize=\tiny](0,0)(-5,-5)(5,5) \pstGeonode[PointName=none](-0,0){O}(-1,0){A}(1,0){B} \pstArcOAB[linecolor=brown,linewidth=2pt]{0}{B}{A} \meinecaption{2}{ Example \ref{exa:upper_semicircle}.}\label{fig:upper_semicircle} \end{minipage} \hfill \begin{minipage}{5cm} \centering \psset{unit=1pc} \psaxes[arrows=->,linewidth=2pt,labelFontSize=\tiny](0,0)(-5,-5)(5,5) \pstGeonode[PointName=none](-3,1){C}(-5,1){A}(-1,1){B} \pstArcOAB[linecolor=brown,linewidth=2pt]{C}{A}{B} \meinecaption{2}{ Example \ref{exa:semicircle1}.}\label{fig:semicircle1} \end{minipage} \hfill \begin{minipage}{5cm} \centering \psset{unit=1pc} \psaxes[arrows=->,linewidth=2pt,labelFontSize=\tiny](0,0)(-5,-5)(5,5) \pstGeonode[PointName=none](-1,1){C}(-1,4){A}(-1,-2){B} \pstArcOAB[linecolor=brown,linewidth=2pt]{C}{A}{B} \meinecaption{2}{ Example \ref{exa:semicircle2}.}\label{fig:semicircle2} \end{minipage} \end{figure} \begin{exa}\label{exa:upper_semicircle} Figure \ref{fig:upper_semicircle} shews the upper semicircle $y=\sqrt{1-x^2}$. \end{exa} \begin{exa}\label{exa:semicircle1} Draw the semicircle of equation $y =1-\sqrt{-x^2-6x-5}$. \end{exa} \begin{solu} Since the square root has a minus sign, the semicircle will be a lower semicircle, lying below the line $y=1$. We must find the centre and the radius of the circle . For this, let us complete the equation of the circle by squaring and rearranging. This leads to $$\begin{array}{lll} y =1-\sqrt{-x^2-6x-5} & \implies & y-1 =-\sqrt{-x^2-6x-5}\\ & \implies & (y-1)^2=-x^2-6x-5 \\ & \implies & x^2+6x+{\bf 9}+(y-1)^2=-5+{\bf 9}\\ & \implies & (x+3)^2+(y-1)^2=4,\\ \end{array}$$whence the semicircle has centre at $(-3,1)$ and radius $2$. Its graph appears in figure \ref{fig:semicircle1}. \end{solu} \begin{exa}\label{exa:semicircle2} Find the equation of the semicircle in figure \ref{fig:semicircle2}. \end{exa} \begin{solu} The semicircle has centre at $(-1,1)$ and radius $3$. The full circle would have equation $$(x+1)^2+(y-1)^2=9. $$Since this is a left semicircle, we must solve for $x$ and take the minus $-$ on the square root: $$(x+1)^2+(y-1)^2=9\implies (x+1)^2=9-(y-1)^2 \implies x+1=-\sqrt{9-(y-1)^2} \implies x=-1-\sqrt{9-(y-1)^2} ,$$ whence the equation sought is $x=-1-\sqrt{9-(y-1)^2}$. \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{pro} Sketch the following curves. \begin{enumerate} \item $y = \sqrt{16 - x^2}$ \item $x = -\sqrt{16 - y^2}$ \item $x = -\sqrt{12 - 4y - y^2}$ \item $x = -5 - \sqrt{12 + 4y - y^2}$ \end{enumerate} \end{pro} \begin{pro}\label{pro:face2} Draw $$(x^2+y^2-100)(y-\sqrt{4-(x+4)^2})(y-\sqrt{4-(x-4)^2})(y+4+\sqrt{4-x^2})=0. $$ \begin{answer}This is asking to draw the circle $x^2+y^2=100$, and the semicircles $y=\sqrt{4-(x+4)^2}$, $y=\sqrt{4-(x-4)^2}$, $y=-4-\sqrt{4-x^2}$, all in the same set of axes.The picture appears in figure \ref{fig:face}. \vspace*{2cm} \begin{figurehere} \centering \psset{unit=.4pc} \psaxes[linewidth=2pt,labels=none](0,0)(-10.5,-10.5)(10.5,10.5) \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(4,0){O1}(-4,0){O2}(0,-4){O3} \pscircle[linewidth=2pt,linecolor=brown](O){10} \psarc[linewidth=2pt,linecolor=blue](O1){2}{0}{180} \psarc[linewidth=2pt,linecolor=blue](O2){2}{0}{180} \psarc[linewidth=2pt,linecolor=brown](O3){2}{180}{360} \meinecaption{2}{Problem \ref{pro:face2}.}\label{fig:face2} \end{figurehere} \end{answer} \end{pro} \section{Lines}\label{sec:lines} In the previous sections we saw the link {\em Algebra to Geometry} by giving the equation of a circle and producing its graph, and conversely, the link {\em Geometry to Algebra} by starting with the graph of a circle and finding its equation. This section will continue establishing these links, but our focus now will be on lines. \bigskip We have already seen equations of vertical and horizontal lines. We give their definition again for the sake of completeness. \begin{df} Let $a$ and $b$ be real number constants. A {\em vertical line} on the plane is a set of the form $$\{(x, y)\in \reals^2: x = a\}. $$Similarly, a {\em horizontal line} on the plane is a set of the form $$\{(x, y)\in \reals^2: y = b\}.$$ \end{df} \vspace{3cm} \begin{figure}[!hptb] \begin{minipage}{4.5cm} $$ \psset{unit=1pc} \psaxes[labels=none, ticks=none,linewidth=1.2pt]{<->}(0,0)(-7,-7)(7,7) \psline[linewidth=2pt,linecolor=brown]{<->}(3,-6.5)(3,6.5)$$ \vspace{2cm}\meinecaption{.25}{A vertical line.} \end{minipage} \hfill \begin{minipage}{4.5cm} $$ \psset{unit=1pc} \psaxes[labels=none, ticks=none,linewidth=1.2pt]{<->}(0,0)(-7,-7)(7,7) \psline[linewidth=2pt,linecolor=brown]{<->}(-6.5,-1)(6.5, -1)$$ \vspace{2cm}\meinecaption{.25}{A horizontal line.} \end{minipage} \hfill \begin{minipage}{4.5cm} $$ \psset{unit=4pc} \rput(-5, -2.5){\psline[linewidth=2pt]{*-*}(3, 1.5)(6,4.5) \psdots[dotstyle=*,dotscale=2](3, 1.5)(4, 2.5)(6,4.5)\uput{8pt}[l](3, 1.5){(x_1,y_1)} \uput{8pt}[l](4, 2.5){(x, y)} \uput{8pt}[l](6,4.5){(x_2,y_2)} \psline[linestyle=dashed,linewidth=2pt](3,1.5)(6,1.5)(6,4.5) \psline[linestyle=dashed, linewidth=2pt](4,2.5)(4,1.5) \pcline[offset=10pt]{|-|}(6.5, 1.5)(6.5, 4.5) \lput*{:U}{y_2 - y_1} \pcline[offset=10pt]{|-|}(4.5, 1.5)(4.5, 2.5)\lput*{:U}{y - y_1} \pcline[offset=10pt]{|-|}(3, 1)(4, 1) \lput*{:U}{x - x_1} \pcline[offset=10pt]{|-|}(3, .7)(6, .7)\lput*{:U}{x_2 - x_1}} $$ \vspace{2cm}\meinecaption{.25}{Theorem \ref{thm:graph_mx+k}.}\label{fig:graph_mx+k} \end{minipage} \end{figure} \begin{thm}\label{thm:graph_mx+k} The equation of any non-vertical line on the plane can be written in the form $y = mx + k$, where $m$ and $k$ are real number constants. Conversely, any equation of the form $y = ax + b$, where $a, b$ are fixed real numbers has as a line as a graph.\index{line!equation of a} \end{thm} \begin{pf} If the line is parallel to the $x$-axis, that is, if it is horizontal, then it is of the form $y = b$, where $b$ is a constant and so we may take $m = 0$ and $k = b$. Consider now a line non-parallel to any of the axes, as in figure \ref{fig:graph_mx+k}, and let $(x,y)$, $(x_1, y_1)$, $(x_2, y_2)$ be three given points on the line. By similar triangles we have $$\dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{y - y_1}{x - x_1}, $$which, upon rearrangement, gives $$y = \left(\dfrac{y_2 - y_1}{x_2 - x_1}\right)x -x_1\left(\dfrac{y_2 - y_1}{x_2 - x_1}\right) + y_1, $$ and so we may take $$m = \dfrac{y_2 - y_1}{x_2 - x_1}, \ \ \ k =-x_1\left(\dfrac{y_2 - y_1}{x_2 - x_1}\right) + y_1. $$ \bigskip Conversely, consider real numbers $x_1 < x_2 < x_3$, and let $P= (x_1, ax_1 + b)$, $Q= (x_2, ax_2 + b)$, and $R= (x_3, ax_3 + b)$ be on the graph of the equation $y=ax+b$. We will shew that $$\distance{P}{Q} + \distance{Q}{R} = \distance{P}{R}. $$ Since the points $P, Q, R$ are arbitrary, this means that any three points on the graph of the equation $y=ax+b$ are collinear, and so this graph is a line. Then $$ \distance{P}{Q} = \sqrt{(x_2 - x_1)^2 + (ax_2 - ax_1)^2} = |x_2 - x_1|\sqrt{1 + a^2} = (x_2 - x_1)\sqrt{1 + a^2}, $$ $$ \distance{Q}{R} = \sqrt{(x_3 - x_2)^2 + (ax_3 - ax_2)^2} = |x_3 - x_2|\sqrt{1 + a^2} = (x_3 - x_2)\sqrt{1 + a^2}, $$ $$ \distance{P}{Q} = \sqrt{(x_3 - x_1)^2 + (ax_3 - ax_1)^2} = |x_3 - x_1|\sqrt{1 + a^2} = (x_3 - x_1)\sqrt{1 + a^2}, $$ from where $$\distance{P}{Q} + \distance{Q}{R} = \distance{P}{R} $$follows. This means that the points $P, Q,$ and $R$ lie on a straight line, which finishes the proof of the theorem. \end{pf} \begin{df} The quantity $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ in Theorem \ref{thm:graph_mx+k} is the {\em slope or gradient of the line passing through $(x_1,y_1)$ and $(x_2,y_2)$}. Since $y = m(0) + k$, the point $(0,k)$ is the {\em $y$-intercept of the line joining $(x_1,y_1)$ and $(x_2,y_2)$}. \index{line!slope of a}\index{line!intercept}\index{line!gradient of a}Figures \ref{fig:m_pos} through \ref{fig:m_infy} shew how the various inclinations change with the sign of $m$. \end{df} \vspace{2cm} \begin{figure}[h] \begin{minipage}{.24\textwidth} \centering\psset{unit=1pc} \psaxes[linewidth=2pt,arrows={->},labels=none,ticks=none](0,0)(-5,-5)(5,5) \psplot[linewidth=2pt,linecolor=brown,algebraic,arrows={<->}]{-5}{5}{1+x/3} \meinecaption{2}{$m>0$}\label{fig:m_pos} \end{minipage}\hfill \begin{minipage}{.24\textwidth} \centering\psset{unit=1pc} \psaxes[linewidth=2pt,arrows={->},labels=none,ticks=none](0,0)(-5,-5)(5,5) \psplot[linewidth=2pt,linecolor=brown,algebraic,arrows={<->}]{-5}{5}{1-x/3} \meinecaption{2}{$m<0$}\label{fig:m_neg} \end{minipage}\hfill \begin{minipage}{.24\textwidth} \centering\psset{unit=1pc} \psaxes[linewidth=2pt,arrows={->},labels=none,ticks=none](0,0)(-5,-5)(5,5) \psplot[linewidth=2pt,linecolor=brown,algebraic,arrows={<->}]{-5}{5}{1} \meinecaption{2}{$m=0$} \label{fig:m_0} \end{minipage}\hfill \begin{minipage}{.24\textwidth} \centering\psset{unit=1pc} \psaxes[linewidth=2pt,arrows={->},labels=none,ticks=none](0,0)(-5,-5)(5,5) \psline[linewidth=2pt,linecolor=brown,arrows={<->}](-3,-5)(-3,5) \meinecaption{2}{$m=\infty$} \label{fig:m_infy} \end{minipage} \end{figure} \begin{exa}\label{ex:y=x} By Theorem \ref{thm:graph_mx+k}, the equation $y=x$ represents a line with slope $1$ and passing through the origin. Since $y=x$, the line makes a $45^\circ$ angle with the $x$-axis, and bisects quadrants I and III. See figure \ref{fig:y=x} \end{exa} \vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{5cm} \centering\psset{unit=1pc} \psline[linewidth=2pt, linecolor=brown]{<->}(-4.5, -4.5)(4.5, 4.5) \psaxes[labels=none,linewidth=2pt]{<->}(0, 0)(-5, -5)(5, 5) \meinecaption{2}{Example \ref{ex:y=x}.}\label{fig:y=x} \end{minipage} \hfill \begin{minipage}{5cm} \centering\psset{unit=1pc}\rput(-1,-2){ \psline[linewidth=2pt, linecolor=brown]{<->}(-1, 6.6667)(4, -1.6667) \psaxes[labels=none,linewidth=2pt]{<->}(0, 0)(-2, -2)(5, 7) \psdots[dotstyle=*,dotscale=1](0, 5)(3, 0)} \meinecaption{2}{Example \ref{ex:line1}.}\label{fig:line1} \end{minipage} \hfill \begin{minipage}{5cm} \centering\psset{unit=1pc} \psaxes[labels=none,linewidth=2pt]{<->}(0, 0)(-5, -5)(5, 5) \psplot[linewidth=2pt, linecolor=brown,algebraic,arrows={<->}]{-5}{6}{.25*x-1.25} \psdots(-3,-2)(1,-1)(4,-.25) \meinecaption{2}{Example \ref{exa:colli_pts}.}\label{fig:colli_pts} \end{minipage} \hfill \end{figure} \begin{exa}\label{ex:line1} A line passes through $(-3, 10)$ and $(6, -5)$. Find its equation and draw it. \label{ex_lin_2}\end{exa} \begin{solu} The equation is of the form $y = mx + k$. We must find the slope and the $y$-intercept. To find $m$ we compute the ratio $$m = \frac{10 - (-5)}{-3 - 6} = -\frac{5}{3}.$$Thus the equation is of the form $y = -\dis{\frac{5}{3}}x + k$ and we must now determine $k$. To do so, we substitute either point, say the first, into $y = -\dis{\frac{5}{3}}x + k$ obtaining $10 = -\dis{\frac{5}{3}}(-3) + k$, whence $k = 5$. The equation sought is thus $y = -\dis{\frac{5}{3}}x + 5$. To draw the graph, first locate the $y$-intercept (at $(0, 5)$). Since the slope is $\dis{-\frac{5}{3}}$, move five units down (to $(0, 0)$) and three to the right (to $(3, 0)$). Connect now the points $(0, 5)$ and $(3, 0)$. The graph appears in figure \ref{fig:line1}. \end{solu} \begin{exa}\label{exa:colli_pts} Three points $(4, u), (1, -1)$ and $(-3, -2)$ lie on the same line. Find $u$. \end{exa} \begin{solu} Since the points lie on the same line, any choice of pairs of points used to compute the gradient must yield the same quantity. Therefore $$\frac{u - (-1)}{4 - 1} = \frac{-1 - (-2)}{1 - (-3)} $$which simplifies to the equation $$\frac{u + 1}{3} = \frac{1}{4}.$$Solving for $u$ we obtain $u = -\frac{1}{4}$. \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro}\label{pro:lines}Assuming that the equations for the lines $l_1$, $l_2$, $l_3$, and $l_4$ in figure \ref{fig:lines} below can be written in the form $y = mx + b$ for suitable real numbers $m$ and $b$, determine which line has the largest value of $m$ and which line has the largest value of $b$. \end{pro} \vspace{2cm} \begin{figurehere} \centering\psset{unit=.5pc} \psline[linewidth=1.5pt,linecolor=brown]{<->}(-2,6.5)(8,-1) \psline[linewidth=1.5pt,linecolor=brown]{<->}(-1,-5)(5,8) \psline[linewidth=1.5pt,linecolor=brown]{<->}(-2,8)(5,-1) \psline[linewidth=1.5pt,linecolor=brown]{<->}(-1,.5)(5,5) \psaxes[labels=none, ticks=none linewidth=1.5pt]{->}(0,0)(-4.5,-4.5)(9.5,9.5) \uput[r](9.5,0){$x$} \uput[u](0, 9.5){y}\uput[r](5,5){$l_1$}\uput[r](8,-1){$l_2$}\uput[r](5,8){$l_3$}\uput[r](5,-1){$l_4$} \meinecaption{.7}{Problem \ref{pro:lines}.}\label{fig:lines} \end{figurehere} \begin{pro}[AHSME 1994] Consider the L-shaped region in the plane, bounded by horizontal and vertical segments with vertices at $(0, 0), (0, 3), (3, 3), (3, 1), (5, 1)$ and $(5, 0)$. Find the gradient of the line that passes through the origin and divides this area exactly in half. \label{ahsme94_1} \begin{answer} $\dfrac{7}{9}$ \end{answer} \end{pro} \vspace{1cm} \begin{figurehere}\centering \psset{unit=.75pc} \psaxes[labelFontSize=\tiny, linewidth=1.5pt]{->}(0, 0)(0, 0)(6, 4) \psline[linewidth=1.4pt, labels=none, showpoints=true](0,0)(0,3)(3,3)(3,1)(5,1)(5,0) \meinecaption{.1}{Problem \ref{ahsme94_1}.}\label{ahsme94_1_1} \end{figurehere} \begin{pro} What is the slope of the line with equation $\dfrac{x}{a} + \dfrac{y}{b} = 1$? \begin{answer} $-\dfrac{b}{a}$ \end{answer} \end{pro} \begin{pro}If the point $(a, -a)$ lies on the line with equation $-2x + 3y = 30$, find the value of $a$. \begin{answer} $-6$\end{answer} \end{pro} \begin{pro} Find the equation of the straight line joining $(3, 1)$ and $(-5, -1)$. \begin{answer}$\dis{y = \frac{x}{4} + \frac{1}{4}}$ \end{answer} \end{pro} \begin{pro}\label{pro:line_ab_to_ba} Let $(a, b)\in\reals^{2}$. Find the equation of the straight line joining $(a, b)$ and $(b, a)$. \begin{answer}$\dis{y = -x + b + a}$ \end{answer} \end{pro} \begin{pro}Find the equation of the line that passes through $(a, a^2)$ and $(b, b^2)$. \begin{answer} $y = (a + b)x - ab$\end{answer} \end{pro} \begin{pro} The points $(1, m), (2, 4)$ lie on a line with gradient $m$. Find $m$. \begin{answer} $m = 2$ \end{answer} \end{pro} \begin{pro} Consider the following regions on the plane. $$R_1 = \{(x, y)\in\reals^2| y \leq 1 - x\},$$ $$R_2 = \{(x, y)\in\reals^2| y \geq x + 2\},$$ $$R_3 = \{(x, y)\in\reals^2| y \leq 1 + x \}.$$ Sketch the following regions. \begin{enumerate} \item $R_1 \setminus R_2$ \item $R_2 \setminus R_1$ \item $R_1 \cap R_2 \cap R_3$ \item $R_2 \setminus (R_1 \cup R_2)$ \end{enumerate} \end{pro} \begin{pro}\label{pro:coord_lines} In figure \ref{fig:coord_lines}, point $M$ has coordinates $(2,2)$, points $A, S$ are on the $x$-axis, point $B$ is on the $y$-axis $\triangle SMA$ is isosceles at $M$, and the line segment $SM$ has slope $ \dfrac{1}{2}$. Find the coordinates of points $A, B, S$. \begin{answer} Let $(x,0)$ be the coordinates of $S$. Since the slope of the line segment $SM$ is $\dfrac{1}{2}$, we have $$\dfrac{2-x}{2-0} =\dfrac{1}{2} \implies x=1, $$whence $S$ is the point $(1,0)$. Let $(a,0)$ be the coordinates of $A$. Since $SM=MA$, we have $$ \sqrt{(a-2)^2+(0-2)^2} = \sqrt{(1-2)^2+(0-2)^2} \implies (a-2)^2+4 =5 \implies a\in\{1,3\}.$$This means that $A$ is the point $(3,0)$. Let $B$ be point $(0,y)$. Since $A, B, M$ are collinear, we may compute the slope in two different ways to obtain, $$ \dfrac{y-2}{0-2}=\dfrac{2-0}{2-3}\implies y-2=4 \implies y=6. $$Thus $B$ is the point $(0,6)$. \end{answer} \end{pro} \vspace*{3cm} \begin{figurehere} \centering \psset{unit=1pc} \psplot[linewidth=2pt,linecolor=brown,arrows={<->},algebraic]{-1}{4}{6-2*x} \psline[linewidth=2pt,linecolor=brown](2,2)(1,0) \psdots(2,2)(0,6)(3,0)(1,0)\uput[ur](2,2){$M$}\uput[d](3,0){$A$}\uput[d](1,0){$S$}\uput[l](0,6){$B$} \psaxes[linewidth=2pt,labels=none, ticks=none,arrows={->}](0,0)(-1,-1)(8,8) \meinecaption{.25}{Problem \ref{pro:coord_lines}.} \label{fig:coord_lines} \end{figurehere} \begin{pro} Which points on the line with equation $y = 6-2x$ are equidistant from the axes? \begin{answer} Let required point be $(x,y)$. The distance of this point to its projection on the $x$-axis is $|y|$ and similarly, the distance of this point to its projection on the $x$-axis is $|x|$. We need $$ |y|=|x| \implies |6-2x|=|x|\implies 6-2x=x\qquad \mathrm{or}\qquad 6-2x=-x. $$ The first case gives $x=2$ and the point is $(2,2)$, and the second case gives $x=6$ and the point is $(6,-6)$. \end{answer} \end{pro} \begin{pro} A vertical line divides the triangle with vertices $(0, 0), (1, 1)$ and $(9, 1)$ in the plane into two regions of equal area. Find the equation of this vertical line. \begin{answer} $x = 3$ \end{answer} \end{pro} \begin{pro}\label{pro:lines_aplenty} Draw $$(x^2-1)(y^2-1)(x^2-y^2)=0. $$ \begin{answer} This is asking to graph the lines $x=-1$, $x=1$, $y=-1$, $y=1$, $y=-x$, and $y=x$, all on the same set of axes. The picture appears in figure \ref{fig:lines_aplenty}. \vspace*{2cm} \begin{figurehere} \centering \psset{unit=.4pc} \psaxes[linewidth=2pt,labels=none](0,0)(-10.5,-10.5)(10.5,10.5) \psline[linecolor=brown,linewidth=2pt](-10,1)(10,1) \psline[linecolor=brown,linewidth=2pt](-10,-1)(10,-1) \psline[linecolor=blue,linewidth=2pt](1,-10)(1,10) \psline[linecolor=blue,linewidth=2pt](-1,-10,-1)(-1,10) \psline[linecolor=brown,linewidth=2pt](-10,10)(10,-10) \psline[linecolor=brown,linewidth=2pt](-10,-10)(10,10) \meinecaption{2}{Problem \ref{pro:lines_aplenty}.}\label{fig:lines_aplenty} \end{figurehere} \end{answer} \end{pro} \end{multicols} \section{Parallel and Perpendicular Lines} \vspace{3cm} \begin{figure}[!hptb] \hfill \begin{minipage}{7cm} \centering\psset{unit=1pc} \pstGeonode[PointName=none](0,-2){A}(4,3){B}(0,0){A'}(4,5){B'} \pstLineAB[arrows={<->},linewidth=2pt,linecolor=blue,nodesep=-2]{A}{B} \pstLineAB[arrows={<->},linewidth=2pt,linecolor=blue,nodesep=-2]{A'}{B'} \uput[r](B){{\tiny $(x_2,y_2)$}}\uput[r](A){{\tiny $(x_1,y_1)$}} \uput[ul](B'){{\tiny $(x_2,y'_2)$}}\uput[ul](A'){{\tiny $(x_1,y'_1)$}} \pstLineAB{A}{A'} \pstLineAB{B}{B'} \psdots(A)(A')(B)(B') \meinecaption{2.7}{Theorem \ref{thm:parallel_line_slope}.}\label{fig:parallel_line_slope} \end{minipage} \hfill \begin{minipage}{7cm} $$\psset{unit=1.7pc} \psaxes[labels=none, ticks=none,linewidth=1.5pt]{<->}(0, 0)(-3, -3)(3, 3) \psline[linewidth=2pt, linecolor=brown, labels=none, showpoints=true]{<->}(-2, -2)(0, 0)(2, 2) \psline[linewidth=2pt, labels=none, showpoints=true, linecolor=blue]{<->}(-2, 2)(0, 0)(2, -2) \psline[linewidth=2pt, labels=none, showpoints=true, linecolor=blue]{<->}(1, 2.5)(1, -2.5) \rput(2.5, 2.3){\qquad y = mx} \rput(2.5, -2.3){\qquad y = m_1x} \rput(1, 1){\bullet} \rput(1, -1){\bullet} \rput(2, 1){\qquad (1, m)} \rput(2, -1){\qquad (1, m_1)} $$ \meinecaption{2}{Theorem \ref{thm:perpendicular_line_slope}.}\label{fig:perpendicular_line_slope}. \end{minipage} \end{figure} \begin{thm} Two lines are parallel if and only if they have the same slope. \label{thm:parallel_line_slope}\end{thm} \begin{pf} Suppose the the lines $L$ and $L'$ are parallel, and that the points $A(x_1,y_1)$ y $B(x_2,y_2)$ lie on $L$ and that the points $A'(x_1,y'_1)$ and $B'(x_2,y'_2)$ lie on $L'$. Observe tha t $ABB'A'$ is a parallelogram, and hence, $y_2-y_1=y'_2-y'_1$, which gives $$ \dfrac{y_2-y_1}{x_2-x_1}=\dfrac{y'_2-y'_1}{x_2-x_1}, $$ demonstrating that the slopes of $L$ and $L'$ are equal. Assume now that $L$ and $L'$ have the same slope. The $$ \dfrac{y_2-y_1}{x_2-x_1}=\dfrac{y'_2-y'_1}{x_2-x_1} \implies y_2-y_1=y'_2-y'_1. $$ Then the sides of $AA'$ and $BB'$ of the quadrilateral $ABB'A'$ are congruent. As these sides are also parallel, since they are on the verticals $x=x_1$ and $x=x_2$, we deduce that $ABB'A'$ is a parallelogram, demonstrating that $L$ and $L'$ are parallel. \end{pf} \begin{exa} Find the equation of the line passing through $(4, 0)$ and parallel to the line joining $(-1, 2)$ and $(2, -4)$. \end{exa} \begin{solu} First we compute the slope of the line joining $(-1, 2)$ and $(2, -4)$: $$ m = \frac{2 - (-4)}{-1 - 2} = -2.$$The line we seek is of the form $y = -2x + k$. We now compute the $y$-intercept, using the fact that the line must pass through $(4, 0)$. This entails solving $0 = -2(4) + k$, whence $k = 8$. The equation sought is finally $y = -2x + 8$. \end{solu} \begin{thm}\label{thm:perpendicular_line_slope} Let $y = mx + k$ be a line non-parallel to the axes. If the line $y = m_1x + k_1$ is {\rm perpendicular} to $y = mx + k$ then $\dis{m_1 = -\frac{1}{m}}.$ Conversely, if $mm_1=-1$, then the lines with equations $y = mx + k$ and $y = m_1x + k_1$ are perpendicular. \end{thm} \begin{pf} Refer to figure \ref{fig:perpendicular_line_slope}. Since we may translate lines without affecting the angle between them, we assume without loss of generality that both $y = mx + k$ and $y = m_1x + k_1$ pass through the origin, giving thus $k = k_1 = 0.$ Now, the line $y = mx$ meets the vertical line $x = 1$ at $(1, m)$ and the line $y = m_1x$ meets this same vertical line at $(1, m_1)$ (see figure \ref{fig:perpendicular_line_slope}). By the Pythagorean Theorem $$ (m - m_1)^2 = (1 + m^2) + (1 + m_1 ^2)\implies m^2-2mm_1+m_1 ^2=2+m^2+m_1 ^2 \implies mm_1=-1 , $$ which proves the assertion. The converse is obtained by retracing the steps and using the converse to the Pythagorean Theorem. \end{pf} \begin{exa} Find the equation of the line passing through $(4, 0)$ and perpendicular to the line joining $(-1, 2)$ and $(2, -4)$. \end{exa} \begin{solu} The slope of the line joining $(-1, 2)$ and $(2, -4)$ is $-2.$ The slope of any line perpendicular to it $$ m_1 = -\dfrac{1}{m} = \dfrac{1}{2}.$$The equation sought has the form $y = \dfrac{x}{2} + k$. We find the $y$-intercept by solving $0 = \dfrac{4}{2} + k,$ whence $k = -2$. The equation of the perpendicular line is thus $y = \dfrac{x}{2} - 2$. \end{solu} \begin{exa} For a given real number $t$, associate the straight line $L_t$ with the equation $$ L_t: \ \ (4 - t)y = (t + 2)x + 6t. $$ \begin{enumerate} \item Determine $t$ so that the point $(1,2)$ lies on the line $L_t$ and find the equation of this line. \item Determine $t$ so that the $L_t$ be parallel to the $x$-axis and determine the equation of the resulting line.\item Determine $t$ so that the $L_t$ be parallel to the $y$-axis and determine the equation of the resulting line. \item Determine $t$ so that the $L_t$ be parallel to the line $-5y = 3x - 1$. \item Determine $t$ so that the $L_t$ be perpendicular to the line $-5y = 3x-1$. \item Is there a point $(a, b)$ belonging to every line $L_t$ regardless of the value of $t$? \end{enumerate}\end{exa} \begin{solu} \begin{enumerate} \item If the point $(1,2)$ lies on the line $L_t$ then we have $$ (4-t)(2)= (t+2)(1) + 6t \implies t=\dfrac{2}{3}. $$ The line sought is thus $$ L_{2/3}: \quad (4-\dfrac{2}{3})y = (\frac{2}{3}+2)x + 6\left(\frac{2}{3}\right) $$ or $y = \dfrac{4}{5}x + \dfrac{6}{5}$. \item We need $t + 2 = 0 \implies t = -2$. In this case $$(4 -(-2))y = -12 \implies y = -2. $$ \item We need $4 - t = 0 \implies t = 4$. In this case $$0 = (4 + 2)x + 24 \implies x = -4. $$ \item The slope of $L_t$ is $$ \dfrac{t + 2}{4 - t}, $$and the slope of the line $-5y = 3x - 1$ is $-\dfrac{3}{5}$. Therefore we need $$\dfrac{t + 2}{4 - t} = -\dfrac{3}{5} \implies -3(4 - t) = 5(t + 2) \implies t = -11. $$ \item In this case we need $$\dfrac{t + 2}{4 - t} = \dfrac{5}{3} \implies 5(4 - t) = 3(t + 2) \implies t = \dfrac{7}{4}. $$ \item Yes. From above, the obvious candidate is $(-4,-2)$. To verify this observe that $$(4-t)(-2) = (t+2)(-4) + 6t, $$regardless of the value of $t$. \end{enumerate} \end{solu} \vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{5cm} \centering \psset{unit=.4pc} \psaxes[labels=none, ticks=none, linewidth=1.7pt, arrows={->}](0,0)(-10,-10)(12,12)\psline[linewidth=1.7pt](2,1.4)(2,0) \psline[linewidth=1.5pt,linecolor=brown]{<->}(6,-1.8)(-5,7) \psline[linewidth=1.5pt,linecolor=brown]{<->}(6,6.4)(-4,-6.1)\psdots[dotstyle=*,dotscale=1](2,1.4)(2,0)(0,3)(-3,5.4) \uput[l](0,3){{\tiny$3$}} \uput[d](2,0){{\tiny$2$}} \uput[dl](-3,5.4){{\tiny$(-3,5.4)$}} \uput[ur](10.5,0){$x$} \uput{2pt}[r](2.1,1.4){$P$} \uput[ur](0, 10.5){$y$}\uput[r](6,-1.8){$L'$}\uput[dr](-4,-6.1){L} \meinecaption{3}{ Example \ref{pro:line_perp}.}\label{fig:line_perp} \end{minipage} \hfill \begin{minipage}{5cm} \centering \psset{unit=.4pc} \psaxes[arrows={->},linewidth=1.7pt,labelFontSize=\tiny](0,0)(-10,-10)(12,12) \pstGeonode[PosAngle={0,0}](1,2){O}(5,5){A} \pstGeonode[PointName=none,PointSymbol=none](8.75,0){B} \pstTranslation[PointName=none,PointSymbol=none]{A}{B}{B}[C] \pstTranslation[PointName=none,PointSymbol=none]{B}{A}{A}[C'] \pstCircleOA[linecolor=brown,linewidth=2pt]{O}{A} \pstLineAB[arrows={<->},linewidth=2pt,linecolor=blue]{C}{C'} \meinecaption{3}{ Example \ref{pro:circle-line}.} \label{fig:circle-line} \end{minipage} \hfill \begin{minipage}{5cm} \centering \psset{unit=.4pc} \psaxes[labels=none, ticks=none, linewidth=1.7pt, arrows={->}](0,0)(-10,-10)(12,12) \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(1,7.5){A}(7.5,1){A'}(-9,-9){M}(9,9){M'} \pstMiddleAB[PointName=none]{A}{A'}{B} \psline[linewidth=2pt,linecolor=blue,arrows={<->}](M)(M') \psline(A)(A')\psdots(A)(A')(B) \uput[r](A'){\tiny$(a,b)$} \uput[ur](A){\tiny$(b,a)$}\uput[ur](M'){$y=x$} \meinecaption{3}{ Theorem \ref{thm:symmetry_ab_ba}.}\label{fig:symmetry_ab_ba} \end{minipage} \end{figure} \begin{exa}\label{pro:line_perp} In figure \ref{fig:line_perp}, the straight lines $L$ y $L'$ are perpendicular and meet at the point $P$. \begin{enumerate} \item Find the equation of $L'.$ \item Find the coordinates of $P.$ \item Find the equation of the line $L.$ \end{enumerate} \end{exa} \begin{solu} \noindent \begin{enumerate} \item Notice that $L'$ passes through $(-3,5.4)$ and through $(0,3)$, hence it must have slope $$ \dfrac{5.4-3}{-3-0}=-0.8. $$The equation of $L'$ has the form $y=-0.8x+k$. Since $L'$ passes through $(0,3)$, we deduce that $L'$ has equation $y=-0.8x+3$. \item Since $P$ if of the form $(2,y)$ and since it lies on $L'$, we deduce that $y=-0.8(2)+3=1.4$. \item $L$ has slope $-\dfrac{1}{-0.8}=1.25$. This means that $L$ has equation of the form $y=1.25x+k$. Since $P(2,1.4)$ lies on $L$, we must have$1.4=1.25(2)+k\implies k=-1.1$. We deduce that $L$ has equation $y=1.25x-1.1$. \end{enumerate} \end{solu} \begin{exa}\label{pro:circle-line} Consider the circle ${\cal C}$ of centre $O(1, 2)$ and passing through $A(5,5)$, as in figure \ref{fig:circle-line}. \begin{enumerate} \item Find the equation of ${\cal C}$. \item Find all the possible values of $a$ for which the point $(2, a)$ lies on the circle ${\cal C}$. \item Find the equation of the line $L$ tangent to ${\cal C}$ at $A$. \end{enumerate} \end{exa} \begin{solu} \noindent \begin{enumerate} \item Let $R>0$ be the radius of the circle . Then equation of the circle has the form $$ (x-1)^2+(y-2)^2=R^2. $$Since $A(5,5)$ lies on the circle, $$ (5-1)^2+(5-2)^2=R^2 \implies 16+9=R^2 \implies 25 =R^2, $$whence the equation sought for ${\cal C}$ is $$ (x-1)^2+(y-2)^2=25. $$ \item If the point $(2, a)$ lies on ${\cal C}$, we will have $$ (2-1)^2+(a-2)^2 = 25\implies 1+(a-2)^2=25 \implies (a-2)^2=24 \implies a-2 = \pm \sqrt{24} \implies a= 2\pm \sqrt{24} = 2\pm 2\sqrt{6}. $$ \item $L$ is perpendicular to the line joining $(1,2)$ and $(5,5)$. As this last line has slope $$\dfrac{5-2}{5-1}=\dfrac{3}{4}, $$ the line $L$ will have slope $-\dfrac{4}{3}$. Thus $L$ has equation of the form $$ y = -\dfrac{4}{3}x +k. $$As $(5,5)$ lies on the line, $$ 5= -\dfrac{4}{3}\cdot 5 +k\implies 5+\dfrac{20}{3}=k\implies k = \dfrac{35}{3},$$from where we gather that $L$ has equation $y=-\dfrac{4}{3}x +\dfrac{35}{3}$. \end{enumerate} \end{solu} We will now demonstrate two results that will be needed later. \begin{thm}[Distance from a Point to a Line] Let $L: y = mx + k$ be a line on the plane and let $P = (x_0, y_0)$ be a point on the plane, not on $L$. The distance $\distance{L}{P}$ from $L$ to $P$ is given by $$ \frac{|x_0m + k- y_0|}{\sqrt{1 + m^2}}.$$ \label{thm:disptline}\end{thm} \begin{pf} If the line had infinite slope, then $L$ would be vertical, and of equation $x = c$, for some constant $c$, and then clearly, $$\distance{L}{P} = |x_0 - c|.$$ If $m = 0$, then $L$ would be horizontal, and then clearly $$\distance{L}{P} = |y_0 - k|,$$agreeing with the theorem. Suppose now that $m \neq 0.$ Refer to figure \ref{fig:disptline}. The line $L$ has slope $m$ and all perpendicular lines to $L$ must have slope $-\frac{1}{m}$. The distance from $P$ to $L$ is the length of the line segment joining $P$ with the point of intersection $(x_1, y_1)$ of the line $L'$ perpendicular to $L$ and passing through $P$. Now, it is easy to see that $L'$ has equation $$L': y = -\frac{1}{m}x + y_0 + \frac{x_0}{m},$$from where $L$ and $L'$ intersect at $$x_1 = \frac{y_0m + x_0 - bm}{1 + m^2}, \ \ \ y_1 = \frac{y_0m^2 + x_0m + k}{1 + m^2}.$$ This gives $${ \everymath{\displaystyle} \renewcommand{\arraystretch}{2.5} \begin{array}{lll} \distance{L}{P} & = & \distance{(x_0, y_0)}{(x_1, y_1)} \\ & = & \sqrt{(x_0 - x_1)^2 + (y_0 - y_1)^2} \\ & = & \sqrt{\left(x_0 - \frac{y_0m + x_0 - km}{1 + m^2}\right)^2 + \left(y_0 - \frac{y_0m^2 + x_0m + k}{1 + m^2}\right)^2} \\ & = & \frac{\sqrt{(x_0m^2 - y_0m + km)^2 + (y_0 - x_0m - k)^2}}{1 + m^2} \\ & = & \frac{\sqrt{(m^2 + 1)(x_0m - y_0 + k)^2 }}{1 + m^2} \\ & = & \frac{|x_0m - y_0 + k|}{\sqrt{1 + m^2}}, \end{array} } \renewcommand{\arraystretch}{1} $$proving the theorem. \bigskip {\em Aliter:} A ``proof without words'' can be obtained by considering the similar right triangles in figure \ref{fig:disptline2}. \end{pf} \vspace*{4cm} \begin{figure}[h] \begin{minipage}{7cm} \centering \psset{unit=.7pc} \pstGeonode[PointName=none,PointSymbol=none](-5,5){B}(5,-5){B'}(-7,0){P} \pstLineAB[nodesep=-2,arrows={<->},linewidth=2pt,linecolor=blue]{B}{B'} \pstProjection[PointName=none]{B}{B'}{P}[A] \pcline[linewidth=2pt,linecolor=brown,nodesep=-4,arrows={<->}](P)(A) \pstRightAngle[RightAngleSize=1.5]{P}{A}{B} \psline[linewidth=2pt](P)(A)\psdots(P)(A)\uput[r](A){\tiny$(x_1,y_1)$}\uput[r](B'){\tiny$L:\ y=mx+k$}\uput[r](P){\tiny$(x_0,y_0)$}\meinecaption{3}{ Theorem \ref{thm:disptline}.} \label{fig:disptline} \end{minipage} \hfill \begin{minipage}{7cm} \centering \psset{unit=.7pc} \pstGeonode[PointName=none,dotscale=1.5](-1,-8){A}(10,12){B}(10,-4){P} \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(1,0){I}(0,1){J} \pstProjection[PointName=none,PointSymbol=none]{A}{B}{P}[C] \psline[linewidth=2pt](A)(B)(P)(C) \uput{8pt}[d](P){$(x_0,y_0)$}\uput[ur](B){$(x_0,mx_0+k)$} \pstRightAngle[RightAngleSize=.8]{B}{C}{P} \pstTranslation[DistCoef=.5,PointName=none]{A}{B}{A}[A1] \pstTranslation[DistCoef=.2,PointName=none]{B}{A}{B}[B1] \pstTranslation[PointName=none,PointSymbol=none]{J}{O}{B1}[S] \pstTranslation[PointName=none,PointSymbol=none]{O}{I}{A1}[T] \pstInterLL[PointName=none]{B1}{S}{A1}{T}{Q} \psline[linewidth=2pt](A1)(Q)(B1) \pstRightAngle[RightAngleSize=.8]{A1}{Q}{B1} \pcline[offset=-12pt]{|-|}(P)(B)\ncput*[nrot=:U]{$|mx_0+k-y_0|$} \pcline[offset=5pt]{|-|}(P)(C)\ncput*[nrot=:D]{$d$} \pcline[offset=12pt]{|-|}(A1)(B1)\ncput*[nrot=:U]{$\sqrt{1+m^2}$} \pcline[offset=-8pt]{|-|}(A1)(Q)\ncput*[nrot=:U]{$1$} \pcline[offset=6pt]{|-|}(B1)(Q)\ncput*[nrot=:D]{$m$} \meinecaption{3}{ Theorem \ref{thm:disptline}.} \label{fig:disptline2} \end{minipage} \end{figure} \begin{exa} Find the distance between the line $L: 2x - 3y = 1$ and the point $(-1, 1)$. \end{exa} \begin{solu} The equation of the line $L$ can be rewritten in the form $L: y = \frac{2}{3}x - \frac{1}{3}$. Using Theorem \ref{thm:disptline}, we have $$\distance{L}{P} = \frac{|-\frac{2}{3} - 1 - \frac{1}{3}|}{\sqrt{1 + (\frac{2}{3})^2}} = \frac{6\sqrt{13}}{13}.$$ \end{solu} \begin{thm}\label{thm:symmetry_ab_ba} The point $(b, a)$ is symmetric to the point $(a, b)$ with respect to the line $y=x$. \end{thm} \begin{pf} The line joining $(b, a)$ to $(a, b)$ has equation $y = -x + a+b$. This line is perpendicular to the line $y = x$ and intersects it when $$x = -x + a + b \implies x = \frac{a+b}{2}.$$ Then, since $y=x=\dfrac{a+b}{2}$, the point of intersection is $(\frac{a+b}{2}, \frac{a+b}{2})$. But this point is the midpoint of the line segment joining $(a, b)$ to $(b, a)$, which means that both $(a, b)$ and $(b,a)$ are equidistant from the line $y=x$, establishing the result. See figure \ref{fig:symmetry_ab_ba}. \end{pf} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro}Find the equation of the straight line parallel to the line $8x - 2y = 6$ and passing through $(5, 6)$. \begin{answer} $y = 4x - 14$ \end{answer} \end{pro} \begin{pro} Let $(a, b)\in(\reals\setminus \{0\})^{2}$. Find the equation of the line passing through $(a, b)$ and parallel to the line $\frac{x}{a} - \frac{y}{b} = 1$. \begin{answer} $\dis{y = \frac{b}{a}x}$ \end{answer} \end{pro} \begin{pro}Find the equation of the straight line normal to the line $8x - 2y = 6$ and passing through $(5, 6)$. \begin{answer} $\dis{y = -\frac{1}{4}x + \frac{29}{4}}$ \end{answer} \end{pro} \begin{pro} Let $a, b$ be strictly positive real numbers. Find the equation of the line passing through $(a, b)$ and perpendicular to the line $\frac{x}{a} - \frac{y}{b} = 1$. \begin{answer} $\dis{y = -\frac{a}{b}x + b + \frac{a^2}{b}}$ \end{answer} \end{pro} \begin{pro}Find the equation of the line passing through $(12, 0)$ and parallel to the line joining $(1, 2)$ and $(-3, -1)$. \begin{answer} $y = \frac{3}{4}x - 9$ \end{answer} \end{pro} \begin{pro}Find the equation of the line passing through $(12, 0)$ and normal to the line joining $(1, 2)$ and $(-3, -1)$. \begin{answer} $y = -\frac{4}{3}x + 16$ \end{answer} \end{pro} \begin{pro} Find the equation of the straight line tangent to the circle $x^2+y^2=1$ at the point $(\frac{1}{2}, \frac{\sqrt{3}}{2})$. \begin{answer} Notice that there is a radius of the circle connecting $(0,0)$ and $(\frac{1}{2}, \frac{\sqrt{3}}{2})$. The line passing through these two points is $y = \sqrt{3}x$. Hence, since the tangent line is perpendicular to the radius at the point of tangency, the line sought is of the form $y = -\dfrac{\sqrt{3}}{3}x+k$. To find $k$ observe that $\dfrac{\sqrt{3}}{2} = -\dfrac{3}{4}+k \implies k = \dfrac{3\sqrt{3}}{4}$. Finally, the desired line is $y = -\dfrac{\sqrt{3}}{3}x+ \dfrac{3\sqrt{3}}{4}$. \end{answer} \end{pro} \begin{pro} Consider the line $L$ passing through $(a, a^2)$ and $(b, b^2)$. Find the equations of the lines $L_1$ parallel to $L$ and $L_2$ normal to $L$, if $L_1$ and $L_2$ must pass through $(1, 1)$. \begin{answer} $L_1: y = (a + b)x + 1 -a-b$, $L_2: y = -\dfrac{x}{a + b} + \dfrac{a + b + 1}{a + b}$ \end{answer} \end{pro} \begin{pro} For any real number $t$, associate the straight line $L_t$ having equation $$(2t - 1)x + (3 - t)y - 7t + 6 = 0.$$In each of the following cases, find an $t$ satisfying the stated conditions. \begin{enumerate} \item $L_t$ passes through $(1, 1)$. \item $L_t$ passes through the origin $(0, 0)$. \item $L_t$ is parallel to the $x$-axis. \item $L_t$ is parallel to the $y$-axis. \item $L_t$ is parallel to the line of equation $3x - 2y - 6 = 0.$ \item $L_t$ is normal to the line of equation $y = 4x - 5$. \item $L_t$ has gradient $-2$. \item Is there a point $(x_0, y_0)$ belonging to $L_t$ no matter which real number $t$ be chosen? \end{enumerate} \begin{answer} (1) $t = 4/3$, (2) $t = 6/7$, (3) $t = 1/2$, (4) $t = 3$, (5) $t = -7$, (6) $t = 7/9$, (7) $t = 7/4$, (8) $(3, -1)$ \end{answer} \end{pro} \begin{pro} For any real number $t$, associate the straight line $L_t$ having equation $$(t - 2)x + (t + 3)y + 10t - 5 = 0.$$ In each of the following cases, find an $t$ and the resulting line satisfying the stated conditions. \begin{enumerate} \item $L_t$ passes through $(-2, 3)$. \item $L_t$ is parallel to the $x$-axis. \item $L_t$ is parallel to the $y$-axis. \item $L_t$ is parallel to the line of equation $x - 2y - 6 = 0.$ \item $L_t$ is normal to the line of equation $y = -\frac{1}{4}x - 5$. \item Is there a point $(x_0, y_0)$ belonging to $L_t$ no matter which real number $t$ be chosen? \end{enumerate} \begin{answer}We have \begin{enumerate} \item If $L_t$ passes through $(-2, 3)$ then $$(t - 2)(-2) + (t + 3)(3) + 10t - 5 = 0,$$from where $t = -\frac{8}{11}$. In this case the line is $$ -\frac{30}{11}x+\frac{25}{11}y-\frac{135}{11} = 0. $$ \item $L_t$ will be parallel to the $x$-axis if the $x$-term disappears, which necessitates $t - 2 = 0$ or $t = 2$. In this case the line is $$y=-3. $$ \item $L_t$ will be parallel to the $y$-axis if the $y$-term disappears, which necessitates $t + 3 = 0$ or $t = -3$. In this case the line is $$ x = -7. $$\item The line $x - 2y - 6 = 0$ has gradient $\frac{1}{2}$ and $L_t$ has gradient $\frac{2 - t}{t + 3}$. The lines will be parallel when $\frac{2 - t}{t + 3} = \frac{1}{2}$ or $t = 1/3$. In this case the line is $$-\frac{5}{3}x+\frac{10}{3}y-\frac{5}{3} = 0. $$\item The line $y = -\frac{1}{4}x - 5$ has gradient $-\frac{1}{4}$ and $L_t$ has gradient $\frac{2 - t}{t + 3}$. The lines will be perpendicular when $\frac{2 - t}{t + 3} = 4$ or $t = -2$. In this case the line is $$-4x + y - 25 = 0. $$\item If such a point existed, it would pass through the horizontal and vertical lines found above, thus $(x_0, y_0) = (-7, -3)$ is a candidate for the point sought. That $(-7, -3)$ passes through every line $L_t$, no matter the choice of $t$ is seen from $$(t - 2)(-7) + (t + 3)(-3) + 10t - 5 = -7t + 14 - 3t - 9 + 10t - 5 = 0.$$ \end{enumerate} \end{answer} \end{pro} \begin{pro} Shew that the four points $A = (-2, 0)$, $B = (4, - 2)$, $C = (5, 1)$, and $D = (-1, 3)$ form the vertices of a rectangle. \end{pro} \begin{pro} Find the distance from the point $(1,1)$ to the line $y=-x$. \begin{answer} $\sqrt{2}$ \end{answer} \end{pro} \begin{pro} Let $a\in\reals$. Find the distance from the point $(a, 0)$ to the line $L: y = ax + 1$. \begin{answer} $\sqrt{1 + a^2}$ \end{answer} \end{pro} \begin{pro} Find the equation of the circle with centre at $(3,4)$ and tangent to the line $x-2y+3=0$. \begin{answer} The radius of the circle is the distance from the centre to the tangent line. This radius is then$$ r = \left|\dfrac{3 - 2\cdot 4 + 3}{\sqrt{1^2 + (-2)^2}}\right| = \dfrac{2}{\sqrt{5}}.$$The desired equation is $$ (x-3)^2 + (y-4)^2 = \dfrac{4}{5}.$$ \end{answer} \end{pro} \begin{pro}\label{pro:concu-medianas} $\triangle ABC$ has vertices at $A(a,0)$, $B(b,0)$ and $C(0,c)$, where $a<0},labelFontSize=\tiny](0,0)(-8,-8)(10,10) \psplot[algebraic=true,arrows={<->},linewidth=2pt,linecolor=brown]{-6}{6}{abs(x)} \meinecaption{3}{$y=|x|$.} \label{fig:absval1} \end{minipage} \hfill \begin{minipage}{5cm} \centering\psset{unit=.5pc} \psaxes[linewidth=1.5pt,labelFontSize=\tiny,arrows={->}](0,0)(-8,-8)(10,10) \psline[linewidth=1.5pt, linecolor=brown]{<->}(-3, 7)(.5,0)(3,5) \meinecaption{3}{Example \ref{exa:abs_val_exa_1}.} \label{fig:abs_val_ex1} \end{minipage} \hfill \begin{minipage}{5cm} \centering \psset{unit=.5pc}\psaxes[linewidth=2pt,arrows={->},labelFontSize=\tiny](0,0)(-8,-8)(10,10) \psplot[algebraic=true,arrows={<->},linewidth=2pt,linecolor=brown]{-6}{6}{abs(x+2)-abs(x-2)}\psdots[dotscale=1.5](-2,-4)(2,4) \meinecaption{3}{Example \ref{exa:absval2}.} \label{fig:absval2} \end{minipage} \end{figure} \begin{exa}\label{exa:absval2} Consider the equation $y= |x+2|-|x-2|$. The terms in absolute values vanish when $x=-2$ or $x=-2$. If $x\leq -2$ then $$ |x+2|-|x-2| = (-x-2)-(-x+2)=-4. $$ For $-2\leq x \leq 2$, we have $$ |x+2|-|x-2| = (x+2)-(-x+2)=2x. $$For $x\geq 2$, we have $$ |x+2|-|x-2| = (x+2)-(x-2)=4. $$ Then, $$ y = |x+2|-|x-2|= \left\{ \begin{array}{ll} -4 & \mathrm{if}\ \ x\leq -2, \\ 2x & \mathrm{if}\ \ -2 \leq x\leq +2, \\ +4 & \mathrm{if}\ \ x\geq +2, \\\end{array}\right. $$ The graph is the union of three lines (or rather, two rays and a line segment), and can be see in figure \ref{fig:absval2}. \end{exa} \vspace*{2cm} \begin{figure}[!hptb] \begin{minipage}{7cm} \centering \psset{unit=.5pc}\psaxes[linewidth=2pt,arrows={->},labelFontSize=\tiny](0,0)(-8,-8)(10,10) \psplot[algebraic=true,arrows={<->},linewidth=2pt,linecolor=brown]{-6}{6}{abs(1-abs(x))} \meinecaption{3}{Example \ref{exa:absval3}.} \label{fig:absval3} \end{minipage} \hfill \begin{minipage}{7cm} \centering \psset{unit=.5pc}\psaxes[linewidth=2pt,arrows={->},labelFontSize=\tiny](0,0)(-8,-8)(10,10) \psline[arrows={<->},linewidth=2pt,linecolor=brown](7,7)(0,0)(7,-7) \meinecaption{3}{Example \ref{exa:absval4}.} \label{fig:absval4} \end{minipage} \end{figure} \begin{exa}\label{exa:absval3}Draw the graph of the curve $y=|1-|x||$. \end{exa} \begin{solu}The expression $1-|x|$ changes sign when $1-|x|=0$, that is, when $x=\pm 1$. The expression $|x|$ changes sign when $x=0$. Thus we puncture the real line at $x=-1$, $x=0$ and $x=1$. When $x\leq -1$ $$ |1-|x|| = |x|-1 = -x-1. $$ When $-1\leq x\leq 0$ $$ |1-|x|| = 1-|x| = 1+x. $$ When $0\leq x\leq 1$ $$ |1-|x|| = 1-|x| = 1-x. $$ When $x\geq 1$ $$ |1-|x|| = |x|-1 = x-1. $$ Hence, $$ y = |1-|x||= \left\{ \begin{array}{ll} -x-1 & \mathrm{if}\ \ x\leq -1, \\ 1+x & \mathrm{if}\ \ -1 \leq x\leq 0, \\ 1-x & \mathrm{if}\ \ 0 \leq x\leq 1, \\ x-1 & \mathrm{if}\ \ x\geq 1, \\\end{array}\right. $$ The graph appears in figure \ref{fig:absval3}. \end{solu} \begin{exa}\label{exa:absval4} Using Theorem \ref{thm:symmetry_ab_ba}, we may deduce that the graph of the curve $x=|y|$ is that which appears in figure \ref{fig:absval4} \end{exa} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro}\label{pro:absval1} Consider the curve $$\mathscr{C}: y = \absval{x-1} - \absval{x} + \absval{x+1}.$$ \begin{enumerate} \item Find an expression without absolute values for $\mathscr{C}$ when $x\leq -1$. \item Find an expression without absolute values for $\mathscr{C}$ when $-1 \leq x\leq 0$. \item Find an expression without absolute values for $\mathscr{C}$ when $0 \leq x\leq 1$. \item Find an expression without absolute values for $\mathscr{C}$ when $x\geq 1$. \item Draw $\mathscr{C}$. \end{enumerate} \begin{answer} \noindent \begin{enumerate} \item This is $y =(-x+1)-(-x)+(-x-1)=-x. $ \item This is $y= (-x+1)-(-x)+(x+1)=x+2. $ \item This is $y =(-x+1)-(+x)+(x+1)=-x+2. $ \item This is $y =(x-1)-(+x)+(x+1)=x. $ \item The graph of $\mathscr{C}$ appears in figure \ref{fig:absval1}. \end{enumerate} \vspace*{2cm} \begin{figurehere} \psset{unit=.5pc}\centering \psaxes[labelFontSize=\tiny]{->}(0,0)(-8,-8)(8,8) \psplot[algebraic=true,linewidth=2pt,linecolor=blue, arrows={<->}]{-7}{7}{abs(x-1)-abs(x)+abs(x+1)} \meinecaption{1}{Problem \ref{pro:absval1}.}\label{fig:absval1} \end{figurehere} \end{answer} \end{pro} \begin{pro}\label{pro:absval2} Draw the graph of the curve of equation $|x|=|y|$. \begin{answer} The graph appears in figure \ref{fig:absval2}.\\ \vspace*{3cm} \begin{figurehere} \psset{unit=.5pc}\centering \psaxes[labelFontSize=\tiny]{->}(0,0)(-8,-8)(8,8) \psplot[algebraic=true,linewidth=2pt,linecolor=blue, arrows={<->}]{-7}{7}{x} \psplot[algebraic=true,linewidth=2pt,linecolor=blue, arrows={<->}]{-7}{7}{-x} \meinecaption{1}{Problem \ref{pro:absval2}.}\label{fig:absval2} \end{figurehere} \end{answer} \end{pro} \begin{pro}\label{pro:absval3} Draw the graph of the curve of equation $y=\dfrac{|x|+x}{2}$. \begin{answer} This is the curve $$ y = \dfrac{|x|+x}{2}= \left\{ \begin{array}{ll} x & \mathrm{if}\ \ x\geq 0, \\ 0 & \mathrm{if}\ x<0. \\ \end{array}\right. $$ The graph appears in figure \ref{fig:absval3}.\\ \vspace*{3cm} \begin{figurehere} \psset{unit=.5pc}\centering \psaxes[labelFontSize=\tiny]{->}(0,0)(-8,-8)(8,8) \psplot[algebraic=true,linewidth=2pt,linecolor=blue, arrows={<->}]{-7}{7}{(x+abs(x))/2} \meinecaption{1}{Problem \ref{pro:absval3}.}\label{fig:absval3} \end{figurehere} \end{answer} \end{pro} \begin{pro}\label{pro:absval4}Draw the plane region $$\{(x,y)\in\reals ^2: x^2+y^2 \leq 16, |x|+|y|\geq 4\}.$$ \begin{answer} The set is composed of four segments of a circle inside the circle of equation $ x^2+y^2 = 16$ and bounded by the lines of equation $y =x+4$, $y =x-4$, $y =-x+4$ and $y =-x-4$. The graph appears in figure \ref{fig:absval4}.\\ \vspace*{3cm} \begin{figurehere}\centering \psset{unit=.5pc}\psaxes[linewidth=2pt,arrows={->},labelFontSize=\tiny](0,0)(-8,-8)(10,10) \pscircle*[linecolor=brown](0,0){4} \pspolygon*[linecolor=yellow](-4,0)(0,-4)(4,0)(0,4) \meinecaption{1}{Problem \ref{pro:absval4}.}\label{fig:absval4} \end{figurehere} \end{answer} \end{pro} \begin{pro} Draw the graphs of the following equations. \begin{enumerate} \item $y = |x + 2|$ \item $y = 3 - |x + 2|$ \item $y = 2|x + 2|$ \item $y = |x - 1| + |x + 1|$ \item $y = |x - 1| - |x + 1|$ \item $y = |x + 1| - |x - 1|$ \item $y = |x - 1| + |x| + |x + 1|$ \item $y = |x - 1| - |x| + |x + 1|$ \item $y = |x - 1| + x + |x + 1|$\item $y = |x + 3| + 2|x - 1 | - |x - 4|$ \end{enumerate} \end{pro} \end{multicols} \section{Parabolas, Hyperbolas, and Ellipses} \begin{df} A {\em parabola}\index{parabola} is the collection of all the points on the plane whose distance from a fixed point $F$ (called the {\em focus}\index{parabola!focus} of the parabola) is equal to the distance to a fixed line $L$ (called the {\em directrix} \index{parabola!directrix} of the parabola). See figure \ref{fig:parabola_definition}, where $FD = DP$. \end{df} We can draw a parabola as follows. Cut a piece of thread as long as the trunk of T-square (see figure \ref{fig:drawing_parabola}). Tie one end to the end of the trunk of the T-square and tie the other end to the focus, say, using a peg. Slide the crosspiece of the T-square along the directrix, while maintaining the thread tight against the ruler with a pencil. \vspace*{2cm} \begin{figure}[!hptb] \hfill \begin{minipage}{5cm} \centering\psset{unit=1pc, algebraic=true} \rput(-1,-2){\psline[linewidth=2pt,linecolor=brown]{<->}(-4,-1)(4,-1) \parabola[linecolor=brown,linewidth=2pt]{<->}(3,2.25)(0,0) \psline[linecolor=blue,linewidth=2pt](0,1)(2,1)(2,-1)\psdots[dotstyle=*,dotscale=1](0,1)(2,1)(2,-1) \uput[u](0,1){F}\uput[r](2,1){D}\uput[d](2,-1){P}\uput[r](4,-1){L} } \meinecaption{2}{Definition of a parabola.} \label{fig:parabola_definition} \end{minipage} \hfill \begin{minipage}{5cm} \centering\psset{unit=1pc, algebraic=true} \rput(-1,-2){\psline[linewidth=2pt,linecolor=brown]{<->}(-4,-1)(4,-1) \parabola[linecolor=brown,linewidth=2pt]{<->}(3,2.25)(0,0) \psline[linecolor=blue,linewidth=2pt](0,1)(2,1)(2,4) \psdots[dotstyle=*,dotscale=1](0,1)(2,1)(2,4) %%%T-square cross-section \pspolygon[fillstyle=solid,fillcolor=black, linewidth=2pt](1,-1.3)(1,-1.1)(3,-1.1)(3,-1.3) %%%T-square trunk \pspolygon[fillstyle=solid,fillcolor=black, linewidth=2pt](2.15,-1)(2.265,-1)(2.265,4)(2.15,4)} \meinecaption{2}{Drawing a parabola.} \label{fig:drawing_parabola} \end{minipage} \hfill \begin{minipage}{5cm} \centering \psset{unit=1pc}\psaxes[linewidth=2pt,labelFontSize=\tiny]{->}(0,0)(-5,-5)(5,5)\psplot[algebraic,linewidth=2pt,linecolor=brown,arrows={<->}]{-2.2361}{2.2361}{x^2} \meinecaption{2}{Example \ref{exa:y=x^2}.}\label{fig:y=x^2} \end{minipage} \end{figure} \begin{thm}\label{thm:parabola_equation} Let $d>0$ be a real number. The equation of a parabola with focus at $(0,d)$ and directrix $y=-d$ is $y=\dfrac{x^2}{4d}$. \end{thm} \begin{pf} Let $(x, y)$ be an arbitrary point on the parabola. Then the distance of $(x, y)$ to the line $y=-d$ is $|y+d|$. The distance of $(x, y)$ to the point $(0,d)$ is $\sqrt{x^2+(y-d)^2}$. We have $$ \begin{array}{lll}|y+d|= \sqrt{x^2+(y-d)^2} & \implies & (|y+d|)^2 = x^2+(y-d)^2\\ & \implies & y^2+2yd+d^2 = x^2+y^2-2yd+d^2\\ & \implies & 4dy = x^2 \\ & \implies & y = \dfrac{x^2}{4d}, \end{array}$$as wanted. \end{pf} \begin{rem} Observe that the midpoint of the perpendicular line segment from the focus to the directrix is on the parabola. We call this point the {\em vertex}\index{parabola!vertex}. For the parabola $y=\dfrac{x^2}{4d}$ of Theorem \ref{thm:parabola_equation}, the vertex is clearly $(0,0)$. \end{rem} \begin{exa}\label{exa:y=x^2} Draw the parabola $y = x^2$. \end{exa}\begin{solu} From Theorem \ref{thm:parabola_equation}, we want $\dfrac{1}{4d}=1$, that is, $d=\dfrac{1}{4}$. Following Theorem \ref{thm:parabola_equation}, we locate the focus at $(0,\frac{1}{4})$ and the directrix at $y=-\dfrac{1}{4}$ and use a T-square with these references. The vertex of the parabola is at $(0,0)$. The graph is in figure \ref{fig:y=x^2}. \end{solu} \vspace*{2cm} \begin{figure}[h] \begin{minipage}{5cm} \centering \psset{unit=1pc}\psaxes[linewidth=2pt, labelFontSize=\tiny]{<->}(0,0)(-5,-5)(5,5) \psplot[linewidth=2pt,linecolor=brown,algebraic=true,arrows={->}]{0}{5}{sqrt(x)}\psplot[linewidth=2pt,linecolor=brown,algebraic=true,arrows={->}]{0}{5}{-sqrt(x)}\meinecaption{3}{$x=y^2$.}\label{fig:x=y^2} \end{minipage} \hfill \begin{minipage}{5cm} \centering \psset{unit=1pc}\psaxes[linewidth=2pt, labelFontSize=\tiny]{<->}(0,0)(-5,-5)(5,5)\psplot[linewidth=2pt,linecolor=brown,algebraic=true,arrows={->}]{0}{5}{sqrt(x)} \meinecaption{3}{$y=\sqrt{x}$.}\label{fig:y=sqrtx} \end{minipage} \hfill \begin{minipage}{5cm} \centering \psset{unit=1pc}\psaxes[linewidth=2pt, labelFontSize=\tiny]{<->}(0,0)(-5,-5)(5,5) \psplot[linewidth=2pt,linecolor=brown,algebraic=true,arrows={->}]{0}{5}{-sqrt(x)}\meinecaption{3}{$y=-\sqrt{x}$.}\label{fig:y=-sqrtx} \end{minipage} \end{figure} \begin{exa}\label{exa:x=y^2} Using Theorem \ref{thm:symmetry_ab_ba}, we may draw the graph of the curve $x=y^2$. Its graph appears in figure \ref{fig:x=y^2}. \end{exa} \begin{exa}\label{exa:y=sqrx} Taking square roots on $x=y^2$, we obtain the graphs of $y=\sqrt{x}$ and of $y=-\sqrt{x}$. Their graphs appear in figures \ref{fig:y=sqrtx} and \ref{fig:y=-sqrtx}. \end{exa} \begin{df} A {\em hyperbola}\index{hyperbola} is the collection of all the points on the plane whose absolute value of the difference of the distances from two distinct fixed points $F_1$ and $F_2$ (called the {\em foci}\footnote{{\em Foci} is the plural of {\em focus}.}\index{hyperbola!foci} of the hyperbola) is a positive constant. See figure \ref{fig:hyperbola_definition}, where $|F_1D-F_2D| = |F_1D'-F_2D'|$. \end{df} We can draw a hyperbola as follows. Put tacks on $F_1$ and $F_2$ and measure the distance $F_1F_2$. Attach piece of thread to one end of the ruler, and the other to $F_2$, while letting the other end of the ruler to pivot around $F_1$. The lengths of the ruler and the thread must satisfy $$\mathrm{length\ of\ the\ ruler} - \mathrm{length\ of\ the\ thread} < F_1F_2.$$Hold the pencil against the side of the rule and tighten the thread, as in figure \ref{fig:drawing_hyperbola}. \vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{5cm} $$\psset{unit=.8pc, algebraic=true} \psline[linestyle=dotted](1.414213562, 1.414213562)(2,.5)(-1.414213562,-1.414213562)\psline[linestyle=dashed](1.414213562, 1.414213562)(-5,-.2)(-1.414213562,-1.414213562) \uput[dl](-1.414213562,-1.414213562){F_1}\uput[ur](1.414213562,1.414213562){F_2}\uput[ur](2,.5){D} \uput[dl](-5,-.2){D'}\psplot[linewidth=2pt,linecolor=brown]{-.166666666}{-6}{1/x} \psplot[linewidth=2pt,linecolor=brown]{.166666666}{6}{1/x}\psdots[dotscale=1,dotstyle=*](1.414213562, 1.414213562)(-1.414213562,-1.414213562)(2,.5)(-5,-.2)$$ \vspace{2cm}\meinecaption{.25}{Definition of a hyperbola.}\label{fig:hyperbola_definition}\end{minipage}\hfill \hfill \begin{minipage}{5cm} $$\psset{unit=.8pc, algebraic=true} \psplot[linewidth=2pt,linecolor=brown, linestyle=dotted]{-.166666666}{-6}{1/x} \psplot[linewidth=2pt,linecolor=brown, linestyle=dotted]{.166666666}{6}{1/x} \rput{60.72238682}(-1.6, -1.2){\pspolygon[fillstyle=solid,fillcolor=black,linewidth=2pt](0,0)(7,0)(7,.5)(0,.5)} \rput{60.72238682}(.5,2){\psline[linecolor=blue,linewidth=2pt](0,0)(3.2,0)\psdots[dotscale=1,dotstyle=*](0,0)(3.2,0)} \psline[linecolor=blue,linewidth=2pt](1.414213562, 1.414213562)(.5,2) \psdots[dotscale=1,dotstyle=*](1.414213562, 1.414213562)(-1.414213562,-1.414213562) $$ \vspace{2cm}\meinecaption{.25}{Drawing a hyperbola.}\label{fig:drawing_hyperbola}\end{minipage} \hfill \begin{minipage}{5cm} $$\psset{unit=.8pc, algebraic=true} \psaxes[linewidth=1.2pt,labels=none,ticks=none]{<->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{-.166666666}{-6}{1/x} \psplot[linewidth=2pt,linecolor=brown]{.166666666}{6}{1/x}$$ \vspace{2cm}\meinecaption{.25}{The hyperbola $y = \dfrac{1}{x}$.}\label{fig:xy=1}\end{minipage} \end{figure} \begin{thm}\label{thm:hyperbola_equation} Let $c>0$ be a real number. The hyperbola with foci at $F_1 = (-c,-c)$ and $F_2=(c,c)$, and whose absolute value of the difference of the distances from its points to the foci is $2c$ has equation $xy = \dfrac{c^2}{2}$. \end{thm} \begin{pf} Let $(x, y)$ be an arbitrary point on the hyperbola. Then $$\begin{array}{l} \left|\distance{(x,y)}{(-c,-c)} - \distance{(x,y)}{(c,c)}\right| = 2c \\ \iff \left|\sqrt{(x+c)^2+(y+c)^2} - \sqrt{(x-c)^2+(y-c)^2}\right| = 2c\\ \iff (x+c)^2+(y+c)^2 +(x-c)^2+(y-c)^2 - 2\sqrt{(x+c)^2+(y+c)^2}\cdot\sqrt{(x-c)^2+(y-c)^2} = 4c^2\\ \iff 2x^2 + 2y^2 = 2\sqrt{(x^2 + y^2 + 2c^2)+(2xc+2yc)}\cdot\sqrt{(x^2 + y^2 + 2c^2)-(2xc+2yc)} \\ \iff 2x^2 + 2y^2 = 2\sqrt{(x^2 + y^2 + 2c^2)^2-(2xc+2yc)^2} \\ \iff (2x^2 + 2y^2)^2 = 4\left((x^2 + y^2 + 2c^2)^2-(2xc+2yc)^2\right) \\ \iff 4x^4 + 8x^2y^2 + 4y^4 = 4((x^4+y^4+4c^4+2x^2y^2 + 4y^2c^2 + 4x^2c^2)-(4x^2c^2 + 8xyc^2 + 4y^2c^2)) \\ \iff xy = \dfrac{c^2}{2}, \end{array} $$ where we have used the identities $$(A + B+C)^2 = A^2+ B^2 + C^2 + 2AB+2AC + 2BC\qquad \mathrm{and} \qquad \sqrt{A-B}\cdot \sqrt{A+B} = \sqrt{A^2-B^2}.$$ \end{pf} \begin{rem} Observe that the points $\left(-\dfrac{c}{\sqrt{2}}, -\dfrac{c}{\sqrt{2}}\right)$ and $\left(\dfrac{c}{\sqrt{2}}, \dfrac{c}{\sqrt{2}}\right)$ are on the hyperbola $xy= \dfrac{c^2}{2}$. We call these points the {\em vertices}\footnote{{\em Vertices} is the plural of {\em vertex}.}\index{hyperbola!vertices} of the hyperbola $xy= \dfrac{c^2}{2}$. \end{rem} \begin{exa}\label{exa:y=1/x}To draw the hyperbola $y = \dfrac{1}{x}$ we proceed as follows. According to Theorem \ref{thm:hyperbola_equation}, its two foci are at $(-\sqrt{2}, -\sqrt{2})$ and $(\sqrt{2}, \sqrt{2})$. Put ${\red \mathrm{length\ of\ the\ ruler} - \mathrm{length\ of\ the\ thread} = 2\sqrt{2}.}$ By alternately pivoting about these points using the procedure above, we get the picture in figure \ref{fig:xy=1}. \end{exa} \begin{df}An {\em ellipse} is the collection of points on the plane whose sum of distances from two fixed points, called the {\em foci}, is constant.\index{ellipse} \end{df} \begin{thm}The equation of an ellipse with foci $F_1 = (h - c, k)$ and $F_2 = ( h + c, k)$ and sum of distances is the constant $t = 2a$ is $$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1, $$ where $b^2 = a^2 - c^2$. \end{thm} \begin{pf} By the triangle inequality, $t > F_1F_2 = 2c$, from where $a > c$. It follows that $$\begin{array}{lll}& & \distance{(x, y)}{(x_1, y_1)} + \distance{(x, y)}{(x_2, y_2)} = t \\ & \iff & \sqrt{(x - h + c)^2 + (y - k)^2} = 2a -\sqrt{(x - h - c)^2 + (y - k)^2}\\ & \iff & (x - h + c)^2 + (y - k)^2 = 4a^2 - 4a\sqrt{(x - h - c)^2 + (y - k)^2} + (x - h - c)^2 + (y - k)^2\\ & \iff & (x - h)^2 + 2c(x - h) + c^2 = 4a^2 - 4a\sqrt{(x - h - c)^2 + (y - k)^2} + (x - h)^2 - 2c(x - h) + c^2\\ & \iff & (x - h)c - a^2 = -a\sqrt{(x - h - c)^2 + (y - k)^2}\\ & \iff & (x - h)^2c^2 - 2a^2c(x - h) + a^2 = a^2(x - h - c)^2 + a^2(y - k)^2\\ & \iff & (x - h)^2c^2 - 2a^2c(x - h) + a^2 = a^2(x - h)^2 - 2a^2c(x - h) + a^2c^2 + a^2(y - k)^2 \\ & \iff & (x - h)^2(c^2 - a^2) - a^2(y - k)^2 = a^2c^2 - a^2 \\ & \iff & \dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{a^2 - c^2} = 1. \end{array}$$ Since $a^2 - c^2 > 0$, we may let $b^2 = a^2 - c^2$, obtaining the result\end{pf} \begin{df}The line joining $(h + a, k)$ and $(h - a, k)$ is called the {\em horizontal axis} of the ellipse and the line joining $(h, k - b)$ and $(h, k + b)$ is called the {\em vertical axis} of the ellipse. $\max(a, b)$ is the {\em semi-major axis} and $\min (a, b)$ the {\em semi-minor axis}. \end{df} \begin{rem}The canonical equation of an ellipse whose semi-axes are parallel to the coordinate axes is thus $$ \dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1.$$ \end{rem} \vspace{1cm} \begin{figure}[h] \centering\psset{unit=1.5pc} \pstLineAB{A}{F_1}\pstLineAB{A}{F_2} \psellipse[linewidth=2pt,linecolor=brown](0,0)(4,1) \pstGeonode[PointName=none](2,.866025404){A}(-1.732050808,0){F_1}(1.732050808,0){F_2} \psline(F_1)(A)(F_2) \meinecaption{1}{Drawing an ellipse.}\label{ellipse_1_1} \end{figure} Figure \ref{ellipse_1_1} shews how to draw an ellipse by putting tags on the foci, tying the ends of a string to them and tightening the string with a pencil. \begin{exa} The curve of equation $9x^2 - 18x + 4y^2 + 8y = 23$ is an ellipse, since, by completing squares, $$9(x^2 - 2x + 1) + 4(y^2 + 2y + 1) = 23 + 9 + 4 \implies 9(x - 1)^2 + 4(y + 1)^2 = 36\implies \frac{(x - 1)^2}{4} + \frac{(y + 1)^2}{9} = 1.$$The centre of the ellipse is $(h, k) = (1, - 1).$ The semi-major axis measures $\sqrt{9} = 3$ units and the semi-minor axis measures $\sqrt{4} = 2$ units. \end{exa} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Let $d>0$ be a real number. Prove that the equation of a parabola with focus at $(d,0)$ and directrix $x=-d$ is $x=\dfrac{y^2}{4d}$. \end{pro} \begin{pro} Find the focus and the directrix of the parabola $x=y^2$. \begin{answer} By the preceding exercise the focus is $(\frac{1}{4}, 0)$ and the directrix is $x=-\frac{1}{4}$. \end{answer} \end{pro} \begin{pro} Find the equation of the parabola with directrix $y = -x$ and vertex at $(1, 1)$. \begin{answer} If $(x, y)$ is an arbitrary point on this parabola we must have $$ \dfrac{|-x-y|}{\sqrt{1+(-1)^2}} = \sqrt{(x-1)^2 + (y-1)^2} $$ Squaring and rearranging, the desired equation is $$x^2 + y^2 -2xy-4x-4y+4=0.$$ \end{answer} \end{pro} \begin{pro} Draw the curve $x^2+2x +4y^2-8y=4$. \end{pro} \begin{pro} The point $(x, y)$ moves on the plane in such a way that it is equidistant from the point $(2, 3)$ and the line $x = -4$. Find the equation of the curve it describes. \begin{answer} The distance of $(x,y)$ to $(2, 3)$ is $\sqrt{(x-2)^2+(y-3)^2}$. The distance of $(x, y)$ to the line $x=-4$ is $\absval{x-(-4)}=\absval{x+4}$. We need $$\sqrt{(x-2)^2+(y-3)^2} = \absval{x+4} \iff (x-2)^2+(y-3)^2 = (x+4)^2\iff x=\dfrac{y^2}{12}-\dfrac{y}{2}-\dfrac{1}{2}, $$ from where the desired curve is a parabola. \end{answer} \end{pro} \begin{pro}\label{exa:parabola-triangulo} The points $A(0,0)$ , $B$, and $C$ lie on the parabola $y=\dfrac{x^2}{2}$ as shewn in figure \ref{fig:parabola-triangulo}. If $\triangle ABC$ is equilateral, determine the coordinates of $B$ and $C$. \vspace{1cm} \begin{figurehere} $$ \psset{unit=.5pc}\psaxes[ticks=none,labels=none]{->}(0,0)(-2,-2)(5,8) \psplot[linewidth=2pt,linecolor=brown, algebraic=true]{-3.6}{3.6}{(x^2)/2} \psline[linewidth=2pt,linecolor=blue](0,0)(3.464101616,6)(-3.464101616,6)(0,0) \uput[dl](0,0){A}\uput[r](3.464101616,6){B}\uput[l](-3.464101616,6){C} $$ \meinecaption{.5}{Problem \ref{exa:parabola-triangulo}.}\label{fig:parabola-triangulo} \end{figurehere} \begin{answer} Put $x>0$ and $A=(0,0)$, $B=\left(x, \dfrac{x^2}{2}\right)$ y $C=\left(-x, \dfrac{x^2}{2}\right)$. Then $$AB=BC\implies \sqrt{(x-0)^2+\left(\dfrac{x^2}{2}-0\right)^2} = \sqrt{(x-(-x))^2+ \left(\dfrac{x^2}{2}-\dfrac{x^2}{2}\right)^2}\implies x^2\left(1+\dfrac{x^2}{4}\right) = 4x^2\implies x=2\sqrt{3}.$$ The points are $A(0,0)$, $B(2\sqrt{3}, 6)$ and $C(-2\sqrt{3},6)$. \end{answer} \end{pro} \end{multicols} \chapter{Functions} This chapter introduces the central concept of a {\em function}. We will only concentrate on functions defined by algebraic formul\ae\ with inputs and outputs belonging to the set of real numbers. We will introduce some basic definitions and will concentrate on the algebraic aspects, as they pertain to formul\ae\ of functions. The subject of {\em graphing} functions will be taken in subsequent chapters. \section{Basic Definitions} \vspace{1cm} \begin{figure}[!hptb]\centering \psset{unit=.7cm}\rput(-2,0){ \pscircle(-2,0){2} \psellipse[fillcolor=yellow, fillstyle=solid](6.5,0)(5,2.2) \pscircle[fillcolor=white, fillstyle=solid](6,0){2} \psline[linewidth=.7pt, labels=none, showpoints=true,arrowsize=.3,nodesep=1]{->}(-1.8,0)(5.9,0) \psline[linewidth=.7pt, labels=none, showpoints=true,arrowsize=.3,nodesep=1]{->}(-1.8,.5)(5.9,.5) \psline[linewidth=.7pt, labels=none, showpoints=true,arrowsize=.3,nodesep=1]{->}(-1.8,-.5)(5.9,-.5) \rput(-2, -1){$\dom{f}$}% \rput(6,1){$\im{f}$}% \rput(1,1){$f$}% \rput(10, 0){$\target{f}$}% \psdots*[linewidth=.5pt](-2,0)(-2, .5)(-2, -.5)(6,0)(6,.5)(6, -.5)(8.3,1)(8.3,.5)} \meinecaption{2}{The main ingredients of a function.} \label{df:function} \end{figure} \begin{df} By a {\em (real-valued) function}\index{function} $\fun{f}{x}{f(x)}{\dom{f}}{\target{f}}$ we mean the collection of the following ingredients: \begin{enumerate} \item a {\em name} for the function. Usually we use the letter $f$. \item a set of real number inputs---usually an interval or a finite union of intervals---called the {\em domain} of the function. The domain of $f$ is denoted by $\dom{f}$. \item an {\em input parameter }, also called {\em independent variable} or {\em dummy variable}. We usually denote a typical input by the letter $x$.\index{function!domain} \item a set of possible real number outputs---usually an interval or a finite union of intervals---of the function, called the {\em target set} of the function. The target set of $f$ is denoted by $\target{f}$.\index{function!target set} \item an {\em assignment rule} or {\em formula}, assigning to {\bf every input} a {\bf unique} output. This assignment rule for $f$ is usually denoted by $x\mapsto f(x)$. The output of $x$ under $f$ is also referred to as the {\em image of $x$ under $f$}, \index{function!image} and is denoted by $f(x)$. \index{function!assingment rule} \end{enumerate}See figure \ref{df:function}. \end{df} \begin{df} Colloquially, we refer to the ``function $f$'' when all the other descriptors of the function are understood. \end{df} \begin{df} The {\em image} of a function $\fun{f}{x}{f(x)}{\dom{f}}{\target{f}}$ is the set $$ \im{f} = \{f(x): x\in \dom{f}\}, $$that is, the collection of all outputs of $f$.\index{function!image of a} \end{df} \begin{rem} Necessarily we have $\im{f} \subseteq \target{f}$, but we will see later on that these two sets may not be equal. \end{rem} \begin{exa}Find all functions with domain $\{a, b\}$ and target set $\{c, d\}$. \end{exa} \begin{solu} Since there are two choices for the output of $a$ and two choices for the output of $b$, there are $2^2 = 4$ such functions, namely: \begin{multicols}{2} \begin{enumerate} \item $f_1$ given by $f_1(a) = f_1(b) = c.$ Observe that $\im{f_1} = \{c\}$. \item $f_2$ given by $f_2(a) = f_2(b) = d.$ Observe that $\im{f_2} = \{d\}$. \item $f_3$ given by $f_3(a) = c, f_3(b) = d.$ Observe that $\im{f_1} = \{c, d\}$. \item $f_4$ given by $f_4(a) = d, f_4(b) = c.$ Observe that $\im{f_1} = \{c, d\}$. \end{enumerate} \end{multicols} \end{solu} \begin{rem} It is easy to see that if $A$ has $n$ elements and $B$ has $m$ elements, then the number of functions from $A$ to $B$ is $m^n$. For, if $a_1, a_2, \ldots , a_n$ are the elements of $A$, then there are $m$ choices for the output of $a_1$, $m$ choices for the output of $a_2$, \ldots , $m$ choices for the output of $a_n$, giving a total of $$ \underbrace{m\cdots m}_{n\ \mathrm{times}}=m^n. $$ possibilities. \end{rem} In some computer programming languages like C, C$++$, and Java, one defines functions by statements like {\tt int f(double)}. This tells the computer that the input set is allocated enough memory to take a double (real number) variable, and that the output will be allocated enough memory to carry an integer variable. \begin{exa}Consider the function $$ \fun{f}{x}{x^2}{\reals}{\reals}. $$ Find the following: \begin{enumerate} \item $f(0)$ \item $f(-\sqrt{2})$ \item $f(1-\sqrt{2})$ \item What is $\im{f}$? \end{enumerate} \end{exa} \begin{solu} We have \begin{enumerate} \item $f(0)=0^2=0$ \item $f(-\sqrt{2})=(-\sqrt{2})^2=2$ \item $f(1-\sqrt{2})=(1-\sqrt{2})^2=1^2-2\cdot 1 \cdot \sqrt{2}+(\sqrt{2})^2=3-2\sqrt{2}$ \item Since the square of every real number is positive, we have $\im{f}\subseteq \co{0;+\infty}$. Now, let $a\in \co{0;+\infty}$. Then $\sqrt{a}\in \reals$ and $f(\sqrt{a})=a$, so $a\in\im{f}$. This means that $\co{0;+\infty}\subseteq \im{f}$. We conclude that $\im{f}=\co{0;+\infty}$. \end{enumerate} \end{solu} In the above example it was relatively easy to determine the image of the function. In most cases, this calculation is in fact very difficult. This is the reason why in the definition of a function we define the target set to be the set of all {\em possible outputs}, not the actual outputs. The target set must be large enough to accommodate all the possible outputs of a function. \begin{exa}Does $$ \fun{f}{x}{x^2}{\reals}{\integers}. $$ define a function? \end{exa} \begin{solu} No. The target set is not large enough to accommodate all the outputs. The above rule is telling us that every output belongs to $\integers$. But this is not true, since for example, $f(1-\sqrt{2})=3-2\sqrt{2}\nin\integers$. \end{solu} Upon consideration of the preceding example, the reader may wonder why not then, select as target set the entire set $\reals$. This is in fact what is done in practice, at least in Calculus. From the point of view of Computer Programming, this is wasteful, as we would be allocating more memory than really needed. When we introduce the concept of {\em surjections} later on in the chapter, we will see the importance of choosing an appropriate target set. \begin{exa}Does $$ \fun{f}{x}{\dfrac{1}{x^2}}{\reals}{\reals}. $$ define a function? \end{exa} \begin{solu} No. In a function, every input must have a defined output. Since $f(0)$ is undefined, this is not a function. \end{solu} \begin{df}[Equality of Functions] Two functions are equal if \begin{enumerate} \item Their domains are identical. \item Their target sets are identical. \item Their assignment rules are identical. \end{enumerate} \end{df} This means that the only two things that can be different are the names of the functions and the name of the input parameter. \begin{exa} Consider the functions $$ \fun{f}{x}{x^2}{\integers}{\integers}\qquad \fun{g}{s}{s^2}{\integers}{\integers}\qquad \fun{h}{x}{x^2}{\integers}{\reals}. $$ Then the functions $f$ and $g$ are the same function. The functions $f$ and $h$ are different functions, as their target sets are different. \end{exa} We must pay special attention to the fact that although a formula may make sense for a ``special input'', the ``input'' may not be part of the domain of the function. \begin{exa} Consider the function $$ \fun{f}{x}{\dfrac{1}{x+\dfrac{1}{x}}}{\naturals \setminus \{0\}}{\rationals}. $$ Determine: \begin{enumerate} \item $f(1)$ \item $f(2)$ \item $f\left(\dfrac{1}{2}\right)$ \item $f(-1)$ \end{enumerate} \end{exa} \begin{solu} \noindent \begin{enumerate} \item $f(1)=\dfrac{1}{1+\dfrac{1}{1}}=\dfrac{1}{2}$ \item $f(2)=\dfrac{1}{2+\dfrac{1}{2}} = \dfrac{1}{\dfrac{5}{2}}=\dfrac{2}{5}$ \item $f\left(\dfrac{1}{2}\right)=\dfrac{1}{\dfrac{1}{2}+\dfrac{1}{\dfrac{1}{2}}}=\dfrac{1}{\dfrac{1}{2}+2}=\dfrac{2}{5}$ \item $f(-1)$ is undefined, as $-1\nin\naturals \setminus \{0\}$, that is $-1$ is not part of the domain. \end{enumerate} \end{solu} It must be emphasised that the exhaustion of the elements of the domain is crucial in the definition of a function. For example, the diagram in figure \ref{mult_ima} {\em does not} represent a function, as some elements of the domain are not assigned. Also important in the definition of a function is the fact that the output must be unique. For example, the diagram in \ref{non-exhau} {\em does not} represent a function, since the last element of the domain is assigned to two outputs. \vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{7cm}\centering \psset{unit=1pc} \pstGeonode[PointName=none](0,0){A}(0,1){B}(0,2){C} \pstGeonode[PointName=none](5,0){A'}(5,1){B'}(5,2){C'} \pcline[nodesep=.2, arrows={->},arrowsize=.5](A)(A') \pcline[nodesep=.2, arrows={->},arrowsize=.5](B)(B') \pstMiddleAB[PointName=none,PointSymbol=none]{A}{C}{O} \pstMiddleAB[PointName=none,PointSymbol=none]{A'}{C'}{O'} \psellipse(O)(.5,2.5)\psellipse(O')(.5,2.5)\meinecaption{2}{Not a function.}\label{mult_ima}\end{minipage}\hfill \begin{minipage}{7cm} \centering\psset{unit=1pc} \pstGeonode[PointName=none](0,0){A}(0,1){B}(0,2){C} \pstGeonode[PointName=none](5,0){A'}(5,1){B'}(5,2){C'} \pcline[nodesep=.2, arrows={->},arrowsize=.5](A)(A') \pcline[nodesep=.2, arrows={->},arrowsize=.5](B)(B') \pcline[nodesep=.2, arrows={->},arrowsize=.5](C)(C') \pcline[nodesep=.2, arrows={->},arrowsize=.5](A)(B') \pstMiddleAB[PointName=none,PointSymbol=none]{A}{C}{O} \pstMiddleAB[PointName=none,PointSymbol=none]{A'}{C'}{O'} \psellipse(O)(.5,2.5)\psellipse(O')(.5,2.5) \meinecaption{2}{Not a function.} \label{non-exhau} \end{minipage} \end{figure} To conclude this section, we will give some miscellaneous examples on evaluation of functions. \begin{exa}[The Identity Function]\index{function!identity} Consider the function $$\fun{\idefun}{x}{x}{\reals}{\reals}.$$ This function assigns to every real its own value. Thus $\idefun (-1) = -1$, $\idefun (0) = 0$, $\idefun (4) = 4$, etc. \end{exa} \begin{rem} In general, if $A\subseteq \reals$, the identity function on the set $A$ is defined and denoted by $$\fun{\idefun _A}{x}{x}{A}{A}.$$ \end{rem} \begin{exa} Let $\fun{\gamma}{x}{x^2 - 2}{\reals}{\reals}.$ Find $\gamma(x^2+1)-\gamma(x^2-1)$. \end{exa} \begin{solu} We have $$ \gamma(x^2+1)-\gamma(x^2-1)=((x^2+1)^2-2)-((x^2-1)^2-2) =(x^4+2x^2+1-2)-(x^4-2x^2+1-2)=4x^2.$$ \end{solu} Sometimes the assignment rule of a function varies through various subsets of its domain. We call any such function a {\em piecewise-defined function}. \index{function!piecewise-defined} \begin{exa} Consider the function $f:\cc{-5;4}\to \reals$ defined by $$ f(x)=\left\{\begin{array}{ll} 1 & \mathrm{if}\ 2x\in \co{-5;1}\\ 2 & \mathrm{if}\ x=1\\ x+1 & \mathrm{if}\ x\in \oc{1;4}\\ \end{array}\right. $$ Determine $f(-3)$, $f(1)$, $f(4)$ and $f(5)$. \end{exa} \begin{solu} Plainly, $f(-3)=2(-3)=-6$, $f(1)=2$, $f(4)=4+1=5$, and $f(5)$ is undefined. \end{solu} \begin{exa} Write $f:\reals\rightarrow \reals$, $f(x) = |2x-1|$ as a piecewise-defined function. \end{exa} \begin{solu} We have $f(x) = 2x -1$ for $2x-1\geq 0$ and $f(x)=-(2x-1)$ for $2x-1<0$. This gives \renewcommand{\arraystretch}{1.5} $$f(x) = \left\{\begin{array}{ll} 2x-1 & \mathrm{if} \ x\leq \frac{1}{2} \\ 1-2x & \mathrm{if} \ x>\frac{1}{2}\\ \end{array}\right. $$ \end{solu} Lest the student think that evaluation of functions is a simple affair, let us consider the following example. \begin{exa} Let $f:\reals\rightarrow \reals$ satisfy $f(2x + 4) = x^2 - 2$. Find \begin{enumerate} \item $f(6)$ \item $f(1)$ \item $f(x)$ \item $f(f(x))$ \end{enumerate} \end{exa} \begin{solu}Since $2x+4$ is what is inside the parentheses in the formula given, we need to make all inputs equal to it. \begin{enumerate} \item We need $2x + 4 = 6 \implies x = 1$. Hence $$f(6) = f(2(1) + 4) = 1^2 - 2 = -1.$$ \item We need $2x + 4 = 1 \implies x = -\frac{3}{2}$. Hence $$f(1) = f\left(2\left(-\frac{3}{2}\right) + 4\right) = \left(-\frac{3}{2}\right)^2 - 2 = \frac{1}{4}.$$ \item Here we confront a problem. If we proceeded blindly as before and set $2x+4=x$, we would get $x=-4$, which does not help us much, because what we are trying to obtain is $f(x)$ {\em for every value of $x$.} The key observation is that the dummy variable has no idea of what one is calling it, hence, we may first rename the dummy variable: say $f(2u + 4) = u^2 - 2$. We need $2u + 4 = x \implies u = \dfrac{x - 4}{2}$. Hence $$f(x) = f\left(2\left(\frac{x - 4}{2} \right)+4\right) = \left(\frac{x - 4}{2}\right)^2 - 2 = \frac{x^2}{4} - 2x + 2.$$ \item Using the above part, $$\begin{array}{lll}f(f(x)) & = & \dfrac{(f(x))^2}{4} - 2f(x) + 2 \\ & = & \dfrac{\left(\dfrac{x^2}{4} - 2x + 2\right)^2}{4} - 2\left(\dfrac{x^2}{4} - 2x + 2\right) + 2 \\ & = & \dfrac{x^4}{64}-\dfrac{x^3}{4}+\dfrac{3x^2}{4}+2x-1 \end{array}$$ \end{enumerate} \end{solu} \begin{exa} $f:\reals \rightarrow \reals$ is a function satisfying $f(3) = 2$ and $f(x + 3) = f(3)f(x)$. Find $f(-3)$. \end{exa} \begin{solu} Since we are interested in $f(-3)$, we first put $x = -3$ in the relation, obtaining $$f(0) = f(3)f(-3). $$Thus we must also know $f(0)$ in order to find $f(-3)$. Letting $x = 0$ in the relation, $$f(3) = f(3)f(0)\implies f(3)=f(3)f(3)f(-3)\implies 2=4f(-3) \implies f(-3) = \dfrac{1}{2}. $$ \end{solu} The following example is a surprising application of the concept of function. \begin{exa} Consider the polynomial $(x^2-2x+2)^{2008}$. Find its constant term. Also, find the sum of its coefficients after the polynomial has been expanded and like terms collected. \end{exa} \begin{solu} The polynomial has degree $2\cdot 2008 = 4016$. This means that after expanding out, it can be written in the form $$(x^2-2x+2)^{2008}=a_0x^{4016}+a_1x^{4015}+ \cdots + a_{4015}x+a_{4016}.$$ Consider now the function $$\fun{p}{x}{a_0x^{4016}+a_1x^{4015}+ \cdots + a_{4015}x+a_{4016}}{\reals}{\reals}. $$ The constant term of the polynomial is $a_{4016}$, which happens to be $p(0)$. Hence the constant term is $$a_{4016}=p(0) = (0^2-2\cdot 0 +2)^{2008}=2^{2008}. $$ The sum of the coefficients of the polynomial is $$a_0+a_1+a_2+\cdots +a_{4016}=p(1) = (1^2-2\cdot 1 +2)^{2008} =1. $$ \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Let $$ \fun{f}{x}{\dfrac{x-1}{x^2+1}}{\reals}{\reals}. $$Find $f(0)+f(1)+f(2)$ and $f(0+1+2)$. Is it true that $$f(0)+f(1)+f(2)=f(0+1+2)\ ?$$Is there a real solution to the equation $f(x)=\dfrac{1}{x}$? Is there a real solution to the equation $f(x)=x$? \begin{answer} We have $$f(0)+f(1)+f(2)=\dfrac{0-1}{0^2+1}+\dfrac{1-1}{1^2+1}+\dfrac{2-1}{2^2+1}=-1+0+\dfrac{1}{5}= -\dfrac{4}{5}. $$ Also, $$ f(0+1+2)=f(3)=\dfrac{3-1}{3^2+1}=\dfrac{1}{5}. $$ Clearly then $f(0)+f(1)+f(2)\neq f(0+1+2)$. \bigskip Now, $$f(x)=\dfrac{1}{x}\implies x^2-x=x^2+1 \implies x=-1. $$ Also, $$f(x)=x\implies x-1=x^3+x \implies x^3=-1 \implies x=-1. $$ \end{answer} \end{pro} \begin{pro} Find all functions from $\{0, 1, 2\}$ to $\{-1, 1\}$. \begin{answer} There are $2^3 = 8$ such functions: \begin{enumerate} \item $f_1$ given by $f_1(0) = f_1(1) = f_1(2) = -1$ \item $f_2$ given by $f_2(0) = 1, f_2(1) = f_2(2) = -1$ \item $f_3$ given by $f_3(0) = f_3(1) = -1, f_3(2) = 1$ \item $f_4$ given by $f_4(0) = -1, f_4(1) = 1, f_4(2) = -1$ \item $f_5$ given by $f_5(0) = f_5(1) = f_5(2) = 1$ \item $f_6$ given by $f_6(0) = -1, f_6(1) = f_6(2) = 1$ \item $f_7$ given by $f_7(0) = f_7(1) = 1, f_7(2) = -1$ \item $f_8$ given by $f_8(0) = 1, f_8(1) = - 1, f_8(2) = 1$ \end{enumerate} \end{answer} \end{pro} \begin{pro} Find all functions from $\{-1, 1\}$ to $\{0, 1, 2\}$ . \begin{answer} There are $3^2 = 9$ such functions: \begin{enumerate} \item $f_1$ given by $f_1(-1) = f_1(1) = 0$ \item $f_2$ given by $f_2(-1) =f_2(1) = 1$ \item $f_3$ given by $f_3(-1) = f_3(1) = 2$ \item $f_4$ given by $f_4(-1) = 0, f_4(1) = 1$ \item $f_5$ given by $f_5(-1) = 0, f_5(1) = 2$ \item $f_6$ given by $f_6(-1) = 1, f_6(1) = 2$ \item $f_7$ given by $f_7(-1) = 1, f_7(1) = 0$ \item $f_8$ given by $f_8(-1) = 2, f_8(1) = 0$ \item $f_9$ given by $f_9(-1) = 2, f_9(1) = 1$ \end{enumerate} \end{answer} \end{pro} \begin{pro} Let $f:\reals \rightarrow \reals$, $x\mapsto x^2 - x$. Find $$\dfrac{f(x + h) - f(x - h)}{h}.$$ \begin{answer} $4x - 2$ \end{answer} \end{pro} \begin{pro} Let $f:\reals \rightarrow \reals$, $x\mapsto x^3 -3x$. Find $$\dfrac{f(x + h) - f(x - h)}{h}.$$ \begin{answer} $6x^2+2h^2-6$ \end{answer} \end{pro} \begin{pro} Consider the function $f:\reals\setminus\{0\}\to \reals$, $f(x)=\dfrac{1}{x}$. Which of the following statements are always true? \begin{enumerate} \item $f\left(\dfrac{a}{b}\right)=\dfrac{f(a)}{f(b)}$. \item $f(a+b)=f(a)+f(b)$. \item $f(a^2)=(f(a))^2$ \end{enumerate} \begin{answer} \noindent \begin{enumerate} \item True. $f\left(\dfrac{a}{b}\right)=\dfrac{1}{\dfrac{a}{b}}=\dfrac{b}{a}=\dfrac{1}{a}\cdot\dfrac{1}{\dfrac{1}{b}}=\dfrac{f(a)}{f(b)}$. \item False. For example, $f(1+1)=f(2)=\dfrac{1}{2}$, but $f(1)+f(1)=\dfrac{1}{1}+\dfrac{1}{1}=2$. \item True. $f(a^2)=\dfrac{1}{a^2}=\left(\dfrac{1}{a}\right)^2=(f(a))^2$. \end{enumerate} \end{answer} \end{pro} \begin{pro} Let $a:\reals \rightarrow \reals$, be given by $a(2 - x) = x^2 - 5x$. Find $a(3)$, $a(x)$ and $a(a(x))$. \begin{answer}$a(3) = 6$; $x^2 + x - 6$; $24-11x-10x^2+2x^3+x^4 $\end{answer} \end{pro} \begin{pro} Let $f:\reals \rightarrow \reals$, $f(1 - x) = x^2 - 2.$ Find $f(-2)$, $f(x)$ and $f(f(x))$. \begin{answer} $7$, $x^2 -2x - 1$, $x^4-4x^3+8x+2$ \end{answer} \end{pro} \begin{pro}Let $f:\dom{f} \rightarrow \reals$ be a function. $f$ is said to have a {\em fixed point} at $t\in\dom{f}$ if $f(t) = t$. Let $s:[0; +\infty[ \rightarrow \reals$, $s(x) = x^5 - 2x^3 + 2x$. Find all fixed points of $s$. \begin{answer} We must look for all $x\in\dom{f}$ such that $s(x) = x$. Thus $$\begin{array}{lll} s(x) = x & \implies & x^5 -2x^3 + 2x = x \\ & \implies & x^5 -2x^3 + x = 0 \\ & \implies & x(x^4 - 2x^2 + 1) = 0 \\ & \implies & x(x^2 - 1)^2 = 0 \\ & \implies & x(x + 1)^2(x - 1)^2 = 0. \end{array}$$ The solutions to this last equation are $\{-1, 0, 1\}$. Since $-1\not\in \dom{s}$, the only fixed points of $s$ are $x = 0$ and $x = 1$. \end{answer} \end{pro} \begin{pro} Let $:\reals \rightarrow \reals$, $h(x + 2) = 1 + x - x^2$. Express $h(x - 1)$, $h(x)$, $h(x + 1)$ as powers of $x$. \begin{answer} $h(x - 1) = -11 + 7x - x^2$; \ $h(x) = -5 + 5x - x^2$; \ $h(x + 1) = -1 + 3x - x^2$ \end{answer} \end{pro} \begin{pro} Let $f:\reals \rightarrow \reals$, $f(x + 1) = x^2.$ Find $f(x), f(x + 2)$ and $f(x - 2)$ as powers of $x$. \begin{answer} $f(x) = x^2 - 2x + 1$; \ $f(x + 2)=x^2 + 2x + 1$; \ $f(x - 2) = x^2 - 6x + 9$ \end{answer} \end{pro} \begin{pro} Let $h:\reals \rightarrow \reals$ be given by $h(1 - x) = 2x$. Find $h(3x)$. \begin{answer} Rename the independent variable, say $h(1 - s) = 2s.$ Now, if $1 - s = 3x$ then $s = 1 - 3x.$ Hence $$h(3x) = h(1 - s) = 2s = 2(1 - 3x) = 2 - 6x.$$ \end{answer} \end{pro} \begin{pro} Consider the polynomial $$(1 - x^2 + x^4)^{2003} = a_0 + a_1x+a_2x^2 + \cdots + a_{8012}x^{8012}. $$Find \begin{enumerate} \item $a_0$ \item $a_0 + a_1+a_2 + \cdots + a_{8012}$ \item $a_0 - a_1+a_2 -a_3 + \cdots -a_{8011}+ a_{8012}$ \item $a_0+a_2 + a_4 + \cdots + a_{8010} + a_{8012}$ \item $a_1+a_3+ \cdots +a_{8009}+a_{8011}$ \end{enumerate} \begin{answer} Consider the function $p:\reals \to \reals$, with $$p(x) = (1 - x^2 + x^4)^{2003} = a_0 + a_1x+a_2x^2 + \cdots + a_{8012}x^{8012}.$$Then \begin{enumerate} \item $a_0 = p(0) = (1 - 0^2 + 0^4)^{2003} = 1.$ \item $a_0 + a_1+a_2 + \cdots + a_{8012} = p(1) = (1 - 1^2 + 1^4)^{2003} = 1.$ \item $$\begin{array}{lll}a_0 - a_1+a_2-a_3 + \cdots -a_{8011} + a_{8012} &= & p(-1)\\ & = & (1 - (-1)^2 + (-1)^4)^{2003}\\ & = & 1. \end{array}$$ \item The required sum is $\dfrac{p(1)+p(-1)}{2} = 1$. \item The required sum is $\dfrac{p(1)-p(-1)}{2} = 0$. \end{enumerate} \end{answer} \end{pro} \begin{pro} Let $f:\reals \rightarrow \reals$, be a function such that $\forall x\in ]0;+\infty[$, $$[f(x^3 + 1)]^{\sqrt{x}} = 5,$$ find the value of $$\left[f\left(\frac{27 + y^3}{y^3}\right)\right]^{\sqrt{\frac{27}{y}}}$$ for $y\in ]0;+\infty[$. \begin{answer}We have $$\begin{array}{lll}\left[f\left(\frac{27 + y^3}{y^3}\right)\right]^{\sqrt{\frac{27}{y}}} & = & \left[f\left( \left(\frac{3}{y}\right)^3 + 1\right)\right]^{3\sqrt{\frac{3}{y}}} \\ & = & \left(\left[f\left( \left(\frac{3}{y}\right)^3 + 1\right)\right]^{\sqrt{\frac{3}{y}}}\right)^3 \\ & = & 5^3 \\ & = & 125. \end{array}$$ \end{answer} \end{pro} \end{multicols} \section{Graphs of Functions and Functions from Graphs} In this section we {\em briefly} describe graphs of functions. The bulk of graphing will be taken up in subsequent chapters, as graphing functions with a given formula is a very tricky matter. \begin{df} The {\em graph}\index{function!graph} of a function $\fun{f}{x}{f(x)}{\dom{f}}{\target{f}}$ is the set $\Gamma _f =\{(x,y)\in\reals^2 : y=f(x)\}$ on the plane. For ellipsis, we usually say {\em the graph of $f$}, or {\em the graph $y=f(x)$} or the {\em the curve $y=f(x)$}. \end{df} By the definition of the graph of a function, the $x$-axis contains the set of inputs and $y$-axis has the set of outputs. Since in the definition of a function every input goes to exactly one output, wee see that {\em if a vertical line crosses two or more points of a graph, the graph does not represent a function}. We will call this the {\em vertical line test} for a function. See figures \ref{fig:not_functions1} and \ref{fig:not_functions2}. \bigskip At this stage there are very few functions with a given formula and infinite domain that we know how to graph. Let us list some of them. \begin{exa}[Identity Function]\index{function!identity} Consider the function $$\fun{\idefun}{x}{x}{\reals}{\reals}.$$ By Theorem \ref{thm:graph_mx+k}, the graph of the identity function is a straight line. \end{exa} \begin{exa}[Absolute Value Function]\index{function!absolute value} Consider the function $$\fun{{\bf AbsVal}}{x}{|x|}{\reals}{\reals}.$$ By Example \ref{exa:absval1}, the graph of the absolute value function is that which appears in figure \ref{fig:absval}.\label{exa:absval_fun} \end{exa} \vspace{1cm} \begin{figure}[h] \begin{minipage}{4cm} $$ \psset{unit=.7pc} \psaxes[labels=none, ](0,0)(-5,-5)(5,5) \psline[linewidth=2pt,linecolor=brown]{<->}(-4,4)(-4,2)(3,2)(3,1)(-3,1)$$ \meinecaption{1}{Fails the vertical line test. Not a function.}\label{fig:not_functions1} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc, algebraic=true} \psaxes[labels=none, ](0,0)(-5,-5)(5,5)\psplot[linewidth=2pt,linecolor=brown]{-1}{2}{sqrt(x^3+1)} \psplot[linewidth=2pt,linecolor=brown]{-1}{2}{-1*sqrt(x^3+1)} $$\meinecaption{1}{Fails the vertical line test. Not a function.} \label{fig:not_functions2} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.8pc,algebraic=true}\psaxes[linewidth=1.2pt,labels=none]{<->}(0,0)(-4.5,-4.5)(4.5,4.5) \psline[linecolor=brown,linewidth=2pt]{<->}(-4.5,-4.5)(4.5,4.5) $$ \meinecaption{1}{$\idefun$}\label{fig:idefun}\end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.8pc,algebraic=true}\psaxes[linewidth=1.2pt,labels=none]{<->}(0,0)(-4.5,-4.5)(4.5,4.5) \psplot[arrows={<->},linecolor=brown,linewidth=2pt,algebraic]{-4}{4}{abs(x)}$$ \meinecaption{1}{${\bf AbsVal}$}\label{fig:absval}\end{minipage} \end{figure} \begin{exa}[The Square Function]\index{function!square} Consider the function $$\fun{{\bf Sq}}{x}{x^2}{\reals}{\reals}.$$ This function assigns to every real its square. By Theorem \ref{thm:parabola_equation}, the graph of the square function is a parabola, and it is presented in in figure \ref{fig:idefun^2}. \label{exa:square_function} \end{exa} \begin{exa}[The Square Root Function]\index{function!square root} Consider the function $$\fun{{\bf Rt}}{x}{\sqrt{x}}{\co{0;+\infty}}{\reals}.$$ By Example \ref{exa:y=sqrx}, the graph of the square root function is the half parabola that appears in figure \ref{fig:sqroot}. \label{exa:square_root_function} \end{exa} \begin{exa}[Semicircle Function]\label{ex:sqrt(1-x^2)} Consider the function\footnote{Since we are concentrating exclusively on real-valued functions, the formula for ${\bf Sc}$ only makes sense in the interval $[-1;1]$. We will examine this more closely in the next section.} $$\fun{{\bf Sc}}{x}{\sqrt{1-x^2}}{[-1;1]}{\reals}.$$ By Example \ref{exa:upper_semicircle}, the graph of ${\bf Sc}$ is the upper unit semicircle, which is shewn in figure \ref{fig:sqrt(1-x^2)}. \end{exa} \begin{exa}[The Reciprocal function]\label{exa:reciprocal_function} Consider the function\footnote{The formula for ${\bf Rec}$ only makes sense when $x\neq 0$.} $$\fun{{\bf Rec}}{x}{\dfrac{1}{x}}{\reals\setminus \{0\}}{\reals}.$$ By Example \ref{exa:y=1/x}, the graph of the reciprocal function is the hyperbola shewn in figure \ref{fig:y=1/x}. \end{exa} \vspace{1cm} \begin{figure}[!hptb] \begin{minipage}{4cm} $$\psset{unit=.8pc,algebraic=true}\psaxes[linewidth=1.2pt,labels=none]{<->}(0,0)(-4.5,-4.5)(4.5,4.5) \parabola[linecolor=brown,linewidth=2pt]{<->}(2.2,4.84)(0,0)$$ \meinecaption{1}{${\bf Sq}$}\label{fig:idefun^2}\end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.8pc,algebraic=true}\psaxes[linewidth=1.2pt,labels=none]{<->}(0,0)(-4.5,-4.5)(4.5,4.5) \psplot[arrows={*->},algebraic,linecolor=brown,linewidth=2pt]{0}{4.5}{sqrt(x)}$$ \meinecaption{1}{${\bf Rt}$}\label{fig:sqroot}\end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.8pc,algebraic=true}\psaxes[linewidth=1.2pt,labels=none]{<->}(0,0)(-4.5,-4.5)(4.5,4.5) \psplot[linecolor=brown,linewidth=2pt]{-1}{1}{sqrt(1-x^2)}\psdots[dotstyle=*,dotscale=1](-1, 0)(1,0)$$ \meinecaption{1}{${\bf Sc}$}\label{fig:sqrt(1-x^2)} \end{minipage} \hfill \hfill \begin{minipage}{4cm} $$\psset{unit=.8pc, algebraic=true} \psaxes[linewidth=1.2pt,labels=none]{<->}(0,0)(-4.5,-4.5)(4.5,4.5) \psplot[linewidth=2pt,linecolor=brown,arrows={<->}]{-.22222222}{-4.5}{1/x} \psplot[linewidth=2pt,linecolor=brown,arrows={<->}]{.22222222}{4.5}{1/x}$$ \meinecaption{1}{${\bf Rec}$}\label{fig:y=1/x}\end{minipage} \end{figure} We can combine pieces of the above curves in order to graph piecewise defined functions. \begin{exa}\label{exa:piecewise_graph} Consider the function $f:\reals\setminus \{-1,1\}\to \reals$ with assignment rule $$ f(x)=\left\{\begin{array}{ll} -x & \mathrm{if}\ x<-1 \\ x^2 & \mathrm{if}\ -11 \\ \end{array}\right. $$ Its graph appears in figure \ref{fig:piecewise_graph}. \end{exa} \vspace{1cm} \begin{figure}[h] \centering \psset{unit=.8pc, algebraic=true} \psaxes[linewidth=1.2pt,labels=none]{<->}(0,0)(-4.5,-4.5)(4.5,4.5) \psplot[linewidth=2pt,linecolor=brown,arrows={<-o}]{-4.5}{-1}{-x} \psplot[linewidth=2pt,linecolor=brown]{-1}{1}{x^2} \psplot[linewidth=2pt,linecolor=brown,arrows={o->}]{1}{4.5}{x} \meinecaption{1.5}{Example \ref{exa:piecewise_graph}.}\label{fig:piecewise_graph} \end{figure} The alert reader will notice that, for example, the two different functions $$\fun{f}{x}{x^2}{\reals}{\reals}\qquad \fun{g}{x}{x^2}{\reals}{\co{0;+\infty}} $$possess the same graph. It is then difficult to recover all the information about a function from its graph, in particular, it is impossible to recover its target set. We will now present a related concept in order to alleviate this problem. \begin{df} A {\em functional curve} on the plane is a curve that passes the vertical line test. The {\em domain of the functional curve} is the ``shadow'' of the graph on the $x$-axis, and the {\em image of the functional curve} is its shadow on the $y$-axis. \end{df} In order to distinguish between finite and infinite sets, we will make the convention that arrow heads in a functional curve indicate that the curve continues to infinity in te direction of the arrow. In order to indicate that a certain value is not part of the domain, we will use a hollow dot. Also, in order to make our graphs readable, we will assume that endpoints and dots fall in lattice points, that is, points with integer coordinates. The following example will elaborate on our conventions. \vspace{2cm} \begin{figure}[h] \begin{minipage}{4cm} \centering \psset{unit=.7pc}\psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5) \pstGeonode[PointName=none,dotscale=1.4,linecolor=brown](-4,-3){A}(4,-3){A'}(-2,-1){B}(2,1){B'} \meinecaption{2}{Example \ref{exa:func_curv1}: $a$.}\label{fig:func_curv1} \end{minipage} \hfill \begin{minipage}{4cm} \centering \psset{unit=.7pc}\pstGeonode[PointName=none,PointSymbol=none](-4,-3){A}(-2,-3){B}(2,1){C}(3,1){D} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5) \psline[arrows={*-o},linecolor=brown,linewidth=2pt,arrowscale=1.2](A)(B)(C)(D) \meinecaption{2}{Example \ref{exa:func_curv1}: $b$.}\label{fig:func_curv2} \end{minipage} \hfill \begin{minipage}{4cm} \centering \psset{unit=.7pc}\pstGeonode[PointName=none,PointSymbol=none](-3,3){A}(2,0){B}(4,4){C} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5) \psline[arrows={o->},linecolor=brown,linewidth=2pt,arrowscale=1.4](A)(B)(C) \psdots[linecolor=brown,dotscale=1.2](B) \meinecaption{2}{Example \ref{exa:func_curv1}: $c$.}\label{fig:func_curv3} \end{minipage} \hfill \begin{minipage}{4cm} \centering \psset{unit=.7pc} \pstGeonode[PointName=none,PointSymbol=none](-4,-2){A}(-3,2){B}(0,4){C}(4,4){D} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5) \psline[arrows={<->},linecolor=brown,linewidth=2pt](A)(B)(C)(D) \psdots[dotscale=1.4,dotstyle=o](B)(C) \meinecaption{2}{Example \ref{exa:func_curv1}: $d$.}\label{fig:func_curv4} \end{minipage} \end{figure} \begin{exa}\label{exa:func_curv1} Determine the domains and images of the functional curves $a, b, c, d$ given in figures \ref{fig:func_curv1} through \ref{fig:func_curv4}. \end{exa} \begin{solu}Figure \ref{fig:func_curv1} consists only a finite number of dots. These dots $x$-coordinates are the set $\{-4,-2,2,4\}$ and hence $\dom{a}=\{-4,-2,2,4\}$. The dots $y$-coordinates are the set $\{-3,-1,1\}$ and so $\im{a}=\{-3.-1,1\}$. \bigskip Figure \ref{fig:func_curv2} has $x$-shadow on the interval $\co{-3;3}$. Notice that $x=3$ is excluded since it has an open dot. We conclude that $\dom{b}=\co{-3;3}$. The $y$-shadow of this set is the interval $\cc{-3;1}$. Notice that we do include $y=1$ since there are points having $y$-coordinate $1$, for example $(2,1)$, which are on the graph. Hence, $\im{b}=\cc{-3;1}$. \bigskip The $x$-shadow of figure \ref{fig:func_curv3} commences just right of $x=-3$ and extends to $+\infty$, as we have put an arrow on the rightmost extreme of the curve. Hence $\dom{c}=\oo{-3:+\infty}$. The $y$-shadow of this curve starts at $y=0$ and continues to $+\infty$, thus $\im{c}=\co{0;+\infty}$. \bigskip We leave to the reader to conclude from figure \ref{fig:func_curv4} that $$\dom{d}=\reals \setminus \{-3,0\}=\oo{-\infty;-3}\cup \oo{-3;0}\cup \oo{0;+\infty},\qquad \im{d}=\oo{-\infty;2}\cup \oc{2;4}.$$ \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Consider the functional curve $d$ shewn in figure \ref{fig:func_curv4}. \begin{enumerate} \item Find consecutive integers $a, b$ such that $d(-2)\in \cc{a;b}$. \item Determine $d(-3)$. \item Determine $d(0)$. \item Determine $d(100)$. \end{enumerate} \begin{answer} \noindent \begin{enumerate} \item $d(-2)\in \cc{2;3}$. \item $d(-3)$ is undefined. \item $d(0)$ is undefined. \item $d(100)=4$. \end{enumerate} \end{answer} \end{pro} \begin{pro}\label{pro:signum} The {\em signum} function is defined as follows: $$\fun{{\bf signum}}{x}{\left\{\begin{array}{ll}+1 & \mathrm{if}\ x>0 \\ 0 & \mathrm{if}\ x=0 \\ -1 & \mathrm{if}\ x<0 \\\end{array}\right.}{\reals}{\{-1,0,1\}}. $$ Graph the signum function. \begin{answer} The graph appears in figure \ref{fig:signum}.\\ \vspace{3cm} \begin{figurehere} \centering \psset{unit=.7pc} \pstGeonode[PointName=none,PointSymbol=none](-4,-1){A}(0,-1){B}(0,0){C}(0,1){D}(4,1){E} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5) \psline[arrows={<-o},linecolor=brown,linewidth=2pt](A)(B) \psline[arrows={o->},linecolor=brown,linewidth=2pt](D)(E) \psdots[dotscale=1.4](C) \meinecaption{2}{Problem \ref{pro:signum}: $d$.}\label{fig:signum} \end{figurehere} \end{answer} \end{pro} \begin{pro} By looking at the graph of the identity function ${\bf Id}$, determine $\dom{{\bf Id}}$ and $\im{{\bf Id}}$. \begin{answer} $\dom{{\bf Id}}=\reals$ and $\im{{\bf Id}}=\reals$. \end{answer} \end{pro} \begin{pro} By looking at the graph of the absolute value function ${\bf AbsVal}$, determine $\dom{{\bf AbsVal}}$ and $\im{{\bf AbsVal}}$. \begin{answer} $\dom{{\bf AbsVal}}=\reals$ and $\im{{\bf AbsVal}}=\co{0;+\infty}$. \end{answer} \end{pro} \begin{pro} By looking at the graph of the square function ${\bf Sq}$, determine $\dom{{\bf Sq}}$ and $\im{{\bf Sq}}$. \begin{answer} $\dom{{\bf Sq}}=\reals$ and $\im{{\bf Sq}}=\co{0;+\infty}$. \end{answer} \end{pro} \begin{pro} By looking at the graph of the square root function ${\bf Rt}$, determine $\dom{{\bf Rt}}$ and $\im{{\bf Rt}}$. \begin{answer} $\dom{{\bf Rt}}=\co{0;+\infty}$ and $\im{{\bf Rt}}=\co{0;+\infty}$. \end{answer} \end{pro} \begin{pro} By looking at the graph of the semicircle function ${\bf Sc}$, determine $\dom{{\bf Sc}}$ and $\im{{\bf Sc}}$. \begin{answer} $\dom{{\bf Sc}}=\cc{-1;1}$ and $\im{{\bf Sc}}=\cc{0;1}$. \end{answer} \end{pro} \begin{pro} By looking at the graph of the reciprocal function ${\bf Rec}$, determine $\dom{{\bf Rec}}$ and $\im{{\bf Rec}}$. \begin{answer} $\dom{{\bf Rec}}=\reals \setminus \{0\}$ and $\im{{\bf Rec}}=\reals \setminus \{0\}$. \end{answer} \end{pro} \begin{pro}\label{pro:piecewise3}Graph the function $g:\reals\rightarrow \reals$ that is piecewise defined by $$g(x) = \left\{\begin{array}{ll} \dfrac{1}{x} & \mathrm{if}\ x \in ]-\infty; -1[ \\ x & \mathrm{if}\ x \in [-1; 1] \\ \dfrac{1}{x} & \mathrm{if}\ x \in ]1;+\infty[ \\ \end{array}\right.$$ \begin{answer} The graph appears in figure \ref{fig:piecewise3}.\\ \vspace*{3cm} \begin{figurehere} \centering\psset{unit=.8pc, algebraic=true} \psaxes[linewidth=1.2pt, labels=none]{<->}(0,0)(-7,-7)(7,7) \psplot[linewidth=2pt,linecolor=brown]{-7}{-1}{1/x} \psplot[linewidth=2pt,linecolor=brown]{1}{7}{1/x} \psdots[dotscale=1.5,dotstyle=*, linecolor=brown](-1,-1)(1,1) \psplot[linewidth=2pt,linecolor=brown]{-1}{1}{x} \meinecaption{2}{Problem \ref{pro:piecewise3}.} \label{fig:piecewise3} \end{figurehere} \end{answer} \end{pro} \begin{pro}\label{pro:piecewise1} Consider the function $f: [-4;4]\to [-5;1] $ whose graph is made of straight lines, as in figure \ref{fig:piecewise1}. Find a piecewise formula for $f$. \begin{answer} The first line segment $\mathscr{L}_1$ has slope $$\mathrm{slope}\ \mathscr{L}_1 = \dfrac{1 - (-3)}{-1 - (-4)} = \dfrac{4}{3}, $$ and so the equation of the line containing this line segment is of the form $y = \dfrac{4}{3}x + k_1$. Since $(-1,1)$ is on the line, $1 = -\dfrac{4}{3} + k_1 \implies k_1 = \dfrac{7}{3}$, so this line segment is contained in the line $y = \dfrac{4}{3}x + \dfrac{7}{3}$. The second line segment $\mathscr{L}_2$ has slope $$\mathrm{slope}\ \mathscr{L}_2 = \dfrac{1 - 1}{2 - (-1)} = 0, $$ and so this line segment is contained in the line $y = 1$. Finally, the third line segment $\mathscr{L}_3$ has slope $$\mathrm{slope}\ \mathscr{L}_3 = \dfrac{-5 - 1}{4-2} = -3, $$and so this line segment is part of the line of the form $y = -3x + k_2$. Since $(1,2)$ is on the line, we have $2 = -3 + k_2 \implies k_2 = 5$, and so the line segment is contained on the line $y = -3x + 5$. Upon assembling all this we see that the piecewise function required is \renewcommand{\arraystretch}{1.5} $$f(x) = \left\{\begin{array}{ll}\dfrac{4}{3}x + \dfrac{7}{3} & \mathrm{if} \ x\in[-4;-1] \\ 1& \mathrm{if} \ x\in[-1;2]\\ -3x + 5 & \mathrm{if} \ x\in[2;4] \end{array}\right. $$ \end{answer} \end{pro} \vspace*{2cm} \begin{figurehere}\centering\psset{unit=.8pc} \psaxes[linewidth=1.2pt,labelFontSize=\tiny ]{<->}(0,0)(-7,-7)(7,7) \psline[linewidth=2pt, linecolor=brown](-4,-3)(-1,1)(2,1)(4,-5)\psdots[dotstyle=*,dotscale=1](-4,-3)(-1,1)(2,1)(4,-5) \uput[l](-4,-3){$\mathscr{L}_1$}\uput[u](2,1){$\mathscr{L}_2$}\uput[r](4,-5){$\mathscr{L}_3$} \meinecaption{2}{Problem \ref{pro:piecewise1}.} \label{fig:piecewise1} \end{figurehere} \end{multicols} \section{Natural Domain of an Assignment Rule} Given a formula, we are now interested in determining which possible subsets of $\reals$ will render the output of the formula also a real number subset. \begin{df} The {\em natural domain of an assignment rule} is the largest set of real number inputs that will give a real number output of a given assignment rule. \end{df} \begin{rem} For the algebraic combinations that we are dealing with, we must then worry about having non-vanishing denominators and taking even-indexed roots of positive real numbers. \end{rem} \begin{exa} Find the natural domain of the rule $x\mapsto \dfrac{1}{x^2 - x - 6}$. \end{exa} \begin{solu} In order for the output to be a real number, the denominator must not vanish. We must have $x^2 -x - 6 = (x + 2)(x - 3) \neq 0$, and so $x \neq -2$ nor $x \neq 3$. Thus the natural domain of this rule is $\reals \setminus \{-2, 3\}$. \end{solu} \begin{exa} Find the natural domain of $x\mapsto \dfrac{1}{x^4 - 16}$. \end{exa}Solution: Since $x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x +2)(x-2)(x^2 + 4)$, the rule is undefined when $x = -2$ or $x = 2$. The natural domain is thus $\reals \setminus \{-2,+2\}$. \begin{exa} Find the natural domain for the rule $f(x) = \dfrac{2}{4 - |x|}$. \end{exa} \begin{solu} The denominator must not vanish, hence $x \neq \pm 4$. The natural domain of this rule is thus $\reals \setminus \{-4,4\}$. \end{solu} \begin{exa} Find the natural domain of the rule $f(x) = \sqrt{x +3}$ \end{exa} \begin{solu} In order for the output to be a real number, the quantity under the square root must be positive, hence $x + 3 \geq 0 \implies x \geq -3$ and the natural domain is the interval $[-3;+\infty[$. \end{solu} \begin{exa} Find the natural domain of the rule $g(x) = \dfrac{2}{\sqrt{x+ 3}}$ \end{exa} \begin{solu} The denominator must not vanish, and hence the quantity under the square root must be positive, therefore $x > -3$ and the natural domain is the interval $]-3+;\infty[$. \end{solu} \begin{exa} Find the natural domain of the rule $x \mapsto \sqrt[4]{x^2}$. \end{exa} \begin{solu} Since for all real numbers $x^2 \geq 0$, the natural domain of this rule is $\reals$. \end{solu} \begin{exa} Find the natural domain of the rule $x \mapsto \sqrt[4]{-x^2}$. \end{exa} \begin{solu} Since for all real numbers $-x^2 \leq 0$, the quantity under the square root is a real number only when $x = 0$, whence the natural domain of this rule is $\{0\}$. \end{solu} \begin{exa} Find the natural domain of the rule $x \mapsto \dfrac{1}{\sqrt{x^2}}$. \end{exa} \begin{solu} The denominator vanishes when $x = 0$. Otherwise for all real numbers, $x\neq 0$, we have $x^2 > 0$. The natural domain of this rule is thus $\reals \setminus \{0\}$. \end{solu} \begin{exa} Find the natural domain of the rule $x \mapsto \dfrac{1}{\sqrt{-x^2}}.$ \end{exa} \begin{solu} The denominator vanishes when $x = 0$. Otherwise for all real numbers, $x\neq 0$, we have $-x^2 < 0$. Thus $\sqrt{-x^2}$ is only a real number when $x = 0$, and in that case, the denominator vanishes. The natural domain of this rule is thus the empty set $\varnothing$. \end{solu} \begin{exa} Find the natural domain of the assignment rule $$x\mapsto \sqrt{1 - x} + \dfrac{1}{\sqrt{1 + x}}.$$ \end{exa} \begin{solu} We need simultaneously $1 - x \geq 0$ (which implies that $x \leq 1$) and $1 + x > 0$ (which implies that $x > -1$), so $x\in ]-1;1]$. \end{solu} \begin{exa} Find the largest subset of real numbers where the assignment rule $x \mapsto \sqrt{x^2 - x - 6}$ gives real number outputs. \end{exa} \begin{solu} The quantity $x^2 - x - 6 = (x + 2)(x - 3)$ under the square root must be positive. Studying the sign diagram $$\begin{array}{|l|c|c|c|} x\in & ]-\infty; -2] & [-2;3] & [3; +\infty[ \\ \hline \sgn{x + 2} = & - & + & + \\ \hline \sgn{x - 3} =& - & - & + \\ \hline \sgn{(x +2)(x - 2)} = & + & - & + \\ \hline \end{array} $$ we conclude that the natural domain of this formula is the set $]-\infty ; -2] \cup [3; +\infty[$. \end{solu} \begin{exa} Find the natural domain for the rule $f(x) = \dfrac{1}{\sqrt{x^2 - x - 6}}$. \end{exa} \begin{solu} The denominator must not vanish, so the quantity under the square root must be positive. By the preceding problem this happens when $x\in]-\infty ; -2[ \ \cup\ ]3; +\infty[$. \end{solu} \begin{exa} Find the natural domain of the rule $x\mapsto \sqrt{x^2 + 1}$. \end{exa} \begin{solu} Since $\forall x\in \reals$ we have $x^2 + 1 \geq 1$, the square root is a real number for all real $x$. Hence the natural domain is $\reals$. \end{solu} \begin{exa} Find the natural domain of the rule $x\mapsto \sqrt{x^2 + x + 1}$. \end{exa} \begin{solu} The discriminant of $x^2 + x + 1 = 0$ is $1^2 - 4(1)(1) < 0$. Since the coefficient of $x^2$ is $1 > 0$, the expression $x^2 + x + 1$ is always positive, meaning that the required natural domain is all of $\reals$. \flushleft{{\em Aliter:}} Observe that since $$x^2 + x + 1 = \left(x + \dfrac{1}{2}\right)^2 + \dfrac{3}{4} \geq \dfrac{3}{4}>0,$$ the square root is a real number for all real $x$. Hence the natural domain is $\reals$. \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Below are given some assignment rules. Verify that the accompanying set is the natural domain of the assignment rule. \renewcommand{\arraystretch}{1.7} \begin{center} \begin{tabular}{|l|l|} \hline Assignment Rule & Natural Domain \\ \hline $x \mapsto \dis{\sqrt{(1 - x)(x + 3)}}$ & $x \in [-3; 1].$ \\ \hline $x \mapsto \dis{\sqrt{\frac{1 - x}{x + 3}}}$ & $x \in ]-3; 1]$ \\ \hline $x \mapsto \dis{\sqrt{\frac{x + 3}{1 - x}}}$ & $x \in [-3; 1[$ \\ \hline $x \mapsto \dis{\sqrt{\frac{1}{(x + 3)(1 - x)}}}$ & $x \in ]-3; 1[$ \\ \hline\end{tabular} \end{center} \end{pro}\begin{pro} Find the natural domain for the given assignment rules. \begin{multicols}{2} \begin{enumerate} \item $x \mapsto \dfrac{1}{\sqrt{1 + |x|}}$ \item $x \mapsto \sqrt[4]{5 - |x|}$\item $x \mapsto \sqrt[3]{5 - |x|}$ \item $x \mapsto \dfrac{1}{x^2 + 2x + 2}$ \item $x \mapsto \dfrac{1}{\sqrt{x^2 -2x - 2}} $ \item $x \mapsto \dfrac{1}{|x - 1| + |x + 1|}$ \item $x \mapsto \dfrac{\sqrt{-x}}{x^2 - 1}$ \item $x \mapsto \dfrac{\sqrt{1 - x^2}}{1 - |x|}$ \item $x \mapsto \sqrt{x} + \sqrt{-x}$ \end{enumerate} \end{multicols} \begin{answer} \begin{multicols}{2} \begin{enumerate} \item $\reals$ \item $[-5;5]$ \item $\reals$ \item $\reals$ \item $]-\infty; 1 - \sqrt{3}[ \cup ]1 + \sqrt{3}; +\infty[$ \item $\reals$ \item $]-\infty; -1[ \cup ]-1; 0]$ \item $]-1;1[$ \item $\{0\}$ \end{enumerate} \end{multicols} \end{answer} \end{pro} \begin{pro} Below are given some assignment rules. Verify that the accompanying set is the natural domain of the assignment rule. \begin{center} \begin{tabular}{|l|l|} \hline Assignment Rule & Natural Domain \\ \hline $x \mapsto \dis{\sqrt{\frac{x}{x^2 - 9}}}$ & $x \in ]-3; 0] \ \bigcup \ ]3; +\infty[$\\ \hline $x \mapsto \dis{\sqrt{-|x|}}$ & $x = 0$ \\ \hline $x \mapsto \sqrt{-||x| - 2|}$ & $x \in \{-2, 2\}$ \\ \hline $x \mapsto \dis{\sqrt{\frac{1}{x}}}$ & $x \in ]0;+\infty[$ \\ \hline $x \mapsto \dis{\sqrt{\frac{1}{x^2}}}$ & $x \in \reals \setminus \{0\}$ \\ \hline $x \mapsto \dis{\sqrt{\frac{1}{-x}}}$ & $x \in ]-\infty; 0[$ \\ \hline $x \mapsto \dis{\sqrt{\frac{1}{-|x|}}}$ & $\varnothing$ (the empty set)\\ \hline $x \mapsto \dis{{\frac{1}{x\sqrt{x + 1}}}}$ & $x \in ]-1; 0[ \ \bigcup \ ]0; +\infty[$ \\ \hline $x \mapsto \sqrt{1 + x} + \sqrt{1 - x}$ & $[-1; 1]$\\ \hline \end{tabular} \end{center} \end{pro} \begin{pro} Find the natural domain for the rule $f(x) = \sqrt{x^3 - 12x}$. \begin{answer} $[-2\sqrt{3}; 0]\cup [2\sqrt{3}; +\infty[$. \end{answer} \end{pro} \begin{pro} Find the natural domain of the rule $x\mapsto \dfrac{1}{\sqrt{x^2 -2x -2}}$. \begin{answer} $x\in ]-\infty; 1 - \sqrt{3}[ \cup ]1 + \sqrt{3}; +\infty[$. \end{answer} \end{pro} \begin{pro} Find the natural domain for the following rules. \begin{multicols}{2} \begin{enumerate} \item $x \mapsto \sqrt{-(x + 1)^2} $, \item $x \mapsto \dfrac{1}{\sqrt{-(x + 1)^2}} $ \item $f(x) = \dfrac{x^{1/2}}{\sqrt{x^4 - 13x^2 + 36}}$ \item $g(x) = \dfrac{\sqrt[4]{3 - x}}{\sqrt{x^4 - 13x^2 + 36}}$ \item $h(x) = \dfrac{1}{\sqrt{x^6 - 13x^4 + 36x^2}}$ \item $j(x) = \dfrac{1}{\sqrt{x^5 - 13x^3 + 36x}}$ \item $k(x) = \dfrac{1}{\sqrt{|x^4 - 13x^2 + 36|}}$ \end{enumerate} \end{multicols} \begin{answer}They are \begin{multicols}{2} \begin{enumerate} \item $\{-1\}$ \item $ \varnothing $ \item $[0;2[ \cup ]3;+\infty[ $ \item $ ]3;+\infty[ $ \item $]-\infty; -3[ \cup ]-2; 0[ \cup ]0; 2[ \cup ]3; +\infty[ $\item $]-3;-2[ \cup ]0;2[ \cup ]3;+\infty[ $ \item $\reals \setminus \{-3,-2,2,3\} $ \end{enumerate} \end{multicols} \end{answer} \end{pro} \end{multicols} \section{Algebra of Functions} \begin{df} Let $f: \dom{f}\rightarrow \target{f}$ and $g: \dom{g}\rightarrow \target{g}$. Then $\dom{f\pm g} = \dom{f}\cap \dom{g}$ and the sum (respectively, difference) function $f + g$ (respectively, $f - g$) is given by $$\fun{f \pm g}{x}{f(x) \pm g(x)}{\dom{f}\cap \dom{g}}{\target{f\pm g}}. $$ In other words, if $x$ belongs both to the domain of $f$ and $g$, then $$(f \pm g)(x) = f(x) \pm g(x).$$ \end{df} \begin{df} Let $f: \dom{f}\rightarrow \reals$ and $g: \dom{g}\rightarrow \reals$. Then $\dom{fg} = \dom{f}\cap \dom{g}$ and the product function $fg$ is given by $$\fun{fg}{x}{f(x)\cdot g(x)}{\dom{f}\cap \dom{g}}{\target{fg}}. $$ In other words, if $x$ belongs both to the domain of $f$ and $g$, then $$(fg)(x) = f(x)\cdot g(x).$$ \end{df} \begin{exa} Let $$\fun{f}{x}{x^2 + 2x}{[-1;1]}{\reals}, \ \ \ \ \fun{g}{x}{3x + 2}{[0;2]}{\reals}. $$ Find \begin{multicols}{2}\begin{enumerate} \item $\dom{f \pm g}$ \item $\dom{fg}$ \item $(f + g)(-1)$ \item $(f + g)(1)$ \item $(fg)(1)$\item $(f - g)(0)$ \item $(f + g)(2)$ \end{enumerate}\end{multicols} \end{exa} \begin{solu} We have \begin{multicols}{2}\begin{enumerate} \item $\dom{f \pm g} = \dom{f}\cap \dom{g} = [-1;1]\cap [0;2] = [0;1]. $ \item $\dom{fg}$ is also $\dom{f}\cap \dom{g} = [0;1]. $ \item Since $-1\not\in [0;1]$, $(f + g)(-1)$ is undefined. \item $(f + g)(1) = f(1) + g(1) = 3 + 5 = 8. $ \item $(fg)(1) = f(1)g(1) = (3)(5) = 15$.\item $(f - g)(0) = f(0) - g(0) = 0 -2 = -2. $ \item Since $2\not\in [0;1]$, $(f + g)(2)$ is undefined. \end{enumerate} \end{multicols} \end{solu} \begin{df} Let $g:\dom{g} \rightarrow \reals$ be a function. The {\em support} of $g$, denoted by $\supp{g}$ is the set of elements in $\dom{g}$ where $g$ does not vanish, that is $$\supp{g} = \{x\in\dom{g}: g(x) \neq 0\}.$$ \end{df} \begin{exa} Let $$\fun{g}{x}{x^3 - 2x}{\reals}{\reals}. $$Then $x^3 - 2x = x(x - \sqrt{2})(x + \sqrt{2})$. Thus $$\supp{g} = \reals\setminus \{-\sqrt{2}, 0 \sqrt{2}\}. $$ \end{exa} \begin{exa} Let $$\fun{g}{x}{x^3 - 2x}{[0;1]}{\reals}. $$Then $x^3 - 2x = x(x - \sqrt{2})(x + \sqrt{2})$. Thus $$\supp{g} = [0;1]\setminus \{-\sqrt{2}, 0 \sqrt{2}\} = ]0;1]. $$ \end{exa} \begin{df} Let $f: \dom{f}\rightarrow \reals$ and $g: \dom{g}\rightarrow \reals$. Then $\dom{\dfrac{f}{g}} = \dom{f}\cap \supp{g}$ and the quotient function $\dfrac{f}{g}$ is given by $$\fun{\dfrac{f}{g}}{x}{\dfrac{f(x)}{g(x)}}{\dom{f}\cap \supp{g}}{\target{\dfrac{f}{g}}}. $$ In other words, if $x$ belongs both to the domain of $f$ and $g$ and $g(x) \neq 0$, then $\left(\dfrac{f}{g}\right)(x) = \dfrac{f(x)}{g(x)}.$ \end{df} \begin{exa} Let $$\fun{f}{x}{x^3 - x}{[-2;3]}{\reals}, \ \ \ \ \fun{g}{x}{x^3 - 2x^2}{[0;5]}{\reals}. $$ Find \begin{multicols}{2}\begin{enumerate} \item $\supp{f}$ \item $\supp{g}$ \item $\dom{\dfrac{f}{g}}$ \item $\dom{\dfrac{g}{f}}$ \item $\left(\dfrac{f}{g}\right)(2)$ \item $\left(\dfrac{g}{f}\right)(2)$ \item $\left(\dfrac{f}{g}\right)(1/3)$\item $\left(\dfrac{g}{f}\right)(1/3)$ \end{enumerate} \end{multicols} \end{exa} \begin{solu} \noindent \begin{enumerate} \item As $x^3 - x = x(x - 1)(x + 1)$, $\supp{f} = [-2; -1[ \cup ]-1; 0[ \cup ]0;3]$ \item As $x^3 -2x^2 = x^2(x -2)$, $\supp{g} = ]0;2[\cup ]2;5]$. \item $\dom{\dfrac{f}{g}} = \dom{f}\cap \supp{g} = [-2;3]\cap (]0;2[\cup ]2;5]) = ]0;2[ \cup ]2;3] $ \item $$\dom{\dfrac{g}{f}}=\dom{g}\cap \supp{f} = [0;5]\cap ([-2; -1[ \cup ]-1; 0[ \cup ]0;3]) = ]0;3]$$ \item $\left(\dfrac{f}{g}\right)(2)$ is undefined, as $2\not\in ]0;2[ \cup ]2;3]$. \item $\left(\dfrac{g}{f}\right)(2) = \dfrac{g(2)}{f(2)} = \dfrac{0}{6} = 0$. \item $\left(\dfrac{f}{g}\right)(1/3) = \dfrac{-\frac{8}{27}}{-\frac{5}{27}} = \dfrac{8}{5}$ \item $\left(\dfrac{g}{f}\right)(1/3) = \dfrac{-\frac{5}{27}}{-\frac{8}{27}} = \dfrac{5}{8}$ \end{enumerate} \end{solu} \bigskip We are now going to consider ``functions of functions.'' \begin{df} Let $f: \dom{f} \rightarrow \target{f}$, $g: \dom{g} \rightarrow \target{g}$ and let $U = \{x\in\dom{g}: g(x) \in \dom{f}\}$. We define the {\em composition} function of $f$ and $g$ as \begin{equation} \fun{f\circ g}{x}{f(g(x))}{U}{\target{f\circ g}} .\end{equation}We read $f\circ g$ as ``$f$ {\em composed with} $g$.'' \end{df} \begin{rem} We have $\dom{f\circ g} = \{x\in\dom{g}: g(x) \in \dom{f}\}$. Thus to find $\dom{f\circ g}$ we find those elements of $\dom{g}$ whose images are in $\dom{f}\cap\im{g}$ \end{rem} \begin{exa} Let $$\fun{f}{x}{2x + 1}{\{-2, -1, 0, 1, 2\}}{\reals}, \ \ \ \fun{g}{x}{x^2 - 4}{\{0, 1, 2, 3\}}{\reals}.$$ \begin{multicols}{2} \begin{enumerate} \item Find $\im{f}$. \item Find $\im{g}$. \item Find $\dom{f\circ g}$. \item Find $\dom{g\circ f}$. \item Find $(f\circ g)(0)$. \item Find $(g\circ f)(0)$. \item Find $(f\circ g)(2)$. \item Find $(g\circ f)(2)$. \end{enumerate}\end{multicols} \begin{solu} \begin{enumerate} \item We have $f(-2) = -3$, $f(-1) = -1$, $f(0) = 1$, $f(1) = 3$, $f(2) = 5$. Hence $\im{f} = \{-3,-1,1,3,5\}$.\item We have $g(0) = -4$, $g(1) = -3$, $g(2) = 0$, $g(3) = 5$. Hence $\im{g} = \{-4,-3,0,5\}$. \item $\dom{f\circ g} = \{x\in\dom{g}: g(x) \in \dom{f}\} = \{2\}$. \item $\dom{g\circ f} = \{x\in\dom{f}: f(x) \in \dom{g}\} = \{0,1\}$. \item $(f\circ g)(0) = f(g(0)) = f(-4)$, but this last is undefined. \item $(g\circ f)(0) = g(f(0)) = g(1) = -3$. \item $(f\circ g)(2) = f(g(2)) = f(0) = 1$. \item $(g\circ f)(2) = g(f(2)) = g(5)$, but this last is undefined. \end{enumerate} \end{solu} \end{exa} \begin{exa} Let $$\fun{f}{x}{2x - 3}{\reals}{\reals}, \ \ \ \fun{g}{x}{5x + 1}{\reals}{\reals}.$$ \begin{enumerate} \item Demonstrate that $\im{f} = \reals$. \item Demonstrate that $\im{g} = \reals$. \item Find $(f\circ g)(x)$. \item Find $(g\circ f)(x).$ \item Is it ever true that $(f\circ g)(x) = (g\circ f)(x)$? \end{enumerate} \end{exa} \begin{solu} \begin{enumerate} \item Take $b\in\reals$. We must shew that $\exists x\in \reals$ such that $f(x) = b$. But $$f(x) = b \implies 2x - 3 = b \implies x = \dfrac{b + 3}{2}. $$ Since $\dfrac{b + 3}{2}$ is a real number satisfying $\dis{f\left(\dfrac{b + 3}{2}\right) = b}$, we have shewn that $\im{f} = \reals$. \item Take $b\in\reals$. We must shew that $\exists x\in \reals$ such that $g(x) = b$. But $$g(x) = b \implies 5x + 1 = b \implies x = \dfrac{b -1}{5}. $$ Since $\dfrac{b -1}{5}$ is a real number satisfying $\dis{g\left(\dfrac{b -1}{5}\right) = b}$, we have shewn that $\im{g} = \reals$. \item We have $$(f\circ g)(x) = f(g(x)) = f(5x + 1) = 2(5x + 1) - 3 = 10x - 1$$\item We have $$(g\circ f)(x) = g(f(x)) = g(2x - 3) = 5(2x - 3) + 1 = 10x - 14.$$ $(g\circ f)(x).$ \item If $$(f\circ g)(x) = (g\circ f)(x)$$ then we would have $$10x - 1 = 10x - 14$$ which entails that $-1 = -14$, absolute nonsense! \end{enumerate} \end{solu} \begin{rem} Composition of functions need not be commutative. \end{rem} \begin{exa} Consider $$\fun{f}{x}{\sqrt{3 - x^2}}{[-\sqrt{3};\sqrt{3}]}{\reals}, \ \ \ \fun{g}{x}{-\sqrt{x + 2}}{[-2; +\infty[}{\reals}. $$ \begin{enumerate} \item Find $\im{f}$. \item Find $\im{g}$. \item Find $\dom{f\circ g}$. \item Find $f\circ g$. \item Find $\dom{g\circ f}$. \item Find $g\circ f$. \end{enumerate} \end{exa}\begin{solu} \begin{enumerate} \item Assume $y = \sqrt{3 - x^2}$. Then $y \geq 0$. Moreover $x = \pm \sqrt{3 - y^2}$. This makes sense only if $-\sqrt{3} \leq y \leq \sqrt{3}$. Hence $\im{f} = [0;\sqrt{3}]$. \item Assume $y = -\sqrt{x + 2}$. Then $y \leq 0$. Moreover, $x=y^2-2$ which makes sense for every real number. This means that $y$ is allowed to be any negative number and so $\im{g} = ]-\infty; 0]$. \item $$\begin{array}{lll}\dom{f\circ g} & = & \{x\in\dom{g}: g(x) \in \dom{f}\} \\ &= & \{x\in [-2; +\infty[: -\sqrt{3}\leq -\sqrt{x + 2}\leq \sqrt{3}\}\\ & = & \{x\in [-2; +\infty[: -\sqrt{3}\leq -\sqrt{x + 2}\leq 0\} \\ & = & \{x\in [-2; +\infty[: x \leq 1\} \\ & = & [-2;1] \end{array}$$ \item $(f\circ g)(x) = f(g(x)) = f(-\sqrt{x + 2}) = \sqrt{1 - x}$. \item $$\begin{array}{lll}\dom{g\circ f} & = & \{x\in\dom{f}: f(x) \in \dom{g}\} \\ &= & \{x\in [-\sqrt{3};\sqrt{3}]: \sqrt{3 - x^2} \geq -2\}\\ &= & \{x\in [-\sqrt{3};\sqrt{3}]: \sqrt{3 - x^2} \geq 0\}\\ &= & [-\sqrt{3};\sqrt{3}]\\ \end{array}$$ \item $(g\circ f)(x) = g(f(x)) = g(\sqrt{3 - x^2}) = -\sqrt{\sqrt{3 - x^2} + 2}$. \end{enumerate} \end{solu} \begin{rem} Notice that $\dom{f\circ g} = [-2;1]$, although the domain of definition of $x\mapsto \sqrt{1-x}$ is $]-\infty;1]$. \end{rem} \begin{exa} Let $$\fun{f}{x}{\dfrac{2x}{x - 1}}{\reals \setminus \{1\}}{\reals} , \ \ \ \fun{g}{x}{\sqrt{2 - x}}{]-\infty; 2]}{\reals} $$ \begin{enumerate} \item Find $\im{f}$. \item Find $\im{g}$. \item Find $\dom{f\circ g}$. \item Find $f\circ g$. \item Find $\dom{g\circ f}$. \item Find $g\circ f$. \end{enumerate} \end{exa}\begin{solu} \begin{enumerate} \item Assume $y = \dfrac{2x}{x - 1}$, $x\in\dom{f}$ is solvable. Then $$y(x - 1) = 2x\implies yx - 2x = y \implies x = \dfrac{y}{y-2}. $$Thus the equation is solvable only when $y \neq 2$. Thus $\im{f} = \reals \setminus \{2\}$. \item Assume that $y = \sqrt{2 - x}$, $x\in\dom{g}$ is solvable. Then $y \geq 0$ since $y = \sqrt{2 - x}$ is the square root of a (positive) real number. All $y \geq 0$ will render $x = 2 - y^2$ in the appropriate range, and so $\im{g} = [0;+\infty[$. \item $$\begin{array}{lll}\dom{f\circ g} & = & \{x\in\dom{g}: g(x) \in \dom{f}\} \\ &= & \{x\in ]-\infty;2]: \sqrt{2 - x} \neq 1\}\\ & = & ]-\infty; 1[ \cup ]1;2]\end{array}$$ \item $(f\circ g)(x) = f(g(x)) = f(\sqrt{2 - x}) = \dfrac{1}{\sqrt{2 - x} - 1}$. \item $$\begin{array}{lll}\dom{g\circ f} & = & \{x\in\dom{f}: f(x) \in \dom{g}\} \\ &= & \{x\in \reals \setminus\{1\}: \dfrac{2x}{x-1} \leq 2\}\\ & = & \{x\in \reals \setminus\{1\}: \dfrac{2}{x-1} \leq 0\}\\ & = & ]-\infty; 1[ \end{array}$$ \item $$(g\circ f)(x) = g(f(x)) = g\left(\dfrac{2x}{x - 1}\right) = \sqrt{2-\dfrac{2x}{x-1}} = \sqrt{\dfrac{2}{1 - x}}$$ \end{enumerate} \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Let $$\fun{f}{x}{x^4 - 16}{[-5;3]}{\reals}, \ \ \ \ \fun{g}{x}{|x| - 4}{[-4;2]}{\reals}. $$ Find \begin{multicols}{2} \begin{enumerate} \item $\dom{f + g}$ \item $\dom{fg}$ \item $\dom{\dfrac{f}{g}}$ \item $\dom{\dfrac{g}{f}}$ \item $(f + g)(2)$ \item $(fg)(2)$ \item $\left(\dfrac{f}{g}\right)(2)$ \item $\left(\dfrac{g}{f}\right)(2)$ \item $\left(\dfrac{f}{g}\right)(1)$\item $\left(\dfrac{g}{f}\right)(1)$ \end{enumerate}\end{multicols} \begin{answer} \begin{enumerate} \item $[-4;2]$ \item $[-4;2]$ \item $]-4;2]$ \item $[-4;-2[\cup ]-2;2[$ \item $-2$ \item $0$ \item $0$ \item undefined \item $5$\item $\dfrac{1}{5}$ \end{enumerate}\end{answer} \end{pro} \begin{pro} Let $$\fun{f}{x}{2x}{\{-2,-1,0,1,2\}}{\integers}, \ \ \ \fun{g}{x}{x^2}{\{0,1,2\}}{\integers}.$$ \begin{enumerate} \item Find $\im{f}$. \item Find $\im{g}$. \item Find $\dom{f\circ g}$. \item Find $\dom{g\circ f}$. \end{enumerate} \begin{answer} 1)$\{-4,-2,0,2,4\}$ 2) $\{0,1,4\}$ 3) $\{0,1\}$. 4) $\{0,2\}$.\end{answer} \end{pro} \begin{pro} Let $f, g, h :\{1, 2, 3, 4\} \rightarrow \{1, 2, 10, 1993\}$ be given by \\ $$f(1) = 1, f(2) = 2, f(3) = 10, f(4) = 1993,$$ $$ g(1) = g(2) = 2, g(3) = g(4) - 1 = 1, $$ $$h(1) = h(2) = h(3) = h(4) + 1 = 2.$$ \begin{enumerate} \item Compute $(f + g + h)(3)$ \\ \item Compute $(fg + gh + hf)(4)$. \\ \item Compute $f(1 + h(3)).$ \\ \item Compute $(f\circ f\circ f\circ f\circ f)(2) + f(g(2) + 2).$ \end{enumerate} \begin{answer} (1) $13$, \ (2) $5981$,\ (3) $10$, \ (4) $1995$ \end{answer} \end{pro} \begin{pro} Two functions $f, g:\reals \to \reals$ are given by $f(x) = ax + b$, $g(x) = bx + a$ with $a$ and $b$ integers. If $f(1) = 8$ and $f(g(50)) - g(f(50)) = 28$, find the product $ab$. \begin{answer} Observe that $a+b=f(1)=8$. We have $f(50)=50a+b$, $g(50)=50b+a$ and $$f(g(50))=f(50b+a)=50ab+a^2+b, \qquad g(f(50))=g(50a+b)=50ab+b^2+a, $$ whence $$28=f(g(50)) - g(f(50)) = a^2-b^2-(a-b)=(a-b)(a+b-1)=7(a-b) \implies a-b=4. $$ Therefore, $$ ab=\dfrac{(a+b)^2-(a-b)^2}{4}=\dfrac{64-16}{4}=12. $$ \end{answer} \end{pro} \begin{pro} If $a, b, c:\reals \rightarrow \reals$ are functions with $a(t) = t - 2, b(t) = t^3, c(t) = 5$ demonstrate that \\ \begin{center} \begin{tabular}{lclclclcl} $(a\circ b)(t)$ & = & $t^3 - 2$ \\ $(b\circ a)(t)$ & = & $(t - 2)^3$ \\ $(b\circ c)(t)$ & = & $125$ \\ $(c\circ b)(t)$ & = & $5$ \\ $(c\circ a)(t)$ & = & $5$\\ $(a\circ b \circ c)(t)$ & = & $123$ \\ $(c\circ b\circ a)(t)$ & = & $5$ \\ $(a\circ c \circ b)(t)$ & = & $3$ \\ \end{tabular}\end{center} \end{pro} \begin{pro} Let $$\fun{f}{x}{\sqrt{x - 2}}{[2;+\infty[}{\reals}, \ \ \ \fun{g}{x}{\sqrt{4 - x^2}}{[-2;2]}{\reals}.$$ \begin{enumerate} \item Find $\im{f}$. \item Find $\im{g}$. \item Find $\dom{f\circ g}$. \item Find $\dom{g\circ f}$. \item Find $(f\circ g)(x)$. \item Find $(g\circ f)(x)$. \end{enumerate} \begin{answer} 1) $[0;+\infty]$ 2) $[0;2]$ 3) $\{0\}$. 4) $[2;6] $. 5) $\sqrt{\sqrt{4 - x^2} - 2}$. 6) $\sqrt{6 - x}$. \end{answer} \end{pro} \begin{pro} Let $$\fun{f}{x}{\sqrt{2 - x^2}}{[-\sqrt{2};+\sqrt{2}[}{\reals}, \ \ \ \fun{g}{x}{-\sqrt{-x}}{]-\infty;0]}{\reals}.$$ \begin{enumerate} \item Find $\im{f}$. \item Find $\im{g}$. \item Find $\dom{f\circ g}$. \item Find $\dom{g\circ f}$. \item Find $(f\circ g)(x)$. \item Find $(g\circ f)(x)$. \end{enumerate} \begin{answer} 1) $[0; \sqrt{2}]$ 2) $]-\infty; 0]$ 3) $[-2;0]$. 4) $\{-\sqrt{2}, \sqrt{2}\} $. 5) $\sqrt{2 + x}$. 6) $-\sqrt{-\sqrt{2 - x^2}}$. \end{answer} \end{pro} \begin{pro} Let $f, g, h: \reals \rightarrow \reals$ be functions. Prove that their composition is associative $$f\circ (g\circ h) = (f\circ g)\circ h$$whenever the given expressions make sense. \end{pro} \begin{pro} Let $f:\reals \rightarrow \reals$ be the function defined by $f(x) = ax^2 - \sqrt{2}$ for some positive $a$. If $(f\circ f)(\sqrt{2}) = -\sqrt{2}$ find the value of $a$. \begin{answer} $\dfrac{\sqrt{2}}{2}$ \end{answer} \end{pro} \begin{pro} Let $f:]0:+\infty[ \rightarrow ]0:+\infty[$, such $f(2x) = \dis{\frac{2}{2 + x}}$. Find $2f(x)$. \begin{answer} $\dfrac{8}{4 + x}$ \end{answer} \end{pro} \begin{pro} Let $f, g:\reals \setminus \{1\} \rightarrow \reals$, with$f(x) = \dis{\frac{4}{x - 1}}, g(x) = 2x$, find all $x$ for which $(g\circ f)(x) = (f\circ g)(x).$ \begin{answer} $x = 1/3.$ \end{answer} \end{pro} \begin{pro} Let $f:\reals \rightarrow \reals$, $f(1 - x) = x^2.$ Find $(f\circ f)(x).$ \begin{answer} $(f\circ f)(x) = 4x^2 - 4x^3 + x^4.$ \end{answer} \end{pro} \begin{pro} Let $f:\reals \setminus \{\dis{-\frac{3}{2}}\} \rightarrow \reals \setminus \{\dis{\frac{c}{2}}\}, x \mapsto \dis{\frac{cx}{2x + 3}}$ be such that $(f\circ f)(x) = x.$ Find the value of $c$. \begin{answer} $c = -3$ \end{answer} \end{pro} \begin{pro} Let $f, g:\reals \to \reals$ be functions satisfying for all real numbers $x$ and $y$ the equality \begin{equation}f(x+g(y)) = 2x+y+5.\label{eq:real-eq-fun}\end{equation} Find an expression for $g(x+f(y))$. \begin{answer} If $y=0$ then $f(x+g(0))=2x+5$. Hence $$f(x)=f(x-g(0)+g(0)) 2(x-g(0)+5)=2x-2g(0)+5. $$We deduce that $f(0)=-2g(0)+5$ and hence, $$-2g(0)+5=f(-g(y)+g(y)) = 2(-g(y))+y+5 \implies g(y)= g(0)+\dfrac{y}{2}. $$ This gives $$ g(x+f(y)) = g(0) + \dfrac{x+f(y)}{2}=g(0) +\dfrac{x+2y-2g(0)+5}{2} = \dfrac{x+2y+5}{2}. $$ \end{answer} \end{pro} \end{multicols} \section{Iteration and Functional Equations} \begin{df} Given an assignment rule $x\mapsto f(x)$, its {\em iterate at $x$} is $f(f(x))$, that is, we use its value as the new input. The iterates at $x$ $$ x, \ f(x), \ f(f(x)), \ f(f(f(x))), \ldots $$ are called {\em $0$-th iterate}, {\em $1$st iterate}, {\em $2$nd iterate}, {\em $3$rd iterate}, etc. We denote the $n$-th iterate by $f^{[n]}$. \end{df} In some particular cases it is easy to find the $n$th iterate of an assignment rule, for example $$a(x)=x^t\implies a^{[n]}(x) =x^{t^n}, $$ $$b(x)=mx\implies b^{[n]}(x) =m^nx, $$ $$c(x)=mx+k\implies c^{[n]}(x) =m^nx+k \left(\dfrac{m^n-1}{m-1}\right). $$ The above examples are more the exception than the rule. Even if its possible to find a closed formula for the $n$-th iterate some cases prove quite truculent. \begin{exa} Let $f(x) = \dfrac{1}{1 - x}$. Find the $n$-th iterate of $f$ at $x$, and determine the set of values of $x$ for which it makes sense. \end{exa} \begin{solu} We have $$f^{[2]}(x) = (f\circ f)(x) = f(f(x)) = \frac{1}{1 - \frac{1}{1 - x}} = \frac{x - 1}{x}, $$ $$f^{[3]}(x) = (f\circ f \circ f)(x) = f(f^{[2]}(x))) = f\left(\frac{x - 1}{x}\right) = \frac{1}{1 - \frac{x - 1}{x}} = x. $$ Notice now that $f^{[4]}(x) = (f\circ f^{[3]})(x) = f(f^{[3]}(x)) = f(x) = f^{[1]}(x)$. We see that $f$ is cyclic of period $3$, that is, $$f^{[1]}(x) = f^{[4]}(x) = f^{[7]}(x) = \ldots = \dfrac{1}{1-x}, $$ $$f^{[2]}(x) = f^{[5]}(x) = f^{[8]}(x) = \ldots = \dfrac{x-1}{x}, $$ $$f^{[3]}(x) = f^{[6]}(x) = f^{[9]}(x) = \ldots =x.$$ The formul\ae\ above hold for $x\not\in\{0,1\}$. \end{solu} \begin{df} A {\em functional equation} is an equation whose variables range over functions, or more often, assignment rules. \end{df} A functional equation problem asks for a formula, or formul\ae\ satisfying certain features. \begin{exa} Find all the functions $g:\reals \to \reals $ satisfying $$g(x + y) + g(x - y) = 2x^2 + 2y^2.$$ \end{exa} \begin{solu} If $y = 0$, then $2g(x) = 2x^2$, that is, $g(x) = x^2.$ Let us verify that $g(x) = x^2$ works. We have $$g(x + y) + g(x - y) = (x + y)^2 + (x - y)^2 = x^2 + 2xy + y^2 + x^2 - 2xy + y^2 = 2x^2 + 2y^2,$$ from where the only solution is $g(x)=x^2$. \end{solu} \begin{exa} Find all functions $f:\reals \to \reals$ such that $$x^2f(x)+f(1-x)=2x-x^4. $$ \end{exa} \begin{solu}From the given equation, $$ f(1-x)=2x-x^4-x^2f(x). $$ Replacing $x$ by $1-x$, we obtain $$ (1-x)^2f(1-x)+f(x)=2(1-x)-(1-x)^4.$$ This implies that $$f(x)=2(1-x)-(1-x)^4-(1-x)^2f(1-x)=2(1-x)-(1-x)^4-(1-x)^2(2x-x^4-x^2f(x)), $$ which in turn, gives $$f(x)=2(1-x)-(1-x)^4-2x(1-x)^2+x^4(1-x)^2+(1-x)^2x^2f(x).$$ Solving now for $f(x)$ we gather that $$\begin{array}{lll} f(x) & = & \dfrac{2(1-x)-(1-x)^4-2x(1-x)^2+x^4(1-x)^2}{1-(1-x)^2x^2}\\ & = &\dfrac{(1-x)(2-(1-x)^3-2x(1-x)+x^4(1-x)}){(1-(1-x)x)(1+(1-x)x)} \\ & = &\dfrac{(1-x)(2-(1-3x+3x^2-x^3)-2x+2x^2+x^4-x^5)}{(1-x+x^2)(1+x-x^2)} \\ & = &\dfrac{(1-x)(1+x-x^2+x^3+x^4-x^5)}{(1-x+x^2)(1+x-x^2)} \\ & = &\dfrac{(1-x)(1+x)(1-x+x^2)(1+x-x^2)}{(1-x+x^2)(1+x-x^2)} \\ & = & 1-x^2.\\ \end{array}$$ We now check. If $f(x)=1-x^2$ then $$x^2f(x)+f(1-x)=x^2(1-x^2)+1-(1-x)^2= x^2-x^4+1-1+2x-x^2=2x-x^4, $$from $f(x)=1-x^2$ is the only such solution. \end{solu} We continue with, perhaps, the most famous functional equation. \begin{exa}[Cauchy's Functional Equation] Suppose $f:\rationals \to \rationals$ satisfies $f(x + y) = f(x) + f(y)$. Prove that $\exists c \in \rationals$ such that $f(x) = cx, \ \forall x \in {\rationals}$. \end{exa}\begin{solu} Letting $y = 0$ we obtain $f(x) = f(x) + f(0)$, and so $f(0) = 0$. If $k$ is a positive integer we obtain $$ \begin{array}{ll} f(kx) & = f(x + (k - 1)x) \\ & = f(x) + f((k - 1)x) \\ & = f(x) + f(x) + f((k - 2)x) = 2f(x) + f((k - 2)x) \\ & = 2f(x) + f(x) + f((k - 3)x) = 3f(x) + f((k - 3)x) \\ & \vdots \\ & = \cdots = kf(x) + f(0) = kf(x). \\ \end{array}$$ Letting $y = -x$ we obtain $0 = f(0) = f(x) + f(-x)$ and so $f(-x) = -f(x).$ Hence $f(nx) = nf(x)$ for $n \in {\integers}$. Let $x \in {\rationals}$, which means that $x = \dfrac{s}{t}$ for integers $s, t$ with $t \neq 0.$ This means that $tx = s\cdot 1$ and so $f(tx) = f(s\cdot 1)$ and by what was just proved for integers, $tf(x) = sf(1).$ Hence $f(x) = \dfrac{s}{t}f(1) = xf(1)$. Since $f(1)$ is a constant, we may put $c = f(1)$. Thus $f(x) = cx$ for rational numbers $x$. \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Let $f^{[1]}(x) = f(x) = x + 1, f^{[n + 1]} = f\circ f^{[n]}, n \geq 1.$ Find a closed formula for $f^{[n]}$ \begin{answer} We have $f^{[2]}(x) = f(x + 1) = (x + 1) + 1 = x + 2, f^{[3]}(x) = f(x + 2) = (x + 2) + 1 = x + 3$ and so, recursively, $f^{[n]}(x) = x + n.$ \end{answer} \end{pro} \begin{pro} Let $f^{[1]}(x) = f(x) = 2x, f^{[n + 1]} = f\circ f^{[n]}, n \geq 1.$ Find a closed formula for $f^{[n]}$ \begin{answer} We have $f^{[2]}(x) = f(2x) = 2^2x, f^{[3]}(x) = f(2^2x) = 2^3x$ and so, recursively, $f^{[n]}(x) = 2^nx.$ \end{answer} \end{pro} \begin{pro} Find all the assignment rules $f$ that satisfy $f(xy) = yf(x)$. \begin{answer} Let $x = 1$. Then $f(y) = yf(1)$. Since $f(1)$ is a constant, we may let $k = f(1).$ So all the functions satisfying the above equation satisfy $f(y) = ky.$ \end{answer} \end{pro} \begin{pro} Find all the assignment rules $f$ for which $$f(x) + 2f\left(\frac{1}{x}\right) = x.$$ \begin{answer} From $\dis{f(x) + 2f(\frac{1}{x}) = x}$ we obtain $f\left(\dfrac{1}{x}\right) = \dfrac{x}{2} - \dfrac{1}{2}f(x)$. Also, substituting $1/x$ for $x$ on the original equation we get $$f(1/x) + 2f(x) = 1/x.$$Hence $$f(x) = \dfrac{1}{2x} - \dfrac{1}{2}f(1/x) = \dfrac{1}{2x} - \dfrac{1}{2}\left(\dfrac{x}{2} - \dfrac{1}{2}f(x)\right),$$ which yields $f(x) = \dfrac{2}{3x} - \dfrac{x}{3}$. \\ \end{answer} \end{pro} \begin{pro}Find all functions $f:\reals\setminus\{-1\}\rightarrow \reals$ such that $$(f(x))^2\cdot f\left(\frac{1 - x}{1 + x}\right) = 64x.$$ \begin{answer} We have $$(f(x))^2\cdot f\left(\frac{1 - x}{1 + x}\right) = 64x, $$whence $$(f(x))^4\cdot \left(f\left(\frac{1 - x}{1 + x}\right))\right)^2 = 64^2x^2 \qquad (I) $$ Substitute $x$ by $\frac{1 - x}{1 + x}$. Then $$f\left(\frac{1 - x}{1 + x}\right) ^2 f(x) = 64\left(\frac{1 - x}{1 + x}\right) . \qquad (II) $$ Divide (I) by (II), $$ f(x)^3 = 64 x^2\left(\frac{1+x}{1-x}\right), $$ from where the result follows. \end{answer} \end{pro} \begin{pro} An assignment rule $f$ is said to be an {\em involution} if for all $x$ for which $f(x)$ and $f(f(x))$ are defined we have $f(f(x))=x$. Prove that $a(x)=\dfrac{1}{x}$ is an involution for $x\neq 0$.\index{involution} \end{pro} \begin{pro} Prove that $f(x)=\sqrt{1-x^2}$ is an involution for $0 \leq x\leq 1$. \end{pro} \begin{pro} Let $f$ satisfy $f(n + 1) = (-1)^{n + 1}n - 2f(n), n \geq 1$ If $f(1) = f(1001)$ find $f(1) + f(2) + f(3) + \cdots + f(1000)$. \begin{answer} We have \\ \begin{tabular}{lllll} $f(2)$ & = & $(-1)^21 - 2f(1)$ & = & $1 - 2f(1)$ \\ $f(3)$ & = & $(-1)^32 - 2f(2)$ & = & $-2 - 2f(2)$ \\ $f(4)$ & = & $(-1)^43 - 2f(3)$ & = & $3 - 2f(3)$ \\ $f(5)$ & = & $(-1)^54 - 2f(4)$ & = & $-4 - 2f(4)$ \\ $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ \\ $f(999)$ & = & $(-1)^{999}998 - 2f(998)$ & = & $-998 - 2f(998)$ \\ $f(1000)$ & = & $(-1)^{1000}999 - 2f(999)$ & = & $999 - 2f(999)$ \\ $f(1001)$ & = & $(-1)^{1001}1000 - 2f(1000)$ & = & $-1000 - 2f(1000)$ \\ \end{tabular} \bigskip Adding columnwise, $$f(2) + f(3) + \cdots + f(1001) = 1 - 2 + 3 - \cdots + 999 - 1000 - 2(f(1) + f(2) + \cdot + f(1000)).$$ This gives $$2f(1) + 3(f(2) + f(3) + \cdots + f(1000)) + f(1001) = - 500.$$ Since $f(1) = f(1001)$ we have $2f(1) + f(1001) = 3f(1).$ Therefore$$f(1) + f(2) + \cdots + f(1000) = - \frac{500}{3}.$$ \end{answer} \end{pro} \begin{pro} Let $f:\reals \to \reals$ satisfy $$ f(1)=1, \quad \forall x\in \reals \quad f(x+3)\geq f(x)+3,\quad f(x+1)\leq f(x)+1.$$ Put $g(x)=f(x)-x+1$. Determine $g(2008)$. \begin{answer} $1$ \end{answer} \end{pro} \begin{pro} If $f(a)f(b) = f(a + b) \ \forall \ a, b \in \reals$ and $f(x) > 0 \ \forall \ x \in \reals$, find $f(0)$. Also, find $f(-a)$ and $f(2a)$ in terms of $f(a).$ \begin{answer} Set $a = b = 0.$ Then $(f(0))^2 = f(0)f(0) = f(0 + 0) = f(0)$. This gives $f(0)^2 = f(0).$ Since $f(0) > 0$ we can divide both sides of this equality to get $f(0) = 1$. \bigskip Further, set $b = -a.$ Then $f(a)f(-a) = f(a - a) = f(0) = 1.$ Since $f(a) \neq 0,$ may divide by $f(a)$ to obtain $f(-a) = \dis{\frac{1}{f(a)}}$. Finally taking $a = b$ we obtain $(f(a))^2 = f(a)f(a) = f(a + a) = f(2a).$ Hence $f(2a) = (f(a))^2$ \end{answer} \end{pro} \end{multicols} \section{Injections and Surjections} \begin{df} A function $$\fun{f}{a}{f(a)}{\dom{f}}{\target{f}}$$ is said to be {\em injective} or {\em one-to-one} if $(a_1, a_2) \in (\dom{f})^2,$ $$ a_1 \neq a_2 \implies f(a_1) \neq f(a_2).$$ That is, $$f(a_1) = f(a_2) \implies a_1 = a_2.$$ $f$ is said to be {\em surjective} or {\em onto} if $\target{f} = \im{f}$. That is, if $(\forall b \in B)\ (\exists a \in A)$ such that $f(a) = b.$ $f$ is {\em bijective} if it is both injective and surjective. The number $a$ is said to the the {\em pre-image} of $b$. \index{function!surjective}\index{function!injective}\index{function!bijective} \end{df} A function is thus injective if different inputs result in different outputs, and it is surjective if every element of the target set is hit. Figures \ref{fig:inj-n-epi} through \ref{fig:biyectiva} present various examples. \vspace*{2cm} \begin{figure}[!hptb] \begin{minipage}{4cm} \centering \centering \psset{unit=1pc} \pstGeonode[PointName=none](0,0){A}(0,1){B}(0,2){C} \pstGeonode[PointName=none](5,0){A'}(5,1){B'}(5,2){C'}(5,-1){D'} \pcline[nodesep=.2, arrows={->},arrowsize=.5](A)(A') \pcline[nodesep=.2, arrows={->},arrowsize=.5](B)(B') \pcline[nodesep=.2, arrows={->},arrowsize=.5](C)(C') \pstMiddleAB[PointName=none,PointSymbol=none]{A}{C}{O} \pstMiddleAB[PointName=none,PointSymbol=none]{A'}{C'}{O'} \psellipse(O)(.5,2.5)\psellipse(O')(.5,2.5) \meinecaption{2}{Injective, not surjective.}\label{fig:inj-n-epi} \end{minipage} \hfill \begin{minipage}{4cm} \centering \psset{unit=1pc} \pstGeonode[PointName=none](0,0){A}(0,1){B}(0,2){C}(0,-1){D} \pstGeonode[PointName=none](5,0){A'}(5,1){B'}(5,2){C'} \pcline[nodesep=.2, arrows={->},arrowsize=.5](A)(A') \pcline[nodesep=.2, arrows={->},arrowsize=.5](B)(B') \pcline[nodesep=.2, arrows={->},arrowsize=.5](C)(C') \pcline[nodesep=.2, arrows={->},arrowsize=.5](D)(C') \pstMiddleAB[PointName=none,PointSymbol=none]{A}{C}{O} \pstMiddleAB[PointName=none,PointSymbol=none]{A'}{C'}{O'} \psellipse(O)(.5,2.5)\psellipse(O')(.5,2.5) \meinecaption{2}{Surjective, not injective.}\label{fig:epi-n-iny} \end{minipage} \hfill \begin{minipage}{4cm} \centering \centering \psset{unit=1pc} \pstGeonode[PointName=none](0,0){A}(0,1){B}(0,2){C} \pstGeonode[PointName=none](5,0){A'}(5,1){B'}(5,2){C'} \pcline[nodesep=.2, arrows={->},arrowsize=.5](B)(B') \pcline[nodesep=.2, arrows={->},arrowsize=.5](C)(C') \pcline[nodesep=.2, arrows={->},arrowsize=.5](A)(B') \pstMiddleAB[PointName=none,PointSymbol=none]{A}{C}{O} \pstMiddleAB[PointName=none,PointSymbol=none]{A'}{C'}{O'} \psellipse(O)(.5,2.5)\psellipse(O')(.5,2.5) \meinecaption{2}{Neither injective nor surjective.}\label{fig:n-iny-n-epi} \end{minipage} \hfill \begin{minipage}{4cm} \centering \psset{unit=1pc} \pstGeonode[PointName=none](0,0){A}(0,1){B}(0,2){C} \pstGeonode[PointName=none](5,0){A'}(5,1){B'}(5,2){C'} \pcline[nodesep=.2, arrows={->},arrowsize=.5](A)(A') \pcline[nodesep=.2, arrows={->},arrowsize=.5](B)(B') \pcline[nodesep=.2, arrows={->},arrowsize=.5](C)(C') \pstMiddleAB[PointName=none,PointSymbol=none]{A}{C}{O} \pstMiddleAB[PointName=none,PointSymbol=none]{A'}{C'}{O'} \psellipse(O)(.5,2.5)\psellipse(O')(.5,2.5) \meinecaption{2}{Bijective.}\label{fig:biyectiva} \end{minipage} \end{figure} It is apparent from figures \ref{fig:inj-n-epi} through \ref{fig:biyectiva} that if the domain and the target set of a function are finite, then there are certain inequalities that must be met in order for the function to be injective, surjective or bijective. We make the precise statement in the following theorem. \begin{thm}\label{thm:size_domain_image_injections_surjections} Let $f:A\rightarrow B$ be a function, and let $A$ and $B$ be finite, with $A$ having $n$ elements, and and $B$ $m$ elements. If $f$ is injective, then $n \leq m$. If $f$ is surjective then $m\leq n$. If $f$ is bijective, then $m = n$. If $n \leq m$, then the number of injections from $A$ to $B$ is $$m(m-1)(m-2)\cdots (m-n+1). $$ \end{thm} \begin{pf} Let $A = \{x_1,x_2, \ldots ,x_n\}$ and $B = \{y_1,y_2, \ldots ,y_m\}$. \bigskip If $f$ were injective then $f(x_1), f(x_2), \ldots , f(x_n)$ are all distinct, and among the $y_k$. Hence $n \leq m$. In this case, there are $m$ choices for $f(x_1)$, $m-1$ choices for $f(x_2)$, \ldots , $m-n+1$ choices for $f(x_n)$. Thus there are $$m(m-1)(m-2)\cdots (m-n+1) $$ injections from $A$ to $B$. \bigskip If $f$ were surjective then each $y_k$ is hit, and for each, there is an $x_i$ with $f(x_i) = y_k$. Thus there are at least $m$ different images, and so $n \geq m$. \end{pf} To find the number of surjections from a finite set to a finite set we need to know about Stirling numbers and inclusion-exclusion, and hence, we refer the reader to any good book in Combinatorics. \begin{exa} Let $A=\{1,2,3\}$ and $B=\{4,5,6,7\}$. How many functions are there from $A$ to $B$? How many functions are there from $B$ to $A$? How many injections are there from $A$ to $B$? How many surjections are there from $B$ to $A$? \end{exa} \begin{solu} There are $4\cdot 4 \cdot 4 =64$ functions from $A$ to $B$, since there are $4$ possibilities for the image of $1$, $4$ for the image of $2$, and $4$ for the image of $3$. Similarly, there are $3\cdot 3 \cdot 3 \cdot 3 = 81$ functions from $B$ to $A$. \bigskip By Theorem \ref{thm:size_domain_image_injections_surjections}, there are $$ 4\cdot 3 \cdot 2 = 24 $$ injections from $A$ to $B$. \bigskip The $3^4$ functions from $B$ to $A$ come in three flavours: (i) those that are surjective, (ii) those that map to exactly two elements of $A$, and (iii) those that map to exactly one element of $A$. \bigskip Take a particular element of $A$, say $1\in A$. There are $2^4$ functions from $B$ to $\{2, 3\}$. Notice that some of these may map to the whole set $\{2, 3\}$ or they may skip an element. Coupling this with the $1\in A$, this means that there are $2^4$ functions from $B$ to $A$ that skip the $1$ and may or may not skip the $2$ or the $3$. Since there is nothing holy about choosing $1\in A$, we conclude that there are $3\cdot 2^4$ from $B$ to $A$ that skip either one or two elements of $A$. \bigskip Now take two particular elements of $A$, say $\{1,2\}\subseteq A$. There are $1^4$ functions from $B$ to $\{3\}$. Since there are three $2$-element subsets in $A$---namely $\{1,2\}$, $\{1,3\}$, and $\{2, 3\}$---this means that there are $3\cdot 1^4$ functions from $B$ to $A$ that map precisely into one element of $A$. \bigskip To find the number of surjections from $B$ to $A$ we weed out the functions that skip elements. In considering the difference $ 3^4- 3\cdot 2^4$, we have taken out all the functions that miss one or two elements of $A$, but in so doing, we have taken out twice those that miss one element. Hence we put those back in and we obtain $$ 3^4- 3\cdot 2^4+3\cdot 1^4=36$$ surjections from $B$ to $A$. \end{solu} \begin{rem} It is easy to see that a graphical criterion for a function to be injective is that every horizontal line crossing the function must meet it at most one point. See figures \ref{fig:injective} and \ref{fig:notinjective}. \end{rem} \vspace*{2cm} \begin{figure}[h] \begin{minipage}{.5\textwidth} \centering \psset{unit=1pc} \psplot[algebraic,linewidth=2pt,linecolor=brown]{-3}{3}{2^x-4} \psaxes[linewidth=2pt,labels=none,ticks=none,arrows={->}](0,0)(-5,-5)(5,5) \meinecaption{2}{Passes horizontal line test: injective.} \label{fig:injective} \end{minipage} \hfill \begin{minipage}{.5\textwidth} \centering \psset{unit=1pc} \psplot[algebraic,linewidth=2pt,linecolor=brown]{-5}{5}{4*sin(sqrt(abs(x)))} \psaxes[linewidth=2pt,labels=none,ticks=none,arrows={->}](0,0)(-5,-5)(5,5) \meinecaption{2}{Fails horizontal line test: not- injective.} \label{fig:notinjective} \end{minipage} \end{figure} \begin{exa} The $\fun{a}{x}{x^2}{\reals }{\reals }$ is neither injective nor surjective. For example, $a(-2)=a(2)=4$ but $-2\neq 2$, and there is no $x\in\reals $ with $a(x)=-1$. The function $\fun{b}{x}{x^2}{\reals }{\co{0;+\infty}}$ is surjective but not injective. The function $\fun{c}{x}{x^2}{\co{0;+\infty}}{\reals }$ is injective but not surjective. The function $\fun{d}{x}{x^2}{\co{0;+\infty}}{\co{0;+\infty}}$ is bijective. \end{exa} Given a formula, it is particularly difficult to know in advance what it set of outputs is going to be. This is why when we talk about function, we specify the target set to be a canister for every possible value. The next few examples shew how to find the image of a formula in a few easy cases. \begin{exa} Let $f: \reals \to \reals$, $f(x)=x^2+2x+3$. Determine $\im{f}$. \end{exa} \begin{solu} Observe that $$x^2+2x+3 = x^2+2x+1 +2 = (x+1)^2+2 \geq 2, $$ since the square of every real number is positive. Since $(x+1)^2$ could be made as arbitrarily close to $0$ as desired (upon taking values of $x$ close to $-1$), and can also be made as large as desired, we conclude that $\im{f} \subseteqq \co{2;+\infty}$. Now, let $a\in \co{2;+\infty}$. Then $$x^2+2x+3 = a \iff (x+1)^2+2=a \iff x=-1\pm \sqrt{a-2}. $$ Since $a\geq 2$, $\sqrt{a-2}\in \reals$ and $x\in \reals$. This means that $\co{2;+\infty }\subseteqq \im{f}$ and so we conclude that \mbox{$\im{f} = \co{2:+\infty}$}. \end{solu} \begin{exa} Let $f: \reals \setminus \{1\} \to \reals$, $f(x)=\dfrac{2x}{x-1}$. Determine $\im{f}$. \end{exa} \begin{solu} Observe that $$\dfrac{2x}{x-1} = 2+\dfrac{2}{x-1} \neq 2 $$since $\dfrac{2}{x-1}$ never vanishes for any real number $x$. We will shew that $\im{f}=\reals \setminus \{2\}$. For let $a\neq 2$. Then $$ \dfrac{2x}{x-1}=a \implies 2x = ax-a \implies x(2-a)=-a \implies x=\dfrac{a}{a-2}. $$ But if $a\neq 2$, then $x\in \reals$ and so we conclude that $\im{f}=\reals \setminus \{2\}$. \end{solu} \begin{exa} Consider the function $\fun{f}{A}{B}{x}{\dfrac{x-1}{x+1}}$ , where $A$ is the domain of definition of $f$. \begin{enumerate} \item Determine $A$. \item Determine $B$ so that $f$ be surjective. \item Demonstrate that $f$ is injective. \end{enumerate} \end{exa} \begin{solu} The formula $f(x)=\dfrac{x-1}{x+1}$ outputs real numbers for all values of $x$ except for $x=-1$, whence $A = \reals \setminus \{-1\}$. Now, $$ \dfrac{x-1}{x+1} = 1+ \dfrac{2}{x-1} \neq 1, $$since $\dfrac{2}{x-1}$ never vanishes. If $a\neq 1$ then $$\dfrac{x-1}{x+1}=a \implies ax-a=x+1 \implies x(a-1)=1+a \implies x=\dfrac{1+a}{1-a}, $$ which is a real number, since $a\neq 1$. It follows that $\im{f} = \reals \setminus \{1\}$. To demonstrate that $f$ is injective, we observe that $$f(a)=f(b) \implies \dfrac{a-1}{a+1} = \dfrac{b-1}{b+1} \implies (a-1)(b+1) = (a+1)(b-1) \implies ab+a-b = ab-a+b \implies 2a = 2b \implies a=b, $$ from where the function is indeed injective. \end{solu} \begin{exa} Prove that $$\fun{h}{x}{x^3}{\reals}{\reals}$$ is a bijection. \end{exa} \begin{solu} Assume $h(b) = h(a).$ Then $$\begin{array}{ccc} h(a) = h(b) & \implies & a^3 = b^3 \\ &\implies & a^3 - b^3 = 0 \\ & \implies & (a - b)(a^2 + ab + b^2) = 0 \\ \end{array} $$ Now, $$b^2 + ab + a^2 = \left(b + \frac{a}{2}\right)^2 + \frac{3a^2}{4}.$$ This shews that $b^2 + ab + a^2$ is positive unless both $a$ and $b$ are zero. Hence $b - a = 0$ in all cases. We have shewn that $h(b) = h(a) \implies b = a$, and the function is thus injective. \bigskip To prove that $h$ is surjective, we must prove that $(\forall \ b \in\reals) \ (\exists a)$ such that $h(a) = b.$ We choose $a$ so that $a = b^{1/3}$. Then $$h(a) = h(b^{1/3}) = (b^{1/3})^3 = b.$$Our choice of $a$ works and hence the function is surjective. \end{solu} \begin{exa} Prove that $\fun{f}{x}{\dfrac{x^{1/3}}{x^{1/3} - 1}}{\reals \setminus \{1\}}{\reals}$ is injective but not surjective. \end{exa} \begin{solu} We have $$\begin{array}{ccccc} f(a) = f(b) & \implies & \dfrac{a^{1/3}}{a^{1/3} - 1} & = & \dfrac{b^{1/3}}{b^{1/3} - 1} \\ & \implies & a^{1/3}b^{1/3} - a^{1/3} & = & a^{1/3}b^{1/3} - b^{1/3} \\ & \implies & -a^{1/3} & = & -b^{1/3} \\ & \implies & a & = & b, \\ \end{array} $$ whence $f$ is injective. To prove that $f$ is not surjective assume that $f(x) = b, b\in \reals$. Then $$f(x) = b \implies \dfrac{x^{1/3}}{x^{1/3} - 1} = b \implies x = \dfrac{b^3}{(b - 1)^3}. $$The expression for $x$ is not a real number when $b = 1$, and so there is no real $x$ such that $f(x) = 1$. \end{solu} \begin{exa} Find the image of the function $$\fun{f}{x}{\dfrac{x-1}{x^2+1}}{\reals}{\reals} $$ \end{exa} \begin{solu} First observe that $f(x)=0$ has the solution $x=1$. Assume $b\in \reals$, $b \neq 0$, with $f(x)=b$. Then $$ \dfrac{x-1}{x^2+1}=b \implies bx^2-x+b+1 =0. $$ Completing squares, $$ bx^2-x+b+1 = b\left(x^2-\dfrac{x}{b}\right) + b+1=b\left(x^2-\dfrac{x}{b}+\dfrac{1}{4b^2}\right) + b+1-\dfrac{1}{4b}= b\left(x-\dfrac{1}{2b}\right)^2 + \dfrac{-1+4b+4b^2}{4b}. $$ Hence $$ bx^2-x+b+1 =0 \iff b\left(x-\dfrac{1}{2b}\right)^2 = \dfrac{1-4b-4b^2}{4b} \iff x=\dfrac{1}{2}\pm \dfrac{\sqrt{1-4b-4b^2}}{2b}.$$ We must in turn investigate the values of $b$ for which $b\neq 0$ and $1-4b-4b^2 \geq 0$. Again, completing squares $$1-4b-4b^2 = -4\left(b^2+b\right) + 1= -4\left(b^2+b+\dfrac{1}{4}\right) + 2 = 2-\left(2b+1\right)^2 = =\left(\sqrt{2}-2b-1\right)\left(\sqrt{2}+2b+1\right).$$A sign diagram then shews that $1-4b-4b^2\geq 0$ for $$b \in \cc{-\dfrac{1}{2}-\dfrac{\sqrt{2}}{2};-\dfrac{1}{2}+\dfrac{\sqrt{2}}{2}}, $$ and so $$\im{f}=\cc{-\dfrac{1}{2}-\dfrac{\sqrt{2}}{2};-\dfrac{1}{2}+\dfrac{\sqrt{2}}{2}}. $$ \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro}\label{exa:g=2x+1} Prove that $$\fun{g}{s}{2s + 1}{\reals}{\reals}$$ is a bijection. \begin{answer} Assume $g(s_1) = g(s_2).$ Then $$\begin{array}{ccccc} g(s_1) = g(s_2) & \implies & 2s_1 + 1 & = & 2s_2 + 1 \\ & \implies & 2s_1 & = & 2s_2 \\ & \implies & s_1 & = & s_2 \\ \end{array} $$ We have shewn that $g(s_1) = g(s_2) \implies s_1 = s_2$, and the function is thus injective. \bigskip To prove that $g$ is surjective, we must prove that $(\forall \ b \in\reals) \ (\exists a)$ such that $g(a) = b.$ We choose $a$ so that $a = \dis{\frac{b - 1}{2}}$. Then $$g(a) = g\left(\frac{b - 1}{2}\right) = 2\left(\frac{b - 1}{2}\right) + 1 = b - 1 + 1 = b.$$Our choice of $a$ works and hence the function is surjective. \end{answer} \end{pro} \begin{pro} Prove that $h:\reals \rightarrow \reals$ given by $h(s) = 3 - s$ is a bijection.\end{pro} \begin{pro} Prove that $g:\reals \rightarrow \reals$ given by $g(x) = x^{1/3}$ is a bijection.\end{pro} \begin{pro} Prove that $\fun{f}{x}{\dfrac{2x}{x + 1}}{\reals \setminus \{1\}}{\reals \setminus \{2\}}$ is surjective but that $\fun{g}{x}{\dfrac{2x}{x + 1}}{\reals \setminus \{1\}}{\reals}$ is not surjective. \begin{answer} We must shew that there is a solution $x$ for the equation $f(x) = b, b\in \reals\setminus \{2\}$. Now $$f(x) = b \implies \dfrac{2x}{x + 1} = b \implies x = \dfrac{b}{2 - b}. $$Thus as long as $b \neq 2$ there is $x\in\reals$ with $f(x) = b$. Since there is no $x$ such that $g(x) = 2$ and $2\in \target{g}$, $g$ is not surjective. \end{answer} \end{pro} \begin{pro} Classify each of the following as injective, surjective, bijective or neither. \\ \begin{enumerate} \item $f: \reals \rightarrow \reals,\ \ \ x \mapsto x^4$ \\ \item $f: \reals \rightarrow \{ 1 \},\ \ \ x \mapsto 1$ \\ \item $f: \{1, 2, 3\} \rightarrow \{ a, b\},\ \ \ f(1) = f(2) = a, f(3) = b$ \\ \item $f: [0;+\infty[ \rightarrow \reals,\ \ \ x \mapsto x^3$ \\ \item $f: \reals \rightarrow \reals, \ \ \ x \mapsto |x|$ \\ \item $f: [0;+\infty[ \rightarrow \reals,\ \ \ x \mapsto -|x|$ \\ \item $f: \reals \rightarrow [0;+\infty[,\ \ \ x \mapsto |x|$ \\ \item $f: [0;+\infty[ \rightarrow [0;+\infty[,\ \ \ x \mapsto x^4$ \\ \end{enumerate} \begin{answer} \begin{enumerate} \item neither, $f(-1) = f(1)$ so not injective. There is no $a$ with $f(a) = -1,$ so not surjective. \item surjective, $f(1) = f(-1)$ so not injective.\\ \item surjective, not injective.\\ \item injective, as proved in text, there is no $a$ with $f(a) = -1,$ so not surjective. \\ \item neither, $|1| = |-1|$ so not injective, there is no $a$ with $|a| = -1$, so not surjective. \\ \item injective, non-surjective since, say, there is no $a$ with $-|a| = 1$. \\ \item surjective, non-injective since, say, $|-1| = |1|$ but $-1\neq 1$. \\ \item bijective. \\ \end{enumerate} \end{answer} \end{pro} \begin{pro} Let $f: E \rightarrow F, g: F \rightarrow G$ be two functions. Prove that if $g\circ f$ is surjective then $g$ is surjective. \end{pro} \begin{pro} Let $f: E \rightarrow F, g: F \rightarrow G$ be two functions. Prove that if $g\circ f$ is injective then $f$ is injective. \end{pro} \end{multicols} \section{Inversion} Let $S \subseteqq \reals$. Recall that $\idefun _S$ is the identity function on $S$, that is , $\idefun _S: S \to S$ with$\idefun _S(x) = x.$ \begin{df}Let $A\times B \subseteq \reals ^2$. A function $f: A \rightarrow B$ is said to be {\em right invertible} if there is a function $g: B\to A$, called the {\em right inverse of $f$} such that $f\circ g = \idefun _B$. In the same fashion, $f$ is said to be {\em left invertible } if there exists a function $h: B \to A$ such that $h\circ f = \idefun _A$. A function is {\em invertible} if it is both right and left invertible. \end{df} \begin{thm} Let $f:A \to B$ be right and left invertible. Then its left inverse coincides with its right inverse. \end{thm} \begin{pf} Let $g, h: B \to A$ be the respective right and left inverses of $f$. Using the associativity of compositions, $$(f\circ g) = (\idefun _B) \implies h\circ (f\circ g) = h\circ\idefun _B \implies (h\circ f)\circ g = h\circ\idefun _B \implies (\idefun _A) \circ g = h\circ \idefun _B\implies g=h. $$ \end{pf} \begin{cor}[Uniqueness of Inverses] If $f: A \to B$ is invertible, then its inverse is unique. \end{cor} \begin{pf} Let $f$ have the two inverses $s, t: B\to A$. In particular, $s$ would be a right inverse and $t$ would be a left inverse. By the preceding theorem, these two must coincide. \end{pf} \begin{df} If $f:A\to B$ is invertible, then its inverse will be denoted by $f^{-1}:B\to A$. \end{df} \begin{rem}We must alert the reader that $f^{-1}$ does not denote the reciprocal (multiplicative inverse) of $f$. \end{rem} \begin{thm} Let $f:A\to B$ and $g: C\to A$ be invertible. Then the composition function $f\circ g: C\to B$ is also invertible and $$ (f\circ g)^{-1}=g^{-1}\circ f^{-1}. $$ \end{thm} \begin{pf}By the uniqueness of inverses, $f\circ g$ may only have one inverse, which is, by definition, $(f\circ g)^{-1}$. This means that any other function that satisfies the conditions of being an inverse of $f\circ g$ must then by default be {\em the} inverse of $f\circ g$. We have, $$(g^{-1}\circ f^{-1})\circ (f\circ g)= g^{-1}\circ (f^{-1}\circ f)\circ g = g^{-1}\circ \idefun _A \circ g = g^{-1}\circ g = \idefun _C. $$ In the same fashion, $$ (f\circ g)\circ (g^{-1}\circ f^{-1}) = f\circ (g\circ g^{-1})\circ f^{-1} = f\circ \idefun _A\circ f^{-1} = f\circ f^{-1} = \idefun _B. $$ The theorem now follows from the uniqueness of inverses.\end{pf} \begin{exa}Let $\fun{f}{\reals \setminus \{1\}}{\reals\setminus \{2\}}{x}{\dfrac{2x}{x-1}}$. Demonstrate that $\fun{g}{\reals \setminus \{2\}}{\reals\setminus \{1\}}{x}{\dfrac{x}{x-2}}$ is the inverse of $f$. \end{exa} \begin{solu}Let $x\in \reals \setminus \{2\}$. We have $$ (f\circ g)(x) = f(g(x)) = \dfrac{2g(x)}{g(x)-1} = \dfrac{\dfrac{2x}{x-2}}{\dfrac{x}{x-2}-1} = \dfrac{2x}{x-(x-2)}=x, $$ from where $g$ is a right inverse of $f$. In a similar manner, $x\in \reals \setminus \{2\}$, $$(g\circ f)(x) =g(f(x)) =\dfrac{f(x)}{f(x)-2} = \dfrac{\dfrac{2x}{x-1}}{\dfrac{2x}{x-1}-2}=\dfrac{2x}{2x-2(x-1)}=x,$$whence $g$ is a left inverse of $f$. \end{solu} Consider the functions $u:\{a, b, c\} \rightarrow \{x, y, z\}$ and $v:\{x, y, z\} \rightarrow \{a, b, c\}$ as given by diagram \ref{inverse_1}. It is clear the $v$ undoes whatever $u$ does. Furthermore, we observe that $u$ and $v$ are bijections and that the domain of $u$ is the image of $v$ and vice-versa. This example motivates the following theorem. \begin{thm} A function $f: A \rightarrow B$ is invertible if and only if it is a bijection. \end{thm} \begin{pf} Assume first that $f$ is invertible. Then there is a function $f^{-1}: B \rightarrow A$ such that \begin{equation} f\circ f^{-1} = \idefun _B \ \ {\rm and} \ \ f^{-1}\circ f = \idefun _A.\end{equation} Let us prove that $f$ is injective and surjective. Let $s, t$ be in the domain of $f$ and such that $f(s) = f(t)$. Applying $f^{-1}$ to both sides of this equality we get $(f^{-1}\circ f)(s) = (f^{-1}\circ f)(t)$. By the definition of inverse function, $(f^{-1}\circ f)(s) = s$ and $(f^{-1}\circ f)(t) = t$. Thus $s = t.$ Hence $f(s) = f(t) \implies s = t$ implying that $f$ is injective. To prove that $f$ is surjective we must shew that for every $b \in f(A) \ \exists a \in A$ such that $f(a) = b.$ We take $a = f^{-1}(b)$ (observe that $f^{-1}(b) \in A$). Then $f(a) = f(f^{-1}(b)) = (f\circ f^{-1})(b) = b$ by definition of inverse function. This shews that $f$ is surjective. We conclude that if $f$ is invertible then it is also a bijection. \bigskip Assume now that $f$ is a bijection. For every $b\in B$ there exists a unique $a$ such that $f(a) = b$. This makes the rule $g: B \rightarrow A$ given by $g(b) = a$ a function. It is clear that $g\circ f = \idefun _A$ and $f\circ g = \idefun _B. $ We may thus take $f^{-1} = g.$ This concludes the proof. \end{pf} \vspace*{1cm} \begin{figure}[!hptb] \centering %NOTE: Beware of leaving blank spaces whenever you are in the PSTRICKS %environment. It does not tolerate blank horizontal lines and it will %give you an error in your LaTeX run $$ \psset{unit=.75cm}\rput(-5,0){ \rput(0,0){a} \rput(0, .5){b} \rput(0, -.5){c} \rput(3,0){x} \rput(3,.5){y} \rput(3,-.5){z} \rput(1.5, .8){u} \psellipse(1,2) \psellipse(3,0)(1,2) \psline[linewidth=.4pt, labels=none, showpoints=true]{*->>}(.2,0)(2.9,0) \psline[linewidth=.4pt, labels=none, showpoints=true]{*->>}(.2,.5)(2.9,.5) \psline[linewidth=.4pt, labels=none, showpoints=true]{*->>}(0.2,-.5)(2.9,-.5)} \rput(5,0){\rput(0,0){x} \rput(0, .5){y} \rput(0, -.5){z} \rput(3,0){a} \rput(3,.5){b} \rput(3,-.5){c} \rput(1.5, .8){v} \psellipse(1,2) \psellipse(3,0)(1,2) \psline[linewidth=.4pt, labels=none, showpoints=true]{*->>}(.2,0)(2.9,0) \psline[linewidth=.4pt, labels=none, showpoints=true]{*->>}(.2,.5)(2.9,.5) \psline[linewidth=.4pt, labels=none, showpoints=true]{*->>}(0.2,-.5)(2.9,-.5)} $$ \meinecaption{1}{A function and its inverse.} \label{inverse_1} \end{figure} We will now give a few examples of how to determine the assignment rule of the inverse of a function. \begin{exa}Assume that the function $$ \fun{f}{x}{\dfrac{x-1}{x+1}}{\reals \setminus \{-1\}}{\reals \setminus \{1\}} $$ is a bijection. Determine its inverse. \end{exa} \begin{solu} Put $$ \dfrac{x-1}{x+1}=y $$ and solve for $x$: $$ \dfrac{x-1}{x+1}=y \implies x-1 = yx+y \implies x-yx = 1+y \implies x(1-y) = 1+y \implies x = \dfrac{1+y}{1-y}. $$ Now, exchange $x$ and $y$: $y= \dfrac{1+x}{1-x}$. The desired inverse is $$ \fun{f^{-1}}{x}{\dfrac{1+x}{1-x}}{\reals \setminus \{1\}}{\reals \setminus \{-1\}}. $$ \end{solu} \begin{exa}Assume that the function $$ \fun{f}{x}{(x-2)^3+1}{\reals}{\reals} $$ is a bijection. Determine its inverse. \end{exa} \begin{solu} Put $$ (x-2)^3+1=y $$ and solve for $x$: $$ (x-2)^3+1=y \implies (x-2)^3=y-1 \implies x-2 = \sqrt[3]{y-1} \implies x= \sqrt[3]{y-1} +2. $$ Now, exchange $x$ and $y$: $y= \sqrt[3]{x-1}+2$. The desired inverse is $$ \fun{f^{-1}}{x}{\sqrt[3]{x-1}+2}{\reals}{\reals}. $$ \end{solu} \begin{rem} Since by Theorem \ref{thm:symmetry_ab_ba}, $(x, f(x))$ and $(f(x), x)$ are symmetric with respect to the line $y = x$, the graph of a function $f$ is symmetric with its inverse with respect to the line $y = x$. See figures \ref{fig:func_inv1} through \ref{fig:func_inv3}. \end{rem} \vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{.3\textwidth} \psset{unit=1pc} \centering \psaxes[labels=none,ticks=none,arrows={->},linewidth=2pt](0,0)(-5,-5)(5,5) \psplot[algebraic=true,linewidth=2pt,linecolor=brown,arrows={*->}]{0}{2.263}{x^2} \psplot[algebraic=true,linewidth=2pt,linecolor=blue,arrows={*->},linestyle=dashed]{0}{5}{sqrt(x)} \psplot[algebraic=true,linestyle=dotted]{-4.5}{4.5}{x} \meinecaption{2}{Function and its inverse.} \label{fig:func_inv1} \end{minipage} \hfill \begin{minipage}{.3\textwidth} \psset{unit=1pc} \centering \psaxes[labels=none,ticks=none,arrows={->},linewidth=2pt](0,0)(-5,-5)(5,5) \psplot[algebraic=true,linewidth=2pt,linecolor=brown,arrows={<->}]{-1.667}{1.667}{x^3} \psplot[algebraic=true,linewidth=2pt,linecolor=blue,arrows={->},linestyle=dashed]{0}{5}{x^(1/3)} \psplot[algebraic=true,linewidth=2pt,linecolor=blue,arrows={<-},linestyle=dashed]{-5}{0}{-abs(x)^(1/3)} \psplot[algebraic=true,linestyle=dotted]{-4.5}{4.5}{x} \meinecaption{2}{Function and its inverse.} \label{fig:func_inv2} \end{minipage} \hfill \begin{minipage}{.3\textwidth} \psset{unit=1pc} \centering \psaxes[labels=none,ticks=none,arrows={->},linewidth=2pt](0,0)(-5,-5)(5,5) \psline[showpoints=true,linecolor=brown,linewidth=2pt](-4,-5)(-3,-2)(0,-1)(2,3)(4,4) \psline[showpoints=true,linecolor=blue,linewidth=2pt,linestyle=dashed](-5,-4)(-2,-3)(-1,0)(3,2)(4,4) \psplot[algebraic=true,linestyle=dotted]{-4.5}{4.5}{x} \meinecaption{2}{Function and its inverse.} \label{fig:func_inv3} \end{minipage} \end{figure} \begin{exa}\label{exa:inverseA}Consider the functional curve in figure \ref{fig:inverseA}. \begin{enumerate} \item Determine $\dom{f}$. \item Determine $\im{f}$. \item Draw the graph of $f^{-1}$. \item Determine $f(+5)$. \item Determine $f^{-1}(-2)$. \item Determine $f^{-1}(-1)$. \end{enumerate} \end{exa} \begin{solu} \noindent \begin{enumerate} \item $\cc{-5;5}$ \item $\cc{-3;3}$ \item To obtain the graph, we look at the endpoints of lines on the graph of $f$ and exchange their coordinates. Thus the endpoints $(-5,-3)$, $(-3,-2)$, $(0,-1)$, $(1,1)$, $(5,3)$ on the graph of $f$ now form the endpoints $(-3,-5)$, $(-2,-3)$, $(-1,0)$, $(1,1)$, and $(3,5)$ on the graph of $f^{-1}$. The graph appears in figure \ref{fig:inverseB} below.\\ \item $f(+5)=3$. \item $f^{-1}(-2)=-3$. \item $f^{-1}(-1)=0$. \end{enumerate} \end{solu} \vspace*{2cm} \begin{figure}[h] \begin{minipage}{.5\textwidth} \centering \psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psline[showpoints=true,linewidth=2pt,linecolor=blue](-5,-3)(-3,-2)(0,-1)(1,1)(5,3) \meinecaption{3}{$f$ for example \ref{exa:inverseA}.}\label{fig:inverseA} \end{minipage} \hfill \begin{minipage}{.5\textwidth} \centering \psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psline[showpoints=true,linewidth=2pt,linecolor=blue](-3,-5)(-2,-3)(-1,0)(1,1)(3,5) \meinecaption{3}{$f^{-1}$ for example \ref{exa:inverseA}.}\label{fig:inverseB} \end{minipage}\end{figure} \begin{exa} Consider the formula $f(x)=x^2+4x+5$. Demonstrate that $f$ is injective in $\co{-2;+\infty}$ and determine $f(\co{-2;+\infty})$. Then, find the inverse of $$\fun{f}{x}{x^2+4x+5}{\co{-2;+\infty}}{f(\co{-2;+\infty})}. $$ \end{exa} \begin{solu} Observe that $x^2+4x+5=(x+2)^2+1$. Now, if $a\in \co{-2;+\infty}$ and $b\in \co{-2;+\infty}$, then $$f(a)=f(b) \implies (a+2)^2+1 = (b+2)^2+1 \implies (a+2)^2=(b+2)^2. $$ As $a+2\geq 0$ and $b+2 \geq 0$, we have $$(a+2)^2=(b+2)^2 \implies a+2 = b+2 \implies a=b, $$whence $f$ is injective in $\co{-2;+\infty}$. We have $f(x) = (x+2)^2+1\geq 1$. We will shew that $f(\co{-2;+\infty}=\co{1;+\infty}$. Let $b\in\co{1;+\infty}$. Solving for $x$: $$f(x)=b \implies (x+2)^2+1=b \implies (x+2)^2=b-1. $$As $b-1\geq 0$, $\sqrt{b-1}$ is a real number and thus $$ x=-2+\sqrt{b-1} $$is a real number with $x\leq -2$. We deduce that $f(\co{-2;+\infty})=\co{1;+\infty}$. Since $$\fun{f}{x}{x^2+4x+5} {\co{-2;+\infty}}{\co{1;+\infty}}$$ is a bijection, it is invertible. To find $f^{-1}$, we solve $$x^2+4x+5 =y \implies (x+2)^2+1=y \implies x=-2+\sqrt{y-1}, $$where we have taken the positive square root, since $x\geq -2$. Exchanging $x$ and $y$ we obtain $y = -2+\sqrt{x-1}$. We deduce that the inverse of $f$ is $$\fun{f^{-1}}{x}{-2+\sqrt{x-1}}{\co{1;+\infty}}{\co{-2;+\infty}}. $$ \end{solu} \begin{rem} In the same fashion it is possible to demonstrate that $$\fun{g}{x}{x^2+4x+5}{\oc{-\infty ; -2}}{\co{1;+\infty}} $$ bijective is, with inverse $$\fun{g^{-1}}{x}{-2-\sqrt{x-1}}{\co{1;+\infty}}{\oc{-\infty; -2}}. $$ \end{rem} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Let $$\fun{c}{x}{\frac{x}{x + 2}}{\reals \setminus \{-2\}}{\reals \setminus \{1\}}.$$ Prove that $c$ is bijective and find the inverse of $c$. \label{rational_1}\begin{answer} Since $c(c^{-1}(x)) = x,$ we have $\dis{\frac{c^{-1}(x)}{c^{-1}(x) + 2} = x}$. Solving for $c^{-1}(x)$ we obtain $c^{-1}(x) = \dis{\frac{2x}{1 - x} = -2 + \frac{2}{1 - x} }$. The inverse of $c$ is therefore $$\fun{c^{-1}}{x}{-2 + \frac{2}{1 - x} }{\reals \setminus \{1\}}{\reals \setminus \{-2\}}.$$ \end{answer} \end{pro}\begin{pro} Assume that $f:\reals\to \reals$ is a bijection, where $f(x) = 2x^3 + 1$. Find $f^{-1}(x)$. \begin{answer} $f^{-1}:\reals\to reals$, $f^{-1}(x) = \sqrt[3]{\dfrac{x-}{2}}$ \end{answer} \end{pro} \begin{pro} Assume that $f:\reals\setminus \{1\} \to \reals\setminus \{1\}$ is a bijection, where $f(x) = \sqrt[3]{\dfrac{x+2}{x-1}}$. Find $f^{-1}$. \begin{answer} $f:\reals\setminus \{1\} \to \reals\setminus \{1\}$, $f^{-1}(x) = \dfrac{x^3+2}{x^3-1}$. \end{answer} \end{pro} \begin{pro} Let $f$ and $g$ be invertible functions satisfying $$f(1) = 2,\quad f(2) = 3,\quad f(3)=1,$$ $$ g(1) = -1,\quad g(2) = 3,\quad g(4) =-2.$$ Find $(f\circ g)^{-1}(1)$. \begin{answer} $(f\circ g)^{-1}(1) = (g^{-1}\circ f^{-1})(1) = g^{-1}(f^{-1}(1)) = g^{-1}(3)=2$. \end{answer} \end{pro} \begin{pro} Consider the formula $f:x\mapsto x^2-4x + 5$. Find two intervals $I_1$ and $I_2$ with $\reals= I_1\cup I_2$ and $I_1 \cap I_2 $ consisting on exactly one point, such that $f$ be injective on the restrictions to each interval $f\Big| _{I_1}$ and $f\Big| _{I_2}$. Then, find the inverse of $f$ on each restriction. \begin{answer} Since $x^2-4x+5=(x-2)^2+1$, consider $I_1=\oc{-\infty;2}$ and $I_2=\co{2;+\infty}$. \end{answer} \end{pro} \begin{pro}\label{pro:dos-rectas-inverse} Consider the function $f:\cc{-5;5}\rightarrow \cc{-3;5}$ whose graph appears in figure \ref{fig:dos-rectas-inverse}, and which is composed of two lines. Observe that $f$ passes the horizontal line test, that it is surjective, and hence invertible. . \begin{enumerate} \item Find a formula for $f$ and $f^{-1}$ in $\cc{-5;0}$. \item Find a formula for $f$ and $f^{-1}$ in $\cc{0;5}$. \item Draw the graph of $f^{-1}$. \end{enumerate} \begin{answer} \noindent \begin{enumerate} \item The first piece of $f$ is a line segment with endpoints at $(-5,5), (0,-1)$, and whose slope is $-\dfrac{6}{5}$. Thus the equation of $f$ os $f(x) =-\dfrac{6}{5}x-1$. Putting $y =-\dfrac{6}{5}x-1$ and solving for $x$ we obtain $x=-\dfrac{5}{6}y-\dfrac{5}{6}$. We deduce that $f^{-1}(x) = -\dfrac{5}{6}x-\dfrac{5}{6}$. For $x\in [-5;0]$, $-1\leq f(x)\leq 5$, and hence the formula for $f^{-1}$ is only valid for $-1\leq x \leq 5$. \item The second piece is a line segment with endpoints at $(0,-1), (5,-3)$, which has slope $-\dfrac{2}{5}$. The equation of $f$ is $f(x) =-\dfrac{2}{5}x-1$. Putting $y =-\dfrac{2}{5}x-1$ and solving for $x$, we obtain $x=-\dfrac{5}{2}y-\dfrac{5}{2}$. We deduce that $f^{-1}(x) = -\dfrac{5}{2}x-\dfrac{5}{2}$. For $x\in [0;5]$, $-3\leq f(x)\leq -1$, and so this formula for $f^{-1}$ is only valid for $-3\leq x \leq -1$. \item The graph of $f^{-1}$ appears in figure \ref{fig:dos-rectas-inverseB}. \vspace{2cm} \begin{figurehere} \centering \psset{unit=.5pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-7,-7)(7,7) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-7,-7)(-7,-7)(7,7)\psline[linewidth=2pt,linecolor=brown]{<->}(-7.5,0)(7.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-7.5)(0,7.5) \psdots[dotscale=1.5,dotstyle=*](5,-5)(-1,0)(-3,5)\psline[linewidth=2pt,linecolor=blue](-3,5)(-1,0)(5,-5) \meinecaption{1cm}{Problem \ref{pro:dos-rectas-inverse}.}\label{fig:dos-rectas-inverseB} \end{figurehere} \end{enumerate} \end{answer} \end{pro} \vspace{1.5cm} \begin{figurehere} \psset{unit=.25cm} \centering\psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-7,-7)(7,7) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-7,-7)(-7,-7)(7,7)\psline[linewidth=2pt,linecolor=brown]{<->}(-7.5,0)(7.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-7.5)(0,7.5) \psdots[dotscale=1.5,dotstyle=*](-5,5)(0,-1)(5,-3)\psline[linewidth=2pt,linecolor=blue](-5,5)(0,-1)(5,-3) \meinecaption{2}{Problem \ref{pro:dos-rectas-inverse}.}\label{fig:dos-rectas-inverse} \end{figurehere} \begin{pro} Consider the rule $$f(x) = \dfrac{1}{\sqrt[3]{x^5 - 1}}. $$ \begin{enumerate} \item Find the natural domain of $f$. \item Find the inverse assignment rule $f^{-1}$. \item Find the image of the natural domain of $f$ and the natural domain of $f^{-1}$. \item Conclude. \end{enumerate} \begin{answer}We have \begin{enumerate} \item The expression under the cubic root must not be $0$. Hence $x^5 \neq 1$ and the natural domain is $\reals \setminus \{1\}$. \item Put $$y = \dfrac{1}{\sqrt[3]{x^5 - 1}}.$$ Now exchange $x$ and $y$ and solve for $y$: $$x = \dfrac{1}{\sqrt[3]{y^5 - 1}} \implies x^3(y^5 - 1) = 1 \implies y = \sqrt[5]{\dfrac{x^3 + 1}{x^3}}.$$ Hence $$f^{-1}(x) = \sqrt[5]{\dfrac{x^3 + 1}{x^3}}.$$ \item As $x$ varies in $ \reals \setminus \{1\}$, the expression $\dfrac{1}{\sqrt[3]{x^5 - 1}}$ assumes all positive and negative values, but it is never $0$. Thus $\im{f} = \reals \setminus \{0\}$. The expression for $f^{-1}(x)$ is undefined when $x = 0$. Hence the natural domain of $f^{-1}$ is $\reals \setminus \{0\}$. \item The function $$\fun{f}{x}{\dfrac{1}{\sqrt[3]{x^5 - 1}}}{\reals\setminus\{1\}}{\reals \setminus \{0\}} $$is a bijection with inverse $$\fun{f^{-1}}{x}{\sqrt[5]{\dfrac{x^3 + 1}{x^3}}}{\reals \setminus \{0\}}{\reals \setminus \{1\}}. $$ \end{enumerate} \end{answer} \end{pro} \begin{pro}Find all the real solutions to the equation $$x^2-\dfrac{1}{4}=\sqrt{x+\dfrac{1}{4}}.$$ \begin{answer} Since $x\geq 0$, $f(x)=x^2-\dfrac{1}{4}$ has inverse $f^{-1}(x)=\sqrt{x+\dfrac{1}{4}}$. The graphs of $f$ and $f^{-1}$ meet on the line $y=x$. Hence we are looking for a positive solution to $$x^2-\dfrac{1}{4}=x\implies x=\dfrac{1+\sqrt{2}}{2}. $$ \end{answer} \end{pro} \begin{pro} Let $f, g, h :\{1, 2, 3, 4\} \rightarrow \{1, 2, 10, 1993\}$ be given by $f(1) = 1, f(2) = 2, f(3) = 10, f(4) = 1993, g(1) = g(2) = 2, g(3) = g(4) - 1 = 1, h(1) = h(2) = h(3) = h(4) + 1 = 2.$ \begin{enumerate} \item Is $f$ invertible? Why? If so, what is $f^{-1}(f(h(4)))$? \item Is $g$ one-to-one? Why? \end{enumerate} \begin{answer} 1) Yes, $f$ is a bijection. $f^{-1}(f(h(4))) = h(4) = 1$, 2) No \end{answer} \end{pro} \begin{pro} Given $g:\reals \rightarrow \reals$, $g(x) = 2x + 8$ and $f:\reals \setminus \{-2\} \rightarrow \reals \setminus \{0\}$, $f(x) = \dis{\frac{1}{x + 2}}$ find $(g\circ f^{-1})(-2)$. \begin{answer} $3$\end{answer} \end{pro} \begin{pro} Prove that $\fun{t}{x}{\sqrt{1 - x}}{]-\infty; 1]}{[0; +\infty[}$ is a bijection and find $t^{-1}.$ \begin{answer} $\fun{t^{-1}}{x}{1 - x^2}{[0; +\infty[}{]-\infty; 1]}$ \end{answer} \end{pro} \begin{pro} Let $f:\reals \rightarrow \reals$, $f(x) = ax + b$. For which parameters $a$ and $b$ is $f = f^{-1}$? \begin{answer} Either $a = 1, b = 0$ or $a = -1$ and $b$ arbitrary. \end{answer} \end{pro} \begin{pro} Prove that if $ab \neq -4$ and $f:\reals \setminus \{2/b\} \rightarrow \reals \setminus \{2/b\}$, $f(x) = \dis{\frac{2x + a}{bx - 2}}$ then $f = f^{-1}$. \end{pro} \begin{pro} Let $f:\co{0;+\infty}\to \co{0;+\infty}$be given by $$f(x)=\sqrt{x+\sqrt{x}}.$$ Demonstrate that $f$ is bijective and that its inverse is $$f^{-1}:\co{0;+\infty}\to \co{0;+\infty}, \quad f^{-1}(x)=\dfrac{1-\sqrt{1+4x^2}}{2}+x^2.$$ \end{pro} \begin{pro} Demonstrate that $$f:\reals\to \cc{-1;1}, \quad f(x)=\dfrac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{\sqrt[3]{1+x}+\sqrt[3]{1-x}},$$ is bijective and that its inverse is $$f^{-1}:\cc{-1;1}\to \reals, \quad f^{-1}(x)=\dfrac{x(x^2+3)}{1+3x^2}.$$ \end{pro} \begin{pro} Demonstrate that $$f:\co{-\frac{1}{4};+\infty}\to \oc{-1;1}, \quad f(x)=\dfrac{1-\sqrt{1+4x}}{1+\sqrt{1+4x}},$$ is bijective and that its inverse is $$f^{-1}: \oc{-1;1}\to \co{-\frac{1}{4};+\infty}, \quad f^{-1}(x)=-\dfrac{x}{(1+x)^2}.$$ \end{pro} \begin{pro} Demonstrate that $$f:\reals\to \reals, \quad f(x)=\sqrt[3]{x+\sqrt{x^2+1}}+\sqrt[3]{x-\sqrt{x^2+1}},$$ is bijective and that its inverse is $$f^{-1}: \reals\to \reals, \quad f^{-1}(x)=\dfrac{x^3+3x}{2}.$$ \end{pro} \begin{pro}\label{pro:inversa-trozos} Consider the function $f:\reals\to \reals$, with $$f(x) = \left\{\begin{array}{ll} 2x & \mathrm{if}\ x\leq 0 \\ x^2 & \mathrm{if}\ x>0 \\ \end{array}\right. $$ whose graph appears in figure \ref{fig:inversa-trozos}. \begin{enumerate} \item Is $f$ invertible? \item If the previous answer is affirmative, draw the graph of $f^{-1}$. \item If $f$ is invertible, find a formula for $f^{-1}$. \end{enumerate} \begin{answer} We have $f^{-1}:\reals \to \reals$, with $$f^{-1}(x) = \left\{\begin{array}{ll} \dfrac{x}{2} & \mathrm{if}\ x\leq 0 \\ \sqrt{x} & \mathrm{if}\ x>0 \\ \end{array}\right. $$ The graph of $f^{-1}$ appears in figure \ref{fig:inversa-trozosB}.\\ \vspace{3cm} \begin{figurehere} \centering \psset{unit=1pc} \psdots[dotstyle=*,dotscale=1](1,1) \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-4,-4)(4,4) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-4,-4)(-4,-4)(4,4)\psline[linewidth=2pt,linecolor=brown]{<->}(-4.5,0)(4.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-4.5)(0,4.5)\psplot[linewidth=1.4pt, algebraic, arrows={<-*},linecolor=blue]{-3.4}{0}{x/2} \psplot[linewidth=1.4pt, algebraic, arrows={*->},linecolor=blue]{0}{3.24}{sqrt(x)} \meinecaption{2}{Problem \ref{pro:inversa-trozos}.}\label{fig:inversa-trozosB} \end{figurehere} \end{answer} \end{pro} \vspace{2cm} \begin{figurehere} \centering \psset{unit=.5cm} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-4,-4)(4,4) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-4,-4)(-4,-4)(4,4)\psline[linewidth=2pt,linecolor=brown]{<->}(-4.5,0)(4.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-4.5)(0,4.5) \psdots[dotstyle=*,dotscale=1](1,1)(-1,-2) \psplot[linewidth=2pt, algebraic=true, arrows={<-*},linecolor=blue]{-1.7}{0}{2*x} \psplot[linewidth=2pt, algebraic=true, arrows={*->},linecolor=blue]{0}{1.8}{x^2} \meinecaption{2}{Problem \ref{pro:inversa-trozos}. }\label{fig:inversa-trozos} \end{figurehere} \begin{pro} Demonstrate that $f:\cc{0;1}\to \cc{0;1}$, with $$f(x)=\left\{\begin{array}{ll} x & \mathrm{if}\ x\in \rationals \cap \cc{0;1} \\ 1- x & \mathrm{if}\ x\in \left(\reals \setminus \rationals\right) \cap \cc{0;1} \\ \end{array}\right. $$ is bijective and that $f=f^{-1}$. \end{pro} \begin{pro} Prove, without using a calculator, that $$ \sum _{k=1} ^{9} \left( \left(\frac{k}{10}\right)^2 + \sqrt{\frac{k}{10}}\right) < 9.5$$ \label{boundingarea_1}\begin{answer} The inverse of $$\fun{f}{x}{x^2}{[0;+\infty[}{[0;+\infty[}$$is$$\fun{f^{-1}}{x}{\sqrt{x}}{[0;+\infty[}{[0;+\infty[}.$$ In diagram \ref{boundingarea_1_1}, each rectangle $V_k$ has its lower left corner at $\dis{(0, \frac{k}{10})}$, base $\sqrt{\frac{k}{10}}$ and height $\frac{1}{10}$. Each rectangle $H_k$ has lower left corner at $\dis{(\frac{k}{10}, 0)}$, base $\frac{1}{10}$ and height $(\frac{k}{10})^2$. The collective area of these rectangles is $$ \frac{1}{10}\left(\left(\frac{1}{10}\right)^2 + \sqrt{\frac{1}{10}} + \left(\frac{2}{10}\right)^2 + \sqrt{\frac{2}{10}} + \left(\frac{3}{10}\right)^2 + \sqrt{\frac{3}{10}} + \cdots + \left(\frac{9}{10}\right)^2 + \sqrt{\frac{9}{10}} \right)$$Since these grey rectangles do not intersect with the green squares on the corners, their collective area is less than the area of the unit square minus these smaller squares: $1 - \dis{\frac{1}{100} - \frac{4}{100} = \frac{95}{100}}$. We thus conclude that $${\small \frac{1}{10}\left(\left(\frac{1}{10}\right)^2 + \sqrt{\frac{1}{10}} + \left(\frac{2}{10}\right)^2 + \sqrt{\frac{2}{10}} + \left(\frac{3}{10}\right)^2 + \sqrt{\frac{3}{10}} + \cdots + \left(\frac{9}{10}\right)^2 + \sqrt{\frac{9}{10}} \right) < \frac{95}{100}. } $$ \vspace*{4cm} \begin{figurehere} $$\psset{unit=1pc} \rput(-3,0){\psset{unit=8pc} \psframe[fillstyle=solid, fillcolor=brown](.1,0)(.2,.01) \psframe[fillstyle=solid, fillcolor=brown](.2,0)(.3,.04) \psframe[fillstyle=solid, fillcolor=brown](.3,0)(.4,.09) \psframe[fillstyle=solid, fillcolor=brown](.4,0)(.5,.16) \psframe[fillstyle=solid, fillcolor=brown](.5,0)(.6,.25) \psframe[fillstyle=solid, fillcolor=brown](.6,0)(.7,.36) \psframe[fillstyle=solid, fillcolor=brown](.7,0)(.8,.49) \psframe[fillstyle=solid, fillcolor=brown](.8,0)(.9,.64) \psframe[fillstyle=solid, fillcolor=brown](.9,0)(1,.81) \psframe[fillstyle=solid, fillcolor=brown](0,.1)(.01,.2) \psframe[fillstyle=solid, fillcolor=brown](0,.2)(.04,.3) \psframe[fillstyle=solid, fillcolor=brown](0,.3)(.09,.4) \psframe[fillstyle=solid, fillcolor=brown](0,.4)(.16,.5) \psframe[fillstyle=solid, fillcolor=brown](0,.5)(.25,.6) \psframe[fillstyle=solid, fillcolor=brown](0,.6)(.36,.7) \psframe[fillstyle=solid, fillcolor=brown](0,.7)(.49,.8) \psframe[fillstyle=solid, fillcolor=brown](0,.8)(.64,.9) \psframe[fillstyle=solid, fillcolor=brown](0,.9)(.81,1) \psplot{0}{1}{x 2 exp} \psplot[linestyle=dashed]{0}{1}{x sqrt} \psaxes*[dx=.1\psxunit,Dx=.1, labels=none, ticks=none](0,0)(-.1,-.1)(1.1,1.1) \rput(.15, -.2){H_1} \rput(.25, -.2){H_2} \rput(.35, -.2){H_3} \rput(.45, -.2){H_4} \rput(.55, -.2){H_5} \rput(.65, -.2){H_6} \rput(.75, -.2){H_7} \rput(.85, -.2){H_8} \rput(.95, -.2){H_9} \rput(-.2,.15){V_1} \rput(-.2,.25){V_2} \rput(-.2,.35){V_3} \rput(-.2,.45){V_4} \rput(-.2,.55){V_5} \rput(-.2,.65){V_6} \rput(-.2,.75){V_7} \rput(-.2,.85){V_8} \rput(-.2,.95){V_9} \psframe[fillstyle=solid, fillcolor=blue](0,0)(.1,.1) \psframe[fillstyle=solid, fillcolor=blue](.8,.8)(1,1)}$$\meinecaption{1}{Problem \ref{boundingarea_1}.} \label{boundingarea_1_1} \end{figurehere} \end{answer} \end{pro} \begin{pro} Verify that the functions below, with their domains and images, have the claimed inverses. \vspace{1cm} \begin{tabular}{|l|l|l|l|} \hline Assignment Rule & Natural Domain & Image & Inverse \\ \hline $x \mapsto \dis{\sqrt{2 - x}}$ & $]-\infty; 2]$ & $[0; +\infty[$ & $x \mapsto 2 - x^2$ \\ \hline $x \mapsto \dis{\frac{1}{\sqrt{2 - x}}}$ & $]-\infty; 2[$ & $]0; +\infty[$ & $x \mapsto 2 - \dis{\frac{1}{x^2}}$ \\ \hline $x \mapsto \dis{\frac{2 + x^3}{{2 - x^3}}}$& $\reals \setminus \{\sqrt[3]{2}\}$ & $\reals \setminus \{-1\}$ & $x \mapsto \dis{\sqrt[3]{\frac{2x - 2}{x + 1}}}$ \\ \hline $x \mapsto \dis{\frac{1}{x^3 - 1}}$& $\reals \setminus \{1\}$& $\reals \setminus \{0\}$& $x \mapsto \dis{\sqrt[3]{1 + \frac{1}{x}}}$ \\ \hline \end{tabular} \end{pro} \end{multicols} \chapter{Transformations of the Graph of Functions} \section{Translations} In this section we study how several rigid transformations affect both the graph of a function and its assignment rule. \begin{thm} \label{thm:translations} Let $f$ be a function and let $v$ and $h$ be real numbers. If $(x_0,y_0)$ is on the graph of $f$, then $(x_0,y_0+v)$ is on the graph of $g$, where $g(x)=f(x)+v$, and if $(x_1,y_1)$ is on the graph of $f$, then $(x_1-h,y_1)$ is on the graph of $j$, where $j(x)=f(x+h)$. \end{thm} \begin{pf} Let $\Gamma_f, \Gamma_g, \Gamma_j$ denote the graphs of $f, g, j$ respectively. $$(x_0, y_0) \in \Gamma_f \iff y_0 = f(x_0) \iff y_0 + v = f(x_0) + v \iff y_0+v = g(x_0) \iff (x_0,y_0+v)\in \Gamma_g. $$ Similarly, $$(x_1, y_1) \in \Gamma_f \iff y_1 = f(x_1) \iff y_1 = f(x_1-h + h) \iff y_1 = j(x_1-h) \iff (x_1-h,y_1)\in \Gamma_j. $$ \end{pf} \begin{df} Let $f$ be a function and let $v$ and $h$ be real numbers. We say that the curve $y = f(x) + v$ is a {\em vertical translation} of the curve $y = f(x)$. If $v>0$ the translation is $v$ up, and if $v<0$, it is $v$ units down. Similarly, we say that the curve $y = f(x + h)$ is a {\em horizontal translation} of the curve $y = f(x)$. If $h > 0$, the translation is $h$ units left, and if $h<0$, then the translation is $h$ units right. \index{translation!vertical} \index{translation!horizontal} \end{df} Given a functional curve, we expect that a translation would somehow affect its domain and image. \vspace*{2cm} \begin{figure}[h] \def\figurinetrans{\pstGeonode[PointName=none,PointSymbol=none](-4,0){A}(-2,0){B}(0,0){C}(1,0){D}(2,0){E}(3,0){F}(4,0){G} \pstArcOAB[linewidth=2pt,linecolor=blue]{B}{A}{C} \pstArcOAB[linewidth=2pt,linecolor=blue]{F}{G}{E} \psline[linewidth=2pt,linecolor=blue](C)(E) \psdots[linewidth=1.5pt](C)(D)(E)(G)(3,1)(-2,-2)(A) } \begin{minipage}{.2\textwidth}\centering\psset{unit=.6pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \figurinetrans \meinecaption{2}{$y=f(x)$. }\label{fig:translation1} \end{minipage} \hfill \begin{minipage}{.2\textwidth}\centering\psset{unit=.6pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \rput(0,1){\figurinetrans} \meinecaption{2}{$y=f(x)+1$. }\label{fig:translation2} \end{minipage} \hfill \begin{minipage}{.2\textwidth}\centering\psset{unit=.6pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \rput(-1,0){\figurinetrans} \meinecaption{2}{$y=f(x+1)$. }\label{fig:translation3} \end{minipage} \hfill \begin{minipage}{.2\textwidth}\centering\psset{unit=.6pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \rput(-1,1){\figurinetrans} \meinecaption{2}{\mbox{$y=f(x+1)+1$}. }\label{fig:translation4} \end{minipage} \end{figure} \begin{exa} Figures \ref{fig:translation2} through \ref{fig:translation4} shew various translations of $f:[-4;4]\to [-2;1]$ in figure \ref{fig:translation1}. Its translation $a: [-4;4]\to [-1;2]$ one unit up is shewn in figure \ref{fig:translation2}. Notice that we have simply increased the $y$-coordinate of every point on the original graph by $1$, without changing the $x$-coordinates. Its translation $b: [-5;3]\to [-2;1]$ one unit left is shewn in figure \ref{fig:translation3}. Its translation $c: [-5;3]\to [-1;2]$ one unit up and one unit left is shewn in figure \ref{fig:translation4}. Notice how the domain and image of the original curve are affected by the various translations. \end{exa} \begin{exa}Consider $$\fun{f}{x}{x^2}{\reals}{\reals}. $$ Figures \ref{fig:y=x^2vertran}, \ref{fig:y=f(x)+3} and \ref{fig:y=f(x)-3} shew the vertical translation $a$ $3$ units up and the vertical translation $b$ $3$ units down, respectively. Observe that $$\fun{a}{x}{x^2+3}{\reals}{\reals}, \qquad \fun{b}{x}{x^2-3}{\reals}{\reals}. $$ Figures \ref{fig:y=f(x-3)} and \ref{fig:y=f(x+3)}, respectively shew the horizontal translation $c$ $3$ units right, and the horizontal translation $d$ $3$ units left. Observe that $$\fun{c}{x}{(x-3)^2}{\reals}{\reals}, \qquad \fun{d}{x}{(x+3)^2}{\reals}{\reals}. $$ Figure \ref{fig:y=f(x+3)-3}, shews $g$, the simultaneous translation $3$ units left and down. Observe that $$\fun{g}{x}{(x+3)^3-3}{\reals}{\reals}. $$ \end{exa} \vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{2cm}$$ \psset{unit=.5pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \parabola[linewidth=2pt,linecolor=brown]{<->}(2.445,6)(0,0) $$ \meinecaption{1}{$y=f(x)=x^2$} \label{fig:y=x^2vertran} \end{minipage} \hfill \begin{minipage}{2cm}$$ \psset{unit=.5pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \rput(0,3){\parabola[linewidth=2pt,linecolor=brown]{<->}(2.445,6)(0,0)} $$ \meinecaption{1}{$y=x^2+3$} \label{fig:y=f(x)+3} \end{minipage} \hfill \begin{minipage}{2cm}$$ \psset{unit=.5pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \rput(0,-3){\parabola[linewidth=2pt,linecolor=brown]{<->}(2.445,6)(0,0)} $$ \meinecaption{1}{$y=x^2-3$} \label{fig:y=f(x)-3} \end{minipage} \hfill \begin{minipage}{2cm}$$ \psset{unit=.5pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \rput(3,0){\parabola[linewidth=2pt,linecolor=brown]{<->}(2.445,6)(0,0)} $$ \meinecaption{1}{$y=(x-3)^2$} \label{fig:y=f(x-3)} \end{minipage} \hfill \begin{minipage}{2cm}$$ \psset{unit=.5pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \rput(-3,0){\parabola[linewidth=2pt,linecolor=brown]{<->}(2.445,6)(0,0)} $$ \meinecaption{1}{$y=(x+3)^2$} \label{fig:y=f(x+3)} \end{minipage} \hfill \begin{minipage}{2cm}$$ \psset{unit=.5pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \rput(-3,-3){\parabola[linewidth=2pt,linecolor=brown]{<->}(2.445,6)(0,0)} $$ \meinecaption{1}{$y=(x+3)^2-3$} \label{fig:y=f(x+3)-3} \end{minipage} \end{figure} \begin{exa} If $g(x) = x$ (figure \ref{fig:y=xvertran}), then figures , \ref{fig:y=g(x)+3} and \ref{fig:y=g(x)-3} shew vertical translations $3$ units up and $3$ units down, respectively. Notice than in this case $g(x+t)= x+t = g(x) + t $, so a vertical translation by $t$ units has exactly the same graph as a horizontal translation $t$ units. \end{exa} \vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{4cm}$$ \psset{unit=.7pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,-5.5)(5.5,5.5) $$ \meinecaption{1.5}{$y=g(x)=x$} \label{fig:y=xvertran} \end{minipage} \hfill \begin{minipage}{4cm}$$ \psset{unit=.7pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \rput(0,3){\psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,-5.5)(5.5,5.5)} $$ \meinecaption{1.5}{$y=g(x)+3=x+3$} \label{fig:y=g(x)+3} \end{minipage} \hfill \begin{minipage}{4cm}$$ \psset{unit=.7pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \rput(0,-3){\psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,-5.5)(5.5,5.5)} $$ \meinecaption{1.5}{$y=g(x)-3=x-3$} \label{fig:y=g(x)-3} \end{minipage} \end{figure} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Graph the following curves: \begin{enumerate} \item $y = |x-2|+3$ \item $y = (x-2)^2+3$ \item $y = \dfrac{1}{x-2}+3$ \item $y = \sqrt{4-x^2} + 1 $ \end{enumerate} \end{pro} \begin{pro} What is the equation of the curve $y = f(x) = x^3 - \dfrac{1}{x}$ after a successive translation one unit down and two units right? \begin{answer} $y = f(x-2)-1 = (x-2)^2- \dfrac{1}{x-2}-1$ \end{answer} \end{pro} \begin{pro} Suppose the curve $y = f(x)$ is translated $a$ units vertically and $b$ units horizontally, in this order. Would that have the same effect as translating the curve $b$ units horizontally first, and then $a$ units vertically? \begin{answer} Yes. \end{answer} \end{pro} \end{multicols} \section{Distortions} \begin{thm}\label{thm:distortions} Let $f$ be a function and let $V\neq 0$ and $H\neq 0$ be real numbers. If $(x_0,y_0)$ is on the graph of $f$, then $(x_0,Vy_0)$ is on the graph of $g$, where $g(x)=Vf(x)$, and if $(x_1,y_1)$ is on the graph of $f$, then $\left(\dfrac{x_1}{H},y_1\right)$ is on the graph of $j$, where $j(x)=f(Hx)$. \end{thm} \begin{pf} Let $\Gamma_f, \Gamma_g, \Gamma_j$ denote the graphs of $f, g, j$ respectively. $$(x_0, y_0) \in \Gamma_f \iff y_0 = f(x_0) \iff Vy_0 = Vf(x_0) \iff Vy_0 = g(x_0) \iff (x_0,Vy_0)\in \Gamma_g. $$ Similarly, $$(x_1, y_1) \in \Gamma_f \iff y_1 = f(x_1) \iff y_1 = f\left(\dfrac{x_1}{H}\cdot H\right) \iff y_1 = j\left(\dfrac{x_1}{H}\right) \iff \left(\dfrac{x_1}{H},y_1\right)\in \Gamma_j. $$ \end{pf} \begin{df}Let $V>0$, $H>0$, and let $f$ be a function. The curve $y = Vf(x)$ is called a {\em vertical distortion} of the curve $y = f(x)$. The graph of $y = Vf(x)$ is a {\em vertical dilatation} of the graph of $y = f(x)$ if $V > 1$ and a {\em vertical contraction} if $0< V < 1.$ The curve $y = f(Hx)$ is called a {\em horizontal distortion} of the curve $y = f(x)$ The graph of $y = f(Hx)$ is a {\em horizontal dilatation} of the graph of $y = f(x)$ if $0< H < 1$ and a {\em horizontal contraction} if $H > 1.$ \index{distortion!vertical}\index{distortion!horizontal} \end{df} \begin{exa}\label{exa:distorsiones} Consider the function $$ \fun{f}{x}{f(x)}{\cc{-4;4}}{\cc{-6;6}} $$whose graph appears in figure \ref{fig:distorsiones1}. If $a(x)=\dfrac{f(x)}{2}$ then $$ \fun{a}{x}{a(x)}{\cc{-4;4}}{\cc{-3;3}}, $$and its graph appears in figure \ref{fig:distorsiones2}. If $b(x)=f(2x)$ then $$ \fun{b}{x}{b(x)}{\cc{-2;2}}{\cc{-6;6}}, $$and its graph appears in figure \ref{fig:distorsiones3}. If $c(x)=\dfrac{f(2x)}{2}$ then $$ \fun{c}{x}{c(x)}{\cc{-2;2}}{\cc{-3;3}}, $$and its graph appears in figure \ref{fig:distorsiones4}. \end{exa} \vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{4cm} \centering\psset{unit=.6pc} \psgrid[subgriddiv=0,griddots=5,gridlabels=0.5](-7,-7)(7,7) \psaxes[labels=none,ticks=none](0,0)(-8,-8)(8,8) \pstGeonode[PointName=none](0,0){O}(2,4){A}(4,4){B}(-2,6){Y}(-4,-6){X} \psline[linecolor=brown,linewidth=2pt](X)(Y)(O)(A)(B) \psdots(X)(Y)(O)(A)(B) \meinecaption{2}{$y=f(x)$} \label{fig:distorsiones1} \end{minipage} \hfill \begin{minipage}{4cm} \centering\psset{unit=.6pc} \psgrid[subgriddiv=0,griddots=5,gridlabels=0.5](-7,-7)(7,7) \psaxes[labels=none,ticks=none](0,0)(-8,-8)(8,8) \pstGeonode[PointName=none](0,0){O}(2,2){A}(4,2){B}(-2,3){Y}(-4,-3){X} \psline[linecolor=brown,linewidth=2pt](X)(Y)(O)(A)(B) \psdots(X)(Y)(O)(A)(B) \meinecaption{2}{$y=\dfrac{f(x)}{2}$} \label{fig:distorsiones2} \end{minipage} \hfill \begin{minipage}{4cm} \centering\psset{unit=.6pc} \psgrid[subgriddiv=0,griddots=5,gridlabels=0.5](-7,-7)(7,7) \psaxes[labels=none,ticks=none](0,0)(-8,-8)(8,8) \pstGeonode[PointName=none](0,0){O}(1,4){A}(2,4){B}(-1,6){Y}(-2,-6){X} \psline[linecolor=brown,linewidth=2pt](X)(Y)(O)(A)(B) \psdots(X)(Y)(O)(A)(B) \meinecaption{2}{$y=f(2x)$} \label{fig:distorsiones3} \end{minipage} \hfill \begin{minipage}{4cm} \centering\psset{unit=.6pc} \psgrid[subgriddiv=0,griddots=5,gridlabels=0.5](-7,-7)(7,7) \psaxes[labels=none,ticks=none](0,0)(-8,-8)(8,8) \pstGeonode[PointName=none](0,0){O}(1,2){A}(2,2){B}(-1,3){Y}(-2,-3){X} \psline[linecolor=brown,linewidth=2pt](X)(Y)(O)(A)(B) \psdots(X)(Y)(O)(A)(B) \meinecaption{2}{$y=\dfrac{f(2x)}{2}$} \label{fig:distorsiones4} \end{minipage} \end{figure} \begin{exa} If $y=\sqrt{4-x^2}$, then $x^2+y^2=4$ gives the equation of a circle with centre at $(0,0)$ and radius $2$ by virtue of \ref{eq:canonical_circle}. Hence $$y = \sqrt{4-x^2}$$ is the upper semicircle of this circle. Figures \ref{fig:y=sqrt(4-x^2)} through \ref{fig:y=2a(2x)+1} shew various transformations of this curve. \end{exa} \vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{2cm}$$ \psset{unit=.5pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \psplot[linewidth=2pt,linecolor=brown]{-2}{2}{sqrt(4-x^2)} $$ \meinecaption{1}{$y=\sqrt{4-x^2}$} \label{fig:y=sqrt(4-x^2)} \end{minipage} \hfill \begin{minipage}{2cm}$$ \psset{unit=.5pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \psplot[linewidth=2pt,linecolor=brown]{-2}{2}{2*sqrt(4-x^2)} $$ \meinecaption{1}{$y=2\sqrt{4-x^2}$} \label{fig:y=2a(x)} \end{minipage} \hfill \begin{minipage}{2cm}$$ \psset{unit=.5pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \psplot[linewidth=2pt,linecolor=brown]{-1}{1}{sqrt(4-4*x^2)} $$ \meinecaption{1}{$y=\sqrt{4-4x^2}$} \label{fig:y=a(2x)} \end{minipage} \hfill \begin{minipage}{2cm}$$ \psset{unit=.5pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \rput(2,0){\psplot[linewidth=2pt,linecolor=brown]{-2}{2}{sqrt(4-x^2)}} $$ \meinecaption{1}{$y=\sqrt{-x^2+4x}$} \label{fig:y=a(x-2)} \end{minipage} \hfill \begin{minipage}{2cm}$$ \psset{unit=.5pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \psplot[linewidth=2pt,linecolor=brown]{-1}{1}{2*sqrt(4-4*x^2)} $$ \meinecaption{1}{$y=2\sqrt{4-4x^2}$} \label{fig:y=2a(2x)} \end{minipage} \hfill \begin{minipage}{2cm}$$ \psset{unit=.5pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \rput(0,1){\psplot[linewidth=2pt,linecolor=brown]{-1}{1}{2*sqrt(4-4*x^2)}} $$ \meinecaption{1}{$y=2\sqrt{4-4x^2}+1$} \label{fig:y=2a(2x)+1} \end{minipage} \end{figure} \begin{exa}\label{exa:y=2x^2-4x+1} Draw the graph of the curve $y = 2x^2 - 4x+1$. \end{exa} \begin{solu} We complete squares. $$ \begin{array}{lll}y = 2x^2 - 4x+1 & \iff & \dfrac{y}{2} = x^2-2x + \dfrac{1}{2} \\ & \iff & \dfrac{y}{2} + 1 = x^2-2x+1 + \dfrac{1}{2} \\ & \iff & \dfrac{y}{2} + 1 = (x-1)^2 + \dfrac{1}{2} \\ & \iff & \dfrac{y}{2} = (x-1)^2 - \dfrac{1}{2} \\ & \iff & y = 2(x-1)^2 - 1, \\ \end{array}$$whence to obtain the graph of $y=2x^2-4x+1$ we (i) translate $y=x^2$ one unit right, (ii) dilate the above graph by factor of two, (iii) translate the above graph one unit down. This succession is seen in figures \ref{fig:y=(x-1)^2} through \ref{fig:y=2c(x-1)-1}. \end{solu} \begin{exa} The curve $y = x^2 + \dfrac{1}{x}$ experiences the following successive transformations: (i) a translation one unit up, (ii) a horizontal shrinkage by a factor of $2$, (iii) a translation one unit left. Find its resulting equation. \end{exa}\begin{solu} After a translation one unit up, the curve becomes $$y = f(x)+1 = x^2 + \dfrac{1}{x}+1 = a(x). $$ After a horizontal shrinkage by a factor of $2$ the curve becomes $$y = a(2x) = 4x^2 + \dfrac{1}{2x}+1 = b(x). $$ After a translation one unit left the curve becomes $$y = b(x+1) = 4(x+1)^2 + \dfrac{1}{2x+2} + 1. $$ The required equation is thus $$y= 4(x+1)^2 + \dfrac{1}{2x+2} + 1 = 4x^2 + 8x + 5+\dfrac{1}{2x+2}. $$ \vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{4cm}$$ \psset{unit=.7pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \psplot[linewidth=2pt,linecolor=brown]{-1}{3}{(x-1)^2} $$ \meinecaption{1}{$y=(x-1)^2$} \label{fig:y=(x-1)^2} \end{minipage} \hfill \begin{minipage}{4cm}$$ \psset{unit=.7pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \psplot[linewidth=2pt,linecolor=brown]{-1}{3}{2*(x-1)^2} $$ \meinecaption{1}{$y=2(x-1)^2$} \label{fig:y=2c(x-1)} \end{minipage} \hfill \begin{minipage}{4cm}$$ \psset{unit=.7pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \psplot[linewidth=2pt,linecolor=brown]{-1}{3}{2*x^2-4*x+1} $$ \meinecaption{1}{$y=2(x-1)^2-1$} \label{fig:y=2c(x-1)-1} \end{minipage} \end{figure} \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Draw the graphs of the following curves: \begin{enumerate} \item $y = \dfrac{x^2}{2}$ \item $y = \dfrac{x^2}{2}-1$ \item $y = 2|x| + 1$ \item $y = \dfrac{2}{x}$ \item $y = x^2 + 4x + 5$ \item $y = 2x^2 + 8x$ \end{enumerate} \end{pro} \begin{pro} The curve $y = \dfrac{1}{x}$ experiences the following successive transformations: (i) a translation one unit left, (ii) a vertical dilatation by a factor of $2$, (iii) a translation one unit down. Find its resulting equation and make a rough sketch of the resulting curve. \begin{answer} The required equation is $y = \dfrac{1}{2x+2}-1$. \end{answer} \end{pro} \begin{pro}\label{pro:distortions} For the functional curve given in figure \ref{fig:distortions}, determine its domain and image and draw the following transformations, also determining their respective domains and images. \begin{enumerate} \item $y = 2f(x)$ \item $y = f(2x)$ \item $y = 2f(2x)$ \end{enumerate} \begin{answer}Observe that $f$ is the function $$ \fun{f}{x}{f(x)}{[-4;4]}{[-2;4]}. $$ Let $a$ be the function with curve $y=2f(x)$. Then $ \fun{a}{x}{a(x)}{[-4;4]}{[-4;8]}.$\\ \vspace*{3cm} \begin{figurehere} \centering\psset{unit=.55pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-8,-8)(8,8) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-8,-8)(-8,-8)(8,8)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-8.5,0)(8.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-8.5)(0,8.5) \psline[linecolor=blue,linewidth=2pt](-4,4)(-2,-4)(0,0)(2,8)(4,8) \psdots(-4,4)(-2,-4)(0,0)(2,8)(4,8) \meinecaption{2}{$y=2f(x)$.}\label{fig:distortions1} \end{figurehere} \bigskip Let $b$ be the function with curve $y=f(2x)$. Then $ \fun{b}{x}{b(x)}{[-2;2]}{[-2;4]}.$\\ \vspace*{3cm} \begin{figurehere} \centering\psset{unit=.55pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-8,-8)(8,8) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-8,-8)(-8,-8)(8,8)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-8.5,0)(8.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-8.5)(0,8.5) \psline[linecolor=blue,linewidth=2pt](-2,2)(-1,-2)(0,0)(1,4)(2,4)\psdots(-2,2)(-1,-2)(0,0)(1,4)(2,4) \meinecaption{2}{$y=f(2x)$.}\label{fig:distortions2} \end{figurehere} \bigskip Let $c$ be the function with curve $y=2f(2x)$. Then $ \fun{c}{x}{c(x)}{[-2;2]}{[-4;8]}. $\\ \vspace*{3cm} \begin{figurehere} \centering\psset{unit=.55pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-8,-8)(8,8) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-8,-8)(-8,-8)(8,8)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-8.5,0)(8.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-8.5)(0,8.5) \psline[linecolor=blue,linewidth=2pt](-2,4)(-1,-4)(0,0)(1,8)(2,8)\psdots(-2,4)(-1,-4)(0,0)(1,8)(2,8) \meinecaption{2}{$y=2f(2x)$.}\label{fig:distortions3} \end{figurehere} \end{answer} \end{pro} \vspace*{1cm} \begin{figurehere} \centering\psset{unit=.55pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psline[linewidth=2pt,linecolor=blue](-4,2)(-2,-2)(0,0)(2,4)(4,4) \psdots(-4,2)(-2,-2)(0,0)(2,4)(4,4) \meinecaption{2}{Problem \ref{pro:distortions}.}\label{fig:distortions} \end{figurehere} \end{multicols} \section{Reflexions} \begin{thm}\label{thm:reflexions} Let $f$ be a function If $(x_0,y_0)$ is on the graph of $f$, then $(x_0,-y_0)$ is on the graph of $g$, where $g(x)=-f(x)$, and if $(x_1,y_1)$ is on the graph of $f$, then $\left(-x_1,y_1\right)$ is on the graph of $j$, where $j(x)=f(-x)$. \end{thm} \begin{pf} Let $\Gamma_f, \Gamma_g, \Gamma_j$ denote the graphs of $f, g, j$ respectively. $$(x_0, y_0) \in \Gamma_f \iff y_0 = f(x_0) \iff -y_0 = -f(x_0) \iff -y_0 = g(x_0) \iff (x_0,-y_0)\in \Gamma_g. $$ Similarly, $$(x_1, y_1) \in \Gamma_f \iff y_1 = f(x_1) \iff y_1 = f\left(-(-x_1)\right) \iff y_1 = j\left(-x_1\right) \iff \left(-x_1,y_1\right)\in \Gamma_j. $$ \end{pf} \begin{df} Let $f$ be a function. The curve $y = -f(x)$ is said to be the {\em reflexion of $f$ about the $x$-axis} and the curve $y = f(-x)$ is said to be the {\em reflexion of $f$ about the $y$-axis}. \end{df} \begin{exa}\label{exa:reflexions} Figure \ref{fig:reflexions1} shews the graph of the function $$ \fun{f}{x}{f(x)}{[-4;4]}{[-2;4]}. $$Figure \ref{fig:reflexions2} shews the graph of its reflexion $a$ about the $x$-axis, $$ \fun{a}{x}{a(x)}{[-4;4]}{[-4;2]}. $$ Figure \ref{fig:reflexions3} shews the graph of its reflexion $b$ about the $y$-axis, $$ \fun{b}{x}{b(x)}{[-4;4]}{[-2;4]}. $$Figure \ref{fig:reflexions4} shews the graph of its reflexion $c$ about the $x$-axis and $y$-axis, $$ \fun{c}{x}{c(x)}{[-4;4]}{[-4;2]}. $$ \end{exa} \vspace*{1cm} \begin{figure}[h] \begin{minipage}{.2\textwidth} \centering\psset{unit=.55pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psline[linewidth=2pt,linecolor=blue](-4,2)(-2,-2)(0,0)(2,4)(4,4) \psdots(-4,2)(-2,-2)(0,0)(2,4)(4,4) \meinecaption{2}{$y=f(x)$.}\label{fig:reflexions1} \end{minipage} \hfill \begin{minipage}{.2\textwidth} \centering\psset{unit=.55pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psline[linewidth=2pt,linecolor=blue](-4,-2)(-2,2)(0,0)(2,-4)(4,-4) \psdots(-4,-2)(-2,2)(0,0)(2,-4)(4,-4) \meinecaption{2}{$y=-f(x)$.}\label{fig:reflexions2} \end{minipage} \hfill \begin{minipage}{.2\textwidth} \centering\psset{unit=.55pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psline[linewidth=2pt,linecolor=blue](4,2)(2,-2)(0,0)(-2,4)(-4,4) \psdots(4,2)(2,-2)(0,0)(-2,4)(-4,4) \meinecaption{2}{$y=f(-x)$.}\label{fig:reflexions3} \end{minipage} \hfill \begin{minipage}{.2\textwidth} \centering\psset{unit=.55pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psline[linewidth=2pt,linecolor=blue](4,-2)(2,2)(0,0)(-2,-4)(-4,-4) \psdots(4,-2)(2,2)(0,0)(-2,-4)(-4,-4) \meinecaption{2}{$y=-f(-x)$.}\label{fig:reflexions4} \end{minipage} \end{figure} \begin{exa} Figures \ref{fig:y=d(x)=(x-1)^2} through \ref{fig:y=-(-x-1)^2} shew various reflexions about the axes for the function $$ \fun{d}{x}{(x-1)^2}{\reals}{\reals}. $$ \end{exa} \vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{4cm}$$ \psset{unit=.5pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \psplot[linewidth=2pt,linecolor=brown]{-1}{3}{(x-1)^2} $$ \meinecaption{1}{$y=d(x)=(x-1)^2$} \label{fig:y=d(x)=(x-1)^2} \end{minipage} \hfill \begin{minipage}{4cm}$$ \psset{unit=.5pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \psplot[linewidth=2pt,linecolor=brown]{-1}{3}{-1*(x-1)^2} $$ \meinecaption{1}{$y=-d(x)=-(x-1)^2$} \label{fig:y=-(x-1)^2} \end{minipage} \hfill \begin{minipage}{4cm}$$ \psset{unit=.5pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \psplot[linewidth=2pt,linecolor=brown]{-3}{1}{(-1*x-1)^2} $$ \meinecaption{1}{$y=d(-x)=(-x-1)^2$} \label{fig:y=(-x-1)^2} \end{minipage} \hfill \begin{minipage}{4cm}$$ \psset{unit=.5pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \psplot[linewidth=2pt,linecolor=brown]{-3}{1}{-1*(-1*x-1)^2} $$ \meinecaption{1}{$y=-d(-x)=-(-x-1)^2$} \label{fig:y=-(-x-1)^2} \end{minipage} \end{figure} \begin{exa} Let $f: \reals \setminus \{0\} \rightarrow \reals$ with $$f(x) = x + \dfrac{2}{x} -1. $$The curve $y = f(x)$ experiences the following successive transformations: \begin{enumerate} \item A reflexion about the $x$-axis. \item A translation $3$ units left. \item A reflexion about the $y$-axis. \item A vertical dilatation by a factor of $2$. \end{enumerate}Find the equation of the resulting curve. Note also how the domain of the function is affected by these transformations. \end{exa} \begin{solu} \begin{enumerate} \item A reflexion about the $x$-axis gives the curve $$y = -f(x) = 1 - \dfrac{2}{x} - x = a(x),$$say, with $\dom{a} = \reals \setminus \{0\}$. \item A translation $3$ units left gives the curve $$y = a(x + 3) = 1 - \dfrac{2}{x + 3} - (x + 3) = -2 -\dfrac{2}{x + 3} - x = b(x),$$say, with $\dom{b} = \reals \setminus \{-3\}$. \item A reflexion about the $y$-axis gives the curve $$y = b(-x) = -2 - \dfrac{2}{-x + 3} + x = c(x),$$say, with $\dom{c} = \reals \setminus \{3\}$. \item A vertical dilatation by a factor of $2$ gives the curve $$y = 2c(x) = -4 + \dfrac{4}{x - 3} + 2x = d(x),$$say, with $\dom{d} = \reals \setminus \{3\}$. Notice that the resulting curve is $$y = d(x) = 2c(x) = 2b(-x) = 2a(-x + 3) = -2f(-x + 3) .$$ \end{enumerate} \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Let $f: \reals \rightarrow \reals$ with $$f(x) = 2 - |x|. $$The curve $y = f(x)$ experiences the following successive transformations: \begin{enumerate} \item A reflexion about the $x$-axis. \item A translation $3$ units up. \item A horizontal stretch by a factor of $\frac{3}{4}$. \end{enumerate}Find the equation of the resulting curve. \begin{answer}Proceeding successively: \begin{enumerate} \item A reflexion about the $x$-axis gives the curve $$y = -f(x) = |x| - 2 = a(x), $$say. \item A translation $3$ units up gives the curve $$y = a(x) + 3 = |x| + 1 = b(x), $$say. \item A horizontal stretch by a factor of $\frac{3}{4}$ gives the curve $$y = b\left(\frac{4}{3}x\right) = \left|\dfrac{4x}{3}\right| + 1 = \dfrac{4}{3}|x| + 1 = c(x), $$say. Observe that the resulting curve is $$y = c(x) = b\left(\frac{4}{3}x\right) = a\left(\frac{4}{3}x\right) + 3 = -f\left(\frac{4}{3}x\right) + 3. $$ \end{enumerate} \end{answer} \end{pro} \begin{pro} The graphs of the following curves suffer the following successive, rigid transformations: \begin{enumerate} \item a vertical translation of $2$ units down, \item a reflexion about the $y$-axis, and finally, \item a horizontal translation of $1$ unit to the left. \end{enumerate} Find the resulting equations after all the transformations have been exerted. \begin{enumerate} \item $y = x(1 - x)$ \item $y = 2x - 3$ \item $y = |x + 2| + 1$ \end{enumerate} \begin{answer} (1) $y = -(x + 1)(x + 2) - 2$ \ (2) $y = -2x - 7$ \ (3) $y = |1 - x| - 1$ \end{answer} \end{pro} \begin{pro}\label{pro:reflexion} For the functional curve $y=f(x)$ in figure \ref{fig:reflexionA}, draw $y=f(x+1) $, $y=f(1-x) $ and $y=-f(1-x)$. \begin{answer} Here is the graph of $x\mapsto f(x+1)$.\\ \vspace{3cm} \begin{figurehere} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psline[linewidth=2pt,linecolor=blue](-5,2)(-3,-2)(-1,0)(0,2)(2,2)(4,-2) \psdots(-1,0)(0,2)(2,2)(4,-2)(3,0)(-3,-2)(-5,2)(-4,0) $$ \meinecaption{2}{$y=f(x+1)$.} \end{figurehere} Here is the graph of $x\mapsto f(-x+1)$.\\ \vspace{3cm} \begin{figurehere} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psline[linewidth=2pt,linecolor=blue](5,2)(3,-2)(1,0)(0,2)(-2,2)(-4,-2) \psdots(1,0)(0,2)(-2,2)(-4,-2)(-3,0)(3,-2)(5,2)(4,0) $$ \meinecaption{2}{$y=f(1-x)$.} \end{figurehere} \columnbreak Here is the graph of $x\mapsto -f(-x+1)$.\\ \vspace{3cm} \begin{figurehere} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psline[linewidth=2pt,linecolor=blue](5,-2)(3,2)(1,0)(0,-2)(-2,-2)(-4,2) \psdots(1,0)(0,-2)(-2,-2)(-4,2)(-3,0)(3,2)(5,-2)(4,0) $$ \meinecaption{2}{$y=-f(1-x)$.} \end{figurehere} \end{answer} \end{pro} \vspace{2cm} \begin{figurehere} \centering\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psline[linewidth=2pt,linecolor=blue](-4,2)(-2,-2)(0,0)(1,2)(3,2)(5,-2) \psdots(0,0)(1,2)(3,2)(5,-2)(4,0)(-2,-2)(-4,2)(-3,0) \meinecaption{2}{Problem \ref{pro:reflexion}.} \label{fig:reflexionA} \end{figurehere} \end{multicols} \section{Symmetry} \begin{df} A function $f$ is {\em even} if for all $x$ it is verified that $f(x) = f(-x)$, that is, if the portion of the graph for $x<0$ is a mirror reflexion of the part of the graph for $x> 0$. This means that the graph of $f$ is symmetric about the $y$-axis. A function $g$ is {\em odd} if for all $x$ it is verified that $g(-x) = -g(x)$, in other words, $g$ is odd if it is symmetric about the origin. This implies that the portion of the graph appearing in quadrant I is a $180^\circ$ rotation of the portion of the graph appearing in quadrant III, and the portion of the graph appearing in quadrant II is a $180^\circ$ rotation of the portion of the graph appearing in quadrant IV. \end{df} \begin{exa} The curve in figure \ref{fig:even} is even. The curve in figure \ref{fig:odd} is odd. \label{exa:even_odd}\end{exa} \vspace{1cm} \begin{figure}[!hptb] \begin{minipage}{7cm} \centering\psset{unit=.8pc, algebraic=true} \psaxes[labels=none,ticks=none,linewidth=1.2pt]{<->}(0, 0)(-5, -5)(5, 5) \psplot[linewidth=2pt,linecolor=brown]{-1.85}{1.85}{abs(x)*abs(x^2-1)} \meinecaption{2}{Example \ref{exa:even_odd}. The graph of an even function.} \label{fig:even} \end{minipage} \hfill \begin{minipage}{7cm} \centering \psset{unit=.8pc,algebraic=true} \psaxes[labels=none,ticks=none,linewidth=1.2pt]{<->}(0, 0)(-5, -5)(5, 5)\psplot[linewidth=2pt,linecolor=brown]{-1.85}{1.85}{x*abs(x^2-1)} \meinecaption{2}{Example \ref{exa:even_odd}. The graph of an odd function.} \label{fig:odd} \end{minipage} \end{figure} \begin{thm} Let $\epsilon_1, \epsilon_2$ be even functions, and let $\omega_1, \omega_2$ be odd functions, all sharing the same common domain. Then \begin{enumerate} \item $\epsilon_1 \pm \epsilon_2$ is an even function. \item $\omega_1 \pm \omega_2$ is an odd function. \item $\epsilon_1 \cdot \epsilon_2$ is an even function. \item $\omega_1 \cdot \omega_2$ is an even function. \item $\epsilon_1 \cdot \omega_1$ is an odd function. \end{enumerate} \end{thm} \begin{pf}We have \begin{enumerate} \item $(\epsilon_1 \pm \epsilon_2) (-x) = \epsilon_1(-x) \pm \epsilon_2(-x) = \epsilon_1(x) \pm \epsilon_2(x)$. \item $(\omega_1 \pm \omega_2)(-x) = \omega_1(-x) \pm \omega_2(-x) = -\omega_1(x) \mp \omega_2(x) = -(\omega_1 \pm \omega_2)(x)$ \item $(\epsilon_1 \epsilon_2)(-x) = \epsilon_1(-x) \epsilon_2(-x) = \epsilon_1(x) \epsilon_2(x)$\item $(\omega_1 \omega_2)(-x) = \omega_1(-x) \omega_2(-x) = (-\omega_1(x)) (-\omega_2(x)) = \omega_1(x)\omega_2(x))$ \item $(\epsilon_1 \omega_1)(-x) = \epsilon_1(-x) \omega_1(-x) = -\epsilon_1(x) \omega_1(x)$ \end{enumerate} \end{pf} \begin{cor} Let $p(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots + a_{n - 1}x^{n - 1} + a_nx^n$ be a polynomial with real coefficients. Then the function $$\fun{p}{x}{p(x)}{\reals}{\reals}$$ is an even function if and only if each of its terms has even degree. \label{cor:evenpoly}\end{cor} \begin{pf} Assume $p$ is even. Then $p(x) = p(-x)$ and so $${\everymath{\displaystyle} \begin{array}{ll} p(x) & = \frac{p(x) + p(-x)}{2} \\ & = \frac{a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots + a_{n - 1}x^{n - 1} + a_nx^n}{2} \\ & \ \ \qquad + \frac{ a_0 - a_1x + a_2x^2 - a_3x^3 + \cdots + (-1)^{n - 1}a_{n - 1}x^{n - 1} + (-1)^na_nx^n }{2} \\ & = a_0 + a_2x^2 + a_4x^4 + \cdots + \\ \end{array}}$$and so the polynomial has only terms of even degree. The converse of this statement is trivial. \end{pf} \begin{exa} Prove that in the product $$(1 - x + x^2 - x^3 + \cdots - x^{99} + x^{100})(1 + x + x^2 + x^3 + \cdots + x^{99} + x^{100})$$ after multiplying and collecting terms, there does not appear a term in $x$ of odd degree. \end{exa} \begin{solu} Let $\fun{f}{x}{f(x)}{\reals}{\reals}$ with $$ f(x) = (1 - x + x^2 - x^3 + \cdots - x^{99} + x^{100})(1 + x + x^2 + x^3 + \cdots + x^{99} + x^{100})$$ Then $$f(-x) = (1 + x + x^2 + x^3 + \cdots + x^{99} + x^{100})(1 - x + x^2 - x^3 + \cdots - x^{99} + x^{100}) = f(x),$$which means that $f$ is an even function. Since $f$ is a polynomial, this means that $f$ does not have a term of odd degree. \end{solu} \bigskip Analogous to Corollary \ref{cor:evenpoly}, we may establish the following. \begin{cor}Let $p(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots + a_{n - 1}x^{n - 1} + a_nx^n$ be a polynomial with real coefficients. Then the function $$\fun{p}{x}{p(x)}{\reals}{\reals}$$ is an odd function if and only if each of its terms has odd degree. \end{cor} \begin{thm} Let $f: \reals \rightarrow \reals$ be an arbitrary function. Then $f$ can be written as the sum of an even function and an odd function. \end{thm} \begin{pf} Given $x\in\reals$, put $E(x) = f(x) + f(-x)$, and $O(x) = f(x) - f(-x)$. We claim that $E$ is an even function and that $O$ is an odd function. First notice that $$E(-x) = f(-x) + f(-(-x)) = f(-x) + f(x) = E(x),$$which proves that $E$ is even. Also, $$O(-x) = f(-x) - f(-(-x)) = -(f(x) - f(-x))) = -O(x),$$which proves that $O$ is an odd function. Clearly $$f(x) = \frac{1}{2}E(x) + \frac{1}{2}O(x),$$which proves the theorem. \end{pf} \begin{exa} Investigate which of the following functions are even, odd, or neither. \begin{enumerate} \item $a: \reals \rightarrow \reals$, $a(x) = \dfrac{x^3}{x^2 + 1}$. \item $b: \reals \rightarrow \reals$, $b(x) = \dfrac{|x|}{x^2 + 1}$. \item $c: \reals \rightarrow \reals$, $c(x) = |x| + 2$. \item $d: \reals \rightarrow \reals$, $d(x) = |x + 2|$. \item $f: [-4; 5] \rightarrow \reals$, $f(x) = |x| + 2$. \end{enumerate} \end{exa} \begin{solu} \begin{enumerate} \item $$a(-x) = \dfrac{(-x)^3}{(-x)^2 + 1} = - \dfrac{x^3}{x^2 + 1} = -a(x),$$whence $a$ is odd, since its domain is also symmetric. \item $$b(-x) = \dfrac{|-x|}{(-x)^2 + 1} = \dfrac{|x|}{x^2 + 1} = b(x),$$whence $b$ is even, since its domain is also symmetric. \item $$c(-x) = |-x| + 2 = |x| + 2 = c(x),$$ whence $c$ is even, since its domain is also symmetric. \item $d(-1) = |-1 + 2| = 1$, but $d(1) = 3$. This function is neither even nor odd. \item The domain of $f$ is not symmetric, so $f$ is neither even nor odd. \end{enumerate} \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro}\label{pro:fragmento} Complete the following fragment of graph so that the completion depicts (i) an even function, (ii) an odd function. \begin{answer} Here is the even completion.\\ \vspace{3cm} \begin{figurehere} \psset{unit=.5cm} \centering \psgrid[subgriddiv=0,griddots=5,gridlabels=0](-3.5,-3.5)(3.5,3.5) \psaxes[labels=none]{->}(0,0)(-4,-4)(4,4) \psline[linewidth=2pt]{o-*}(0,2)(2,2)(3,-1)(4,-1) \psline[linewidth=2pt](0,2)(-2,2)(-3,-1)(-4,-1) \meinecaption{2}{Even completion.} \end{figurehere} Here is the odd completion.\\ \vspace{3cm} \begin{figurehere} \psset{unit=.5cm} \centering \psgrid[subgriddiv=0,griddots=5,gridlabels=0](-3.5,-3.5)(3.5,3.5) \psaxes[labels=none]{->}(0,0)(-4,-4)(4,4) \psline[linewidth=2pt]{o-*}(0,2)(2,2)(3,-1)(4,-1) \psline[linewidth=2pt]{o-*}(0,-2)(-2,-2)(-3,1)(-4,1) \meinecaption{2}{Odd Completion.} \end{figurehere} \end{answer} \end{pro} \vspace{1cm} \begin{figurehere}\psset{unit=.25cm} \centering \psgrid[subgriddiv=0,griddots=5,gridlabels=0](-3.5,-3.5)(3.5,3.5) \psaxes[labels=none]{->}(0,0)(-4,-4)(5,5) \psline[linewidth=2pt]{o-*}(0,2)(2,2)(3,-1)(4,-1) \meinecaption{1}{Problem \ref{pro:fragmento}.} \end{figurehere} \begin{pro} Let $f:\reals \to \reals$ be an even function and let $g:\reals \to \reals$ be an odd function. If $f(-2)=3$, $f(3)=2$ and $g(-2)=2$, $g(3)=4$, find $$ (f+g)(2), \qquad (g\circ f)(2). $$ \begin{answer} Since $f$ is even, $f(2)=3$, $f(-3)=2$. Since $g$ is odd, $g(2)=-2$, $g(-3)=-4$. Thus $$ (f+g)(2)=f(2)+g(2)=3+(-2)=1, \qquad (g\circ f)(2) = g(f(2)) = g(3)=4. $$ \end{answer} \end{pro} \begin{pro} Let $f$ be an odd function and assume that $f$ is defined at $x=0$. Prove that $f(0)=0$. \begin{answer} Since $f$ is odd, $f(-0) = -f(0)$. But $f(-0) = f(0)$, giving $f(0) = -f(0)$, that is, $2f(0) = 0$ which implies that $f(0) = 0$. \end{answer} \end{pro} \begin{pro} Can a function be simultaneously even and odd? What would the graph of such a function look like? \begin{answer} The constant function $\reals \rightarrow \{0\}$ with assignment rule $f:x\mapsto 0$ is both even and odd. It is the only such function, for if $g$ were both even and odd and $g(x) = a \neq 0$ for some real number $x$, then we would have $a=g(x)=g(-x) = -g(x) = -a$, implying that $a=0$. \end{answer} \end{pro} \begin{pro} Let $A\times B \subseteqq \reals^2$ and suppose that $f:A \to B$ is invertible and even. Determine the sets $A$ and $B$. \begin{answer} We will shew that $A=\{0\}$ and consequently, $B=\{f(0)\}$. Let $x\in A$. If $x\neq 0$ then $-x$ must also be in $A$ because $f$ is even. Thus then $x\neq -x$ and $f(x)=f(-x)$, which means that $f$ in not injective and hence not invertible, a contradiction. This means that the only element of $A$ is $x=0$. In turn, since $f$ is surjective, $B$ must have exactly one element, which perforce must be $f(0)$. \end{answer} \end{pro} \end{multicols} \section{Transformations Involving Absolute Values} \begin{thm} Let $f$ be a function. Then both $x\mapsto f(|x|)$ and $x\mapsto f(-|x|)$ are even functions. \end{thm} \begin{pf} Put $a(x) = f(|x|)$. Then $a(-x) = f(|-x|) = f(|x|) = a(x)$, whence $x\mapsto a(x)$ is even. Similarly, if $b(x) = f(-|x|)$, then $b(-x) = f(-|-x|) = f(-|x|) = b(x)$ proving that $x\mapsto b(x)$ is even. \end{pf} Notice that $f(x) = f(|x|)$ for $x>0$. Since $x\mapsto f(|x|)$ is even, the graph of $x\mapsto f(|x|)$ is thus obtained by erasing the portion of the graph of $x \mapsto f(x)$ for $x<0$ and reflecting the part for $x>0$. Similarly, since $f(x) = f(-|x|)$ for $x<0$, the graph of $x\mapsto f(-|x|)$ is obtained by erasing the portion of the graph of $x \mapsto f(x)$ for $x>0$ and reflecting the part for $x<0$. \begin{thm} Let $f$ be a function If $(x_0,y_0)$ is on the graph of $f$, then $(x_0,|y_0|)$ is on the graph of $g$, where $g(x)=|f(x)|$. \end{thm} \begin{pf} Let $\Gamma_f, \Gamma_g$ denote the graphs of $f, g$, respectively. $$(x_0, y_0) \in \Gamma_f \implies y_0 = f(x_0) \implies |y_0| = |f(x_0)| \implies |y_0| = g(x_0) \implies (x_0,|y_0|)\in \Gamma_g. $$ \end{pf} \begin{exa}The graph of $y=f(x)$ is given in figure \ref{fig:absvaltrans1}. The transformation $y = |f(x)|$ is given in figure \ref{fig:absvaltrans2}. The transformation $y = f(|x|)$ is given in figure \ref{fig:absvaltrans3}. The transformation $y = f(-|x|)$ is given in figure \ref{fig:absvaltrans4}. The transformation $y = |f(|x|)|$ is given in figure \ref{fig:absvaltrans5}. \end{exa} \vspace{1cm} \begin{figure}[h] \begin{minipage}{.15\textwidth} $$\psset{unit=.6pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.4pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psline[linewidth=2.2pt,linecolor=blue](-4,2)(-2,2)(0,0)(1,-1)(5,-1) \psdots(-4,2)(-2,2)(0,0)(1,-1)(5,-1) $$ \meinecaption{1}{$y=f(x)$.} \label{fig:absvaltrans1} \end{minipage} \hfill \begin{minipage}{.15\textwidth} $$\psset{unit=.6pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.4pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psline[linewidth=2.2pt,linecolor=blue](-4,2)(-2,2)(0,0)(1,1)(5,1) \psdots(-4,2)(-2,2)(0,0)(1,1)(5,1) $$ \meinecaption{1}{$y=|f(x)|$.} \label{fig:absvaltrans2} \end{minipage} \hfill \begin{minipage}{.15\textwidth} $$\psset{unit=.6pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.4pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psline[linewidth=2.2pt,linecolor=blue](-5,-1)(-1,-1)(0,0)(1,-1)(5,-1) \psdots(-5,-1)(-1,-1)(0,0)(1,-1)(5,-1) $$ \meinecaption{1}{$y=f(|x|)$.} \label{fig:absvaltrans3} \end{minipage} \hfill \begin{minipage}{.15\textwidth} $$\psset{unit=.6pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.4pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psline[linewidth=2.2pt,linecolor=blue](-4,2)(-2,2)(0,0)(2,2)(4,2) \psdots(-4,2)(-2,2)(0,0)(2,2)(4,2) $$ \meinecaption{1}{$y=f(-|x|)$.} \label{fig:absvaltrans4} \end{minipage} \hfill \begin{minipage}{.15\textwidth} $$\psset{unit=.6pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.4pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psline[linewidth=2.2pt,linecolor=blue](-5,1)(-1,1)(0,0)(1,1)(5,1) \psdots(-5,1)(-1,1)(0,0)(1,1)(5,1) $$ \meinecaption{1}{$y=|f(|x|)|$.}\label{fig:absvaltrans5} \end{minipage} \end{figure} \begin{exa}Figures \ref{fig:y=f(x)=(x-1)^2-3} through \ref{fig:y=|(|x|-1)^2-3|} exhibit various transformations of $f:x\mapsto (x-1)^2-3$. \end{exa} \vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{4cm}$$ \psset{unit=.7pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \psplot[linewidth=2pt,linecolor=brown]{-2}{4}{(x-1)^2-3} $$ \meinecaption{1}{$y=f(x)=(x-1)^2-3$} \label{fig:y=f(x)=(x-1)^2-3} \end{minipage} \hfill \begin{minipage}{4cm}$$ \psset{unit=.7pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \psplot[linewidth=2pt,linecolor=brown]{-4}{4}{(abs(x)-1)^2-3} $$ \meinecaption{1}{$y=f(|x|)|=(|x|-1)^2-3$} \label{fig:y=(|x|-1)^2-3} \end{minipage} \hfill \begin{minipage}{4cm}$$ \psset{unit=.7pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \psplot[linewidth=2pt,linecolor=brown]{-2}{2}{(-1*abs(x)-1)^2-3} $$ \meinecaption{1}{$y=f(-|x|)=(-|x|-1)^2-3$} \label{fig:y=(-|x|-1)^2-3} \end{minipage} \hfill \begin{minipage}{4cm}$$ \psset{unit=.7pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \psplot[linewidth=2pt,linecolor=brown]{-4}{4}{abs((abs(x)-1)^2-3)} $$ \meinecaption{1}{$y=|f(|x|)|=|(|x|-1)^2-3|$} \label{fig:y=|(|x|-1)^2-3|} \end{minipage} \end{figure} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Use the graph of $f$ in figure \ref{fig:m165_s05_function 8} in order to draw \begin{multicols}{2} \begin{enumerate} \item $y = 2f(x)$ \item $y = f(2x)$ \item $y = f(-x)$ \item $y = -f(x)$ \item $y = -f(-x)$ \item $y = f(|x|)$ \item $y = |f(x)|$ \item $y = f(-|x|)$ \end{enumerate} \end{multicols} \begin{answer} Here are the graphs of $x\mapsto 2f(x)$ and $x\mapsto f(2x)$.\\ \vspace{3cm} \begin{figurehere} \begin{minipage}{3cm} \centering\psset{unit=.6pc} \psgrid[subgriddiv=0,griddots=5,gridlabels=0.5](-7,-7)(7,7) \psaxes[labels=none,ticks=none](0,0)(-8,-8)(8,8) \psline[linewidth=2pt, linecolor=brown](-4,-4)(-3,-4)(-1,2)(0,-4)(3,2)(4,-6) \psdots[dotstyle=*, dotscale=1](-4,-4)(-3,-4)(-1,2)(0,-4)(3,2)(4,-6) \meinecaption{2}{$y=2f(x)$} \end{minipage} \hfill \begin{minipage}{3cm} \centering\psset{unit=.6pc} \psgrid[subgriddiv=0,griddots=5,gridlabels=0.5](-7,-7)(7,7) \psaxes[labels=none,ticks=none](0,0)(-8,-8)(8,8) \psline[linewidth=2pt, linecolor=brown](-2,-2)(-1.5,-2)(-.5,1)(0,-2)(1.5,1)(2,-3) \psdots(-2,-2)(-1.5,-2)(-.5,1)(0,-2)(1.5,1)(2,-3) \meinecaption{2}{$y=f(2x)$} \end{minipage} \hfill \end{figurehere} Here are the graphs of $x\mapsto f(-x)$ and $x\mapsto -f(x)$.\\ \vspace{3cm} \begin{figurehere} \begin{minipage}{3cm} \centering\psset{unit=.6pc} \psgrid[subgriddiv=0,griddots=5,gridlabels=0.5](-7,-7)(7,7) \psaxes[labels=none,ticks=none](0,0)(-8,-8)(8,8) \psline[linewidth=2pt, linecolor=brown](4,-2)(3,-2)(1,1)(0,-2)(-3,1)(-4,-3) \psdots(4,-2)(3,-2)(1,1)(0,-2)(-3,1)(-4,-3) \meinecaption{2}{$y=f(-x)$} \end{minipage} \hfill \begin{minipage}{3cm} \centering\psset{unit=.6pc} \psgrid[subgriddiv=0,griddots=5,gridlabels=0.5](-7,-7)(7,7) \psaxes[labels=none,ticks=none](0,0)(-8,-8)(8,8) \psline[linewidth=2pt, linecolor=brown](-4,2)(-3,2)(-1,-1)(0,2)(3,-1)(4,3) \psdots(-4,2)(-3,2)(-1,-1)(0,2)(3,-1)(4,3) \meinecaption{2}{$y=-f(x)$} \end{minipage} \hfill \end{figurehere} Here are the graphs of $x\mapsto -f(-x)$ and $x\mapsto f(|x|)$.\\ \vspace{3cm} \begin{figurehere} \begin{minipage}{3cm} \centering\psset{unit=.6pc} \psgrid[subgriddiv=0,griddots=5,gridlabels=0.5](-7,-7)(7,7) \psaxes[labels=none,ticks=none](0,0)(-8,-8)(8,8) \psline[linewidth=2pt, linecolor=brown](4,2)(3,2)(1,-1)(0,2)(-3,-1)(-4,3) \psdots(4,2)(3,2)(1,-1)(0,2)(-3,-1)(-4,3) \meinecaption{2}{$y=-f(-x)$} \end{minipage} \hfill \begin{minipage}{3cm} \centering\psset{unit=.6pc} \psgrid[subgriddiv=0,griddots=5,gridlabels=0.5](-7,-7)(7,7) \psaxes[labels=none,ticks=none](0,0)(-8,-8)(8,8) \psline[linewidth=2pt, linecolor=brown](-4,-3)(-3.25,0)(-3,1)(-2,0)(0,-2)(2,0)(3,1)(3.25,0)(4,-3) \psdots(-4,-3)(-3.25,0)(-3,1)(0,-2)(3,1)(3.25,0)(4,-3) \meinecaption{2}{$y=f(|x|)$} \end{minipage} \hfill \end{figurehere} \columnbreak Here are the graphs of $x\mapsto |f(x)|$ and $x\mapsto f(-|x|)$.\\ \vspace{3cm} \begin{figurehere} \hfill \begin{minipage}{3cm} \centering\psset{unit=.6pc} \psgrid[subgriddiv=0,griddots=5,gridlabels=0.5](-7,-7)(7,7) \psaxes[labels=none,ticks=none](0,0)(-8,-8)(8,8) \psline[linewidth=2pt, linecolor=brown](-4,2)(-3,2)(-1.6667,0)(-1,1)(-.6667,0)(0,2)(2,0)(3,1)(3.25,0)(4,3) \psdots(-4,2)(-3,2)(-1.6667,0)(-1,1)(-.6667,0)(0,2)(2,0)(3,1)(3.25,0)(4,3) \meinecaption{2}{$y=|f(x)|$} \end{minipage} \hfill \begin{minipage}{3cm} \centering\psset{unit=.6pc} \psgrid[subgriddiv=0,griddots=5,gridlabels=0.5](-7,-7)(7,7) \psaxes[labels=none,ticks=none](0,0)(-8,-8)(8,8) \psline[linewidth=2pt, linecolor=brown](-4,-2)(-3,-2)(-1.6667,0)(-1,1)(-.6667,0)(0,-2)(.6667,0)(1,1)(1.6667,0)(3,-2)(4,-2) \psdots(-4,-2)(-3,-2)(-1.6667,0)(-1,1)(-.6667,0)(0,-2)(.6667,0)(1,1)(1.6667,0)(3,-2)(4,-2) \meinecaption{2}{$y=f(-|x|)$} \end{minipage} \hfill \end{figurehere} \end{answer} \end{pro} \vspace{1cm} \begin{figurehere} $$\psset{unit=.75pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.4pt, gridlabels=0](0,0)(-4,-4)(4,4) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-4,-4)(-4,-4)(4,4)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-4.5,0)(4.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-4.5)(0,4.5) \psline[linewidth=2pt, linecolor=blue](-4,-2)(-3,-2)(-1.6667,0)(-1,1)(-.6667,0)(0,-2)(2,0)(3,1)(4,-3) \psdots[dotstyle=*, dotscale=1](-4,-2)(-3,-2)(-1.6667,0)(-1,1)(-.6667,0)(0,-2)(2,0)(3,1)(3.25,0)(4,-3) $$\meinecaption{1}{$y = f(x)$} \label{fig:m165_s05_function 8} \end{figurehere} \begin{pro} Draw the curves $y=x^2-1$ and $y=|x^2-1|$ in succession. \begin{answer} The graphs appear below.\\ \vspace{3cm} \begin{figurehere}\hfill \begin{minipage}{2cm}$$ \psset{unit=.5pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \psplot[linewidth=2pt,linecolor=brown]{-2.5}{2.5}{x^2-1} $$ \meinecaption{1}{$y=g(x)=x^2-1$} \label{fig:y=x^2-1} \end{minipage} \hfill \begin{minipage}{2cm}$$ \psset{unit=.5pc, algebraic=true} \psaxes[labels=none,tickstyle=bottom]{<->}(0,0)(-6, -6)(6, 6) \psplot[linewidth=2pt,linecolor=brown]{-2.5}{2.5}{abs(x^2-1)} $$ \meinecaption{1}{$y=|g(x)|=|x^2-1|$} \label{fig:y=|x^2-1|} \end{minipage} \hfill \end{figurehere} \end{answer} \end{pro} \begin{pro}\label{pro:sqrtabsval} Draw the graph of the curve $y=\sqrt{|x|}$. \begin{answer} The graph appears in figure \ref{fig:sqrtabsval}. \vspace{3cm} \begin{figurehere} $$\psset{unit=1pc} \psaxes[arrows={->},linewidth=2pt,labelFontSize=\tiny](0,0)(-5,-5)(5,5) \psplot[linecolor=brown,linewidth=2pt,algebraic=true,arrows={<->}]{-5}{5}{sqrt(abs(x))} $$ \meinecaption{1}{Problem \ref{pro:sqrtabsval}.}\label{fig:sqrtabsval} \end{figurehere} \end{answer} \end{pro} \begin{pro}\label{pro:semicircle4} Draw the graphs of the curves $$y=\sqrt{-x^2+2|x|+3}, \qquad y=\sqrt{-x^2-2|x|+3}.$$ \begin{answer} Observe that $y=\sqrt{x^2+2x+3}$ is an upper semicircle and that $$ y=\sqrt{-x^2+2x+3} \implies x^2-2x+y^2 =3 \implies (x-1)^2+y^2 =4, $$ from where the semicircle has radius $2$ and centre at $(1,0)$, as appears in figure \ref{fig:semicircle4}. \\ \vspace{3cm} \begin{figurehere} \centering \psset{unit=1pc} \psaxes[arrows={->},linewidth=2pt,labelFontSize=\tiny](0,0)(-5,-5)(5,5) \psplot[linecolor=brown,linewidth=2pt,algebraic=true]{-1}{3}{sqrt(-x^2+2*x+3)} \meinecaption{2}{Problem \ref{pro:semicircle4}.}\label{fig:semicircle4} \end{figurehere} The graph of $y=\sqrt{-x^2+2|x|+3}$ appears in figure \ref{fig:semicircle5}. \\ \vspace{3cm} \begin{figurehere} \centering \psset{unit=1pc} \psaxes[arrows={->},linewidth=2pt,labelFontSize=\tiny](0,0)(-5,-5)(5,5) \psplot[linecolor=brown,linewidth=2pt,algebraic=true]{-3}{3}{sqrt(-x^2+2*abs(x)+3)} \meinecaption{2}{Problem \ref{pro:semicircle4}.}\label{fig:semicircle5} \end{figurehere} The graph of of $y=\sqrt{-x^2-2|x|+3}$ appears in figure \ref{fig:semicircle6}.\\ \vspace{3cm} \begin{figurehere} \centering \psset{unit=1pc} \psaxes[arrows={->},linewidth=2pt,labelFontSize=\tiny](0,0)(-5,-5)(5,5) \psplot[linecolor=brown,linewidth=2pt,algebraic=true]{-1}{1}{sqrt(-x^2-2*abs(x)+3)} \meinecaption{2}{Problem \ref{pro:semicircle4}.}\label{fig:semicircle6} \end{figurehere} \end{answer} \end{pro} \begin{pro} Draw the following graphs in succession. \begin{enumerate} \item $y = (x-1)^2-2$ \item $y = |(x-1)^2-2|$ \item $y = (|x|-1)^2-2$ \item $y = (1+|x|)^2-2$ \end{enumerate} \begin{answer} Here is the graph of $y=(x-1)^2-2$.\\ \vspace{3cm} \begin{figurehere} $$\psset{unit=.6pc} \psaxes[linewidth=2pt,labelFontSize=\tiny](0,0)(-5,-5)(5,5) \psplot[linewidth=2pt,linecolor=brown,arrows={<->},algebraic=true]{-2}{4}{(x-1)^2-2}$$ \meinecaption{1}{$y=(x-1)^2-2$.} \end{figurehere} \bigskip Here is the graph of $y=|(x-1)^2-2|$.\\ \vspace{2cm} \begin{figurehere} $$\psset{unit=.6pc} \psaxes[linewidth=2pt,labelFontSize=\tiny](0,0)(-5,-5)(5,5) \psplot[linewidth=2pt,linecolor=brown,arrows={<->},algebraic=true]{-2}{4}{abs((x-1)^2-2)} $$ \meinecaption{1}{$y=|(x-1)^2-2|$.} \end{figurehere} \bigskip Here is the graph of $y=(|x|-1)^2-2$.\\ \vspace{2cm} \begin{figurehere}$$\psset{unit=.6pc} \psaxes[linewidth=2pt,labelFontSize=\tiny](0,0)(-5,-5)(5,5) \psplot[linewidth=2pt,linecolor=brown,arrows={<->},algebraic=true]{-4}{4}{(abs(x)-1)^2-2}$$ \meinecaption{1}{$y=(|x|-1)^2-2$.} \end{figurehere} \bigskip Observe that $(-|x|-1)^2= (-1)^2(|x|+1)^2= (|x|+1)^2$. Here is the graph of $y=(|x|+1)^2-2$.\\ \vspace{2cm} \begin{figurehere} $$\psset{unit=.6pc} \psaxes[linewidth=2pt,labelFontSize=\tiny](0,0)(-5,-5)(5,5) \psplot[linewidth=2pt,linecolor=brown,arrows={<->},algebraic=true]{-2}{2}{(abs(x)+1)^2-2}$$ \meinecaption{1}{$y=(|x|+1)^2-2$.} \end{figurehere} \end{answer} \end{pro} \begin{pro} Draw the graph of $f:\reals\rightarrow \reals$, with assignment rule $f(x)=x|x|$. \end{pro} \begin{pro} Draw the following curves in succession: \begin{enumerate} \item $y = x^2$ \item $y = (x -1)^2$ \item $y = (|x|-1)^2$ \end{enumerate} \end{pro} \begin{pro} Draw the following curves in succession: \begin{enumerate} \item $y = x^2$ \item $y = x^2-1$ \item $y = |x^2-1|$ \end{enumerate} \end{pro} \begin{pro} Draw the following curves in succession: \begin{enumerate} \item $y = x^2 + 2x + 3$ \item $y = x^2+ 2|x| + 3$ \item $y = |x^2 + 2x + 3|$ \item $y = |x^2 + 2|x| + 3|$ \end{enumerate} \end{pro} \begin{pro} Draw the following curves in succession: \begin{enumerate} \item $y = 1-x$ \item $y = |1-x|$ \item $y = 1-|1-x|$ \item $y = |1-|1-x||$ \item $y = 1-|1-|1-x||$ \item $y = |1-|1-|1-x|||$ \item $y = 1-|1-|1-|1-x|||$ \item $y = |1-|1-|1-|1-x||||$ \end{enumerate} \begin{answer} Here is the graph of $y=1-x$.\\ \vspace{2cm} \begin{figurehere} $$\psset{unit=.6pc} \psaxes[linewidth=2pt,labelFontSize=\tiny](0,0)(-5,-5)(5,5) \psplot[linewidth=2pt,linecolor=brown,arrows={<->},algebraic=true]{-5}{5}{1-x}$$ \meinecaption{1}{$y=1-x$.} \end{figurehere} \bigskip Here is the graph of $y=|1-x|$.\\ \vspace{2cm} \begin{figurehere} $$\psset{unit=.6pc} \psaxes[linewidth=2pt,labelFontSize=\tiny](0,0)(-5,-5)(5,5) \psplot[linewidth=2pt,linecolor=brown,arrows={<->},algebraic=true]{-5}{6}{abs(1-x)}$$ \meinecaption{1}{$y=|1-x|$.} \end{figurehere} \bigskip Here is the graph of $y=1-|1-x|$.\\ \vspace{2cm} \begin{figurehere} $$\psset{unit=.6pc} \psaxes[linewidth=2pt,labelFontSize=\tiny](0,0)(-5,-5)(5,5) \psplot[linewidth=2pt,linecolor=brown,arrows={<->},algebraic=true]{-5}{7}{1-abs(1-x)}$$ \meinecaption{1}{$y=1-|1-x|$.} \end{figurehere} \bigskip Here is the graph of $y=|1-|1-x||$.\\ \vspace{2cm} \begin{figurehere} $$\psset{unit=.6pc} \psaxes[linewidth=2pt,labelFontSize=\tiny](0,0)(-5,-5)(5,5) \psplot[linewidth=2pt,linecolor=brown,arrows={<->},algebraic=true]{-5}{7}{abs(1-abs(1-x))}$$ \meinecaption{1}{$y=|1-|1-x||$.} \end{figurehere} \columnbreak Here is the graph of $y=1-|1-|1-x||$.\\ \vspace{2cm} \begin{figurehere} $$\psset{unit=.6pc} \psaxes[linewidth=2pt,labelFontSize=\tiny](0,0)(-5,-5)(5,5) \psplot[linewidth=2pt,linecolor=brown,arrows={<->},algebraic=true]{-5}{7}{1-abs(1-abs(1-x))}$$ \meinecaption{1}{$y=1-|1-|1-x||$.} \end{figurehere} \bigskip Here is the graph of $y=|1-|1-|1-x|||$.\\ \vspace{2cm} \begin{figurehere} $$\psset{unit=.6pc} \psaxes[linewidth=2pt,labelFontSize=\tiny](0,0)(-5,-5)(5,5) \psplot[linewidth=2pt,linecolor=brown,arrows={<->},algebraic=true]{-5}{7}{abs(1-abs(1-abs(1-x)))}$$ \meinecaption{1}{$y=|1-|1-|1-x|||$.} \end{figurehere} \bigskip Here is the graph of $y=1-|1-|1-|1-x|||$.\\ \vspace{2cm} \begin{figurehere} $$\psset{unit=.6pc} \psaxes[linewidth=2pt,labelFontSize=\tiny](0,0)(-5,-5)(5,5) \psplot[linewidth=2pt,linecolor=brown,arrows={<->},algebraic=true]{-5}{7}{1-abs(1-abs(1-abs(1-x)))}$$ \meinecaption{1}{$y=1-|1-|1-|1-x|||$.} \end{figurehere} \bigskip Here is the graph of $y=|1-|1-|1-|1-x||||$.\\ \vspace{2cm} \begin{figurehere} $$\psset{unit=.6pc} \psaxes[linewidth=2pt,labelFontSize=\tiny](0,0)(-5,-5)(5,5) \psplot[linewidth=2pt,linecolor=brown,arrows={<->},algebraic=true]{-5}{7}{abs(1-abs(1-abs(1-abs(1-x))))}$$ \meinecaption{1}{$y=|1-|1-|1-|1-x||||$.} \end{figurehere} \end{answer} \end{pro} \begin{pro} Put $f_1(x) = x$; $f_2(x) = |1-f_1(x)|$; $f_3(x) = |1-f_2(x)|$; \ldots $f_n(x) = |1-f_{n-1}(x)|$. Prove that the solutions of the equation $f_n(x) = 0$ are $\{\pm 1, \pm 3, \ldots, \pm (n-3), (n-1)\}$ if $n$ is even and $\{0, \pm 2, \ldots , \pm (n-3), (n-1) \}$ if $n$ is odd. \end{pro} \begin{pro} \label{pro:distortion_4} Given in figures \ref{fig:distortion_4_1} and \ref{fig:distortion_4_2} are the graphs of two curves, $y = f(x)$ and $y = f(ax)$ for some real constant $a<0$. \\ \begin{enumerate} \item Determine the value of the constant $a$. \item Determine the value of $C$. \end{enumerate} \begin{answer} Notice that the graph of $ y = f(ax)$ is a horizontal shrinking of the graph of $y = f(x)$. Put $g(x) = f(ax)$. Since $g(4/3) = 0$ we must have $4a/3=-2 \implies a = -3/2$, so the point $(-2,0)$ on the original graph was mapped to the point $(4/3,0)$ on the new graph. Hence the point $(3,0)$ in the old graph gets mapped to $(-2,0)$ and so $C=-2$. \end{answer} \end{pro} \vspace{2cm} \begin{figurehere} \begin{minipage}{3cm} $$\psset{unit=.7pc}\psaxes[labels=none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \uput[r](7,0){x} \uput[u](0, 7.5){y} \psplot[linewidth=1.5pt, linecolor=brown]{-3.3}{3.3}{x -3 add -.01 mul x 2 exp 10 add mul x mul x 2 add 2 exp mul}$$ \vspace{1.5cm} \meinecaption{.25}{Problem \ref{pro:distortion_4}. $y = f(x)$} \label{fig:distortion_4_1} \end{minipage} \hfill \begin{minipage}{3cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks=none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \uput[r](7,0){x} \uput[u](0, 7.5){y} \psplot[linewidth=1.5pt, linecolor=brown]{-2.2}{2}{x -1.5 mul -3 add -.01 mul x -1.5 mul 2 exp 10 add mul x -1.5 mul mul x -1.5 mul 2 add 2 exp mul} \psdots[dotstyle=*,dotscale=1](-2,0)(1.33333333333333,0)\uput[d](2.5,0){\frac{4}{3}}\uput[d](-2.5,0){C} $$ \meinecaption{1.5}{Problem \ref{pro:distortion_4}. $y = f(ax)$} \label{fig:distortion_4_2} \end{minipage} \hfill \end{figurehere} \end{multicols} \section{Behaviour of the Graphs of Functions} So far we have limited our study of functions to those families of functions whose graphs are known to us: lines, parabolas, hyperbolas, or semicircles. Through some arguments involving symmetry we have been able to extend this collection to compositions of the above listed functions with the absolute value function. We would now like to increase our repertoire of functions that we can graph. For that we need the machinery of Calculus, which will be studied in subsequent courses. We will content ourselves with {\em informally} introducing various terms useful when describing curves and with proving that these properties hold for some simple curves. \subsection{Continuity} \begin{df} We write $x\to a+$ to indicate the fact that $x$ is progressively getting closer and closer to $a$ through values greater (to the right) of $a$. Similarly, we write $x\to a-$ to indicate the fact that $x$ is progressively getting closer and closer to $a$ through values smaller (to the left) of $a$. Finally, we write $x\to a$ to indicate the fact that $x$ is progressively getting closer and closer to $a$ through values left and right of $a$. \end{df} \begin{df} Given a function $f$, we write $f(a+)$ for the value that $f(x)$ approaches as $x\to a+$. In other words, we consider the values of a dextral neighbourhood of $a$, progressively decrease the length of this neighbourhood, and see which value $f$ approaches in this neighbourhood. Similarly, we write $f(a-)$ for the value that $f(x)$ approaches as $x\to a-$. In other words, we consider the values of a sinistral neighbourhood of $a$, progressively decrease the length of this neighbourhood, and see which value $f$ approaches in this neighbourhood. \end{df} \begin{exa} Let $f: [-4;4]\to \reals$ be defined as follows: $$f(x) = \left\{\begin{array}{ll}x^2+1 & \mathrm{if}\ -4 \leq x < -2 \\ 2 & \mathrm{if}\ x=-2 \\ 2+2x & \mathrm{if}\ -2}](0,0)(-5,-5)(5,5) \psplot[algebraic=true,linewidth=2pt,linecolor=brown,arrows={<->}]{-4.5}{4.5}{abs(x)} \psdots[dotstyle=o,linecolor=brown,dotscale=1.3](0,0) \meinecaption{2}{$a: x\mapsto |x|$, $x\neq 0$.} \label{fig:cont1} \end{minipage}\hfill \begin{minipage}{5cm} \centering\psset{unit=1pc} \psaxes[labels=none,linewidth=2pt,arrows={->}](0,0)(-5,-5)(5,5) \psplot[algebraic=true,linewidth=2pt,linecolor=brown,arrows={<-o}]{-4.5}{0}{-1} \psplot[algebraic=true,linewidth=2pt,linecolor=brown,arrows={o->}]{0}{4.5}{1} \psdots[dotstyle=o,linecolor=brown,dotscale=1.3](0,0) \meinecaption{2}{$b: x\mapsto \dfrac{x}{|x|}$, $x\neq 0$.} \label{fig:cont2} \end{minipage} \hfill \begin{minipage}{5cm} \centering\psset{unit=1pc} \psaxes[labels=none,linewidth=2pt,arrows={->}](0,0)(-5,-5)(5,5) \psplot[algebraic=true,linewidth=2pt,linecolor=brown,arrows={<->}]{-4.5}{-.22222}{1/x} \psplot[algebraic=true,linewidth=2pt,linecolor=brown,arrows={<->}]{.22222}{4.5}{1/x} \psdots[dotstyle=o,linecolor=brown,dotscale=1.3](0,0) \meinecaption{2}{$c: x\mapsto \dfrac{1}{x}$, $x\neq 0$.} \label{fig:cont3} \end{minipage} \end{figure} \begin{df}A function $f$ is said to be {\em left continuous}\index{function!continuous} at the point $x=a$ if $f(a-) = f(a)$. A function $f$ is said to be {\em right continuous}\index{function!continuous} at the point $x=a$ if $f(a) = f(a+)$. A function $f$ is said to be {\em continuous}\index{function!continuous} at the point $x=a$ if $f(a-) = f(a) = f(a+)$. It is continuous on the interval $I$ if it is continuous on every point of $I$. \index{continuity} \end{df} Heuristically speaking, a continuous function is one whose graph has no ``breaks.'' \begin{exa}\label{exa:continuous1} Given that $$ f(x) = \left\{ \begin{array}{lr}6 + x & \mathrm{if}\ x \in ]-\infty; -2] \\ 3x^2 + xa &\mathrm{if}\ x\in ]-2; +\infty[ \\\end{array} \right. $$ is continuous, find $a$. \end{exa} \begin{solu} Since $f(-2-) =f(-2)= 6-2 = 4$ and $f(-2+) = 3(-2)^2 -2a = 12-2a$ we need $$f(-2-) = f(-2+) \implies 4 = 12-2a \implies a = 4. $$ \end{solu} \subsection{Monotonicity} \begin{df} A function $f$ is said to be {\em increasing} (respectively, {\em strictly increasing}) if $a 0).$$ Similarly, a function $g$ is (strictly) decreasing if for all $a1 \\\end{array} \right. $$ is continuous, find $a$. \begin{answer} We have $f(1-) = 0$ and $f(1+) = 2+3a$. We need then $0 = 2+3a$ or $a= -\dfrac{2}{3}$. \end{answer} \end{pro} \begin{pro} Let $n$ be a strictly positive integer. Given that $$ f(x) = \left\{ \begin{array}{lr}\dfrac{x^n-1}{x-1} & \mathrm{if}\ x\neq 1 \\ a &\mathrm{if}\ x=1 \\\end{array} \right. $$ is continuous, find $a$. \begin{answer} For $x\neq 1$ we have $f(x) = \dfrac{x^n-1}{x-1} = x^{n-1} + x^{n-2} + \cdots + x^2 + x+1$. Since $f(1-) =n$ and $f(1+) = n$ we need $a= f(1) = n$. \end{answer} \end{pro} \begin{pro} Give an example of a function discontinuous at the points $\pm \sqrt[3]{1}, \pm \sqrt[3]{2}, \pm \sqrt[3]{3}, \pm \sqrt[3]{4}, \pm \sqrt[3]{5}, \ldots$. \begin{answer} Examine the assignment rule $x \mapsto \lfloor x^3 \rfloor$. \end{answer} \end{pro} \end{multicols} \section{The functions $x\mapsto \floor{x}$, $x\mapsto \ceil{x}$, $x\mapsto \deci{x}$} \begin{df}The {\em floor} $\floor{x}$ of a real number $x$ is the unique integer defined by the inequality $$ \floor{x}\leq x < \floor{x}+1. $$ \end{df} In other words, $\floor{x}$ is $x$ if $x$ is an integer, or the integer just to the left, if $x$ is not an integer. For example $$\floor{3}=3, \qquad \floor{3.9}=3, \qquad \floor{-\pi}=-4. $$ If $n\in \integers$ and if $$n \leq x < n+1,$$ then $\floor{x}=n$. This means that the function $x\mapsto \floor{x}$ is constant between two consecutive integers. For example, between $0$ and $1$ it will have output $0$; between $1$ and $2$, it will have output $1$, etc., always taking the smaller of the two consecutive integers. Its graph has the staircase shape found in figure \ref{fig:floor_function }. \begin{df} The {\em ceiling} $\ceil{x}$ of a real number $x$ is the unique integer defined by the inequality $$ \ceil{x}-1< x \leq \ceil{x}. $$ \end{df} In other words, $\ceil{x}$ is $x$ if $x$ is an integer, or the integer just to the right, if $x$ is not an integer. For example $$\ceil{3}=3, \qquad \ceil{3.9}=4, \qquad \floor{-\pi}=-3. $$ If $n\in \integers$ and if $$n < x \leq n+1,$$ then $\ceil{x}=n+1$. This means that the function $x\mapsto \ceil{x}$ is constant between two consecutive integers. For example, between $0$ and $1$ it will have output $1$; between $1$ and $2$, it will have output $2$, etc., always taking the larger of the two consecutive integers. Its graph has the staircase shape found in figure \ref{fig:ceil_function }. \vspace*{2cm} \begin{figure}[!hptb] \begin{minipage}{.25\textwidth} $$\psset{unit=1pc} \multips(-4,-4)(1,1){9}{\psline[linecolor=brown,linewidth=2pt,arrows={*-o}](0,0)(1,0)} \psaxes[linewidth=1.5pt, labelFontSize=\tiny]{->}(0,0)(-5,-5)(5,5) $$ \meinecaption{2}{$x\mapsto \floor{ x }$.} \label{fig:floor_function } \end{minipage} \hfill \begin{minipage}{.25\textwidth} $$ \psset{unit=1pc} \psaxes[linewidth=1.5pt, labelFontSize=\tiny ]{->}(0,0)(-5,-5)(5,5) \multips(-4,-3)(1,1){9}{\psline[linecolor=brown,linewidth=2pt,arrows={o-*}](0,0)(1,0)} $$ \meinecaption{2}{$x\mapsto \ceil{x}$.} \label{fig:ceil_function } \end{minipage} \hfill \begin{minipage}{.25\textwidth} $$ \psset{unit=1pc} \psaxes[linewidth=1.5pt, labelFontSize=\tiny ]{->}(0,0)(-5,-5)(5,5) \multips(-4,0)(1,0){9}{\psline[linecolor=brown,linewidth=2pt,arrows={*-o}](0,0)(1,1)} $$ \meinecaption{2}{$x\mapsto x-\floor{x}$.} \label{fig:deci_function } \end{minipage} \end{figure} \begin{df} A function $f$ is said to be {\em periodic of period $P$} if there a real number $P>0$ such that $$ x\in \dom{f}\implies (x+P)\in\dom{f}, \quad f(x+P)=f(x). $$ \end{df} That is, if $f$ is periodic of period $P$ then once $f$ is defined on an interval of period $P$, then it will be defined for all other values of its domain. The discussion below will make use of the following lemma. \begin{lem}\label{lem:parte-entera} Let $x\in \reals$ and $z\in \integers$. Then $$ \floor{x+z} = \floor{x}+z. $$ \end{lem} \begin{pf} Recall that $ \floor{x} $ is the unique integer with the property $$ \floor{x} \leq x < \floor{x}+1. $$In turn, this means that $\floor{x+z}-z$ also satisfies this inequality. By definition, $$ \floor{x+z} \leq x+z < \floor{x+z}+1, $$ and so we have, $$ \floor{x+z}-z \leq x < \floor{x+z}-z+1, $$ from where $\floor{x+z}-z$ satisfies the desired inequality and we conclude that e $\floor{x+z}-z=\floor{x}$, demonstrating theorem. \end{pf} \begin{exa}Put $\deci{x}=x-\floor{x}$. Consider the function $f:\reals \to \co{0;1}$, $f(x)=\deci{x}$, the decimal part {\em decimal part of $x$}. We have $$ \floor{x}\leq x < \floor{x}+1 \implies 0 \leq x-\floor{x} <1. $$ Also, by virtue of lemma \ref{lem:parte-entera}, $$ f(x+1)= \deci{x+1}=(x+1)-\floor{x+1} = (x+1)-(\floor{x}+1)=x-\floor{x}=\deci{x}=f(x),$$which means that $f$ is periodic of period $1$. Now, $$x\in \co{0;1} \implies \deci{x}=x, $$from where we gather that between $0$ and $1$, $f$ behaves like the identity function. The graph of $x \mapsto \deci{x}$ appears in figure \ref{fig:deci_function } . \end{exa} \subsection*{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Give an example of a function $r$ discontinuous at the reciprocal of every non-zero integer. \begin{answer} Examine the assignment rule $r(x) = \dis{\lfloor\frac{1}{x}\rfloor}, x \neq 0$. \end{answer} \end{pro} \begin{pro} Give an example of a function discontinuous at the odd integers. \begin{answer} Examine the assignment rule $x \mapsto \dis{\lfloor \frac{x - 1}{2} \rfloor}$. \end{answer} \end{pro} \begin{pro} Give an example of a function discontinuous at the square of every integer. \begin{answer} Examine the assignment rule $x \mapsto \lfloor \sqrt{|x|} \rfloor$. \end{answer} \end{pro} \begin{pro}\label{pro:distance_to_Z} Let $||x|| = \min _{n \in {\integers}} |x - n|$. Prove that $x \mapsto ||x||$ is periodic and find its period. Also, graph this function. Notice that this function measures the distance of a real number to its nearest integer. \begin{answer} The function is periodic of period $1$. If $x\in \cc{0;1}$ then $$ ||x||=\min (x,1-x). $$ Its graph appears in figure \ref{fig:distance_to_Z}. \vspace*{3cm} \begin{figurehere} $$ \psset{unit=1pc} \psaxes[linewidth=1.5pt, labelFontSize=\tiny ]{->}(0,0)(-5,-5)(5,5) \multips(-5,0)(1,0){9}{\psline[linecolor=brown,linewidth=1.2pt,arrows={*-*}](0,0)(.5,.5)(1,0)}$$ \meinecaption{2}{Problem \ref{pro:distance_to_Z}.}\label{fig:distance_to_Z} \end{figurehere} \end{answer} \end{pro} \begin{pro}\label{pro:floor1} Investigate the graph of $x\mapsto \floor{2x}$. \begin{answer} Its graph will jump each time $2x=n$, an integer, that is, when $x=\dfrac{n}{2}$, which means it jumps at every fraction with denominator $2$. Its graph appears in figure \ref{pro:floor1}. \vspace{3cm} \begin{figurehere} $$ \psset{unit=1pc} \psaxes[linewidth=1.5pt, labelFontSize=\tiny ]{->}(0,0)(-5,-5)(5,5) \multips(-2.5,-5)(.5,1){12}{\psline[linecolor=brown,linewidth=1.2pt,arrows={*-o}](0,0)(.5,0)}$$ \meinecaption{2}{Problem \ref{pro:floor1}.}\label{fig:floor1} \end{figurehere} \end{answer} \end{pro} \begin{pro} Is it true that for all real numbers $x$ we have $\deci{x^2}=\deci{x}^2$? \begin{answer}The assertion is false. For example, if $x=2.1$ then $\deci{2.1}^2=0.1^2=0.01$ but $\deci{2.1^2}=\deci{4.41}=0.41$ \end{answer} \end{pro} \begin{pro} Demonstrate that the function $f:\reals \to \{-1,1\}$ given by $f(x)=(-1)^{\floor{x}}$ is periodic of period $2$ and draw its graph. \end{pro} \begin{pro} Discuss the graph of $x\mapsto \dfrac{1}{\ceil{x}-\floor{x}}$. \begin{answer} The formula $x\mapsto \dfrac{1}{\ceil{x}-\floor{x}}$ is not defined for $x\in \integers$. If $x\in \reals \setminus \integers$ then $\dfrac{1}{\ceil{x}-\floor{x}}=1$. Thus the graph consists of the horizontal line of equation $y=1$ but with punctures at the points $(n,1)$, $n\in \integers$. \end{answer} \end{pro} \begin{pro} Find the points of discontinuity of the function $f:\reals\rightarrow \reals$, $f:x\mapsto \floor{x}+ \sqrt{x-\floor{x}}$. \begin{answer} First consider $n\in\integers $. We have $$ \floor{x}\rightarrow \left\{\begin{array}{ll}n-1 & \mathrm{as}\ x\to n- \\ n& \mathrm{as}\ x\to n+ \\ \end{array}\right. $$ Then $$ \floor{x} + \sqrt{x-\floor{x}}\rightarrow \left\{\begin{array}{llll}n-1 + \sqrt{n-(n-1)}& = & n & \mathrm{as}\ x\to n- \\ n + \sqrt{n-n}& = & n & \mathrm{as}\ x\to n+ \\ \end{array}\right. $$ We deduce that $f$ is continuous at the integers. Since $f$ is clearly continuous at non-integral points, we conclude that $f$ is everywhere continuous. \end{answer} \end{pro} \begin{pro} Find the points of discontinuity of the function $\fun{f}{\reals}{\reals}{x}{\left\{\begin{array}{ll} x & \mathrm{if}\ x\in \rationals\\ 0 & \mathrm{if}\ x\in \reals\setminus \rationals\\ \end{array}\right.}$. \end{pro} \begin{pro} Find the points of discontinuity of the function $\fun{f}{\reals}{\reals}{x}{\left\{\begin{array}{ll} 0 & \mathrm{if}\ x\in \rationals\\ x & \mathrm{if}\ x\in \reals\setminus \rationals\\ \end{array}\right.}$. \end{pro} \begin{pro} Find the points of discontinuity of the function $\fun{f}{\reals}{\reals}{x}{\left\{\begin{array}{ll} 0 & \mathrm{if}\ x\in \rationals\\ 1 & \mathrm{if}\ x\in \reals\setminus \rationals\\ \end{array}\right.}$. \end{pro} \begin{pro} Prove that $f:\reals \rightarrow \reals,\ \dis{f(t + 1) = \frac{1}{2} + \sqrt{f(t) - (f(t))^2}}$ has period $2.$ \end{pro} \end{multicols} \chapter{Polynomial Functions} \begin{df} A {\em polynomial} $p(x)$ of degree $n\in\naturals$ is an expression of the form $$p(x) = a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0, \qquad a_n \neq 0,\qquad a_k\in \reals,$$where the $a_k$ are constants. If the $a_k$ are all integers then we say that $p$ has integer coefficients, and we write $p(x)\in\integers[x]$; if the $a_k$ are real numbers then we say that $p$ has real coefficients and we write $p(x)\in \reals[x]$; etc. The degree $n$ of the polynomial $p$ is denoted by $\deg p$. The coefficient $a_n$ is called the {\em leading coefficient} of $p(x)$. \index{polynomial}\index{polynomial!leading coefficient}\index{polynomial!degree of a} A {\em root} of $p$ is a solution to the equation $p(x) = 0$.\index{root!polynomial}\index{polynomial!root}\index{function!polynomial} \end{df} In this chapter we learn how to graph polynomials all whose roots are real numbers. \begin{exa}Here are a few examples of polynomials. \begin{itemize} \item $a(x) = 2x+1 \in \integers[x]$, is a polynomial of degree $1$, and leading coefficient $2$. It has $x=-\dfrac{1}{2}$ as its only root. A polynomial of degree $1$ is also known as an {\em affine function}. \item $b(x) = \pi x^2+x-\sqrt{3}\in\reals[x]$, is a polynomial of degree $2$ and leading coefficient $\pi$. By the quadratic formula $b$ has the two roots $$ x= \dfrac{-1+\sqrt{1+4\pi\sqrt{3}}}{2\pi}\qquad\mathrm{and}\qquad x= \dfrac{-1-\sqrt{1+4\pi\sqrt{3}}}{2\pi}. $$ A polynomial of degree $2$ is also called a {\em quadratic polynomial} or {\em quadratic function}. \item $C(x) = 1\cdot x^0:=1$, is a constant polynomial, of degree $0$. It has no roots, since it is never zero. \end{itemize} \end{exa} \begin{thm}The degree of the product of two polynomials is the sum of their degrees. In symbols, if $p, q$ are polynomials, $\deg pq = \deg p + \deg q$. \label{thm:degree_of_product} \end{thm} \begin{pf} If $p(x) = a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0,$ and $q(x) = b_mx^m + b_{m - 1}x^{m - 1} + \cdots + b_1x + b_0$, with $a_n\neq 0$ and $b_m\neq 0$ then upon multiplication, $$ p(x)q(x) = (a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0)(b_mx^m + b_{m - 1}x^{m - 1} + \cdots + b_1x + b_0) = a_nb_mx^{m+n} + \cdots + ,$$ with non-vanishing leading coefficient $a_nb_m$. \end{pf} \begin{exa} The polynomial $p(x) = (1+2x+3x^3)^{4}(1-2x^2)^5$ has leading coefficient $3^4(-2)^5 =-2592$ and degree $3\cdot 4 + 2\cdot 5 = 22$. \end{exa} \begin{exa} What is the degree of the polynomial identically equal to $0$? Put $p(x) \equiv 0$ and, say, $q(x) = x+1$. Then by Theorem \ref{thm:degree_of_product} we must have $\deg pq = \deg p + \deg q = \deg p + 1$. But $pq$ is identically $0$, and hence $\deg pq = \deg p$. But if $\deg p$ were finite then $$\deg p = \deg pq = \deg p + 1 \implies 0 = 1, $$nonsense. Thus the $0$-polynomial does not have any finite degree. We attach to it, by convention, degree $-\infty$. \end{exa} \section{Power Functions }\label{sec:power_functions} \begin{df} A {\em power function} is a function whose formula is of the form $x\mapsto x^\alpha$, where $\alpha \in \reals$. In this chapter we will only study the case when $\alpha$ is a positive integer. \index{function!power} \end{df} \bigskip If $n$ is a positive integer, we are interested in how to graph $x\mapsto x^n$. We have already encountered a few instances of power functions. For $n = 0$, the function $x\mapsto 1$ is a constant function, whose graph is the straight line $y=1$ parallel to the $x$-axis. For $n=1$, the function $x\mapsto x$ is the identity function, whose graph is the straight line $y=x$, which bisects the first and third quadrant. These graphs were not obtained by fiat, we demonstrated that the graphs are indeed straight lines in Theorem \ref{thm:graph_mx+k}. Also, for $n=2$, we have the square function $x\mapsto x^2$ whose graph is the parabola $y=x^2$ encountered in example \ref{exa:y=x^2}. We reproduce their graphs below in figures \ref{fig:poly_y=1} through \ref{fig:poly_y=x^2} for easy reference. \vspace{1cm} \begin{figure}[!hptb] \begin{minipage}{4cm} $$\psset{algebraic=true, unit=1pc} \psaxes[ labels=none, linewidth=1.5pt]{->}(0,0)(-2.5,-2.5)(2.5,2.5) \psline[linewidth=2pt, linecolor=brown]{<->}(-2.5,1)(2.5,1) $$ \meinecaption{1}{$x\mapsto 1$.} \label{fig:poly_y=1} \end{minipage}\hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=1pc} \psaxes[ labels=none, linewidth=1.5pt]{->}(0,0)(-2.5,-2.5)(2.5,2.5) \psline[linewidth=2pt, linecolor=brown]{<->}(-2.5,-2.5)(2.5,2.5) $$ \meinecaption{1}{$x\mapsto x$.} \label{fig:poly_y=x} \end{minipage}\hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=1pc} \psaxes[ labels=none, linewidth=1.5pt]{->}(0,0)(-2.5,-2.5)(2.5,2.5) \parabola[linewidth=2pt, linecolor=brown]{<->}(1.58,2.5)(0,0) $$ \meinecaption{1}{$x\mapsto x^2$.} \label{fig:poly_y=x^2} \end{minipage} \end{figure} \bigskip The graphs above were obtained by geometrical arguments using similar triangles and the distance formula. This method of obtaining graphs of functions is quite limited, and hence, as a view of introducing a more general method that argues from the angles of continuity, monotonicity, and convexity, we will derive the shape of their graphs once more. \section{Affine Functions}\label{sec:affine_functions} \begin{df} Let $m, k$ be real number constants. A function of the form $x\mapsto mx + k$ is called an {\em affine function}. In the particular case that $m = 0$, we call $x\mapsto k$ a {\em constant function}. If, however, $k = 0$ and $m\neq 0$, then we call the function $x\mapsto mx$ a {\em linear function}. \index{function!linear}\index{function!affine}\index{function!constant} \end{df} \begin{thm}[Graph of an Affine Function] The graph of an affine function $$\fun{f}{x}{mx+k}{\reals}{\reals} $$is a continuous straight line. It is strictly increasing if $m>0$ and strictly decreasing if $m<0$. If $m\neq 0$ then $x\mapsto mx+k$ has a unique zero $x=-\dfrac{k}{m}$. If $m\neq 0$ then $\im{f} = \reals$. \end{thm} \begin{pf} Since for any $a\in \reals$, $f(a+)=f(a)=f(a-)=ma+k$, an affine function is everywhere continuous. Let $\lambda \in [0;1]$. Since $$ f (\lambda a+ (1-\lambda)b) =m(\lambda a+ (1-\lambda)b) + k=m\lambda a+mb-mb\lambda +k =\lambda mf(a) + (1-\lambda)m f(b), $$ an affine function is both convex and concave. This means that it does not bend upwards or downwards (or that it bends upwards and downwards!) always, and hence, it must be a straight line. Let $a < b$. Then $$\dfrac{f(b) - f(a)}{b - a} = \dfrac{mb+k - ma-k}{b - a} = m ,$$ which is strictly positive for $m>0$ and strictly negative for $m<0$. This means that $f$ is a strictly increasing function for $m>0$ and strictly decreasing for $m<0$. Also given any $a\in\reals $ we have $$f(x) = a\implies mx+k=a \implies x=\dfrac{a-k}{m}, $$ which is a real number as long as $m\neq 0$. Hence every real number is an image of $f$ meaning that $\im{f} = \reals$. In particular, if $a=0$, then $x=-\dfrac{k}{m}$ is the only solution to the equation $f(x)=0$. Clearly, if $m=0$, then $\im{f}=\{k\}$.\end{pf} This information is summarised in the following tables. \begin{figure}[h] \begin{minipage}{.4\textwidth} $$ \begin{array}{|c|ccccc|} \hline x& -\infty & & -\dfrac{k}{m} & & +\infty \\ \hline & & & & & \\ & & & & \nearrow & \\ f(x)=mx+k & & & 0 & & \\ & & \nearrow & & & \\ \hline \end{array}$$ \meinecaption{1}{Variation chart for $x\mapsto mx+k$, with $m>0$.} \end{minipage} \hfill \begin{minipage}{.4\textwidth} $$\psset{unit=1pc} \psplot[linewidth=2pt, linecolor=brown,arrows={<->},algebraic]{-5}{5}{x/2+1} $$\meinecaption{1}{Graph of $x\mapsto mk+k$, $m>0$.} \end{minipage} \end{figure} \begin{figure}[h] \begin{minipage}{.4\textwidth} $$ \begin{array}{|c|ccccc|} \hline x& -\infty & & -\dfrac{k}{m} & & +\infty \\ \hline & & & & & \\ & & \searrow & & & \\ f(x)=mx+k & & & 0 & & \\ & & & & \searrow & \\ \hline \end{array}$$ \meinecaption{1}{Variation chart for $x\mapsto mx+k$, with $m<0$.} \end{minipage} \hfill \begin{minipage}{.4\textwidth} $$\psset{unit=1pc}\psplot[linewidth=2pt, linecolor=brown,arrows={<->},algebraic]{-5}{5}{-x/2+1} $$\meinecaption{1}{Graph of $x\mapsto mk+k$, $m<0$.} \end{minipage} \end{figure} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro}[Graph of the Absolute Value Function] Prove that the graph of the absolute value function $$\fun{{\bf AbsVal}}{x}{|x|}{\reals}{\reals} $$is convex. Prove that $x \mapsto |x|$ is an even function, decreasing for $x < 0$ and increasing for $x > 0$. Moreover, prove that $\im{{\bf AbsVal}} = [0;+\infty[$. \begin{answer} To prove that $x \mapsto |x|$ is convex, we use the triangle inequality theorem \ref{tri_ineq} and the fact that $|\lambda| = \lambda$, $|1 - \lambda| = 1 - \lambda$ for $\lambda \in [0; 1]$. We have $$\begin{array}{lll} {\bf AbsVal} (\lambda a + (1 - \lambda)b) & = & |\lambda a + (1 - \lambda)b| \\ & \leq & |\lambda a| + |(1 - \lambda)b| \\ & = & \lambda |a| + (1 - \lambda)|b| \\ & = & \lambda {\bf AbsVal} (a) + (1 - \lambda) {\bf AbsVal} (b), \end{array}$$whence $x \mapsto |x| $ is convex. As ${\bf AbsVal} (-x) = |-x| = |x| = {\bf AbsVal} (x)$, the absolute value function is an even function. For $ a < b< 0$, $$\dfrac{{\bf AbsVal}(b) - {\bf AbsVal} (a)}{b - a} = \dfrac{|b| - |a|}{b - a} = \dfrac{-b - (-a)}{b - a} = -1 < 0,$$ $x \mapsto |x|$ is a strictly decreasing function for $x < 0$. Similarly, for $0< a < b $ $$\dfrac{{\bf AbsVal}(b) - {\bf AbsVal}(a)}{b - a} = \dfrac{|b| - |a|}{b - a} = \dfrac{b - a}{b - a} = 1 > 0,$$ and so $x \mapsto |x|$ is a strictly increasing function for $x > 0$. \bigskip Also, assume that $y\in\im{{\bf AbsVal}}$. Then $\exists x\in \reals$ with $y = {\bf AbsVal} (x) = |x| $, which means that $y \geq 0$ and so $\im{{\bf AbsVal}} = [0;+\infty[$. \bigskip To obtain the graph of $x \mapsto |x|$ we graph the line $y = -x$ for $x < 0$ and the line $y = x$ for $x \geq 0$. \end{answer} \end{pro} \end{multicols} \section{The Square Function} In this section we study the shape of the graph of the square function $x\mapsto x^2$. \begin{thm}[Graph of the Square Function] \label{thm:graph_square_fun} The graph of the square function $$\fun{{\bf Sq}}{x}{x^2}{\reals}{\reals} $$is a convex curve which is strictly decreasing for $x < 0$ and strictly increasing for $x>0$. Moreover, $x \mapsto x^2$ is an even function and $\im{{\bf Sq}} = [0;+\infty[$. \end{thm} \begin{pf} As ${\bf Sq} (-x) = (-x)^2 = x^2 = {\bf Sq} (x)$, the square function is an even function. Now, for $a < b$ $$\dfrac{{\bf Sq} (b) - {\bf Sq} (a)}{b - a} = \dfrac{b^2 - a^2}{b - a} = b + a.$$ If $a < b < 0$ the sum $a + b$ is negative and $x \mapsto x^2$ is a strictly decreasing function. If $0 < a < b$ the sum $a + b$ is positive and $x \mapsto x^2$ is a strictly increasing function. To prove that $x \mapsto x^2$ is convex we observe that $${\bf Sq}(\lambda a + (1 - \lambda )b) \leq \lambda {\bf Sq}(a) + (1 - \lambda ){\bf Sq}(b) $$ $$ \begin{array}{lll} & \iff & \lambda ^2a^2 + 2\lambda(1 - \lambda)ab + (1 - \lambda)^2 b^2 \leq \lambda a^2 + (1 - \lambda)b^2 \\ & \iff & 0 \leq \lambda(1 - \lambda) a^2 - 2\lambda(1 - \lambda)ab + ((1 - \lambda) - (1 - \lambda)^2) b^2\\ & \iff & 0 \leq \lambda(1 - \lambda) a^2 - 2\lambda(1 - \lambda)ab + \lambda (1 - \lambda)b^2 \\ & \iff & 0 \leq \lambda(1 - \lambda) (a^2 - 2ab + b^2) \\ & \iff & 0 \leq \lambda(1 - \lambda) (a - b)^2. \end{array}$$This last inequality is clearly true for $\lambda \in [0; 1]$, establishing the claim. Also suppose that $y\in \im{{\bf Sq}}.$ Thus there is $x\in\reals $ such that $ {\bf Sq}(x) = y \implies x^2=y$. But the equation $y = x^2$ is solvable only for $y \geq 0$ and so only positive numbers appear as the image of $x \mapsto x^2$. Since for $x\in [0;+\infty[$ we have ${\bf Sq}(\sqrt{x})=x$, we conclude that $\im{{\bf Sq}} = [0;+\infty[$. The graph of the $x \mapsto x^2$ is called a {\em parabola}. We summarise this information by means of the following diagram. \begin{figure}[h] \begin{minipage}{.4\textwidth} $$ \begin{array}{|c|ccccc|} \hline x& -\infty & & 0 & & +\infty \\ \hline & & & & & \\ f(x) =x^2 & & \searrow & & \nearrow & \\ & & & 0 & & \\ \hline \end{array} $$ \meinecaption{1}{Variation chart for $x\mapsto x^2$.} \end{minipage} \hfill \begin{minipage}{.4\textwidth} $$\psset{unit=1pc} \psplot[algebraic,linewidth=2pt,linecolor=brown,arrows={<->}]{-2.1}{2.1}{x^2} $$ \meinecaption{1}{Graph of $x\mapsto x^2$.} \end{minipage} \end{figure} \end{pf} \section{Quadratic Functions}\label{sec:quadratic_functions} \begin{df}Let $a, b, c$ be real numbers, with $a\neq 0$. A function of the form $$\fun{f}{x}{ax^2+bx+c}{\reals}{\reals} $$ is called a {\em quadratic function} with leading coefficient $a$. \end{df} \begin{thm}\label{thm:behaviour_of_a_quadratic}Let $a\neq 0, b, c$ be real numbers and let $x \mapsto ax^2 + bx + c$ be a quadratic function. Then its graph is a parabola. If $a>0$ the parabola has a local minimum at $x=-\dfrac{b}{2a}$ and it is convex. If $a < 0$ the parabola has a local maximum at $x=-\dfrac{b}{2a}$ and it is concave. \end{thm} \begin{pf}Put $f(x) =ax^2+bx + c$. Completing squares, $$\begin{array}{lll} ax^2 + bx + c & = & a\left(x^2 + 2\dfrac{b}{2a}x + \dfrac{b^2}{4a^2}\right) + c-\dfrac{b^2}{4a} \\ & = & a\left(x + \dfrac{b}{2a}\right)^2 + \dfrac{4ac - b^2}{4a}, \end{array}$$ and hence this is a horizontal translation $-\dfrac{b}{2a}$ units and a vertical translation $\dfrac{4ac - b^2}{4a}$ units of the square function $x\mapsto x^2$ and so it follows from Theorems \ref{thm:graph_square_fun}, \ref{thm:translations} and \ref{thm:distortions}, that the graph of $f$ is a parabola. \bigskip Assume first that $a>0$. Then $f$ is convex, decreases if $x<-\dfrac{b}{2a}$ and increases if $x>-\dfrac{b}{2a}$, and so it has a minimum at $x=-\dfrac{b}{2a}$. The analysis of $-f$ yields the case for $a<0$, and the Theorem is proved. \end{pf}The information of Theorem \ref{thm:behaviour_of_a_quadratic} is summarised in the following tables. \begin{figure}[!h]\centering \begin{minipage}{.4\textwidth} $$ \begin{array}{|c|ccccc|} \hline x& -\infty & & -\dfrac{b}{2a} & & +\infty \\ \hline & & & & & \\ & & \searrow & & \nearrow & \\ f(x)=ax^2+bx+c & & & 0 & & \\ \hline \end{array}$$ \meinecaption{.25}{$x\mapsto ax^2+bx+c$, with $a>0$.} \end{minipage} \hfill \begin{minipage}{.4\textwidth} $$\psset{unit=1pc} \psplot[algebraic,linewidth=2pt,linecolor=brown,arrows={<->}]{-2.1}{2.1}{x^2-4} $$ \meinecaption{4}{Graph of $x\mapsto ax^2+bx+c$, $a>0$.} \end{minipage} \end{figure} \begin{figure}[h] \centering \begin{minipage}{.4\textwidth} $$ \begin{array}{|c|ccccc|} \hline x& -\infty & & -\dfrac{b}{2a} & & +\infty \\ \hline & & & & & \\ f(x)=ax^2+bx+c & & & 0 & & \\ & & \nearrow & & \searrow & \\ \hline \end{array}$$ \meinecaption{1}{$x\mapsto ax^2+bx+c$, with $a<0$.} \end{minipage} \hfill \begin{minipage}{.4\textwidth} $$\psset{unit=1pc} \psplot[algebraic,linewidth=2pt,linecolor=brown,arrows={<->}]{-2.1}{2.1}{-x^2-4} $$ \meinecaption{4}{Graph of $x\mapsto ax^2+bx+c$, $a<0$.} \end{minipage} \end{figure} \begin{df} The point $\left(-\dfrac{b}{2a}, \dfrac{4ac - b^2}{4a}\right)$ lies on the parabola and it is called the {\em vertex} of the parabola $y=ax^2+bx+c$. The quantity $b^2-4ac$ is called the {\em discriminant} of $ax^2+bx+c$. The equation $$ y = a\left(x + \dfrac{b}{2a}\right)^2 + \dfrac{4ac - b^2}{4a} $$ is called the {\em canonical equation of the parabola} $y=ax^2+bx+c$. \end{df} \begin{rem}The parabola $x\mapsto ax^2+bx+c$ is symmetric about the vertical line $x=-\dfrac{b}{2a}$ passing through its vertex. Notice that the axis of symmetry is parallel to the $y$-axis. If $(h, k)$ is the vertex of the parabola, by completing squares, the equation of a parabola with axis of symmetry parallel to the $y$-axis can be written in the form $y=a(x-h)^2+k$. Using Theorem \ref{thm:symmetry_ab_ba}, the equation of a parabola with axis of symmetry parallel to the $x$-axis can be written in the form $x=a(y-k)^2+h$. \end{rem} \begin{exa} A parabola with axis of symmetry parallel to the $y$-axis and vertex at $(1,2)$. If the parabola passes through $(3,4)$, find its equation. \end{exa} \begin{solu} The parabola has equation of the form $y=a(x-h)^2+k=a(x-1)^2+2$. Since when $x=3$ we get $y=4$, we have, $$4=a(3-1)^2+2 \implies 4=4a+2 \implies a = \dfrac{1}{2}. $$The equation sought is thus $$y = \dfrac{1}{2}\left(x-1\right)^2+2. $$ \end{solu} \subsection{Zeros and Quadratic Formula} \vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{4.5cm}$$ \psset{unit=1pc} \psaxes[labels=none, ticks=none](0,0)(-4,-4)(4,4) \rput(-1,1){\parabola[linewidth=2pt,linecolor=brown]{<->}(2,4)(0,0)} \rput{180}(1,-1){\parabola[linewidth=2pt,linecolor=blue]{<->}(2,4)(0,0)}$$ \meinecaption{1}{No real zeroes.} \label{fig:no_real_zero} \end{minipage} \hfill \begin{minipage}{4.5cm}$$ \psset{unit=1pc} \psaxes[labels=none, ticks=none](0,0)(-4,-4)(4,4) \rput(-1,0){\parabola[linewidth=2pt,linecolor=brown]{<->}(2,4)(0,0)} \rput{180}(1,0){\parabola[linewidth=2pt,linecolor=blue]{<->}(2,4)(0,0)}$$ \meinecaption{1}{One real zero.}\label{fig:one_real_zero} \end{minipage} \hfill \begin{minipage}{4.5cm}$$ \psset{unit=1pc} \psaxes[labels=none, ticks=none](0,0)(-4,-4)(4,4) \rput(-2,-2){\parabola[linewidth=2pt,linecolor=brown]{<->}(2,4)(0,0)} \rput{180}(2,2){\parabola[linewidth=2pt,linecolor=blue]{<->}(2,4)(0,0)}$$ \meinecaption{1}{Two real zeros.}\label{fig:two_real_zeros} \end{minipage} \end{figure} \begin{df} In the quadratic equation $ax^2+bx+c = 0$, $a\neq 0$, the quantity $b^2-4ac$ is called the {\em discriminant}. \end{df} \begin{cor}[Quadratic Formula] \label{cor:quadratic_formula} The roots of the equation $ ax^2+bx+c=0$ are given by the formula \begin{equation} ax^2 + bx + c = 0 \iff x = \dfrac{-b\pm\sqrt{b^2 -4ac}}{2a} \label{eq:quadratic_formula}\index{Formula!Quadratic}\end{equation} If $a\neq 0, b, c$ are real numbers and $b^2-4ac= 0$, the parabola $x\mapsto ax^2+bx+c$ is tangent to the $x$-axis and has one (repeated) real root. If $b^2-4ac > 0$ then the parabola has two distinct real roots. Finally, if $b^2-4ac < 0$ the parabola has two complex roots. \end{cor} \begin{pf} By Theorem \ref{thm:behaviour_of_a_quadratic} we have $$ax^2 + bx + c = a\left(x + \dfrac{b}{2a}\right)^2 + \dfrac{4ac - b^2}{4a}, $$ and so $$\begin{array}{lll} ax^2 + bx + c = 0 & \iff & \left(x + \dfrac{b}{2a}\right)^2 =\dfrac{b^2 -4ac}{4a^2} \\ & \iff & x + \dfrac{b}{2a} =\pm\dfrac{\sqrt{b^2 -4ac}}{2|a|} \\ & \iff & x = \dfrac{-b\pm\sqrt{b^2 -4ac}}{2a}, \\ \end{array}$$ where we have dropped the absolute values on the last line because the only effect of having $a<0$ is to change from $\pm$ to $\mp$. \bigskip If $b^2-4ac=0$ then the vertex of the parabola is at $\left(-\dfrac{b}{2a}, 0\right)$ on the $x$-axis, and so the parabola is tangent there. Also, $x=-\dfrac{b}{2a}$ would be the only root of this equation. This is illustrated in figure \ref{fig:one_real_zero}. \bigskip If $b^2-4ac>0$, then $\sqrt{b^2-4ac}$ is a real number $\neq 0$ and so $\dfrac{-b-\sqrt{b^2 -4ac}}{2a}$ and $\dfrac{-b+\sqrt{b^2 -4ac}}{2a}$ are distinct numbers. This is illustrated in figure \ref{fig:two_real_zeros}. \bigskip If $b^2-4ac<0$, then $\sqrt{b^2-4ac}$ is a complex number $\neq 0$ and so $\dfrac{-b-\sqrt{b^2 -4ac}}{2a}$ and $\dfrac{-b+\sqrt{b^2 -4ac}}{2a}$ are distinct complex numbers. This is illustrated in figure \ref{fig:no_real_zero}. \end{pf} \begin{rem} If a quadratic has real roots, then the vertex lies on a line crossing the midpoint between the roots. \end{rem} \vspace{2.5cm} \begin{figure}[!hptb] \begin{minipage}{5cm} $$\psset{algebraic=true, unit=.5pc} \psaxes[ labels=none, linewidth=1.5pt]{->}(0,0)(-10,-10)(10,10) \psplot[linewidth=2pt, linecolor=brown]{-1}{6}{x^2-5*x+3}$$ \vspace{2cm}\meinecaption{.25}{$y = x^2-5x + 3$} \label{fig:parabola4} \end{minipage} \hfill \begin{minipage}{5cm} $$\psset{algebraic=true, unit=.5pc} \psaxes[ labels=none, linewidth=1.5pt]{->}(0,0)(-10,-10)(10,10) \psplot[linewidth=2pt, linecolor=brown]{-1}{6}{abs(x^2-5*x+3)}$$ \vspace{2cm}\meinecaption{.25}{$y = |x^2-5x + 3|$} \label{fig:parabola4|f|} \end{minipage} \hfill \begin{minipage}{5cm} $$\psset{algebraic=true, unit=.5pc} \psaxes[ labels=none, linewidth=1.5pt]{->}(0,0)(-10,-10)(10,10) \psplot[linewidth=2pt, linecolor=brown]{-6}{6}{x^2-5*abs(x)+3}$$ \vspace{2cm}\meinecaption{.25}{$y = |x|^2-5|x| + 3$} \label{fig:parabola4f(|.|)} \end{minipage} \end{figure} \begin{exa}\label{exa:parabola4} Consider the quadratic function $f:\reals \to \reals$, $f(x) = x^2 - 5x + 3$. \begin{multicols}{2} \begin{enumerate} \item Write this parabola in canonical form and hence find the vertex of $f$. Determine the intervals of monotonicity of $f$ and its convexity. \item Find the $x$-intercepts and $y$-intercepts of $f$. \item Graph $y = f(x)$, $y = |f(x)|$, and $y = f(|x|)$. \item Determine the set of real numbers $x$ for which $f(x) > 0$. \end{enumerate}\end{multicols} \end{exa}\begin{solu} \begin{enumerate} \item Completing squares $$ y = x^2-5x+3 = \left(x-\dfrac{5}{2}\right)^2 -\dfrac{13}{4}. $$ From this the vertex is at $\left(\dfrac{5}{2}, -\dfrac{13}{4}\right)$. Since the leading coefficient of $f$ is positive, $f$ will be increasing for $x>\dfrac{5}{2}$ and it will be decreasing for $x < \dfrac{5}{2}$ and $f$ is concave for all real values of $x$. \item For $x = 0$, $f(0) = 0^2 - 5\cdot 0 + 3 = 3$, and hence $y = f(0) = 3$ is the $y$-intercept. By the quadratic formula, $$f(x) = 0 \iff x^2-5x+3 = 0 \iff x = \dfrac{-(-5)\pm\sqrt{(-5)^2-4(1)(3)}}{2(1)} = \dfrac{5\pm \sqrt{13}}{2}. $$ Observe that $\dfrac{5- \sqrt{13}}{2} \approx 0.697224362$ and $\dfrac{5+ \sqrt{13}}{2} \approx 4.302775638$. \item The graphs appear in figures \ref{fig:parabola4} through \ref{fig:parabola4f(|.|)}. \item From the graph in figure \ref{fig:parabola4}, $x^2 -5x + 3 > 0$ for values $x\in \oo{-\infty; \dfrac{5- \sqrt{13}}{2}}$ or $x\in \oo{\dfrac{5+ \sqrt{13}}{2}; +\infty }$. \end{enumerate} \end{solu} \begin{cor}\label{cor:sign_of_a_quadratic} If $a\neq 0, b, c$ are real numbers and if $b^2-4ac<0$, then $ax^2 + bx + c$ has the same sign as $a$. \end{cor} \begin{pf} Since $$ax^2 + bx + c = a\left(\left(x + \dfrac{b}{2a}\right)^2 + \dfrac{4ac - b^2}{4a^2}\right), $$and $4ac-b^2>0$, $\left(\left(x + \dfrac{b}{2a}\right)^2 + \dfrac{4ac - b^2}{4a^2}\right)>0$ and so $ax^2+bx+c$ has the same sign as $a$. \end{pf} \begin{exa} Prove that the quantity $q(x) = 2x^2+x+1$ is positive regardless of the value of $x$. \end{exa}\begin{solu} The discriminant is $1^2 - 4(2)(1) = -7 < 0$, hence the roots are complex. By Corollary \ref{cor:sign_of_a_quadratic}, since its leading coefficient is $2>0$, $q(x)> 0$ regardless of the value of $x$. Another way of seeing this is to complete squares and notice the inequality $$ 2x^2+x+1 = 2\left(x + \dfrac{1}{4}\right)^2 + \dfrac{7}{8} \geq \dfrac{7}{8}, $$ since $\left(x + \dfrac{1}{4}\right)^2$ being the square of a real number, is $\geq 0$. \end{solu} \bigskip By Corollary \ref{cor:quadratic_formula}, if $a\neq 0, b, c$ are real numbers and if $b^2-4ac\neq 0$ then the numbers $$r_1 = \dfrac{-b-\sqrt{b^2 -4ac}}{2a}\qquad \mathrm{and}\qquad r_2 = \dfrac{-b+\sqrt{b^2 -4ac}}{2a}$$ are distinct solutions of the equation $ax^2+bx+c=0$. Since $$ r_1+r_2 = -\dfrac{b}{a},\qquad \mathrm{and}\qquad r_1r_2 = \dfrac{c}{a}, $$any quadratic can be written in the form $$ ax^2+bx+c =a\left(x^2+\dfrac{bx}{a} + \dfrac{c}{a}\right) = a\left(x^2 - (r_1+r_2)x + r_1r_2\right) = a(x-r_1)(x-r_2). $$ We call $a(x-r_1)(x-r_2)$ a {\em factorisation} of the quadratic $ax^2+bx+c$.\index{factorisation!of a quadratic}\index{quadratic equation!factorisation of a} \begin{exa} A quadratic polynomial $p$ has $1 \pm \sqrt{5}$ as roots and it satisfies $p(1) = 2$. Find its equation. \end{exa}\begin{solu} Observe that the sum of the roots is $$r_1+r_2 = 1-\sqrt{5} + 1+\sqrt{5} = 2$$ and the product of the roots is $$r_1r_2 = (1-\sqrt{5})(1+\sqrt{5}) = 1-(\sqrt{5})^2 = 1-5=-4.\footnote{As a shortcut for this multiplication you may wish to recall the {\em difference of squares identity}: $(a-b)(a+b) = a^2-b^2$.}$$ Hence $p$ has the form $$p(x) = a\left(x^2-(r_1+r_2)x+r_1r_2\right) = a(x^2-2x-4). $$Since $$2= p(1)\implies 2 = a(1^2-2(1)-4) \implies a = -\dfrac{2}{5},$$ the polynomial sought is $$ p(x) = -\dfrac{2}{5}\left(x^2-2x-4\right). $$ \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Let $$R_1 = \{(x, y)\in\reals^2| y \geq x^2 - 1\},$$ $$R_2 = \{(x, y)\in\reals^2| x^2 + y^2 \leq 4 \},$$ $$R_3 = \{(x, y)\in\reals^2| y \leq -x^2 + 4\}.$$ Sketch the following regions. \begin{enumerate} \item $R_1 \setminus R_2$ \item $R_1 \cap R_3$ \item $R_2 \setminus R_1$ \item $R_1 \cap R_2$ \end{enumerate} \end{pro} \begin{pro} Write the following parabolas in canonical form, determine their vertices and graph them: (i) $y = x^2 + 6x + 9$, (ii) $y = x^2 + 12x + 35$, (iii) $y = (x - 3)(x + 5)$, (iv) $y = x(1 - x)$, (v) $y = 2x^2 - 12x + 23$, (vi) $y = 3x^2 - 2x + \frac{8}{9}$, (vii) $y = \frac{1}{5}x^2 + 2x + 13$ \\ \begin{answer} (i) $y = (x + 3)^2$ vertex at $(-3, 0)$, (ii) $y = (x + 6)^2 - 1$ vertex at $(-6, -1)$, (iii) $y = (x + 1)^2 - 16$, vertex at $(-1, -16)$ (iv) $y = -(x - \frac{1}{2})^2 + \frac{1}{4}$, vertex at $(\frac{1}{2}, \frac{1}{4})$ (v) $y = 2(x - 3)^2 + 5$, vertex at $(3, 5)$, (vi) $3(x - \frac{1}{3})^2 + \frac{5}{9}$, vertex at $(\frac{1}{3}, \frac{5}{9})$ (vii) $y = \frac{1}{5}(x + 5)^2 + 8$, vertex at $(-5, 8)$ \end{answer} \end{pro} \begin{pro} Find the vertex of the parabola $y = (3x - 9)^2 - 9$. \begin{answer} $(3, - 9)$ \end{answer} \end{pro} \begin{pro} Find the equation of the parabola whose axis of symmetry is parallel to the $y$-axis, with vertex at $(0, -1)$ and passing through $(3, 17)$. \begin{answer} $y = 2x^2 - 1$ \end{answer} \end{pro} \begin{pro} Find the equation of the parabola having roots at $x = -3$ and $x = 4$ and passing through $(0, 24).$ \begin{answer} $y = -2(x + 3)(x - 4)$ \end{answer} \end{pro} \begin{pro} Let $0 \leq a, b, c \leq 1.$ Prove that at least one of the products $a(1 - b), b(1 - c), c(1 - a)$ is smaller than or equal to $\frac{1}{4}$. \begin{answer} Observe that $\dis{x(1 - x) = \frac{1}{4} - (x - \frac{1}{2})^2 \leq \frac{1}{4}}$ and that for $x \in [0, 1], 0 \leq x(1 - x).$ Thus if all these products are $> \frac{1}{4}$ we obtain $\dis{\frac{1}{4^3} < a(1 - b)b(1 - c)c(1 - a) = a(1 - a)b(1 - b)c(1 - c) \leq \frac{1}{4^3}}$, a contradiction. Thus one of the products must be $\leq \frac{1}{4}$. \end{answer} \end{pro} \begin{pro} An apartment building has $30$ units. If all the units are inhabited, the rent for each unit is $\$ 700$ per unit. For every empty unit, management increases the rent of the remaining tenants by $\$ 25$. What will be the profit $P(x)$ that management gains when $x$ units are empty? What is the maximum profit? \begin{answer} $P(x) = 21025 - 25(x - 1)^2$; \ \ $\$ 21025$ \end{answer} \end{pro} \begin{pro} Find all real solutions to $|x^2 - 2x| = |x^2 + 1|$. \begin{answer} We have $$\begin{array}{lll} |x^2 - 2x| = |x^2 + 1| & \iff & (x^2 -2x = x^2 + 1) \ \mathrm{or}\ (x^2 + 2x = -x^2 - 1) \\ & \iff & (-2x - 1= 0) \ \mathrm{or}\ (2x^2 + 2x + 1 = 0) \\ & \iff & \left(x = -\dfrac{1}{2}\right) \ \mathrm{or}\ \left(x = -\dfrac{1}{2} \pm \dfrac{i}{2}\right),\end{array}$$ whence the solution set is $ \left\{-\dfrac{1}{2}\right\}.$ \end{answer} \end{pro} \begin{pro} Find all the real solutions to $$(x^2 + 2x -3)^2 = 2.$$ \begin{answer} We have $$\begin{array}{lll} (x^2 + 2x - 3)^2 = 2 & \iff & (x^2 + 2x - 3 = \sqrt{2}) \ \mathrm{or}\ (x^2 + 2x - 3= -\sqrt{2}) \\ & \iff & (x^2 + 2x - 3 - \sqrt{2} = 0) \ \mathrm{or}\ (x^2 + 2x - 3 + \sqrt{2} = 0) \\ & \iff & \left(x = \dfrac{-2\pm \sqrt{4 - 4(-3 - \sqrt{2})}}{2}\right)\\ & & \qquad \ \mathrm{or}\ \left(x = \dfrac{-2\pm \sqrt{4 - 4(-3 + \sqrt{2})}}{2}\right) \\ & \iff & \left(x = \dfrac{-2\pm \sqrt{16 + 4\sqrt{2}}}{2}\right) \\ & & \qquad\ \mathrm{or}\ \left(x = \dfrac{-2\pm \sqrt{16 - 4\sqrt{2}}}{2}\right) \\ & \iff & (x = -1 \pm \sqrt{4 + \sqrt{2}}) \ \mathrm{or}\ (x = -1 \pm \sqrt{4 - \sqrt{2}}).\end{array}$$ Since each of $4\pm \sqrt{2} > 0$, all four solutions found are real. The set of solutions is $ \left\{-1 \pm \sqrt{4\pm \sqrt{2}}\right\}.$ \end{answer} \end{pro} \begin{pro} Solve $x^3-x^2-9x+9 = 0$. \begin{answer} $$\begin{array}{lll} x^3-x^2-9x+9 = 0 & \iff & x^2(x-1)-9(x-1) = 0 \\ & \iff & (x - 1)(x^2 - 9) = 0 \\ & \iff & (x - 1)(x - 3)(x + 3)=0 \\ & \iff & x\in\{-3,1,3\}.\end{array}$$ \end{answer} \end{pro} \begin{pro} Solve $x^3-2x^2-11x+12 = 0$. \begin{answer} $$\begin{array}{lll} x^3-2x^2-11x+12 = 0 & \iff & x^3 - x^2 - x^2 + x - 12x + 12 = 0 \\ & \iff & x^2(x - 1) - x(x - 1) - 12(x - 1) = 0 \\ & \iff & (x - 1)(x^2 - x - 12)=0 \\ & \iff & (x - 1)(x + 3)(x - 4)=0 \\ & \iff & x\in\{-3,1,4\}.\end{array}$$ \end{answer} \end{pro} \begin{pro} Find all real solutions to $x^3 - 1 = 0$. \begin{answer}$x^3 - 1 = (x - 1)(x^2 + x + 1)$. If $x \neq 1$, the two solutions to $x^2 + x + 1 = 0$ can be obtained using the quadratic formula, getting $\dis{x = 1/2 \pm i\sqrt{3}/2}$. There is only one real solution, namely $x = 1.$ \end{answer} \end{pro} \begin{pro} A parabola with axis of symmetry parallel to the $x$-axis and vertex at $(1,2)$. If the parabola passes through $(3,4)$, find its equation. \begin{answer} The parabola has equation of the form $x=a(y-k)^2+h=a(y-2)^2+1$. Since when $x=3$ we get $y=4$, we have, $$3=a(4-2)^2+1 \implies 3=4a+1 \implies a = \dfrac{1}{2}. $$The equation sought is thus $$x = \dfrac{1}{2}\left(y-2\right)^2+1. $$ \end{answer} \end{pro} \begin{pro} Solve $9 + x^{-4} = 10x^{-2}.$ \begin{answer}Observe that $$x^{-4} - 10x^{-2} + 9 = (x^{-2} - 9)(x^{-2} - 1).$$ Thus $\dfrac{1}{x^2} = 9$ and $\dfrac{1}{x^2} =1$, whence $x = \pm \frac{1}{3}$ and $x = \pm 1.$ \end{answer} \end{pro} \begin{pro} Find all the real values of the parameter $t$ for which the equation in $x$ $$t^2x - 3t = 81x - 27 $$has a solution. \begin{answer} Rearranging, \begin{equation}\label{eq:equivalent_equation} (t^2 - 81)x = 3(t - 9) \implies (t - 9)(t + 9)x = 3(t - 9). \end{equation} If $t = 9$, (\ref{eq:equivalent_equation}) becomes $0 = 0$, which will be true for all values of $x$. If $t = -9$, (\ref{eq:equivalent_equation}) becomes $0 = -54$, which is clearly nonsense. If $t\in\reals \setminus \{-9,9\}$, then $$ x = \dfrac{3}{t + 9} $$is the unique solution to the equation. \end{answer} \end{pro} \begin{pro} The sum of two positive numbers is $50$. Find the largest value of their product. \begin{answer} Let $x$ and $50 - x$ be the numbers. We seek to maximise the product $P(x) = x(50 - x).$ But $P(x) = 50x - x^2 = -(x^2 - 50x) = -(x^2 - 50x + 625) + 625 = 625 - (x - 25)^2.$ We deduce that $P(x) \leq 625,$ as the square of any real number is always positive. The maximum product is thus 625 occurring when $x = 25$. \end{answer} \end{pro} \begin{pro} Of all rectangles having perimeter $20$ shew that the square has the largest area. \begin{answer} If $b, h$ are the base and height, respectively, of the rectangle, then we have $20 = 2b + 2h$ or $10 = b + h.$ The area of the rectangle is then $A(h) = bh = h(10 - h) = 10h - h^2 = 25 - (h - 5)^2.$ This shows that $A(h) \leq 25$, and equality occurs when $h = 5.$ In this case $b = 10 - h = 5.$ The height is the same as the base, and so the rectangle yielding maximum area is a square. \end{answer} \end{pro} \begin{pro} An orchard currently has $25$ trees, which produce $600$ fruits each. It is known that for each additional tree planted, the production of each tree diminishes by $15$ fruits. Find: \begin{enumerate} \item the current fruit production of the orchard, \item a formula for the production obtained from each tree upon planting $x$ more trees, \item a formula $P(x)$ for the production obtained from the orchard upon planting $x$ more trees. \item How many trees should be planted in order to yield maximum production?\end{enumerate} \begin{answer} \begin{enumerate} \item The current production is $25\times 600 = 15 000$ fruits. \\ \item If $x$ more trees are planted, the production of each tree will be $600 - 15x.$ \\ \item Let $P(x)$ be the total production after planting $x$ more trees. Then $P(x) = (25 + x)(600 - 15x) = -15x^2 + 225x + 15 000.$ A good function modelling this problem is $$\fun{P}{x}{-15x^2 + 225x + 15000 }{\{x\in\naturals|x \geq 25\}}{\naturals}.$$ This model assumes that the amount of trees is never fewer than $25$. \item We maximise $P(x) = -15x^2 + 225x + 15000 = 15000 -15(x^2 - 15x) = 15843.75 - 15(x - 7.5)^2$. The production is maximised if either $7$ or $8$ more trees are added, in which case the production will be $ 15843.75 - 15(7 - 7.5)^2 = 15840$ fruits. \end{enumerate} \end{answer} \end{pro} \end{multicols} \section{$x\mapsto x^{2n+2}$, $n\in \naturals$} The graphs of $y = x^2$, $y = x^4$, $y = x^6$, etc., resemble one other. For $-1 \leq x \leq 1$, the higher the exponent, the flatter the graph (closer to the $x$-axis) will be, since $$|x|<1 \implies \cdots < x^6 < x^4 < x^2 < 1. $$ For $|x| \geq 1$, the higher the exponent, the steeper the graph will be since $$|x|>1 \implies \cdots > x^6 > x^4 > x^2 > 1. $$ We collect this information in the following theorem, of which we omit the proof. \begin{thm}\label{thm:power_functions} Let $n \geq 2$ be an integer and $f(x) = x^n$. Then if $n$ is even, $f$ is convex, $f$ is decreasing for $x<0$, and $f$ is increasing for $x>0$. Also, $f(-\infty) = f(+\infty) = +\infty$. \end{thm} \vspace*{1cm} \begin{figure}[!h] \begin{minipage}{.2\textwidth} $$\psset{algebraic=true, unit=1pc} \psaxes[ labels=none]{->}(0,0)(-2.5,-2.5)(2.5,2.5) \psplot[linewidth=2pt, linecolor=brown]{-1.2}{1.2}{x^2} $$ \meinecaption{2}{$y = x^2$.} \label{fig:xmapsx^2} \end{minipage}\hfill \begin{minipage}{.2\textwidth} $$\psset{algebraic=true, unit=1pc} \psaxes[ labels=none]{->}(0,0)(-2.5,-2.5)(2.5,2.5) \psplot[linewidth=2pt, linecolor=brown]{-1.2}{1.2}{x^4} $$ \meinecaption{2}{$y = x^4$.} \label{fig:xmapsx^4} \end{minipage}\hfill \begin{minipage}{.2\textwidth} $$\psset{algebraic=true, unit=1pc} \psaxes[ labels=none]{->}(0,0)(-2.5,-2.5)(2.5,2.5) \psplot[linewidth=2pt, linecolor=brown]{-1.2}{1.2}{x^6} $$ \meinecaption{2}{$y = x^6$.} \label{fig:xmapsx^6} \end{minipage} \hfill \begin{minipage}{.2\textwidth} $$ \begin{array}{|c|ccccc|} \hline x& -\infty & & 0 & & +\infty \\ \hline & & & & & \\ & & \searrow & & \nearrow & \\ f(x) = x^n & & & 0 & & \\ \hline \end{array}$$ \meinecaption{.25}{$x\mapsto x^n$, with $n>0$ integer and even.} \end{minipage} \hfill \end{figure} \section{The Cubic Function} We now deduce properties for the cube function. \begin{thm}[Graph of the Cubic Function]\label{thm:cubic_graph} The graph of the cubic function $$\fun{{\bf Cube}}{x}{x^3}{\reals}{\reals} $$is concave for $x < 0$ and convex for $x > 0$. $x \mapsto x^3$ is an increasing odd function and $\im{{\bf Cube}} = \reals$. \end{thm} \begin{pf} Consider $${\bf Cube}(\lambda a + (1 - \lambda )b) - \lambda {\bf Cube}(a) - (1 - \lambda ){\bf Cube}(b), $$which is equivalent to $$(\lambda a + (1 - \lambda )b)^3 - \lambda a^3 - (1 - \lambda)b^3, $$which is equivalent to $$(\lambda^3 - \lambda)a^3 + ((1 - \lambda)^3 - (1 - \lambda))b^3 + 3\lambda (1 - \lambda)ab(\lambda a + (1 - \lambda )b),$$ which is equivalent to $$-(1 - \lambda)(1 + \lambda)\lambda a^3 + (-\lambda^3 + 3\lambda ^2 - 2\lambda)b^3 + 3\lambda (1 - \lambda)ab(\lambda a + (1 - \lambda )b),$$ which in turn is equivalent to $$(1 - \lambda)\lambda (-(1 + \lambda)a^3 + (\lambda - 2)b^3 + 3ab(\lambda a + (1 - \lambda)b)).$$ This last expression factorises as $$-\lambda(1 - \lambda)(a-b)^2(\lambda (a - b) + 2b + a).$$Since $\lambda(1 - \lambda)(a-b)^2 \geq 0$ for $\lambda \in [0; 1]$, $${\bf Cube}(\lambda a + (1 - \lambda )b) - \lambda {\bf Cube}(a) - (1 - \lambda ){\bf Cube}(b)$$has the same sign as $$-(\lambda (a - b) + 2b + a) = -(\lambda a + (1 - \lambda)b + b + a).$$If $(a, b)\in]0;+\infty[^2$ then $\lambda a + (1 - \lambda)b \geq 0$ by lemma \ref{lem:betwixt} and so $$-(\lambda a + (1 - \lambda)b + b + a) \leq 0$$ meaning that ${\bf Cube}$ is convex for $x \geq 0.$ Similarly, if $(a, b)\in]-\infty; 0[^2$ then $$-(\lambda a + (1 - \lambda)b + b + a) \geq 0$$ and so $x \mapsto x^3$ is concave for $x \geq 0.$ This proves the claim. \bigskip As ${\bf Cube} (-x) = (-x)^3 = -x^3 = -{\bf Cube} (x)$, the cubic function is an odd function. Since for $a < b$ $$\dfrac{{\bf Cube} (b) - {\bf Cube} (a)}{b - a} = \dfrac{b^3 - a^3}{b - a} = b^2 + ab + b^2 = \left(b + \dfrac{a}{2}\right)^2 + \dfrac{3a^2}{4} > 0,$$ ${\bf Cube}$ is a strictly increasing function. Also if $y\in \im{{\bf Cube}}$ then there is $x\in \reals$ such that $x^3 = {\bf Cube} (x) = y$. The equation $y = x^3$ has a solution for every $y\in \reals$ and so $\im{{\bf Cube}} = \reals$. The graph of $x \mapsto x^3$ appears in figure \ref{fig:xmapsx^3}. \end{pf} \vspace*{1cm} \begin{figure}[!h] \begin{minipage}{.2\textwidth} $$ \begin{array}{|c|ccccc|} \hline x& -\infty & & 0 & & +\infty \\ \hline & & & & & \\ & & & & \nearrow & \\ f(x) = x^n & & & 0 & & \\ & & \nearrow & & & \\ \hline \end{array}$$ \meinecaption{.25}{$x\mapsto x^n$, with $n\geq 3$ odd.} \end{minipage} \hfill \begin{minipage}{.2\textwidth} $$\psset{algebraic=true, unit=1pc} \psaxes[ labels=none]{->}(0,0)(-2.5,-2.5)(2.5,2.5) \psplot[linewidth=2pt, linecolor=brown]{-1.2}{1.2}{x^3} $$ \meinecaption{2}{$y = x^3$.} \label{fig:xmapsx^3} \end{minipage}\hfill \begin{minipage}{.2\textwidth} $$\psset{algebraic=true, unit=1pc} \psaxes[ labels=none]{->}(0,0)(-2.5,-2.5)(2.5,2.5) \psplot[linewidth=2pt, linecolor=brown]{-1.2}{1.2}{x^5} $$ \meinecaption{2}{$y = x^5$.} \label{fig:xmapsx^5} \end{minipage}\hfill \begin{minipage}{.2\textwidth} $$\psset{algebraic=true, unit=1pc} \psaxes[ labels=none]{->}(0,0)(-2.5,-2.5)(2.5,2.5) \psplot[linewidth=2pt, linecolor=brown]{-1.2}{1.2}{x^7} $$ \meinecaption{2}{$y = x^7$.} \label{fig:xmapsx^7} \end{minipage} \hfill \end{figure} \section{$x\mapsto x^{2n+3}$, $n\in \naturals$} The graphs of $y = x^3$, $y = x^5$, $y = x^7$, etc., resemble one other. For $-1 \leq x \leq 1$, the higher the exponent, the flatter the graph (closer to the $x$-axis) will be, since $$|x|<1 \implies \cdots < |x^7| < |x^5| < |x^3| < 1. $$ For $|x| \geq 1$, the higher the exponent, the steeper the graph will be since $$|x|>1 \implies \cdots > |x^7| > |x^5| > |x^3| > 1. $$ We collect this information in the following theorem, of which we omit the proof. \begin{thm}\label{thm:power_functions2} Let $n \geq 3$ be an integer and $f(x) = x^n$. Then if $n$ is odd, $f$ is increasing, $f$ is concave for $x<0$, and $f$ is convex for $x>0$. Also, $f(-\infty)= -\infty$ and $ f(+\infty) = +\infty$. \end{thm} \section{Graphs of Polynomials} Recall that the zeroes of a polynomial $p(x)\in \reals [x]$ are the solutions to the equation $p(x)=0$, and that the polynomial is said to {\em split in $\reals$} if all the solutions to the equation $p(x)=0$ are real. \bigskip In this section we study how to graph polynomials that split in $\reals$, that is, we study how to graph polynomials of the form $$p(x) = a(x-r_1)^{m_1}(x-r_2)^{m_2} \cdots (x-r_k)^{m_k}, $$where $a\in \reals \setminus \{0\}$ and the $r_i$ are real numbers and the $m_i \geq 1$ are integers. \bigskip To graph such polynomials, we must investigate the {\em global behaviour} of the polynomial, that is, what happens as $x\to \pm \infty$, and we must also investigate the {\em local behaviour} around each of the roots $r_i$. \bigskip We start with the following theorem, which we will state without proof. \begin{thm} A polynomial function $x\mapsto p(x)$ is an everywhere continuous function. \end{thm} \begin{thm}\label{thm:behaviour_polynomials_at_infty} Let $p(x) = a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0 \ \ \ a_n \neq 0$, be a polynomial with real number coefficients. Then $$p(-\infty) = (\sgn{a_n})(-1)^n\infty,\qquad p(+\infty) = (\sgn{a_n})\infty.$$ Thus a polynomial of odd degree will have opposite signs for values of large magnitude and different sign, and a polynomial of even degree will have the same sign for values of large magnitude and different sign. \end{thm} \begin{pf} If $x\neq 0$ then $$p(x) = a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0 = a_nx^n\left(1 + \dfrac{a_{n - 1}}{x} + \cdots + \dfrac{a_1}{x^{n-1}} + \dfrac{a_0}{x^n}\right) \sim a_nx^n,$$since as $\tends{x}{\pm \infty}$, the quantity in parenthesis tends to $1$ and so the eventual sign of $p(x)$ is determined by $a_nx^n$, which gives the result. \end{pf} We now state the basic result that we will use to graph polynomials. \begin{thm}\label{thm:graph_splitting_polynomials} Let $a\neq 0$ and the $r_i$ are real numbers and the $m_i$ be positive integers. Then the graph of the polynomial $$p(x) = a(x-r_1)^{m_1}(x-r_2)^{m_2} \cdots (x-r_k)^{m_k}, $$ \begin{itemize} \item crosses the $x$-axis at $x=r_i$ if $m_i$ is odd. \item is tangent to the $x$-axis at $x=r_i$ if $m_i$ is even. \item has a convexity change at $x=r_i$ if $m_i\geq 3$ and $m_i$ is odd. \end{itemize} \end{thm} \begin{pf} Since the local behaviour of $p(x)$ is that of $c(x-r_i)^{m_i}$ (where $c$ is a real number constant) near $r_i$, the theorem follows at once from our work in section \ref{sec:power_functions}. \end{pf} \vspace*{3cm} \begin{figure}[!hptb] \begin{minipage}{.2\textwidth} $$\psset{algebraic=true, unit=.8pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.4pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psplot[linewidth=2pt, linecolor=brown,arrows={<->}]{-2.5}{1.7}{(x+2)*x*(x-1)} $$ \meinecaption{2}{Example \ref{exa:polygraph1}.} \label{y=(x+2)x(x-1)} \end{minipage} \hfill \begin{minipage}{.2\textwidth} $$\psset{algebraic=true, unit=.8pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.4pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psplot[linewidth=2pt, linecolor=brown,arrows={<->}]{-2.5}{.75}{(x+2)^3*x^2*(1-2*x)} $$ \meinecaption{2}{Example \ref{exa:polygraph2}.} \label{y=(x+2)^3x^2(1-2x)} \end{minipage} \hfill \begin{minipage}{.2\textwidth} $$\psset{algebraic=true, unit=.8pc}\psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.4pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psplot[linewidth=2pt, linecolor=brown,arrows={<->}]{-2.45}{1.5}{(x+2)^2*x*(1-x)^2} $$ \meinecaption{2}{Example \ref{exa:polygraph3}.} \label{y=(x+2)^2x(1-x)^2} \end{minipage} \hfill \begin{minipage}{.2\textwidth} $$\psset{unit=.8pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.4pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psplot[linewidth=1.5pt, linecolor=brown, algebraic=true, arrows={<->}]{-2.12}{3.1}{x*(x-2)^3*(x+2)/3} \psdots[dotscale=1.4,dotstyle=*](0,0)(2,0)(-2,0)(1,-1) $$ \meinecaption{2}{Example \ref{exa:quintic-f06},.} \label{fig:quintic-f06} \end{minipage} \hfill \end{figure} \begin{exa}\label{exa:polygraph1} Make a rough sketch of the graph of $y = (x+2)x(x-1)$. Determine where it achieves its local extrema and their values. Determine where it changes convexity. \end{exa}\begin{solu}We have $p(x) = (x+2)x(x-1)\sim (x)\cdot x(x) = x^3$, as $\tends{x}{+\infty}$. Hence $p(-\infty) = (-\infty)^3 = -\infty$ and $p(+\infty) = (+\infty)^3 = +\infty$. This means that for large negative values of $x$ the graph will be on the negative side of the $y$-axis and that for large positive values of $x$ the graph will be on the positive side of the $y$-axis. By Theorem \ref{thm:graph_splitting_polynomials}, the graph crosses the $x$-axis at $x=-2$, $x=0$, and $x=1$. The graph is shewn in figure \ref{y=(x+2)x(x-1)}. \end{solu} \begin{exa}\label{exa:polygraph2} Make a rough sketch of the graph of $y=(x+2)^3x^2(1-2x)$. \end{exa}\begin{solu} We have $(x+2)^3x^2(1-2x) \sim x^3\cdot x^2 (-2x) = -2x^6$. Hence if $p(x) = (x+2)^3x^2(1-2x)$ then $p(-\infty) = -2(-\infty)^6 = -\infty$ and $p(+\infty) = -2(+\infty)^6 = -\infty$, which means that for both large positive and negative values of $x$ the graph will be on the negative side of the $y$-axis. By Theorem \ref{thm:graph_splitting_polynomials}, in a neighbourhood of $x=-2$, $p(x)\sim 20(x+2)^3$, so the graph crosses the $x$-axis changing convexity at $x=-2$. In a neighbourhood of $0$, $p(x)\sim 8x^2$ and the graph is tangent to the $x$-axis at $x=0$. In a neighbourhood of $x=\dfrac{1}{2}$, $p(x)\sim \dfrac{25}{16}(1-2x)$, and so the graph crosses the $x$-axis at $x= \frac{1}{2}$. The graph is shewn in figure \ref{y=(x+2)^3x^2(1-2x)}. \end{solu} \begin{exa}\label{exa:polygraph3} Make a rough sketch of the graph of $y=(x+2)^2x(1-x)^2$. \end{exa}\begin{solu} The dominant term of $(x+2)^2x(1-x)^2$ is $x^2\cdot x (-x)^2 = x^5$. Hence if $p(x) = (x+2)^2x(1-x)^2$ then $p(-\infty) = (-\infty)^5 = -\infty$ and $p(+\infty) = (+\infty)^5 = +\infty$, which means that for large negative values of $x$ the graph will be on the negative side of the $y$-axis and for large positive values of $x$ the graph will be on the positive side of the $y$-axis. By Theorem \ref{thm:graph_splitting_polynomials}, the graph crosses the $x$-axis changing convexity at $x=-2$, it is tangent to the $x$-axis at $x=0$ and it crosses the $x$-axis at $x= \frac{1}{2}$. The graph is shewn in figure \ref{y=(x+2)^2x(1-x)^2}. \end{solu} \begin{exa}\label{exa:quintic-f06}, The polynomial in figure \ref{fig:quintic-f06}, has degree $5$. You may assume that the points marked below with a dot through which the polynomial passes have have integer coordinates. You may also assume that the graph of the polynomial changes concavity at $x=2$.\\ \noindent \begin{enumerate} \item Determine $p(1)$. \item Find the general formula for $p(x)$. \item Determine $p(3)$. \end{enumerate} \end{exa} \begin{solu} \noindent \begin{enumerate} \item From the graph $p(1)=-1$. \item $p$ has roots at $x=-2$, $x=0$, $x=+2$. Moreover, $p$ has a zero of multiplicity at $x=2$, and so it must have an equation of the form $p(x)=A(x+2)(x)(x-2)^3$. Now $$ -1=p(1) = A(1+2)(1)(1-2)^3\implies A=\dfrac{1}{3} \implies p(x)=\dfrac{(x+2)(x)(x-2)^3}{3}. $$ \item $p(3)=\dfrac{(3+2)(3)(3-2)^3}{3}=5$. \end{enumerate} \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro}Make a rough sketch of the following curves. \begin{enumerate} \item $y = x^3-x$ \item $y = x^3-x^2$ \item $y = x^2(x-1)(x+1)$ \item $y = x(x-1)^2(x+1)^2$ \item $y = x^3(x-1)(x+1)$ \item $y = -x^2(x-1)^2(x+1)^3$ \item $y = x^4-8x^2+16$ \end{enumerate} \end{pro} \begin{pro}\label{pro:quartic1} The polynomial in figure \ref{fig:quartic1} has degree $4$. \begin{enumerate} \item Determine $p(0)$. \item Find the equation of $p(x)$. \item Find $p(-3)$. \item Find $p(2)$. \end{enumerate} \end{pro} \vspace{1cm} \begin{figurehere} $$\psset{unit=.7pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.4pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psplot[linewidth=1.5pt, linecolor=brown]{-2.7}{2.1}{ x 2 add x 1 add mul x -1 add 2 exp mul 0.5 mul} \psdots[dotscale=1,dotstyle=*](0,1)(1,0)(-1,0)(-2,0) $$\meinecaption{2}{Problem \ref{pro:quartic1}.} \label{fig:quartic1} \end{figurehere} \end{multicols} \section{Polynomials} \subsection{Roots} In sections \ref{sec:affine_functions} and \ref{sec:quadratic_functions} we learned how to find the roots of equations (in the unknown $x$) of the type $ax+b = 0$ and $ax^2+bx+c = 0$, respectively. We would like to see what can be done for equations where the power of $x$ is higher than $2$. We recall that \begin{df}\label{df:splitting_ring} If all the roots of a polynomial are in $\integers$ (integer roots), then we say that the {\em polynomial splits or factors over $\integers$}. If all the roots of a polynomial are in $\rationals$ (rational roots), then we say that the {\em polynomial splits or factors over $\rationals$}. If all the roots of a polynomial are in $\complex$ (complex roots), then we say that the {\em polynomial splits (factors) over $\complex$}. \index{polynomial!splitting} \end{df} \begin{rem} Since $\integers \subset \rationals \subset\reals \subset \complex$, any polynomial splitting on a smaller set immediately splits over a larger set. \end{rem} \begin{exa} The polynomial $l(x)= x^2-1 = (x-1)(x+1)$ splits over $\integers$. The polynomial $p(x) = 4x^2-1=(2x-1)(2x+1)$ splits over $\rationals$ but not over $\integers$. The polynomial $q(x) = x^2-2=(x-\sqrt{2})(x+\sqrt{2})$ splits over $\reals$ but not over $\rationals$. The polynomial $r(x) = x^2 + 1 = (x-i)(x+i)$ splits over $\complex$ but not over $\reals$. Here $i = \sqrt{-1}$ is the imaginary unit. \end{exa} \subsection{Ruffini's Factor Theorem} \begin{thm}[Division Algorithm] If the polynomial $p(x)$ is divided by $a(x)$ then there exist polynomials $q(x), r(x)$ with \begin{equation} p(x) = a(x)q(x) + r(x)\end{equation} and $0 \leq$ degree $r(x) <$ degree $a(x)$. \end{thm} \begin{exa}If $x^5 + x^4 + 1$ is divided by $x^2 + 1$ we obtain $$x^5 + x^4 + 1 = (x^3 + x^2 - x - 1)(x^2 + 1) + x + 2,$$and so the quotient is $q(x) = x^3 + x^2 - x - 1$ and the remainder is $r(x) = x + 2.$\end{exa} \begin{exa} Find the remainder when $(x + 3)^5 + (x + 2)^8 + (5x + 9)^{1997}$ is divided by $x + 2.$ \end{exa} \begin{solu} As we are dividing by a polynomial of degree $1$, the remainder is a polynomial of degree 0, that is, a constant. Therefore, there is a polynomial $q(x)$ and a constant $r$ with $$ (x + 3)^5 + (x + 2)^8 + (5x + 9)^{1997} = q(x)(x + 2) + r $$Letting $x = - 2$ we obtain $$(-2 + 3)^5 + (-2 + 2)^8 + (5(-2) + 9)^{1997} = q(-2)(-2 + 2) + r = r. $$As the sinistral side is 0 we deduce that the remainder $r = 0$. \end{solu} \begin{exa} A polynomial leaves remainder $-2$ upon division by $x - 1$ and remainder $-4$ upon division by $x + 2.$ Find the remainder when this polynomial is divided by $x^2 + x - 2$. \end{exa} \begin{solu} From the given information, there exist polynomials $q_1(x), q_2(x)$ with $p(x) = q_1(x)(x - 1) - 2$ and $p(x) = q_2(x)(x + 2) - 4.$ Thus $p(1) = -2$ and $p(-2) = - 4.$ As $x^2 + x - 2 = (x - 1)(x + 2)$ is a polynomial of degree 2, the remainder $r(x)$ upon dividing $p(x)$ by $x^2 + x - 1$ is of degree 1 or smaller, that is $r(x) = ax + b$ for some constants $a, b$ which we must determine. By the Division Algorithm, $$p(x) = q(x)(x^2 + x - 1) + ax + b.$$Hence $$- 2 = p(1) = a + b$$and $$- 4 = p(-2) = -2a + b.$$From these equations we deduce that $a = 2/3, b = -8/3.$ The remainder sought is $$r(x) = \frac{2}{3}x - \frac{8}{3}.$$ \end{solu} \begin{thm}[Ruffini's Factor Theorem] The polynomial $p(x)$ is divisible by $x - a$ if and only if $p(a) = 0.$ Thus if $p$ is a polynomial of degree $n$, then $p(a) = 0$ if and only if $p(x) = (x-a)q(x)$ for some polynomial $q$ of degree $n-1$. \label{thm:ruffini}\index{Theorem!Ruffini}\end{thm} \begin{pf} As $x - a$ is a polynomial of degree $1$, the remainder after diving $p(x)$ by $x - a$ is a polynomial of degree $0$, that is, a constant. Therefore $$p(x) = q(x)(x - a) + r.$$From this we gather that $p(a) = q(a)(a - a) + r = r,$ from where the theorem easily follows. \end{pf} \begin{exa} Find the value of $a$ so that the polynomial $$t(x) = x^3 - 3ax^2 + 2$$ be divisible by $x + 1$. \end{exa} \begin{solu} By Ruffini's Theorem \ref{thm:ruffini}, we must have $$0 = t(-1) = (-1)^3 - 3a(-1)^2 + 2 \implies a = \frac{1}{3}.$$ \end{solu} \begin{df} Let $a$ be a root of a polynomial $p$. We say that $a$ is a root of {\em multiplicity} $m$ if $p(x)$ is divisible by $(x-a)^m$ but not by $(x-a)^{m+1}$. This means that $p$ can be written in the form $p(x) = (x-a)^mq(x)$ for some polynomial $q$ with $q(a)\neq 0$. \end{df} \begin{cor}\label{cor:ruffini2}If a polynomial of degree $n$ had any roots at all, then it has at most $n$ roots. \end{cor} \begin{pf} If it had at least $n+1$ roots then it would have at least $n+1$ factors of degree $1$ and hence degree $n+1$ at least, a contradiction. \end{pf} Notice that the above theorem only says that if a polynomial has any roots, then it must have at most its degree number of roots. It does not say that a polynomial must possess a root. That all polynomials have at least one root is much more difficult to prove. We will quote the theorem, without a proof. \begin{thm}[Fundamental Theorem of Algebra]\index{Theorem!Fundamental of Algebra} A polynomial of degree at least one with complex number coefficients has at least one complex root. \end{thm} \begin{rem} The Fundamental Theorem of Algebra implies then that a polynomial of degree $n$ has {\em exactly} $n$ roots (counting multiplicity). \end{rem} A more useful form of Ruffini's Theorem is given in the following corollary. \begin{cor} If the polynomial $p$ with integer coefficients, $$p(x) = a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0.$$ has a rational root $\frac{s}{t}\in\rationals$ (here $\frac{s}{t}$ is assumed to be in lowest terms), then $s$ divides $a_0$ and $t$ divides $a_n$. \label{cor:ruffini}\end{cor} \begin{pf} We are given that $$0 = p\left(\frac{s}{t}\right) = a_n\left(\frac{s^n}{t^n}\right) + a_{n - 1}\left(\frac{s^{n - 1}}{t^{n - 1}}\right) + \cdots + a_1\left(\frac{s}{t}\right) + a_0.$$Clearing denominators, $$0 = a_ns^n + a_{n - 1}s^{n - 1}t + \cdots + a_1st^{n - 1} + a_0t^n.$$ This last equality implies that $$-a_0t^n = s(a_ns^{n - 1} + a_{n - 1}s^{n - 2}t + \cdots + a_1t^{n - 1}).$$Since both sides are integers, and since $s$ and $t$ have no factors in common, then $s$ must divide $a_0.$ We also gather that $$-a_ns^n = t(a_{n - 1}s^{n - 1} + \cdots + a_1st^{n - 2} + a_0t^{n - 1}),$$from where we deduce that $t$ divides $a_n$, concluding the proof. \end{pf} \begin{exa} Factorise $a(x) = x^3 - 3x - 5x^2 + 15$ over $\integers[x]$ and over $\reals[x]$. \end{exa} \begin{solu} By Corollary \ref{cor:ruffini}, if $a(x)$ has integer roots then they must be in the set $\{-1, 1, -3, 3, -5, 5\}$. We test $a(\pm 1), a(\pm 3), a(\pm 5)$ to see which ones vanish. We find that $a(5) = 0$. By the Factor Theorem, $x - 5$ divides $a(x).$ Using long division, $$\polylongdiv{x^3-3x-5x^2+15}{x-5} $$we find $$a(x) = x^3 - 3x - 5x^2 + 15 = (x - 5)(x^2 - 3),$$which is the required factorisation over $\integers[x]$. The factorisation over $\reals[x]$ is then $$a(x) = x^3 - 3x - 5x^2 + 15 = (x - 5)(x - \sqrt{3})(x + \sqrt{3}).$$ \end{solu} \begin{exa} Factorise $b(x) = x^5 - x^4 - 4x + 4$ over $\integers[x]$ and over $\reals[x]$. \end{exa} \begin{solu} By Corollary \ref{cor:ruffini}, if $b(x)$ has integer roots then they must be in the set $\{-1, 1, -2, 2, -4, 4\}$. We quickly see that $b(1) = 0$, and so, by the Factor Theorem, $x - 1 $ divides $b(x)$. By long division $$\polylongdiv{x^5 - x^4 - 4x + 4}{x-1} $$ we see that $$b(x) = (x - 1)(x^4 - 4) = (x - 1)(x^2 - 2)(x^2 + 2),$$which is the desired factorisation over $\integers[x]$. The factorisation over $\reals$ is seen to be $$b(x) = (x - 1)(x - \sqrt{2})(x + \sqrt{2})(x^2 + 2).$$Since the discriminant of $x^2+2$ is $-8<0$, $x^2+2$ does not split over $\reals$. \end{solu} \begin{lem}Complex roots of a polynomial with real coefficients occur in conjugate pairs, that is, if $p$ is a polynomial with real coefficients and if $u+vi$ is a root of $p$, then its conjugate $u-vi$ is also a root for $p$. Here $i=\sqrt{-1}$ is the imaginary unit. \end{lem} \begin{pf} Assume $$p(x) =a_0 + a_1x+ \cdots + a_nx^n $$ and that $p(u+vi) = 0$. Since the conjugate of a real number is itself, and conjugation is multiplicative (Theorem \ref{thm:multiplicative_conjugation}), we have $$\begin{array}{lll} 0 & = & \overline{0}\\ & = & \overline{p(u+vi)} \\ & = &\overline{a_0 + a_1(u+vi)+ \cdots + a_n(u+vi)^n} \\ & = &\overline{a_0} +\overline{ a_1(u+vi)}+ \cdots + \overline{a_n(u+vi)^n} \\ & = & a_0 + a_1(u-vi) + \cdots + a_n(u-vi)^n\\ & = & p(u-vi),\\ \end{array}$$ whence $u-vi$ is also a root. \end{pf} Since the complex pair root $u\pm vi$ would give the polynomial with real coefficients $$ (x-u-vi)(x-u+vi) = x^2 - 2ux + (u^2+v^2), $$ we deduce the following theorem. \begin{thm} Any polynomial with real coefficients can be factored in the form $$ A(x-r_1)^{m_1}(x-r_2)^{m_2}\cdots (x-r_k)^{m_k}(x^2+a_1x+b_1)^{n_1}(x^2+a_2x+b_2)^{n_2} \cdots (x^2+a_lx+b_l)^{n_l}, $$ where each factor is distinct, the $m_i, l_k$ are positive integers and $A, r_i, a_i, b_i$ are real numbers.\end{thm} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Find the cubic polynomial $p$ having zeroes at $x = -1, 2, 3$ and satisfying $p(1) = -24.$ \begin{answer}Such polynomial must have the form $p(x) = a(x + 1)(x - 2)(x - 3)$, and so we must determine $a$. But $-24 = p(1) = a(2)(-1)(-2) = 4a.$ Hence $a = -6.$ We thus find $p(x) = -6(x + 1)(x - 2)(x - 3)$. \end{answer} \end{pro} \begin{pro} How many cubic polynomials with leading coefficient $-2$ are there splitting in the set $\{1, 2, 3\}$? \begin{answer} There are ten such polynomials. They are $p_1(x) = -2(x - 1)^3$, $p_2(x) = -2(x - 2)^3$, $p_3(x) = -2(x - 3)^3$, $p_4(x) = -2(x - 1)(x - 2)^2$, $p_5(x) = -2(x - 1)^2(x - 2)$, $p_6(x) = -2(x - 1)(x - 3)^2$, $p_7(x) = -2(x - 1)^2(x - 3)$, $p_8(x) = -2(x - 2)(x - 3)^2$, $p_9(x) = -2(x - 2)^2(x - 3)$, $p_{10}(x) = -2(x - 1)(x - 2)(x - 3)$. \end{answer} \end{pro} \begin{pro} Find the cubic polynomial $c$ having a root of $x = 1$, a root of multiplicity $2$ at $x = -3$ and satisfying $c(2) = 10$. \begin{answer} This polynomial must have the form $c(x) = a(x - 1)(x + 3)^2$. Now $10 = c(2) = a(2 - 1)(2 + 3)^2 = 25a$, whence $\dis{a = \frac{2}{5}}$. The required polynomial is thus $\dis{c(x) = \frac{2}{5}(x - 1)(x + 3)^2}$. \end{answer} \end{pro} \begin{pro} A cubic polynomial $p$ with leading coefficient $1$ satisfies $p(1) = 1, p(2) = 4, p(3) = 9$. Find the value of $p(4)$. \begin{answer} Put $g(x) = p(x) - x^2.$ Observe that $g$ is also a cubic polynomial with leading coefficient $1$ and that $g(x) = 0$ for $x = 1, 2, 3$. This means that $g(x) = (x - 1)(x - 2)(x - 3)$ and hence $p(x) = (x - 1)(x - 2)(x - 3) + x^2$. This yields $p(4) = (3)(2)(1) + 4^2 = 22.$ \end{answer} \end{pro} \begin{pro} The polynomial $p(x)$ has integral coefficients and $p(x) = 7$ for four different values of $x$. Shew that $p(x)$ never equals $14$. \begin{answer}The polynomial $g(x) = p(x) - 7$ vanishes at the $4$ different integer values $a, b, c, d$. In virtue of the Factor Theorem, $$g(x) = (x - a)(x - b)(x - c)(x - d)q(x),$$ where $q(x)$ is a polynomial with integral coefficients. Suppose that $p(t) = 14$ for some integer $t$. Then $g(t) = p(t) - 7 = 14 - 7 = 7.$ It follows that $$7 = g(t) = (t - a)(t - b)(t - c)(t - d)q(t),$$that is, we have factorised $7$ as the product of at least $4$ different factors, which is impossible since $7$ can be factorised as $7(-1)1$, the product of at most $3$ distinct integral factors. From this contradiction we deduce that such an integer $t$ does not exist. \end{answer} \end{pro} \begin{pro} Find the value of $a$ so that the polynomial $$t(x) = x^3 - 3ax^2 + 12$$ be divisible by $x + 4.$. \begin{answer} By the Factor Theorem, we must have $$0 = t(-4) = (-4)^3 - 3a(-4)^2 + 40$$ $$\iff 0 = -24 - 48a$$ $$\iff a = -\frac{1}{2}.$$ \end{answer} \end{pro} \begin{pro} Let $f(x) = x^4 + x^3 + x^2 + x + 1.$ Find the remainder when $f(x^5)$ is divided by $f(x).$ \begin{answer} Observe that $f(x)(x - 1) = x^5 - 1$ and $$f(x^5) = x^{20} + x^{15} + x^{10} + x^5 + 1 = (x^{20} - 1) + (x^{15} - 1) + (x^{10} - 1) + (x^5 - 1) + 5.$$ Each of the summands in parentheses is divisible by $x^5 - 1$ and, a fortiori, by $f(x).$ The remainder sought is thus $5$. \end{answer} \end{pro} \begin{pro} If $p(x)$ is a cubic polynomial with $p(1) = 1, p(2) = 2, p(3) = 3, p(4) = 5,$ find $p(6)$. \begin{answer} Put $g(x) = p(x) - x$, then $p(6) = 16 $. \end{answer} \end{pro} \begin{pro} The polynomial $p(x)$ satisfies $p(-x) = -p(x).$ When $p(x)$ is divided by $x - 3$ the remainder is $6$. Find the remainder when $p(x)$ is divided by $x^2 - 9$. \end{pro} \begin{pro} Factorise $x^3 + 3x^2 - 4x + 12$ over $\integers[x]$. \begin{answer} $(x - 2)(x + 2)(x - 3)$ \end{answer} \end{pro} \begin{pro} Factorise $3x^4 + 13x^3 - 37x^2 - 117x + 90$ over $\integers[x]$. \begin{answer} $(x - 3)(x + 3)(x + 5)(3x - 2)$ \end{answer} \end{pro} \begin{pro} Find $a, b$ such that the polynomial $x^3 + 6x^2 + ax + b$ be divisible by the polynomial $x^2 + x - 12$. \begin{answer} $a = -7, b = -60$ \end{answer} \end{pro} \begin{pro} How many polynomials $p(x)$ of degree at least one and integer coefficients satisfy $$16p(x^2) = (p(2x)) ^2,$$for all real numbers $x$? \begin{answer} Let $p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots + a_1x+a_0$ with $a_n\neq 0$, $n \geq 1$. Then $$16p(x^2) = (p(2x)) ^2 \implies 16(a_nx^{2n}+a_{n-1}x^{2n-2}+\cdots + a_1x^2+a_0)=\left(2^na_nx^n+2^{n-1}a_{n-1}x^{n-1}+\cdots + 2a_1x+a_0\right)^2 $$ Since the coefficients on both sides of the equality must agree, we must have $$16a_n=2^{2n}a_n ^2 \implies 2^4=2^{2n}a_n $$ since $a_n\neq 0$. As $a_n$ is an integer, we must have the following cases: $n=1, a_n=4$, $n=2, a_n=1$. Clearly we may not have $n\geq 3$. Thus such polynomials are either linear or quadratic. Also, for $x=0$, $16p(0)=(p(0))^2$ and therefore either $p(0)=0$ or $p(0)=16$. For $n=1$ we seek $p(x)=4x+a$. Solving $$ 16(4x^2+a)=(8x+a)^2 \implies a=0, $$whence $p(x)=4x$. For $n=2$, let $p(x)=x^2+ax+b$. Solving $$ 16(x^4+ax^2+b)=(4x^2+2ax+b)^2 \implies a=0. $$ Since $p(0)=0$ or $p(0)=16$, we must test $p(x)=x^2$ and $p(x)=x^2+16$. It is easy to see that only $p(x) =x^2$ satisfies the desired properties. In conclusion, $ 4x$ and $x^2$ are the only two such polynomials.. \end{answer} \end{pro} \end{multicols} \chapter{Rational Functions and Algebraic Functions} \section{The Reciprocal Function} \begin{df} Given a function $f$ we write $f(-\infty)$ for the value that $f$ may eventually approach for large (in absolute value) and negative inputs and $f(+\infty)$ for the value that $f$ may eventually approach for large (in absolute value) and positive input. The line $y = b$ is a (horizontal) {\em asymptote} for the function $f$ if either $$f(-\infty) = b \qquad \mathrm{or} \qquad f(+\infty) = b. $$ \end{df}\index{asymptote!horizontal} \begin{df} Let $k>0$ be an integer. A function $f$ has a {\em pole of order $k$} at the point $x = a$ if $ (x-a)^{k-1}f(x) \rightarrow \pm \infty $ as $\tends{x}{a}$, but $ (x-a)^kf(x) $ as $\tends{x}{a}$ is finite. Some authors prefer to use the term {\em vertical asymptote}, rather than pole. \index{asymptote!vertical} \index{pole} \end{df} \begin{exa} Since $xf(x) = 1$, $f(0-) = -\infty$, $f(0+) = +\infty$ for $ \fun{f}{x}{\dfrac{1}{x}}{\reals\setminus\{0\}}{\reals \setminus\{0\}}$, $f$ has a pole of order $1$ at $x=0$. \end{exa} \begin{thm}[Graph of the Reciprocal Function] The graph of the reciprocal function $$\fun{{\bf Rec}}{x}{\dfrac{1}{x}}{\reals \setminus \{0\}}{\reals} $$is concave for $x < 0$ and convex for $x > 0$. $x \mapsto \dfrac{1}{x}$ is decreasing for $x < 0$ and $x > 0$. $x \mapsto \dfrac{1}{x}$ is an odd function and $\im{{\bf Rec}} = \reals \setminus \{0\}$. \end{thm} \begin{pf} Assume first that $0 < a < b$ and that $\lambda \in [0;1]$. By the Arithmetic-Mean-Geometric-Mean Inequality, Theorem \ref{thm:arith_mean_geo_mean}, we deduce that $$ \dfrac{a}{b} + \dfrac{b}{a} \geq 2. $$ Hence the product $$ \begin{array}{lll}(\lambda a + (1 - \lambda)b)\left(\dfrac{\lambda}{a} + \dfrac{1 - \lambda}{b}\right) & = & \lambda^2 + (1 - \lambda)^2 + \lambda(1-\lambda)\left(\dfrac{a}{b} + \dfrac{b}{a}\right) \\ & \geq & \lambda^2 + (1 - \lambda)^2 + 2\lambda(1-\lambda) \\ & = & (\lambda + 1 - \lambda)^2 \\ & = & 1. \end{array}$$ Thus for $0 < a< b$ we have $$ \dfrac{1}{\lambda a + (1 - \lambda)b} \leq \left(\dfrac{\lambda}{a} + \dfrac{1 - \lambda}{b}\right) \implies {\bf Rec} (\lambda a + (1 - \lambda)b) \leq \lambda {\bf Rec} (a) + (1 - \lambda){\bf Rec} (b), $$from where $x \mapsto \dfrac{1}{x}$ is convex for $x > 0$. If we replace $a, b$ by $-a, -b$ then the inequality above is reversed and we obtain that $x \mapsto \dfrac{1}{x}$ is concave for $x<0$. \bigskip As ${\bf Rec} (-x) = \dfrac{1}{-x} = -\dfrac{1}{x} = -{\bf Rec} (x)$, the reciprocal function is an odd function. Assume $a < b$ are non-zero and have the same sign. Then $$\dfrac{{\bf Rec} (b) - {\bf Rec} (a)}{b - a} = \dfrac{\dfrac{1}{b} - \dfrac{1}{a}}{b - a} = -\dfrac{1}{ab} < 0,$$ since we are assuming that $a, b$ have the same sign, whence $x \mapsto \dfrac{1}{x}$ is a strictly decreasing function whenever arguments have the same sign. Also given any $y\in\im{{\bf Rec}} $ we have $y = {\bf Rec} (x) = \dfrac{1}{x} $, but this equation is solvable only if $y \neq 0$. and so every real number is an image of $\idefun$ meaning that $\im{{\bf Rec}} = \reals \setminus \{0\}$. \end{pf} \begin{exa} Figures \ref{fig:a(x)=1/x} through \ref{fig:a(x+2)+3=1/(x+2)+3} exhibit various transformations of $y = a(x) = \dfrac{1}{x}$. Notice how the poles and the asymptotes move with the various transformations. \end{exa} \vspace{1cm} \begin{figure}[!hptb] \begin{minipage}{4.5cm}$$\psset{unit=.7pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{-6}{-.166666667}{1/x} \psplot[linewidth=2pt,linecolor=brown]{.166666667}{6}{1/x} $$\meinecaption{1}{$x\mapsto \dfrac{1}{x}$} \label{fig:a(x)=1/x} \end{minipage} \hfill \begin{minipage}{4.5cm}$$\psset{unit=.7pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \rput(1,-1){ \psline[linestyle=dashed](0,-6)(0,6)\psline[linestyle=dashed](-6,0)(6,0) \psplot[linewidth=2pt,linecolor=brown]{-6}{-.166666667}{1/x} \psplot[linewidth=2pt,linecolor=brown]{.166666667}{6}{1/x}} $$\meinecaption{1}{$x\mapsto \dfrac{1}{x-1} -1$} \label{fig:a(x-1)-1=1/(x-1)-1} \end{minipage} \hfill \begin{minipage}{4.5cm}$$\psset{unit=.7pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \rput(-2,3){ \psline[linestyle=dashed](0,-6)(0,6)\psline[linestyle=dashed](-6,0)(6,0) \psplot[linewidth=2pt,linecolor=brown]{-6}{-.166666667}{1/x} \psplot[linewidth=2pt,linecolor=brown]{.166666667}{6}{1/x}} $$\meinecaption{1}{$x\mapsto \dfrac{1}{x+2} +3$} \label{fig:a(x+2)+3=1/(x+2)+3} \end{minipage} \end{figure} \vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{4.5cm}$$\psset{unit=.7pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \rput(1,-1){ \psline[linestyle=dashed](0,-6)(0,6)\psline[linestyle=dashed](-6,0)(6,0) \psplot[linewidth=2pt,linecolor=brown]{-6}{-.166666667}{1/x} \psplot[linewidth=2pt,linecolor=brown]{.166666667}{6}{1/x}} $$\meinecaption{1}{$x\mapsto \dfrac{1}{x-1} -1$} \label{fig:1/(x-1)-1} \end{minipage} \hfill \begin{minipage}{4.5cm}$$\psset{unit=.7pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \rput(1,1){\psplot[linewidth=2pt,linecolor=brown]{-6}{-.25}{1/abs(x)}}\rput(1,-1){ \psline[linestyle=dashed](0,1)(0,6)\psline[linestyle=dashed](-6,2)(6,2) \psplot[linewidth=2pt,linecolor=brown]{.166666667}{1}{1/x}} \rput(1,1){\psplot[linewidth=2pt,linecolor=brown]{1}{6}{-1/x}} $$\meinecaption{1}{$x\mapsto \left|\dfrac{1}{x-1} -1\right|$} \label{fig:1/|x-1|-1} \end{minipage} \hfill \begin{minipage}{4.5cm}$$\psset{unit=.7pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \rput(1,-1){ \psline[linestyle=dashed](0,-6)(0,6)\psline[linestyle=dashed](-6,0)(6,0) \psplot[linewidth=2pt,linecolor=brown]{-1}{-.166666667}{1/x} \psplot[linewidth=2pt,linecolor=brown]{.166666667}{6}{1/x}} \rput(-1,-1){ \psline[linestyle=dashed](0,-6)(0,6)\psline[linestyle=dashed](-6,0)(6,0) \psplot[linewidth=2pt,linecolor=brown]{-6}{-.166666667}{-1/x} \psplot[linewidth=2pt,linecolor=brown]{.166666667}{1}{-1/x}} $$\meinecaption{1}{$x\mapsto \dfrac{1}{|x|-1}-1$} \label{fig:1/(|x|-1)-1} \end{minipage} \hfill \end{figure} \section{Inverse Power Functions} We now proceed to investigate the behaviour of functions of the type $x\mapsto \dfrac{1}{x^n}$, where $n>0$ is an integer. \begin{thm}\label{thm:graph_inverse_powers} Let $n>0$ be an integer. Then \begin{itemize}\item if $n$ is even, $x\mapsto \dfrac{1}{x^n}$ is increasing for $x<0$, decreasing for $x>0$ and convex for all $x\neq 0$. \item if $n$ is odd, $x\mapsto \dfrac{1}{x^n}$ is decreasing for all $x\neq 0$, concave for $x<0$, and convex for $x>0$. \end{itemize} Thus $x\mapsto \dfrac{1}{x^n}$ has a pole of order $n$ at $x = 0$ and a horizontal asymptote at $y=0$. \end{thm} \begin{exa}A few functions $x \mapsto \dfrac{1}{x^n}$ are shewn in figures \ref{fig:reciprocal_function} through \ref{fig:1/x^6}. \bigskip \end{exa} \vspace{1cm} \begin{figure}[!hptb] \begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{-6}{-.166666667}{1/x} \psplot[linewidth=2pt,linecolor=brown]{.166666667}{6}{1/x} $$\meinecaption{1}{$x\mapsto \dfrac{1}{x}$} \label{fig:reciprocal_function} \end{minipage} \hfill \begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{-6}{-.4082482905}{1/x^2} \psplot[linewidth=2pt,linecolor=brown]{.4082482905}{6}{1/x^2} $$\meinecaption{1}{$x\mapsto \dfrac{1}{x^2}$} \label{fig:1/x^2} \end{minipage} \hfill \begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{-6}{-.5503212082}{1/x^3} \psplot[linewidth=2pt,linecolor=brown]{.5503212082}{6}{1/x^3} $$\meinecaption{1}{$x\mapsto \dfrac{1}{x^3}$} \label{fig:1/x^3} \end{minipage} \hfill \begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{-6}{-.6389431042}{1/x^4} \psplot[linewidth=2pt,linecolor=brown]{.6389431042}{6}{1/x^4} $$\meinecaption{1}{$x\mapsto \dfrac{1}{x^4}$} \label{fig:1/x^4} \end{minipage} \hfill \begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{-6}{-.6988271188}{1/x^5} \psplot[linewidth=2pt,linecolor=brown]{.6988271188}{6}{1/x^5} $$\meinecaption{1}{$x\mapsto \dfrac{1}{x^5}$} \label{fig:1/x^5} \end{minipage} \hfill \begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{-6}{-.7418363756}{1/x^6} \psplot[linewidth=2pt,linecolor=brown]{.7418363756}{6}{1/x^6} $$\meinecaption{1}{$x\mapsto \dfrac{1}{x^6}$} \label{fig:1/x^6} \end{minipage} \hfill \end{figure} \section{Rational Functions} \begin{df} By a {\em rational function} $x\mapsto r(x)$ we mean a function $r$ whose assignment rule is of the $r(x) = \dfrac{p(x)}{q(x)}$, where $p(x)$ and $q(x) \neq 0$ are polynomials. \index{function!rational} \end{df} We now provide a few examples of graphing rational functions. Analogous to theorem \ref{thm:graph_splitting_polynomials}, we now consider rational functions $x\mapsto r(x) = \dfrac{p(x)}{q(x)}$ where $p$ and $q$ are polynomials with no factors in common and splitting in $\reals$. \begin{thm}\label{thm:graph_splitting_rational_functions} Let $a\neq 0$ and the $r_i$ are real numbers and the $m_i$ be positive integers. Then the rational function with assignment rule $$r(x) = K\dfrac{(x-a_1)^{m_1}(x-a_2)^{m_2} \cdots (x-a_k)^{m_k}}{(x-b_1)^{n_1}(x-b_2)^{n_2} \cdots (x-b_l)^{n_l}}, $$ \begin{itemize} \item has zeroes at $x=a_i$ and poles at $x=b_j$. \item crosses the $x$-axis at $x=a_i$ if $m_i$ is odd. \item is tangent to the $x$-axis at $x=a_i$ if $m_i$ is even. \item has a convexity change at $x=a_i$ if $m_i\geq 3$ and $m_i$ is odd. \item both $r(b_j-)$ and $r(b_j+)$ blow to infinity. If $n_i$ is even, then they have the same sign infinity: $r(b_i+) = r(b_i-) = +\infty$ or $r(b_i+) = r(b_i-) = -\infty$. If $n_i$ is odd, then they have different sign infinity: $r(b_i+) = -r(b_i-) = +\infty$ or $r(b_i+) = -r(b_i-) = -\infty$. \end{itemize} \end{thm} \begin{pf} Since the local behaviour of $r(x)$ is that of $c(x-r_i)^{t_i}$ (where $c$ is a real number constant) near $r_i$, the theorem follows at once from Theorem \ref{thm:power_functions} and \ref{thm:graph_inverse_powers}. \end{pf} \begin{exa}\label{exa:(x-1)^2(x+2)/(x+1)(x-2)^2} Draw a rough sketch of $x\mapsto \dfrac{(x-1)^2(x+2)}{(x+1)(x-2)^2}$. \end{exa}\begin{solu} Put $r(x) = \dfrac{(x-1)^2(x+2)}{(x+1)(x-2)^2}$. By Theorem \ref{thm:graph_splitting_rational_functions}, $r$ has zeroes at $x=1$, and $x=-2$, and poles at $x=-1$ and $x=2$. As $\tends{x}{1}$, $r(x)\sim \dfrac{3}{2}(x-1)^2$, hence the graph of $r$ is tangent to the axes, and positive, around $x=2$. As $\tends{x}{-2}$, $r(x)\sim -\dfrac{9}{16}(x+2)$, hence the graph of $r$ crosses the $x$-axis at $x=-2$, coming from positive $y$-values on the left of $x=-2$ and going to negative $y$=values on the right of $x=-2$. As $\tends{x}{-1}$, $r(x)\sim \dfrac{4}{9(x+1)}$, hence the graph of $r$ blows to $-\infty$ to the left of $x=-1$ and to $+\infty$ to the right of $x=-1$. As $\tends{x}{2}$, $r(x)\sim \dfrac{4}{3(x-2)^2}$, hence the graph of $r$ blows to $+\infty$ both from the left and the right of $x=2$. Also we observe that $$r(x) \sim \dfrac{(x)^2(x)}{(x)(x)^2} = \dfrac{x^3}{x^3} = 1, $$ and hence $r$ has the horizontal asymptote $y=1$. A sign diagram for $\dfrac{(x-1)^2(x+2)}{(x+1)(x-2)^2}$ follows: $$\begin{array}{|c|c|c|c|c|}\hline ]-\infty;-2[ & ]-2;-1[ & ]-1;1[ & ]1;2[ & ]2;+\infty[ \\ \hline \plus & \minus & \plus & \plus & \plus \\ \hline \end{array}$$ The graph of $r$ can be found in figure \ref{fig:(x-1)^2(x+2)/(x+1)(x-2)^2}. \end{solu} \vspace{3cm} \begin{figure}[!hptb] \begin{minipage}{.4\textwidth}$$\psset{unit=1pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{-6}{-1.083676189}{((x+2)*(x-1)^2)/((x+1)*(x-2)^2)} \psplot[linewidth=2pt,linecolor=brown]{-.9058688457}{1.655868846}{((x+2)*(x-1)^2)/((x+1)*(x-2)^2)} \psplot[linewidth=2pt,linecolor=brown]{3}{6}{((x+2)*(x-1)^2)/((x+1)*(x-2)^2)} \psline[linestyle=dotted](-6,1)(6,1) \psline[linestyle=dotted](-1,-6)(-1,6) \psline[linestyle=dotted](2,-6)(2,6) $$\vspace{2cm}\meinecaption{.25}{$x\mapsto \dfrac{(x-1)^2(x+2)}{(x+1)(x-2)^2}$} \label{fig:(x-1)^2(x+2)/(x+1)(x-2)^2} \end{minipage} \hfill \begin{minipage}{.4\textwidth}$$\psset{unit=2pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-3,-3)(3,3) \psplot[linewidth=2pt,linecolor=brown]{-1.740903741}{-1.046066042}{(((x-.75)^2)*((x+.75)^2))/((x+1)*(x-1))} \psplot[linewidth=2pt,linecolor=brown]{-.9746582010}{.9746582010}{(((x-.75)^2)*((x+.75)^2))/((x+1)*(x-1))} \psplot[linewidth=2pt,linecolor=brown]{1.046066042}{1.740903741}{(((x-.75)^2)*((x+.75)^2))/((x+1)*(x-1))} \psline[linestyle=dotted](-3,1)(3,1) \psline[linestyle=dotted](-1,-3)(-1,3) \psline[linestyle=dotted](1,-3)(1,3) $$\vspace{2cm}\meinecaption{.25}{$x\mapsto \dfrac{(x-3/4)^2(x+3/4)^2}{(x+1)(x-1)}$} \label{fig:(x-3/4)^2(x+3/4)^2)/(x+1)(x-1)} \end{minipage} \end{figure} \begin{exa}\label{exa:(x-3/4)^2(x+3/4)^2)/(x+1)(x-1)} Draw a rough sketch of $x\mapsto \dfrac{(x-3/4)^2(x+3/4)^2}{(x+1)(x-1)}$. \end{exa}\begin{solu} Put $r(x) = \dfrac{(x-3/4)^2(x+3/4)^2}{(x+1)(x-1)}$. First observe that $r(x) = r(-x)$, and so $r$ is even. By Theorem \ref{thm:graph_splitting_rational_functions}, $r$ has zeroes at $x=\pm\dfrac{3}{4}$, and poles at $x=\pm 1$. As $\tends{x}{\dfrac{3}{4}}$, $r(x)\sim -\dfrac{36}{7}(x-3/4)^2$, hence the graph of $r$ is tangent to the axes, and negative, around $x=3/4$, and similar behaviour occurs around $x=-\dfrac{3}{4}$. As $\tends{x}{1}$, $r(x)\sim \dfrac{49}{512(x-1)}$, hence the graph of $r$ blows to $-\infty$ to the left of $x=1$ and to $+\infty$ to the right of $x=1$. As $\tends{x}{-1}$, $r(x)\sim -\dfrac{49}{512(x-1)}$, hence the graph of $r$ blows to $+\infty$ to the left of $x=-1$ and to $-\infty$ to the right of $x=-1$. Also, as $\tends{x}{+\infty}$, $$ r(x) \sim \dfrac{(x)^2(x)^2}{(x)(x)} = x^2, $$so $r(+\infty) = +\infty$ and $r(-\infty) = +\infty$. A sign diagram for $\dfrac{(x-3/4)^2(x+3/4)^2}{(x+1)(x-1)}$ follows: $$\begin{array}{|c|c|c|c|c|}\hline ]-\infty;-1[ & \oo{-1;-\dfrac{3}{4}} & \oo{-\dfrac{3}{4};\dfrac{3}{4}} & \oo{\dfrac{3}{4};1} & ]1;+\infty[ \\ \hline \plus & \minus & \minus & \minus & \plus \\ \hline \end{array}$$ The graph of $r$ can be found in figure \ref{fig:(x-3/4)^2(x+3/4)^2)/(x+1)(x-1)}. \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Find the condition on the distinct real numbers $a, b, c$ such that the function $x\mapsto \dfrac{(x-a)(x-b)}{x-c}$ takes all real values for real values of $x$. Sketch two scenarios to illustrate a case when the condition is satisfied and a case when the condition is not satisfied. \end{pro} \begin{pro} Make a rough sketch of the following curves. \begin{enumerate} \item $y = \dfrac{x}{x^2-1}$ \item $y = \dfrac{x^2}{x^2-1}$ \item $y = \dfrac{x^2-1}{x}$ \item $y = \dfrac{x^2-x-6}{x^2+x-6}$ \item $y = \dfrac{x^2+x-6}{x^2-x-6}$ \item $y = \dfrac{x}{(x+1)^2(x-1)^2}$ \item $y = \dfrac{x^2}{(x+1)^2(x-1)^2}$ \end{enumerate} \end{pro} \begin{pro} \label{pro:rationalA} The rational function $q$ in figure \ref{fig:rational1A} has only two simple poles and satisfies $q(x)\rightarrow 1$ as $x\rightarrow \pm \infty$. You may assume that the poles and zeroes of $q$ are located at integer points. \begin{enumerate} \item Find $q(0)$. \\ \item Find $q(x)$ for arbitrary $x$. \item Find $q(-3)$. \item To which value does $q(x)$ approach as $x\to -2+$? \end{enumerate} \end{pro} \vspace{2cm} \begin{figurehere} $$\psset{unit=.7pc} \renewcommand{\pshlabel}[1]{{\tiny #1}} \renewcommand{\psvlabel}[1]{{\tiny #1}} \psgrid[subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-8,-8)(8,8) \psline[linewidth=1.3pt](0,-8)(0,8)\psline[linewidth=1.3pt](-8, 0 )(8, 0) \psplot[linewidth=1.5pt, linecolor=brown,algebraic=true,arrows={<->}]{-7}{-2.230138587}{((x+1)*(x-2))/((x-1)*(x+2))} \psplot[linewidth=1.5pt, linecolor=brown,algebraic=true,arrows={<->}]{-1.838087489}{.8968052533}{((x+1)*(x-2))/((x-1)*(x+2))} \psplot[linewidth=1.5pt, linecolor=brown,algebraic=true,arrows={<->}]{1.088087489}{6}{((x+1)*(x-2))/((x-1)*(x+2))} \psdots[dotstyle=*,dotscale=1](-1,0)(2,0)(0,1) $$\meinecaption{2}{Problem \ref{pro:rationalA}.} \label{fig:rational1A} \end{figurehere} \end{multicols} \section{Algebraic Functions} \begin{df} We will call {\em algebraic function} a function whose assignment rule can be obtained from a rational function by a finite combination of additions, subtractions, multiplications, divisions, exponentiations to a rational power. \index{function!algebraic} \end{df} \begin{thm}\label{thm:graph_rational_powers} Let $|q|\geq 2$ be an integer. If \begin{itemize}\item if $q$ is even then $x\mapsto x^{1/q}$ is increasing and concave for $q \geq 2$ and decreasing and convex for $q \leq -2$ for all $x>0$ and it is undefined for $x<0$. \item if $q$ is odd then $x\mapsto x^{1/q}$ is everywhere increasing and convex for $x<0$ but concave for $x>0$ if $q\geq 3$. If $q\leq -3$ then $x\mapsto x^{1/q}$ is decreasing and concave for $x<0$ and increasing and convex for $x>0$. \end{itemize} \end{thm} \clearpage A few of the functions $x\mapsto x^{1/q}$ are shewn in figures \ref{fig:x^1/2} through \ref{fig:x^-1/7}. \vspace{1cm} \begin{figure}[!hptb] \begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{0}{6}{sqrt(x)} $$\meinecaption{1}{$x\mapsto x^{1/2}$} \label{fig:x^1/2} \end{minipage} \hfill \begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{.02777777778}{6}{x^(-1/2)} $$\meinecaption{1}{$x\mapsto x^{-1/2}$} \label{fig:x^-1/2} \end{minipage} \hfill \begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{0}{6}{x^(1/4)} $$\meinecaption{1}{$x\mapsto x^{1/4}$} \label{fig:x^1/4} \end{minipage} \hfill \begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{.0007716049383}{6}{x^(-1/4)} $$\meinecaption{1}{$x\mapsto x^{-1/4}$} \label{fig:x^-1/4} \end{minipage} \hfill \begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{0}{6}{x^(1/6)} $$\meinecaption{1}{$x\mapsto x^{1/6}$} \label{fig:x^1/6} \end{minipage} \hfill \begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{.00002143347051}{6}{x^(-1/6)} $$\meinecaption{1}{$x\mapsto x^{-1/6}$} \label{fig:x^-1/6} \end{minipage} \hfill \end{figure} \vspace{1cm} \begin{figure}[!hptb] \begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{0}{6}{x^(1/3)} \psplot[linewidth=2pt,linecolor=brown]{-6}{0}{-1*abs(x)^(1/3)} $$\meinecaption{1}{$x\mapsto x^{1/3}$} \label{fig:x^1/3} \end{minipage} \hfill \begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{.004629629630}{6}{x^(-1/3)} \psplot[linewidth=2pt,linecolor=brown]{-.004629629630}{-6}{-1*abs(x)^(-1/3)} $$\meinecaption{1}{$x\mapsto x^{-1/3}$} \label{fig:x^-1/3} \end{minipage} \hfill \begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{0}{6}{x^(1/5)} \psplot[linewidth=2pt,linecolor=brown]{0}{-6}{-1*abs(x)^(1/5)} $$\meinecaption{1}{$x\mapsto x^{1/5}$} \label{fig:x^1/5} \end{minipage} \hfill \begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{.0001286008230}{6}{x^(-1/5)} \psplot[linewidth=2pt,linecolor=brown]{-.0001286008230}{-6}{-1*abs(x)^(-1/5)} $$\meinecaption{1}{$x\mapsto x^{-1/5}$} \label{fig:x^-1/5} \end{minipage} \hfill \begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{0}{6}{x^(1/7)} \psplot[linewidth=2pt,linecolor=brown]{0}{-6}{-1*abs(x)^(1/7)} $$\meinecaption{1}{$x\mapsto x^{1/7}$} \label{fig:x^1/7} \end{minipage} \hfill \begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt, labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown]{.000003572245085}{6}{x^(-1/7)} \psplot[linewidth=2pt,linecolor=brown]{-.000003572245085}{-6}{-1*abs(x)^(-1/7)} $$\meinecaption{1}{$x\mapsto x^{-1/7}$} \label{fig:x^-1/7} \end{minipage} \hfill \end{figure} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Draw the graph of each of the following curves. \begin{enumerate} \item $x\mapsto (1+x)^{1/2} $ \item $x\mapsto (1-x)^{1/2} $ \item $x\mapsto 1+(1+x)^{1/3} $ \item $x\mapsto 1-(1-x)^{1/3} $ \item $x\mapsto \sqrt{x}+\sqrt{-x} $ \end{enumerate} \end{pro} \end{multicols} \chapter{Exponential Functions} \section{Exponential Functions} \begin{df} Let $a > 0, a \neq 1$ be a fixed real number. The function $$\funnoname{x}{a^x}{\reals}{]0;+\infty[},$$ is called the {\em exponential function} of {\em base} $a$. \end{df} \vspace{2cm} \begin{figure}[!hptb]\begin{minipage}{.4\textwidth}\psset{unit=1pc}\centering \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.4pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psplot[algebraic,linewidth=2pt,linecolor=brown,arrows={<->}]{-5.5}{5.5}{1.3^x} \meinecaption{3}{$x \mapsto a^x, \ a > 1.$}\label{fig:expbasemore1}\end{minipage}\hfill \begin{minipage}{.4\textwidth}\psset{unit=1pc} \centering \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.4pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psplot[algebraic,linewidth=2pt,linecolor=brown,arrows={<->}]{-5.5}{5.5}{1/1.3^x} \meinecaption{3}{$x \mapsto a^x, \ 0 < a < 1.$} \label{fig:expbaseless1} \end{minipage} \hfill\end{figure} \vspace{2cm} We will now prove that the generic graphs of the exponential function resemble those in figures \ref{fig:expbasemore1} and \ref{fig:expbaseless1}. \begin{thm} If $a > 1, \ x \mapsto a^x$ is strictly increasing and convex, and if $0 < a < 1$ then $x \mapsto a^x$ is strictly decreasing and convex. \end{thm} \begin{pf} Put $f(x) = a^x$. Recall that a function $f$ is strictly increasing or decreasing depending on whether the ratio $$\dfrac{f(t) - f(s)}{t - s} >0 \ \ \ \mathrm{or}\ \ \ <0 $$for $t\neq s$. Now, $$ \dfrac{f(t) - f(s)}{t - s} = \dfrac{a^t - a^s}{t-s} = (a^{s})\cdot \dfrac{a^{t-s} - 1}{t-s}. $$ If $a>1$, and $t-s>0$ then also $a^{t-s}>1$.\footnote{The alert reader will find this argument circular! I have tried to prove this theorem from first principles without introducing too many tools. Alas, I feel tired\ldots } If $a>1$, and $t-s<0$ then also $a^{t-s}<1$. Thus regardless on whether $t-s > 0$ or $<0$ the ratio $$ \dfrac{a^{t-s} - 1}{t-s}>0, $$whence $f$ is increasing for $a > 1$. A similar argument proves that for $0 < a < 1$, $f$ would be decreasing. \bigskip To prove convexity will be somewhat more arduous. Recall that $f$ is convex if for arbitrary $0 \leq \lambda \leq 1$ we have $$f(\lambda s + (1 - \lambda)t) \leq \lambda f(s) + (1 - \lambda)f(t),$$ that is, a straight line joining any two points of the curve lies above the curve. We will not be able to prove this quickly, we will just content with proving midpoint convexity: we will prove that $$ f\left(\dfrac{s + t}{2}\right) \leq \dfrac{1}{2}f(s) + \dfrac{1}{2}f(t). $$This is equivalent to $$ a^{\frac{s+t}{2}} \leq \dfrac{1}{2}a^s + \dfrac{1}{2}a^t ,$$ which in turn is equivalent to $$2 \leq a^{\frac{s-t}{2}} + a^{\frac{t-s}{2}}. $$But the square of a real number is always non-negative, hence $$ \left(a^{\frac{s-t}{4}} - a^{\frac{t-s}{4}}\right)^2 \geq 0 \implies a^{\frac{s-t}{2}} + a^{\frac{t-s}{2}} \geq 2, $$ proving midpoint convexity. \end{pf} \begin{rem} The line $y = 0$ is an asymptote for $x \mapsto a^x.$ If $a > 1,$ then $a^x \rightarrow 0$ as $x \rightarrow -\infty$ and $a^x \rightarrow +\infty$ as $x \rightarrow + \infty$. If $0 < a < 1,$ then $a^x \rightarrow +\infty$ as $x \rightarrow -\infty$ and $a^x \rightarrow 0$ as $x \rightarrow + \infty$. \end{rem} \subsection*{Homework} \addcontentsline{toc}{section}{Homework} \markright{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\begin{pro} Make rough sketches of the following curves. \begin{enumerate} \item $x\mapsto 2^x$ \item $x\mapsto 2^{|x|}$ \item $x\mapsto 2^{-|x|}$ \item $x\mapsto 2^{x} + 3$ \item $x\mapsto 2^{x+3}$ \end{enumerate} \end{pro} \end{multicols} \section{The number $e$} Consider now the following problem, first studied by the Swiss mathematician Jakob Bernoulli around the 1700s: Query: If a creditor lends money at interest under the condition that during each individual moment the proportional part of the annual interest be added to the principal, what is the balance at the end of a full year?\footnote{``Qu\ae ritur, si creditor aliquis pecuniam suam f\oe nori exponat, ea lege, ut singulis momentis pars proportionalis usur\ae \ annu\ae\ sorti annumeretur; quantum ipsi finito anno debeatur?'' } Suppose $a$ dollars are deposited, and the interest is added $n$ times a year at a rate of $x$. After the first time period, the balance is $$b_1 = \left(1 + \frac{x}{n}\right)a. $$After the second time period, the balance is $$b_2 = \left(1 + \frac{x}{n}\right)b_1 = \left(1 + \frac{x}{n}\right)^2a.$$ Proceeding recursively, after the $n$-th time period, the balance will be $$b_n = \left(1 + \frac{x}{n}\right)^na.$$ \bigskip The study of the sequence $$e_n = \left(1 + \frac{1}{n}\right)^n$$thus becomes important. It was Bernoulli's pupil, Leonhard Euler, who shewed that the sequence $\left(1 + \frac{1}{n}\right)^n, n = 1, 2, 3, \ldots$ converges to a finite number, which he called $e$. In other words, Euler shewed that \begin{equation} e = \lim _{n \rightarrow \infty} \left(1 + \frac{1}{n}\right) ^n. \label{eq:def_of_e}\end{equation} It must be said, in passing, that Euler did not rigourously shewed the existence of the above limit. He, however, gave other formulations of the irrational number $$e = 2.718281828459045235360287471352...,$$ among others, the infinite series \begin{equation} e = 2 + \frac{1}{2!} + + \frac{1}{3!} + + \frac{1}{4!} + + \frac{1}{5!} + \cdots, \end{equation}and the infinite continued fraction \begin{equation} e = 2 + \cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{4 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{6 + \cfrac{1}{\dotsb }}}}}}}}}. \end{equation} We will now establish a series of results in order to prove that the limit in \ref{eq:def_of_e} exists. \begin{lem} Let $n$ be a positive integer. Then $$x^n - y^n = (x - y)(x^{n - 1} + x^{n - 2}y + x^{n - 3}y^2 + \cdots + x^2y^{n - 3} + xy^{n - 2} + y^{n - 1}).$$ \label{lem:diffbinom}\end{lem} \begin{pf} The lemma follows by direct multiplication of the dextral side. \end{pf} \begin{lem} If $0 \leq a < b, \ n \in \naturals $ $$na^{n - 1} < \frac{b^{n} - a^{n}}{b - a} < nb^{n - 1}.$$ \label{lem:ineq_bin}\end{lem} \begin{pf} By Lemma \ref{lem:diffbinom} $$\begin{array}{lll} \dfrac{b^{n} - a^{n}}{b - a} & = & b^{n - 1} + b^{n - 2}a + b^{n - 3}a^2 + \cdots + b^2a^{n - 3} + ba^{n - 2} + a^{n - 1} \\ & < & b^{n - 1} + b^{n - 1} + \cdots + b^{n - 1} + b^{n - 1} \\ & = & nb^{n - 1}, \end{array} $$from where the dextral inequality follows. The sinistral inequality can be established similarly. \end{pf} \begin{thm}\label{thm:e} The sequence $$e_n = \left(1 + \frac{1}{n}\right)^n, n = 1, 2, \ldots $$ is a bounded increasing sequence, and hence it converges to a limit, which we call $e$. \end{thm} \begin{pf} By Lemma \ref{lem:ineq_bin} $$ \dfrac{b^{n + 1} - a^{n + 1}}{b - a} \leq (n + 1)b^{n}\implies b^n[(n + 1)a - nb] < a^{n + 1}. $$Putting $a = 1 + \dfrac{1}{n + 1}$, $b = 1 + \dfrac{1}{n}$ we obtain $$ e_n = \left(1 + \frac{1}{n}\right)^n < \left(1 + \frac{1}{n+1}\right)^{n + 1} = e_{n + 1}, $$ whence the sequence $e_n, n = 1, 2, \ldots$ increases. Again, by putting $a = 1$, $b = 1 + \dfrac{1}{2n}$ we obtain $$ \left(1 + \frac{1}{2n}\right)^n < 2 \implies \left(1 + \frac{1}{2n}\right)^{2n} <4 \implies e_{2n} < 4. $$ Since $e_n < e_{2n} < 4$ for all $n$, the sequence is bounded. In view of Theorem \ref{thm:conve_seq} the sequence converges to a limit. We call this limit $e$. \end{pf} \begin{rem} Since the sequence increases towards $e$ we have $$ 2 = \left(1 + \frac{1}{1}\right)^1< e. $$From the proof of Theorem \ref{thm:e} it stems that $2 < e < 4$. In fact, in can be shewn that $e\approx 2.718281828459045235360287471352\ldots$ and so $2 < e < 3$. \end{rem} \begin{rem} $e$ is called the {\em natural} exponential base. The function $x \mapsto e^x$ has the property that any tangent drawn to the curve at the point $x$ has slope $e^x$. The notation $\exp (x) = \exp x = e^x$ is often used. \end{rem} \vspace*{3cm} \begin{figure}[!hptb] \psset{unit=1pc}\centering \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.4pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psplot[algebraic,linewidth=2pt,linecolor=brown,arrows={<->}]{-5}{1.609437912}{2.71828^x} \psplot[algebraic,linewidth=2pt,linecolor=blue,arrows={<->}]{-5}{4}{1+x} \psdot(0,1) \meinecaption{3}{$1 + x \leq e^x \ \forall x\in\reals$.} \label{fig:linelessexp} \end{figure} \begin{thm}\label{thm:line_exponential} If $x\in\reals$ then $$1 + x \leq e^x,$$with equality only for $x = 0$. \end{thm} \begin{pf} From figure \ref{fig:linelessexp}, the line $y= 1 + x$ lies below the graph of $y = e^x$, proving the theorem.\end{pf} \bigskip Replacing $x$ by $x - 1$ we obtain, \begin{cor} $$x \leq e^{x - 1}, \ \forall x\in\reals .$$ Equality occurs if and only if $x = 1$. \label{cor:linelessexp_1}\end{cor} \subsection*{Homework} \addcontentsline{toc}{section}{Homework} \markright{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro}True or False. \begin{multicols}{2} \begin{enumerate} \item $\exists t\in\reals$ such that $e^t = -9$. \item As $x \rightarrow -\infty$, $2^x \rightarrow -\infty$. \item $\forall x\in\reals$, $10 + x^2 + x^4 > 2^x.$ \item $x \mapsto e^x$ is increasing over $\reals$. \item $x \mapsto \frac{e^x}{\pi^x}$ is increasing over $\reals$. \item $ex \leq e^x, \ \ \forall x\in\reals$. \end{enumerate}\end{multicols} \begin{answer} F; F; F; T; F; T \end{answer} \end{pro} \begin{pro} By using Theorem \ref{thm:line_exponential}, and the fact that $\pi > e,$ prove that $e^\pi > \pi^e$. \end{pro}(Hint: Put $x = \frac{\pi}{e} - 1$.) \begin{pro} Make a rough sketch of each of the following. \begin{multicols}{2}\begin{enumerate} \item $x \mapsto 2^x$ \item $x \mapsto e^x$ \item $x \mapsto \dis{\left(\frac{1}{2}\right)^x}$ \item $x \mapsto -1 + 2^x$ \item $x \mapsto e^{|x|}$ \item $x \mapsto e^{-|x|}$ \end{enumerate} \end{multicols} \end{pro} \begin{pro} Let $n\in\naturals, n > 1$. Prove that $$n! < \left(\frac{n + 1}{2}\right)^n.$$ \end{pro} \begin{pro} Put $$\cosh x = \frac{e^x + e^{-x}}{2}$$and $$\sinh x = \frac{e^x - e^{-x} }{2}.$$Prove that $$\cosh^2x - \sinh^2x = 1.$$The function $x \mapsto \cosh x$ is known as the {\em hyperbolic cosine}. The function $x \mapsto \sinh x$ is known as the {\em hyperbolic sine}. \end{pro} \begin{pro}Prove that for $n\in\naturals$, $$\left(1 + \frac{1}{n}\right)^n < \left(1 + \frac{1}{n + 1}\right)^{n + 1}.$$ and $$\left(1 + \frac{1}{n}\right)^{n + 1} > \left(1 + \frac{1}{n + 1}\right)^{n + 2}.$$ \end{pro} (Hint: Use a suitable choice of $a$ and $b$ in Lemma \ref{lem:ineq_bin}.) \begin{pro} Prove that the function $x\mapsto \dfrac{x}{e^x-1}+\dfrac{x}{2}$ is even. \end{pro} \end{multicols} \section{Arithmetic Mean-Geometric Mean Inequality} Using Corollary \ref{cor:linelessexp_1}, we may prove, \`{a} la P\'{o}lya, the Arithmetic-Mean-Geometric-Mean Inequality (AM-GM Inequality, for short). \begin{thm}[Arithmetic-Mean-Geometric-Mean Inequality] Let $$a_1, a_2, \ldots , a_n$$ be non-negative real numbers. Then $$(a_1a_2\cdots a_n)^{1/n} \leq \frac{a_1 + a_2 + \cdots + a_n}{n}. $$ Equality occurs if and only if $a_1 = a_2 = \ldots = a_n.$\end{thm} \begin{pf} Put $$A_k = \frac{na_k}{a_1 + a_2 + \cdots + a_n},$$ and $G_n = a_1a_2\cdots a_n$. Observe that $$A_1A_2\cdots A_n = \frac{n^nG_n}{(a_1 + a_2 + \cdots + a_n )^n},$$ and that $$A_1 + A_2 + \cdots + A_n = n.$$ By Corollary \ref{cor:linelessexp_1}, we have $$A_1 \leq \exp (A_1 - 1),$$ $$A_2 \leq \exp (A_2 - 1),$$ $$\vdots$$ $$A_n \leq \exp (A_n - 1).$$Since all the quantities involved are non-negative, we may multiply all these inequalities together, to obtain, $$A_1A_2\cdots A_n \leq \exp (A_1 + A_2 + \cdots + A_n - n).$$In view of the observations above, the preceding inequality is equivalent to $$\frac{n^nG_n}{(a_1 + a_2 + \cdots + a_n )^n} \leq \exp (n - n) = e^0 = 1.$$We deduce that $$G_n \leq \left(\frac{a_1 + a_2 + \cdots + a_n}{n}\right)^n,$$which is equivalent to $$(a_1a_2\cdots a_n)^{1/n} \leq \frac{a_1 + a_2 + \cdots + a_n}{n}. $$ Now, for equality to occur, we need each of the inequalities $A_k \leq \exp (A_k - 1)$ to hold. This occurs, in view of Corollary \ref{cor:linelessexp_1} if and only if $A_k = 1, \ \forall k$, which translates into $a_1 = a_2 = \ldots = a_n.$. This completes the proof. \end{pf} \begin{cor}[Harmonic-Mean-Geometric-Mean Inequality] If $$a_1, a_2, \ldots , a_n$$ are positive real numbers, then $$\frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}}\leq \sqrt[n]{a_1a_2\cdots a_n}.$$ \end{cor} \begin{pf} By the AM-GM Inequality, $$\sqrt[n]{\frac{1}{a_1}\cdot\frac{1}{a_2}\cdots\frac{1}{a_n}} \leq \frac{\frac{1}{a_1}+ \frac{1}{a_2} + \cdots + \frac{1}{a_n}}{n},$$ from where the result follows by rearranging. \end{pf} \begin{exa} The sum of two positive real numbers is $100$. Find their maximum product. \end{exa} \begin{solu} Let $x$ and $y$ be the numbers. We use the AM-GM Inequality for $n = 2$. Then $$\sqrt{xy} \leq \frac{x + y}{2}.$$ In our case, $x + y = 100,$ and so $$\sqrt{xy} \leq 50,$$ which means that the maximum product is $xy \leq 50^2 = 2500.$ If we take $x = y = 50,$ we see that the maximum product is achieved for this choice of $x$ and $y$. \end{solu} \begin{exa} From a rectangular cardboard piece measuring $75\times 45$ a square of side $x$ is cut from each of its corners in order to make an open box. See figure \ref{fig:openbox_1}. Find the function $x \mapsto V(x)$ that gives the volume of the box as a function of $x$, and obtain an upper bound for the volume of this box. \label{exa:openbox} \end{exa} \begin{solu} From the diagram shewn, the height of the box is $x$, its length $75 - 2x$ and its width $45 - 2x$. Hence $$V(x) = x(75 - 2x)(45 - 2x).$$ Now, if we used the AM-GM Inequality for the three quantities $x, \ 80 - 2x,$ and $50 - 2x$, we would obtain $$\begin{array}{lll}V(x) & = & x(75 - 2x)(45 - 2x) \\ & < & \left(\dfrac{x + 75 - 2x + 45 - 2x }{3}\right)^3 \\ & = & \left(\dfrac{120 - 3x}{3}\right)^3 \\ & = & (40 - x )^3. \end{array}$$ (We use the strict inequality sign because we know that equality will never be achieved: $75 - 2x$ never equals $45 - 2x$.) This has the disadvantage of depending on $x$. In order to overcome this, we use the following trick. Consider, rather, the three quantities $4x, \ 75 - 2x,$ and $45 - 2x$. Then $$\begin{array}{lll} 4V(x) & = & 4x(75 - 2x)(45 - 2x)\\ & < & \left(\dfrac{4x + 75 - 2x + 45 - 2x }{3}\right)^3 \\ & = & \left(\dfrac{120}{3}\right)^3 \\ & = & 64000. \end{array}$$ This means that $$V(x) < \frac{64000}{4} = 16000.$$ \end{solu} \vspace{2cm} \begin{figure}[!hptb] $$ \psset{unit=1.8pc} \rput(-1,-.5){\psline(0,0)(5,0)(5,3)(0,3)(0,0) \psline(0,1)(1,1)(1,0) \psline(0,2)(1,2)(1,3) \psline(4,0)(4,1)(5,1) \psline(4,3)(4,2)(5,2) \uput[180](0,.5){x}\uput[270](.5,0){x} \uput[180](0,2.5){x}\uput[90](.5,3){x} \uput[270](4.5,0){x} \uput[0](5,.5){x} \uput[0](5,2.5){x} \uput[90](4.5,3){x}} $$ \meinecaption{1}{Example \ref{exa:openbox}.} \label{fig:openbox_1} \end{figure} \begin{rem} Later in calculus, you will see that the volume is $$V(x) \leq (20 - \frac{5}{2}\sqrt{19})(35 + 5\sqrt{19})(5 + 5\sqrt{19}),$$ and that the maximum is achieved when $$x = 20 - \frac{5}{2}\sqrt{19}.$$\end{rem} \begin{exa} Find the maximum value of the function $\fun{f}{x}{x(1 - x )^2}{[0;1]}{\reals}$. \end{exa} \begin{solu} Observe that for $x\in[0; 1]$ both $x$ and $1 - x$ are non-negative. We maximise, rather, $2f(x)$ via the AM-GM Inequality. $$ 2f(x) = 2x(1 - x)^2 = 2x(1 - x)(1 - x) \leq \left(\frac{2x + 1 - x + 1 - x}{3}\right)^3 = \frac{8}{27}.$$ Thus $$f(x) \leq \frac{1}{2}\cdot\frac{8}{27} = \frac{4}{27}.$$ The maximum value is attained when $2x = 1 - x$, that is, when $x = \frac{1}{3}.$ \end{solu} \subsection*{Homework} \addcontentsline{toc}{section}{Homework} \markright{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\begin{pro} Let $x, y, z$ be any real numbers. Prove that $$3x^2y^2z^2 \leq x^6 + y^6 + z^6.$$ \end{pro} \begin{pro} The sum of $5$ positive real numbers is $S$. What is their maximum product? \begin{answer} $\frac{S^5}{3125}.$ \end{answer} \end{pro} \begin{pro} Use AM-GM to prove that $\cosh x \geq 1, \ \ \forall x\in\reals.$ \end{pro} \begin{pro} Maximise the following functions over $[0; 1]$. \begin{enumerate} \item $a: x \mapsto x(1 - x)^3$. \item $b: x \mapsto x^2(1 - x)^2$. \item $c: x \mapsto x^2(1 - x)^3$. \end{enumerate} \begin{answer} (1) $a(x) \leq \frac{27}{256}$ achieved at $x = \frac{1}{4}$,\\ (2) $b(x) \leq \frac{1}{16}$ achieved at $x = \frac{1}{2}$, \\ (3) $c(x) \leq \frac{108}{3125}$ achieved at $x = \frac{2}{5}$, (Hint: Consider $\frac{9}{4}c(x) = (\frac{3}{2}x)(\frac{3}{2}x)(1 - x)^3$.) \end{answer} \end{pro} \begin{pro} Prove that of all rectangular boxes with a given surface area, the cube has the largest volume. \end{pro} \end{multicols} \chapter{Logarithmic Functions} \section{Logarithms} Recall that if $a > 0, a \neq 1$ is a fixed real number, $\funnoname{x}{a^x}{\reals}{]0;+\infty[}$ maps a real number $x$ to a positive number $y,$ i.e., $a^x = y$. We call $x$ the {\em logarithm} of $y$ to the base $a$, and we write $x = \log _a\ y$. In other words, the function $\funnoname{x}{a^x}{\reals}{]0;+\infty[}$ has inverse $\funnoname{x}{\log_ax}{]0;+\infty[}{\reals}.$\footnote{ In higher mathematics, and in many computer algebra programmes like Maple \circledR, the notation ``$\log$'' without indicating the base, is used for the natural logarithm of base $e$. Misguided authors, enemies of the State, communists,Al-Qaeda members, vegetarians and other vile criminals use ``$\log$'' in calculators and in lower mathematics to denote the logarithm of base $10$, and use ``$\ln$'' to denote the natural logarithm. This makes things somewhat confusing. In these notes we will denote the logarithm base 10 by ``$\log _{10}$'' and the natural logarithm by ``$\log _e$'', which is hardly original but avoids confusion.} \begin{exa} $\log _5 \ 25 = 2$ since $5^2 = 25.$ \end{exa} \begin{exa} $\log _2\ 1024 = 10$ since $2^{10} = 1024.$ \end{exa} \begin{exa} $\log _3\ 27 = 3$ since $3^3 = 27.$ \end{exa} \begin{exa} $\log _{190123456}\ 1 = 0$ as $190123456^0 = 1.$ \end{exa} \begin{rem} If $a > 0, a \neq 1$, it should be clear that $\log _a 1 = 0, \ \log _a a = 1,$ and in general $\log _a a^t = t$, where $t$ is any real number. \end{rem} \begin{exa} $\log _{\sqrt{2}}\ 8 = \log _{2^{1/2}}\ (2^{1/2})^6= 6$. \end{exa} \begin{exa} $\log _{\sqrt{2}}\ 32 = \log _{2^{1/2}}\ (2^{1/2})^{10} = 10.$ \end{exa} \begin{exa} $\dis{\log _{3\sqrt{3}}\ 81\sqrt[8]{27} = \log _{3^{3/2}}\ (3^{3/2})^{(2/3)(35/8)} = \frac{2}{3}\cdot\frac{35}{8} = \frac{35}{12}}$. \end{exa} \flushleft{\em Aliter:} We seek a solution $x$ to $$(3\sqrt{3})^x = 81\sqrt[8]{27}$$ Expressing the sinistral side as powers of 3, we have $$ \begin{array}{lll} (3\sqrt{3})^x & = & (3\cdot 3^{1/2})^x \\ & = & (3^{1 + 1/2})^x \\ & = & (3^{3/2})^x \\ & = & 3^{3x/2} \end{array} $$ Also, the dextral side equals $$ \begin{array}{lll} 81\sqrt[8]{27} & = & 3^4\cdot (3^3)^{1/8} \\ & = & 3^{4 + 3/8} \\ & = & 3^{35/8} \end{array} $$ Thus $(3\sqrt{3})^x = 81\sqrt[8]{27}$ implies that $3^{3x/2} = 3^{35/8}$ or $\dis{\frac{3x}{2} = \frac{35}{8}}$ from where $x = \frac{35}{12}$. \vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{.4\textwidth}\psset{unit=1pc}\centering \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.4pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psplot[algebraic,linewidth=2pt,linecolor=brown,arrows={<->}]{.006737946999}{5}{ln(x)} \meinecaption{3}{$x \mapsto \log _a x, a > 1$} \label{logbaseg1} \end{minipage} \hfill\begin{minipage}{.4\textwidth}\psset{unit=1pc}\centering \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.4pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=1.2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psplot[algebraic,linewidth=2pt,linecolor=brown,arrows={<->}]{.006737946999}{5}{(ln(x))/(ln(1/2.718281828))} \meinecaption{3}{$x \mapsto \log _a x, 0 < a < 1$.} \label{logbases1} \end{minipage} \end{figure} Since $x \mapsto a^x$ and $x \mapsto \log_ax$ are inverses, the graph of $x \mapsto \log_ax$ is symmetric with respect to the line $y = x$ to the graph of $x \mapsto a^x$. For $a > 1, x \mapsto a^x$ is increasing and convex, $x \mapsto \log _a\ x, \ a > 1$ will be increasing and concave, as in figure \ref{logbaseg1}. Also, for $0 < a < 1, x \mapsto a^x$ is decreasing and convex, $x \mapsto \log _a\ x, \ 0 < a < 1$ will be decreasing and concave, as in figure \ref{logbases1}. \begin{exa} Between which two integers does $\log _2\ 1000$ lie? \end{exa} \begin{solu} Observe that $2^9 = 512 < 1000 < 1024 = 2^{10}$. Since $x \mapsto \log _2 x$ is increasing, we deduce that $\log _2 1000$ lies between $9$ and $10$. \end{solu} \begin{exa} Find $\lfloor \log _3\ 201 \rfloor$. \end{exa} \begin{solu} $3^{4} = 81 < 201 < 243 = 3^5.$ Hence $\lfloor \log _3 201 \rfloor = 4$. \end{solu} \begin{exa} Which is greater $\log _5 7$ or $\log _{8} 3$?\end{exa} \begin{solu} Clearly $\log _5 7 > 1 > \log _8 3$. \end{solu} \begin{exa} Find the integer that equals $$\lfloor \log _2 1 \rfloor + \lfloor \log _2 2\rfloor + \lfloor \log _2 3\rfloor + \lfloor \log _2 4\rfloor + \cdots + \lfloor \log _2 66 \rfloor .$$ \end{exa} \begin{solu} Firstly, $\log _2\ 1 = 0$. We may decompose the interval $[2; 66]$ into dyadic blocks, as $$ [2; 66] = [2; 4[ \cup [4; 8[ \cup [8; 16[ \cup [16,; 32[ \cup [32,; 64[ \cup [64; 66].$$ On the first interval there are $4 - 2 = 2$ integers with $\lfloor\log _2\ x \rfloor = 1, x \in [2; 4[$. On the second interval there are $8 - 4 = 4$ integers with $\lfloor\log _2\ x \rfloor = 2, x \in [4; 8[$. On the third interval there are $16 - 8 = 8$ integers with $\lfloor \log _2\ x \rfloor = 3, x \in [8; 16[$. On the fourth interval there are $32 - 16 = 16$ integers with $\lfloor\log _2\ x\rfloor = 4, x \in [16; 32[$. On the fifth interval there are $64 - 32 = 32$ integers with $\lfloor \log _2\ x \rfloor = 5, x \in [32; 64[$. On the sixth interval there are $66 - 64 + 1= 3$ integers with $\lfloor\log _2\ x \rfloor = 6, x \in [64; 66]$. Thus $$ \begin{array}{lll} \lfloor \log _2\ 1 \rfloor + \lfloor \log _2\ 2\rfloor + \lfloor \log _2\ 3\rfloor + & & \\ \qquad + \lfloor \log _2\ 4\rfloor + \cdots + \lfloor \log _2\ 66 \rfloor & &\\ & = & 2(1) + 4(2) + 8(3) + \\ & & \qquad + 16(4) + 32(5) + 3(6) \\ & = & 276. \end{array}$$ \end{solu} \begin{exa} What is the natural domain of definition of $x\mapsto \log_2 (x^2 - 3x - 4)$? \end{exa} \begin{solu} We need $x^2 - 3x -4 = (x-4)(x+1) > 0$. By making a sign diagram, or looking at the graph of the parabola $y = (x-4)(x+1)$ we see that this occurs when $x\in ]-\infty;-1[\cup ]4;+\infty[$. \end{solu} \begin{exa} What is the natural domain of definition of $x\mapsto \log_{|x|-4} (2 - x)$? \end{exa} \begin{solu} We need $2-x > 0$ and $|x|-4\neq 1$. Thus $x<2$ and $x\neq 5, x\neq -5$. We must have $x\in ]-\infty;-5[\cup ]-5;2[$. \end{solu} \subsection*{Homework} \addcontentsline{toc}{section}{Homework} \markright{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} True or False. \begin{enumerate} \begin{multicols}{2}\columnseprule 1pt \columnsep 5pt \multicoltolerance=500 \item $\exists x \in \reals$ such that $\log_4 x = 2.$ \item $\exists x \in \reals$ such that $\log_4 x = -2.$ \item $\log_2 1 = 0.$ \item $\log_2 0 = 1.$ \item $\log_2 2 = 1.$ \item $x \mapsto \log_{1/5} x$ is increasing over $\reals_+^*$. \item $\forall x > 0, (\log_5 x)^2 = \log_5 x^2.$ \item $\log _3\ 201 = 4$. \end{multicols}\end{enumerate} \begin{answer} T; T; T; F; T; T; F; F\end{answer} \end{pro} \begin{pro} Compute the following. \begin{enumerate} \item $\dis{\log _{1/3}\ 243}$ \item $\dis{\log _{10}\ .00001}$ \item $\dis{\log _{.001}\ 100000}$ \item $\dis{\log _{9}\ \frac{1}{3}}$ \item $\dis{\log _{1024}\ 64}$ \item $\dis{\log _{5^{2/3}}\ 625}$ \item $\dis{\log _{2\sqrt{2}}\ 32\sqrt[5]{2}}$ \item $\dis{\log _{2}\ .0625}$ \item $\dis{\log _{.0625}\ 2}$ \item $\dis{\log _{3}\ \sqrt[4]{729\sqrt[3]{9^{-1}27^{-4/3}}}}$ \end{enumerate} \begin{answer} (1) $-5$, (2) $-5$, (3) $-\frac{5}{3}$, (4) $-\frac{1}{2}$, (5) $\frac{3}{5}$, (6) $6$, (7) $\frac{52}{15}$, (8) $-4$ , (9) $-\frac{1}{4}$, (10) $1$ \end{answer} \end{pro} \begin{pro} Let $a > 0, a \neq 1.$ Compute the following. \begin{enumerate} \item $\log _{a}\ \sqrt[4]{a^{8/5}}$ \item $\log _{a}\ \sqrt[3]{a^{-15/2}}$ \item $\log _{a}\ \dis{\frac{1}{a^{1/2}}}$ \item $\log _{a^3}\ a^6$ \item $\log _{a^2}\ a^3$ \item $\log _{a^{5/6}}\ a^{7/25} $ \end{enumerate} \begin{answer} (1) $\frac{2}{5}$, (2) $-\frac{5}{2}$, (3) $-\frac{1}{2}$, (4) $2$, (5) $\frac{42}{125}$ \end{answer} \end{pro} \begin{pro} Make a rough sketch of the following. \begin{enumerate} \item $x \mapsto \log _2\ x$ \item $x \mapsto \log _2\ |x|$ \item $x \mapsto 4 + \log _{1/2}\ x$ \item $x \mapsto 5 - \log _3\ x$ \item $x \mapsto 2 - \log _{1/4}\ x$ \item $x \mapsto \log _{5}\ x$ \item $x \mapsto \log _{5}\ |x|$ \item $x \mapsto |\log _{5}\ x|$ \item $x \mapsto |\log _{5}\ |x||$ \item $x \mapsto 2 + \log_e |x|$ \item $x \mapsto -3 + \log_{1/2} |x|$ \item $x \mapsto 5 - |\log_4 x|$ \end{enumerate} \end{pro} \begin{pro} Prove that for $x > 0$, $$1 - x \leq -\log_e x.$$ \end{pro} \begin{pro} Prove that for $x > 0$ we have $$x^e \leq e^x.$$ Use this to prove that for $x > 0$, $$\log_e x \leq \frac{x}{e}.$$ \end{pro} \begin{pro}Find the natural domain of definition of the following. \begin{enumerate} \item $x\mapsto \log_2 (x^2 - 4)$ \item $x\mapsto \log_2 (x^2 +4)$ \item $x\mapsto \log_2 (4 - x^2)$ \item $x\mapsto \log_2 (\frac{x + 1}{x - 2})$ \item $x\mapsto \log_{x^2 + 1} (x^2 +1)$ \item $x\mapsto \log_{1-x^2} x$ \end{enumerate} \end{pro} \end{multicols} \section{Simple Exponential and Logarithmic Equations} Recall that for $a > 0, \ a \neq 1,\ b > 0$ the relation $a^x = b$ entails $x = \log _a\ b.$ This proves useful in solving the following equations. \begin{exa} Solve the equation $$\log _4 x = -3.$$ \end{exa} \begin{solu} $x = 4^{-3} = \dis{\frac{1}{64}}$. \end{solu} \begin{exa} Solve the equation $$\log _2 x = 5.$$ \end{exa} \begin{solu} $x = 2^{5} = 32$. \end{solu} \begin{exa} Solve the equation $$\log _x\ 16 = 2.$$ \end{exa} \begin{solu} $16 = x^2$. Since the base must be positive, we have $x = 4$. \end{solu} \begin{exa} Solve the equation $3^x = 2$. \end{exa}\begin{solu} By definition, $x = \log _3\ 2$. \end{solu} \begin{exa} Solve the equation $9^x - 5\cdot 3^x + 6 = 0.$ \end{exa} \begin{solu} We have $$9^x - 5\cdot 3^x + 6 = (3^x)^2 - 5\cdot 3^x + 6 = (3^x - 2)(3^x - 3).$$Thus either $3^x - 2 = 0$ or $3^x - 3 = 0$. This implies that $x = \log _3\ 2$ or $x = 1$. \end{solu} \begin{exa} Solve the equation $25^x - 5^x - 6 = 0.$ \end{exa} \begin{solu} We have $$25^x - 5^x - 6 = (5^x)^2 - 5^x - 6 = (5^x + 2)(5^x - 3),$$whence $5^x - 3 = 0$ or $x = \log_5\ 3$ as $5^x + 2 = 0$ does not have a real solution. (Why?) \end{solu} \par\bigskip Since $x \mapsto a^x$ and $x \mapsto \log _a x$ are inverses, we have \begin{equation} x = a^{\log _a x} \ \forall a > 0, \ a \neq 1, \ \forall x > 0 \end{equation} Thus for example, $5^{\log _5 4} = 4, \ 26^{\log _{26} 8} = 8$. This relation will prove useful in solving some simple equations. \begin{exa} Solve the equation $$\log _2\log _4 x = -1.$$ \begin{solu} As $\log _2\log _4 x = -1,$ we have $$\log _4 x = 2^{\log _2\log _4 x} = 2^{-1} = \frac{1}{2}.$$ Hence $\dis{x = 4^{\log _4 x} = 4^{1/2} = \sqrt{4} = 2.}$ \end{solu} \end{exa} \begin{exa} Solve the equation $$\log _2 \log _3 \log _5 x = 0$$ \end{exa} \begin{solu} Since $\log _2 \log _3 \log _5 x = 0$ we have $$\log _3\log_5 x = 2^{\log _2\log _3\log _5 x} = 2^0 = 1.$$ Hence $$\log _5 x = 3^{\log _3\log _5 x} = 3^1 = 3.$$ Finally $x = 5^{\log _5 x} = 5^3 = 125.$ \end{solu} \begin{exa} Solve the equation $$\log _2 x(x - 1) = 1.$$ \end{exa}\begin{solu} We have $x(x - 1) = 2^1 = 2.$ Hence $x^2 - x - 2 = 0.$ This gives $x = 2$ or $x = -1$. Check that both are indeed solutions! \end{solu} \begin{exa} Solve the equation $\log_{e + x} e^8 = 2$. \end{exa} \begin{solu} We have $(e + x)^2 = e^8$ or $e + x = \pm e^4.$ Now the base $e + x$ cannot be negative, so we discard the minus sign alternative. The only solution is when $e + x = e^4,$ that is, $x = e^4 - e$. \end{solu} \subsection*{Homework} \addcontentsline{toc}{section}{Homework} \markright{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Find real solutions to the following equations for $x$. \begin{enumerate} \item $\log _{x}\ 3 = 4$ \item $\log _{3}\ x = 4$ \item $\log _{4}\ x = 3$ \item $\log _{x - 2}\ 9 = 2$ \item $\log _{|x|}\ 16 = 4$ \item $23^x - 2 = 0$ \item $(2^x - 3)(3^x - 2)(6^x - 1) = 0$ \item $4^x - 9\cdot 2^x + 14 = 0$ \item $49^x - 2\cdot 7^x + 1 = 0$ \item $36^x - 2\cdot 6^x = 0$ \item $36^x + 6^x - 6 = 0$ \item $5^x + 12\cdot 5^{-x} = 7$ \item $\log_2\log_3 x = 2$ \item $\log _3\log _5 x = -1 $ \end{enumerate} \begin{answer} (1) $\sqrt[4]{3}$, (2) $81$, (3) $64$ (4) $5$, (5) $\pm 2$, (6) $\log_{23}2$, (7) $\log_23, \ \log_32,\ 0$, (8) $\log_2 7, 1,$ (9) $0$, (10) $\log_62$, (11) $\log_62,$ (12) $\log_5 4, \log_53,$ (13) $81$, (14) $\sqrt[3]{5}$ \end{answer} \end{pro} \end{multicols} \section{Properties of Logarithms} A few properties of logarithms that will simplify operations with them will now be deduced. \begin{thm} If If $a > 0, a \neq 1,\ M > 0,$ and $\alpha$ is any real number, then \begin{equation} \log _a\ M^\alpha = \alpha\log _a\ M \end{equation} \end{thm} \begin{pf} Let $x = \log _a\ M$. Then $a^x = M$. Raising both sides of this equality to the exponent $\alpha,$ one gathers $a^{\alpha x} = M^\alpha$. But this entails that $\log _a M^\alpha = \alpha x = \alpha (\log _a\ M)$, which proves the theorem. \end{pf} \begin{exa} How many digits does $8^{330}$ have? \end{exa} \begin{solu} Let $n$ be the integer such that $\dis{10^n < 8^{330} < 10^{n + 1}}$. Clearly then $8^{330}$ has $n + 1$ digits. Since $x \mapsto \log _{10} x$ is increasing, taking logarithms base 10 one has $n < 330\log _{10} 8 < n + 1$. Using a calculator, we see that $298.001 < 330\log _{10} 8 < 298.02$, whence $n = 298$ and so $8^{330}$ has $299$ digits. \end{solu} \begin{exa} If $\log _a t = 2,$ then $\log _a t^3 = 3\log _a t = 3(2) = 6.$ \end{exa} \begin{exa} $\log _5\ 125 = \log _5 \ 5^3 = 3\log _5\ 5 = 3(1) = 3.$ \end{exa} \begin{thm} Let $a > 0, a \neq 1,\ M > 0,$ and let $\beta \neq 0$ be a real number. Then \begin{equation} \log _{ a^\beta}\ M = \frac{1}{\beta} \log _a\ M. \end{equation} \end{thm} \begin{pf} Let $x = \log _a\ M$. Then $a^x = M$. Raising both sides of this equality to the power $\dis{\frac{1}{\beta}}$ we gather $a^{x/\beta} = M^{1/\beta}$. But this entails that $$ \log _a M^{1/\beta} = \frac{x}{\beta} = \frac{1}{\beta}(\log _a\ M), $$ which proves the theorem. \end{pf} \begin{exa} Given that $\dis{\log _{8\sqrt{2}} 1024}$ is a rational number, find it. \end{exa} \begin{solu} We have $$\log _{8\sqrt{2}} 1024 = \log _{2^{7/2}} 1024 = \frac{2}{7}\log _{2} 2^{10} = \frac{2}{7}\cdot 10\log _2\ 2 = \frac{20}{7}. $$ \end{solu} \begin{thm} If $a > 0, a \neq 1,\ M > 0,\ N > 0$ then \begin{equation} \log _a\ MN = \log _a\ M + \log _a\ N \end{equation} In words, the logarithm of a product is the sum of the logarithms. \end{thm} \begin{pf} Let $x = \log _a\ M$ and let $y = \log _a\ N$. Then $a^x = M$ and $a^{y} = N$. This entails that $a^{x + y} = a^xa^y = MN$. But $a^{x + y} = MN$ entails $x + y = \log _a\ MN$, that is $$\log _a\ M + \log _a\ N = x + y = \log _a\ MN,$$as required. \end{pf} \begin{exa} If $\log _a\ t = 2, \ \log _a\ p = 3$ and $\log _a\ u^3 = 21$, find $\log _a\ t^3pu.$ \end{exa} \begin{solu} First observe that $\log _a\ t^3pu = \log _a\ t^3 + \log _a\ p + \log _a u.$ Now, $\log _a t^3 = 3\log _a\ t = 6. $ Also, $21 = \log _a u^3 = 3\log _a\ u,$ from where $\log _a\ u = 7.$ Hence $$\log _a\ t^3pu = \log _a\ t^3 + \log _a\ p + \log _a u = 6 + 3 + 7 = 16.$$ \end{solu} \begin{exa} Solve the equation $$\log _2 x + \log _2 (x - 1) = 1.$$ \end{exa}\begin{solu} If $x > 1$ then $$\log _2 x + \log _2 (x - 1) = \log_2x(x - 1).$$This entails $x(x - 1) = 2,$ from where $x = -1$ or $x = 2$. The solution $x = -1$ must be discarded, as we need $x > 1$. \end{solu} \begin{thm} If $a > 0, a \neq 1,\ M > 0,\ N > 0$ then \begin{equation} \log _a \frac{M}{N} = \log _a M - \log _a N \end{equation} \end{thm} \begin{pf} Let $x = \log _a\ M$ and let $y = \log _a\ N$. Then $a^x = M$ and $a^{y} = N$. This entails that $\dis{a^{x - y} = \frac{a^x}{a^y} = \frac{M}{N}}$. But $\dis{a^{x - y} = \frac{M}{N}}$ entails $x - y = \log _a\ \frac{M}{N}$, that is $$\log _a\ M - \log _a\ N = x - y = \log _a\ \frac{M}{N},$$as required. \end{pf} \begin{exa} Let $\log _a\ t = 2, \ \log _a\ p = 3$ and $\log _a\ u^3 = 21$, find $\dis{\log _a\ \frac{p^2}{tu} }.$ \end{exa} \begin{solu} First observe that $$\log _a\ \frac{p^2}{tu} = \log _a\ p^2 - \log _a\ tu = 2\log _a p - (\log _a\ t + \log _a\ u).$$ This entails that $$\log _a \frac{p^2}{tu} = 2(3) - (2 + 21) = -17.$$ \end{solu} \begin{thm} If $a > 0, a \neq 1, \ b > 0, b \neq 1$ and $M > 0$ then \begin{equation} \log _a\ M = \frac{\log _b\ M}{\log _b\ a}. \end{equation} \end{thm} \begin{pf} From the identity $\dis{b^{\log _b\ M} = M}$, we obtain, upon taking logarithms base $a$ on both sides $$\log _a\left(b^{\log _b\ M}\right) = \log _a\ M.$$By Theorem 3.4.1 $$\log _a\ \left(b^{\log _b\ M}\right) = (\log _b\ M)(\log _a\ b),$$ whence the theorem follows. \end{pf} \begin{exa} Given that $$ (\log _2 3)\cdot (\log _3 4)\cdot (\log _4 5) \cdots (\log _{511} 512)$$is an integer, find it. \end{exa} \begin{solu} Choose $a > 0, a \neq 1$. Then $$ \begin{array}{lll} (\log _2 3)\cdot (\log _3 4)\cdot (\log _4 5) \cdots (\log _{511} 512) & = & \dfrac{\log _a 3}{\log _a 2}\cdot\dfrac{\log _a 4}{\log _a 3}\cdot \dfrac{\log _a 5}{\log _a 4}\cdots \dfrac{\log _a 512}{\log _a 511}\\ & = & \dfrac{\log _a 512}{\log _a 2}. \\ \end{array}$$But $$\dfrac{\log _a 512}{\log _a 2} = \log_2 512 = \log _2 2^9 = 9,$$so the integer sought is $9$. \end{solu} \begin{cor} If $a > 0, a \neq 1, \ b > 0, b \neq 1$ then \begin{equation} \log _a\ b = \frac{1}{\log _b\ a}. \end{equation} \end{cor} \begin{pf} Let $M = b$ in the preceding theorem. \end{pf} \begin{exa} Given that $\log _n\ t = 2a, \ \log _s n = 3a^2$, find $\log _t s$ in terms of $a$. \end{exa}\begin{solu} We have $$\log_t s = \frac{\log_n s}{\log_n t}.$$Now, $\dis{\log_n s = \frac{1}{\log_s n} = \frac{1}{3a^2}}$. Hence $$\log_t s = \frac{\log_n s}{\log_n t} = \frac{\frac{1}{3a^2}}{2a} = \frac{1}{6a^3}.$$ \end{solu} \begin{exa} Given that $\log _a 3 = s^{-3},\ \log _{\sqrt{3}} b = s^2 + 2, \ \log _9 c = s^3$, write $\dis{\log _3\ \frac{a^2b^5}{c^4}}$ as a polynomial in $s$. \end{exa}\begin{solu} Observe that $$\log _3\ \frac{a^2b^5}{c^4} = 2\log_3a + 5\log_3b - 4\log_3c,$$so we seek information about $\log_3a, \log_3b$ and $\log_3c$. Now, $$\log_3a = \frac{1}{\log_a3} = s^3,\ \ \ \log_3b = \frac{1}{2}\log_{\sqrt{3}}b = \frac{1}{2}s^2 + 1$$ and $\log_3c = 2\log_9c = 2s^3.$ Hence $$\log _3\ \frac{a^2b^5}{c^4} = 2s^3 + \frac{5}{2}s^2 + 5 - 8s^3 = -6s^3 + \frac{5}{2}s^2 + 5.$$ \end{solu} \begin{exa} Given that $.63 < \log _3\ 2 < .631$, find the smallest positive integer $a$ such that $3^a > 2^{102}$. \end{exa} \begin{solu} Since $x \mapsto \log _3\ x$ is an increasing function, we have $a \log _3 \ 3 > 102 \log _3\ 2$, that is, $a > 102\log _3\ 2.$ Using the given information, $64.26 < 102 \log _3\ 2 < 64.362$, which means that $a = 65$ is the smallest such integer. \end{solu} \begin{exa} Assume that there is a positive real number $x$ such that $$x^{x^{x^{.^{.^{.}}}}} = 2,$$where there is an infinite number of $x$'s. What is the value of $x$? \end{exa} \begin{solu} Since $\dis{x^{x^{x^{.^{.^{.}}}}} = 2},$ one has $$2 = x^{x^{x^{.^{.^{.}}}}} = x^2,$$whence, as $x$ is positive, $x = \sqrt{2}.$ \end{solu} \begin{rem} Euler shewed that the equation $$a^{x^{x^{.^{.^{.}}}}} = x$$has real roots only for $a\in [e^{-e}; e^{1/e}]$. \end{rem} \begin{exa} How many real positive solutions does the equation $$x^{(x^x)} = (x^x)^x$$have? \end{exa}\begin{solu} Assuming $x > 0$ we have $x^x\log_e x = x\log_e x^x$ or $x^x\log_e x = x^2\log_e x.$ Thus $(\log_ex)(x^x - x^2) = 0$. Thus either $\log_ex = 0$, in which case $x = 1,$ or $x^x = x^2,$ in which case $x = 2$. The equation has therefore only two positive solutions. \end{solu} \begin{exa} The non-negative integers smaller than $10^n$ are split into two subsets $A$ and $B$. The subset $A$ contains all those integers whose decimal expansion does not contain a $5$, and the set $B$ contains all those integers whose decimal expansion contains at least one $5$. Given $n$, which subset, $A$ or $B$ is the larger set? One may use the fact that $\log _{10}\ 2 := .3010$ and that $\log _{10}\ 3 := .4771.$ \end{exa} \begin{solu} The set $B$ contains $10^n - 9^n$ elements and the set $A$ contains $9^n$ elements. Now if $10^n - 9^n > 9^n$ then $10^n > 2\cdot 9^n$ and taking logarithms base 10 we deduce $$n > \log _{10} \ 2 + 2n \log _{10}\ 3.$$Thus $$ n > \frac{\log _{10}\ 2}{1 - 2 \log _{10}\ 3} := 6.57...$$Therefore, if $n \leq 6$, $A$ has more elements than $B$ and if $n > 6,$ $B$ has more elements than $A$. \end{solu} \begin{exa} Shew that if $a, b, c,$ are real numbers with $a^2 = b^2 + c^2, a + b > 0, a + b \neq 1, a - b > 0, a - b \neq 1,$ then $$\log _{a - b}\ c + \log _{a + b}\ c = 2(\log _{a - b}\ c)(\log _{a + b}\ c).$$ \end{exa} \begin{solu} As $c^2 = a^2 - b^2 = (a - b)(a + b)$, upon taking logarithms base $a + b$ we have \begin{equation} 2\log _{a + b}\ c = \log _{a + b}\ (a - b)(a + b) = 1 + \log _{a + b}\ (a - b) \end{equation} Similarly, taking logarithms base $a - b$ on the identity $c^2 = (a - b)(a + b)$ we obtain \begin{equation} 2\log _{a - b}\ c = \log _{a - b}\ (a - b)(a + b) = 1 + \log _{a - b}\ (a + b) \end{equation} Multiplying these last two identities, $$ \begin{array}{lll} 4(\log _{a - b}\ c)(\log _{a + b}\ c) & = & (1 + \log _{a + b}\ (a - b))(1 + \log _{a - b}\ (a + b)) \\ & = & 1 + \log _{a - b}\ (a + b) + \log _{a + b}\ (a - b)\\ & & \qquad + (\log _{a - b}\ (a + b)) (\log _{a + b}\ (a - b)) \\ & = & 2 + \log _{a - b}\ (a + b) + \log _{a + b}\ (a - b) \\ & = & 2 + \log _{a - b}\ \frac{c}{a - b} + \log _{a + b}\ \frac{c}{a + b} \\ & = & \log _{a - b}\ c + \log _{a + b}\ c, \end{array} $$as we wanted to shew. \end{solu} \begin{exa} If $\log _{12} \ 27 = a$ prove that $\dis{\log _6\ 16 = \frac{4(3 - a)}{3 + a}.}$ \end{exa} \begin{solu} First notice that $\dis{a = \log_{12} \ 27 = 3\log_{12}\ 3 = \frac{3}{\log_{3}\ 12} = \frac{3}{1 + 2\log_{3}\ 2}}$, whence $\dis{\log_{3}\ 2 = \frac{3 - a}{2a}}$ or $\dis{\log_{2}\ 3 = \frac{2a}{3 - a}}$. Also $$\begin{array}{lll} \log _6 \ 16 & = & 4 \log_{6}\ 2 \\ & = & \frac{4}{\log_{2}\ 6} \\ & = & \frac{4}{1 + \log_{2}\ 3} \\ & = & \frac{4}{1 + \frac{2a}{3 - a}} \\ & = & \frac{4(3 - a)}{3 + a}, \end{array} $$as required. \end{solu} \begin{exa} Solve the system $$5\left(\log _x y + \log _y x\right) = 26$$ $$xy = 64$$ \end{exa} \begin{solu} Clearly we need $x > 0, y > 0, x \neq 1, y \neq 1.$ The first equation may be written as $5\left(\log _x y + \dfrac{1}{\log _x y}\right) = 26$ which is the same as $(\log _x y - 5)(\log _y x - \dfrac{1}{5}) = 0$. Thus the system splits into the two equivalent systems (I) $\log _x y = 5, xy = 64$ and (II) $\log _x y = 1/5, xy = 64.$ Using the conditions $x > 0, y > 0, x \neq 1, y \neq 1$ we obtain the two sets of solutions $x = 2, y = 32$ or $x = 32, y = 2.$ \end{solu} \subsection*{Homework} \addcontentsline{toc}{section}{Homework} \markright{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Find the exact value of $$\frac{1}{\log _2 1996!} + \frac{1}{\log _3 1996!} +\frac{1}{\log _4 1996!} + \cdots + \frac{1}{\log _{1996} 1996!}.$$ \begin{answer} $1$ \end{answer} \end{pro} \begin{pro} \begin{enumerate} \item $\log_4 MN = \log_4 M + \log N \ \forall M, N\in \reals$. \item $\log_5M^2 = 2\log_5M \forall M\in\reals$. \item $\exists \ M\in\reals$ such that $\log_5M^2 = 2\log_5M$. \end{enumerate} \begin{answer} F; F; T \end{answer} \end{pro} \begin{pro} Given that $\log _a p = 2, \ \log _a m = 9, \ \log _a n = -1$ find \begin{enumerate} \item $\log _a p^7$ \item $\log _{a^7} p$ \item $\log _{a^4} p^2n^3$ \item $\dis{\log _{a^6} \frac{m^3n}{p^6}}$ \end{enumerate} \begin{answer} (1) $14$, (2) $\frac{2}{7}$, (3) $\frac{1}{4}$, (4) $\frac{7}{3}$ \end{answer} \end{pro} \begin{pro} Which number is larger, $3^{1000}$ or $5^{600}$? \begin{answer} $3^{1000}$ \end{answer} \end{pro} \begin{pro} Find $(\log _3\ 169)(\log _{13}\ 243)$ without recourse of a calculator or tables. \begin{answer} $10$ \end{answer} \end{pro} \begin{pro} Find $\dis{\frac{1}{\log _{2} \ 36} + \frac{1}{\log _{3} \ 36}}$ without recourse of a calculator or tables. \begin{answer} $\frac{1}{2}$ \end{answer} \end{pro} \begin{pro} Given that $\log _a\ p = b, \ \log _q a = 3b^{-2}$, find $\log _p q$ in terms of $b$. \begin{answer} $\frac{b}{3}$ \end{answer} \end{pro} \begin{pro} Given that $\log _2 a = s,\ \log _4 b = s^2, \ \log _{c^2} 8 = \frac{2}{s^3 + 1}$, write $\dis{\log _2\ \frac{a^2b^5}{c^4}}$ as a function of $s$. \begin{answer} $-3s^3 + 10s^2 + 2s - 3$ \end{answer} \end{pro} \begin{pro} Given that $\log _{a^2} (a^2 + 1) = 16$, find the value of $$\log _{a^{32}\ \left(a + \frac{1}{a}\right)}.$$ \begin{answer} $\frac{31}{32}$ \end{answer} \end{pro} \begin{pro} Write without logarithms. Assume the proper restrictions on the variables wherever necessary. \begin{enumerate}\renewcommand{\arraystretch}{1.7} \item $\dis{(a^\alpha)^{-\beta\log _{\alpha ^S}\ N^\gamma}}$ \item $-\log _8\ \log _4\ \log _2\ 16$ \item $\log_{0.75}\ \log_{2}\ \sqrt{\sqrt[-2]{0.125}}$ \item $\dis{\left({5^{(\log_{7}\ 5)^{-1}} + {(-\log_{10}\ 0.1}})^{-1/2}\right)^{1/3}}$ \item $\dis{b^{a^{({\log_{b}\log_{b}N})/{(\log_{b}a)}}}}$ \item $\dis{2^{(\log _3\ 5)} - 5^{(\log _3\ 2)}}$ \item $\dis{\left(\frac{1}{49}\right)^{1 + (\log _7\ 2)} + 5^{-(\log _{1/5}\ 7)}}$ \end{enumerate} \begin{answer} (1) $N^{-\alpha\beta\gamma /s}$, (2) $0$, (3) $1$ (4) $2$, (5) $N$, (6) $0$, (7) $\frac{1373}{196}$ \end{answer} \end{pro} \begin{pro} A sheet of paper has approximately $0.1$ mm of thickness. Suppose you fold the sheet by halves, thirty times consecutively. (1) What is the thickness of the folded paper?, (2) How many times should you fold the sheet in order to obtain the distance from Earth to the Moon? (the distance from Earth to the Moon is about 384 000 km.) \begin{answer} (1) About 107.37 km (2) 42 times. \end{answer} \end{pro} \begin{pro} How many digits does $11^{2000}$ have? \begin{answer} $2083$ \end{answer} \end{pro} \begin{pro} Let $A = \log _6 16, B = \log _{12} 27$. Find integers $a, b, c$ such that $(A + a)(B + b) = c$. \begin{answer} $a = 4, b = 3, c = 24$ \end{answer} \end{pro} \begin{pro} Given that $\log _{ab} a = 4$, find $$\log _{ab} \frac{\sqrt[3]{a}}{\sqrt{b}}.$$ \begin{answer} $\frac{17}{6}$ \end{answer} \end{pro} \begin{pro} The number $5^{100}$ is written in binary (base-$2$) notation. How many binary digits does it have? \end{pro} \begin{pro} Prove that if $x > 0, a > 0, a \neq 1$ then $\dis{x^{{1}/{\log_{a}\ x}} = a.}$ \end{pro} \begin{pro} Let $a, b, x$ be positive real numbers distinct from $1$. When is it true that $$4(\log _a\ x)^2 + 3(\log _b\ x)^2 = 8(\log _a x)(\log _b\ x)\ ?$$ \end{pro} \begin{pro} Prove that $\log _3 \pi + \log _\pi 3 > 2.$ \end{pro} \begin{pro} Solve the equation $$4\cdot 9^{x - 1} = 3\sqrt{2^{2x + 1}}$$ \end{pro} \begin{pro} Solve the equation $$5^{x - 1} + 5\left(0.2\right)^{x - 2} = 26$$ \end{pro} \begin{pro} Solve the equation $$25^x - 12\cdot 2^x - (6.25)(0.16)^x = 0$$ \end{pro} \begin{pro} Solve the equation $$\log _3(3^x - 8) = 2 - x$$ \end{pro} \begin{pro} Solve the equation $$\log_4 (x^2 - 6x + 7) = \log_4 (x - 3)$$ \end{pro} \begin{pro} Solve the equation $$\log _3 (2 - x) - \log _3 (2 + x) - \log _3 x + 1 = 0$$ \end{pro} \begin{pro} Solve the equation $$2\log_4 (2x) = \log_4 (x^2 + 75)$$ \end{pro} \begin{pro} Solve the equation $$\log_2 (2x) = \frac{1}{4}\log_2 (x - 15)^4$$ \end{pro} \begin{pro} Solve the equation $$\frac{\log _2 x}{\log _4 2x} = \frac{\log _8 4x}{\log _{16} 8x}$$ \end{pro} \begin{pro} Solve the equation $$\log _3 x = 1 + \log _x 9$$ \end{pro} \begin{pro} Solve the equation $$25^{\log_2 x} = 5 + 4x^{\log_2 5}$$ \end{pro} \begin{pro} Solve the equation $$x^{\log_{10} 2x} = 5$$ \end{pro} \begin{pro} Solve the equation $$|x- 3|^{(x^2 - 8x + 15)/(x - 2)} = 1$$ \end{pro} \begin{pro} Solve the equation $$\log _{2x - 1} \frac{x^4 + 2}{2x + 1} = 1$$ \end{pro} \begin{pro} Solve the equation $$\log _{3x} x = \log _{9x} x$$ \end{pro} \begin{pro} Solve $$\log _2 x + \log _4 y + \log _4 z = 2,$$ $$\log _3 x + \log _9 y + \log _9 z = 2,$$ $$\log _4 x + \log _{16} y + \log _{16} z = 2.$$ \end{pro} \begin{pro} Solve the equation $$x^{0.5\log _{\sqrt{x}} (x^2 - x)} = 3^{\log _9 4}.$$\end{pro} \end{multicols} \chapter{Goniometric Functions} \section{The Winding Function} Recall that a circle of radius $r$ has a circumference of $2\pi r$ units of length. Hence a unit circle, i.e., one with $r = 1$, has circumference $2\pi$. \begin{df} A {\em radian} is a $\dis{\frac{1}{2\pi}}$th part of the circumference of a unit circle. \end{df} \vspace{1cm} \begin{figure}[!hptb]\psset{unit=3pc}\centering \pstGeonode[PointName=none](0,0){O}(1,0){A} \pstCircleOA[linewidth=2pt]{O}{A} \pstCurvAbsNode[PointName=none]{O}{A}{B}{\pstDistVal{1}} \pstArcOAB[linecolor=blue,linewidth=2pt]{O}{A}{B} \pcline(O)(A)\lput*{:U}{$1$}\pcline(O)(B) \meinecaption{2}{A radian.} \label{radian} \end{figure} Since $\frac{1}{2\pi} \approx 0.16$, a radian is about $\dis{\frac{4}{25}}$ of the circumference of the unit circle. A quadrant or quarter part of a circle has arc length of $\frac{\pi}{4}$ radians. A semicircle has arc length $\frac{2\pi}{2} = \pi$ radians. \begin{rem} \begin{enumerate} \item A radian is simply a real number! \item If a central angle of a unit circle cuts an arc of $x$ radians, then the central angle measures $x$ radians. \item The sum of the internal angles of a triangle is $\pi$ radians. \end{enumerate} \end{rem} Suppose now that we cut a unit circle into a ``string'' and use this string to mark intervals of length $2\pi$ on the real line. We put an endpoint $0$, mark off intervals to the right of $0$ with endpoints at $2\pi, 4\pi, 6\pi, \ldots,$ etc. We start again, this time going to the left and marking off intervals with endpoints at $-2\pi, -4\pi, -6\pi, \ldots,$ etc., as shewn in figure \ref{relnm2pi}. \vspace{1cm} \begin{figure}[!h] \centering\psset{xunit=3.141592654} \multips(0,0)(.5,0){4}{\pscurve{->}(0, .2)(.25, .8)(.5, .2)} \multips(0,0)(-.5,0){4}{\pscurve{<-}(-.5, .2)(-.25, .8)(0, .2)} \psaxes[linewidth=2pt,yAxis=false,trigLabels,Dx=2,dx=.5,labelsep=-.8](0,0)(-2,0)(2,0) \meinecaption{1}{The Real Line modulo $2\pi$.} \label{relnm2pi} \end{figure} \vspace{1cm} We have decomposed the real line into the union of disjoint intervals $$\ldots \cup [-6\pi; -4\pi[\cup[-4\pi; -2\pi[\cup[-2\pi; 0[\cup[0; 2\pi[\cup[2\pi; 4\pi[\cup[4\pi; 6\pi[\cup\ldots$$ Observe that each real number belongs to one, and only one of these intervals, that is, there is a unique integer $k$ such that if $x\in\reals$ then $x\in[2\pi k; (2k + 2)\pi[$. For example $100 \in [30\pi; 32\pi[$ and $-9\in[-4\pi; -2\pi[.$ \begin{df} Given two real numbers $a$ and $b$, we say that $a$ {\em is congruent to} $b$ modulo $2\pi$, written $a \equiv b \mod 2\pi$, if $\dfrac{a - b}{2\pi}$ is an integer. If $\dis{\frac{a - b}{2\pi}}$ is not an integer, we say that $a$ and $b$ are {\em incongruent} modulo $2\pi$ and we write $a \not\equiv b \mod 2\pi$. \end{df} \bigskip For example, $5\pi \equiv -7\pi \mod 2\pi$, since $\dfrac{5\pi - (-7\pi) }{2\pi} = \dfrac{12\pi}{2\pi} = 6,$ an integer. However, $5\pi \not\equiv 2\pi \mod 2\pi$ as $\dfrac{5\pi - 2\pi }{2\pi} = \dfrac{3\pi}{2\pi} = \dfrac{3}{2},$ which is not an integer. \begin{df} If $a \equiv b \mod 2\pi$, we say that $a$ and $b$ belong to the same {\em residue class} $\mod 2\pi$. We also say that $a$ and $b$ are {\em representatives} of the same residue class modulo $2\pi$. \end{df} \begin{thm} Given a real number $a$, all the numbers of the form $a + 2\pi k, \ k\in\integers$ belong to the same residue class modulo $2\pi$. \end{thm} \begin{pf} Take two numbers of this form, $a + 2\pi k_1$ and $a + 2\pi k_2$, say, with integers $k_1, k_2$. Then $$\frac{(a + 2\pi k_1) - (a + 2\pi k_2)}{2\pi} = k_1 - k_2,$$ which being the difference of two integers is an integer. This shews that $a + 2\pi k_1 \equiv a + 2\pi k_2 \mod 2\pi$. \end{pf} \begin{exa} Take $x = \frac{\pi}{3}$. Then $$ \begin{array}{lllll} \frac{\pi}{3} & \equiv & \frac{\pi}{3} + 2\pi & \equiv &\frac{7\pi}{3} \ \mod 2\pi \\ & \equiv & \frac{\pi}{3} - 2\pi & \equiv &-\frac{5\pi}{3} \ \mod 2\pi \\ & \equiv & \frac{\pi}{3} + 4\pi & \equiv &\frac{13\pi}{3} \ \mod 2\pi \\ & \equiv & \frac{\pi}{3} - 4\pi & \equiv &-\frac{11\pi}{3} \ \mod 2\pi \\ \end{array} $$ Thus all of $$\frac{\pi}{3}, \frac{7\pi}{3}, -\frac{5\pi}{3}, \frac{13\pi}{3}, -\frac{11\pi}{3} $$belong to the same residue class $\mod 2\pi$. \end{exa} \begin{rem} If $a \equiv b \mod 2\pi$ then there exists an integer $k$ such that $a = b + 2\pi k$. \end{rem} \bigskip Given a real number $x$, it is clear that there are infinitely many representatives of the class to which $x$ belongs, as we can add any integral multiple of $2\pi$ to $x$ and still lie in the same class. However, exactly one representative $x_0$ lies in the interval $[0, 2\pi[$, as we saw above. We call $x_0$ the {\em canonical} representative of the class (to which $x$ belongs modulo $2\pi$). \bigskip To find the canonical representative of the class of $x$, we simply look for the integer $k$ such that $2k\pi \leq x < (2k + 2)\pi$. Then then $0 \leq x - 2k\pi < 2\pi$ and so $x - 2\pi k$ is the canonical representative of the class of $x$. \begin{df} We will call the procedure of finding a canonical representative for the class of $x$, {\em reduction} modulo $2\pi$. \end{df} \begin{exa} Reduce $5\pi \mod 2\pi$. \end{exa} \begin{solu} Since $4\pi < 5\pi < 6\pi$, we have $5\pi \equiv 5\pi - 4\pi \equiv \pi \mod 2\pi$. Thus $\pi$ is the canonical representative of the class to which $5\pi$ belongs, modulo $2\pi.$ \end{solu} \bigskip To speed up the computations, we may avail of the fact that $2\pi k \equiv 0 \mod 2\pi$, that is, any integral multiple of $2\pi$ is congruent to $0$ $\mod 2\pi.$ \vspace{4cm} \begin{figure}[!hptb] \begin{minipage}{.4\textwidth} \centering \psset{unit=1.5pc} \psaxes[labels=none,arrows={->},linewidth=2pt,arrowscale=1.5,ticks=none](0,0)(-4,-4)(4,4) \pstGeonode[PointName=none](0,0){O}(2,0){A}(2,2){I}(-2,2){II}(-2,-2){III}(2,-2){IV} \pstCircleOA[linewidth=2pt,linecolor=brown]{O}{A} \uput[r](I){Quadrant I $(+,+)$} \uput[r](IV){Quadrant IV $(+,-)$} \uput[l](II){Quadrant II $(-,+)$}\uput[l](III){Quadrant III $(-,-)$} \pstextpath[c]{\psarc[linecolor=blue]{->}(0,0){7.2}{330}{60}}{\textbf{Positive\ direction}} \meinecaption{4}{The unit circle on the Cartesian Plane.}\label{fig:sign_sin_cos} \end{minipage} \hfill \begin{minipage}{.4\textwidth} \centering \psset{unit=1.5pc} \psaxes[labels=none,linewidth=2pt,ticks=none](0,0)(-4,-4)(4,4) \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(4,0){A} \pstCircleOA[linewidth=2pt,linecolor=brown]{O}{A} \multido{\iA=0+30}{12}{\psline[linewidth=0.1pt](3.6;\iA)(4;\iA)} \multido{\iA=0+45}{8}{\psline[linewidth=0.1pt](3.6;\iA)(4;\iA)} \rput(5.5;0){$0$ radians} \rput{30}(5.5;30){$\frac{\pi}{6}$ radians} \rput{45}(5.5;45){$\frac{\pi}{4}$ radians} \rput{60}(5.5;60){$\frac{\pi}{3}$ radians} \rput{90}(5.5;90){$\frac{\pi}{2}$ radians} \rput{120}(5.5;120){$\frac{2\pi}{3}$ radians} \rput{135}(5.5;135){$\frac{3\pi}{4}$ radians} \rput{150}(5.5;150){$\frac{5\pi}{6}$ radians} \uput[l](5.5;180){$\pi$ radians} \rput{210}(5.5;210){$\frac{7\pi}{6}$ radians} \rput{225}(5.5;225){$\frac{5\pi}{4}$ radians} \rput{240}(5.5;240){$\frac{4\pi}{3}$ radians} \rput{270}(5.5;270){$\frac{3\pi}{2}$ radians} \rput{300}(5.5;300){$\frac{5\pi}{3}$ radians} \rput{315}(5.5;315){$\frac{7\pi}{4}$ radians} \rput{330}(5.5;330){$\frac{11\pi}{6}$ radians} \meinecaption{4}{The unit circle on the Cartesian Plane.}\label{circart2} \end{minipage} \end{figure} \vspace{1cm} \begin{exa} Reduce $\dis{\frac{200\pi}{7}}$ modulo $2\pi$. \end{exa} \begin{solu} $\dis{\frac{200\pi}{7} \equiv \frac{196\pi + 4\pi}{7} \equiv 28\pi + \frac{4\pi}{7} \equiv \frac{4\pi}{7} \mod 2\pi}$. \end{solu} \begin{exa} Reduce $\dis{-\frac{5\pi}{7}}$ modulo $2\pi$. \end{exa} \begin{solu} $ \dis{-\frac{5\pi}{7} \equiv 2\pi - \frac{5\pi}{7}\equiv \frac{9\pi}{7} \mod 2\pi}$. \end{solu} \begin{exa} Reduce $7 \mod 2\pi$.\end{exa}\begin{solu} Since $2\pi < 6.29 < 7 < 4\pi$, the largest even multiple of $\pi$ smaller than $7$ is $2\pi$, whence $7 \equiv 7 - 2\pi \mod 2\pi. $. \end{solu} Place now the centre of a unit circle at the origin of the Cartesian Plane. Choosing the point $(1, 0)$ as our departing point (a completely arbitrary choice), we traverse the circumference of the unit circle counterclockwise (again, the choice is completely arbitrary). If we traverse 0 units, we are still at $(1, 0)$, on the positive portion of the $x$-axis. If we traverse a number of units in the interval $\oo{0; \frac{\pi}{2}}$, we are in the first quadrant. \bigskip If we have traversed exactly $\dfrac{\pi}{2}$ units, we are at $(0, 1)$, on the positive portion of the $y$-axis. Traversing a number of units in the interval $\oo{\frac{\pi}{2}; \pi}$, puts us in the second quadrant. If we travel exactly $\pi$ units, we are at $(-1, 0)$, the negative portion of the $x$-axis. Traversing a number of units in the interval $\oo{\pi; \frac{3\pi}{2}}$, puts us in the third quadrant. Traversing exactly $\frac{3\pi}{2}$ units puts us at the point $(0, -1)$, the negative portion of the $y$-axis. Travelling a number of units in the interval $\oo{\frac{3\pi}{2}; 2\pi}$, puts us in the fourth quadrant. Finally, travelling exactly $2\pi$ units brings us back to $(1, 0)$. So, after one revolution around the unit circle, we are back in already travelled territory. See figure \ref{fig:sign_sin_cos}. \vspace*{3cm} \begin{figure}[!hptb]\centering \psset{unit=2pc} \pstGeonode[PosAngle={225,0}](0,0){O}(4,3){x} \pstGeonode[PointName=none](1,0){A}(1.3,0){D} \pstGeonode[PointName=none,PointSymbol=none](4,-4){B}(4,4){B'} \psaxes[linewidth=2pt,labels=none,ticks=none,arrows={->}](0,0)(-3,-3)(3,3) \pstCircleOA[linewidth=2pt,linecolor=brown]{O}{A} \pstCurvAbsNode[PosAngle=110]{O}{A}{M}{\pstDistVal{1}}\pstCurvAbsNode[PointName=none,PointSymbol=none]{O}{D}{D'}{\pstDistVal{1.3}} \pstCurvAbsNode[PosAngle=20,PointSymbol=none]{O}{D}{x_0}{\pstDistVal{.65}} \pstLineAB[arrows={<->},arrowscale=2]{B}{B'} \pstArcOAB[arrows=|-|]{O}{D}{D'} \pscurve{->}(x)(3.5,3.5)(3.3,2.4)(3,4)(M) \meinecaption{3}{$\winding:\reals \rightarrow \reals^2, \ \winding (x) = M$.} \label{camelmap} \end{figure} \begin{rem} If we traverse the unit circle {\em clockwise}, then the arc length is measured negatively. \end{rem} We now define a function ${\cal C}: \reals \rightarrow \reals^2$ in the following fashion. Given a real number $x$, let $x_0$ be its canonical representative modulo $2\pi$. Starting at $(1, 0)$, traverse the circumference of the unit circle $x_0$ units counterclockwise. Your final destination is a point on the Cartesian Plane, call it $M$. We let $\winding (x) = M.$ See figure \ref{camelmap}. The function $\winding$ is called the {\em winding function}. \begin{exa} In what quadrant does $\winding \left(-\frac{283\pi}{5}\right)$ lie? \end{exa} \begin{solu} Observe that $$\begin{array}{lll} -\frac{283\pi}{5} & \equiv & \frac{-280\pi - 3\pi}{5}\\ & \equiv & -56\pi - \frac{3\pi}{5}\\ & \equiv & -\frac{3\pi}{5} \\ & \equiv & 2\pi - \frac{3\pi}{5} \\ & \equiv & \frac{7\pi}{5} \ \mod 2\pi.\end{array}$$Since $\frac{7\pi}{5} \in ]\pi; \frac{3\pi}{2}[$, $\winding \left(-\frac{283\pi}{5}\right)$ lies in quadrant III. \end{solu} \begin{exa} In what quadrant does $\winding (451)$ lie? \end{exa} \begin{solu} Since $71 < \frac{451}{2\pi} < 71.8$, $142\pi < 451 < 144\pi$, and hence $451 \equiv 451 - 142\pi \mod 2\pi$. Now, $451 - 142\pi \approx 4.89 \in \oo{\frac{3\pi}{2}; 2\pi}$, and so $\dis{{\cal C}(451)}$ lies in the fourth quadrant. \end{solu} \begin{exa} In which quadrant does $\dis{{\cal C}(\pi^2)}$ lie? \end{exa}\begin{solu} We multiply the inequality $2 < \pi < 4$ through by $\pi$, obtaining $2\pi < \pi^2 < 4\pi$, whence the largest even multiple of $\pi$ less than $\pi^2$ is $2\pi$. Therefore $\pi^2 \equiv \pi^2 - 2\pi \mod 2\pi$. Now we claim that $$ \pi < \pi^2 - 2\pi < \frac{3\pi}{2}.$$The sinistral inequality is easily deduced from the obvious inequality $3\pi < \pi^2.$ The dextral inequality is deduced from the fact that $\pi^2 < 3.5\pi.$ The inequality $\pi < \pi^2 - 2\pi < \frac{3\pi}{2}$ is thus proven, which means that $\dis{{\cal C}(\pi^2)}$ lies in the third quadrant. \end{solu} \begin{exa} Find the members of the set $\left\{\frac{\pi}{2} + \frac{k\pi}{3}: k \in \integers\}\right\}$ that belong to the interval $[8\pi; 10\pi[$. \end{exa} \begin{solu} The problem is asking for all integers $k$ such that $$8\pi \leq \frac{\pi}{2} + \frac{k\pi}{3} < 10\pi.$$Now, $$\begin{array}{lll} 8\pi \leq \frac{\pi}{2} + \frac{k\pi}{3} < 10\pi & \Longleftrightarrow & 8\pi - \frac{\pi}{2} \leq \frac{k\pi}{3} < 10\pi - \frac{\pi}{2} \\ & \Longleftrightarrow & \frac{15\pi}{2} \leq \frac{k\pi}{3} < \frac{19\pi}{2} \\ & \Longleftrightarrow & 22.5 \leq k < 28.5. \end{array} $$ Since $k$ is an integer, $k \in \{23, 24, 25, 26, 27, 28\}$. The required elements are thus \begin{multicols}{2} $$\frac{\pi}{2} + \frac{23\pi}{3} = \frac{49\pi}{6},$$ $$\frac{\pi}{2} + \frac{24\pi}{3} = \frac{17\pi}{2},$$ $$\frac{\pi}{2} + \frac{25\pi}{3} = \frac{53\pi}{6},$$ $$\frac{\pi}{2} + \frac{26\pi}{3} = \frac{55\pi}{6},$$ $$\frac{\pi}{2} + \frac{27\pi}{3} = \frac{19\pi}{2},$$ $$\frac{\pi}{2} + \frac{28\pi}{3} = \frac{59\pi}{6}.$$ \end{multicols} \end{solu} \begin{exa} Is $\dfrac{275\pi}{6}\in \left\{\frac{\pi}{2} + \frac{k\pi}{3}: k \in \integers\right\}$? \end{exa}\begin{solu} The problem is asking whether there is an integer $k$ such that $$\frac{275\pi}{6} = \frac{\pi}{2} + \frac{k\pi}{3}.$$ Solving for $k$ we find $k = 136,$ which is an integer. The answer is affirmative and indeed, $$\frac{275\pi}{6} = \frac{\pi}{2} + \frac{136\pi}{3}.$$ \end{solu} \subsection*{Homework} \addcontentsline{toc}{section}{Homework} \markright{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro}True or False. \begin{enumerate} \item $10 \equiv 8 \mod 2\pi$. \item $-\frac{9\pi}{7} \equiv \frac{5\pi}{7} \mod 2\pi$. \item $\frac{1}{\pi} \equiv \frac{2}{\pi} \mod 2\pi$. \item $\frac{7\pi}{6} \equiv \frac{\pi}{6} \mod 2\pi$. \item $\frac{-8\pi}{41} \equiv -\frac{500\pi}{41} \mod 2\pi$. \item $x \in [-1; 0[$ then $\winding(x)$ is in quadrant IV. \end{enumerate} \begin{answer} F; T; F; F; T; T \end{answer} \end{pro} \begin{pro} Reduce the following real numbers $\mod 2\pi$. Determine the quadrant in which their image under $\dis{{\cal C}}$ would lie.\\ \begin{multicols}{3} \begin{enumerate} \item $\dis{\frac{3\pi}{5}}$; \item $\dis{-\frac{3\pi}{5}}$; \item $\dis{\frac{7\pi}{5}}$,; \item $\dis{\frac{8\pi}{57}}$; \item $\dis{\frac{57\pi}{8}}$; \item $\dis{\frac{6\pi}{79}}$; \item $\dis{\frac{790\pi}{7}}$; \item $1$;\item $2$; \item $3$; \item $4$; \item $5$; \item $6$,; \item $100$,; \item $-3.14$; \item $-3.15$ \end{enumerate} \end{multicols}\begin{answer} \begin{multicols}{2} \begin{enumerate} \item $\frac{3\pi}{5}$, quadrant II ; \item $\frac{7\pi}{5}$, quadrant III ; \item $\frac{7\pi}{5}$, quadrant III; \item $\dis{\frac{8\pi}{57}}$, quadrant I; \item $\dis{\frac{9\pi}{8}}$, quadrant III; \item $\dis{\frac{6\pi}{79}}$, quadrant I; \item $\dis{\frac{6\pi}{7}}$, quadrant II; \item $1$, quadrant I; \item $2$, quadrant II; \item $3$, quadrant II; \item $4$, quadrant III; (xii) $5$, quadrant IV; \item $6$, quadrant IV; \item $100 - 30\pi$, quadrant IV; \item $\dis{2\pi - 3.14}$, quadrant III; \item $2\pi - 3.15$, quadrant II \end{enumerate}\end{multicols} \end{answer} \end{pro} \begin{pro} Find all the members of the set $\dis{\{\frac{3\pi}{4} + \frac{k\pi}{5}:k\in\integers\}}$ that lie in the interval (i) $[0; \pi[$; (ii) $[-\pi; 0[$. \begin{answer} (i) $\frac{3\pi}{20}$, $\frac{7\pi}{20}$, $\frac{11\pi}{20}$, $\frac{3\pi}{4}$, $\frac{19\pi}{20}$; (ii) $-\frac{17\pi}{20}$, $-\frac{13\pi}{20}$, $-\frac{9\pi}{20}$, $-\frac{\pi}{4}$, $\frac{-\pi}{20}$. \end{answer} \end{pro} \begin{pro} Is $\dis{\frac{279\pi}{20}\in\{\frac{3\pi}{4} + \frac{3k\pi}{5}|k\in\integers\}}$? Is $\dis{-\frac{251\pi}{20}\in \{ \frac{3\pi}{4} + \frac{3k\pi}{5}|k\in\integers\}}$? \begin{answer} Yes; No. \end{answer} \end{pro} \begin{pro} Prove that congruence modulo $2\pi$ is reflexive, that is, if $a\in\reals$, then $a \equiv a \mod 2\pi$. \end{pro} \begin{pro} Prove that congruence modulo $2\pi$ is symmetric, that is, if $a, b\in\reals$, and if $a \equiv b \mod 2\pi$ then $b \equiv a \mod 2\pi$. \end{pro} \begin{pro} Prove that congruence modulo $2\pi$ is transitive, that is if if $a, b, c\in\reals$, then $a \equiv b \mod 2\pi$ and $b \equiv c \mod 2\pi$ imply $a \equiv c \mod 2\pi$. \end{pro} \end{multicols} \section{Cosines and Sines: Definitions}Consider any real number $x$. We find its canonical representative $x_0$ $\mod 2\pi$ and use this to find ${\cal C}(x) = M$, as in figure \ref{fig:sin_cos}. We now project the point $M$ so obtained onto $C$ and $S$ on the axes. The {\em cosine} function $\funnoname{x}{\cos x }{\reals}{[-1; 1]}$ is given by $\cos (x) = \cos x = {OC}$ (the algebraic length of the segment $OC$, that is, the signed distance from $O$ to $C$) and the {\em sine} function $\funnoname{x}{\sin x }{\reals}{[-1; 1]}$ is given by $\sin (x) = \sin x = {OS}$ (the algebraic length of the segment $OS$). \vspace{3cm} \begin{figure}[!hptb]\centering \psset{unit=2pc} \pstGeonode[PosAngle={225,0}](0,0){O}(6,3){x} \pstGeonode[PointName=none](2,0){A}(2.3,0){D} \pstGeonode[PointName=none,PointSymbol=none](6,-4){B}(6,4){B'}(1,0){XP}(-1,0){XM}(0,1){YP}(0,-1){YM} \psaxes[linewidth=2pt,labels=none,ticks=none,arrows={->}](0,0)(-3,-3)(3,3) \pstCircleOA[linewidth=2pt,linecolor=brown]{O}{A} \pstCurvAbsNode[PointName=none]{O}{A}{M}{\pstDistVal{2}}\pstCurvAbsNode[PointName=none,PointSymbol=none]{O}{D}{D'}{\pstDistVal{2.3}} \pstCurvAbsNode[PosAngle=-20,PointSymbol=none]{O}{D}{x_0}{\pstDistVal{1.65}} \pstLineAB[arrows={<->},arrowscale=2]{B}{B'} \pstArcOAB[arrows=|-|]{O}{D}{D'} \uput[ur](M){ $M=(\cos x, \sin x)$} \pstProjection[PosAngle=180]{YP}{YM}{M}[S]\pstProjection[PosAngle=-90]{XP}{XM}{M}[C] \psline[linewidth=2pt,linecolor=blue](S)(M)(C) \pscurve{->}(x)(3.5,3.5)(3.3,2.4)(3,4)(x_0) \psdots(M)(O)(S)(C) \meinecaption{3}{Geometric construction of the cosine and sine functions.}\label{fig:sin_cos} \end{figure} \begin{rem} \begin{enumerate} \item The farthest right $M$ can go is to $(1, 0)$ and the farthest left is to $(-1,0).$ Thus $-1 \leq \cos x \leq 1.$ Similarly, the farthest up $M$ can go is to $(0, 1)$ and the farthest down it can go is to $(0, -1)$. Hence $-1 \leq \sin x \leq 1$. \item The sine and cosine functions are defined for all real numbers. \item If $a \equiv b \mod 2\pi$ then $\cos a = \cos b$ and $\sin a = \sin b.$ In other words, the cosine and sine functions are periodic with period $2\pi,$ that is \begin{equation} \sin (2\pi + x) = \sin x \ \forall x \in \reals, \end{equation} \begin{equation} \cos (2\pi + x) = \cos x \ \forall x \in \reals. \end{equation} \item The point $M$ has abscissa $\cos x$ and ordinate $\sin x$, that is, $M = (\cos x, \sin x)$. \item The functions $\funnoname{x}{\cos x }{\reals}{[-1; 1]}$ and $\funnoname{x}{\sin x }{\reals}{[-1; 1]}$ are surjective (onto) but not injective (one-to-one). \end{enumerate} \end{rem} We may now compute some simple sines and cosines. \begin{exa}From figure \ref{fig:sin_cos_cardinalpts}, if $x = 0$ then the point $M$ is $(1, 0)$. Thus $\cos 0 = 1,\ \sin 0 = 0.$ If $x = \dfrac{\pi}{2}$ the point $M$ is $(0, 1)$. From this we gather that $\cos \frac{\pi}{2} = 0$ and $\sin \frac{\pi}{2} = 1$. If $x = \pi$ then the point $M$ is $(-1, 0)$. Thus $\cos \pi = -1,\ \sin \pi = 0.$ If $x = \frac{3\pi}{2}$ the point $M$ is $(0, 1)$. From this we gather that $\cos \frac{3\pi}{2} = 0$ and $\sin \frac{3\pi}{2} = -1$. \end{exa}\vspace*{3cm} \begin{figure}[!hptb] \begin{minipage}{.4\textwidth} \centering \psset{unit=1pc} \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(4,0){A} \pstCircleOA[linewidth=2pt,linecolor=brown]{O}{A} \multido{\iA=0+90}{4}{\psline[linewidth=0.1pt](3.6;\iA)(4;\iA)} \rput(3;0){$0$ } \rput{90}(3;90){$\frac{\pi}{2}$} \uput[r](3;180){$\pi$ }\rput{270}(3;270){$\frac{3\pi}{2}$} \rput(5;0){$(0,1)$} \rput{90}(5;90){$(0,1)$} \uput[l](5;180){$(-1,0)$ }\rput{270}(5;270){\quad $(0,-1)$} \psaxes[labels=none,linewidth=2pt,ticks=none](0,0)(-2,-2)(2,2) \meinecaption{3}{Some values of $\sin$ and $\cos$.}\label{fig:sin_cos_cardinalpts} \end{minipage} \hfill \begin{minipage}{.4\textwidth} \centering \psset{unit=1pc} \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(4,0){A}(4;60){I}(4;-60){IV}(4;120){II}(4;240){III} \pstCircleOA[linewidth=2pt,linecolor=brown]{O}{A} \psline(I)(O)(A) \pstMarkAngle[MarkAngleRadius=.8]{A}{O}{I}{$x$} \uput[ur](I){$(\cos x, \sin x)$} \uput[dr](IV){$(\cos (-x), \sin (-x))$} \uput[ul](II){$(\cos (\pi -x), \sin (\pi- x))$}\uput[dl](III){$(\cos (\pi +x), \sin (\pi+ x))$}\psdots(O)(A)(I)(II)(III)(IV)\pspolygon(I)(II)(III)(IV) \meinecaption{3}{Symmetry Identities.}\label{fig:symm_identities} \end{minipage} \end{figure} \begin{df}If $K \neq -1$, we write $\sin ^K x, \cos^K x$ to denote $(\sin x)^K, (\cos x)^K$, respectively. $\sin ^{-1} x, \cos^{-1}x$, a are reserved for when we study inversion later in these notes. \end{df} The following relation, known as the {\em Pythagorean Relation} is fundamental in the study of circular functions. \begin{thm}[Pythagorean Relation] Let $x$ be any real number. Then \begin{equation} \cos^2 x + \sin^2x = 1.\label{pytrel}\end{equation} \end{thm} \begin{pf} Let ${\cal C}(x) = M = (\cos x, \sin x)$, as in figure \ref{fig:sin_cos}., where $O = (0,0)$, and $S, C$ are the projections of $M$ onto the axes. In $\triangle OCM$, $\cos x = {OC},$ and $\sin x = {OS} = {CM}.$ As $\triangle OCM$ is a right triangle and ${OM} = 1$, by the Pythagorean Theorem, we have $$\cos^2x + \sin^2x = {OC}^2 + {CM}^2 = {OM}^2 = 1^2 = 1,$$ which completes the proof. \end{pf} \begin{rem} Pay attention to the notation $\cos^2x$ for $(\cos x)^2$ and respectively to $\sin^2x$ for $(\sin x)^2$. Do not confuse these with $\cos x^2$ and $\sin x^2$. For example, if $x = \pi$ then $\cos^2\pi = (-1)^2 = 1$ and $\sin^2\pi = 0^2 = 0.$ Since ${\cal C }(\pi^2)$ lies in the third quadrant, $\cos \pi^2 < 0$ and $\sin \pi^2 < 0$. Hence $\cos ^2 \pi \neq \cos \pi^2$ and $\sin^2\pi \neq \sin \pi^2.$ \end{rem} From the Pythagorean Relation, $$\cos x = \pm\sqrt{1 - \sin^2x}$$and $$\sin x = \pm\sqrt{1 - \cos^2x}.$$ The ambiguity in sign is resolved by specifying in which quadrant $\winding(x)$ lies, see figure \ref{fig:sign_sin_cos}. \begin{exa} Let $\dis{\frac{3\pi}{2} < x < 2\pi}$ and $\dis{\cos x = \frac{1}{3}}$. Find $\sin x$. \end{exa} \begin{solu} $\winding(x)$ lies in the fourth quadrant, where $\sin x < 0$. We have $$\sin x = -\sqrt{1 - \cos^2x} = -\sqrt{\frac{8}{9}} = -\frac{2\sqrt{2}}{3}.$$ \end{solu} \begin{exa} Given that $\frac{\pi}{2} < x < \pi$, and that $\sin x = \frac{3}{5}$, find $\cos x$.\end{exa} \begin{solu} Since $\winding(x)$ lies in the second quadrant, the cosine is negative. Hence $$\cos x = -\sqrt{1 - \sin ^2x} = -\sqrt{1 - \left(\frac{3}{5}\right)^2} = -\frac{4}{5}.$$ \end{solu} \begin{thm}[Symmetry Identities] Let $x\in \reals$. Then the following are identities. \begin{equation} \cos (-x) = \cos x, \end{equation} \begin{equation} \sin(-x) = -\sin x, \end{equation} \begin{equation} \cos (\pi - x) = -\cos x, \end{equation} \begin{equation} \sin (\pi -x) = \sin x, \end{equation} \begin{equation} \cos (\pi + x) = -\cos x, \end{equation} \begin{equation} \sin (\pi + x) = -\sin x, \end{equation} \label{symidth}\end{thm} \begin{pf} The first identity says that the cosine is an even function; the second that the sine is an odd function. The third and fourth identities are ``supplementary angle'' identities. The fifth and the sixth identities are a ``reflexion about the origin.'' All of these identities can be derived at once from figure \ref{fig:symm_identities}. \end{pf} \bigskip By the $2\pi$-periodicity of the cosine and sine we have \begin{equation} \cos (2\pi k + x) = \cos x, \ \forall x \in \reals \ \forall k \in \integers \end{equation} \begin{equation} \sin (2\pi k + x) = \sin x, \ \forall x \in \reals \ \forall k \in \integers. \end{equation} Now, $$\cos ((2k + 1)\pi + x) = \cos (2\pi k + \pi + x) = \cos (\pi + x) = -\cos x$$ and $$\sin ((2k + 1)\pi + x) = \sin (2\pi k + \pi + x) = \sin (\pi + x) = -\sin x,$$ whence the following corollary is proved. \begin{cor} Let $x \in \reals$ and $k \in \integers$. Then \begin{equation}\cos ((2k + 1)\pi + x) = -\cos x \end{equation} and \begin{equation}\sin ((2k + 1)\pi + x) = -\sin x \end{equation} \end{cor} In other words, if we add even multiples of $\pi$ to a real number, we get back the same cosine and the sine of the real number. If we add odd multiples of $\pi$ to a real number, we get minus the cosine or sine of the real number. \begin{exa} Write $$\sin (32\pi + x) - 18\cos (19\pi - x) + \cos (56\pi + x) - 9\sin (x + 17\pi)$$in the form $a\sin x + b\cos x.$ \end{exa} \begin{solu} The even multiples of $\pi$ addends give $$\sin (32\pi + x) = \sin x$$ and $$\cos (56\pi + x) = \cos x.$$ Examining the odd multiples of $\pi$ addends we see that $\cos (19\pi - x) = -\cos (-x)$. But $\cos (-x) = \cos x$, as the cosine is an even function and so $$\cos (19\pi - x) = -\cos x.$$ Similarly, $$\sin (17\pi + x) = -\sin x.$$ Upon gathering all of these equalities, we deduce that $$\begin{array}{ll} \sin (32\pi + x) - 18\cos (19\pi - x) & \\ \qquad + \cos (56\pi + x) - 9\sin (x + 17\pi) & \\ & = \qquad \sin x - 18(-\cos x)\\ & \quad\qquad + \cos x - 9\sin x \\ & = \qquad -8\sin x + 19\cos x. \end{array}$$ \end{solu} \begin{exa} Prove that $\cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$. \end{exa} \begin{solu} $\dis{\winding(\frac{\pi}{4})}$ is half-way between $\winding(0)$ and $\dis{\winding(\frac{\pi}{2}})$. Thus $\triangle OCM$ in figure \ref{fig:sin_cos_pi_4} is an isosceles right triangle. As ${OC} = {CM}$, we have $$\cos \frac{\pi}{4} = \sin \frac{\pi}{4}. $$By the Pythagorean Relation, $$\cos^2\frac{\pi}{4} + \sin^2\frac{\pi}{4} = 1,$$ and so $2\cos^2\frac{\pi}{4} = 1$. Since $\winding(\frac{\pi}{4})$ lies in the first quadrant, we take the positive square root. We deduce $\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$. This implies that $\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$. \end{solu} \begin{exa} Prove that $\dis{\cos \frac{\pi}{3} = \frac{1}{2}}$ and that $\dis{\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}}$. \end{exa}\begin{solu} In figure \ref{fig:sin_cos_pi_3} , $A = (\cos\frac{\pi}{3}, \sin\frac{\pi}{3}), \ B = (0, 0)$ and $C = (1, 0)$. Since ${BA} = {BC} = 1$, $\triangle BAC$ is isosceles. Thus $\angle A = \angle C$. Moreover, since the sum of the angles of a triangle is $\pi$ radians and central $\angle B$ measures $\dis{\frac{\pi}{3}}$ radians, the triangle is equilateral. Let $D$ denote the foot of the perpendicular from $A$ to the side $BC$. Since $\triangle BAC$ is equilateral, $D$ is halfway of the distance between $B$ and $C$, which means that $ = \dis{\frac{1}{2}}$. Thus $$\cos \frac{\pi}{3} = \frac{1}{2}.$$ Also, taking the positive square root (why?) $$\sin\frac{\pi}{3} = \sqrt{1 - \cos ^2 \frac{\pi}{3}} = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \frac{\sqrt{3}}{2},$$as we wanted to shew. \end{solu} \begin{exa} Prove that $\dis{\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}}$ and that $\dis{\sin \frac{\pi}{6} = \frac{1}{2}} $. \end{exa} \begin{solu} Reflect the point $A = (\cos \frac{\pi}{6}, \sin \frac{\pi}{6})$ about the $x$-axis to the point $C = (\cos \frac{\pi}{6}, -\sin \frac{\pi}{6})$, as in figure \ref{fig:sin_cos_pi_6}. Observe that since $\angle DBA = \angle CBD = \dis{\frac{\pi}{6}} $ then $\angle CBA = \dis{\frac{\pi}{3}}$. Thus $\triangle ABC$ is equilateral, and so ${AD} = \frac{1}{2}$, which implies that $$\sin \frac{\pi}{6} = \frac{1}{2}.$$ We deduce that $$\cos\frac{\pi}{6} = \sqrt{1 - \sin ^2 \frac{\pi}{6}} = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \frac{\sqrt{3}}{2}.$$ \end{solu} \vspace{2cm} \begin{figure}[!h] \begin{minipage}{.3\textwidth} \centering\psset{unit=3pc} \psaxes[labels=none,linewidth=2pt,arrows=->,arrowscale=1.4,ticks=none](0,0)(-1.5,-1.5)(1.5,1.5) \pstGeonode[PointName=none,PointSymbol=none](1,0){A} \pstGeonode[PosAngle={90,225}](1;45){M}(0,0){O} \pstProjection[PosAngle=-90]{O}{A}{M}[C] \pstCircleOA[linewidth=2pt,linecolor=brown]{O}{A} \psline[linewidth=2pt,linecolor=red](C)(M)(O) \psdots(O)(M)(C) \meinecaption{2}{$\sin\frac{\pi}{4}$ and $\cos\frac{\pi}{4}$}\label{fig:sin_cos_pi_4} \end{minipage} \hfill \begin{minipage}{.3\textwidth} \centering\psset{unit=3pc}\psaxes[labels=none,linewidth=2pt,arrows=->,arrowscale=1.4,ticks=none](0,0)(-1.5,-1.5)(1.5,1.5)\pstGeonode[PosAngle={225,-45}](0,0){B}(1,0){C} \pstGeonode[PosAngle={90}](1;60){A} \pstProjection[PosAngle=-90]{B}{C}{A}[D] \pstCircleOA[linewidth=2pt,linecolor=brown]{B}{C} \psline[linewidth=2pt,linecolor=red](C)(A)(B)\psline[linewidth=2pt,linecolor=red](D)(A) \psdots(A)(B)(C)(D) \meinecaption{2}{$\sin\frac{\pi}{3}$ and $\cos\frac{\pi}{3}$}\label{fig:sin_cos_pi_3} \end{minipage} \hfill \begin{minipage}{.3\textwidth} \centering\psset{unit=3pc}\psaxes[labels=none,ticks=none,linewidth=2pt,arrows=->,arrowscale=1.4](0,0)(-1.5,-1.5)(1.5,1.5)\pstGeonode[PointName=none,PointSymbol=none](1,0){T} \pstGeonode[PosAngle={45,-45,225}](1;30){A}(1;-30){C}(0,0){B} \pstProjection[PosAngle=225]{B}{T}{A}[D] \pstCircleOA[linewidth=2pt,linecolor=brown]{B}{T} \psline[linewidth=2pt,linecolor=red](A)(B)(C)(A) \psdots(A)(B)(C)(D) \meinecaption{2}{$\sin\frac{\pi}{6}$ and $\cos\frac{\pi}{6}$}\label{fig:sin_cos_pi_6} \end{minipage} \end{figure} The student will do well in memorising the special values deduced above, which are conveniently gathered in the table below. \\ $$ \begin{array}{|c|c|c|} \hline x & \sin x & \cos x \\ \hline 0 & 0 & 1 \\ \hline \frac{\pi}{6} & \frac{1}{2} & \frac{\sqrt{3}}{2} \\ \hline \frac{\pi}{4} & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ \hline \frac{\pi}{3} & \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \hline \frac{\pi}{2} & 1 & 0 \\ \hline \end{array} $$ \begin{exa} Find $\dis{\cos(-\frac{\pi}{6})}$ and $\dis{\sin(-\frac{\pi}{6})}$. \end{exa} \begin{solu} Since $x \mapsto \cos x$ is an even function, we have $$\cos(-\frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}.$$ Since $x \mapsto \sin x$ is an odd function, we have $$\sin(-\frac{\pi}{6}) = -\sin(\frac{\pi}{6}) = -\frac{1}{2}.$$ \end{solu} \begin{exa} Find $\dis{\cos{\frac{7\pi}{6}}}$ and $\dis{\sin{\frac{7\pi}{6}}}$. \end{exa} \begin{solu} By the reflexion about the origin identities $$ \cos\frac{7\pi}{6} = \cos (\pi + \frac{\pi}{6}) = -\cos\frac{\pi}{6} = -\frac{\sqrt{3}}{2}$$ and $$\sin\frac{7\pi}{6} = \sin (\pi + \frac{\pi}{6}) = -\sin\frac{\pi}{6} = -\frac{1}{2}.$$ \end{solu} \begin{exa} $$\cos \frac{2\pi}{3} = \cos \left(\pi - \frac{\pi}{3}\right) = -\cos \left(-\frac{\pi}{3}\right) = -\cos \frac{\pi}{3} = -\frac{1}{2} $$ and $$\sin \frac{2\pi}{3} = \sin \left(\pi - \frac{\pi}{3}\right) = -\sin \left(-\frac{\pi}{3}\right) = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$$ \end{exa} \begin{exa}Find the exact value of $$\cos \left(-\frac{32}{3}\pi\right)$$ \end{exa} \begin{solu} $$ \begin{array}{lllr} \cos \left(-\frac{32}{3}\pi\right) & = & \cos\left(\frac{32\pi}{3}\right) \\ & = & \cos(10\pi + \frac{2\pi}{3}) \\ & = & \cos(\frac{2\pi}{3}) \\ & = & -\frac{1}{2} \\ \end{array} $$ \flushleft{{\em Aliter:}} $$ \begin{array}{lllr} \cos \left(-\frac{32}{3}\pi\right) & = & \cos\left(\frac{32\pi}{3}\right) \\ & = & \cos(11\pi - \frac{\pi}{3}) \\ & = & -\cos(-\frac{\pi}{3}) \\ & = & -\cos\left(\frac{\pi}{3}\right) \\ & = & -\frac{1}{2} \\ \end{array} $$ \end{solu} \begin{exa} Find the exact value of $$\sin \left(-\frac{31}{3}\pi\right)$$ \end{exa} \begin{solu} $$ \begin{array}{lll} \sin \left(-\frac{31}{3}\pi\right) & = & -\sin\left(\frac{31\pi}{3}\right) \\ & = & -\sin(10\pi + \frac{\pi}{3}) \\ & = & -\sin(\frac{\pi}{3}) \\ & = & -\frac{\sqrt{3}}{2} \\ \end{array} $$ \end{solu} \begin{thm}[Complementary Angle Identities] The following identities hold: \begin{equation}\cos (\frac{\pi}{2} - x) = \sin x \ \forall x\in\reals\end{equation} \begin{equation}\sin (\frac{\pi}{2} - x) = \cos x \ \forall x\in\reals\end{equation} \end{thm} \begin{pf} We will prove the result for $x\in[0; \frac{\pi}{2}[$. The extension of these identities to all real numbers depends on Theorem \ref{symidth} and we leave it as an exercise. In figure \ref{fig:complementary_angle_iden} assume that arc $\arc{MA}$ (read counterclockwise) measures $x$ and that $x\in[0; \frac{\pi}{4}[$. Reflect point $A = (\cos x, \sin x)$ about the line $y = x$, to point $B =(\sin x, \cos x)$ as in figure \ref{fig:complementary_angle_iden}. Arc $\arc{BT}$ (read counterclockwise) measures $x$, and so arc $\arc{MAB}$ measures $\frac{\pi}{2} - x$. This means that $B = (\cos (\frac{\pi}{2} - x), \sin (\frac{\pi}{2} - x))$, from where the Theorem follows for $x \in ]0; \frac{\pi}{4}[$. Assume now that $x\in [\frac{\pi}{4}; \frac{\pi}{2}[$. Then $\frac{\pi}{2} - x \in [0; \frac{\pi}{4}[$, and so we apply the result just obtained to $\frac{\pi}{2} - x$: $$\cos \left(\frac{\pi}{2} - x \right) = \sin\left(\frac{\pi}{2} - \left(\frac{\pi}{2} - x \right) \right) = \sin x,$$ and $$\sin \left(\frac{\pi}{2} - x \right) = \cos\left(\frac{\pi}{2} - \left(\frac{\pi}{2} - x \right) \right) = \cos x.$$ So, we have established the result for $x\in [0; \frac{\pi}{2}[$. \end{pf} \vspace{1cm} \begin{figure}[!hptb] \centering\psset{unit=3pc}\psaxes[labels=none,linewidth=2pt,arrows=->,arrowscale=1.4,ticks=none](0,0)(-1.5,-1.5)(1.5,1.5) \psline[linewidth=2pt,linecolor=red,arrows=<->,arrowscale=1.4](-1.4,-1.4)(1.4,1.4)\uput[r](1.3,1.3){$y=x$} \pstGeonode[PosAngle={135,-45}](0,0){O}(1,0){M} \pstGeonode[PosAngle={0,90}](1;30){A}(1;60){B} \pstGeonode[PointName=none](1;45){T} \pstCircleOA[linewidth=2pt,linecolor=brown]{O}{M} \psline[linewidth=2pt,linecolor=red](B)(O)(A) \psdots(M)(O)(A)(B) \meinecaption{2}{Complementary Angle Identities.}\label{fig:complementary_angle_iden} \end{figure} \begin{rem} Using the complementary angle identities, $$\sin \frac{\pi}{6} = \cos\left( \frac{\pi}{2} - \frac{\pi}{6} \right) = \cos \frac{\pi}{3} = \frac{1}{2},$$and $$\cos \frac{\pi}{6} = \sin\left( \frac{\pi}{2} - \frac{\pi}{6} \right) = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2},$$for instance. \end{rem} \begin{exa} Prove that $$\sin x = \cos \left(x - \frac{\pi}{2}\right), \ \forall x \in \reals.$$ \end{exa}\begin{solu} Since the cosine is an even function, $$\sin x = \cos \left(\frac{\pi}{2} - x\right) = \cos \left(-\left(\frac{\pi}{2} - x\right)\right) = \cos \left(x - \frac{\pi}{2}\right).$$ \vspace{2cm} \end{solu} \begin{exa} Prove that the following hold identically. $$\cos x = \sin \left(x + \frac{\pi}{2}\right), \ \forall x \in \reals.$$ $$-\sin x = \cos \left(x + \frac{\pi}{2}\right), \ \forall x \in \reals.$$ \end{exa} \begin{solu} Using the fact the fact that the cosine is an even function, and using the complementary angle identity for the cosine, $$\cos x = \cos (-x) = \sin \left(\frac{\pi}{2} - (-x)\right) = \sin \left(\frac{\pi}{2} + x\right) .$$ Since the sine is an odd function, $$\sin x = -\sin (-x) = -\cos \left(\frac{\pi}{2} - (-x)\right) = -\cos \left(\frac{\pi}{2} + x\right) .$$ \end{solu} \begin{exa} Let $0 < \theta < \frac{\pi}{2}$. Given that $\sin 2\theta = \cos 3\theta$ find $\sin 5\theta$. \end{exa} \begin{solu} Since $\sin 2\theta = \cos 3\theta$, these two quantities have the same sign. Since $0 < 2\theta < \pi$, then both $\winding(2\theta)$ and $\winding(3\theta)$ must be in quadrant I. By the complementary angle identities, we have $\sin 2\theta = \cos (\frac{\pi}{2} - 2\theta)$. Thus $\cos (\frac{\pi}{2} - 2\theta) = \cos 3\theta$, and so, $\frac{\pi}{2} - 2\theta = 3\theta$ or $5\theta = \frac{\pi}{2}$. Hence $\sin 5\theta = 1.$ \end{solu} \begin{exa} Write in the form $a\sin\alpha + b\sin\alpha$: $$\sin (\pi - \alpha) + \cos \left(\frac{\pi}{2} + \alpha\right) - \cos (\pi + \alpha)$$ \end{exa}\begin{solu} By reflexion about the origin, $\sin (\pi - \alpha) = -\sin ( -\alpha)$. Since the sine is an odd function, $-\sin (-\alpha) = -(-\sin \alpha) = \sin\alpha$. By the complementary angle identities, and since the sine is an odd function $$\cos \left(\frac{\pi}{2} + \alpha\right) = \cos \left(\frac{\pi}{2} - (-\alpha)\right) = \sin (-\alpha) = -\sin\alpha.$$ Finally, by reflexion about the origin, $\cos (\pi + \alpha) = -\cos\alpha$. Upon collecting all of these equalities, $$ \sin (\pi - \alpha) + \cos \left(\frac{\pi}{2} + \alpha\right) - \cos (\pi + \alpha) = \cos\alpha . $$ \end{solu} \begin{exa} Given that $$3\sin x + 4\cos x = 5,$$find $\sin x$ and $\cos x$. \end{exa} \begin{solu} We have $$3\sin x + 4\cos x = 5 \Longleftrightarrow \sin x = \frac{5 - 4\cos x}{3}.$$Putting this in the identity $\cos^2x + \sin^2x = 1$ we obtain $$\cos^2x + \left(\frac{5 - 4\cos x}{3}\right)^2 = 1$$ $$\cos^2x + \frac{25 - 40\cos x + 16\cos^2x}{9} = 1$$ $$9\cos^2x + 25 - 40\cos x + 16\cos^2x = 9$$ $$25\cos^2x - 40\cos x + 16 = 0$$ $$(5\cos x - 4)^2 = 0$$ $$\cos x = \frac{4}{5}$$ Substituting this value we obtain $$\sin x = \frac{5 - 4\cos x}{3} = \frac{5 - \frac{16}{5}}{3} = \frac{3}{5}.$$ \end{solu} \begin{exa} Find $k$ such that the expression $$(\sin x + \cos x)^2 + k\sin x\cos x = 1$$becomes an identity. \end{exa} \begin{solu} We have $$ \begin{array}{lll} 1 & = & (\sin x + \cos x)^2 + k\sin x\cos x\\ & = & \sin^2x + 2\sin x\cos x + \cos ^2x + k\sin x\cos x \\ & = & 1 + (k + 2)\sin x\cos x \end{array} $$We thus have $(k + 2)\sin x\cos x = 0$. This will hold for all real numbers $x$ if $k = -2.$ \end{solu} \subsection*{Homework} \addcontentsline{toc}{section}{Homework} \markright{Homework} \begin{pro} Write in the form $a\sin x + b\cos x$, with real constants $a, b$. $$A(x) = \sin\left(\frac{\pi}{2} - x\right) + \cos(5\pi - x) + \cos\left(\frac{3\pi}{2} - x\right) + \sin\left(\frac{3\pi}{2} + x\right)$$ \end{pro} \begin{pro}True or False. \begin{multicols}{2} \begin{enumerate} \item $\sin \frac{7\pi}{6} = 1/2$. \item $\cos (\frac{\pi}{2} + 99) = \sin 99$. \item $\cos (-1993) = \cos 1993 $. \item $\sin (-1993) = -\sin 1993$. \item If $\sin x = 1$, then $x = \pi /2.$ \item $\cos (\cos \pi) = \cos (\cos 0) $. \item $\forall x\in\reals, \ \sin 2x = 2\sin x$. \item $\exists x\in\reals$ such that $\cos x = 2.$ \item $\exists x\in \reals$ such that $\cos^2x = \cos x^2$ \item $(\sin x + \cos x)^2 = 1, \ \forall x\in \reals$. \item $\cos x = \sin (x + \frac{\pi}{2}), \ \forall x \in \reals$. \item $\sin x = \cos (x - \frac{\pi}{2}), \ \forall x \in \reals$. \item $\sin x = \cos (x + \frac{\pi}{2}), \ \forall x \in \reals$. \item $-\frac{1}{2} \leq \cos\frac{x}{2} \leq \frac{1}{2}, \ \forall x\in\reals$. \item $1 \leq -2\cos \frac{x}{2} + 3 \leq 5, \ \forall x\in\reals$. \item $\exists A\in\reals$ such that the equation $\cos x = A$ has exactly $7$ real solutions. \item $\cos^2x - \sin^2x = -1, \forall x\in\reals$. \end{enumerate} \end{multicols} \begin{answer} F; F; T; T; F; T; F; F; T; F; T; T; F; T; F; F\end{answer} \end{pro} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Given that $\sin t = -0.8$ and $\winding(t)$ lies in the fourth quadrant, find $\cos t$. \begin{answer} $\cos t = 0.6$ \end{answer} \end{pro} \begin{pro} Given that $\cos u = -0.9$ and $\winding(u)$ lies in the second quadrant, find $\sin u$. \begin{answer} $\sin u = \sqrt{.19}$ \end{answer} \end{pro}\begin{pro} Given that $\sin t = \frac{\sqrt{7}}{5}$ and $\winding(t)$ lies in the first quadrant, find $\cos t$. \begin{answer} $\cos t = \dfrac{3\sqrt{2}}{5}$ \end{answer} \end{pro} \begin{pro} Given that $\cos u = \frac{\sqrt{13}}{4}$ and $\winding(u)$ lies in the third quadrant, find $\sin u$. \begin{answer} $\sin u = -\dfrac{\sqrt{3}}{4}$ \end{answer} \end{pro}\begin{pro} Using the fact that $\frac{5\pi}{6} = \pi - \frac{\pi}{6}$, find $\cos \frac{5\pi}{6}$ and $\sin \frac{5\pi}{6}$. \begin{answer} $\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$, $\sin \frac{5\pi}{6} = \frac{1}{2}$ \end{answer} \end{pro} \begin{pro} Using the fact that $\frac{3\pi}{4} = \pi - \frac{\pi}{4}$, find $\cos \frac{3\pi}{4}$ and $\sin \frac{3\pi}{4}$. \begin{answer} $\cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$ and $\sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2}$ \end{answer} \end{pro} \begin{pro} Find $\dis{\sin (\frac{31\pi}{6})}$ and $\dis{\cos (\frac{31\pi}{6})}$. \begin{answer} $\sin (\frac{31\pi}{6}) = -\frac{1}{2}$ and $\cos (\frac{31\pi}{6}) = -\frac{\sqrt{3}}{2}$ \end{answer} \end{pro} \begin{pro} Find $\dis{\sin (\frac{20\pi}{3})}$ and $\dis{\cos (\frac{20\pi}{3})}$. \begin{answer} $\sin (\frac{20\pi}{3}) = \frac{\sqrt{3}}{2}$ and $\cos (\frac{20\pi}{3}) = -\frac{1}{2}$ \end{answer} \end{pro} \begin{pro} Find $\dis{\sin (\frac{17\pi}{4})}$ and $\dis{\cos (\frac{17\pi}{4})}$. \begin{answer} $\sin (\frac{17\pi}{4}) = \frac{\sqrt{2}}{2}$ and $\cos (\frac{17\pi}{4}) = \frac{\sqrt{2}}{2}$ \end{answer} \end{pro}\begin{pro} Find $\dis{\sin (\frac{-15\pi}{4})}$ and $\dis{\cos (-\frac{15\pi}{4})}$. \begin{answer} $\sin (\frac{-15\pi}{4}) = \frac{\sqrt{2}}{2}$ and $\cos (\frac{-15\pi}{4}) = \frac{\sqrt{2}}{2}$ \end{answer} \end{pro}\begin{pro} Find $\dis{\sin (\frac{202\pi}{3})}$ and $\dis{\cos (\frac{202\pi}{3})}$. \begin{answer} $\sin (\frac{202\pi}{3}) = -\frac{\sqrt{3}}{2}$ and $\cos (\frac{202\pi}{3}) = -\frac{1}{2}$ \end{answer} \end{pro}\begin{pro} Find $\dis{\sin (\frac{171\pi}{4})}$ and $\dis{\cos (\frac{171\pi}{4})}$. \begin{answer} $\sin (\frac{171\pi}{4}) = \frac{\sqrt{2}}{2}$ and $\cos (\frac{171\pi}{4}) = -\frac{\sqrt{2}}{2}$ \end{answer} \end{pro} \begin{pro} If $|\sin\theta| < 1$ and $|\cos\theta| > 0$, prove that $$\frac{\cos\theta}{1 - \sin\theta} + \frac{\cos\theta}{1 + \sin\theta} = \frac{2}{\cos\theta}$$holds identically. \end{pro} \begin{pro} Given that $$\cos \frac{2\pi}{5} = \frac{\sqrt{5} - 1}{4},$$find $\sin \frac{2\pi}{5}$, $\cos \frac{3\pi}{5}$ and $\sin\frac{3\pi}{5} $ \end{pro} \begin{pro} Given that $\cos \alpha + \sin \alpha = A$ and $\sin\alpha \cos \alpha = B$, prove that $A^2 - 2B = 1$ \end{pro} \begin{pro} Given that $\cos \alpha + \sin \alpha = A$ and $\sin\alpha \cos \alpha = B$, prove that $\sin ^3 \alpha + \cos ^3 \alpha = A - AB.$ \end{pro} \begin{pro} Demonstrate that for all real numbers $x$, the following is an identity $$(\sin x + 4\cos x)^2 + (4\sin x - \cos x)^2 = 17$$ \end{pro} \begin{pro} Prove that $\dis{\cos ^4x - \sin ^4 x = \cos ^2x - \sin ^2 x}$ is an identity.\end{pro} \begin{pro} Prove that $$\sqrt{1 + 2\sin x\cos x} =|\sin x + \cos x|,\ \ \forall x\in \reals .$$ \end{pro} \begin{pro} Prove that $\forall x\in\reals$, $$\sin^4x + \cos^4x + 2(\sin x\cos x)^2 = 1.$$ \end{pro} \begin{pro} Prove, by recurrence, that $$\sin (x + n\pi) = (-1)^n\sin x,$$and $$\cos (x + n\pi) = (-1)^n\cos x.$$ \end{pro} \begin{pro} Prove that $\forall x \in \reals$, $$\sin^6x + \cos^6x + 3(\sin x\cos x)^2 = 1.$$ \end{pro} \begin{pro} Prove that $$\frac{\sin x - \cos x + 1}{\sin x + \cos x - 1} = \frac{\sin x + 1}{\cos x}$$ $\forall x\in\reals$ such that $\sin x + \cos x \neq 1$ and $\cos x \neq 0.$ \end{pro} \begin{pro}[AHSME 1976] If $\sin x + \cos x = \frac{1}{5}$ and $x\in ]0; \pi[$, find $\cos x$ and $\sin x.$ \end{pro} \begin{pro}[AIME 1983] Find the minimum value of the function $$x \mapsto \frac{9x^2\sin^2x + 4}{x\sin x}$$over the interval $]0; \pi[$. \begin{answer} Hint: Use the Arithmetic-Geometric-Mean Inequality $\frac{a + b}{2} \geq \sqrt{ab}$, for non-negative real numbers $a, b$. \end{answer} \end{pro} \end{multicols} \clearpage \section{The Graphs of Sine and Cosine} To obtain the graph of $x\mapsto \sin x$, we traverse the circumference of the unit circle, starting from $(1, 0)$, in a levogyrate (counterclockwise) sense, recording each time the abscissa of the point visited. See figure \ref{fig:graph_of_sine}.\vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{.4\textwidth} \centering \psaxes[linewidth=2pt,labels=none](0,0)(-1.2,-1.2)(1.2,1.2) \pstGeonode[PointName=none](0,0){O'}(0,1){L}(1;60){M}(1;45){N}(1;30){P}(0,-1){L'}(1;-60){M'}(1;-45){N'}(1,0){X} \pstCircleOA[linewidth=2pt]{O'}{L} \pstProjection[PointName=none]{X}{O'}{M}[PM] \pstProjection[PointName=none]{X}{O'}{N}[PN] \pstProjection[PointName=none]{X}{O'}{P}[PP] \psline[linestyle=dashed](M)(PM) \psline[linestyle=dashed](M')(PM) \psline[linestyle=dashed](P)(PP) \psline[linestyle=dashed](N)(PN) \psline[linestyle=dashed](N')(PN) \end{minipage} \begin{minipage}{.4\textwidth} \centering \psaxes[linewidth=2pt, dx=3.09, showorigin=false,ticks=all,trigLabels,labels=x](0,0)(-.5 ,-1.2)(6.5,1.2) \psplot[algebraic=true,linewidth=2pt,linecolor=brown]{0}{6.29}{sin(x)} \pstGeonode[PointName=none](!PI 2 div 1){A}(!PI 4 div .7071067810){B}(!PI 3 div .8660254040){C}(!PI 6 div .5){D}(!PI 1.5 mul -1){E}(!PI 4 div 5 mul -.7071067810){B'}(!PI 3 div 4 mul -.8660254040){C'} \pstGeonode[PointName=none](!PI 2 div 0){PA}(!PI 4 div 0){PB}(!PI 3 div 0){PC}(!PI 6 div 0){PD}(!PI 1.5 mul 0){PE}(!PI 4 div 5 mul 0){PB'}(!PI 3 div 4 mul 0){PC'} \psline[linestyle=dashed](L)(A)\psline[linestyle=dashed](B)(N) \psline[linestyle=dashed](P)(D) \psline[linestyle=dashed](M)(C) \psline[linestyle=dashed](L')(E) \psline[linestyle=dashed](M')(C') \psline[linestyle=dashed](N')(B') \psline[linestyle=dashed](A)(PA) \psline[linestyle=dashed](B)(PB) \psline[linestyle=dashed](C)(PC) \psline[linestyle=dashed](D)(PD) \psline[linestyle=dashed](E)(PE) \psline[linestyle=dashed](B')(PB') \psline[linestyle=dashed](C')(PC') \end{minipage} \meinecaption{3}{The graph $y = \sin x$ for $x\in [0; 2\pi [.$} \label{fig:graph_of_sine} \end{figure} Since $x \mapsto \sin x$ is periodic with period $2\pi$ and an odd function, we may now graph $x \mapsto \sin x$ for all values of $x$. See figure \ref{fig:graph_sine_2}. \vspace*{1cm} \begin{figure}[!hptb] \centering \psaxes[trigLabels=true, trigLabelBase=2,dx=\psPiH,xunit=\psPi,linewidth=2pt,arrows={->}](0,0)(-3,-1.5)(3,1.5) \psplot[linecolor=red,linewidth=1.5pt]{-7}{7}{x RadtoDeg sin} \meinecaption{1}{The graph of $x \mapsto \sin x$.} \label{fig:graph_sine_2} \end{figure} \begin{exa}[Jordan's Inequality] Give a graphical argument justifying the inequality $\frac{2}{\pi}x \leq \sin x$ for $0 \leq x \leq \frac{\pi}{2}$. \end{exa} \begin{solu} The equation of the straight line joining $(0, 0)$ and $(\frac{\pi}{2}, 1)$ is $y = \frac{2}{\pi}x$. From the graphs below, the graph of $y = \frac{2}{\pi}x$ lies below that of $y = \sin x $ in the interval $[0; \frac{\pi}{2}]$. See figure \ref{fig:xlesssin}. \end{solu} \vspace*{2cm} \begin{figure}[!hptb] \begin{minipage}{.4\textwidth}\def\psPi{3.14159265} \def\psPiH{1.570796327} \newdimen\pstRadUnit \newdimen\pstRadUnitInv \pstRadUnit=1.047198cm % this is pi/3 \pstRadUnitInv=0.95493cm % \centering \psaxes[trigLabels=true, trigLabelBase=2,dx=\psPiH,xunit=\psPi,linewidth=2pt,arrows={->}](0,0)(-.5,0)(1.2,1.5) \psplot[linecolor=red,linewidth=1.5pt]{0}{3.1416}{x RadtoDeg sin} \pstGeonode[PointName=none](!PI 2 div 1){A}(0,0){O} \pstLineAB[linewidth=2pt,linecolor=magenta,nodesepB=-.4]{O}{A}\psdots(A)(O) \meinecaption{2}{Jordan's Inequality.} \label{fig:xlesssin} \end{minipage} \hfill \begin{minipage}{.4\textwidth}\def\psPi{3.14159265} \def\psPiH{1.570796327} \newdimen\pstRadUnit \newdimen\pstRadUnitInv \pstRadUnit=1.047198cm % this is pi/3 \pstRadUnitInv=0.95493cm % \centering \psaxes[trigLabels=true, trigLabelBase=2,dx=\psPiH,xunit=\psPi,linewidth=2pt,arrows={->}](0,0)(-1.2,-2.1)(1.2,2.1) \psplot[linecolor=brown,linewidth=1.5pt]{-3.14}{3.14}{x RadtoDeg sin 2 mul} \meinecaption{2}{The graph of $x \mapsto 2\sin x$.} \label{fig:2sinx} \end{minipage} \end{figure} \begin{exa} Graph $x \mapsto 2\sin x$. \end{exa} \begin{solu} Recall that if $y = f(x),$ then $y = 2f(x)$ is a distortion of the graph of $y = f(x)$, in which the $y$-coordinate is doubled. The graph of $x \mapsto 2\sin x$ is shewn in figure \ref{fig:2sinx}. Observe that $-2 \leq 2\sin x \leq 2,$ so the least value that $x \mapsto 2\sin x$ could attain is $-2$ and the largest value is $2$. \end{solu} \begin{df} The average between the least value and largest value of of a periodic function its {\em amplitude.} \end{df} \begin{thm} Let $A\in\reals\setminus\{0\}$. Then $\funnoname{x}{\sin Ax}{\reals}{[-1; 1]}$ and $\funnoname{x}{\cos Ax}{\reals}{[-1; 1]}$ have period $\dfrac{2\pi}{|A|}$. \end{thm} \begin{pf} Since $x \mapsto \sin x$ and $x \mapsto \cos x$ have period $2\pi$, then, if $A\in\reals\setminus\{0\}$ is constant, we have $$\sin A\left(x + \frac{2\pi}{|A|}\right) = \sin (Ax \pm 2\pi) = \sin Ax,$$ and $$\cos A\left(x + \frac{2\pi}{|A|}\right) = \cos (Ax \pm 2\pi) = \cos Ax,$$ whence $x \mapsto \sin Ax$ and $x \mapsto \cos Ax$ have period at most $\frac{2\pi}{|A|}$. Could the period of $x \mapsto \sin Ax, A \neq 0$ and $x \mapsto \cos Ax, A \neq 0$ be {\em smaller} than $\dfrac{2\pi}{|A|}$? Assume $0 < P < \dfrac{2\pi}{|A|}$ is a period for these functions. Then $0 < P|A| < 2\pi$ and $\sin Ax = \sin A(x \pm P)$ and $\cos Ax = \cos A(x \pm P).$ In particular, $$0 = \sin 0 = \sin \pm AP.$$This means that $|A|P$ is a zero of $x \mapsto \sin x.$ Since $0 < |A|P < 2\pi$, we must have $|A|P = \pi$. Now $$1 = \cos 0 = \cos \pm AP = \cos |A|P = \cos \pi = -1,$$a contradiction. Thus the period of $x \mapsto \sin Ax, A \neq 0$ and $x \mapsto \cos Ax, A \neq 0$ is precisely $\dfrac{2\pi}{|A|}$, as we wanted to shew. \end{pf} \begin{exa} Graph $x \mapsto \sin 2x$. \end{exa} \begin{solu} Since $-1 \leq \sin 2x \leq 1,$ the amplitude of $x \mapsto \sin 2x$ is $\frac{1 - (-1)}{2} = 1.$ The period of $x \mapsto \sin 2x$ is $2\pi\div 2 = \pi$. Recall that if $y = f(2x),$ then $y = f(2x)$ is a distortion of the graph of $y = f(x)$, in which the $x$-coordinate is halved. The graph of $x \mapsto \sin 2x$ is shewn in figure \ref{fig:sin2x}. \end{solu} \vspace{2cm} \begin{figure}[!h] \begin{minipage}{.4\textwidth}\def\psPi{3.14159265} \def\psPiH{1.570796327} \newdimen\pstRadUnit \newdimen\pstRadUnitInv \pstRadUnit=1.047198cm % this is pi/3 \pstRadUnitInv=0.95493cm % \centering \psaxes[trigLabels=true, trigLabelBase=2,dx=\psPiH,xunit=\psPi,linewidth=2pt,arrows={->}](0,0)(-1.2,-2.1)(1.2,2.1) \psplot[linecolor=brown,linewidth=1.5pt]{-3.14}{3.14}{x 2 mul RadtoDeg sin } \meinecaption{2}{The graph of $x \mapsto \sin 2x$.} \label{fig:sin2x} \end{minipage} \hfill \begin{minipage}{.4\textwidth}\def\psPi{3.14159265} \def\psPiH{1.570796327} \newdimen\pstRadUnit \newdimen\pstRadUnitInv \pstRadUnit=1.047198cm % this is pi/3 \pstRadUnitInv=0.95493cm % \centering \psaxes[trigLabels=true, trigLabelBase=2,dx=\psPiH,xunit=\psPi,linewidth=2pt,arrows={->}](0,0)(-1.2,-2.1)(1.2,2.1) \psplot[linecolor=brown,linewidth=1.5pt]{-3.14}{3.14}{x RadtoDeg cos} \meinecaption{2}{The graph of $x \mapsto \cos x$.} \label{fig:cosx} \end{minipage} \end{figure} \begin{exa} Graph $\dis{x \mapsto \sin \left(x + \frac{\pi}{2}\right)}$ \end{exa} \begin{solu} Recall that if $a>0$ the graph of $x \mapsto f(x + a)$ is a translation $a$ units to the left of the graph $x \mapsto f(x)$. Now, the cosine is an even function, and by the complementary angle identities, we have $$\cos x = \cos (-x) = \sin\left(\frac{\pi}{2} - (-x)\right) = \sin\left(\frac{\pi}{2} + x\right),$$ and so this graph is the same as that of the cosine function. The graph of $y = \sin(x + \frac{\pi}{2}) = \cos x$ is shewn in in figure \ref{fig:cosx}. \end{solu} \begin{exa} Give a purely graphical argument (no calculators allowed!) justifying $\cos 1 < \sin 1.$ \end{exa} \begin{solu} At $x = \frac{\pi}{4}$, the graphs of the sine and the cosine coincide. For $x \in [\frac{\pi}{4}; \frac{\pi}{2}]$, the values of the sine increase from $\frac{\sqrt{2}}{2}$ to 1, whereas the values of the cosine decrease from $\frac{\sqrt{2}}{2}$ to $0$. Since $\frac{\pi}{4} < 1 < \frac{\pi}{2}$, we have $\cos 1 < \sin 1.$ \end{solu} \begin{exa} Graph $\dis{x \mapsto -2\cos \frac{x}{2} + 3}$ \end{exa} \begin{solu} Since $-1 \leq \cos \frac{x}{2} \leq 1$, we have $1 \leq -2\cos \frac{x}{2} + 3 \leq 5$. The amplitude of $x \mapsto -2\cos \frac{x}{2} + 3$ is therefore $\frac{5 - 1}{2} = 2.$ The period of $\dis{x \mapsto -2\cos \frac{x}{2} + 3}$ is $\dis{\frac{2\pi}{\frac{1}{2}} = 4\pi}$. The graph is shewn in figure \ref{fig:2cos.5x3}. \end{solu} \begin{exa}Draw the graph of $x \mapsto -3\sin \frac{x}{4}$. What is the amplitude, period, and where is the first positive real zero of this function? \end{exa} \begin{solu} Since $-3 \leq -3\sin x \leq 3,$ the amplitude of $x \mapsto -3\sin \frac{x}{4}$ is $\frac{3 - (-3)}{2} = 3.$ The period is $2\pi \div \frac{1}{4} = 8\pi$, and the first positive zero occurs when $\frac{x}{4} = \pi$, i.e., at $x = 4\pi$. A portion of the graph is shewn in figure \ref{fig:3sin.25x}. \end{solu} \vspace{2cm} \begin{figure}[!h] \begin{minipage}{.4\textwidth}\centering\psset{unit=1pc} \psaxes[linewidth=2pt,labelFontSize=\tiny](0,0)(-8,-5)(8,5) \psplot[algebraic,linewidth=2pt,linecolor=red]{-8}{8}{-2*cos(x/2)+3} \meinecaption{2}{The graph of $x \mapsto -2\cos \frac{x}{2} + 3$.} \label{fig:2cos.5x3} \end{minipage} \hfill \begin{minipage}{.4\textwidth}\centering\psset{unit=1pc} \psaxes[linewidth=2pt,labelFontSize=\tiny](0,0)(-8,-5)(8,5) \psplot[algebraic,linewidth=2pt,linecolor=red]{-8}{8}{-3*sin(x/4)} \meinecaption{2}{The graph of $x \mapsto -3\sin \frac{x}{4}$.} \label{fig:3sin.25x} \end{minipage} \end{figure} \begin{exa}For which real numbers $x$ is $\log _{\cos x} x$ a real number? \end{exa} \begin{solu} If $\log _a t$ is defined and real, then $a > 0, a \neq 1$ and $t > 0.$ Hence one must have $\cos x > 0, \cos x \neq 1$ and $x > 0.$ All this happens when $$x\in \left. \right]0; \frac{\pi}{2}\left[ \ \cup \ \right]\frac{3\pi}{2} + 2\pi n; 2\pi (n + 1) \left[ \right. \cup \left. \right]2\pi (n + 1); \frac{5\pi}{2} + 2\pi n\left[ \right. , $$ for $n \geq 0, n\in \integers .$ \end{solu} \begin{exa} For which real numbers $x$ is $\log _x {\cos x}$ a real number? \end{exa} \begin{solu} In this case one must have $x > 0, x \neq 1$ and $\cos x > 0.$ Hence $$x\in \left. \right]0; 1\left[ \ \cup \ \right]1; \frac{\pi}{2}\left[ \ \cup \ \right]\frac{3\pi}{2} + 2\pi n; \frac{5\pi}{2} + 2\pi n\left[ \right. ,$$ for $n \geq 0, n\in\integers$. \end{solu} \begin{exa} Find the period of $x \mapsto \sin 2x + \cos 3x.$ \end{exa} \begin{solu} Let $P$ be the period of $x \mapsto \sin 2x + \cos 3x.$ The period of $x \mapsto \sin 2x$ is $\pi$ and the period of $x \mapsto \cos 3x$ is $\frac{2\pi}{3}$. In one full period of length $P$, both $x \mapsto \sin 2x$ and $x \mapsto \cos 3x$ must go through an integral number of periods. Thus $P = s\pi = \frac{2\pi t}{3}$, for some positive integers $s$ and $t$. But then $3s = 2t.$ The smallest positive solutions of this is $s = 2, t = 3.$ The period sought is then $P = s\pi = 2\pi$. \end{solu} \begin{exa} How many real numbers $x$ satisfy $$\sin x = \frac{x}{100} ?$$\end{exa} \begin{solu} Plainly $x = 0$ is a solution. Also, if $x > 0$ is a solution, so is $-x < 0$. So, we can restrict ourselves to positive solutions. \bigskip If $x$ is a solution then $|x| = 100|\sin x| \leq 100$. So one can further restrict $x$ to the interval $]0; 100].$ Decompose $]0; 100]$ into $2\pi$-long intervals (the last interval is shorter): $$ ]0; 100] = ]0; 2\pi ] \cup ]2\pi ; 4\pi ] \cup ]4\pi ; 6\pi ] \cup \cdots \cup ]28\pi ; 30\pi] \cup ]30\pi ; 100].$$ From the graphs of $y = \sin x, y = x/100$ we see that that the interval $]0; 2\pi]$ contains only one solution. Each interval of the form $]2\pi k;\ 2(k+ 1)\pi ], k = 1, 2, \ldots , 14$ contains two solutions. As $31\pi < 100$, the interval $]30\pi ; 100]$ contains a full wave, hence it contains two solutions. Consequently, there are $1 + 2\cdot 14 + 2 = 31$ positive solutions, and hence, 31 negative solutions. Therefore, there is a total of $31 + 31 + 1 = 63$ solutions. \end{solu} \subsection*{Homework} \addcontentsline{toc}{section}{Homework} \markright{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro}True or False. Use graphical arguments for the numerical premises. No calculators! \begin{enumerate} \item $x \mapsto \cos 3x$ has period $3$. \item $\cos 3 > \sin 1$. \item The first real zero of $x \mapsto 2\sin x + 8$ occurs at $x = \pi$ \item There is a real number $x$ for which the graph of $x \mapsto 8 + \cos\frac{x}{10}$ touches the $x$-axis. \end{enumerate} \begin{answer} F; F; F; F \end{answer} \end{pro} \begin{pro} Graph portions of the following. Find the amplitude, period, and the location of the first positive real zero, if there is one, of each function. \begin{multicols}{2} \begin{enumerate} \item $x \mapsto 3\sin x$ \item $x \mapsto \sin 3x$ \item $x \mapsto \sin (-3x)$ \item $x \mapsto 3\sin 3x$ \item $x \mapsto 3\cos x$ \item $x \mapsto \cos 3x$ \item $x \mapsto \frac{1}{3}\cos x$ \item $x \mapsto \cos \frac{1}{3}x$ \item $x \mapsto -2\cos \frac{1}{3}x + 13$ \item $x \mapsto \frac{1}{4}\cos \frac{1}{3}x - 10$ \item $x \mapsto |\sin x|$ \item $x \mapsto \sin |x|$ \end{enumerate} \end{multicols} \end{pro} \begin{pro} Find the period of $x \mapsto \sin 3x + \cos 5x$ \end{pro} \begin{pro} Find the period of $x \mapsto \sin x + \cos 5x$ \end{pro} \begin{pro} How many real solutions are there to $$\sin x = \log_e x\ ?$$ \end{pro} \begin{pro} Let $x \geq 0.$ Justify graphically that $$\sin x \leq x.$$ Your argument must make no appeal to graphing software. \end{pro} \begin{pro} Let $x \in \reals$. Justify graphically that $$1 - \frac{x^2}{2} \leq \cos x.$$ Your argument must make no appeal to graphing software. \end{pro} \end{multicols} \clearpage \section{Inversion} Since $\funnoname{x}{\sin x}{\reals}{[-1; 1]}$ is periodic, it is not injective, and hence it does not have an inverse. We can, however, restrict the domain and in this way obtain an inverse of sorts. The choice of the restriction of the domain is arbitrary, but the interval $\cc{-\frac{\pi}{2}; \frac{\pi}{2}}$ is customarily used. \vspace*{3cm} \begin{figure}[!hptb] \begin{minipage}{.4\textwidth}\centering \psaxes[trigLabels=true, trigLabelBase=2,dx=\psPiH,xunit=\psPi,linewidth=2pt,arrows={->}](0,0)(-.7,-1.1)(.7,1.5) \psplot[linecolor=red,linewidth=1.5pt]{-1.570796327}{1.570796327}{x RadtoDeg sin} \meinecaption{2}{$y = \Sin x$} \label{fig:Sin} \end{minipage} \hfill \begin{minipage}{.4\textwidth}\centering \SpecialCoor % For label positionning \psaxes[labels=x,Dy= \pstPI2,linewidth=2pt]{->}% (0,0)(-1.1,-1.6)(1.5,2) \uput[0](!0 PI 2 div){$\frac{\pi}2$} \uput[0](!0 PI 2 div neg)% {$-\frac{\pi}2$} \psplot[linecolor=red,linewidth=2pt]{-1}{1}% {x ASIN }\meinecaption{2}{$y = \arcsin x$} \label{fig:arcsine} \end{minipage} \end{figure} \begin{df} The {\em Principal Sine Function}, $\funnoname{x}{\Sin x}{[-\frac{\pi}{2}; \frac{\pi}{2}]}{[-1; +1]}$ is the restriction of the function $x \mapsto \sin x$ to the interval $\dis{[-\frac{\pi}{2}; \frac{\pi}{2}]}$. With such restriction $$\funnoname{x}{\Sin x}{[-\frac{\pi}{2}; \frac{\pi}{2}]}{[-1; +1]}$$ is bijective with inverse $$\funnoname{x}{\arcsin x}{[-1; +1]}{[-\frac{\pi}{2}; \frac{\pi}{2}]}$$ \end{df} The graph of $\funnoname{x}{\arcsin x}{[-1; +1]}{[-\frac{\pi}{2}; \frac{\pi}{2}]}$ is thus symmetric with the graph of $\funnoname{x}{\Sin x}{[-\frac{\pi}{2}; \frac{\pi}{2}]}{[-1; +1]}$ with respect to the line $y = x$. See figures \ref{fig:Sin} and \ref{fig:arcsine} for the graph of $y = \arcsin x$. The notation $\sin ^{-1}$ is often used to represent $\arcsin$. The function $x \mapsto \arcsin x$ is an odd function, that is, $$\arcsin (-x) = -\arcsin x, \ \forall x\in[-1;1].$$Also, $[-\frac{\pi}{2}; \frac{\pi}{2}]$ is the smallest interval containing $0$ where all the values of $x \mapsto \Sin x$ in the interval $[-1; 1]$ are attained. Moreover, $\forall (x, y)\in[-1;1]\times[-\frac{\pi}{2};\frac{\pi}{2}]$, $y = \arcsin x \Longleftrightarrow x = \sin y.$ \begin{rem} \begin{enumerate} \item Whilst it is true that $\sin \arcsin x = x, \forall x\in [-1; 1]$, the relation $\arcsin \sin x = x$ is not always true. For example, $\arcsin\sin \frac{7\pi}{6} = \arcsin (-\frac{1}{2}) = -\frac{\pi}{6} \neq \frac{7\pi}{6}.$ \item $\funnoname{x}{(\arcsin\circ\sin)(x)}{\reals}{\reals}$ is a $2\pi$-periodic odd function with $$(\arcsin \circ \sin)(x) = \left\{ \begin{array}{ll} x & {\rm if} \ x\in\left[0; \frac{\pi}{2}\right] \\ \pi - x & {\rm if} \ x\in\left[\frac{\pi}{2}; \pi\right]. \end{array} \right. $$ The graph of $x \mapsto (\arcsin\circ\sin)(x)$ is shewn in figure \ref{arcsinsin}.\vspace{2cm} \end{enumerate} \end{rem} \vspace*{2cm} \begin{figure}[!hptb] \begin{minipage}{.4\textwidth} \centering \psset{unit=1.7pc} \psaxes[labels=x,Dy= \pstPI2,linewidth=2pt]{->}% (0,0)(-7,-1.6)(7,2) \uput[0](!0 PI 2 div){$\frac{\pi}2$} \uput[0](!0 PI 2 div neg)% {$-\frac{\pi}2$} \psplot[linecolor=red,linewidth=2pt,plotpoints=5001]{-6.3}{6.3}% {x SIN ASIN} \meinecaption{2}{$y = (\arcsin\circ\sin)(x)$} \label{arcsinsin} \end{minipage} \hfill \begin{minipage}{.4\textwidth} \centering %\psset{unit=1.5pc} \psaxes[ticks=none,labels=none,linewidth=2pt]{->}(0,0)(-4,-2)(4,2) %% Must convert degrees into radians. \psplot[linecolor=brown,linewidth=2pt,algebraic]{-3.1416}{3.5}{sin(x)} \rput(1.571,0){|} \rput(1.571,-.6){$\frac{\pi}{2}$} \rput(-1.571,0){|} \rput(-1.571,-.6){$-\frac{\pi}{2}$} \psline[linewidth=2pt]{<->}(-4, .5)(4, .5) \rput(3.1416, 0){|}\rput(3.1416, -.4){$\pi$} \rput(4.5, .5){$y = A$} \rput{270}(.5236,-1){$\arcsin\ A$}\rput{270}(2.6178,-1.3){$\pi - \arcsin \ A$}\rput(.5236, 0){|} \rput(2.6178, 0){|}\meinecaption{2}{The equation $\sin x = A$} \label{fig:sinequa} \end{minipage} \end{figure} \begin{thm} The equation $$\sin x = A$$ has (i) no real solutions if $|A| > 1$, (ii) the infinity of solutions $$x = (-1)^n\arcsin A + n\pi , \ n\in\integers,$$if $|A| \leq 1.$ \end{thm} \begin{pf} Since $-1 \leq \sin x \leq 1$ for $x\in\reals$, the first assertion is clear. Now, let $|A| \leq 1$. In figure \ref{fig:sinequa} (where we have chosen $0 \leq A \leq 1,$ the argument for $-1 \leq A < 0$ being similar), the first two positive intersections of $y = A$ with $y = \sin x$ occur at $x = \arcsin A$ and $x = \pi - \arcsin A$. Since the sine function is periodic with period $2\pi$, this means that $$x = \arcsin A + 2\pi n, \ n\in\integers$$ and $$x = \pi - \arcsin A + 2\pi n = -\arcsin A + (2n + 1)\pi , \ n\in\integers$$ are the real solutions of this equation. Both relations can be summarised by writing $$x = (-1)^n\arcsin A + n\pi , \ n\in\integers .$$ This proves the theorem. \end{pf} \begin{exa} Find all real solutions to $\sin x = -\frac{1}{2}$, and then find all solutions in the interval $[12\pi ; \frac{27\pi}{2}]$. \end{exa} \begin{solu} The general solution to $\sin x = -\frac{1}{2}$ is given by $$\begin{array}{lll} x & = & (-1)^n\arcsin\left(-\frac{1}{2}\right) + n\pi \\ & = & (-1)^n\left(-\frac{\pi}{6}\right) + n\pi \\ & = & (-1)^{n + 1}\frac{\pi}{6} + n\pi \end{array} $$Now, if $$12\pi \leq (-1)^{n + 1}\frac{\pi}{6} + n\pi \leq \frac{27\pi}{2}$$ then $$12 - (-1)^{n + 1}\frac{1}{6} \leq n \leq \frac{27}{2} - (-1)^{n + 1}\frac{1}{6}.$$ The smallest $12 - (-1)^{n + 1}\frac{1}{6}$ can be is $12 - \frac{1}{6} = \frac{71}{6} > 11$. The largest $\frac{27}{2} - (-1)^{n + 1}\frac{1}{6}$ can be is $\frac{27}{2} + \frac{1}{6} = \frac{41}{3} < 14$. So possibly, $$11 < n < 14,$$which means that $n = 12$ or $n = 13.$ Testing $n = 12,$ $x = -\frac{\pi}{6} + 12\pi = \frac{71\pi}{6}$, which falls outside the interval and $x = \frac{\pi}{6} + 13\pi = \frac{79\pi}{6},$ which falls in the interval. So the only solution in the interval $[12\pi ; \frac{27\pi}{2}]$ is $\frac{79\pi}{6}$. \end{solu} \begin{exa} Find the set of all solutions of $$\sin \frac{\pi}{x^2} = \frac{1}{2}.$$Are there any solutions in the interval $]1; 3[$ ? \end{exa}\begin{solu} We have $$\frac{\pi}{x^2} = (-1)^n\arcsin\frac{1}{2} + n\pi = (-1)^n\frac{\pi}{6} + n\pi $$ $$\frac{1}{x^2} = (-1)^n\frac{1}{6} + n$$ $$x^2 = \frac{1}{(-1)^n\frac{1}{6} + n}$$ $$x^2 = \frac{6}{(-1)^n + 6n}.$$ The expression on the right is negative for integers $n \leq -1.$ Therefore $$x = \pm\sqrt{\frac{6}{(-1)^n + 6n}}, n = 0, 1, 2, 3, \ldots .$$ The set of all solutions is thus $$\left\{-\sqrt{\frac{6}{(-1)^n + 6n}},\ \sqrt{\frac{6}{(-1)^n + 6n}} \ n = 0, 1, 2, 3, \ldots \right\} .$$ If $$1 < \sqrt{\frac{6}{(-1)^n + 6n}} < 3,$$ then $$1 < \frac{6}{(-1)^n + 6n} < 9,$$ $$\frac{1}{6} < \frac{1}{(-1)^n + 6n} < \frac{3}{2},$$ $$\frac{2}{3} < 6n + (-1)^n < 6,$$ $$\frac{2}{3} - (-1)^n < 6n < 6 - (-1)^n.$$The smallest $\frac{2}{3} - (-1)^n$ can be is $-\frac{1}{3}$ and the largest $6 - (-1)^n$ can be is $7$. Hence we must test $n$ such that $-\frac{1}{3} < 6n < 7,$ that is, $n =0$ and $n = 1.$ If $n = 0,$ then $x = \sqrt{6} \in ]1; 3[$. If $n = 1$, then $x = \sqrt{\frac{6}{5}} \in ]1; 3[.$ So the solutions belonging to $]1; 3[$ are $x = \sqrt{6}$ and $x = \sqrt{\frac{6}{5}}$. \end{solu} \begin{exa} Find the set of all real solutions to $$\sin \frac{2}{2x + 1} = \frac{\sqrt{2}}{2}$$ \end{exa} \begin{solu} We have $$\frac{2}{2x + 1} = (-1)^n\arcsin\left(\frac{\sqrt{2}}{2}\right) + \pi n, n \in\integers,$$ which is equivalent to each of the following equations $$\frac{2}{2x + 1} = (-1)^n\frac{\pi}{4} + \pi n,$$ $$\frac{2x + 1}{2} = \frac{1}{(-1)^n\frac{\pi}{4} + \pi n},$$ $$x + \frac{1}{2} = \frac{1}{(-1)^n\frac{\pi}{4} + \pi n},$$ whence the solution set is $$\left\{-\frac{1}{2} + \frac{4}{(-1)^n\pi + 4n\pi} , \ n\in\integers\right\}.$$ \end{solu} \begin{exa} Find the set of all real solutions to $$2\sin^2x - \sin x - 1 = 0.$$ \end{exa} \begin{solu} Factoring, $$0 = 2\sin^2x - \sin x - 1 = (2\sin x + 1)(\sin x - 1) $$ Hence either $\sin x = -\frac{1}{2}$ and so $$x = (-1)^n\arcsin \frac{1}{2} + \pi n = (-1)^n(\frac{-\pi}{6}) + \pi n = (-1)^{n + 1}\frac{\pi}{6} + \pi n,$$ or $\sin x = 1$ and so $$x = (-1)^n\arcsin1 + \pi n = (-1)^n\frac{\pi}{2} + \pi n.$$ The solution set is therefore $$\left\{(-1)^{n + 1}\frac{\pi}{6} + \pi n, \ (-1)^n\frac{\pi}{2} + \pi n, n\in\integers\right\}.$$ \end{solu} \begin{df} The {\em Principal Cosine Function}, $\funnoname{x}{\Cos\ x}{[0; \pi]}{[-1; 1]}$ is the restriction of the function $x \mapsto \cos x$ to the interval $\dis{[0; \pi]}$. With such restriction $$\funnoname{x}{\Cos\ x}{[0; \pi]}{[-1; 1]}$$ is bijective with inverse $$\funnoname{x}{\arccos x}{[-1; 1]}{[0; \pi]}.$$ \end{df} \begin{rem}\begin{enumerate}\item The notation $\cos ^{-1}$ is often used to represent $\arccos$. \item Whilst it is true that $\cos \arccos x = x, \forall x\in [-1; 1]$, the relation $\arccos \cos x = x$ is not always true. For example, $\arccos\cos \frac{7\pi}{6} = \arccos (-\frac{\sqrt{3}}{2}) = \frac{5\pi}{6} \neq \frac{7\pi}{6}.$ \item $x \mapsto \arccos x$ is neither an even nor an odd even function. \item $\funnoname{x}{(\arccos\circ\cos)(x)}{\reals}{\reals}$ is a $2\pi$-periodic even function with $$(\arccos \circ \cos)(x) = \left\{ \begin{array}{ll} x & {\rm if} \ x\in\left[0; \pi\right] \\ -x & {\rm if} \ x\in\left[-\pi; 0\right]. \end{array} \right. $$ \item $\forall (x, y)\in[-1;1]\times[0;\pi]$, $y = \arccos x \Longleftrightarrow x = \cos y.$ \item The graphs of $x \mapsto \Cos x$ and $x \mapsto \arccos x$ are symmetric with respect to the line $y = x$. \end{enumerate} \end{rem} The graph of $x \mapsto \arccos x$ is shewn in figure \ref{arccos}. \bigskip For convenience, we provide the following table. \\ $$\begin{array}{|c|c|c|c|c|c|} \hline x & \arcsin x & \arccos x & x & \arcsin x & \arccos x \\ \hline 0 & 0 & \frac{\pi}{2} & & & \\ \hline 1 & \frac{\pi}{2} & 0 & -1 & -\frac{\pi}{2} & \pi \\ \hline \frac{1}{2} & \frac{\pi}{6} & \frac{\pi}{3} & -\frac{1}{2} & -\frac{\pi}{6}& \frac{2\pi}{3} \\ \hline \frac{\sqrt{2}}{2} & \frac{\pi}{4}& \frac{\pi}{4} & -\frac{\sqrt{2}}{2} & -\frac{\pi}{4}& \frac{3\pi}{4} \\ \hline \frac{\sqrt{3}}{2} & \frac{\pi}{3}& \frac{\pi}{6} & -\frac{\sqrt{3}}{2} & -\frac{\pi}{3}& \frac{5\pi}{6} \\ \hline \end{array} $$ \begin{thm} The equation $$\cos x = A$$ has (i) no real solutions if $|A| > 1$, (ii) the infinity of solutions $$x = \pm\arccos A + 2n\pi , \ n\in\integers,$$if $|A| \leq 1.$ \end{thm} \begin{pf} Since $-1 \leq \cos x \leq 1$ for $x\in\reals$, the first assertion is clear. \bigskip Now, let $|A| \leq 1$. In figure \ref{cosequa} (where we have chosen $0 \leq A \leq 1,$ the argument for $-1 \leq A < 0$ being similar), the two intersections of $y = A$ with $y = \cos x$ closest to $x = 0$ occur at $x = \arccos A$ and $x = -\arccos A$. \bigskip Since the cosine function is periodic with period $2\pi$, this means that $$x = \arccos A + 2\pi n, \ n\in\integers$$ and $$x = -\arccos A + 2\pi n, \ n\in\integers$$ are the real solutions of this equation. Both relations can be summarised by writing $$x = \pm\arccos A + 2n\pi , \ n\in\integers .$$ This proves the theorem. \end{pf} \vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{.4\textwidth} \centering\psset{unit=1.5pc} \rput(0,3.1416){-}\rput(.4,3.1416){{\tiny $\pi$}}\rput(1, 0){|}\rput(1, -0.3){+1} \rput(1, -0.3){+1} \rput(-1, -0.3){-1} \rput(-1, 0){|} \rput(.4,1.571){{\tiny$\frac{\pi}{2}$}} \rput(0,1.571){-} \psaxes[ticks=none,labels=none,linewidth=2pt]{->}(0,0)(-4, -4)(4, 4) \psplot[linewidth=2pt,algebraic]{-1}{1}{acos(x)} \psplot[linestyle=dashed, linecolor=brown,linewidth=2pt ]{0}{3.1416}{x 57.2958 mul cos} \psplot[linestyle=dotted, linecolor=blue]{-1}{3.14}{x} \rput(5.2,3){{\rm solid}\ $y=\arccos x$} \rput(5.2,2){{\rm dashed}\ $y = \cos x$} \meinecaption{3}{$y = \arccos x$} \label{arccos} \end{minipage}\hfill \begin{minipage}{.4\textwidth} \centering \psset{unit=3pc} \psaxes[ticks=none,labels=none,linewidth=2pt]{->}(0,0)(-3,-2)(3,2) %% Must convert degrees into radians. \psplot[linewidth=2pt,linecolor=brown,algebraic]{-2}{2}{cos(x)} \rput(1.571,0){|} \rput(1.571,-.6){$\frac{\pi}{2}$} \rput(-1.571,0){|} \rput(-1.571,-.6){$-\frac{\pi}{2}$} \psline[linewidth=2pt,linecolor=red]{<->}(-2, .5)(2, .5) \rput(3.5, .5){$y = A$} \rput{270}(1.0472,-1){$\arccos A$}\rput{270}(-1.0472,-1.1){ $-\arccos A$}\rput(1.0472, 0){|} \rput(-1.0472, 0){|}\meinecaption{3}{The equation $\cos x = A$} \label{cosequa} \end{minipage} \end{figure} \begin{exa} Find the set of all real solutions to $$2\sin^2x + 3\cos x - 3 = 0.$$ \end{exa} \begin{solu} Since the equation has a cosine to the first power, we write the equation in terms of cosine only, obtaining $$\begin{array}{lll} & 0 & = 2\sin^2x + 3\cos x - 3 \\ & = & 2(1 - \cos^2x) + 3\cos x - 3\\ & = & 2\cos^2 x - 3\cos x + 1\\ & = & (2\cos x - 1)(\cos x - 1)\\ \end{array} $$ Thus either $\cos x = \frac{1}{2}$, in which case $$x = \pm \arccos \frac{1}{2} + 2\pi n = \pm \frac{\pi}{3} + 2\pi n$$ or $\cos x = 1$ in which case $$x = \pm \arccos 1 + 2\pi n = 2\pi n.$$ The solution set is $$\left\{\pm \frac{\pi}{3} + 2\pi n, 2\pi n,\ n\in\integers\right\}.$$ \end{solu} \begin{exa} Find the solutions of the equation$$ \log _{\sqrt{2}\sin x}(1 + \cos x) = 2$$ in the interval $[0; 2\pi]$. \end{exa} \begin{solu} If the logarithmic expression is to make sense, then $\sqrt{2}\sin x > 0,$ $\sqrt{2}\sin x \neq 1$ and $1 + \cos x > 0.$ For this we must have $$x\in\left]0; \frac{\pi}{4}\right[ \cup \left]\frac{\pi}{4}; \frac{3\pi}{4}\right[ \cup \left]\frac{3\pi}{4}; \pi\right[.$$ Now, if $x$ belongs to this set $$\log _{\sqrt{2}\sin x}(1 + \cos x) = 2 \Longleftrightarrow 2\sin^2x = 1 + \cos x.$$Using $\sin^2x = 1 - \cos^2x$, the last equality occurs if and only if $$(2\cos x - 1)(\cos x + 1) = 0.$$If $\cos x + 1 = 0$, then $x = \pi$, a value that must be discarded (why?). If $\cos x = \frac{1}{2}$, then $x = \frac{\pi}{3}$, which is the only solution in $[0; 2\pi]$ \end{solu}. \begin{exa} Find the set of all the real solutions to $$2^{\sin^2x} + 5(2^{\cos^2x}) = 7$$ \end{exa} \begin{solu} Observe that $$\everymath{\displaystyle}{\begin{array}{lll} 2^{\sin^2x} + 5(2^{\cos^2x}) - 7 & = & 2^{\sin^2x} + 5(2^{1 - \sin^2x}) - 7 \\ & = & 2^{\sin^2x} + 5(2^1 \cdot 2^{-\sin^2x}) - 7 \\ & = & 2^{\sin^2x} + \left(\frac{10}{2^{\sin^2x}}\right) - 7 \\ & = & u + \frac{10}{u} - 7. \end{array}}$$ with $u = 2^{\sin^2 x}$. From this, $0 = u^2 - 7u + 10 = (u - 5)(u - 2)$. Thus either $u = 2,$, meaning $2^{\sin^2x} = 2$ which is to say $\sin x = \pm 1$ or $x = (-1)^n(\frac{\pm\pi}{2}) + n\pi$. When $2^{\sin^2x} = 5$ one sees that $\sin^2x = \log _2 5$. Since the sinistral side of the last equality is at most 1 and its dextral side is greater than 1, there are no real roots in this instance. The solution set is thus $$\left\{(-1)^n(\frac{\pm\pi}{2}) + n\pi , \ n\in\integers\right\}.$$ \end{solu} \begin{exa} Find all the real solutions of the equation $$\cos^{2000}x - \sin^{2000}x = 1.$$ \end{exa} \begin{solu} Transposing $$\cos^{2000}x = \sin^{2000}x + 1.$$The dextral side is $\geq 1$ and the sinistral side is $\leq 1.$ Thus equality is only possible if both sides are equal to $1$, which entails that $\cos x = 1$ or $\cos x = -1$, whence $x = \pi n, n \in\integers$. \end{solu} \begin{exa} Find all the real solutions of the equation $$\cos^{2001}x - \sin^{2001}x = 1.$$ \end{exa}\begin{solu} Since $|\cos x| \leq 1$ and $|\sin x| \leq 1$, we have $$ \begin{array}{lll} 1 & = & \cos^{2001}x - \sin^{2001}x \\ & = & \cos^{2001}(-x) + \sin^{2001} (-x) \\ & \leq & | \cos^{2001}(-x)| + | \sin^{2001}(-x)| \\ & = & |\cos^{1999}(-x)|\cos^2(-x) + |\sin^{1999}(-x)|\sin^2(-x) \\ & \leq & \cos^2(-x) + \sin^2(-x) \\ & = & 1. \end{array} $$The inequalities are tight, and so equality holds throughout. The first inequality above is true if and only if $\cos (-x) \geq 0$ and $\sin (-x) \geq 0$. The second inequality is true if and only if $|\cos (-x)| = 1$ or $|\sin (-x)| = 1$. Hence we must have either $\cos (-x) = 1$ or $\sin (-x) = 1$.This means $x = 2n\pi$ or $x = -\frac{\pi}{2} + 2n\pi$ where $n\in\integers$. \end{solu} \begin{exa} What is $\sin\arccos\frac{3}{4}$? \end{exa} \begin{solu} Put $t = \arccos \frac{3}{4}$. Then $\frac{3}{4} = \cos t$ with $t \in [0; \frac{\pi}{2}]$. In the interval $[0; \frac{\pi}{2}]$, $\sin t$ is positive. Hence $$\sin t = \sqrt{1 - \cos ^2 y} = \sqrt{1 - \left(\frac{3}{4}\right)^2} = \frac{\sqrt{7}}{4}.$$ \end{solu} \begin{exa} What is $\sin\arccos (-\frac{3}{7})$? \end{exa} \begin{solu} Put $t = \arccos (-\frac{3}{7})$. Then $-\frac{3}{7} = \cos t$ with $y \in [\frac{\pi}{2}; \pi]$. In the interval $[\frac{\pi}{2}; \pi]$, $\sin t$ is positive. Hence $$\sin t = \sqrt{1 - \cos ^2 t} = \sqrt{1 - \left(-\frac{3}{7}\right)^2} = \frac{2\sqrt{10}}{7}.$$ \end{solu} \begin{exa} Let $x\in ]-\frac{1}{5}; 0[$. Express $\sin \arccos 5x$ as a function of $x$. \end{exa}\begin{solu} First notice that $5x \in ]-1; 0[,$ which means that $\arccos 5x \in ]\frac{\pi}{2}; \pi[$, an interval where the sine is positive. Put $t = \arccos 5x$. Then $5x = \cos t$. Finally, $$\sin t = \sqrt{1 - \cos^2t} = \sqrt{1 - 25x^2}.$$ \end{solu} \begin{exa} Prove that $$\arcsin x + \arccos x = \frac{\pi}{2}, \forall x\in [-1; 1]. $$ \end{exa} \begin{solu} By the complementary angle identity for the cosine, $$\cos \left(\frac{\pi}{2} - \arcsin x\right) = \sin (\arcsin x) = x.$$Since $-\frac{\pi}{2} \leq \arcsin x \leq \frac{\pi}{2}$, we have $\frac{\pi}{2} - \arcsin x \in [0; \pi]$. This means that $$\cos \left(\frac{\pi}{2} - \arcsin x\right) = x \Longleftrightarrow \frac{\pi}{2} - \arcsin x = \arccos x,$$whence the desired result follows. \end{solu} \subsection*{Homework} \addcontentsline{toc}{section}{Homework} \markright{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} True or False. \begin{enumerate} \item $\arcsin\frac{\pi}{2} = 1$. \item If $\arccos x = -\frac{1}{2}$, then $x = -\frac{\pi}{3}$. \item If $\arcsin x \geq 0$ then $x\in [0; \frac{\pi}{2}]$. \item $\arccos\ \cos (-\frac{\pi}{3}) = \frac{\pi}{3}$. \item $\arccos\ \cos (-\frac{\pi}{6}) = -\frac{\pi}{6}$. \item $\arcsin\frac{1}{2000} + \arccos\frac{1}{2000} = \frac{\pi}{2}$. \item $\exists x\in\reals$ such that $\arcsin x > 1.$ \item $-1 \leq \arccos x \leq 1, \ \forall x\in\reals$. \item $\sin\arcsin x = x, \ \forall x\in\reals$. \item $\arccos (\cos x) = x, \ \forall x\in [0; \pi]$. \end{enumerate} \begin{answer} F; F; T; T; F; T; T; F; T; F \end{answer} \end{pro}\begin{pro} Find all the real solutions to $2\sin x + 1 = 0$ in the interval $[-\pi ; \pi]$. \begin{answer} $\{-\frac{5\pi}{6}, -\frac{\pi}{6}\}$ \end{answer} \end{pro} \begin{pro}Find the set of all real solutions to $$\sin\left(3x - \frac{\pi}{4}\right) = 0.$$ \begin{answer} $\{\frac{\pi}{12} + \frac{n\pi}{3}, n\in\integers\}$ \end{answer} \end{pro} \begin{pro} Find the set of all real solutions of the equation $$-2\sin^2 x - \cos x + 1 = 0.$$ \begin{answer} $\{\pm\frac{2\pi}{3} + 2\pi n, 2\pi n, n \in\integers\}$. \end{answer} \end{pro} \begin{pro} Find all the real solutions to $\sin 3x = -1.$ Find all the solutions belonging to the interval $[98\pi ; 100\pi]$. \begin{answer} $\{(-1)^{n + 1}\frac{\pi}{6} + \frac{n\pi}{3}, n \in\integers\}$; $\{(-1)^{n + 1}\frac{\pi}{6} + \frac{n\pi}{3}, n = 295, 296, 297, 298, 299, 300\}$ \end{answer} \end{pro} \begin{pro} Find the set of all real solutions to $$5\cos^2x - 2\cos x - 7 = 0.$$ \begin{answer} $\{(2n + 1)\pi , n\in\integers\}$ \end{answer} \end{pro}\begin{pro} Find the set of all real solutions to $$\sin x\cos x = 0.$$ \begin{answer} $\{\frac{n\pi}{2}, \ n\in\integers\}$ \end{answer} \end{pro} \begin{pro} Find the set of all real solutions to $$\cos 3x = \frac{4}{3}.$$ \begin{answer} $\emptyset$ \end{answer} \end{pro} \begin{pro} Find the set of all real solutions to $$4\sin^2 2x - 3 = 0.$$ \begin{answer} $\{-\frac{\pi}{6} + n\pi , \frac{2\pi}{3} + n\pi\}$ \end{answer} \end{pro} \begin{pro} Find all real solutions belonging to the interval $[-2; 2]$, if any, to the following equations. \begin{enumerate} \item $4\sin^2x - 3 = 0$ \item $2\sin^2x = 1$ \item $\cos \frac{2x}{3} = -\frac{\sqrt{3}}{2}$ \item $\sin \frac{3}{x} = 1$ \item $\dis{\frac{1 + \sin x}{1 - \cos x} = 0}$ \end{enumerate} \begin{answer} (1) $\{-\frac{\pi}{3}, \frac{\pi}{3}\}$ ; (2) $\{-\frac{\pi}{6}, \frac{\pi}{6}\}$; (3) No solutions in this interval; (4) All the solutions belong to this interval $\{\frac{6}{(-1)^n\pi + 2n\pi}, \ n\in\integers\}$ ; (5) $\{-\frac{\pi}{2}, \frac{\pi}{2}\}$ \end{answer} \end{pro} \begin{pro} Find $\sin\arccos\frac{1}{3}$. \begin{answer} $\frac{2\sqrt{2}}{3}$ \end{answer} \end{pro} \begin{pro} Find $\cos\arcsin (-\frac{2}{3})$. \begin{answer} $\frac{\sqrt{5}}{3}$ \end{answer} \end{pro} \begin{pro} Find $\sin\arccos (-\frac{2}{3})$. \begin{answer} $\frac{\sqrt{5}}{3}$ \end{answer} \end{pro} \begin{pro} Find $\arcsin (\sin 5)$; $\arccos (\cos 10)$ \begin{answer} $5 - 2\pi$; $4\pi - 10$ \end{answer} \end{pro} \begin{pro} Find all the real solutions of the following equations. \begin{enumerate} \item $\cos x + \dfrac{1}{\cos x} = \frac{3}{2}$. \item $2\cos^3x + \cos^2x - 2\cos x - 1 = 0$. \item $\dis{6\cos^2\left(5x - \frac{\pi}{3}\right) - \cos\left(5x - \frac{\pi}{3}\right) = 2.}$ \item $4\cos^2x - 2(\sqrt{2} + 1)\cos x + \sqrt{2} = 0$. \item $4\cos^4x - 17\cos^2x + 4 = 0.$ \item $(2\cos x + 1)^2 - 4\cos^2x + (\sin x)(2\cos x + 1) + 1 = 0$. \item $4\sin^2x - 2(\sqrt{3} - \sqrt{2})\sin x = \sqrt{6}.$ \item $-2\sin^2x + 19|\sin x| + 10 = 0.$ \end{enumerate} \end{pro} \begin{pro} Demonstrate that $$\arccos \ x + \arccos (-x) = \pi , \ \ \forall x\in [-1; 1],$$ $$\arcsin x = -\arcsin (-x), \ \ \forall x\in [-1; 1].$$ \end{pro} \begin{pro} Shew that $$\arcsin x = \arccos\sqrt{1 - x^2}, \ \forall x\in [0; 1 ],$$ $$\arccos x = \arcsin\sqrt{1 - x^2}, \ \forall x\in [0; 1 ].$$ \end{pro} \begin{pro} Let $0 < x < \frac{1}{3}$. Find $\cos\ \arcsin 3x$ and $\cos \ \arccos 3x$ as functions of $x$. \end{pro} \begin{pro} Let $-\frac{1}{2} < x < 0$. Find $\sin\ \arcsin 2x$ and $\sin\ \arccos 2x$ as functions of $x$. \end{pro} \begin{pro} Find real constants $a, b$ such that $$(\arcsin\circ\sin)(x) = ax + b, \ \forall x\in [\frac{99\pi}{2};\frac{101\pi}{2}].$$ \end{pro} \begin{pro} Prove that $\funnoname{x}{(\arccos\circ\cos )(x)}{\reals}{\reals}$ is a $2\pi$-periodic even function and graph a portion of this function for $x\in[-2\pi; 2\pi].$ \end{pro} \end{multicols} \section{The Goniometric Functions} We define the {\em tangent, secant, cosecant} and {\em cotangent} of $x\in\reals$ as follows. \begin{equation} \tan x = \frac{\sin x}{\cos x}, \ x \neq \frac{\pi}{2} + \pi n, \ n \in \integers, \end{equation} \begin{equation} \sec x = \frac{1}{\cos x}, \ x \neq \frac{\pi}{2} + \pi n, \ n \in \integers, \end{equation} \begin{equation} \csc x = \frac{1}{\sin x}, \ x \neq \pi n, \ n \in \integers, \end{equation} \begin{equation} \cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x}, \ x \neq \pi n, \ n \in \integers. \end{equation} The circles below have all radius $1$. \begin{center} \begin{tabular}{ll||ll} cosine & \pscircle[fillstyle=solid,fillcolor=brown]{0.15} & sine & \pscircle[fillstyle=solid,fillcolor=blue]{0.15} \\ secant & \pscircle[fillstyle=solid,fillcolor=magenta]{0.15} & tangent & \pscircle[fillstyle=solid,fillcolor=blue]{0.15} \\ cosecant & \pscircle[fillstyle=solid,fillcolor=orange]{0.15} & cotangent& \pscircle[fillstyle=solid,fillcolor=yellow]{0.15} \\ \end{tabular} \end{center} \vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{4cm} $$ \psset{unit=2.5pc} \pscircle[linewidth=2pt](0,0){1} \psaxes[labels=none,linewidth=2pt]{->}(0,0)(-2,-2)(2,2) \psline(0,0)(0.86602540378443864676372317075294,.5)\psdots[dotstyle=*,dotscale=1](0.86602540378443864676372317075294,.5) \psline[linecolor=brown,linewidth=2pt](0,0)(0.86602540378443864676372317075294,0) \psline[linecolor=blue,linewidth=2pt](0.86602540378443864676372317075294,0)(0.86602540378443864676372317075294,.5) $$ \end{minipage} \hfill \begin{minipage}{4cm} $$ \psset{unit=2.5pc} \pscircle[linewidth=2pt](0,0){1} \psaxes[labels=none,linewidth=2pt]{->}(0,0)(-2,-2)(2,2) \psdots[dotstyle=*,dotscale=1](0.86602540378443864676372317075294,.5) \psline[linecolor=blue,linewidth=2pt](1,0)(1,0.57735026918962576450914878050196) \psline[linecolor=magenta,linewidth=2pt](0,0)(1,0.57735026918962576450914878050196) \psline[linestyle=dotted](1,-2)(1,0) \psline[linestyle=dotted](1,0.57735026918962576450914878050196)(1,2) $$ \end{minipage} \hfill \begin{minipage}{4cm} $$ \psset{unit=2.5pc} \pscircle[linewidth=2pt](0,0){1} \psaxes[labels=none,linewidth=2pt]{->}(0,0)(-2,-2)(2,2) \psdots[dotstyle=*,dotscale=1](0.86602540378443864676372317075294,.5) \psline(1.7320508075688772935274463415059,0)(1.7320508075688772935274463415059,1) \psline[linecolor=yellow,linewidth=2pt](0,0)(1.7320508075688772935274463415059,0) \psline[linecolor=orange,linewidth=2pt](0,0)(1.7320508075688772935274463415059,1) \psline[linestyle=dotted](-2,1)(2,1) $$ \end{minipage} \end{figure} \vspace{2cm} \begin{rem} \begin{enumerate} \item The image of $x \mapsto \tan x$ over its domain $\reals - \{\frac{\pi}{2} + \pi n, \ n \in \integers\}$ is $\reals$. \item The image of $x \mapsto \cot x$ over its domain $\reals - \{\pi n, \ n \in \integers\}$ is $\reals$. \item The image of $x \mapsto \sec x$ over its domain $\reals - \{\frac{\pi}{2} + \pi n, \ n \in \integers\}$ is $]-\infty; -1] \cup [1; +\infty[$. \item The image of $x \mapsto \csc x$ over its domain $\reals - \{\pi n, \ n \in \integers\}$ is $]-\infty; -1] \cup [1; +\infty[$. \end{enumerate} \end{rem} \begin{exa} Given that $\tan x = -3$ and $\winding(x)$ lies in the fourth quadrant, find $\sin x$ and $\cos x$. \end{exa} \begin{solu} In the fourth quadrant $\sin x < 0$ and $\cos x > 0.$ Now, $-3 = \tan x = \frac{\sin x}{\cos x}$ entails $\sin x = -3\cos x$. As $\sin ^2 x + \cos^2 x = 1$, One gathers $9\cos ^2 x + \cos ^2 x = 1$ or $\cos ^2 x = \frac{1}{10}$. Choosing the positive root, $\cos x = \frac{\sqrt{10}}{10}.$ Finally, $$\sin x = -3\cos x = -\frac{3\sqrt{10}}{10}.$$ \end{solu} \begin{exa} Given that $\cot x = 4$ and $\winding(x)$ lies in the third quadrant, find the values of $\tan x, \sin x, \cos x, \csc x, \sec x.$ \end{exa} \begin{solu} From $\cot x = 4$, we have $\cos x = 4\sin x$. Using this and $\sin^2x + \cos^2x = 1,$ we gather $\sin^2x + 16\sin^2x = 1,$ and since $\winding(x)$ lies in the third quadrant, $\sin x = -\frac{\sqrt{17}}{17}$. Moreover, $\cos x = 4\sin x = -\frac{4\sqrt{17}}{17}$. Finally, $\tan x = \frac{1}{\cot x} = \frac{1}{4}$, $\csc x = \frac{1}{\sin x} = -\sqrt{17}$ and $\sec x = \frac{1}{\cos x} = -\frac{\sqrt{17}}{4}$. \end{solu} \begin{thm} The function $\funnoname{x}{\tan x}{\reals - \{\frac{\pi}{2} + \pi n, \ n \in \integers\}}{\reals}$ is an odd function. \label{tanodd}\end{thm} \begin{pf} If $\ x \neq \frac{\pi}{2} + \pi n, \ n \in \integers$ $$\tan (-x) = \frac{\sin (-x)}{\cos (-x)} = -\frac{\sin x}{\cos x} = -\tan x,$$ which proves the assertion. \end{pf} \begin{thm} The function $\funnoname{x}{\tan x}{\reals - \{\frac{\pi}{2} + \pi n, \ n \in \integers\}}{\reals}$ is periodic with period $\pi$. \label{tanper}\end{thm} \begin{pf} Since $$\tan (x + \pi) = \frac{\sin (x + \pi)}{\cos(x + \pi)} = \frac{-\sin x}{-\cos x} = \tan x,$$the period is at most $\pi$. Assume now that $0 < P < \pi$ is a period for $x \mapsto \tan x$. Then $\tan x = \tan (x + P) \ \forall x \in \reals$ and in particular, $$ 0 = \tan 0 = \tan P = \frac{\sin P}{\cos P},$$ which entails that $\sin P = 0$. But then $P$ would be a positive zero of $x \mapsto \sin x$ smaller than $\pi$, a contradiction. Hence the period of $x \mapsto \tan x$ is exactly $\pi$, which completes the proof. \end{pf} How to graph $x \mapsto \tan x$? We start with $x \in[0;\frac{\pi}{2}[$ and then appeal to theorem \ref{tanodd} and theorem \ref{tanper} to extend this construction for all $x\in\reals$. \bigskip In figure \ref{fig:tangent_graph_construction}, choose $B$ such that the measure of arc $\arc{AB}$ (measured counterclockwise) be $x$. Point $A = (1, 0)$, and point $B = (\sin x, \cos x)$. Since points $B$ and $(1, t )$ are collinear, the gradient (slope) of the line joining $(0, 0)$ and $B$ is the same as that joining $(0, 0)$ and $(1, t)$. Computing gradients, we have $$\frac{\sin x - 0}{\cos x - 0} = \frac{t - 0}{1 - 0},$$whence $t = \tan x.$ We have thus produced a line segment measuring $\tan x.$ If we let $x$ vary from $0$ to $\pi/2$ we obtain the graph of $x \mapsto \tan x$ for $x\in[0; \frac{\pi}{2}[$. \bigskip Since $\cos x = 0$ at $x = \frac{\pi}{2}(2n + 1), \ n\in\integers$, $x \mapsto \tan x $ has poles at the points $x = \frac{\pi}{2}(2n + 1), \ n \in \integers$. The graph of $x \mapsto \tan x$ is shewn in figure \ref{fig:graph_tangent}. \bigskip \vspace*{3cm} \begin{figure}[!hptb] $$ \psset{unit=2pc} \psline[linewidth=0.2mm, linestyle=dashed]{<->}(-8,0)(-2,0) \psline(-5,0)(-3.5,1.5)(-3.5,0) \psline[linewidth=0.2mm,linestyle=dashed]{<->}(-5, -3)(-5, 3) \pscircle[linewidth= 0.2mm](-5,0){1.5} %\psarc[linewidth=.4mm]{|-|}(-5,0){4}{0}{45} %\rput(-3, .7){x} \rput(-3.5,-.4){A} \psdots[dotstyle=*,dotscale=1](-3.93934,1.06066)\psdots[dotstyle=*,dotscale=1](-3.5,0) \psdots[dotstyle=*,dotscale=1](-3.5,1.5)\rput(-3, 1.8){(1, t)} \psline[linestyle=dotted](-3.5,1.5)(5.81,1.5) \rput(-5.4,-.4){O}\rput(-5,0){\rput(1.1,1.4){B}} \psaxes[labels=none, ticks=none ](5,0)(7,4)(3,-4)\rput(5,0){\psplot[linewidth=2pt,linecolor=brown]{-1.3}{1.3}{x 57.2958 mul sin x 57.2958 mul cos div}}\psline[linestyle=dotted](6.5708,-4)(6.5708,4) \psline[linestyle=dotted](3.4292,-4)(3.4292,4) \rput(3.4292,-.6){-\frac{\pi}{2}} \rput(6.571,-.6){\frac{\pi}{2}} $$ \meinecaption{3}{Construction of the graph of $x \mapsto \tan x$ for $x\in[0; \frac{\pi}{2}[$.} \label{fig:tangent_graph_construction} \end{figure} \vspace*{2cm} \begin{figure}[h] \begin{minipage}{.4\textwidth} \centering $$\psset{unit=1pc} \psplot[linewidth=2pt,linecolor=brown,arrows={<->}]{-4.44}{-1.85}{x 57.2958 mul sin x 57.2958 mul cos div} \psplot[linewidth=2pt,linecolor=brown,arrows={<->}]{-1.3}{1.3}{x 57.2958 mul sin x 57.2958 mul cos div} \psplot[linewidth=2pt,linecolor=brown,arrows={<->}]{1.85}{4.44}{x 57.2958 mul sin x 57.2958 mul cos div} \uput[d](-1.571,-4){-\frac{\pi}{2}} \uput[d](1.571,-4){\frac{\pi}{2}} \uput[d](4.7124,-4){\frac{3\pi}{2}} \uput[d](-4.7124,-4){-\frac{3\pi}{2}} \psline[linestyle=dotted](-4.7124,-4)(-4.7124,4) \psline[linestyle=dotted](-1.5708,-4)(-1.5708,4) \psline[linestyle=dotted](1.5708,-4)(1.5708,4) \psline[linestyle=dotted](4.7124,-4)(4.7124,4) \psaxes[ticks=none,labels=none,linewidth=2pt]{->}(0,0)(-7,-4)(7,4) $$ \meinecaption{3}{$y = \tan x$} \label{fig:graph_tangent} \end{minipage}\hfill \begin{minipage}{.4\textwidth} $$ \psset{unit=1pc} \psaxes[ticks=none,labels=none,linewidth=2pt]{->}(0,0)(-6, -2)(6, 2) \rput(.4,-1.571){-\frac{\pi}{2}}\rput(0,-1.571){-} \rput(0,1.571){-} \rput(.4,1.571){\frac{\pi}{2}} \psplot[linewidth=2pt,linecolor=brown,arrows={<-}]{-5}{-.01}{x abs 1 atan -57.2958 div} \psplot[linewidth=2pt,linecolor=brown,arrows={->}]{.01}{5}{x 1 atan 57.2958 div} \psline[linestyle=dotted](-6,1.571)(6,1.571) \psline[linestyle=dotted](-6,-1.571)(6,-1.571)$$ \meinecaption{3}{$y = \arctan x$} \label{fig:graph_arctan} \end{minipage} \end{figure} We now define the Principal Tangent function and the $\arctan$ function. \begin{df} The {\em Principal Tangent Function}, $x \mapsto \Tan\ x$ is the restriction of the function $x \mapsto \tan x$ to the interval $\dis{]-\frac{\pi}{2}; \frac{\pi}{2}[}$. With such restriction $$\funnoname{x}{\Tan\ x}{]-\frac{\pi}{2}; \frac{\pi}{2}[}{\reals}$$ is bijective with inverse $$\funnoname{x}{\arctan x}{\reals}{]-\frac{\pi}{2}; \frac{\pi}{2}[} $$ \end{df} The graph of $x \mapsto \arctan x$ is shewn in figure \ref{fig:graph_arctan}. Observe that the lines $y = \pm\frac{\pi}{2}$ are asymptotes to $x \mapsto \arctan x.$ \begin{rem} \begin{enumerate} \item $\forall x\in\reals, \ \tan (\arctan(x)) = x.$ \item $\funnoname{x}{(\arctan \circ\tan)(x)}{\reals - \{\frac{\pi}{2} + n\pi, n\in \integers\}}{\reals}$ is an odd $\pi$-periodic function. \end{enumerate} \end{rem} \begin{thm} The equation $$\tan x = A, \ A \in \reals$$ has the infinitely many solutions $$x = \arctan A + n\pi , n \in\integers .$$ \end{thm} \begin{pf} Since the graph of $x \mapsto \tan x$ is increasing in $]-\frac{\pi}{2}; \frac{\pi}{2}[$, it intersects the graph of $y = A$ at exactly one point, $$\tan x = A \Longrightarrow x = \arctan A$$ if $x\in]-\frac{\pi}{2}; \frac{\pi}{2}[.$ Since $x \mapsto \tan x$ is periodic with period $\pi$, each of the points $$x = \arctan A + n\pi , n\in\integers$$ is also a solution. \end{pf} \begin{exa} Solve the equation $$\tan ^2 x = 3$$ \end{exa} \begin{solu} Either $\tan x = \sqrt{3}$ or $\tan x = -\sqrt{3}$. This means that $x = \arctan \sqrt{3} + \pi n = \frac{\pi}{3} + \pi n$ or $x = \arctan (-\sqrt{3}) + \pi n = -\frac{\pi}{3} + \pi n$. We may condense this by writing $x = \pm\frac{\pi}{3} + \pi n, n \in \integers$. \end{solu} \begin{exa} Solve the equation $(\tan x)^{\sin x} = (\cot x)^{\cos x}$. \end{exa} \begin{solu} For the tangent and cotangent to be defined, we must have $x \neq \frac{n\pi}{2}, n\in\integers$. Then $$(\tan x)^{\sin x} = (\cot x)^{\cos x} = \frac{1}{(\tan x)^{\cos x}}$$implies $$(\tan x)^{\sin x + \cos x} = 1.$$ Thus either $\tan x = 1,$ in which case $x = \frac{\pi}{4} + n\pi, \ n\in\integers$ or $\sin x + \cos x = 0,$ which implies $\tan x = -1$, but this does not give real values for the expressions in the original equation. The solution is thus $$x = \frac{\pi}{4} + n\pi, \ n\in\integers.$$ \end{solu} \begin{exa} Find $\sin \arctan \frac{2}{3}$. \end{exa} \begin{solu} Put $t = \arctan \frac{2}{3}$. Then $\frac{2}{3} = \tan t, t \in ]0; \frac{\pi}{2}[$ and thus $\sin t > 0.$ We have $\frac{3}{2}\sin t = \cos t$. As $$1 = \cos ^2 t + \sin ^2 t = \frac{9}{4}\sin ^2 t + \sin ^2 t,$$ we gather that $\sin ^2 t = \frac{4}{13}$. Taking the positive square root $\sin t = \frac{2}{13}$. \end{solu} \begin{exa} Find the exact value of $\tan \arccos (-\frac{1}{5})$.\end{exa} \begin{solu} Put $t = \arccos (-\frac{1}{5})$. As the arccosine of a negative number, $t \in [\frac{\pi}{2}, \pi]$. Now, $\cos t = -\frac{1}{5}$, and so $$\sin t = \sqrt{1 - \left(-\frac{1}{5}\right)^2} = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5}.$$ We deduce that $\tan t = \frac{\sin t}{\cos t} = -2\sqrt{6}.$ \end{solu} \begin{exa} Let $x \in [0; 1[$. Prove that $$\arcsin x = \arctan\frac{x}{\sqrt{1 - x^2 }}.$$ \end{exa} \begin{solu} Since $x \in [0; 1[,$ $\arcsin x \in [0; \frac{\pi}{2}[$. Put $t = \arcsin x,$ then $\sin t = x$, and $\cos t > 0$ since $t\in [0; \frac{\pi}{2}[$. Now, $\cos t = \sqrt{1 - \sin^2t} = \sqrt{1 - x^2}$, and $$\tan t = \frac{\sin t}{\cos t} = \frac{x}{\sqrt{1 - x^2}}.$$Since $t\in [0; \frac{\pi}{2}[$ this implies that $$t = \arctan\frac{x}{\sqrt{1 - x^2}},$$from where the desired equality follows. \end{solu} \begin{thm} The following Pythagorean-like Relation holds. \begin{equation} \tan ^2 x + 1 = \sec ^2 x, \ \forall x\in\reals\setminus\{(2n + 1 )\frac{\pi}{2}, \ n\in\integers\} . \end{equation} \end{thm}\begin{pf} This immediately follows from $\sin ^2 x + \cos ^2 x = 1$ upon dividing through by $\cos ^2 x$. \end{pf} \begin{exa} Given that $\tan x + \cot x = a$, write $\tan ^3x + \cot ^3 x$ as a polynomial in $a$. \end{exa} \begin{solu} Using the fact that $\tan x\cot x = 1$, and the Binomial Theorem: $${\everymath{\displaystyle} \begin{array}{lll} (\tan x + \cot x)^3 & = & \tan^3x + 3\tan^2x\cot x + 3\tan x\cot^2x + \cot^3x \\ & = & \tan^3x + \sin^3x + 3\tan x\cot x(\tan x + \cot x) \\ & = & \tan^3x + \sin^3x + 3(\tan x + \cot x) \\ \end{array}}$$ It follows that $$\tan ^3 x + \cot ^3 x = (\tan x + \cot x)^3 - 3(\tan x + \cot x) = a^3 - 3a.$$ \flushleft{{\bf Aliter:}} Observe that $a^2 = (\tan x + \cot x)^2 = \tan ^2x + \cot ^2x + 2$, hence $\tan ^2x + \cot ^2x = a^2 - 2.$ Factorising the sum of cubes $$\tan ^3 x + \cot ^3 x = (\tan x + \cot x)(\tan ^2x - 1 + \cot ^2x) = a(a^2 - 1 - 2) $$ which equals $ a^3 - 3a,$ as before. \end{solu} \begin{exa} Prove that $$\frac{2\sin y + 3}{2\tan y + 3\sec y} = \cos y,$$whenever the expression on the sinistral side be defined. \end{exa} \begin{solu} Decomposing the tangent and the secant as cosines we obtain, $$\renewcommand{\arraystretch}{1.5}{\everymath{\displaystyle} \begin{array}{lll} \frac{2\sin y + 3}{2\tan y + 3\sec y} & = & \frac{2\sin y + 3}{2\frac{\sin y}{\cos y} + \frac{3}{\cos y}} \\ & = & \frac{2\sin y\cos y + 3 \cos y}{2\sin y + 3} \\ & = & \frac{(\cos y)(2\sin y + 3)}{2\sin y + 3} \\ & = & \cos y, \end{array}} $$as we wished to shew. \end{solu} \begin{exa} Prove the identity $$\frac{\tan A + \tan B}{\sec A - \sec B} =\frac{\sec A + \sec B}{\tan A - \tan B},$$ whenever the expressions involved be defined. \end{exa} \begin{solu} We have $$\renewcommand{\arraystretch}{1.5}{\everymath{\displaystyle}\begin{array}{lll} \frac{\tan A + \tan B}{\sec A - \sec B} & = & \left(\frac{\tan A + \tan B}{\sec A - \sec B}\right) \left(\frac{\tan A - \tan B}{\sec A + \sec B}\right) \left(\frac{\sec A + \sec B}{\tan A - \tan B}\right) \\ & = & \left(\frac{\tan ^2 A - \tan ^2 B}{\sec ^2 A - \sec ^2 B}\right) \left(\frac{\sec A + \sec B}{\tan A - \tan B}\right) \\ & = & \left(\frac{(\sec ^2 A - 1) - (\sec ^2 B - 1)}{\sec ^2 A - \sec ^2 B}\right) \left(\frac{\sec A + \sec B}{\tan A - \tan B}\right) \\ & = & \frac{\sec A + \sec B}{\tan A - \tan B}, \\ \end{array}} $$as we wished to shew. \end{solu} \begin{exa}Given that $\sin A + \csc A = T$, express $\sin^4A + \csc^4A$ as a polynomial in $T$. \end{exa} \begin{solu} First observe that $$T^2 = (\sin A + \csc A)^2 = \sin^2A + \csc^2A + 2\sin A\csc A,$$ hence $$\sin^2A + \csc^2A = T^2 - 2.$$ By the Binomial Theorem $$ \begin{array}{lll} T^4 & = & (\sin A + \csc A)^4\\ & = & \sin ^4A + 4\sin^3A\csc A + 6\sin^2A\csc^2A + 4\sin A\csc^3A + \csc^4A\\ & = & \sin^4A + \csc^4A + 6 + 4(\sin^2 A + \csc^2 A) \\ & = & \sin^4A + \csc^4A + 6 + 4(T^2 - 2), \\ \end{array} $$whence $\sin^4A + \csc^4A = T^4 - 4T + 2.$ \end{solu} \subsection*{Homework} \addcontentsline{toc}{section}{Homework} \markright{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} True or False. \begin{enumerate} \item $\tan x = \cot \frac{1}{x}, \ \forall x\in\reals\setminus\{0\}$. \item $\exists x\in\reals$ such that $\sec x = \frac{1}{2}$. \item $\arctan 1 = \frac{\arcsin 1}{\arccos 1}$. \item $x \mapsto \tan 2x$ has period $\pi$. \end{enumerate} \begin{answer} F; F; F; F \end{answer} \end{pro} \begin{pro} Given that $\csc x = -1.5$ and $\winding(x)$ lies on the fourth quadrant, find $\sin x, \cos x$ and $\tan x$. \begin{answer} $\sin x = -\frac{2}{3}, \cos x = \frac{\sqrt{5}}{3}, \tan x = -\frac{-2\sqrt{5}}{5}.$ \end{answer} \end{pro}\begin{pro} Given that $\tan x = 2$ and $\winding(x)$ lies on the third quadrant, find $\sin x$ and $\cos x$. \begin{answer} $\sin x = -\frac{2\sqrt{5}}{5}, \cos x = -\frac{\sqrt{5}}{5}$ \end{answer} \end{pro} \begin{pro} Given that $\sin x = t^2$ and $\winding(x)$ lies in the second quadrant, find $\cos x$ and $\tan x$. \begin{answer} $\cos x = -\sqrt{1 - t^4}, \tan x = -\frac{t^2}{\sqrt{1 - t^4}}$ \end{answer} \end{pro} \begin{pro} Let $x < -1.$ Find $\sin \arcsec x$ as a function of $x$. \begin{answer} $\sin\arcsec x = -\sqrt{1 - \frac{1}{x^2}}$ \end{answer} \end{pro} \begin{pro} Find $\cos\arctan (-\frac{1}{3})$. \begin{answer} $\frac{3\sqrt{10}}{10}$ \end{answer} \end{pro} \begin{pro} Find $\arctan (\tan (-6))$, $\arccot (\cot (-10))$. \begin{answer} $2\pi - 6$; $4\pi - 10$ \end{answer} \end{pro} \begin{pro} Give a sensible definition of the Principal Cotangent, Secant, and Cosecant functions, and their inverses. Graph each of these functions. \end{pro} \begin{pro} Solve the following equations. \begin{enumerate} \item $\sec^2x - \sec x - 2 = 0$ \item $\tan x + \cot x = 2$ \item $\tan 4x = 1$ \item $2\sec^2x + \tan^2x - 3 = 0$ \item $2\cos x - \sin x = 0$ \item $\tan (x + \frac{\pi}{3}) = 1$ \item $3\cot ^2x + 5\csc x + 1 = 0$ \item $2\sec^2 x = 5\tan x$ \item $\tan^2x + \sec^2x = 17$ \item $6\cos^2x + \sin x - 5 = 0$ \end{enumerate} \end{pro} \begin{pro} Prove that $$\tan x = \cot \left(\frac{\pi}{2} - x\right),$$ $$\cot x = \tan \left(\frac{\pi}{2} - x\right).$$ \end{pro} \begin{pro} Prove that if $x\in\reals$ then $$\arctan x + \arccot \frac{1}{x} = \frac{\pi}{2}{\rm sgn}(x), $$where ${\rm sgn}(x) = -1$ if $x < 0,$ ${\rm sgn}(x) = 1$ if $x > 0,$ and ${\rm sgn}(0) = 0.$ \end{pro} \begin{pro} Graph $x \mapsto (\arctan\circ\tan)(x)$ \end{pro} \begin{pro} Let $x \in ]0; 1[$. Prove that $$\arcsin x = \arccot\frac{\sqrt{1 - x^2 }}{x}.$$ \end{pro} \begin{pro} Let $x \in ]0; 1[$. Prove that $$\arccos x = \arctan\frac{\sqrt{1 - x^2 }}{x} = \arccot \frac{x}{\sqrt{1 - x^2}}.$$ \end{pro} \begin{pro} Let $x > 0$. Prove that $$\arctan x = \arcsin\frac{x}{\sqrt{1 + x^2 }} = \arccos \frac{1}{\sqrt{1 + x^2}}.$$ \end{pro} \begin{pro} Let $x > 0$. Prove that $$\arccot x = \arcsin\frac{1}{\sqrt{1 + x^2 }} = \arccos \frac{x}{\sqrt{1 + x^2}}.$$ \end{pro} \begin{pro} Prove the following identities. Assume, whenever necessary, that the given expressions are defined. \begin{enumerate} \item $$\sin x\tan x = \sec x - \cos x$$ \item $\dis{\tan ^3 x + 1 = (\tan x + 1)(\sec^2x - \tan x)}$ \item $\dis{1 + \tan ^2 x = \frac{1}{2 - 2\sin x} + \frac{1}{2 + 2\sin x}}$ \item $\dis{\frac{\sec\alpha\sin\alpha}{\tan\alpha + \cot\alpha} = \sin^2\alpha}$ \item $\dis{\frac{1 - \sin\alpha}{\cos\alpha} =\frac{\cos\alpha}{1 + \sin\alpha}}$ \item $\dis{7\sec ^2x - 6\tan^2x + 9\cos^2x = \frac{(1 + 3\cos^2x)^2}{\cos^2x}}$ \item $\dis{\frac{1 - \tan ^2 t}{1 + \tan^2 t} = \cos^2 t - \sin ^2 t}$ \item $\dis{\frac{1 + \tan B + \sec B}{1 + \tan B - \sec B} = (1 + \sec B)(1 + \csc B)}$ \end{enumerate} \end{pro} \end{multicols} \section{ Addition Formulae} We will now derive the following formul\ae. \begin{equation} \cos (\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta \end{equation} \begin{equation} \sin (\alpha \pm \beta) = \sin\alpha\cos\beta \pm \sin\beta\cos\alpha \end{equation} \begin{equation} \tan (\alpha \pm \beta) = \frac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha\tan\beta} \end{equation}\vspace*{2cm} \begin{figure}[!hptb] \begin{minipage}{7cm} \centering \psset{unit=6pc} \pstGeonode[PointName=none,PointSymbol=none](1,0){T} \pstGeonode[PointName=none,PointSymbol=none](0,0){O} \pstGeonode[PosAngle={45,45}](1;25){A}(1;75){B} \pstLineAB{O}{A}\pstLineAB{O}{B}\pstLineAB{O}{T} \pstCircleOA{O}{T} \pstMarkAngle[arrows=->,linecolor=brown,LabelSep=.45]{T}{O}{A}{$b$} \pstMarkAngle[arrows=->,linecolor=blue,LabelSep=1.4, MarkAngleRadius= 1.2]{A}{O}{B}{$a-b$} \pstMarkAngle[arrows=->,linecolor=blue,LabelSep=.75, MarkAngleRadius= .65]{T}{O}{B}{$a$} \vspace{3cm} \meinecaption{.25}{Theorem \ref{thm:cosine-of-difference}.} \label{fig:cosine-of-difference-1} \end{minipage} \hfill \begin{minipage}{7cm} \centering \psset{unit=6pc} \pstGeonode[PointName=none,PointSymbol=none](0,0){O} \pstGeonode[PosAngle={0,45}](1;0){A'}(1;50){B'} \pstLineAB{O}{A'}\pstLineAB{O}{B'} \pstCircleOA{O}{A'} \pstMarkAngle[arrows=->,linecolor=blue, LabelSep=.65]{A'}{O}{B'}{$a-b$} \vspace{3cm} \meinecaption{.25}{Theorem \ref{thm:cosine-of-difference}.} \label{fig:cosine-of-difference-2} \end{minipage} \end{figure} We begin by proving \begin{thm}\label{thm:cosine-of-difference} Let $(a, b)\in \reals^2$. Then $\cos (a-b) = \cos a \cos b + \sin a \sin b$.\end{thm} \begin{pf} Consider the points $A(\cos b, \sin b)$ and $B(\cos a, \sin a)$ in figure \ref{fig:cosine-of-difference-1}. Their distance is $$\begin{array}{lll}\sqrt{(\cos b-\cos a)^2+(\sin b - \sin a)^2}& = & \sqrt{\cos^2 b-2\cos b\cos a+\cos^2 a+\sin^2 b -2\sin b\sin a+ \sin^2a} \\ & = & \sqrt{2-2(\cos a\cos b+\sin a\sin b)} . \end{array}$$If we rotate $A$ $b$ radians clockwise to $A'(1,0)$, and $B$ $b$ radians clockwise to $B'(\cos (a-b), \sin (a-b))$ as in figure \ref{fig:cosine-of-difference-2}, the distance is preserved, that is, the distance of $A'$ to $B'$, which is $$ \sqrt{(\cos (a-b)-1)+\sin^2 (a-b)} = \sqrt{1-2\cos (a-b)+\cos^2(a-b) + \sin^2(a-b)} = \sqrt{2-2\cos (a-b)}, $$then equals the distance of $A$ to $B$. Therefore we have $$\begin{array}{lll}\sqrt{2-2(\cos a\cos b+\sin a\sin b)}=\sqrt{2-2\cos (a-b)} & \implies & 2-2(\cos a\cos b+\sin a\sin b) =2-2\cos (a-b) \\ & \implies & \cos (a-b) = \cos a\cos b+\sin a\sin b. \end{array}$$ \end{pf} \begin{cor} $\cos (a+b)=\cos a\cos b-\sin a\sin b$. \end{cor} \begin{pf} This follows by replacing $b$ by $-b$ in Theorem \ref{thm:cosine-of-difference}, using the fact that $x\mapsto \cos x$ is an even function and so $\cos (-b) =\cos b$, and that $x\mapsto \sin x$ is an odd function and so $\sin (-b) =-\sin b$: $$ \cos (a+b) = \cos (a-(-b))=\cos a\cos (-b)+\sin a\sin (-b) = \cos a \cos b-\sin a\sin b.$$ \end{pf} \begin{thm}\label{thm:sine-of-diff-sum} Let $(a, b)\in \reals^2$. Then $\sin (a\pm b) = \sin a \cos b \pm \sin b \cos a$.\end{thm} \begin{pf} We use the fact that $\sin x = \cos \left(\dfrac{\pi}{2}-x\right)$ and that $\cos x = \sin \left(\dfrac{\pi}{2}-x\right)$. Thus $$\begin{array}{lll}\sin (a+b) & = & \cos \left(\dfrac{\pi}{2}-\left(a+b\right)\right) \\ & = & \cos \left(\left(\dfrac{\pi}{2}-a\right)-b\right) \\ & = & \cos \left(\dfrac{\pi}{2}-a\right)\cos b+\sin\left(\dfrac{\pi}{2}-a\right)\sin b \\ & = & \sin a\cos b + \cos a\sin b, \end{array}$$proving the addition formula. For the difference formula, we have $$ \sin (a-b) = \sin (a+(-b)) = \sin a\cos (-b) + \sin (-b)\cos a = \sin a\cos b - \sin b \cos a. $$ \end{pf} \begin{thm}\label{thm:tangent-of-diff-sum} Let $(a, b)\in \reals^2$. Then $\tan (a\pm b) = \dfrac{\tan a\pm \tan b}{1\mp \tan a\tan b}$.\end{thm} \begin{pf} Using the formul\ae\ derived above, $$\begin{array}{lll}\tan (a \pm b) & = & \dfrac{\sin (a \pm b)}{\cos (a \pm b)} \\ & = & \dfrac{\sin a\cos b\pm \sin b\cos a}{\cos a\cos b \mp \sin a\sin b}.\\ \end{array}$$Dividing numerator and denominator by $\cos a\cos b$ we obtain the result. \end{pf} By letting $a + b = A, a - b = B$ in the above results we obtain the following corollary. \begin{cor} \begin{equation} \cos A + \cos B = 2\cos\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right) \end{equation} \begin{equation} \cos A - \cos B = -2\sin\left(\frac{A + B}{2}\right)\sin\left(\frac{A - B}{2}\right) \end{equation} \begin{equation} \sin A + \sin B = 2\sin\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right) \end{equation} \begin{equation} \sin A - \sin B = 2\sin\left(\frac{A - B}{2}\right)\cos\left(\frac{A + B}{2}\right) \end{equation} \label{sumtoprod}\end{cor} \begin{exa} Given that $\cos a = -.1$ and $\pi < a < \frac{3\pi}{2}$, and that $\sin b = .2$ and $0 < b < \frac{\pi}{2}$, find $\cos (a + b)$. \end{exa} \begin{solu} Since $\winding(a)$ is in the third quadrant, $\sin a = -\sqrt{1 - (.1)^2} = -\sqrt{0.99}$. As $\winding(b)$ is in the first quadrant, $\cos b = \sqrt{1 - (.2)^2} = \sqrt{0.96}$. By the addition formula for the cosine $$ \begin{array}{lll}\cos(a + b) & = & \cos a\cos b - \sin a\sin b\\ & = & (-.1)(\sqrt{0.96}) - (-\sqrt{0.99})(.2)\\ & = & .2\sqrt{.99} - .1\sqrt{.96}. \end{array}$$ \end{solu} \begin{exa} Write $\sin 5x\cos x$ as a sum of sines. \end{exa} \begin{solu} We have $$\sin 6x = \sin (5x + x) = \sin 5x\cos x + \sin x\cos 5x$$ $$\sin 4x = \sin (5x - x) = \sin 5x\cos x - \sin x\cos 5x$$ Adding both equalities and dividing by $2$, we gather, $$\sin 5x\cos x = \frac{1}{2}\sin 6x + \frac{1}{2}\sin 4x .$$ \end{solu} \begin{exa} Solve the equation $$\sin 6x + \sin 4x = 0.$$ \end{exa} \begin{solu} As $\sin 6x + \sin 4x = 2\sin 5x\cos x$ we must have either $\sin 5x = 0$ or $\cos x = 0$. Thus $$x = \frac{\pi n}{5}, \ \ x = \pm\frac{\pi}{2} + \pi n, n \in \integers .$$ \end{solu} \begin{exa} Write $\sin x\sin 2x$ as a sum of cosines. \end{exa} \begin{solu} We have $$\cos 3x = \cos (2x + x) = \cos 2x\cos x - \sin 2x\sin x,$$ $$\cos x = \cos (2x - x) = \cos 2x\cos x + \sin 2x\sin x.$$ Subtracting both equalities $\cos 3x - \cos x = -2\sin 2x\sin x$, whence $$\sin 2x\sin x = -\frac{1}{2}\cos 3x + \frac{1}{2}\cos x. $$ \end{solu} \begin{exa} Find the exact value of $\cos \frac{7\pi}{12}$. \end{exa} \begin{solu} Observe that $\frac{7}{12} = \frac{1}{3} + \frac{1}{4}$. Using the addition formul\ae $$ \begin{array}{lll} \cos \frac{7\pi}{12} & = & \cos \left(\frac{\pi}{3} + \frac{\pi}{4}\right) \\ & = & \cos\frac{\pi}{3}\cos\frac{\pi}{4} - \sin\frac{\pi}{3}\sin\frac{\pi}{4} \\ & = & (\frac{1}{2})(\frac{\sqrt{2}}{2}) - (\frac{\sqrt{3}}{2})(\frac{\sqrt{2}}{2}) \\ & = & \frac{\sqrt{2} - \sqrt{6}}{4} \end{array} . $$ \end{solu} \begin{exa} (i) Write $\sqrt{3}\cos x + \sin x$ in the form $A\cos (x - \theta)$, with $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$. (ii) Use the preceding identity in order to solve the equation $$\sqrt{3}\cos x + \sin x = -1.$$ (iii) Find all the solutions in the interval $[0; 2\pi]$. \end{exa} \begin{solu} First observe that $A \neq 0$, since $\sqrt{3}\cos x + \sin x$ is not identically $0$. We have $$A\cos (x - \theta) = A\cos x\cos\theta + A\sin x\sin\theta . $$If the expression on the dextral side of the above equality is to be equal to $\sqrt{3}\cos x + \sin x$ then $A\cos\theta = \sqrt{3}$ and $A\sin\theta = 1.$ This entails that $\tan\theta = \frac{\sqrt{3}}{3}$ and so $\theta = \frac{\pi}{6}$. This in turn yields $A = 2.$ Hence $$\sqrt{3}\cos x + \sin x = 2\cos\left(x - \frac{\pi}{6}\right).$$ Now, if $2\cos\left(x - \frac{\pi}{6}\right) = -1,$ then $$x - \frac{\pi}{6} = \pm\arccos (-\frac{1}{2}) + 2n\pi, \ n\in\integers,$$ $$x = \frac{\pi}{6} \pm \frac{2\pi}{3} + 2n\pi, \ n\in\integers,$$ which is the same family as $x = \frac{5\pi}{6} + 2n\pi, x = -\frac{\pi}{2} + 2n\pi$ and the solutions in $[0; 2\pi]$ are clearly $x = \frac{5\pi}{6}$ and $x = \frac{3\pi}{2}$. \flushleft{{\bf Aliter:}} Write the equation as $\sqrt{3}\cos x + 1 = -\sin x$ and square $$3\cos^2x + 2\sqrt{3}\cos x + 1 = \sin^2x.$$Using $\sin^2x = 1 - \cos^2x$ we obtain $$3\cos^2x + 2\sqrt{3}\cos x + 1 = 1 - \cos^2x,$$or $$(\cos x)(4\cos x + 2\sqrt{3}) = 0.$$This equation has solutions $x = \pm\frac{\pi}{2} + 2n\pi$ and $x = \pm \frac{5\pi}{6} + 2n\pi$. Testing $x = \frac{\pi}{2}$ in the original equation $\sqrt{3}\cos x + \sin x = -1$ we see that it is not a solution, hence the family $x = \frac{\pi}{2} + 2n\pi$ is not part of the solution set of the original equation. The same happens when we test $x = -\frac{5\pi}{6}$, so we must also discard this family. The two remaining families, $x = \frac{5\pi}{6} + 2n\pi,$ $x = -\frac{\pi}{2} + 2n\pi$ agree with our previous solution. \end{solu} \begin{exa} Obtain a formula for $\cos(a + b + c)$ in terms of cosines and sines of $a, b,$ and $c.$ \end{exa} \begin{solu} Using the addition formula twice $$ \begin{array}{lll} \cos(a + b + c) & = & \cos a\cos(b + c) - \sin a \sin(b + c) \\ & = & \cos a(\cos b \cos c - \sin b\sin c) - \\ & & \qquad - \sin a(\sin b\cos c + \sin c\cos b) \\ & = & \cos a\cos b\cos c - \cos a\sin b\sin c - \\ & & \qquad - \sin a\sin b\cos c - \sin a\cos b\sin c \\ \end{array} $$ \end{solu} \begin{exa}[Canadian Mathematical Olympiad 1984] Given any $7$ real numbers, prove that there are two of them, say, $x$ and $y$, such that $$0 \leq \frac{x - y}{1 + xy} \leq \frac{1}{\sqrt{3}}.$$ \end{exa}\begin{solu} Let the numbers be $a_k, k = 1, 2, \ldots, 7.$ There exists $b_k$ such $a_k = \tan b_k$, since $\funnoname{x}{\tan x}{]-\frac{\pi}{2};\frac{\pi}{2}[}{\reals}$ is a bijection. Divide the interval $]-\frac{\pi}{2};\frac{\pi}{2}[$ into six subintervals, each of length $\frac{\pi}{6}$. Since we have $7\ b_k$'s and $6$ subintervals, two of the $b_k$'s, say $b_s$ and $b_t$, must lie in the same subinterval. Assuming $b_s \geq b_t$ we then have $0 \leq b_s - b_t \leq \frac{\pi}{6}$. Since $x \mapsto \tan x$ is an increasing function, $$\tan 0 \leq \tan (b_s - b_t) \leq \tan\frac{\pi}{6},$$which is to say, $$0 \leq \frac{\tan b_s - \tan b_t}{1 + \tan b_s\tan b_t} \leq \frac{1}{\sqrt{3}}.$$ This implies that $$ 0 \leq \frac{a_s - a_t}{1 + a_sa_t} \leq \frac{1}{\sqrt{3}},$$which completes the proof. \end{solu} \begin{exa} Prove that if $$\frac{a - b}{1 + ab} + \frac{b - c}{1 + bc} + \frac{c - a}{1 + ca} = 0,$$ for real numbers $a, b, c$, then at least two of the numbers $a, b, c$ are equal. \end{exa} \begin{solu} $\exists u, v, w$ with $-\frac{\pi}{2} < u, v, w < \frac{\pi}{2}$ such that $a = \tan u, b = \tan v, c = \tan w$ (why?). The given equation becomes $$\frac{\tan u - \tan v}{1 + \tan u\tan v} + \frac{\tan v - \tan w}{1 + \tan v\tan w} + \frac{\tan w - \tan u}{1 + \tan w\tan u} = 0.$$ Using the addition for the tangents, the preceding relation is equivalent to $$\tan (u - v) + \tan (v - w) + \tan (w - u) = 0.$$ Applying $\tan X + \tan Y = (\tan (X + Y))(1 - \tan X\tan Y)$ with $X = u - v$ and $Y = v - w,$ we obtain $$(\tan (u - w))(1 - \tan (u - v)\tan (v - w)) + \tan (w - u) = 0.$$ Factorising the above expression, $$(\tan (u - w))(\tan (u - v))(\tan (v - w)) = 0.$$This implies that one of the tangents in this product must be $0$. Since $$-\pi < u - w, u - v, v - w < \pi,$$ this means that one of these differences must be exactly $0$, which in turn implies that two of the numbers $a, b, c$ are equal. \end{solu} \begin{exa} Prove that $$\arctan a + \arctan b = \left\{\begin{array}{ll} \arctan\frac{a + b}{1 - ab} & {\rm if} \ ab < 1, \\ \frac{\pi}{2}({\rm sgn}(a)) & {\rm if}\ ab = 1, \\ \arctan\frac{a + b}{1 - ab} + \frac{\pi}{2}({\rm sgn}(a)) & {\rm if }\ ab > 1. \end{array}\right. .$$ \end{exa} \begin{solu} Put $x = \arctan a, y = \arctan b.$ If $(x, y) \in ]-\frac{\pi}{2}; \frac{\pi}{2}[^2$ and $x + y \neq \frac{(2n + 1)\pi}{2} , n\in\integers$, then $$\tan (x + y) = \frac{\tan x + \tan y}{1 - \tan x\tan y} = \frac{a + b}{1 - ab}.$$Now, $-\pi < x + y < \pi$. Conditioning on $x$ we have, $$-\frac{\pi}{2} < x + y < \frac{\pi}{2} \Longleftrightarrow \left|\begin{array}{l} x = 0 \\ {\rm or}\ x > 0\ {\rm and}\ y < \frac{\pi}{2} - x \\ {\rm or}\ x < 0\ {\rm and}\ y > - \frac{\pi}{2} - x \\ \end{array} \right. $$ The above choices hold if and only if $$ \begin{array}{l} a = 0 \\ {\rm or}\ a > 0\ {\rm and}\ b < \frac{1}{a} \\ {\rm or}\ a < 0\ {\rm and}\ b > \frac{1}{a} \\ \end{array} . $$ Hence, if $ab < 1,$ then $x + y\in ]-\frac{\pi}{2}; \frac{\pi}{2}[$ and thus $$x + y = \arctan (\tan (x + y)) = \arctan\frac{a + b}{1 - ab}.$$ If $ab > 1$ and $a > 0$ then $x + y \in]\frac{\pi}{2}; \pi[$ and thus $$x + y = \arctan\frac{a + b}{1 - ab} + \pi .$$ If $ab > 1$ and $a < 0$, then $x + y \in]-\pi; -\frac{\pi}{2}[$ and thus $$x + y = \arctan\frac{a + b}{1 - ab} - \pi .$$ The case $ab = 1$ is left as an exercise. \end{solu} \begin{exa} Solve the equation $\arccos x = \arcsin\frac{1}{3} + \arccos\frac{1}{4}$. \end{exa} \begin{solu} Observe that $\arccos x \in [0;\pi]$ and that since both $0 \leq \arcsin\frac{1}{3} \leq \frac{\pi}{2}$ and $0 \leq \arccos\frac{1}{4} \leq \frac{\pi}{2}$, we have $0 \leq \arcsin\frac{1}{3} + \arccos\frac{1}{4} \leq \pi$. Hence, we may take cosines on both sides of the equation and obtain $$ \begin{array}{lll}x & = & \cos (\arccos x)\\ & = & \cos (\arcsin\frac{1}{3} + \arccos\frac{1}{4}) \\ & = & (\cos\arcsin\frac{1}{3})(\cos\arccos\frac{1}{4}) - (\sin\arcsin\frac{1}{3})(\sin\arccos\frac{1}{4}) \\ & = & \frac{\sqrt{2}}{6} - \frac{\sqrt{15}}{12} \end{array} .$$ \end{solu} \begin{exa}[Machin's Formula] Prove that $$\frac{\pi}{4} = 4\arctan \frac{1}{5} - \arctan \frac{1}{239}. $$ \end{exa} \begin{solu} Observe that $$\begin{array}{lll} 4\arctan \frac{1}{5} & = & 2\arctan \frac{1}{5} + 2\arctan\frac{1}{5} \\ & = & 2\arctan \frac{\frac{1}{5} + \frac{1}{5}}{1 - \frac{1}{5}\cdot\frac{1}{5}} \\ & = & 2\arctan \frac{5}{12} \\ & = & \arctan\frac{5}{12} + \arctan\frac{5}{12} \\ & = & \arctan\frac{\frac{5}{12} + \frac{5}{12}}{1 - \frac{5}{12}\cdot\frac{5}{12}} \\ & = & \arctan\frac{120}{119}. \end{array} $$ Also $$ \begin{array}{lll} \arctan\frac{120}{119} - \arctan\frac{1}{239} & = & \arctan\frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119}\cdot\frac{1}{239}}\\ & = & \arctan 1\\ & = & \frac{\pi}{4}. \end{array} $$ Upon assembling the equalities, we obtain the result. \end{solu} \subsection*{Homework} \addcontentsline{toc}{section}{Homework} \markright{Homework} \begin{pro} Demonstrate the identity $$\sin(a + b)\sin(a - b) = \sin^2a - \sin^2b =\cos^2b - \cos^2a$$ \end{pro} \begin{pro} Prove that for all real numbers $x$, $$\cos\left(2x - \frac{4\pi}{3}\right) + \cos 2x + \cos\left(2x + \frac{4\pi}{3}\right) = 0.$$ \end{pro} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Using the fact that $\frac{1}{12} = \frac{1}{3} - \frac{1}{4}$, find the exact value of the following. \begin{enumerate} \item $\dis{\cos {\pi}/{12}}$ \item $\dis{\sin {\pi}/{12}}$ \end{enumerate} \begin{answer} $\cos (\pi/12) = \frac{\sqrt{2}}{4}(\sqrt{3} + 1)$, $\sin (\pi/12) = \frac{\sqrt{2}}{4}(\sqrt{3} - 1)$. \end{answer} \end{pro} \begin{pro} Write $\cot (a + b)$ in terms of $\cot a$ and $\cot b$. \begin{answer} $\frac{\cot a\cot b - 1}{\cot a + \cot b}$ \end{answer} \end{pro} \begin{pro} Write $\sin x\sin 2x$ as a sum of cosines. \begin{answer} $\frac{1}{2}\cos x - \frac{1}{2}\cos 3x$ \end{answer} \end{pro} \begin{pro} Write $\cos x\cos 4x$ as a sum of cosines. \begin{answer} $\frac{1}{2}\cos 3x + \frac{1}{2}\cos 5x$ \end{answer} \end{pro} \begin{pro} Write using only one arcsine: $\arccos \frac{4}{5} - \arccos\frac{1}{4}$. \begin{answer} $-\arcsin\frac{4\sqrt{15} - 3}{20}$ \end{answer} \end{pro} \begin{pro} Write using only one arctangent: $\arctan\frac{1}{3} - \arctan\frac{1}{4}$. \begin{answer} $\arctan\frac{1}{13}$. \end{answer} \end{pro} \begin{pro} Write using only one arctangent: $\arccot (-2) - \arctan (-\frac{2}{3}).$ \begin{answer} $\pi + \arctan\frac{1}{8}$ \end{answer} \end{pro} \begin{pro} Write $\sin x\cos 2x$ as a sum of sines. \begin{answer} $\frac{1}{2}\sin 3x - \frac{1}{2}\sin x$ \end{answer} \end{pro} \begin{pro} Write $\sin x\sin 2x\sin 3x$ as a sum of sines. \begin{answer} $\frac{1}{4}\sin 2x + \frac{1}{4}\sin 4x - \frac{1}{4}\sin 6x$ \end{answer} \end{pro} \begin{pro} Given real numbers $a, b$ with $0 < a < \pi /2$ and $\pi < b < 3\pi /2$ and given that $\sin a = 1/3$ and $ \cos b = -1/2$, find $\cos (a - b)$. \begin{answer} $-(\frac{\sqrt{2}}{6 } + \frac{\sqrt{3}}{6})$ \end{answer} \end{pro} \begin{pro} Solve the equation $\cos x + \cos 3x = 0.$. \begin{answer} $x = \pm\frac{\pi}{4} + n\pi,\ x = \pm\frac{\pi}{2} + 2n\pi, \ n\in \integers$ \end{answer} \end{pro} \begin{pro} Solve the equation $$\arcsin (\tan x) = x.$$ \begin{answer} $x = 0.$ \end{answer} \end{pro} \begin{pro} Solve the equation $$\arccos x = \arcsin (1 - x).$$ \begin{answer} $x = 0$ or $x = 1.$ \end{answer} \end{pro}\begin{pro} Solve the equation $$\arctan x + \arctan 2x = \frac{\pi}{4}.$$ \begin{answer} $x = \frac{\sqrt{17} - 3}{4}$ \end{answer} \end{pro} \begin{pro} Prove the identity $$\cos^4x = \frac{1}{8}(\cos 4x + 4\cos 2x + 3).$$ \end{pro} \begin{pro} Prove the identities $$\tan a + \tan b = \frac{\sin (a + b)}{(\cos a)(\cos b)},$$ $$\cot a + \cot b = \frac{\sin (a + b)}{(\sin a)(\sin b)}.$$ \end{pro} \begin{pro} Given that $0 \leq \alpha , \beta , \gamma \leq \frac{\pi}{2}$ and satisfy $\sin\alpha = 12/13, \cos\beta = 8/17, \sin\gamma = 4/5, $ find the value of $\sin (\alpha + \beta - \gamma )$ and $\cos (\alpha - \beta + 2\gamma ).$ \end{pro} \begin{pro} Establish the identity $$\frac{\sin (a - b)\sin (a + b)}{1 - \tan ^2 a\cot ^2 b} = -\cos ^2 a\sin ^2 b.$$ \end{pro} \begin{pro} Find real constants $a, b, c$ such that $$\sin 3x - \sqrt{3}\cos 3x = a\sin (bx + c).$$Use this to solve the equation $$\sin 3x - \sqrt{3}\cos 3x = -\sqrt{2}.$$ \end{pro} \begin{pro} Solve the equation $$\sin 2x + \cos 2x = -1$$ \end{pro} \begin{pro} Simplify: $\dis{\sin({\rm arcsec}\ \frac{17}{8} - \arctan (-\frac{2}{3}))}$. \end{pro} \begin{pro} Shew that if $\cot (a + b) = 0$ then $\sin (a + 2b) = \sin a.$ \end{pro} \begin{pro} Let $a + b + c = \frac{\pi}{2}$. Write $\cos a\cos b\cos c$ as a sum of sines. \end{pro} \begin{pro}Shew that the amplitude of $x \mapsto a\sin Ax + b\cos Ax$ is $\sqrt{a^2 + b^2}$. \end{pro} \begin{pro} Solve the equation $$\cos x - \sin x = 1.$$ \end{pro} \begin{pro} Let $a + b + c = \pi$. Simplify $$\sin^2a + \sin^2b + \sin^2c - 2\cos a\cos b\cos c.$$ \end{pro} \begin{pro} Prove that if $$\cot a + \csc a\cos b\sec c= \cot b + \cos a\csc b\sec c, $$then either $a - b = k\pi$, or $a + b + c = \pi + 2m\pi$ or $a + b - c = \pi + 2n\pi$ for some integers $k,m, n$. \end{pro} \begin{pro} Prove that if $$\tan a + \tan b + \tan c = \tan a\tan b\tan c,$$ then $a + b + c = k\pi$ for some integer $k$. \end{pro} \begin{pro} Prove that if any of $a + b + c, \ a + b - c, \ a - b + c$ or $a - b - c$ is equal to an odd multiple of $\pi$, then $$\cos^2a + \cos^2b + \cos^2c + 2\cos a\cos b\cos c = 1,$$and that the converse is also true. \end{pro} \end{multicols} \appendix \renewcommand{\chaptername}{Appendix} \chapter{Complex Numbers} \section{Arithmetic of Complex Numbers} One uses the symbol $i$ to denote the {\em imaginary unit} $i = \sqrt{-1}$. Then $i^2 = -1.$ \begin{exa} Find $\sqrt{-25}$. \end{exa} \begin{solu} $\sqrt{-25} = 5i.$ \end{solu} \indent Since $i^0 = 1, i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1, i^5 = i,$ etc., the powers of $i$ repeat themselves cyclically in a cycle of period $4$. \begin{exa} Find $i^{1934}$. \end{exa} \begin{solu} Observe that $1934 = 4(483) + 2$ and so $i^{1934} = i^2 = -1$. \end{solu} \begin{exa} For any integral $\alpha$ one has $$i^\alpha + i^{\alpha + 1} + i^{\alpha + 2} + i^{\alpha + 3} = i^{\alpha}(1 + i + i^2 + i^3) = i^{\alpha}(1 + i - 1 - i) = 0.$$ \end{exa} \indent If $a, b$ are real numbers then the object $a + bi$ is called a {\em complex number}. One uses the symbol $\complex$ to denote the set of all complex numbers. If $a, b, c, d \in {\reals}$, then the sum of the complex numbers $a + bi$ and $c + di$ is naturally defined as \begin{equation} (a + bi) + (c + di) = (a + c) + (b + d)i \end{equation} The product of $a + bi$ and $c + di$ is obtained by multiplying the binomials: \begin{equation} (a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i \end{equation} \begin{exa} Find the sum $(4 + 3i) + (5 - 2i)$ and the product $(4 + 3i)(5 - 2i)$. \end{exa} \begin{solu} One has $$(4 + 3i) + (5 - 2i) = 9 + i$$and $$(4 + 3i)(5 - 2i) = 20 - 8i + 15i - 6i^2 = 20 + 7i + 6 = 26 + 7i.$$ \end{solu} \begin{df} Let $z\in\complex, (a, b) \in \reals^2$ with $z = a + bi$. The {\em conjugate} $\overline{z}$ of $z$ is defined by \begin{equation} \overline{z} = \overline{a + bi} = a - bi \end{equation} \end{df} \begin{exa} The conjugate of $5 + 3i$ is $\overline{5 + 3i} = 5 - 3i$. The conjugate of $2 - 4i$ is $\overline{2 - 4i} = 2 + 4i.$ \end{exa} \begin{rem} The conjugate of a real number is itself, that is, if $a\in \reals$, then $\overline{a} = a$. Also, the conjugate of the conjugate of a number is the number, that is, $\overline{\overline{z}} = z.$ \end{rem} \begin{thm}\label{thm:multiplicative_conjugation} The function $z: {\complex} \rightarrow {\complex}, \ z \mapsto \overline{z}$ is multiplicative, that is, if $z_1, z_2$ are complex numbers, then \begin{equation} \overline{z_1z_2} = \overline{z_1} \cdot \overline{z_2} \end{equation} \end{thm} \begin{pf} Let $z_1 = a + bi, z_2 = c + di$ where $a, b, c, d$ are real numbers. Then $$ \begin{array}{lll} \overline{z_1z_2} & = & \overline{(a + bi)(c + di)}\\ & = & \overline{(ac - bd) + (ad + bc)i} \\ & = & (ac - bd) - (ad + bc)i\\ \end{array}$$ Also, $$ \begin{array}{lll} \overline{z_1}\cdot\overline{z_2} & = & (\overline{a + bi})(\overline{c + di}) \\ & = & (a - bi)(c - di) \\ & = & ac - adi - bci + bdi^2 \\ & = & (ac - bd) - (ad + bc)i, \end{array} $$which establishes the equality between the two quantities. \end{pf} \begin{exa} Express the quotient $\dis{\frac{2 + 3i}{3 - 5i}}$ in the form $a + bi$. \end{exa} \begin{solu} One has $$ \frac{2 + 3i}{3 - 5i} = \frac{2 + 3i}{3 - 5i}\cdot \frac{3 + 5i}{3 + 5i} = \frac{-9 + 19i}{34} = \frac{-9}{34} + \frac{19i}{34}$$ \end{solu} \begin{df} The {\em modulus} $|a + bi|$ of $a + bi$ is defined by \begin{equation} |a + bi| = \sqrt{(a + bi)(\overline{a + bi})} = \sqrt{a^2 + b^2} \end{equation} \end{df} Observe that $z \mapsto |z|$ is a function mapping ${\complex}$ to $[0;+\infty[$. \begin{exa} Find $|7 + 3i|$. \end{exa} \begin{solu} $|7 + 3i| = \sqrt{(7 + 3i)(7 - 3i)} = \sqrt{7^2 + 3^2} = \sqrt{58}$. \end{solu} \begin{exa} Find $|\sqrt{7} + 3i|$. \end{exa} \begin{solu} $|\sqrt{7} + 3i| = \sqrt{(\sqrt{7} + 3i)(\sqrt{7} - 3i)} = \sqrt{7 + 3^2} = 4$. \end{solu} \begin{thm} The function $z \mapsto |z|,\ {\complex} \rightarrow {\reals}_+$ is multiplicative. That is, if $z_1, z_2$ are complex numbers then \begin{equation} |z_1z_2| = |z_1||z_2|\end{equation} \end{thm} \begin{pf} By Theorem \ref{thm:multiplicative_conjugation}, conjugation is multiplicative, hence $$ \begin{array}{lll} |z_1z_2| & = & \sqrt{z_1z_2\overline{z_1z_2}}\\ & = & \sqrt{z_1z_2\overline{z_1}\cdot\overline{z_2}} \\ & = & \sqrt{z_1\overline{z_1}z_2\overline{z_2}} \\ & = & \sqrt{z_1\overline{z_1}}\sqrt{z_2\overline{z_2}} \\ & = & |z_1||z_2| \end{array}$$whence the assertion follows. \end{pf} \begin{exa} Write $(2^2 + 3^2)(5^2 + 7^2)$ as the sum of two squares. \end{exa} \begin{solu} The idea is to write $2^2 + 3^2 = |2 + 3i|^2,\ 5^2 + 7^2 = |5 + 7i|^2$ and use the multiplicativity of the modulus. Now \\ \begin{center} \begin{tabular}{lll} $(2^2 + 3^2)(5^2 + 7^2)$ & $=$ & $|2 + 3i|^2|5 + 7i|^2$ \\ & $=$ & $|(2 + 3i)(5 + 7i)|^2$ \\ & $=$ & $|-11 + 29i|^2$ \\ & $=$ & $11^2 + 29^2$ \end{tabular} \end{center} \end{solu} \section{Equations involving Complex Numbers} Recall that if $ux^2 + vx + w = 0$ with $u \neq 0$, then the roots of this equation are given by the {\em Quadratic Formula} \begin{equation} x = -\frac{v}{2u} \pm \frac{\sqrt{v^2 - 4uw}}{2u} \end{equation} The quantity $v^2 - 4uw$ under the square root is called the {\em discriminant} of the quadratic equation $ux^2 + vx + w = 0$. If $u, v, w$ are real numbers and this discriminant is negative, one obtains complex roots. Complex numbers thus occur naturally in the solution of quadratic equations. Since $i^2 = -1$, one sees that $x = i$ is a root of the equation $x^2 + 1 = 0$. Similary, $x = -i$ is also a root of $x^2 + 1.$ \begin{exa} Solve $2x^2 + 6x + 5 = 0$ \end{exa} \begin{solu} Using the quadratic formula $$ x = -\frac{6}{4} \pm \frac{\sqrt{-4}}{4} = -\frac{3}{{2}} \pm i\frac{1}{{2}}$$ \end{solu} In solving the problems that follow, the student might profit from the following identities. \begin{equation} s^2 - t^2 = (s - t)(s + t) \end{equation} \begin{equation} s^{2k} - t^{2k} = (s^k - t^k)(s^k + t^k), \ k\in\naturals \end{equation} \begin{equation} s^3 - t^3 = (s - t)(s^2 + st + t^2) \end{equation} \begin{equation} s^3 + t^3 = (s + t)(s^2 - st + t^2) \end{equation} \begin{exa} Solve the equation $x^4 - 16 = 0$. \end{exa} \begin{solu} One has $x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$. Thus either $x = -2, x = 2$ or $x^2 + 4 = 0.$ This last equation has roots $\pm 2i$. The four roots of $x^4 - 16 = 0$ are thus $x = -2, x = 2, x = -2i, x = 2i.$ \end{solu} \begin{exa} Find the roots of $x^3 - 1 = 0$. \end{exa} \begin{solu} $x^3 - 1 = (x - 1)(x^2 + x + 1)$. If $x \neq 1$, the two solutions to $x^2 + x + 1 = 0$ can be obtained using the quadratic formula, getting $\dis{x = -\frac{1}{2} \pm i\frac{\sqrt{3}}{2}}$. \end{solu} \begin{exa} Find the roots of $x^3 + 8 = 0$. \end{exa} \begin{solu} $x^3 + 8 = (x + 2)(x^2 - 2x + 4)$. Thus either $x = -2$ or $x^2 - 2x + 4 = 0$. Using the quadratic formula, one sees that the solutions of this last equation are $\dis{x = 1 \pm i\sqrt{3}}$. \end{solu} \begin{exa} Solve the equation $x^4 + 9x^2 + 20 = 0$. \end{exa} \begin{solu} One sees that $$x^4 + 9x^2 + 20 = (x^2 + 4)(x^2 + 5) = 0$$Thus either $x^2 + 4 = 0$, in which case $x = \pm 2i$ or $x^2 + 5 = 0$ in which case $x = \pm i\sqrt{5}$. The four roots are $x = \pm 2i, \pm i\sqrt{5}$ \end{solu} \subsection*{Homework} \addcontentsline{toc}{subsection}{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Perform the following operations. Write your result in the form $a + bi$, with $(a, b) \in \reals^2$. \begin{enumerate} \item $\sqrt{36} + \sqrt{-36}$ \item $(4 + 8i) - (9 - 3i) + 5(2 + i) - 8i$ \item $4 + 5i + 6i^2 + 7i^3$ \item $i(1 + i) + 2i^2(3 - 4i)$ \item $(8 - 9i)(10 + 11i)$ \item $i^{1990} + i^{1991} + i^{1992} + i^{1993}$ \item $\dis{\frac{2 - i}{2 + i}}$ \item $\dis{\frac{1 - i}{1 + 2i} + \frac{1 + i}{1 + 2i}}$ \item $(5 + 2i)^2 + (5 - 2i)^2$ \item $(1 + i)^3$ \end{enumerate} \end{pro} \begin{pro} Find real numbers $a, b$ such that $$(a - 2) + (5b + 3)i = 4 - 2i$$ \end{pro} \begin{pro} Write $(2^2 + 3^2)(3^2 + 7^2)$ as the sum of two squares. \end{pro} \begin{pro} Prove that $(1 + i)^2 = 2i$ and that $(1 - i)^2 = -2i$. Use this to write $$\frac{(1 + i)^{2004}}{(1 - i)^{2000}}$$ in the form $a + bi, \ (a, b) \in \reals^2$.\end{pro} \begin{pro} Prove that $(1 + i\sqrt{3})^3 = 8$. Use this to prove that $$(1 + i\sqrt{3})^{30} = 2^{30}.$$ \end{pro} \begin{pro} Find $|5 + 7i|$, $|\sqrt{5} + 7i|$, $|5 + i\sqrt{7}|$ and $|\sqrt{5} + i\sqrt{7}|$. \end{pro} \begin{pro} Prove that if $k$ is an integer then $$(4k + 1)i^{4k} + (4k + 2)i^{4k + 1} + (4k + 3)i^{4k + 2} + (4k + 4)i^{4k + 3} = -2 - 2i.$$ Use this to prove that $$1 + 2i + 3i^2 + 4i^{3} + \cdots + 1995i^{1994} + 1996i^{1995} = -998 - 998i.$$ \end{pro} \begin{pro} If $z$ and $z'$ are complex numbers with either $|z| = 1$ or $|z'| = 1,$ prove that $$\left|\frac{z - z'}{1 - \overline{z}z'}\right| = 1.$$ \end{pro} \begin{pro} Prove that if $z, z'$ and $w$ are complex numbers with $|z| = |z'| = |w| = 1$, then $$|zz' + zw + z'w| = |z + z' + w|$$ \end{pro} \begin{pro} Prove that if $n$ is an integer which is not a multiple of $4$ then $$1^n + i^n + i^{2n} + i^{3n} = 0.$$ Now let $$f(x) = (1 + x + x^2)^{1000} = a_0 + a_1x + \cdots + a_{2000}x^{2000}.$$ By considering $f(1) + f(i) + f(i^2) + f(i^3)$, find $$a_0 + a_4 + a_8 + \cdots + a_{2000}.$$ \end{pro} \begin{pro} Find all the roots of the following equations. \begin{enumerate} \item $x^2 + 8 = 0$ \item $x^2 + 49 = 0$ \item $x^2 - 4x + 5 = 0$ \item $x^2 - 3x + 6 = 0$ \item $x^4 - 1 = 0$ \item $x^4 + 2x^2 - 3 = 0$ \item $x^3 - 27 = 0$ \item $x^6 - 1 = 0$ \item $x^6 - 64 = 0$ \end{enumerate} \end{pro} \end{multicols} \section{Polar Form of Complex Numbers} Complex numbers can be given a geometric representation in the {\em Argand diagram} (see figure \ref{fig:argand}), where the horizontal axis carries the real parts and the vertical axis the imaginary ones. \vspace{3cm} \begin{figure}[h] \begin{minipage}{.4\textwidth} \centering \psset{unit=1pc} \psaxes[linewidth=2pt,labels=none,ticks=none]{->}(0,0)(-5,-5)(5,5) \pstGeonode[PointName=none,linecolor=blue](4;60){A} \uput[ur](A){$a+bi$} \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(0,5){B}(5,0){C} \pstLineAB[linecolor=blue]{O}{A} \pstProjection[PointName=none,linecolor=magenta]{C}{O}{A}[D] \pstProjection[PointName=none,linecolor=magenta]{B}{O}{A}[E] \pstLineAB[linecolor=blue]{A}{D} \pstMarkAngle{C}{O}{A}{$\theta$} \pscircle[linestyle=dashed,linecolor=red](O){4} \uput[d](D){$a$} \uput[l](E){$b$} \uput[u](B){$\Im$}\uput[u](C){$\Re$} \psdots[linecolor=blue](A) \meinecaption{2}{Argand's diagram.}\label{fig:argand} \end{minipage} \begin{minipage}{.4\textwidth} \centering \psset{unit=1pc} \psaxes[linewidth=2pt,labels=none,ticks=none]{->}(0,0)(-5,-5)(5,5) \pstGeonode[PointName=none,linecolor=blue](6;55){A} \uput[ur](A){$z$} \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(0,5){B}(5,0){C} \pstLineAB[linecolor=blue]{O}{A}\aput{:U}{$|z|$} \pstProjection[PointName=none,linecolor=magenta]{C}{O}{A}[D] \pstLineAB[linecolor=blue]{A}{D} \aput{:D}{$|z|\sin\theta$} \pstLineAB[linecolor=blue]{O}{D} \bput{:U}{$|z|\cos\theta$} \pstMarkAngle{C}{O}{A}{$\theta$} \uput[u](B){$\Im$}\uput[dr](C){$\Re$} \psdots[linecolor=blue](A) \meinecaption{2}{Polar Form of a Complex Number.}\label{fig:polar-form} \end{minipage} \end{figure} Given a complex number $z=a+bi$ on the Argand diagram, consider the angle $\theta \in ]-\pi;\pi ]$ that a straight line segment passing through the origin and through $z$ makes with the positive real axis. Considering the polar coordinates of $z$ we gather \begin{equation} z = |z|(\cos \theta + i\sin\theta), \qquad \theta \in ]-\pi; \pi], \end{equation} which we call the {\em polar form} of the complex number $z$. The angle $\theta$ is called the {\em argument} of the complex number $z$. \begin{exa} Find the polar form of $\sqrt{3} - i$. \end{exa} \begin{solu} First observe that $|\sqrt{3}-i|=\sqrt{\sqrt{3}^2+1^2}= 2$. Now, if $$\sqrt{3}-i=2(\cos \theta+i\sin\theta),$$we need $\cos \theta = \dfrac{\sqrt{3}}{2}$, $\sin \theta = -\dfrac{1}{2}$. This happens for $\theta \in ]-\pi; \pi]$ when $\theta = -\dfrac{\pi}{6}$. Therefore, $$ \sqrt{3}-i=2(\cos\left(-\dfrac{\pi}{6}\right) +i\sin\left(-\dfrac{\pi}{6}\right) $$is the required polar form. \end{solu} We now present some identities involving complex numbers. Let us start with the following classic result. The proof requires Calculus and can be omitted. If we allow complex numbers in our MacLaurin expansions, we readily obtain Euler's Formula. \begin{thm}[Euler's Formula] Let $x\in\reals$. Then $$ e^{ix} = \cos x + i \sin x. $$\label{thm:euler} \end{thm} \begin{pf} Using the MacLaurin expansion's of $x\mapsto e^x$, $x\mapsto \cos x$, and $x\mapsto \sin x$, we gather $$\begin{array}{lll} e^{ix} & = & \sum _{k=0} ^{+\infty} \dfrac{(ix)^n}{n!} \\ & = & \sum _{k=0} ^{+\infty} \dfrac{(ix)^{2n}}{(2n)!}+ \sum _{k=0} ^{+\infty} \dfrac{(ix)^{2n+1}}{(2n+1)!} \\ & = & \sum _{k=0} ^{+\infty} \dfrac{(-1)^nx^{2n}}{(2n)!}+ i\sum _{k=0} ^{+\infty} \dfrac{(-1)^nx^{2n+1}}{(2n+1)!} \\ & = & \cos x + i\sin x. \end{array}$$ \end{pf} Taking complex conjugates, $$ e^{-ix}=\overline{e^{ix}}=\overline{\cos x + i\sin x} = \cos x -i\sin x. $$ Solving for $\sin x$ we obtain \begin{equation} \sin x = \frac{e^{ix} - e^{-ix}}{2i} \end{equation} Similarly, \begin{equation} \cos x = \frac{e^{ix} + e^{-ix}}{2} \end{equation} \begin{cor}[De Moivre's Theorem] Let $n\in\integers$ and $x\in\reals$. Then $$(\cos x+ i\sin x)^n=\cos nx + i\sin nx$$ \label{cor:demoivre}\end{cor} \begin{pf} We have $$ (\cos x+ i\sin x)^n = (e^{ix})^n = e^{ixn} = \cos nx + i\sin nx, $$by theorem \ref{thm:euler}. \bigskip {\em Aliter:} An alternative proof without appealing to Euler's identity follows. We first assume that $n>0$ and give a proof by induction. For $n=1$ the assertion is obvious, as $$ (\cos x+ i\sin x)^1 = \cos 1\cdot x + i\sin 1\cdot x. $$ Assume the assertion is true for $n-1>1$, that is, assume that $$(\cos x+ i\sin x)^{n-1}=\cos (n-1)x + i\sin (n-1)x.$$Using the addition identities for the sine and cosine, $$ \begin{array}{lll} (\cos x+ i\sin x)^n & = & (\cos x+ i\sin x)(\cos x+ i\sin x)^{n-1} \\ & = & (\cos x+ i\sin x)(\cos (n-1)x + i\sin (n-1)x).\\ & = & (\cos x)(\cos (n-1)x)-(\sin x)(\sin (n-1)x) + i((\cos x)(\sin (n-1)x) + (\cos (n-1)x)(\sin x)).\\ & = & \cos (n-1+1)x + i\sin (n-1+1)x\\ & = & \cos nx + i\sin nx, \end{array}$$proving the theorem for $n>0$. \bigskip Assume now that $n<0$. Then $-n>0$ and we may used what we just have proved for positive integers we have $$\begin{array}{lll} (\cos x+ i\sin x)^n & = & \dfrac{1}{(\cos x+ i\sin x)^{-n}}\\ & = & \dfrac{1}{\cos (-nx)+i\sin (-nx)}\\ & = & \dfrac{1}{\cos nx -i\sin nx}\\ & = & \dfrac{\cos nx + i\sin nx}{(\cos nx + i\sin nx)(\cos nx - i\sin nx)} \\ & = & \dfrac{\cos nx + i\sin nx}{\cos^2 nx + \sin^2 nx}\\ & = & \cos nx + i\sin nx, \end{array}$$proving the theorem for $n<0$. If $n=0$, then since $\sin$ and $\cos$ are not simultaneously zero, we get $1=(\cos x+ i\sin x)^0 = \cos 0x+i\sin 0x=\cos 0x=1$, proving the theorem for $n=0$. \end{pf} \begin{exa}\label{exa:trisector} Prove that $$ \cos 3x = 4\cos ^3x-3\cos x, \qquad \sin 3x = 3\sin x - 4\sin^3x. $$ \end{exa} \begin{solu} Using Euler's identity and the Binomial Theorem, $$\begin{array}{lll}\cos 3x + i\sin 3x & = & e^{3ix}\\ & = & (e^{ix})^3=(\cos x + i\sin x)^3\\ & = & \cos^3x+3i\cos^2x\sin x-3\cos x\sin^2x-i\sin^3x\\ & = & \cos^3x+3i(1-\sin^2x)\sin x-3\cos x(1-\cos^2x)-i\sin^3x,\end{array} $$we gather the required identities. \end{solu} The following corollary is immediate. \begin{cor}[Roots of Unity] If $n>0$ is an integer, the $n$ numbers $e^{2\pi i k/n}=\cos \dfrac{2\pi k}{n}+i\sin \dfrac{2\pi k}{n}$, $0 \leq k \leq n-1$, are all different and satisfy $(e^{2\pi i k/n}) ^n =1$. \end{cor} \vspace{2cm} \begin{figure}[h] \begin{minipage}{5cm} \centering \psset{unit=1pc} \psaxes[linewidth=2pt,labels=none,ticks=none]{->}(0,0)(-5,-5)(5,5) \pstGeonode[PointName=none,linecolor=blue,linewidth=2pt](4;0){A}(4;120){B}(4;240){C} \pspolygon[linewidth=1.2pt,linecolor=red](A)(B)(C) \pstGeonode[PointName=none,PointSymbol=none](0,0){O} \pscircle[linestyle=dashed,linecolor=red](O){4} \meinecaption{2}{Cubic Roots of $1$.}\label{fig:cbc-rt1} \end{minipage} \hfill \begin{minipage}{5cm} \centering \psset{unit=1pc} \psaxes[linewidth=2pt,labels=none,ticks=none]{->}(0,0)(-5,-5)(5,5) \pstGeonode[PointName=none,linecolor=blue,linewidth=2pt](4;0){A}(4;90){B}(4;180){C}(4;270){D} \pspolygon[linewidth=1.2pt,linecolor=red](A)(B)(C)(D) \pstGeonode[PointName=none,PointSymbol=none](0,0){O} \pscircle[linestyle=dashed,linecolor=red](O){4} \meinecaption{2}{Quartic Roots of $1$.}\label{fig:qt-rt1} \end{minipage} \hfill \begin{minipage}{5cm} \centering \psset{unit=1pc} \psaxes[linewidth=2pt,labels=none,ticks=none]{->}(0,0)(-5,-5)(5,5) \pstGeonode[PointName=none,linecolor=blue,linewidth=2pt](4;0){A}(4;72){B}(4;144){C}(4;216){D}(4;288){E} \pspolygon[linewidth=1.2pt,linecolor=red](A)(B)(C)(D)(E) \pstGeonode[PointName=none,PointSymbol=none](0,0){O} \pscircle[linestyle=dashed,linecolor=red](O){4} \meinecaption{2}{Quintic Roots of $1$.}\label{fig:qu-rt1} \end{minipage} \end{figure} \begin{exa} For $n=2$, the square roots of unity are the roots of $$x^2-1=0 \implies x\in \{-1,1\}.$$ \bigskip For $n=3$ we have $x^3-1=(x-1)(x^2+x+1)=0$ hence if $x\neq 1$ then $x^2+x+1=0 \implies x=\dfrac{-1\pm i\sqrt{3}}{2}$. Hence the cubic roots of unity are $$\left\{-1, \dfrac{-1- i\sqrt{3}}{2}, \dfrac{-1+ i\sqrt{3}}{2}\right\}.$$ Or, we may find them trigonometrically, $$\begin{array}{lllll} e^{2\pi i\cdot 0 /3} & = & \cos \dfrac{2\pi \cdot 0}{3} + i\sin \dfrac{2\pi \cdot 0}{3} & = & 1, \\ e^{2\pi i\cdot 1 /3} & = & \cos \dfrac{2\pi \cdot 1}{3} + i\sin \dfrac{2\pi \cdot 1}{3} & = & -\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \\ e^{2\pi i\cdot 2 /3} & = & \cos \dfrac{2\pi \cdot 2}{3} + i\sin \dfrac{2\pi \cdot 2}{3} & = & -\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2} \\ \end{array} $$ \bigskip For $n=4$ they are the roots of $x^4-1=(x-1)(x^3+x^2+x+1)=(x-1)(x+1)(x^2+1)=0$, which are clearly $$ \{-1,1,-i,i\}. $$Or, we may find them trigonometrically, $$\begin{array}{lllll} e^{2\pi i\cdot 0 /4} & = & \cos \dfrac{2\pi \cdot 0}{4} + i\sin \dfrac{2\pi \cdot 0}{4} & = & 1, \\ e^{2\pi i\cdot 1 /4} & = & \cos \dfrac{2\pi \cdot 1}{4} + i\sin \dfrac{2\pi \cdot 1}{4} & = & i \\ e^{2\pi i\cdot 2 /4} & = & \cos \dfrac{2\pi \cdot 2}{4} + i\sin \dfrac{2\pi \cdot 2}{4} & = & -1\\ e^{2\pi i\cdot 3 /4} & = & \cos \dfrac{2\pi \cdot 3}{4} + i\sin \dfrac{2\pi \cdot 3}{4} & = & -i \\ \end{array} $$ \bigskip For $n=5$ they are the roots of $x^5-1=(x-1)(x^4+x^3+x^2+x+1)=0$. To solve $x^4+x^3+x^2+x+1=0$ observe that since clearly $x\neq 0$, by dividing through by $x^2$, we can transform the equation into $$ x^2 + \dfrac{1}{x^2} + x+ \dfrac{1}{x}+1 =0. $$ Put now $u =x+ \dfrac{1}{x} $. Then $u^2-2=x^2+\dfrac{1}{x^2}$, and so $$ x^2 + \dfrac{1}{x^2} + x+ \dfrac{1}{x}+1 =0\implies u^2-2+u+1=0 \implies u=\dfrac{-1\pm \sqrt{5}}{2}.$$ Solving both equations$$ x+ \dfrac{1}{x}= \dfrac{-1- \sqrt{5}}{2}, \qquad x+ \dfrac{1}{x}= \dfrac{-1+ \sqrt{5}}{2},$$ we get the four roots $$ \left\{\dfrac{-1-\sqrt{5}}{4}-i\dfrac{\sqrt{10-2\sqrt{5}}}{4}, \quad \dfrac{-1-\sqrt{5}}{4}+i\dfrac{\sqrt{10-2\sqrt{5}}}{4}, \quad \dfrac{\sqrt{5}-1}{4}-i\dfrac{\sqrt{2\sqrt{5}+10}}{4}, \quad \dfrac{\sqrt{5}-1}{4}+i\dfrac{\sqrt{2\sqrt{5}+10}}{4} \right\}, $$ or, we may find, trigonometrically, $$\begin{array}{lllll} e^{2\pi i\cdot 0 /5} & = & \cos \dfrac{2\pi \cdot 0}{5} + i\sin \dfrac{2\pi \cdot 0}{5} & = & 1, \\ e^{2\pi i\cdot 1 /5} & = & \cos \dfrac{2\pi \cdot 1}{5} + i\sin \dfrac{2\pi \cdot 1}{5} & = & \left(\dfrac{\sqrt{5}-1}{4}\right)+i\left(\dfrac{\sqrt{2}\cdot \sqrt{5+\sqrt{5}}}{4}\right), \\ e^{2\pi i\cdot 2 /5} & = & \cos \dfrac{2\pi \cdot 2}{5} + i\sin \dfrac{2\pi \cdot 2}{5} & = & \left(\dfrac{-\sqrt{5}-1}{4}\right)+i\left(\dfrac{\sqrt{2}\cdot \sqrt{5-\sqrt{5}}}{4}\right), \\ e^{2\pi i\cdot 3 /5} & = & \cos \dfrac{2\pi \cdot 3}{5} + i\sin \dfrac{2\pi \cdot 3}{5} & = & \left(\dfrac{-\sqrt{5}-1}{4}\right)-i\left(\dfrac{\sqrt{2}\cdot \sqrt{5-\sqrt{5}}}{4}\right), \\ e^{2\pi i\cdot 4 /5} & = & \cos \dfrac{2\pi \cdot 4}{5} + i\sin \dfrac{2\pi \cdot 4}{5} & = & \left(\dfrac{\sqrt{5}-1}{4}\right)-i\left(\dfrac{\sqrt{2}\cdot \sqrt{5+\sqrt{5}}}{4}\right), \\ \end{array} $$ \bigskip See figures \ref{fig:cbc-rt1} through \ref{fig:qu-rt1}. \end{exa} By the Fundamental Theorem of Algebra the equation $x^n-1 = 0$ has exactly $n$ complex roots, which gives the following result. \begin{cor}\label{cor:decom-of-root-1} Let $n>0$ be an integer. Then $$x^n-1 = \prod _{k=0} ^{n-1} (x-e^{2\pi i k/n}). $$ \end{cor} \begin{thm}\label{thm:series-multisection} We have, $$ 1 + x + x^2 + \cdots + x^{n - 1} = \left\{\begin{array}{ll} 0 & x = e^{\frac{2\pi ik}{n}}, \quad 1 \leq k \leq n - 1, \\ n & x = 1. \end{array} \right.$$ \end{thm} \begin{pf} Since $x^n-1 =(x-1)(x^{n-1}+x^{n-2}+\cdots + x +1)$, from Corollary \ref{cor:decom-of-root-1}, if $x\neq 1$, $$ x^{n-1}+x^{n-2}+\cdots + x +1 = \prod _{k=1} ^{n-1} (x-e^{2\pi i k/n}).$$ If $\epsilon $ is a root of unity different from $1$, then $\epsilon =e^{2\pi i k/n}$ for some $k\in [1;n-1]$, and this proves the theorem. Alternatively, $$ 1 + \epsilon + \epsilon ^2 + \epsilon ^3 + \cdots + \epsilon ^{n - 1} = \frac{\epsilon ^n - 1}{\epsilon - 1} = 0.$$ This gives the result. \end{pf} We may use complex numbers to select certain sums of coefficients of polynomials. The following problem uses the fact that if $k$ is an integer \begin{equation} i^k + i^{k + 1} + i^{k + 2} + i^{k + 3} = i^k(1 + i + i^2 + i^3) = 0 \end{equation} \begin{exa} Let $$(1 + x^4 + x^8)^{100} = a_0 + a_1x + a_2x^2 + \cdots + a_{800}x^{800}.$$ Find: \begin{dingautolist}{202} \item $a_0 + a_1 + a_2 + a_3 + \cdots + a_{800}.$ \\ \item $a_0 + a_2 + a_4 + a_6 + \cdots + a_{800}.$ \\ \item $a_1 + a_3 + a_5 + a_7 + \cdots + a_{799}.$ \\ \item $a_0 + a_4 + a_8 + a_{12} + \cdots + a_{800}.$ \\ \item $a_1 + a_5 + a_9 + a_{13} + \cdots + a_{797}.$ \end{dingautolist} \end{exa} \begin{solu} Put $$p(x) = (1 + x^4 + x^8)^{100} = a_0 + a_1x + a_2x^2 + \cdots + a_{800}x^{800}.$$ Then \begin{dingautolist}{202} \item $$ a_0 + a_1 + a_2 + a_3 + \cdots + a_{800} = p(1) = 3^{100}. $$ \item $$a_0 + a_2 + a_4 + a_6 + \cdots + a_{800} = \frac{p(1) + p(-1)}{2} = 3^{100}. $$ \item $$a_1 + a_3 + a_5 + a_7 + \cdots + a_{799} = \frac{p(1) - p(-1)}{2} = 0. $$ \item $$a_0 + a_4 + a_8 + a_{12} + \cdots + a_{800} = \frac{p(1) + p(-1) + p(i) + p(-i)}{4} = 2\cdot 3^{100}.$$ \item $$a_1 + a_5 + a_9 + a_{13} + \cdots + a_{797} = \frac{p(1) - p(-1) -ip(i) + ip(-i)}{4} = 0.$$ \end{dingautolist} \end{solu} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} Prove that $$\cos^62x = \frac{1}{32}\cos 12x + \frac{3}{16}\cos 8x + \frac{15}{32}\cos 4x + \frac{5}{16}.$$ \begin{answer} Using the binomial theorem and Euler's formula, $$\begin{array}{lll} 32\cos^6 2x & = & \left(e^{2ix} + e^{-2ix}\right)^6\\ & = & \binom{6}{0}e^{12ix} + \binom{6}{1}e^{10ix}e^{-2ix}+ \binom{6}{2}e^{8ix}e^{-4ix}+ \binom{6}{3}e^{6ix}e^{-6ix}+ \binom{6}{4}e^{4ix}e^{-8ix} + \binom{6}{5}e^{2ix}e^{-10ix} + \binom{6}{6}e^{-12ix}\\ & = & e^{12ix} + 6e^{8ix} + 15e^{4ix} + 20 + 15e^{-4ix} + 6e^{-8ix} + e^{-12ix}\\ & = & (e^{12ix}+e^{-12ix})+6(e^{8ix}+e^{-8ix})+15(e^{4ix} +e^{-4ix}) + 20 \\ & = & 2\cos 12x+ 12\cos 8x + 30\cos 4x + 20, \end{array} $$ from where we deduce the result. \end{answer} \end{pro} \begin{pro} Prove that $$ \sqrt{3}=\tan\dfrac{\pi}{9}+4\sin \dfrac{\pi}{9}.$$ \begin{answer} From $$ \cos 3x = 4\cos ^3x-3\cos x, \qquad \sin 3x = 3\sin x - 4\sin^3x, $$we gather, upon using the double angle and the sum identities, $$ \begin{array}{lll} \renewcommand{\arraystretch}{3.5} \tan 3x & = & \dfrac{3\sin x-4\sin ^3x}{4\cos^3 x-3\cos x}\\ & = & \tan x \left(\dfrac{3-4\sin^2x}{4\cos^2x-3}\right)\\ & = & \tan x \left(\dfrac{3-4\sin^2x}{1-4\sin^2x}\right)\\ & = & \tan x \left(1 + \dfrac{2}{1-4\sin^2x}\right)\\ & = & \tan x+ \dfrac{2\sin x}{\cos x - 4\sin^2x\cos x}.\\ & = & \tan x+ \dfrac{2\sin x}{\cos x - 2\sin x \sin 2x}\\ & = & \tan x+ \dfrac{2\sin x}{\cos x - 2\left(\dfrac{\cos x}{2}-\dfrac{\cos 3x}{2}\right)}\\ & = & \tan x + \dfrac{2\sin x}{\cos 3x}. \end{array}$$ Finally, upon letting $x=\dfrac{\pi}{9}$ we gather, $$ \sqrt{3}=\tan \dfrac{\pi}{3}= \tan\dfrac{\pi}{9}+\dfrac{2\sin \dfrac{\pi}{9}}{\cos \dfrac{\pi}{3}} = \tan\dfrac{\pi}{9}+4\sin \dfrac{\pi}{9},$$ as it was to be shewn. \end{answer} \end{pro} \end{multicols} \chapter{Binomial Theorem}\label{chap:bino_mio} \section{Pascal's Triangle} It is well known that \begin{equation}(a + b)^2 = a^2 + 2ab + b^2\end{equation}Multiplying this last equality by $a + b$ one obtains $$(a + b)^3 = (a + b)^2(a + b) = a^3 + 3a^2b + 3ab^2 + b^3$$ Again, multiplying \begin{equation} (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \end{equation}by $a + b$ one obtains $$(a + b)^4 = (a + b)^3(a + b) = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$$ Dropping the variables, a pattern with the coefficients emerges, a pattern called {\em Pascal's Triangle.} \flushleft{{\bf Pascal's Triangle}} \begin{center}\tiny{1 \\ 1 \hspace{3mm} 1\\ 1 \hspace{3mm} 2 \hspace{3mm} 1 \\ 1 \hspace{3mm} 3 \hspace{3mm} 3 \hspace{3mm} 1 \\ 1 \hspace{3mm} 4 \hspace{3mm} 6 \hspace{3mm} 4 \hspace{3mm} 1 \\ 1 \hspace{3mm} 5 \hspace{3mm} 10 \hspace{3mm} 10 \hspace{3mm} 5 \hspace{3mm} 1 \\ 1 \hspace{3mm} 6 \hspace{3mm} 15 \hspace{3mm} 20 \hspace{3mm} 15 \hspace{3mm} 6 \hspace{3mm} 1 \\ 1 \hspace{3mm} 7 \hspace{3mm} 21 \hspace{3mm} 35 \hspace{3mm} 35 \hspace{3mm} 21 \hspace{3mm} 7 \hspace{3mm} 1 \\ 1 \hspace{3mm} 8 \hspace{3mm} 28 \hspace{3mm} 56 \hspace{3mm} 70 \hspace{3mm} 56 \hspace{3mm} 28 \hspace{3mm} 8 \hspace{3mm} 1\\ 1 \hspace{3mm} 9 \hspace{3mm} 36 \hspace{3mm} 84 \hspace{3mm} 126 \hspace{3mm} 126 \hspace{3mm} 84 \hspace{3mm} 36 \hspace{3mm} 9 \hspace{3mm} 1 \\ 1 \hspace{3mm} 10 \hspace{3mm} 45 \hspace{3mm} 120 \hspace{3mm} 210 \hspace{3mm} 252 \hspace{3mm} 210 \hspace{3mm} 120 \hspace{3mm} 45 \hspace{3mm} 10 \hspace{3mm} 1 \\ ................................................................................................................................. \\ ...........................................................................................................................................}\end{center} Notice that each entry different from $1$ is the sum of the two neighbours just above it. Pascal's Triangle can be used to expand binomials to various powers, as the following examples shew. \begin{exa} $$ \begin{array}{lll} (4x + 5)^3 & = & (4x)^3 + 3(4x)^2(5) + 3(4x)(5)^2 + 5^3 \\ & = & 64x^3 + 240x^2 + 300x + 125 \end{array} $$ \end{exa} \begin{exa} $$\begin{array}{lll} (2x - y^2)^4 & = & (2x)^4 + 4(2x)^3(-y^2) + 6(2x)^2(-y^2)^2 + \\ & & \qquad + 4(2x)(-y^2)^3 + (-y^2)^4 \\ & = & 16x^4 - 32x^3y^2 + 24x^2y^4 - 8xy^6 + y^8 \end{array} $$ \end{exa} \begin{exa} $$ \begin{array}{lll} (2 + i)^5 & = & 2^5 + 5(2)^4(i) + 10(2)^3(i)^2 + \\ & & + 10(2)^2(i)^3 + 5(2)(i)^4 + i^5 \\ & = & 32 + 80i - 80 - 40i + 10 + i \\ & = & -38 + 39i \end{array} $$ \end{exa} \begin{exa} $$ \begin{array}{lll} (\sqrt{3} + \sqrt{5})^4 & = & (\sqrt{3})^4 + 4(\sqrt{3})^3(\sqrt{5}) \\ & & \qquad + 6(\sqrt{3})^2(\sqrt{5})^2 + 4(\sqrt{3})(\sqrt{5})^3 + (\sqrt{5})^4 \\ & = & 9 + 12\sqrt{15} + 90 + 20\sqrt{15} + 25 \\ & = & 124 + 32\sqrt{15} \end{array} $$ \end{exa} \begin{exa} Given that $a - b = 2, ab = 3$ find $a^3 - b^3.$ \end{exa} \begin{solu} One has $$ \begin{array}{lll} 8 & = & 2^3 \\ & = & (a - b)^3 \\ & = & a^3 - 3a^2b + 3ab^2 - b^3 = a^3 - b^3 - 3ab(a - b)\\ & = & a^3 - b^3 - 18, \end{array}$$ whence $a^3 - b^3 = 26.$ \\ {\em Aliter:} Observe that $4 = 2^2 = (a - b)^2 = a^2 + b^2 - 2ab = a^2 - b^2 - 6$, whence $a^2 + b^2 = 10.$ This entails that $$a^3 - b^3 = (a - b)(a^2 + ab + b^2) = (2)(10 + 3) = 26,$$as before. \end{solu} \section{Homework} \addcontentsline{toc}{section}{Homework} \markright{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Expand \begin{enumerate} \item $(x - 4y)^3$ \item $(x^3 + y^2)^4$ \item $(2 + 3x)^3$ \item $(2i - 3)^4$ \item $(2i + 3)^4 + (2i - 3)^4$ \item $(2i + 3)^4 - (2i - 3)^4$ \item $(\sqrt{3} - \sqrt{2})^3$ \item $(\sqrt{3} + \sqrt{2})^3 + (\sqrt{3} - \sqrt{2})^3$ \item $(\sqrt{3} + \sqrt{2})^3 - (\sqrt{3} - \sqrt{2})^3$ \end{enumerate} \end{pro} \begin{pro} Prove that $$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$$ Prove that $$(a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd )$$Generalise. \end{pro} \begin{pro} Compute $(x + 2y + 3z)^2$. \end{pro} \begin{pro} Given that $a + 2b = -8, \ ab = 4,$ find (i) $a^2 + 4b^2$, (ii) $a^3 + 8b^3$, (iii) $\dis{\frac{1}{a} + \frac{1}{2b}}$. \end{pro} \begin{pro} The sum of the squares of three consecutive positive integers is $21170$. Find the sum of the cubes of those three consecutive positive integers. \end{pro} \begin{pro} What is the coefficient of $x^4 y^6$ in $$(x\sqrt{2} - y)^{10}?$$\end{pro} Answer: 840. \begin{pro} Expand and simplify $$ (\sqrt{1 - x^2} + 1)^7 - (\sqrt{1 - x^2} - 1)^7.$$\end{pro} \end{multicols} \chapter{Sequences and Series} \section{Sequences} \begin{df} A sequence of real numbers is a function whose domain is the set of natural numbers and whose output is a subset of the real numbers. We usually denote a sequence by one of the notations $$a_0, a_1, a_2, \ldots , $$ or$$ \{a_n\}_{n=0} ^{+\infty}\ . $$ \end{df} \begin{rem} Sometimes we may not start at $n = 0$. In that case we may write $$a_m, a_{m+1}, a_{m+2}, \ldots , $$ or$$ \{a_n\}_{n=m} ^{+\infty}\ , $$where $m$ is a non-negative integer. \end{rem} We will be mostly interested in two types of sequences: sequences that have an explicit formula for their $n$-th term and sequences that are defined recursively. \begin{exa} Let $a_n = 1 - \frac{1}{2^n}, n = 0, 1, \ldots$. Then $\{a_n\}_{n = 0} ^{+\infty}$ is a sequence for which we have an explicit formula for the $n$-th term. The first five terms are $$\begin{array}{lllll} a_0 & = & 1 -\frac{1}{2^0} & = & 0, \\ a_1 & = & 1 -\frac{1}{2^1} & = & \frac{1}{2}, \\ a_2 & = & 1 -\frac{1}{2^2} & = & \frac{3}{4}, \\ a_3 & = & 1 -\frac{1}{2^3} & = & \frac{7}{8}, \\ a_4 & = & 1 -\frac{1}{2^4} & = & \frac{15}{16}. \\ \end{array} $$ \end{exa} \begin{exa} Let $$x_0 = 1,\ \ x_n = \left(1 + \frac{1}{n}\right)x_{n-1},\ \ \ n = 1, 2, \ldots .$$ Then $\{x_n\}_{n = 0} ^{+\infty}$ is a sequence recursively defined. The terms $x_1, x_2, \ldots , x_5$ are $$\begin{array}{lllll} x_1 & = & \left(1 +\frac{1}{1}\right)x_0 & = & 2, \\ x_2 & = & \left(1 +\frac{1}{2}\right)x_1 & = & 3, \\ x_3 & = & \left(1 +\frac{1}{3}\right)x_2 & = & 4, \\ x_4 & = & \left(1 +\frac{1}{4}\right)x_3 & = & 5, \\ x_5 & = & \left(1 +\frac{1}{5}\right)x_4 & = & 6. \\ \end{array} $$You might conjecture that an explicit formula for $x_n$ is $x_n = n + 1$, and you would be right! \end{exa} \begin{df} A sequence $\{a_n\} _{n=0} ^{+\infty}$ is said to be {\em increasing} if $a_{n}\leq a_{n + 1} \ \forall n\in\naturals$\footnote{Some people call these sequences {\em non-decreasing.}} and {\em strictly increasing} if $a_n < a_{n + 1} \ \forall n\in\naturals$\footnote{Some people call these sequences {\em increasing.}} \bigskip Similarly, a sequence $\{a_n\} _{n=0} ^{+\infty}$ is said to be {\em decreasing} if $a_{n}\geq a_{n + 1} \ \forall n\in\naturals$\footnote{Some people call these sequences {\em non-increasing.}} and {\em strictly decreasing} if $a_n > a_{n + 1} \ \forall n\in\naturals$\footnote{Some people call these sequences {\em decreasing.}} \bigskip A sequence is {\em monotonic} if is either increasing, strictly increasing, decreasing, or strictly decreasing. \end{df} \begin{exa} Recall that $0! = 1$, $1! = 1$, $2! = 1\cdot 2 = 2$, $3! = 1\cdot 2 \cdot 3 = 6$, etc. Prove that the sequence $x_n = n!, n = 0, 1, 2, \ldots$ is strictly increasing for $n \geq 1$. \end{exa} \begin{solu} For $n > 1$ we have $$x_n = n! = n(n - 1)! = nx_{n - 1} > x_{n - 1}, $$since $n>1$. This proves that the sequence is strictly increasing. \end{solu} \begin{exa} Prove that the sequence $x_n = 2 + \dfrac{1}{2^n}$, $n =0, 1, 2, \ldots$ is strictly decreasing. \end{exa} \begin{solu} We have $$\begin{array}{lll}x_{n + 1} - x_n & = & \left(2 + \dfrac{1}{2^{n+1}} \right) - \left(2 + \dfrac{1}{2^n}\right) \\ & = & \dfrac{1}{2^{n+1}} -\dfrac{1}{2^n} \\ & = & -\dfrac{1}{2^{n+1}} \\ & < & 0, \end{array} $$whence $$x_{n + 1} - x_n < 0 \implies x_{n + 1} < x_n,$$i.e., the sequence is strictly decreasing. \end{solu} \begin{exa} Prove that the sequence $x_n = \dfrac{n^2 + 1}{n}$, $n = 1, 2, \ldots$ is strictly increasing. \end{exa} \begin{solu} First notice that $\dfrac{n^2 + 1}{n} = n + \dfrac{1}{n}$. Now, $$ \begin{array}{lll}x_{n + 1} - x_n & = & \left(n + 1 + \dfrac{1}{n + 1}\right) -\left(n + \dfrac{1}{n }\right) \\ & = & 1 + \dfrac{1}{n + 1} - \dfrac{1}{n} \\ & = & 1 -\dfrac{1}{n(n +1)} \\ & > & 0,\\ \end{array}$$ since from $1$ we are subtracting a proper fraction less than $1$. Hence $$x_{n + 1} - x_n > 0 \implies x_{n + 1} > x_n,$$i.e., the sequence is strictly increasing. \end{solu} \begin{df} A sequence $\{x_n\} _{n=0} ^{+\infty}$ is said to be {\em bounded} if eventually the absolute value of every term is smaller than a certain positive constant. The sequence is {\em unbounded} if given an arbitrarily large positive real number we can always find a term whose absolute value is greater than this real number. \end{df} \begin{exa} Prove that the sequence $x_n = n!, n = 0, 1, 2, \ldots$ is unbounded.\end{exa} \begin{solu} Let $M > 0$ be a large real number. Then its integral part $\lfloor M\rfloor$ satisfies the inequality $M-1<\lfloor M\rfloor \leq M$ and so $\lfloor M\rfloor + 1 > M$. We have $$x_{\lfloor M\rfloor+1} = (\lfloor M\rfloor+1)! = (\lfloor M\rfloor+1)(\lfloor M\rfloor)(\lfloor M\rfloor-1)\cdots 2\cdot 1 > M, $$ since the first factor is greater than $M$ and the remaining factors are positive integers. \end{solu} \begin{exa} Prove that the sequence $a_n = \dfrac{n + 1}{n},$ $n = 1, 2, \ldots,$ is bounded. \end{exa} \begin{solu} Observe that $a_n = \dfrac{n + 1}{n} = 1 + \dfrac{1}{n}$. Since $\dfrac{1}{n}$ strictly decreases, each term of $a_n$ becomes smaller and smaller. This means that each term is smaller that $a_1 = 1 + \dfrac{1}{2}$. Thus $a_n < 2$ for $n \geq 2$ and the sequence is bounded. \end{solu} \subsection*{Homework} \addcontentsline{toc}{section}{Homework} \markright{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Find the first five terms of the following sequences. \begin{multicols}{2} \begin{enumerate} \item $x_n = 1 + (-2)^n, n = 0, 1, 2, \ldots$ \item $x_n = 1 + (-\frac{1}{2})^n, n = 0, 1, 2, \ldots$ \item $x_n = n! + 1, n = 0, 1, 2, \ldots$ \item $x_n = \dfrac{1}{n! + (-1)^n}, n = 2, 3, 4, \ldots$ \item $x_n = \left(1 +\dfrac{1}{n}\right)^{n}, n = 1, 2, \ldots, $ \end{enumerate} \end{multicols} \begin{answer} (1) $2$, $-1$, $5$, $-7$, $17$; (2) $2$, $1/2$, $5/4$, $7/8$, $17/16$; (3) $2$, $2$, $3$, $7$, $25$; (4) $1/3$, $1/5$, $1/25$, $1/119$, $1/721$; (5) $2$, $9/4$, $64/27$, $625/256$, $7776/3125$ \end{answer} \end{pro} \begin{pro} Decide whether the following sequences are eventually monotonic or non-monotonic. Determine whether they are bounded or unbounded. \begin{multicols}{2}\begin{enumerate} \item $x_n = n,$ $n = 0, 1, 2, \ldots$ \item $x_n = (-1)^nn,$ $n = 0, 1, 2, \ldots$ \item $x_n = \dfrac{1}{n!},$ $n = 0, 1, 2, \ldots$ \item $x_n = \dfrac{n}{n + 1},$ $n = 0, 1, 2, \ldots$ \item $x_n = n^2 - n,$ $n = 0, 1, 2, \ldots$ \item $x_n = (-1)^n,$ $n = 0, 1, 2, \ldots$ \item $x_n = 1 - \dfrac{1}{2^n},$ $n = 0, 1, 2, \ldots$ \item $x_n = 1 + \dfrac{1}{2^n},$ $n = 0, 1, 2, \ldots$ \end{enumerate} \end{multicols} \begin{answer} (1) Strictly increasing, unbounded (2) non-monotonic, unbounded (3) strictly decreasing, bounded (4) strictly increasing, bounded (5) strictly increasing, unbounded, (6) non-monotonic, bounded, (7) strictly increasing, bounded, (8) strictly decreasing, bounded \end{answer} \end{pro} \end{multicols} \section{Convergence and Divergence} We are primarily interested in the behaviour that a sequence $\{a_n\} _{n=0} ^{+\infty}$ exhibits as $n$ gets larger and larger. First some shorthand. \begin{df} The notation $n\rightarrow +\infty$ means that the natural number $n$ increases or tends towards $+\infty$, that is, that it becomes bigger and bigger. \end{df} \begin{df} We say that the sequence $\{x_n\}_{n=0} ^{+\infty}$ {\em converges}\footnote{This definition is necessarily imprecise, as we want to keep matters simple. A more precise definition is the following: we say that a sequence $c_ n, n = 0, 1, 2, \ldots$ {\em converges} to $L$ (written $c_n \rightarrow L$) as $n \rightarrow +\infty$, if $\forall \varepsilon > 0 \ \exists N\in\naturals$ such that $|c_n - L| < \varepsilon \ \forall n > N.$ We say that a sequence $d_ n, n = 0, 1, 2, \ldots$ diverges to $+\infty$ (written $d_n \rightarrow +\infty$) as $n \rightarrow +\infty$, if $\forall M > 0 \ \exists N\in\naturals$ such that $d_n > M \ \forall n > N.$ A sequence $f_n, n = 0, 1, 2, \ldots$ diverges to $-\infty$ if the sequence $-f_n, n = 0, 1, 2, \ldots$ converges to $+\infty$. } to a limit $L$, written $x_n \rightarrow L$ as $n\rightarrow +\infty$, if eventually all terms after a certain term are closer to $L$ by any preassigned distance. A sequence which does not converge is said to {\em diverge.} \end{df} To illustrate the above definition, some examples are in order. \begin{exa} The constant sequence $$1,1,1,1, \ldots $$converges to $1$. \end{exa} \begin{exa} Consider the sequence $$1, \dfrac{1}{2}, \dfrac{1}{3}, \ldots, \frac{1}{n}, \ldots, $$ We claim that $\dfrac{1}{n}\rightarrow 0$ as $n\rightarrow+\infty$. Suppose we wanted terms that get closer to $0$ by at least $.00001 = \dfrac{1}{10^5}$. We only need to look at the $100000$-term of the sequence: $\dfrac{1}{100000} = \dfrac{1}{10^5}$. Since the terms of the sequence get smaller and smaller, any term after this one will be within $.00001$ of $0$. We had to wait a long time---till after the $100000$-th term---but the sequence eventually did get closer than $.00001$ to $0$. The same argument works for any distance, no matter how small, so we can eventually get arbitrarily close to $0$.\footnote{A rigorous proof is as follows. If $\varepsilon > 0$ is no matter how small, we need only to look at the terms after $N = \lfloor \frac{1}{\varepsilon} + 1\rfloor$ to see that, indeed, if $n > N$, then $$s_n = \frac{1}{n} < \frac{1}{N} = \frac{1}{\lfloor\frac{1}{\varepsilon} + 1\rfloor } < \varepsilon.$$ Here we have used the inequality $$ t - 1 < \lfloor t \rfloor \leq t, \ \ \forall t\in \reals.$$ }. \end{exa} \begin{exa} The sequence $$0, 1,4,9,16, \ldots ,n^2, \ldots$$ diverges to $+\infty$, as the sequence gets arbitrarily large.\footnote{A rigorous proof is as follows. If $M > 0$ is no matter how large, then the terms after $N = \lfloor \sqrt{M}\rfloor + 1$ satisfy ($n > N$) $$t_n = n^2 > N^2 = (\lfloor \sqrt{M}\rfloor + 1)^2 > M. $$} \end{exa} \begin{exa} The sequence $$1,-1,1,-1,1,-1,\ldots, (-1)^n,\ldots$$ has no limit (diverges), as it bounces back and forth from $-1$ to $+1$ infinitely many times. \end{exa} \begin{exa} The sequence $$0, -1,2,-3,4,-5,\ldots, (-1)^nn, \ldots , $$ has no limit (diverges), as it is unbounded and alternates back and forth positive and negative values.. \end{exa} \vspace{1cm} \begin{figure}[!hptb] $$ \psset{unit=1pc} \psline[linewidth=1.5pt]{<->}(-10,0)(10,0) \rput(-7,0){|} \rput(-3,0){|} \rput(0,0){|} \rput(2.9,0){|} \rput(4,0){|} \rput(5,0){|}\rput(6.5,0){|}\uput[d](-7,0){x_0} \uput[d](-3,0){x_1}\uput[d](0,0){{\tiny x_2}}\uput[d](2.9,0){\ddots }\uput[d](4,0){{\tiny x_n}}\uput[d](5,0){\ddots}\uput[d](6.5,0){{\tiny s}} $$ \meinecaption{1}{Theorem \ref{thm:conve_seq}.} \label{fig:thm_conve_seq} \end{figure} \vspace{2cm} When is it guaranteed that a sequence of real numbers has a limit? We have the following result. \begin{thm}\label{thm:conve_seq}Every bounded increasing sequence $\{a_n\}_{n = 0} ^{+\infty}$ of real numbers converges to its supremum. Similarly, every bounded decreasing sequence of real numbers converges to its infimum. \end{thm} \begin{pf} The idea of the proof is sketched in figure \ref{fig:thm_conve_seq}. By virtue of Axiom \ref{axi:completeness}, the sequence has a supremum $s$. Every term of the sequence satisfies $a_n \leq s$. We claim that eventually all the terms of the sequence are closer to $s$ than a preassigned small distance $\epsilon > 0$. Since $s - \epsilon$ is not an upper bound for the sequence, there must be a term of the sequence, say $a_{n_0}$ with $s - \epsilon \leq a_{n_0}$ by virtue of the Approximation Property Theorem \ref{thm:approx_sup}. Since the sequence is increasing, we then have $$ s-\epsilon \leq a_{n_0} \leq a_{n_0+1} \leq a_{n_0+2} \leq a_{n_0+2} \leq \ldots \leq s, $$ which means that after the $n_0$-th term, we get to within $\epsilon$ of $s$. \bigskip To obtain the second half of the theorem, we simply apply the first half to the sequence $\{-a_n\}_{n=0} ^{+\infty}$. \end{pf} \subsection*{Homework} \addcontentsline{toc}{section}{Homework} \markright{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Give plausible arguments to convince yourself that \begin{enumerate} \item $\frac{1}{2^n} \rightarrow 0$ as $n\rightarrow +\infty$ \item $2^n \rightarrow +\infty$ as $n\rightarrow +\infty$ \item $\frac{1}{n!} \rightarrow 0$ as $n\rightarrow +\infty$ \item $\frac{n +1}{n} \rightarrow 1$ as $n\rightarrow +\infty$ \item $(\frac{2}{3})^n \rightarrow 0$ as $n\rightarrow +\infty$ \item $(\frac{3}{2})^n \rightarrow +\infty$ as $n\rightarrow +\infty$ \item the sequence $(-2)^n, n = 0, 1, \ldots$ diverges as $n\rightarrow +\infty$ \item $\frac{n}{2^n} \rightarrow 0$ as $n\rightarrow +\infty$ \item $\frac{2^n}{n} \rightarrow +\infty$ as $n\rightarrow +\infty$ \item the sequence $1 + (-1)^n, n = 0, 1, \ldots$ diverges as $n\rightarrow +\infty$ \end{enumerate} \end{pro} \end{multicols} \section{Finite Geometric Series} \begin{df} A {\em geometric sequence} or {\em progression} is a sequence of the form $$ a, \ ar, ar^2,\ ar^3, \ ar^4, \ldots,$$that is, every term is produced from the preceding one by multiplying a fixed number. The number $r$ is called the {\em common ratio}. \end{df} \begin{rem} \begin{center} \begin{enumerate} \item Trivially, if $a = 0,$ then every term of the progression is $0$, a rather uninteresting case. \item If $ar \neq 0$, then the common ratio can be found by dividing any term by that which immediately precedes it. \item The $n$-th term of the progression $$ a, \ ar, ar^2,\ ar^3, \ ar^4, \ldots,$$ is $ar^{n - 1}$. \end{enumerate}\end{center} \end{rem} \begin{exa} Find the $35$-th term of the geometric progression $$\frac{1}{\sqrt{2}},\ -2, \ \frac{8}{\sqrt{2}}, \ldots . $$ \end{exa} \begin{solu} The common ratio is $-2 \div \frac{1}{\sqrt{2}} = -2\sqrt{2}$. Hence the $35$-th term is $\frac{1}{\sqrt{2}}(-2\sqrt{2})^{34} = \frac{2^{51}}{\sqrt{2}} = 1125899906842624\sqrt{2}.$ \end{solu} \begin{exa} The fourth term of a geometric progression is $24$ and its seventh term is $192.$ Find its second term. \end{exa}\begin{solu} We are given that $ar^3 = 24$ and $ar^6 = 192,$ for some $a$ and $r$. Clearly, $ar \neq 0,$ and so we find $$\frac{ar^6}{ar^3} = r^3 = \frac{192}{24} = 8,$$ whence $r = 2$. Now, $a(2)^3 = 24,$ giving $a = 3.$ The second term is thus $ar = 6.$ \end{solu} \begin{exa} Find the sum $$2 + 2^2 + 2^3 + 2^4 + \cdots + 2^{64}.$$Estimate (without a calculator!) how big this sum is. \end{exa} \begin{solu} Let $$S = 2 + 2^2 + 2^3 + 2^4 + \cdots + 2^{64}.$$Observe that the common ratio is $2$. We multiply $S$ by $2$ and notice that every term, with the exception of the last, appearing on this new sum also appears on the first sum. We subtract $S$ from $2S$: $$\begin{array}{lllllllllllllll} S & = & 2 & + & 2^2 & + & 2^3 & + & 2^4 & + & \cdots & + & 2^{64} & & \\ 2S & = & & & 2^2 & + & 2^3 & + & 2^4 & + & \cdots & + & 2^{64} & + & 2^{65}\\ \hline 2S - S & = & -2 + 2^{65} & & & & & & & & & & & & \\ \end{array} $$ Thus $S = 2^{65} - 2$. To estimate this sum observe that $2^{10} = 1024 \approx 10^3.$ Therefore $$2^{65} = (2^{10})^6\cdot (2^5) = 32(2^{10})^6 \approx 32 (10^3)^6 = 32\times 10^{18} = 3.2\times 10^{19}.$$ The exact answer (obtained via Maple \circledR ), is $$ 2^{65} - 2 = 36893488147419103230.$$ My pocket calculator yields $3.689348815 \times 10^{19}.$ Our estimate gives the right order of decimal places. \end{solu} \begin{rem} \begin{enumerate} \item If a chess player is paid $\$2$ for the first square of a chess board, $\$4$ for the second square, $\$8$ for the third square, etc., after reaching the $64$-th square he would be paid $\$36893488147419103230$. Query: After which square is his total more than $\$1000000$? \item From the above example, the sum of a geometric progression with positive terms and common ratio $r > 1$ grows rather fast rather quickly. \end{enumerate}\end{rem} \begin{exa} Sum $$\frac{2}{3} + \frac{2}{3^2} + \frac{2}{3^3} + \cdots + \frac{2}{3^{99}}.$$ \label{geomser2thi}\end{exa} \begin{solu} Put $$S = \frac{2}{3} + \frac{2}{3^2} + \frac{2}{3^3} + \cdots + \frac{2}{3^{99}}.$$ Then $$\frac{1}{3}S = \frac{2}{3^2} + \frac{2}{3^3} + \frac{2}{3^4} + \cdots + \frac{2}{3^{100}}.$$ Subtracting, $$S - \frac{1}{3}S = \frac{2}{3}S = \frac{2}{3} - \frac{2}{3^{100}}.$$ It follows that $$S = \frac{3}{2}\left( \frac{2}{3} - \frac{2}{3^{100}}\right) = 1 - \frac{1}{3^{99}}. $$ \end{solu} \begin{rem} The sum of the first two terms of the series in example \ref{geomser2thi} is $\frac{2}{3} + \frac{2}{3^2} = \frac{8}{9}$, which, though close to $1$ is not as close as the sum of the first $99$ terms. A geometric progression with positive terms and common ratio $0 < r < 1$ has a sum that grows rather slowly. \end{rem} To close this section we remark that the approximation $2^{10} \approx 1000$ is a useful one. It is nowadays used in computer lingo, where a kilobyte is $1024$ bytes---``kilo'' is a Greek prefix meaning ``thousand.'' \begin{exa} Without using a calculator, determine which number is larger: $2^{900}$ or $3^{500}$. \label{approexa1}\end{exa} \begin{solu} The idea is to find a power of $2$ close to a power of $3$. One readily sees that $2^3 = 8 < 9 = 3^2.$ Now, raising both sides to the $250$-th power, $$2^{750} = (2^3)^{250} < (3^2)^{250} = 3^{500}.$$The inequality just obtained is completely useless, it does not answer the question addressed in the problem. However, we may go around this with a similar idea. Observe that $9 < 8\sqrt{2}$: for, if $9 \geq 8\sqrt{2}$, squaring both sides we would obtain $81 > 128,$ a contradiction. Raising $9 < 8\sqrt{2}$ to the $250$-th power we obtain $$3^{500} = (3^2)^{250} < (8\sqrt{2})^{250} = 2^{875} < 2^{900},$$ whence $ 2^{900}$ is greater. \end{solu} \begin{rem} You couldn't solve example \ref{approexa1} using most pockets calculators and the mathematical tools you have at your disposal (unless you were {\em really} clever!). Later in this chapter we will see how to solve this problem using logarithms. \end{rem} \subsection*{Homework} \addcontentsline{toc}{section}{Homework} \markright{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Find the $17$-th term of the geometric sequence\ $$-\frac{2}{3^{17}},\ \frac{2}{3^{16}},\ -\frac{2}{3^{15}}, \cdots . $$ \begin{answer} $-\dfrac{2}{3}$ \end{answer} \end{pro} \begin{pro} The $6$-th term of a geometric progression is $20$ and the $10$-th is $320$. Find the absolute value of its third term. \begin{answer} One is given that $ar^5 = 20$ and $ar^9 = 320$. Hence $|ar^2| = \frac{5}{2}$ \end{answer} \end{pro} \begin{pro} Find the sum of the following geometric series. \begin{enumerate} \item $$1 + 3 + 3^2 + 3^3 + \cdots + 3^{49}.$$ \item If $y \neq 1,$ $$1 + y + y^2 + y^3 + \cdots + y^{100}.$$ \item If $y \neq 1,$ $$1 - y + y^2 - y^3 + y^4 - y^5 + \cdots - y^{99}+ y^{100}.$$ \item If $y \neq 1,$ $$1 + y^2 + y^4 + y^6 + \cdots + y^{100}.$$ \end{enumerate} \begin{answer} (1) $\frac{3^{50} - 1}{2} = 358948993845926294385124$, (2) $\frac{1 - y^{101}}{1 - y}$, (3) $\frac{1 + y^{101}}{1 + y}$, (4) $\frac{1 - y^{102}}{1 - y^2}$ \end{answer} \end{pro} \begin{pro} A colony of amoebas\footnote{Why are amoebas bad mathematicians? Because they divide to multiply!} is put in a glass at $2:00$ PM. One second later each amoeba divides in two. The next second, the present generation divides in two again, etc.. After one minute, the glass is full. When was the glass half-full? \begin{answer} At $2:00:59$ PM (the second just before $2:01$ PM.) \end{answer} \end{pro} \begin{pro} Without using a calculator: which number is greater $2^{30}$ or $30^2$?\begin{answer} $2^{30}$ \end{answer} \end{pro} \begin{pro} In this problem you may use a calculator. Legend says that the inventor of the game of chess asked the Emperor of China to place a grain of wheat on the first square of the chessboard, $2$ on the second square, $4$ on the third square, $8$ on the fourth square, etc.. (1) How many grains of wheat are to be put on the last ($64$-th) square?, (2) How many grains, total, are needed in order to satisfy the greedy inventor?, (3) Given that $15$ grains of wheat weigh approximately one gramme, what is the approximate weight, in kg, of wheat needed?, (4) Given that the annual production of wheat is $350$ million tonnes, how many years, approximately, are needed in order to satisfy the inventor (assume that production of wheat stays constant)\footnote{ Depending on your ethnic preference, the ruler in this problem might be an Indian maharajah or a Persian shah, but never an American businessman!!!}. \begin{answer} (1) $2^{63} = 9223372036854775808$, (2) $2^{64} - 1 = 18446744073709551614$, (3) $1.2\times 10^{15}$ kg, or $1200$ billion tonnes (4) 3500 years \end{answer} \end{pro} \begin{pro} Prove that $$1 + 2\cdot 5 + 3\cdot 5^2 + 4\cdot 5^3 + \cdots + 99\cdot 5^{100} = \frac{99\cdot 5^{101}}{4} - \frac{5^{101} - 1}{16}. $$ \end{pro} \begin{pro} Shew that $$1 + x + x^2 + \cdots + x^{1023} = (1 + x)(1 + x^2)(1 + x^4)\cdots (1 + x^{256})(1 + x^{512}).$$ \end{pro} \begin{pro} Prove that $$1 + x + x^2 + \cdots + x^{80} = (x^{54} + x^{27} + 1)(x^{18} + x^9 + 1)(x^6 + x^3 + 1)(x^2 + x + 1).$$ \end{pro} \end{multicols} \section{Infinite Geometric Series} \begin{df} Let$$s_n = a + ar + ar^2 + \cdots + ar^{n - 1}$$ be the sequence of partial sums of a geometric progression. We say that the {\em infinite} geometric sum $$a + ar + ar^2 + \cdots + ar^{n - 1} + ar^n + \cdots$${\em converges} to a finite number $s$ if $|s_n - s| \rightarrow 0$ as $n\rightarrow +\infty.$ We say that infinite sum $$a + ar + ar^2 + \cdots + ar^{n - 1} + ar^n + \cdots$${\em diverges} if there is no finite number to which the sequence of partial sums converges. \end{df} \begin{lem}\label{lem:a^ngoesto0} If $0 < a<1$ then $a^n \rightarrow 0$ as $n\rightarrow 0$. \end{lem} \begin{pf} Observe that by multiplying through by $a$ we obtain $$0 < a<1 \implies 0 < a^2 1,$ the infinite sum diverges. \end{enumerate} \label{geomsum}\end{thm} \begin{pf} Put $$S = a + ar + ar^2 + \cdots + ar^{n - 1}.$$Then $$rS = ar + ar^2 + ar^3 + \cdots + ar^{n}.$$ Subtracting, $$S - rS = S(1 - r) = a - ar^n.$$Since $|r| \neq 1$ we may divide both sides of the preceding equality in order to obtain $$S = \frac{a - ar^n}{1 - r},$$proving the first statement of the theorem. Now, if $|r| < 1,$ then $|r|^n \rightarrow 0$ as $n \rightarrow +\infty$ by virtue of Lemma \ref{lem:a^ngoesto0}, and if $|r| > 1,$ then $|r|^n \rightarrow + \infty$ as $n \rightarrow +\infty$. The second and third statements of the theorem follow from this. \end{pf} \begin{rem} We have thus created a dichotomy amongst infinite geometric sums. If their common ratio is smaller than $1$ in absolute value, the infinite geometric sum converges. Otherwise, the sum diverges. \end{rem} \begin{exa} Find the sum of the infinite geometric series $$\frac{3}{5^3} - \frac{3}{5^4} + \frac{3}{5^5} - \frac{3}{5^6} + \cdots .$$ \end{exa} \begin{solu} We have $a = \frac{3}{5^3}, r = -\frac{1}{5}$ in Theorem \ref{geomsum}. Therefore $$\frac{3}{5^3} - \frac{3}{5^4} + \frac{3}{5^5} - \frac{3}{5^6} + \cdots = \frac{\frac{3}{5^3}}{1 - \left(-\frac{1}{5}\right)} = \frac{1}{50}.$$ \end{solu} \begin{exa} Find the rational number which is equivalent to the repeating decimal $0.23\overline{45}$. \end{exa} \begin{solu} $$0.23\overline{45} = \frac{23}{10^2} + \frac{45}{10^4} + \frac{45}{10^6} + \cdots = \frac{23}{10^2} + \frac{\frac{45}{10^4}}{1 - \frac{1}{10^2}} = \frac{23}{100} + \frac{1}{220} = \frac{129}{550}. $$ \end{solu} \begin{rem} The geometric series above did not start till the second term of the sum. \end{rem} \begin{exa} A celestial camel is originally at the point $(0,0)$ on the Cartesian Plane. The camel is told by a Djinn that if it wanders $1$ unit right, $1/2$ unit up, $1/4$ unit left, $1/8$ unit down, $1/16$ unit right, and so, ad infinitum, then it will find a houris. What are the coordinate points of the houris? \end{exa} \begin{solu} Let the coordinates of the houris be $(X, Y)$. Then $$X = \frac{1}{4} + \frac{1}{4^2} - \frac{1}{4^3} + \cdots = \frac{1}{1 - \left(-\frac{1}{4}\right)} = \frac{4}{5},$$ and $$Y = \frac{1}{2} - \frac{1}{2^3} + \frac{1}{2^5} - \frac{1}{2^7} \cdots = \frac{\frac{1}{2}}{1 - \left(-\frac{1}{4}\right)} = \frac{2}{5}.$$ \end{solu} \begin{exa} What is wrong with the statement $$1 + 2 + 2^2 + 2^3 + \cdots = \frac{1}{1 - 2} = -1?$$Notice that the sinistral side is positive and the dextral side is negative. \end{exa} \begin{solu} The geometric sum diverges, as the common ratio $2$ is $> 1$, so we may not apply the formula for an infinite geometric sum. There is an interpretation (called {\em convergence in the sense of Abel}), where statements like the one above do make sense. \end{solu} \subsection*{Homework} \addcontentsline{toc}{section}{Homework} \markright{Homework} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small \begin{pro} Find the sum of the given infinite geometric series. \begin{enumerate} \item $$\frac{8}{5} + 1 + \frac{5}{8}+ \cdots$$ \item $$0.9 + 0.03 + 0.001 + \cdots$$ \item $$\frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}} + 1 + \frac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} + \cdots$$ \item $$\frac{\sqrt{3}}{\sqrt{2}} + \frac{\sqrt{2}}{3} + \frac{2\sqrt{2}}{9\sqrt{3}} + \cdots$$ \item $$1 + \frac{\sqrt{5} - 1}{2} + \left(\frac{\sqrt{5} - 1}{2}\right)^2 + \cdots $$ \item $$1 + 10 + 10^2 + 10^3 + \cdots$$ \item $$1 - x + x^2 - x^3 + \cdots, \ \ |x| < 1.$$ \item $$\frac{\sqrt{3}}{\sqrt{3} + 1} + \frac{\sqrt{3}}{\sqrt{3} + 3} + \cdots $$ \item $$x - y + \frac{y^2}{x} - \frac{y^3}{x^2} + \frac{y^4}{x^3} - \frac{y^5}{x^4} + \cdots, $$with $|y| < |x|$.\end{enumerate} \begin{answer} (1) $\frac{64}{15}$ , (2) $\frac{27}{29}$ , (3) $\frac{140 + 99\sqrt{2}}{8}$, (4) $\frac{27\sqrt{6} + 18\sqrt{2}}{46}$, (5) $\frac{3 + \sqrt{5}}{2}$, (6) diverges, (7) $\frac{1}{1 + x}$, (8) $\frac{3}{2}$, (9) $\frac{x^2}{x - y}$ \end{answer} \end{pro} \begin{pro} Give rational numbers (that is, the quotient of two integers), equivalent to the repeating decimals below. \begin{enumerate} \item $0.\overline{3}$ \item $0.\overline{6}$ \item $0.2\overline{5}$ \item $2.12\overline{35}$ \item $0.\overline{428571}$ \end{enumerate} \begin{answer} (1) $\frac{1}{3}$ , (2) $\frac{2}{3}$, (3) $\frac{23}{90}$ , (4) $\frac{21023}{9900}$ , (5) $\frac{3}{7}$ \end{answer} \end{pro} \begin{pro} Give an example of an infinite series with all positive terms, adding to $666$. \end{pro} \end{multicols} \chapter{Old Exam Questions} \section{Multiple-Choice} \begin{enumerate} \subsection{Real Numbers} \item The infinite repeating decimal $0.102102\ldots = 0.\overline{102}$ as a quotient of two integers is\\ \pscirclebox{A} $\dfrac{15019}{147098}$ \hfill \pscirclebox{B} $\dfrac{34}{333}$ \hfill \pscirclebox{C} $\dfrac{51}{500}$ \hfill \pscirclebox{D} $\dfrac{101}{999}$ \hfill \pscirclebox{E} none of these \item Express the infinite repeating decimal $0.424242\ldots = 0.\overline{42}$ as a fraction. \\ \psovalbox{A} $\dfrac{21}{50}$ \hfill \psovalbox{B} $\dfrac{14}{33}$ \hfill \psovalbox{C} $\dfrac{7}{15}$ \hfill \psovalbox{D} $\dfrac{14}{333}$ \hfill \psovalbox{E} none of these \item Write the infinite repeating decimal as a fraction: $0.121212\ldots = 0.\overline{12}$.\\ \psovalbox{A} $\dfrac{4}{33}$ \hfill \psovalbox{B} $\dfrac{3}{25}$ \hfill \psovalbox{C} $\dfrac{1}{2}$ \hfill \psovalbox{D} $\dfrac{102}{333}$ \hfill \psovalbox{E} none of these \item Let $a\in\rationals$ and $b\in\reals \setminus \rationals$. How many of the following are {\em necessarily} irrational numbers? $$I: a+b, \qquad II: ab, \qquad III: 1+a+b, \qquad IV: 1+a^2+b^2 $$ \psovalbox{A} exactly one \hfill \psovalbox{B} exactly two \hfill \psovalbox{C} exactly three \hfill \psovalbox{D} all four \hfill \psovalbox{E} none \item Let $a\in\integers$. How many of the following are {\em necessarily} true? $$I: \sqrt{|a|}\in\reals \setminus \rationals, \qquad II: \sqrt{a^2}\in\integers, \qquad III: \dfrac{a}{1+|a|}\in \rationals, \qquad IV: \sqrt{1+a^2}\in \reals \setminus \rationals $$ \psovalbox{A} exactly one \hfill \psovalbox{B} exactly two \hfill \psovalbox{C} exactly three \hfill \psovalbox{D} all four \hfill \psovalbox{E} none \subsection{Sets on the Line} \item $\oo{-3;2}\ \cap\ \cc{1;3}=$ \\ \pscirclebox{A} $\oo{-3;1} $ \hfill \pscirclebox{B} $ \oc{-3;1}$ \hfill \pscirclebox{C} $\co{1;2} $ \hfill \pscirclebox{D} $ \oc{-3;3}$ \hfill \pscirclebox{E} none of these \item Determine the set of all real numbers $x$ satisfying the inequality $\dfrac{x+2}{x-1}<1$. \\ \pscirclebox{A} $\oo{1;+\infty}$ \hfill \pscirclebox{B} $\oo{-2;1}$ \hfill \pscirclebox{C} $\oo{-\infty;1}$ \hfill \pscirclebox{D} $\oc{-\infty;1}$ \hfill \pscirclebox{E} none of these \item $\oc{-3;8}\quad \cap \quad \co{-8;-3}=$. \\ \psovalbox{A} $\{-3\}$ \hfill \psovalbox{B} $\varnothing$ \hfill \psovalbox{C} $\oc{-8;8}$ \hfill \psovalbox{D} $\oo{-8;8}$ \hfill \psovalbox{E} none of these \item Write as a single interval: $\oc{-2;4}\cup \co{1;5}$. \pscirclebox{A} $\oo{-2;1}$ \hfill \pscirclebox{B} $\oo{1;4}$ \hfill \pscirclebox{C} $\oo{-2;5}$\hfill \pscirclebox{D} $\cc{1;4}$ \hfill \pscirclebox{E} none of these \item Write as a single interval the following interval difference: \quad $\oo{-5;2}\quad \setminus\quad \cc{-3;3}$. \\ \psovalbox{A} $\oo{-5;-3}$ \hfill \psovalbox{B} $\co{-5;-3}$ \hfill \psovalbox{C} $\cc{-5;-3}$ \hfill \psovalbox{D} $\oc{-5;-3}$ \hfill \psovalbox{E} none of these \clearpage \item If $\dfrac{x+1}{x(x-1)}\geq 0$ then $x\in $\\ \psovalbox{A} $\oc{-\infty;0}\cup \co{1;+\infty} $ \\ \psovalbox{B} $\co{-1;0}\cup \oo{1;+\infty}$ \\ \psovalbox{C} $\co{-1;1}\cup \oo{1;+\infty}$ \\ \psovalbox{D} $\oo{-\infty;0}\cup \oo{0;1} $ \\ \psovalbox{E} none of these \item If $\dfrac{3}{x-1} - \dfrac{1}{x}\leq \dfrac{1}{x}$ then $x\in $\\ \psovalbox{A} $\oc{-\infty;-2} \cup \oo{0;1} $ \hfill \psovalbox{B} $ \oo{-2;1}$ \hfill \psovalbox{C} $\co{-2;0}\cup \oo{1;+\infty}$ \hfill \psovalbox{D} $\oo{-\infty;+\infty} $ \hfill \psovalbox{E} none of these \subsection{Absolute Values} \item[] {\bf Situation: Consider the absolute value expression $\absval{x+2} +\absval{x}-x$. Problems \ref{pro:y=|x+2|+|x|-x-begin} through \ref{pro:y=|x+2|+|x|-x-end} refer to it.} \item \label{pro:y=|x+2|+|x|-x-begin} Write $\absval{x+2}+\absval{x}-x$ without absolute values in the interval $\oc{-\infty;-2}$.\\ \psovalbox{A} $-x-2$ \hfill \psovalbox{B} $x+2$ \hfill \psovalbox{C} $-3x-2$ \hfill \psovalbox{D} $2-x$ \hfill \psovalbox{E} none of these \item Write $\absval{x+2}+\absval{x}-x$ without absolute values in the interval $\cc{-2;0}$.\\ \psovalbox{A} $-x-2$ \hfill \psovalbox{B} $x+2$ \hfill \psovalbox{C} $-3x-2$ \hfill \psovalbox{D} $2-x$ \hfill \psovalbox{E} none of these \item Write $\absval{x+2}+\absval{x}-x$ without absolute values in the interval $\co{0;+\infty}$.\\ \psovalbox{A} $-x-2$ \hfill \psovalbox{B} $x+2$ \hfill \psovalbox{C} $-3x-2$ \hfill \psovalbox{D} $2-x$ \hfill \psovalbox{E} none of these \item If $\absval{x+2}+\absval{x}-x=2$, then $x\in$ \psovalbox{A} $\varnothing$ \hfill \psovalbox{B} $\{-2\}$\hfill \psovalbox{C} $\cc{-2;0}$\hfill \psovalbox{D} $\{0\}$ \hfill \psovalbox{E} none of these \item \label{pro:y=|x+2|+|x|-x-end} If $\absval{x+2}+\absval{x}-x=3$, then $x\in$ \psovalbox{A} $\{0,1\}$ \hfill \psovalbox{B} $\{-1,0\}$\hfill \psovalbox{C} $\cc{-1;1}$\hfill \psovalbox{D} $\{-1,1\}$ \hfill \psovalbox{E} none of these \item $||\sqrt{2}-2|-2|=$ \\ \pscirclebox{A} $\sqrt{2}$ \hfill \pscirclebox{B} $\sqrt{2}-4$ \hfill \pscirclebox{C} $4-\sqrt{2}$ \hfill \pscirclebox{D} $1+\sqrt{2}$ \hfill \pscirclebox{E} none of these \item If $|x+1|=4$ then\\ \pscirclebox{A} $x\in\{-5,3\}$ \hfill \pscirclebox{B} $x\in\{-4,4\}$ \hfill \pscirclebox{C} $x\in\{-3,5\}$ \hfill \pscirclebox{D} $x\in\{-5,5\}$ \hfill \pscirclebox{E} none of these \item If $-1},linewidth=1.7pt,labels=none,ticks=none](0,0)(-5,-5)(5,5) \pstCircleOA{O}{A} \pscustom[fillstyle=solid,fillcolor=brown]{\pstLineAB{B}{B'}\pstArcOAB{O}{B'}{B}} \pscustom[fillstyle=solid,fillcolor=brown]{\pstLineAB{C}{C'}\pstArcOAB{O}{C}{C'}} \meinecaption{2}{A} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=1pc} \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(2,0){A}(1.732,1){B}(-1.732,1){B'}(1.732,-1){C}(-1.732,-1){C'} \psaxes[arrows={->},linewidth=1.7pt,labels=none,ticks=none](0,0)(-5,-5)(5,5) \pstCircleOA{O}{A} \pscustom[fillstyle=solid,fillcolor=brown]{\pstLineAB{B}{B'}\pstArcOAB{O}{B}{B'}} \pscustom[fillstyle=solid,fillcolor=brown]{\pstLineAB{C}{C'}\pstArcnOAB{O}{C}{C'}} \meinecaption{2}{B} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=1pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none,ticks=none](0,0)(-5,-5)(5,5) \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(2,0){A}(1,1.732){B}(1,-1.732){B'}(-1,1.732){C}(-1,-1.732){C'}(1,0){D} \psaxes[arrows={->},linewidth=1.7pt,labels=none,ticks=none](0,0)(-5,-5)(5,5) \pstCircleOA[fillstyle=solid,fillcolor=brown]{O}{A}\pstCircleOA[fillstyle=solid,fillcolor=white]{O}{D} \meinecaption{2}{C} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=1pc} \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(2,0){A}(1,1.732){B}(1,-1.732){B'}(-1,1.732){C}(-1,-1.732){C'} \psaxes[arrows={->},linewidth=1.7pt,labels=none,ticks=none](0,0)(-5,-5)(5,5) \pstCircleOA{O}{A} \pscustom[fillstyle=solid,fillcolor=brown]{\psline(B)(B')\pstArcnOAB{O}{B'}{C'}\psline(C')(C)\pstArcnOAB{O}{C}{B}} \meinecaption{2}{D} \end{minipage} \end{figure} \vspace{1cm} \pscirclebox{A} A \hfill \pscirclebox{B} B \hfill \pscirclebox{C} C\hfill \pscirclebox{D} D \hfill \pscirclebox{E} none of these \item Which one of the following graphs best represents the set $$ \{(x,y)\in\reals^2: x^2+y^2 \geq 1, \quad (x-1)^2+y^2\leq 1\}\ ? $$Notice that there are four graphs, but five choices. \vspace{3cm} \begin{figure}[hptb] \begin{minipage}{4cm} \centering \psset{unit=1pc} \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(1,0){A} \psaxes[arrows={->},linewidth=1.7pt,labels=none,ticks=none](0,0)(-5,-5)(5,5) \pstCircleOA{O}{A}\pstCircleOA{A}{O}\pstInterCC[PointNameA=none,PointNameB=none]{O}{A}{A}{O}{B}{B'} \pscustom[fillstyle=solid,fillcolor=brown]{\pstArcOAB{O}{B'}{B}\pstArcOAB{A}{B}{B'}} \meinecaption{2}{A} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=1pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none,ticks=none](0,0)(-5,-5)(5,5) \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(1,0){A} \pstCircleOA{O}{A}\pstCircleOA{A}{O}\pstInterCC[PointNameA=none,PointNameB=none]{O}{A}{A}{O}{B}{B'} \pscustom[fillstyle=solid,fillcolor=brown]{\pstArcOAB{O}{B'}{B}\pstArcnOAB{A}{B}{B'}} \meinecaption{2}{B} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=1pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none,ticks=none](0,0)(-5,-5)(5,5) \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(1,0){A} \pstCircleOA{O}{A}\pstCircleOA{A}{O}\pstInterCC[PointNameA=none,PointNameB=none]{O}{A}{A}{O}{B}{B'} \pscustom[fillstyle=solid,fillcolor=brown]{\pstArcnOAB{A}{B'}{B}\pstArcOAB{O}{B}{B'}} \meinecaption{2}{C} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=1pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none,ticks=none](0,0)(-5,-5)(5,5) \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(-1,0){A} \pstCircleOA{O}{A}\pstCircleOA{A}{O}\pstInterCC[PointNameA=none,PointNameB=none]{O}{A}{A}{O}{B}{B'} \pscustom[fillstyle=solid,fillcolor=brown]{\pstArcOAB{O}{B'}{B}\pstArcnOAB{A}{B}{B'}} \meinecaption{2}{D} \end{minipage} \end{figure} \vfill \vspace{1cm} \pscirclebox{A} A \hfill \pscirclebox{B} B \hfill \pscirclebox{C} C\hfill \pscirclebox{D} D \hfill \pscirclebox{E} none of these \clearpage \item Which one of the following graphs best represents the set $$ \{(x,y)\in\reals^2: x^2+y^2 \leq 16, \quad y \geq -x\}\ ? $$Notice that there are four graphs, but five choices. \vfill\vspace{2cm} \vfill \begin{figure}[hptb] \begin{minipage}{4cm} \centering \psset{unit=1pc} \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(4,0){A}(4;-45){C}(4;135){D} \psaxes[arrows={->},linewidth=1.7pt,labels=none,ticks=none](0,0)(-5,-5)(5,5) \pstCircleOA{O}{A} \pscustom[fillstyle=solid,fillcolor=brown]{\psline(D)(C)\pstArcOAB{O}{D}{C}} \meinecaption{2}{A} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=1pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none,ticks=none](0,0)(-5,-5)(5,5) \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(4,0){A}(4;-45){C}(4;135){D} \pstCircleOA{O}{A} \pscustom[fillstyle=solid,fillcolor=brown]{\psline(D)(C)\pstArcOAB{O}{C}{D}} \meinecaption{2}{B} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=1pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none,ticks=none](0,0)(-5,-5)(5,5) \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(4,0){A}(4;45){C}(4;225){D} \pstCircleOA{O}{A} \pscustom[fillstyle=solid,fillcolor=brown]{\psline(D)(C)\pstArcOAB{O}{C}{D}} \meinecaption{2}{C} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=1pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none,ticks=none](0,0)(-5,-5)(5,5) \pstGeonode[PointName=none,PointSymbol=none](0,0){O}(4,0){A}(4;45){C}(4;225){D} \pstCircleOA{O}{A} \pscustom[fillstyle=solid,fillcolor=brown]{\psline(D)(C)\pstArcOAB{O}{D}{C}} \meinecaption{2}{D} \end{minipage} \end{figure} \vspace{1cm} \pscirclebox{A} A \hfill \pscirclebox{B} B \hfill \pscirclebox{C} C\hfill \pscirclebox{D} D \hfill \pscirclebox{E} none of these \item Which of the following graphs represents the set $$ \{(x,y)\in\reals ^2: x^2 + y^2 \leq 4, \quad \absval{x}\geq 1 \} ?$$ \vspace{2cm} \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7.5,7.5) \psgrid[subgriddiv=0,linewidth=.7pt, gridlabels=0](0,0)(-6,-6)(6,6) \pscircle[linewidth=2pt](0,0){2} \pscustom[fillcolor=brown,fillstyle=solid]{\psline(1,1.732050808)(1,-1.732050808)\psarc(0,0){2}{-60}{60}} $$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7.5,7.5) \psgrid[subgriddiv=0,linewidth=.7pt, gridlabels=0](0,0)(-6,-6)(6,6) \pscircle[linewidth=2pt](0,0){2} \pscustom[fillcolor=brown,fillstyle=solid]{\psline(1,1.732050808)(1,-1.732050808)\psarc(0,0){2}{-60}{60}} \pscustom[fillcolor=brown,fillstyle=solid]{\psline(-1,1.732050808)(-1,-1.732050808)\psarc(0,0){2}{120}{240}} $$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7.5,7.5) \psgrid[subgriddiv=0,linewidth=.7pt, gridlabels=0](0,0)(-6,-6)(6,6) \pscircle[linewidth=2pt](0,0){2} \pscustom[fillcolor=brown,fillstyle=solid]{\psline(-1,1.732050808)(-1,-1.732050808)\psarc(0,0){2}{120}{240}} $$ \meinecaption{1}{C} \end{minipage} \hfill \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7.5,7.5) \psgrid[subgriddiv=0,linewidth=.7pt, gridlabels=0](0,0)(-6,-6)(6,6) \pscircle[linewidth=2pt](0,0){2} \pscustom[fillcolor=brown,fillstyle=solid]{\psline(-1.732050808,1)(1.732050808,1)\psarc(0,0){2}{60}{120}} \pscustom[fillcolor=brown,fillstyle=solid]{\psline(1.732050808,-1)(-1.732050808,-1)\psarc(0,0){2}{240}{330}} $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \psovalbox{A} A\hfill \psovalbox{B} B \hfill \psovalbox{C} C\hfill \psovalbox{D} D \hfill \psovalbox{E} none of these \clearpage \item Which of the following graphs represents the set $$ \{(x,y)\in\reals ^2: 0\leq x \leq 2, \ \ \ 3\leq y\leq 4\} ?$$ \vspace{2cm} \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7.5,7.5) \psgrid[subgriddiv=0,linewidth=.7pt, gridlabels=0](0,0)(-6,-6)(6,6) \pscustom[fillcolor=brown,fillstyle=solid]{\psline(3,0)(4,0)(4,2)(3,2)(3,0)} $$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7.5,7.5) \psgrid[subgriddiv=0,linewidth=.7pt, gridlabels=0](0,0)(-6,-6)(6,6) \pscustom[fillcolor=brown,fillstyle=solid]{\psline(-6,3)(0,3)(0,4)(-6,4)(-6,3)} \pscustom[fillcolor=brown,fillstyle=solid]{\psline(6,3)(6,4)(2,4)(2,3)(6,3)} $$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7.5,7.5) \psgrid[subgriddiv=0,linewidth=.7pt, gridlabels=0](0,0)(-6,-6)(6,6) \pscustom[fillcolor=brown,fillstyle=solid]{\psline(0,3)(0,4)(2,4)(2,3)(0,3)} $$ \meinecaption{1}{C} \end{minipage} \hfill \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7.5,7.5) \psgrid[subgriddiv=0,linewidth=.7pt, gridlabels=0](0,0)(-6,-6)(6,6) \pscustom[fillcolor=brown,fillstyle=solid]{\psline(-2,-4)(-2,-3)(3,4)(3,3)(-2,-4)} $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \psovalbox{A} A\hfill \psovalbox{B} B \hfill \psovalbox{C} C\hfill \psovalbox{D} D \hfill \psovalbox{E} none of these \subsection{Lines} \item The lines with equations $ax + by = c$ and $dx + ey = f$ are perpendicular, where $a, b, c, d, e, f$ are non-zero constants. Which of the following must be true? \\ \psovalbox{A} $ad - be = 0$\hfill \psovalbox{B} $ad + be = -1$ \hfill \psovalbox{C}$ae + bd = -1$ \hfill \psovalbox{D} $ae + bd = 0$ \hfill \psovalbox{E} $ad + be = 0 $ \item If $a, b$ are non-zero real constants, find the equation of the line passing through $(a, b)$ and parallel to the line $L: \dfrac{x}{a}-\dfrac{y}{b}=1$.\\ \psovalbox{A} $y = \dfrac{b}{a}x-a$ \hfill \psovalbox{B} $y = -\dfrac{a}{b}x-b$ \hfill \psovalbox{C} $y = \dfrac{a}{b}x+a$\hfill \psovalbox{D} $y = \dfrac{b}{a}x$ \hfill \psovalbox{E} none of these \item If $a, b$ are non-zero real constants, find the equation of the line passing through $(a, b)$ and perpendicular to the line $L: \dfrac{x}{a}-\dfrac{y}{b}=1$.\\ \psovalbox{A} $y = -\dfrac{a}{b}x+b+\dfrac{a^2}{b}$ \hfill \psovalbox{B} $y = -\dfrac{a}{b}x-b$ \hfill \psovalbox{C} $y = \dfrac{a}{b}x+a$\hfill \psovalbox{D} $y = \dfrac{b}{a}x+a$ \hfill \psovalbox{E} none of these \item If the points $(1,1)$, $(2,3)$, and $(4,a)$ are on the same line, find the value of $a$.\\ \pscirclebox{A} $7$ \hfill \pscirclebox{B} $-7$ \hfill \pscirclebox{C} $6$\hfill \pscirclebox{D} $2$ \hfill \pscirclebox{E} none of these \item If the lines $L:\quad ax-2y=c$ and $L':\quad by-x=a$ are parallel, then \\ \psovalbox{A} $\dfrac{a}{2}=\dfrac{1}{b}$\hfill \psovalbox{B} $\dfrac{a}{2}=-\dfrac{1}{b}$ \hfill \psovalbox{C} $\dfrac{a}{2}=b$\hfill \psovalbox{D} $\dfrac{a}{2}=-b$ \hfill \psovalbox{E} none of these \item If the lines $L:\quad ax-2y=c$ and $L':\quad by-x=a$ are perpendicular, then\\ \psovalbox{A} $\dfrac{a}{2}=\dfrac{1}{b}$\hfill \psovalbox{B} $\dfrac{a}{2}=-\dfrac{1}{b}$ \hfill \psovalbox{C} $\dfrac{a}{2}=b$\hfill \psovalbox{D} $\dfrac{a}{2}=-b$ \hfill \psovalbox{E} none of these \item Find the equation of the line parallel to $y=mx+k$ and passing through $(1,1)$. \\ \pscirclebox{A} $y=mx+1$ \hfill \pscirclebox{B} $y=mx+1-m$ \hfill \pscirclebox{C} $y=mx+m-1$\hfill \pscirclebox{D} $y=mx$ \hfill \pscirclebox{E} none of these \clearpage \item Find the equation of the line perpendicular to $y=mx+k$ and passing through $(1,1)$.\\ \pscirclebox{A} $y=-\dfrac{x}{m}-1+\dfrac{1}{m} $ \\ \pscirclebox{B} $y=-\dfrac{x}{m}+1+\dfrac{1}{m} $ \\ \pscirclebox{C} $y=-\dfrac{x}{m}+1-\dfrac{1}{m} $\\ \pscirclebox{D} $y=-\dfrac{x}{m}-1-\dfrac{1}{m} $ \\ \pscirclebox{E} none of these \item[] {\bf Problems \ref{pro:two-points-begin} through \ref{pro:two-points-end} refer to the two points $(a, -a)$ and $(1,1)$.} \item \label{pro:two-points-begin} Find the slope of the line joining $(a, -a)$ and $(1,1)$.\\ \pscirclebox{A} $\dfrac{1-a}{1+a}$ \hfill \pscirclebox{B} $\dfrac{1+a}{1-a}$ \hfill \pscirclebox{C} $\dfrac{1+a}{a-1}$ \hfill \pscirclebox{D} $-1$ \hfill \pscirclebox{E} none of these \item Find the equation of the line passing through $(a, -a)$ and $(1,1)$.\\ \pscirclebox{A} $y = \left(\dfrac{1+a}{1-a}\right)x +\dfrac{2a}{1-a}$ \\ \pscirclebox{B} $y=\left(\dfrac{1+a}{1-a}\right)x$ \\ \pscirclebox{C} $y = \left(\dfrac{1+a}{1-a}\right)x +\dfrac{2a}{a-1}$\\ \pscirclebox{D} $y=\left(\dfrac{a-1}{a+1}\right)x$ \\ \pscirclebox{E} none of these \item Find the equation of the line passing through $(0,0)$ and parallel to the line passing through $(a, -a)$ and $(1,1)$.\\ \pscirclebox{A} $y = \left(\dfrac{1+a}{1-a}\right)x +\dfrac{2a}{1-a}$ \\ \pscirclebox{B} $y=\left(\dfrac{1+a}{1-a}\right)x$ \\ \pscirclebox{C} $y = \left(\dfrac{1+a}{1-a}\right)x +\dfrac{2a}{a-1}$\\ \pscirclebox{D} $y=\left(\dfrac{a-1}{a+1}\right)x$ \\ \pscirclebox{E} none of these \item \label{pro:two-points-end} Find the equation of the line passing through $(0,0)$ and perpendicular to the line passing through $(a, -a)$ and $(1,1)$.\\ \pscirclebox{A} $y = \left(\dfrac{1-a}{1+a}\right)x $ \vfill \pscirclebox{B} $y=\left(\dfrac{1+a}{1-a}\right)x$ \\ \pscirclebox{C} $y = \left(\dfrac{1+a}{1-a}\right)x +\dfrac{2a}{a-1}$\\ \pscirclebox{D} $y=\left(\dfrac{a-1}{a+1}\right)x$ \\ \pscirclebox{E} none of these \item[] Problems \ref{pro:line-with-parameter1} through \ref{pro:line-with-parameter2} refer to the following. For a given real parameter $ u$, consider the family of lines $L_u$ given by $$ L_u: \quad (u+1)y+(u-2)x=u. $$ \item \label{pro:line-with-parameter1} For which value of $u$ is $L_u$ horizontal? \\ \pscirclebox{A} $u=-1$ \hfill \pscirclebox{B} $u=2$ \hfill \pscirclebox{C} $u=\dfrac{1}{3}$\hfill \pscirclebox{D} $u=\dfrac{2}{3}$ \hfill \pscirclebox{E} none of these \item For which value of $u$ is $L_u$ vertical? \\ \pscirclebox{A} $u=-1$ \hfill \pscirclebox{B} $u=2$ \hfill \pscirclebox{C} $u=\dfrac{1}{3}$\hfill \pscirclebox{D} $u=\dfrac{2}{3}$ \hfill \pscirclebox{E} none of these \item For which value of $u$ is $L_u$ parallel to the line $y=2x-1$? \\ \pscirclebox{A} $u=0$ \hfill \pscirclebox{B} $u=2$ \hfill \pscirclebox{C} $u=5$\hfill \pscirclebox{D} $u=\dfrac{2}{3}$ \hfill \pscirclebox{E} none of these \item \label{pro:line-with-parameter2} For which value of $u$ is $L_u$ perpendicular to the line $y=2x-1$? \\ \pscirclebox{A} $u=-5$ \hfill \pscirclebox{B} $u=0$ \hfill \pscirclebox{C} $u=-\dfrac{1}{2}$\hfill \pscirclebox{D} $u=5$ \hfill \pscirclebox{E} none of these \item[] For a real number parameter $u$ consider the line $L_u$ given by the equation $$L_u: \ \ (u - 2)y = (u + 1)x + u. $$ Questions \ref{pro:line_begin} to \ref{pro:line_end} refer to $L_u$. \item \label{pro:line_begin} For which value of $u$ does $L_u$ pass through the point $(-1,1)$?\\ \vfill A) $1$ \hfill B) $-1$ \hfill C) $2$ \hfill D) $3$ \hfill E) none of these \item For which value of $u$ is $L_u$ parallel to the $x$-axis? \\ \vfill A) $-2$\hfill B) $2$ \hfill C) $-1$ \hfill D) $1$\hfill E) none of these \item For which value of $u$ is $L_u$ parallel to the $y$-axis? \\ \vfill A) $-2$\hfill B) $2$ \hfill C) $-1$ \hfill D) $1$\hfill E) none of these \item For which value of $u$ is $L_u$ parallel to the line $2x - y = 2$? \\ \vfill A) $5$\hfill B) $0$ \hfill C) $-3$ \hfill D) $\dfrac{1}{3}$\hfill E) none of these \item For which value of $u$ is $L_u$ perpendicular to the line $2x - y = 2$? \\ \vfill A) $5$\hfill B) $0$ \hfill C) $ \dfrac{1}{3}$ \hfill D) $-\dfrac{1}{3}$\hfill E) none of these \item \label{pro:line_end} Which of the following points is on every line $L_u$ regardless the value of $u$?\\ \vfill A) $(-1,2)$ \hfill B) $(2,-1)$\hfill C) $(\frac{1}{3}, -\frac{2}{3})$ \hfill D) $(-\frac{2}{3}, \frac{1}{3})$ \hfill E) none of these \subsection{Absolute Value Curves} \item[] {\bf Situation:} {\em Problems \ref{pro:absval-begin} and \ref{pro:absval-end} refer to the curve} \mbox{$y = |x-2|+|x+1|$}. \item \label{pro:absval-begin} Write $y=|x-2|+|x+1|$ without absolute values.\\ \psovalbox{A} $y = \left\{\begin{array}{lll}-2x+1 & \mathrm{if}\ x\leq -1 \\ 3 & \mathrm{if}\ -1\leq x\leq 2 \\ 2x-1 & \mathrm{if}\ x\geq 2 \\ \end{array}\right.$ \vfill \psovalbox{B} $y = \left\{\begin{array}{lll}-2x+3 & \mathrm{if}\ x\leq -1 \\ 1 & \mathrm{if}\ -1\leq x\leq 2 \\ 2x-3 & \mathrm{if}\ x\geq 2 \\ \end{array}\right.$ \vfill \psovalbox{C} $y = \left\{\begin{array}{lll}-2x-3 & \mathrm{if}\ x\leq -1 \\ 3 & \mathrm{if}\ -1\leq x\leq 2 \\ 2x+3 & \mathrm{if}\ x\geq 2 \\ \end{array}\right.$ \vfill \psovalbox{D} $y = \left\{\begin{array}{lll}-2x-3 & \mathrm{if}\ x\leq -1 \\ 1 & \mathrm{if}\ -1\leq x\leq 2 \\ 2x+3 & \mathrm{if}\ x\geq 2 \\ \end{array}\right.$ \vfill \psovalbox{E} none of these \item \label{pro:absval-end}Which graph most resembles the curve $y=|x-2|+|x+1|$?\\ \vspace{3cm} \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-5}{4}{abs(x-1)+abs(x+2)} $$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-6}{6}{-abs(x+1)+abs(x-2)} $$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc}\psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-4}{5}{abs(x+1)+abs(x-2)} $$ \meinecaption{1}{C} \end{minipage} \hfill \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-4}{5}{abs(x+1)+abs(x-2)-5} $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \vfill \psovalbox{A} \hfill \psovalbox{B} \hfill \psovalbox{C} \hfill \psovalbox{D} \hfill \psovalbox{E} none of these \clearpage \item Which graph most resembles the curve $y=|x-2|-|x+1|$?\\ \vspace{2cm} \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-6}{6}{abs(x+1)-abs(x-2)} $$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-6}{6}{-abs(x+1)+abs(x-2)} $$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc}\psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-4}{5}{abs(x+1)+abs(x-2)} $$ \meinecaption{1}{C} \end{minipage} \hfill \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-4}{5}{5-abs(x+1)-abs(x-2)} $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \psovalbox{A} \hfill \psovalbox{B} \hfill \psovalbox{C} \hfill \psovalbox{D} \hfill \psovalbox{E} none of these \subsection{Circles and Semicircles} \item The point $A(1,2)$ lies on the circle ${\cal C}: (x+1)^2+(y-1)^2=5$. Which of the following points is diametrically opposite to $A$ on ${\cal C}$? \vfill \pscirclebox{A} $(-1,-2)$ \hfill \pscirclebox{B} $(-3,0)$ \hfill \pscirclebox{C} $(0,3)$\hfill \pscirclebox{D} $(0,\sqrt{5}+1)$ \hfill \pscirclebox{E} none of these \item A circle has a diameter with endpoints at $(-2,3)$ and $(6,5)$. Find its equation. \vfill \pscirclebox{A} $\left(x+2\right)^2+\left(y-3\right)^2=68$ \vfill \pscirclebox{B} $\left(x-4\right)^2+\left(y-8\right)^2=61$ \vfill \pscirclebox{C} $\left(x-2\right)^2+\left(y-4\right)^2=17$\vfill \pscirclebox{D} $\left(x-2\right)^2+\left(y-4\right)^2=\sqrt{17}$ \vfill \pscirclebox{E} none of these \item Which figure represents the circle with equation $$ x^2-2x+y^2+6y=-6 \ ?$$ Again, notice that there are four figures, but five choices. \\ \vspace{2cm} \begin{figure}[hptb] \begin{minipage}{4cm} \centering \psset{unit=.5pc} \pstGeonode[PointName=none](0,0){O}(1,-3){T} \psaxes[arrows={->},linewidth=1.7pt,labels=none](0,0)(-10,-10)(10,10) \rput(T){\pscircle[linewidth=2pt,linecolor=brown](0,0){2}\psdots(2,0)(-2,0)(0,2)(0,-2)} \meinecaption{2}{A} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=.5pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none](0,0)(-10,-10)(10,10) \pstGeonode[PointName=none](-2,6){T} \rput(T){\pscircle[linewidth=2pt,linecolor=brown](0,0){2}\psdots(2,0)(-2,0)(0,2)(0,-2)} \meinecaption{2}{B} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=.5pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none](0,0)(-10,-10)(10,10) \pstGeonode[PointName=none](-1,3){T} \rput(T){\pscircle[linewidth=2pt,linecolor=brown](0,0){2}\psdots(2,0)(-2,0)(0,2)(0,-2)}\meinecaption{2}{C} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=.5pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none](0,0)(-10,-10)(10,10) \pstGeonode[PointName=none](1,3){T} \rput(T){\pscircle[linewidth=2pt,linecolor=brown](0,0){2}\psdots(2,0)(-2,0)(0,2)(0,-2)} \meinecaption{2}{D} \end{minipage} \end{figure} \vspace{1cm} \pscirclebox{A} A \hfill \pscirclebox{B} B \hfill \pscirclebox{C} C\hfill \pscirclebox{D} D \hfill \pscirclebox{E} none of these \item Which figure represents the semicircle with equation $$ x = 1-\sqrt{-y^2-6y-5}?$$ Again, notice that there are four figures, but five choices.\\ \vspace{2cm} \begin{figure}[hptb] \begin{minipage}{4cm} \centering \psset{unit=.5pc} \pstGeonode[PointName=none](0,0){O}(1,-3){T} \psarc[linecolor=brown,linewidth=2pt](T){2}{-90}{90} \psaxes[arrows={->},linewidth=1.7pt,labels=none](0,0)(-10,-10)(10,10) \meinecaption{2}{A} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=.5pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none](0,0)(-10,-10)(10,10) \pstGeonode[PointName=none](-1,3){T} \psarc[linecolor=brown,linewidth=2pt](T){2}{-90}{90} \meinecaption{2}{B} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=.5pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none](0,0)(-10,-10)(10,10) \pstGeonode[PointName=none](1,-3){T} \psarc[linecolor=brown,linewidth=2pt](T){2}{90}{-90} \meinecaption{2}{C} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=.5pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none](0,0)(-10,-10)(10,10) \pstGeonode[PointName=none](-1,3){T} \psarc[linecolor=brown,linewidth=2pt](T){2}{90}{-90} \meinecaption{2}{D} \end{minipage} \end{figure} \vspace{1cm} \pscirclebox{A} A \hfill \pscirclebox{B} B \hfill \pscirclebox{C} C\hfill \pscirclebox{D} D \hfill \pscirclebox{E} none of these \item Find the equation of the circle with centre at $(-1,2)$ and passing through $(0,1)$.\\ \psovalbox{A} $(x-1)^2+(y+2)^2=10$ \\ \psovalbox{B} $(x+1)^2+(y-2)^2=2$ \\ \psovalbox{C} $(x+1)^2+(y-2)^2=10$ \\ \psovalbox{D} $(x-1)^2+(y+2)^2=2$ \\ \psovalbox{E} none of these \item Let $a$ and $b$ be real constants. Find the centre and the radius of the circle with equation $$x^2+2ax+y^2-4by=1. $$ \\ \pscirclebox{A} Centre: $(-a,2b)$, Radius: $\sqrt{a^2+4b^2}$ \\ \pscirclebox{B} Centre: $(a,2b)$, Radius: $\sqrt{1+a^2+4b^2}$ \\ \pscirclebox{C} Centre: $(a,-2b)$, Radius: $\sqrt{1+a^2+4b^2}$\\ \pscirclebox{D} Centre: $(-a,2b)$, Radius: $\sqrt{1+a^2+4b^2}$ \\ \pscirclebox{E} none of these \item A circle has a diameter with endpoints $A(b,-a)$ and $B(-b,a)$. Find its equation.\\ \psovalbox{A} $(x-b)^2+(y+a)^2=a^2+b^2$\\ \psovalbox{B} $(x-b)^2+(y-a)^2=a^2+b^2$ \\ \psovalbox{C} $x^2+y^2=a^2+b^2$\\ \psovalbox{D} $x^2+y^2=\sqrt{a^2+b^2}$ \\ \psovalbox{E} none of these \clearpage \item Find the centre $C$ and the radius $R$ of the circle with equation $x^2+y^2=2ax-b$. \\ \psovalbox{A} $C(0,0), R=\sqrt{2a-b}$ \\ \psovalbox{B} $C\left(a, -\dfrac{b}{2}\right), R=\sqrt{a^2+\frac{b^2}{4}}$ \\ \psovalbox{C} $C(-a,0), R=\sqrt{a^2-b}$ \\ \psovalbox{D} $C(a,0), R=\sqrt{a^2-b}$ \\ \psovalbox{E} none of these \subsection{Functions: Definition} \item Which one of the the following represents a function?\\ \vspace{2.5cm} \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \rput{-90}(-1,1){\parabola[linewidth=1.9pt, linecolor=brown]{<->}(3,9)(0,0)}$$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \rput{-45}(-1,1){\parabola[linewidth=1.9pt, linecolor=brown]{<->}(3,9)(0,0)}$$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \rput(-1,3){\parabola[linewidth=1.9pt, linecolor=brown]{<->}(3,-9)(0,0)} $$ \meinecaption{1}{C} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \pscircle[linewidth=1.9pt,linecolor=brown](-2,2){2} $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \vspace{.5cm} \psovalbox{A}A\hfill\psovalbox{B} B \hfill\psovalbox{C} C\hfill\psovalbox{D} D \hfill\psovalbox{E} none of these \item How many functions are there from the set $\{a, b, c\}$ to the set $\{1,2\}$? \\ \psovalbox{A} $ 9 $ \hfill \psovalbox{B} $ 8 $ \hfill \psovalbox{C} $ 6 $ \hfill \psovalbox{D} $ 1$ \hfill \psovalbox{E} none of these \subsection{Evaluation of Formul\ae} \item[] Figure \ref{fig:functional_curve1} shews a functional curve $y =f(x)$, and refers to problems \ref{pro:func_curve1_begin1} to \ref{pro:func_curve1_end1}.\\ \vspace{2.3cm} \begin{center}\begin{figure}[h] $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5)\psgrid[subgriddiv=0,linewidth=.7pt, gridlabels=0](0,0)(-6,-6)(6,6) \psline[linewidth=2pt,linecolor=brown]{*-o}(-5,-5)(-1,-3)\psline[linewidth=2pt,linecolor=brown]{o-*}(2,2)(5,5) $$ \meinecaption{1}{Problems \ref{pro:func_curve1_begin1} to \ref{pro:func_curve1_end1}.} \label{fig:functional_curve1} \end{figure}\end{center} \item \label{pro:func_curve1_begin1}The domain of the functional curve in figure \ref{fig:functional_curve1} is \\ \psovalbox{A}$[-5;5]$ \hfill\psovalbox{B} $[-5;-1[\cup ]2;5]$ \hfill\psovalbox{C} $[-5;-1]\cup [2;5]$\hfill\psovalbox{D} $[-5;-1[\cup [2;5[$ \hfill\psovalbox{E} none of these \item The image of the functional curve in figure \ref{fig:functional_curve1} is \\ \psovalbox{A}$[-5;5]$ \hfill\psovalbox{B} $[-5;-3] \cup ]2;5]$ \hfill\psovalbox{C} $[-5;-3[ \cup ]2;5[$ \hfill\psovalbox{D} $[-5;-3[ \cup ]2;5[$ \hfill\psovalbox{E} none of these \hfill \\ \item $f(3) =$ \\ \psovalbox{A}$1$ \hfill\psovalbox{B} $2$ \hfill\psovalbox{C} $3$ \hfill\psovalbox{D} $5$ \hfill\psovalbox{E} none of these \\ \item \label{pro:func_curve1_end1} $f$ \ \ \ is \\ \psovalbox{A}an even function \hfill\psovalbox{B} increasing \hfill\psovalbox{C} an odd function \hfill\psovalbox{D} decreasing \hfill\psovalbox{E} none of these \item[] Problems \ref{pro:funtzi-1} through \ref{pro:funtzi-1} refer to the functional curve in figure \ref{fig:funtzi}. \vspace{3cm} \begin{figure}[h!] $$\psset{unit=.75pc} \psgrid[subgriddiv=0,gridlabels=.5,gridcolor=darkgray](-10,-10)(10,10) \psline[linewidth=1.5pt](0,-10)(0,10)\psline[linewidth=1.5pt](-10, 0 )(10, 0) \psline[linewidth=1.6pt](-7,-3)(-4,-3)(-2,2)(2,5)(5,4)\psdots[dotstyle=o,dotscale=1.5](-2, 2)\psdots[dotstyle=*,dotscale=1.5](-7, -3)(-4,-3)(2,5)(5,4) $$ \meinecaption{3}{Problems \ref{pro:funtzi-1} through \ref{pro:funtzi-1}.} \label{fig:funtzi}\end{figure} \item \label{pro:funtzi-1}The domain of the function $f$ is \\ \psovalbox{A} $[-7; 5]$ \hfill \psovalbox{B} $[-7; -2[ \cup ]-2;5]$\hfill (C) $]-7;5[$\hfill (D) $]-7; 5]$\hfill (E) none of these \item The image of the function $f$ is \\ \psovalbox{A} $[-3; 4]$ \hfill \psovalbox{B} $[-3; 4] \setminus \{2\}$\hfill \psovalbox{C} $[-3; 5]$\hfill (D) $[-3;2[\cup ]2;5]$\hfill (\psovalbox{E} none of these \item $f(2) = $\\ \psovalbox{A} $2$\hfill \psovalbox{B} $3$ \hfill \psovalbox{C} $4$\hfill \psovalbox{D} $5$\hfill \psovalbox{E} none of these \item \label{pro:funtzi-2} $f(-2) = $ \\ \psovalbox{A} $2$\hfill \psovalbox{B} $3$\hfill \psovalbox{C} $5$\hfill \psovalbox{D} undefined\hfill \psovalbox{E} none of these \item Let $f(x)= 1+x+x^2$. What is $f(0)+f(1)+f(2)$? \\ \pscirclebox{A} $10$ \hfill \pscirclebox{B} $11$ \hfill \pscirclebox{C} $7$ \hfill \pscirclebox{D} $3$ \hfill \pscirclebox{E} none of these \item Let $f:\reals \rightarrow \reals$ with the assignment rule $x\mapsto (x-(x-(x-1)^2)^2)^2$. Find $f(2)$.\\ \psovalbox{A} $1$\hfill \psovalbox{B} $4$ \hfill \psovalbox{C} $16$\hfill \psovalbox{D} $0$ \hfill \psovalbox{E} none of these \item Let $f(x)=\dfrac{x-1}{x+1}$. Find $f(2)$. \\ \pscirclebox{A} $0$ \hfill \pscirclebox{B} $\dfrac{1}{3}$ \hfill \pscirclebox{C} $\dfrac{2}{3}$\hfill \pscirclebox{D} $\dfrac{1}{2}$ \hfill \pscirclebox{E} none of these \item Consider a function $f:\reals \to \reals$ such that $f\left(\dfrac{x}{3}\right) = 9x$. Find $f(x)$. \\ \pscirclebox{A} $3x$ \hfill \pscirclebox{B} $\dfrac{x}{3}$ \hfill \pscirclebox{C} $\dfrac{x}{9}$\hfill \pscirclebox{D} $27x$ \hfill \pscirclebox{E} none of these \item Consider $f(x)=\dfrac{1}{x}$, for $x\neq 0$. How many of the following assertions are necessarily true? $$I: f(ab)=f(a)f(b), \quad II: f\left(\dfrac{a}{b}\right)=\dfrac{f(a)}{f(b)}, \quad III: f(a+b) =f(a)+f(b), \quad IV: f\left(\dfrac{1}{a}\right)=\dfrac{1}{f(a)} $$ \pscirclebox{A} exactly one \hfill \pscirclebox{B} exactly two \hfill \pscirclebox{C} exactly three \hfill \pscirclebox{D} all four \hfill \pscirclebox{E} none of them \subsection{Algebra of Functions} \item Let $f(x)=2x+1$. Find $(f\circ f\circ f)(1)$.\\ \vfill \psovalbox{A} $8$\hfill \psovalbox{B} $3$ \hfill \psovalbox{C} $9$ \hfill \psovalbox{D} $15$ \hfill \psovalbox{E} none of these \item Let $f(x)=x-2$ and $g(x)=2x+1$. Find $$ (f\circ g)(1) +(g\circ f)(1).$$ \pscirclebox{A} $-1$ \hfill \pscirclebox{B} $1$ \hfill \pscirclebox{C} $0$\hfill \pscirclebox{D} $2$ \hfill \pscirclebox{E} none of these \item Let $f:\reals \to \reals$ be such that $f(2x-1)=x+1$. Find $f(-3)$.\\ \psovalbox{A} $-2$\hfill \psovalbox{B} $1$ \hfill \psovalbox{C} $-1$\hfill \psovalbox{D} $0$ \hfill \psovalbox{E} none of these \item Let $f(x)=x+1$. What is $\underbrace{(f\circ \cdots \circ f)}_{100\ f\mathrm{'s}}(x)$? \\ \psovalbox{A} $x+100$\hfill \psovalbox{B} $x^{100}+1$ \hfill \psovalbox{C} $x^{100}+100$ \hfill \psovalbox{D} $x+99$\hfill \psovalbox{E} none of these \item Let $f:\reals \to \reals$ satisfy $f(1-x)=x-2$. Find $f(x)$. \\ \pscirclebox{A} $-1-x$ \hfill \pscirclebox{B} $x+1$ \hfill \pscirclebox{C} $x-1$\hfill \pscirclebox{D} $1-x$ \hfill \pscirclebox{E} none of these \item[] Questions \ref{pro:func1a} through \ref{pro:func1b} refer to the assignment rules given by $f(x)=\dfrac{x}{x-1}$ and $g(x) = 1-x$. \item \label{pro:func1a} Determine $(f\circ g)(2)$. \\ \psovalbox{A} $0$\hfill \psovalbox{B} $-2$\hfill \psovalbox{C} $-1$\hfill \psovalbox{D} $\dfrac{1}{2}$\hfill \psovalbox{E} none of these \item Determine $(g\circ f)(2)$. \\ \psovalbox{A} $0$\hfill \psovalbox{B} $-2$\hfill \psovalbox{C} $-1$\hfill \psovalbox{D} $\dfrac{1}{2}$\hfill \psovalbox{E} none of these \item Determine $(gf)(2)$. \\ \psovalbox{A} $0$\hfill \psovalbox{B} $-2$\hfill \psovalbox{C} $-1$\hfill \psovalbox{D} $\dfrac{1}{2}$\hfill \psovalbox{E} none of these \item Determine $(g+f)(2)$. \\ \psovalbox{A} $1$\hfill \psovalbox{B} $-2$\hfill \psovalbox{C} $-1$\hfill \psovalbox{D} $\dfrac{1}{2}$\hfill \psovalbox{E} none of these \item \label{pro:func1b} If $(f+g)(x)=(g\circ f)(x)$ then $x\in$ \\ \psovalbox{A} $\{-1,1\}$\hfill \psovalbox{B} $\{-3,0\}$\hfill \psovalbox{C} $\{-3,3\}$\hfill \psovalbox{D} $\{0,3\}$\hfill \psovalbox{E} none of these \item[] Problems \ref{pro:composi-a} through \ref{pro:composi-b} refer to the functions $f$ and $g$ with $$f(x) = \dfrac{2}{2-x}, \qquad g(x) = \dfrac{x-2}{x-1}, \qquad h(x) = \dfrac{2x-2}{x}.$$ \item \label{pro:composi-a} $f(-1)=$ \\ \psovalbox{A} $4$\hfill \psovalbox{B} $\dfrac{2}{3}$ \hfill \psovalbox{C} $1$ \hfill \psovalbox{D} $\dfrac{3}{2}$\hfill \psovalbox{E} none of these \item Find $(fgh)(-1)$.\\ \psovalbox{A} $\dfrac{37}{6}$\hfill \psovalbox{B} $\dfrac{2}{3}$ \hfill \psovalbox{C} $\dfrac{3}{2}$ \hfill \psovalbox{D} $4$ \hfill \psovalbox{E} none of these \item Find $(f+g+h)(-1)$. \\ \psovalbox{A} $\dfrac{37}{6}$\hfill \psovalbox{B} $\dfrac{2}{3}$ \hfill \psovalbox{C} $\dfrac{3}{2}$ \hfill \psovalbox{D} $4$ \hfill \psovalbox{E} none of these \item $(f\circ g)(x)=$ \\ \psovalbox{A} $f(x)$\hfill \psovalbox{B} $g(x)$ \hfill \psovalbox{C} $h(x)$ \hfill \psovalbox{D} $x$\hfill \psovalbox{E} none of these \item $(g\circ h)(x)=$ \\ \psovalbox{A} $f(x)$\hfill \psovalbox{B} $g(x)$ \hfill \psovalbox{C} $h(x)$ \hfill \psovalbox{D} $x$\hfill \psovalbox{E} none of these \item \label{pro:composi-b} $(h\circ f)(x)=$ \\ \psovalbox{A} $f(x)$\hfill \psovalbox{B} $g(x)$ \hfill \psovalbox{C} $h(x)$ \hfill \psovalbox{D} $x$\hfill \psovalbox{E} none of these \item[] Problems \ref{pro:composi-a} through \ref{pro:composi-b} refer to the functions $f$ and $g$ with $f(x) = \sqrt{x^2+1}$ and $g(x) = \sqrt{x^2-1}$. \item \label{pro:composi-a} Find $(fg)(2)$. \\ \psovalbox{A} $4$\hfill \psovalbox{B} $2$ \hfill \psovalbox{C} $\sqrt{5}+\sqrt{3}$ \hfill \psovalbox{D} $\sqrt{15}$\hfill \psovalbox{E} none of these % \item Find $(f+g)(2)$. \\ \psovalbox{A} $4$\hfill \psovalbox{B} $2$ \hfill \psovalbox{C} $\sqrt{5}+\sqrt{3}$ \hfill \psovalbox{D} $\sqrt{15}$\hfill \psovalbox{E} none of these % \item Find $(f\circ g)(2)$. \\ \psovalbox{A} $4$\hfill \psovalbox{B} $2$ \hfill \psovalbox{C} $\sqrt{5}+\sqrt{3}$ \hfill \psovalbox{D} $\sqrt{15}$\hfill \psovalbox{E} none of these % \item Find $(g\circ f)(2)$. \\ \psovalbox{A} $4$\hfill \psovalbox{B} $2$ \hfill \psovalbox{C} $\sqrt{5}+\sqrt{3}$ \hfill \psovalbox{D} $\sqrt{15}$\hfill \psovalbox{E} none of these % \item \label{pro:composi-b} Find $(g\circ f\circ g \circ f \circ g \circ f \circ g \circ f)(2)$. \\ \psovalbox{A} $4$\hfill \psovalbox{B} $2$ \hfill \psovalbox{C} $\sqrt{3}$ \hfill \psovalbox{D} $\sqrt{5}$\hfill \psovalbox{E} none of these % \item A function $f:\reals\rightarrow \reals$ satisfies $f(2x)=x^2$. Find $(f\circ f)(x)$. \\ \psovalbox{A} $x^4$ \hfill \psovalbox{B} $\dfrac{x^4}{4}$ \hfill \psovalbox{C} $\dfrac{x^4}{16}$\hfill \psovalbox{D} $\dfrac{x^4}{64}$ \hfill \psovalbox{E} none of these \subsection{Domain of Definition of a Formula} \item What is the natural domain of definition of the assignment rule $x\mapsto \dfrac{\sqrt{x^2-1}}{|x|-1}$?\\ \psovalbox{A}$[-1;1]$ \hfill\psovalbox{B} $]-\infty; -1] \cup [1;+\infty[$\hfill\psovalbox{C} $]-\infty; -1[ \cup ]1;+\infty[ $\hfill\psovalbox{D} $\reals \setminus \{\pm 1\}$ \hfill\psovalbox{E} none of these \item What is the natural domain of definition of the assignment rule $x\mapsto \dfrac{\sqrt{x-2}}{x^3-8}$?\\ \psovalbox{A}$]2;+\infty[$\hfill\psovalbox{B} $\reals \setminus \{2\}$ \hfill\psovalbox{C} $]-\infty; -2[ $\hfill\psovalbox{D} $[2;+\infty[$ \hfill\psovalbox{E} none of these \clearpage \item[] Questions \ref{pro:dom-def1} through \ref{pro:dom-def2} are related. \item \label{pro:dom-def1} Consider the assignment rule $x\mapsto \dfrac{1+x}{1-x}$. Find its domain of definition. \vfill \pscirclebox{A} $\reals \setminus \{1\}$ \vfill \pscirclebox{B} $\co{-1;1}$ \vfill \pscirclebox{C} $\reals \setminus \{-1,1\}$ \vfill \pscirclebox{D} $\reals \setminus \{-1\}$ \vfill \pscirclebox{E} none of these \item Consider the assignment rule $x\mapsto \sqrt{\dfrac{1+x}{1-x}}$. Find its domain of definition. \vfill \pscirclebox{A} $\oo{-\infty; -1}\ \cup\ \oo{1;+\infty}$ \vfill \pscirclebox{B} $\co{-1;1}$ \vfill \pscirclebox{C} $\oc{-\infty; -1}\ \cup\ \oo{1;+\infty}$ \vfill \pscirclebox{D} $\cc{-1;1}$ \vfill \pscirclebox{E} none of these \item Consider the assignment rule $x\mapsto \sqrt{1+x}+\sqrt{1-x}$. Find its domain of definition. \vfill \pscirclebox{A} $\oo{-\infty; -1}\ \cup\ \oo{1;+\infty}$ \vfill \pscirclebox{B} $\co{-1;1}$ \vfill \pscirclebox{C} $\oc{-\infty; -1}\ \cup\ \oo{1;+\infty}$ \vfill \pscirclebox{D} $\cc{-1;1}$ \vfill \pscirclebox{E} none of these \item \label{pro:dom-def2}Consider the assignment rule $x\mapsto \sqrt{\dfrac{1+x}{1-x}-1}$. Find its domain of definition. \vfill \pscirclebox{A} $\oo{0;1}$ \vfill \pscirclebox{B} $\cc{0;1}$ \vfill \pscirclebox{C} $\co{-1;1}$ \vfill \pscirclebox{D} $\co{0;1}$ \vfill \pscirclebox{E} none of these \item What is the domain of definition of the formula $x\mapsto \sqrt{1-x^2}$\ ? \vfill \pscirclebox{A} $\cc{-1;1}$ \hfill \pscirclebox{B} $\oc{-\infty;-1}$ \hfill \pscirclebox{C} $\oc{-\infty;1}$\hfill \pscirclebox{D} $\co{1;+\infty}$ \hfill \pscirclebox{E} none of these \item Find the natural domain of definition of $x\mapsto \sqrt{-x} + \sqrt{1+x}$.\\ \psovalbox{A} $[-1;0]$ \hfill \psovalbox{B} $[0;1]$ \hfill \psovalbox{C} $[-1;1]$ \hfill \psovalbox{D} $\reals \setminus [-1;1]$ \hfill \psovalbox{E} none of these \item Find the natural domain of definition of $x \mapsto \sqrt{\dfrac{x}{x^2-x-6}}$.\\ \psovalbox{A} $[-2;3]$ \hfill \psovalbox{B} $[-2;0[ \cup [3;+\infty[$ \hfill \psovalbox{C} $]-2;0] \cup ]3;+\infty[$ \hfill \psovalbox{D} $]-3;+\infty[$ \hfill \psovalbox{E} none of these \subsection{Piecewise-defined Functions} \item Which one most resembles the graph of $y = f(x) = \left\{ \begin{array}{ll} \dfrac{1}{x}+1 & \mathrm{if}\ x\in ]-\infty; -1] \\ 1-x^2 & \mathrm{if}\ x\in ]-1;1[ \\ \dfrac{1}{x}-1 & [1;+\infty[ \end{array} \right. ?$\\ \vspace{2cm} \begin{figure}[!h] \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psplot[linecolor=blue,linewidth=1.4pt]{-6}{-1}{2-2/x^2} \psplot[linecolor=brown,linewidth=1.4pt]{-1}{1}{1-x^2} \psplot[linecolor=blue,linewidth=1.4pt]{1}{6}{2-2/x^2} $$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psplot[linecolor=blue,linewidth=1.4pt]{-6}{-1}{1/x+1} \psplot[linecolor=brown,linewidth=1.4pt]{-1}{1}{1-x^2} \psplot[linecolor=blue,linewidth=1.4pt]{1}{6}{1/x-1}$$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psplot[linecolor=blue,linewidth=1.4pt]{-6}{6}{abs(1-abs(1-abs(x)))} $$ \meinecaption{1}{C} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psplot[linecolor=blue,linewidth=1.4pt]{-6}{-2}{2/x-2} \psplot[linecolor=brown,linewidth=1.4pt]{-2}{2}{1-x^2} \psplot[linecolor=blue,linewidth=1.4pt]{2}{6}{2/x-4} $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \psovalbox{A} A\hfill\psovalbox{B} B \hfill\psovalbox{C} C\hfill\psovalbox{D} D \hfill\psovalbox{E} none of these \item Which one most resembles the graph of \ \ \ $y = f(x) = \left\{ \begin{array}{ll} (x+3)^2-5 & \mathrm{if}\ x\in ]-\infty; -1] \\ x^3 & \mathrm{if}\ x\in ]-1;1[ \\ 5-(x-3)^2 & [1;+\infty[ \end{array} \right. ?$\\ \vspace{2cm} \begin{figure}[h] \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psplot[linecolor=blue,linewidth=1.4pt]{-6}{-1}{1/x} \psplot[linecolor=brown,linewidth=1.4pt]{-1}{1}{x^3} \psplot[linecolor=blue,linewidth=1.4pt]{1}{6}{1/x} $$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psplot[linecolor=blue,linewidth=1.4pt]{-6}{-1}{x} \psplot[linecolor=brown,linewidth=1.4pt]{-1}{1}{x^3} \psplot[linecolor=blue,linewidth=1.4pt]{1}{6}{x}$$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psplot[linecolor=blue,linewidth=1.4pt]{-6}{-1}{(x+3)^2-5} \psplot[linecolor=brown,linewidth=1.4pt]{-1}{1}{x^3} \psplot[linecolor=blue,linewidth=1.4pt]{1}{6}{5-(x-3)^2} $$ \meinecaption{1}{C} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psplot[linecolor=blue,linewidth=1.4pt]{-6}{-1}{(x+3)^2-5} \psplot[linecolor=brown,linewidth=1.4pt]{-1}{1}{x} \psplot[linecolor=blue,linewidth=1.4pt]{1}{6}{5-(x-3)^2} $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \psovalbox{A}A\hfill\psovalbox{B} B \hfill\psovalbox{C} C\hfill\psovalbox{D} D \hfill\psovalbox{E} none of these \subsection{Parity of Functions} \item Which one of the following functions $f:\reals \rightarrow \reals $ with the assignment rules given below, represents an even function?\\ \psovalbox{A} $f(x)=x\absval{x}$\hfill \psovalbox{B} $f(x)=\absval{x-x^2}$ \hfill \psovalbox{C} $f(x)=x^2 - x^4 + 1-x$\hfill \psovalbox{D} $f(x)=\absval{x}^3$ \hfill \psovalbox{E} none of these \item How many of the following are assignment rules of even functions? $$I: a(x)=|x|^3, \qquad II: b(x) = x^2|x|, \qquad III: c(x)=x^3-x, \qquad IV: d(x)=|x+1| $$ \pscirclebox{A} exactly one \hfill \pscirclebox{B} exactly two \hfill \pscirclebox{C} exactly three \hfill \pscirclebox{D} all four \hfill \pscirclebox{E} none \item Let $f$ be an odd function and let $g$ be an even function, both with the same domain. How many of the following functions are necessarily even? $$I: x\mapsto f(x)g(x) \qquad II: x\mapsto f(x)+g(x) \qquad III: x\mapsto (f(x))^2+(g(x))^2 \qquad IV: x\mapsto f(x)|g(x)|$$ \pscirclebox{A} exactly one \hfill \pscirclebox{B} exactly two \hfill \pscirclebox{C} exactly three \hfill \pscirclebox{D} all four \hfill \pscirclebox{E} none of them \item Let $f$ be an even function and let $g$ be an odd function, with $f(2)=3$ and $g(2)=5$. Find the value of \mbox{$f(-2)+g(-2)+(fg)(-2)$}.\\ \psovalbox{A} $-17$\hfill \psovalbox{B} $23$\hfill \psovalbox{C} $13$\hfill \psovalbox{D} $7$\hfill \psovalbox{E} none of these \item Let $f$ be an even function and let $g$ be an odd function, both defined over all reals. How many of the following functions are necessarily even? $$I: x\mapsto (f+g)(x)\quad II: x\mapsto (f\circ g)(x) \quad III: x\mapsto (g\circ f)(x) \quad IV: x\mapsto |f(x)|+|g(x)| $$ \psovalbox{A} none\hfill \psovalbox{B} exactly one\hfill \psovalbox{C} exactly two\hfill \psovalbox{D} exactly three\hfill \psovalbox{E} all four \item Let $f$ be an odd function defined over all real numbers. How many of the following are necessarily even? \\ $$I: 2f; \qquad II: |f|; \qquad III: f^2; \qquad IV: f\circ f. $$ \\ \psovalbox{A} Exactly one \hfill \psovalbox{B} Exactly two \hfill \psovalbox{C} Exactly three \hfill \psovalbox{D} All four\hfill \psovalbox{E} none is even \item Let $f$ be an odd function such that $f(-a)=b$ and let $g$ be an even function such that $g(c)=a$. What is $(f\circ g)(-c)$? \\ \psovalbox{A} $b$\hfill \psovalbox{B} $-b$ \hfill \psovalbox{C} $-a$ \hfill \psovalbox{D} $a$\hfill \psovalbox{E} none of these % \subsection{Transformations of Graphs} \item The curve $y = \dfrac{x-1}{x+1}$ experiences the following successive transformations: (1) a reflexion about the $y$ axis, (2) a translation $1$ unit down, (3) a reflexion about the $x$-axis. Find the equation of the resulting curve.\\ \pscirclebox{A} $y=\dfrac{2}{1-x}$ \hfill \pscirclebox{B} $y=\dfrac{x}{2-x}$ \hfill \pscirclebox{C} $y=\dfrac{2}{x-1}$\hfill \pscirclebox{D} $y=\dfrac{x-2}{x}$ \hfill \pscirclebox{E} none of these \item What is the equation of the resulting curve after $y=x^2-x$ has been, successively, translated one unit up and reflected about the $y$-axis? \\ \psovalbox{A} $y= x^2-x+1$\hfill \psovalbox{B} $y = x^2+x+1$ \hfill \psovalbox{C} $y=-x^2+x-1$\hfill \psovalbox{D} $y = (x+1)^2-x-1$ \hfill \psovalbox{E} none of these \item What is the equation of the curve symmetric to the curve $y = \dfrac{1}{x^3}+1$ with respect to the line $y=0$ ?\\ \psovalbox{A} $y = -\dfrac{1}{x^3}+1$ \hfill \psovalbox{B} $y = -\dfrac{1}{x^3}-1$\hfill \psovalbox{C}$y = \dfrac{1}{(x-1)^3}$ \hfill \psovalbox{D} $y = \dfrac{1}{(x-1)^{1/3}}$ \hfill \psovalbox{E} $y = \dfrac{1}{(1-x)^{1/3}}$ \item What is the equation of the resulting curve after the curve $y=x|x+1|$ has been successively translated one unit right and reflected about the $y$-axis? \\ \pscirclebox{A} $y=(x-1)|x|$ \hfill \pscirclebox{B} $y=-(x+1)|x|$ \hfill \pscirclebox{C} $y=-x|x|$\hfill \pscirclebox{D} $y=-x|x|-1$ \hfill \pscirclebox{E} none of these \item The curve $y = |x|+x$ undergoes the following successive transformations: (1) a translation $1$ unit down, (2) a reflexion about the $y$-axis, (3) a translation $2$ units right. Find the equation of the resulting curve.\\ \psovalbox{A} $y = |x-2|-x+1$ \hfill \psovalbox{B} $y = |x-2|-x-1$ \hfill \psovalbox{C} $y = |x+2|-x-3$ \hfill \psovalbox{D} $y = |x-2|+x-1$ \hfill \psovalbox{E} none of these \clearpage \item[] There are six graphs shewn below. The first graph is that of the original curve $y=f(x)$, and the other five are various transformations of the original graph. You are to match each graph letter below with the appropriate equation in \ref{pro:transform-a1} through \ref{pro:transform-b1} below. \vspace{2cm} \begin{figure}[!h] \begin{minipage}[c]{4cm} \centering\psset{unit=.5pc} \psaxes[linewidth=2pt, labels=none,ticks=none,arrows={->}](0,0)(-5,-5)(5,5) \psline[linecolor=blue,linewidth=2pt](-5,3)(-3,3)(0,0)(2,-2)(4,2) \meinecaption{1}{$y=f(x)$.} \end{minipage} \hfill \begin{minipage}[c]{4cm} \centering\psset{unit=.5pc} \psaxes[linewidth=2pt, labels=none,ticks=none,arrows={->}](0,0)(-5,-5)(5,5) \psline[linecolor=blue,linewidth=2pt](-4,2)(-2,-2)(0,0)(2,-2)(4,2) \meinecaption{1}{A.} \end{minipage} \hfill \begin{minipage}[c]{4cm} \centering\psset{unit=.5pc} \psaxes[linewidth=2pt, labels=none,ticks=none,arrows={->}](0,0)(-5,-5)(5,5) \psline[linecolor=blue,linewidth=2pt](-5,-3)(-3,-3)(0,0)(2,2)(4,-2) \meinecaption{1}{B.} \end{minipage} \end{figure} \vspace{2cm} \begin{figure}[!h] \begin{minipage}[c]{4cm} \centering\psset{unit=.5pc} \psaxes[linewidth=2pt, labels=none,ticks=none,arrows={->}](0,0)(-5,-5)(5,5) \psline[linecolor=blue,linewidth=2pt](-4,2)(-2,-2)(0,0)(3,3)(5,3) \meinecaption{1}{C.} \end{minipage} \hfill \begin{minipage}[c]{4cm} \centering\psset{unit=.5pc} \psaxes[linewidth=2pt, labels=none,ticks=none,arrows={->}](0,0)(-5,-5)(5,5) \psline[linecolor=blue,linewidth=2pt](-5,3)(-3,3)(0,0)(3,3)(5,3) \meinecaption{1}{D.} \end{minipage} \hfill \begin{minipage}[c]{4cm} \centering\psset{unit=.5pc} \psaxes[linewidth=2pt, labels=none,ticks=none,arrows={->}](0,0)(-5,-5)(5,5) \psline[linecolor=blue,linewidth=2pt](-5,3)(-3,3)(0,0)(2,2)(3,0)(4,2) \meinecaption{1}{E.} \end{minipage} \end{figure} \vfill \item \label{pro:transform-a1} $y = f(-x) $ is \\ \psovalbox{A}\hfill \psovalbox{B} \hfill \psovalbox{C}\hfill \psovalbox{D} \hfill \psovalbox{E} \vfill \item $y = -f(x) $ is \\ \psovalbox{A}\hfill \psovalbox{B} \hfill \psovalbox{C}\hfill \psovalbox{D} \hfill \psovalbox{E} \vfill \item $y = f(|x|) $ is \\ \psovalbox{A}\hfill \psovalbox{B} \hfill \psovalbox{C}\hfill \psovalbox{D} \hfill \psovalbox{E} \vfill \item $y = |f(x)| $ is \\ \psovalbox{A}\hfill \psovalbox{B} \hfill \psovalbox{C}\hfill \psovalbox{D} \hfill \psovalbox{E} \vfill \item \label{pro:transform-b1} $y = f(-|x|) $ is \\ \psovalbox{A}\hfill \psovalbox{B} \hfill \psovalbox{C}\hfill \psovalbox{D} \hfill \psovalbox{E} \clearpage \item[] There are six graphs shewn below. The first graph is that of the original curve $f:\reals \to \reals$, where $f(x)=\sqrt[3]{x}$, and the other five are various transformations of the original graph. You are to match each graph letter below with the appropriate equation in \ref{pro:transform-a} through \ref{pro:transform-b} below. \vspace{2cm} \vfill \begin{figure}[!h] \begin{minipage}[c]{4cm} \centering\psset{unit=.7pc}\psgrid[subgriddiv=0,gridlabels=0,gridcolor=darkgray](-5,-5)(5,5) \psaxes[linewidth=2pt, labels=none,ticks=none,arrows={->}](0,0)(-5.5,-5.5)(5.5,5.5) \psplot[linecolor=brown,linewidth=2pt,algebraic]{0}{5}{x^(1/3)} \psplot[linecolor=brown,linewidth=2pt,algebraic]{0}{-5}{-abs(x)^(1/3)} \meinecaption{1.5}{$y=f(x)$.} \end{minipage} \hfill \begin{minipage}[c]{4cm} \centering\psset{unit=.7pc}\psgrid[subgriddiv=0,gridlabels=0,gridcolor=darkgray](-5,-5)(5,5) \psaxes[linewidth=2pt, labels=none,ticks=none,arrows={->}](0,0)(-5.5,-5.5)(5.5,5.5) \psplot[linecolor=brown,linewidth=2pt,algebraic]{-5}{5}{abs(x)^(1/3)} \meinecaption{1.5}{A.} \end{minipage} \hfill \begin{minipage}[c]{4cm} \centering\psset{unit=.7pc}\psgrid[subgriddiv=0,gridlabels=0,gridcolor=darkgray](-5,-5)(5,5) \psaxes[linewidth=2pt, labels=none,ticks=none,arrows={->}](0,0)(-5.5,-5.5)(5.5,5.5) \psplot[linecolor=brown,linewidth=2pt,algebraic]{0}{5}{x^(1/3)+2} \psplot[linecolor=brown,linewidth=2pt,algebraic]{0}{-5}{-abs(x)^(1/3)+2} \meinecaption{1.5}{B.} \end{minipage} \end{figure} \vfill \vspace*{1cm} \begin{figure}[!h] \begin{minipage}[c]{4cm} \centering\psset{unit=.7pc}\psgrid[subgriddiv=0,gridlabels=0,gridcolor=darkgray](-5,-5)(5,5) \psaxes[linewidth=2pt, labels=none,ticks=none,arrows={->}](0,0)(-5.5,-5.5)(5.5,5.5) \psplot[linecolor=brown,linewidth=2pt,algebraic]{-1.709975947}{1.709975947}{x^3} \meinecaption{1.5}{C.} \end{minipage} \hfill \begin{minipage}[c]{4cm} \centering\psset{unit=.7pc}\psgrid[subgriddiv=0,gridlabels=0,gridcolor=darkgray](-5,-5)(5,5) \psaxes[linewidth=2pt, labels=none,ticks=none,arrows={->}](0,0)(-5.5,-5.5)(5.5,5.5) \psplot[linecolor=brown,linewidth=2pt,algebraic]{0}{5}{-(x)^(1/3)} \psplot[linecolor=brown,linewidth=2pt,algebraic]{0}{-5}{abs(x)^(1/3)} \meinecaption{1.5}{D.} \end{minipage} \hfill \begin{minipage}[c]{4cm} \centering\psset{unit=.7pc}\psgrid[subgriddiv=0,gridlabels=0,gridcolor=darkgray](-5,-5)(5,5) \psaxes[linewidth=2pt, labels=none,ticks=none,arrows={->}](0,0)(-5.5,-5.5)(5.5,5.5) \psplot[linecolor=brown,linewidth=2pt,algebraic]{0}{5}{-(x)^(1/3)+1} \psplot[linecolor=brown,linewidth=2pt,algebraic]{0}{-5}{abs(x)^(1/3)+1} \meinecaption{1.5}{E.} \end{minipage} \end{figure} \item \label{pro:transform-a} $y = f(x)+1 $ is \\ \psovalbox{A}\hfill \psovalbox{B} \hfill \psovalbox{C}\hfill \psovalbox{D} \hfill \psovalbox{E} \vfill \item $y = f^{-1}(x) $ is \\ \psovalbox{A}\hfill \psovalbox{B} \hfill \psovalbox{C}\hfill \psovalbox{D} \hfill \psovalbox{E} \vfill \item $y = -f(x)+1 $ is \\ \psovalbox{A}\hfill \psovalbox{B} \hfill \psovalbox{C}\hfill \psovalbox{D} \hfill \psovalbox{E} \vfill \item $y = |f(x)| $ is \\ \psovalbox{A}\hfill \psovalbox{B} \hfill \psovalbox{C}\hfill \psovalbox{D} \hfill \psovalbox{E} \vfill \item \label{pro:transform-b} $y = f(-x) $ is \\ \psovalbox{A}\hfill \psovalbox{B} \hfill \psovalbox{C}\hfill \psovalbox{D} \hfill \psovalbox{E} \vfill \subsection{Quadratic Functions} \item Find the vertex of the parabola with equation $y=x^2-6x+1$.\\ \pscirclebox{A} $(3,10)$ \hfill \pscirclebox{B} $(-3,10)$ \hfill \pscirclebox{C} $(-3,-8)$\hfill \pscirclebox{D} $(3,-8)$ \hfill \pscirclebox{E} none of these \item Find the equation of the parabola whose axis of symmetry is parallel to the $y$-axis, passes through $(2,1)$, and has vertex at $(-1,2)$.\\ \psovalbox{A} $x= 3(y-2)^2-1$ \vfill \psovalbox{B} $y=-9(x+1)^2+2$ \vfill \psovalbox{C} $y=-(x-1)^2+2$ \vfill \psovalbox{D} $y=-\dfrac{1}{9}(x+1)^2+2$ \vfill \psovalbox{E} none of these \item Let $a, b, c$ be real constants. Find the vertex of the parabola $y =cx^2+2bx+a$.\\ \psovalbox{A} $\left(-\dfrac{b}{2c}, a-\dfrac{3b^2}{4c}\right)$ \hfill \psovalbox{B} $\left(-\dfrac{b}{c}, a-\dfrac{b^2}{c}\right)$ \hfill \psovalbox{C} $\left(-\dfrac{b}{c}, a+\dfrac{b^2}{c}\right)$ \hfill \psovalbox{D} $\left(\dfrac{b}{c}, a+3\dfrac{b^2}{c}\right)$ \hfill \psovalbox{E} none of these \item A parabola has vertex at $(1,2)$, symmetry axis parallel to the $x$-axis, and passes through $(-1,0)$. Find its equation.\\ \psovalbox{A} $x=-\dfrac{(y-2)^2}{2}+1$\vfill \psovalbox{B} $x=-2(y-2)^2+1$\vfill \psovalbox{C} $y=-\dfrac{(x-1)^2}{2}+2$\vfill \psovalbox{D} $y=-2(x-1)^2+2$\vfill \psovalbox{E} none of these \item \label{pro:parabola} The graph in figure \ref{fig:parabola} below belongs to a curve with equation of the form $y = A(x+1)^2+4$. Find $A$.\\ \vspace{3cm} \begin{figure}[h] \psset{unit=1pc}\centering \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psplot[algebraic,linewidth=2pt,linecolor=blue, arrows={<->}]{-5}{3}{-((x+1)^2)/2+4}\psdots[dotscale=1.5](-1,4)(-3,2)(1,2) \meinecaption{2.5}{Problem \ref{pro:parabola}.}\label{fig:parabola} \end{figure} \vspace{1cm} \pscirclebox{A} $A=\dfrac{1}{2}$ \hfill \pscirclebox{B} $A=-1$ \hfill \pscirclebox{C} $A=-\dfrac{1}{2}$\hfill \pscirclebox{D} $A=-2$ \hfill \pscirclebox{E} none of these \clearpage \item[] Problems \ref{pro:quada} through \ref{pro:quadb} refer to the quadratic function $q:\reals \to \reals$ with assignment rule given by $$q(x)= x^2-6x+5.$$ \vfill \item \label{pro:quada} How many of the following assertions is (are) true? \begin{enumerate} \item $q$ is convex. \item $q$ is invertible over $\reals$. \item the graph $q$ has vertex $(-3,-4)$. \item the graph of $q$ has $y$-intercept $(0,5)$ and $x$-intercepts $(-1,0)$ and $(5,0)$. \end{enumerate} \psovalbox{A} none\hfill \psovalbox{B} exactly one\hfill \psovalbox{C} exactly two\hfill \psovalbox{D} exactly three\hfill \psovalbox{E} all four \vfill \item Which one most resembles the graph of $q$? Notice that there are four graphs but five choices.\\ \vspace{1cm} \vfill \begin{figure}[!hptb] \begin{minipage}{4cm} \psset{unit=.75pc}\centering \psaxes[linewidth=2pt,labels=none,ticks=none,arrows={->}](0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown,algebraic,arrows={<->}]{0}{6}{x^2-6*x+5} \meinecaption{2}{A} \end{minipage} \hfill \begin{minipage}{4cm} \psset{unit=.75pc}\centering \psaxes[linewidth=2pt,labels=none,ticks=none,arrows={->}](0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown,algebraic,arrows={<->}]{0}{6}{-x^2+6*x-5} \meinecaption{2}{B} \end{minipage} \hfill \begin{minipage}{4cm} \psset{unit=.75pc}\centering \psaxes[linewidth=2pt,labels=none,ticks=none,arrows={->}](0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown,algebraic,arrows={<->}]{-6}{0}{x^2+6*x+5} \meinecaption{2}{C} \end{minipage} \hfill \begin{minipage}{4cm} \psset{unit=.75pc}\centering \psaxes[linewidth=2pt,labels=none,ticks=none,arrows={->}](0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown,algebraic,arrows={<->}]{-3}{3}{5-x^2} \meinecaption{2}{D} \end{minipage} \end{figure} \psovalbox{A} A\hfill \psovalbox{B} B\hfill \psovalbox{C} C\hfill \psovalbox{D} D\hfill \psovalbox{E} none of these \vfill \item \label{pro:quadb} Which one most resembles the graph of $y=q(|x|)$? Notice that there are four graphs but five choices.\\ \vspace{1cm} \vfill \begin{figure}[!hptb] \begin{minipage}{4cm} \psset{unit=.75pc}\centering \psaxes[linewidth=2pt,labels=none,ticks=none,arrows={->}](0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown,algebraic,arrows={<->}]{0}{6}{abs(x^2-6*x+5)} \meinecaption{2}{A} \end{minipage} \hfill \begin{minipage}{4cm} \psset{unit=.75pc}\centering \psaxes[linewidth=2pt,labels=none,ticks=none,arrows={->}](0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown,algebraic,arrows={<->}]{-6}{6}{x^2-6*abs(x)+5} \meinecaption{2}{B} \end{minipage} \hfill \begin{minipage}{4cm} \psset{unit=.75pc}\centering \psaxes[linewidth=2pt,labels=none,ticks=none,arrows={->}](0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown,algebraic,arrows={<->}]{-1}{1}{x^2+6*abs(x)-3} \meinecaption{2}{C} \end{minipage} \hfill \begin{minipage}{4cm} \psset{unit=.75pc}\centering \psaxes[linewidth=2pt,labels=none,ticks=none,arrows={->}](0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown,algebraic,arrows={<->}]{-6}{6}{abs(x^2-6*abs(x)+5)} \meinecaption{2}{D} \end{minipage} \end{figure} \psovalbox{A} A\hfill \psovalbox{B} B\hfill \psovalbox{C} C\hfill \psovalbox{D} D\hfill \psovalbox{E} none of these \vfill \clearpage \item \label{pro:parabola}Find the equation of the parabola shewn below. You may assume that the points marked with a dot have integer coordinates.\\ \vspace{2.5cm} \begin{figure}[!h] \centering \psset{unit=.9pc} \psaxes[linewidth=2pt,arrows={->},labelFontSize=\tiny](0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown,arrows={<->},algebraic=true]{-6}{2}{-(x+2)^2/2+1} \psdots[dotstyle=*,dotscale=1.2](-2,1)(0,-1)(-4,-1) \meinecaption{2.5}{Problem \ref{pro:parabola}. } \end{figure} \psovalbox{A} $y= -\dfrac{-(x+2)^2}{2}+1$\\ \psovalbox{B} $y = -2(x+2)^2+1$ \\ \psovalbox{C} $y=(x+2)^2+1$\\ \psovalbox{D} $y = -(x+2)^2+1$ \\ \psovalbox{E} none of these \subsection{Injections and Surjections} \item How many injective functions are there from the set $\{a, b, c\}$ to the set $\{1, 2\}$? \\ \psovalbox{A} $ 6 $ \hfill \psovalbox{B} $ 9 $ \hfill \psovalbox{C} $ 8 $ \hfill \psovalbox{D} $0 $ \hfill \psovalbox{E} none of these \item How many surjective functions are there from the set $\{a, b, c\}$ to the set $\{1, 2\}$? \\ \psovalbox{A} $ 0 $ \hfill \psovalbox{B} $ 6 $ \hfill \psovalbox{C} $ 9 $ \hfill \psovalbox{D} $ 8$ \hfill \psovalbox{E} none of these \item How many invertible functions are there from the set $\{a, b, c\}$ to the set $\{1, 2\}$? \\ \psovalbox{A} $ 0 $ \hfill \psovalbox{B} $ 6 $ \hfill \psovalbox{C} $ 9 $ \hfill \psovalbox{D} $8 $ \hfill \psovalbox{E} none of these \subsection{Inversion of Functions} \item What is the equation of the curve symmetric to the curve $y = \dfrac{1}{x^3}+1$ with respect to the line $y=x$ ?\\ \psovalbox{A} $y = -\dfrac{1}{x^3}+1$ \hfill \psovalbox{B} $y = -\dfrac{1}{x^3}-1$\hfill \psovalbox{C}$y = \dfrac{1}{(x-1)^3}$ \hfill \psovalbox{D} $y = \dfrac{1}{(x-1)^{1/3}}$ \hfill \psovalbox{E} $y = \dfrac{1}{(1-x)^{1/3}}$ \clearpage \item[] Figure \ref{fig:functional_curve2} shews a functional curve $$f:[-5;5]\rightarrow [-3;6], \qquad y =f(x),$$ and refers to problems \ref{pro:func_curve2_begin} to \ref{pro:func_curve2_end}.\\ \vspace{3cm} \begin{figure}[h!] $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \uput[r](7,0){x} \uput[u](0, 7.5){y}\psgrid[subgriddiv=0,linewidth=.7pt, gridlabels=0](0,0)(-6,-6)(6,6) \psline[linewidth=1.5pt, linecolor=brown](-5,-3)(-2,-1)(0,1)(2,3)(3,5)(5,6) \psdots[dotstyle=*,dotscale=1](-5,-3)(-2,-1)(0,1)(2,3)(3,5)(5,6) $$\meinecaption{1}{Problems \ref{pro:func_curve2_begin} to \ref{pro:func_curve2_end}.} \label{fig:functional_curve2} \end{figure} \item \label{pro:func_curve2_begin}$f(-2) + f(2) = $\\ \psovalbox{A}$0$ \hfill\psovalbox{B}$1$ \hfill\psovalbox{C}$2$\hfill\psovalbox{D}$3$ \hfill\psovalbox{E}none of these \item $f(-3)$ belongs to the interval \\ \psovalbox{A}$[-1;0]$ \hfill\psovalbox{B}$[-2;-1]$ \hfill\psovalbox{C} $[-3;-2]$ \hfill\psovalbox{D}$[0;1]$ \hfill\psovalbox{E} none of these \item $f^{-1}(3) =$ \\ \psovalbox{A}$-3$ \hfill\psovalbox{B}$-\dfrac{1}{3}$ \hfill\psovalbox{C} $2$ \hfill\psovalbox{D}$5$ \hfill\psovalbox{E}none of these \item $(f\circ f)(2) =$ \\ \psovalbox{A}$4$ \hfill\psovalbox{B}$5$ \hfill\psovalbox{C}$6$ \hfill\psovalbox{D}undefined \hfill\psovalbox{E}none of these \item\label{pro:func_curve2_end} The graph of $f^{-1}$ is \\ \vspace{2cm} \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \uput[r](7,0){x} \uput[u](0, 7.5){y} \psgrid[subgriddiv=0,linewidth=.7pt, gridlabels=0](0,0)(-6,-6)(6,6) \psline[linewidth=1.5pt, linecolor=brown](-5,-3)(-2,-1)(0,1)(2,3)(3,5)(5,6) \psdots[dotstyle=*,dotscale=1](-5,-3)(-2,-1)(0,1)(2,3)(3,5)(5,6) $$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \uput[r](7,0){x} \uput[u](0, 7.5){y}\psgrid[subgriddiv=0,linewidth=.7pt, gridlabels=0](0,0)(-6,-6)(6,6) \psline[linewidth=1.5pt, linecolor=brown](-5,-3)(-2,-1)(0,1)(2,2)(3,5)(5,6) \psdots[dotstyle=*,dotscale=1](-5,-3)(-2,-1)(0,1)(2,2)(3,5)(5,6)$$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \uput[r](7,0){x} \uput[u](0, 7.5){y}\psgrid[subgriddiv=0,linewidth=.7pt, gridlabels=0](0,0)(-6,-6)(6,6) \psline[linewidth=1.5pt, linecolor=brown](-3,-5)(-1,-2)(1,0)(3,2)(5,3)(6,5) \psdots[dotstyle=*,dotscale=1](-3,-5)(-1,-2)(1,0)(3,2)(5,3)(6,5) $$ \meinecaption{1}{C} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \uput[r](7,0){x} \uput[u](0, 7.5){y}\psgrid[subgriddiv=0,linewidth=.7pt, gridlabels=0](0,0)(-6,-6)(6,6) \psline[linewidth=1.5pt, linecolor=brown](-3,-5)(-1,-2)(1,0)(3,3)(5,4)(6,5) \psdots[dotstyle=*,dotscale=1](-3,-5)(-1,-2)(1,0)(3,3)(5,4)(6,5) $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \psovalbox{A}A\hfill\psovalbox{B}B \hfill\psovalbox{C}C\hfill\psovalbox{D}D \hfill\psovalbox{E}none of these \item Let $f(x)= \dfrac{x}{x+1}$. Find $g(x)$ such that $(f\circ g)(x)=x$. \\ \psovalbox{A} $g(x)=\dfrac{x}{x-1}$\hfill \psovalbox{B} $g(x)=\dfrac{x}{1+x}$ \hfill \psovalbox{C} $g(x)=\dfrac{x}{1-x}$ \hfill \psovalbox{D} $g(x)=-\dfrac{x}{1+x}$\hfill \psovalbox{E} none of these \item Let $f(x) = \dfrac{x+1}{1-2x}$. Then $f^{-1}(x)=$\\ \psovalbox{A} $\dfrac{1-x}{1+2x}$ \hfill \psovalbox{B} $\dfrac{1+x}{1-2x}$ \hfill \psovalbox{C} $\dfrac{x-1}{1+2x}$ \hfill \psovalbox{D} $\dfrac{1-x}{1-2x}$ \hfill \psovalbox{E} none of these \item[] Problems \ref{pro:piecewise_linear_begin1} through \ref{pro:piecewise_linear_end1} refer to the function $f$ with assignment rule $$y = f(x) = \left\{\begin{array}{ll} \dfrac{x}{3} - \dfrac{10}{3} & \mathrm{if}\ x\in [-5;-2[ \\ 2x & \mathrm{if}\ x\in [-2;2] \\ \dfrac{x}{3} + \dfrac{10}{3} & \mathrm{if}\ x\in ]2;5] \end{array}\right.$$ \item \label{pro:piecewise_linear_begin1}Which one most resembles the graph of $f$? \\ \vspace{2cm} \begin{figure}[h] \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psplot[linecolor=brown,linewidth=1.4pt]{-5}{-2}{(-10+x)/3} \psplot[linecolor=blue,linewidth=1.4pt]{-2}{2}{2*x} \psplot[linecolor=blue,linewidth=1.4pt]{2}{5}{(x+10)/3} \psdots[dotscale=1,dotstyle=*](-5,-5)(5,5)(-2,-4)(2,4) $$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psline[linewidth=1.4pt,linecolor=brown](-5,-5)(-2,-3)(2,3)(5,5) \psdots[dotscale=1,dotstyle=*](-5,-5)(-2,-3)(2,3)(5,5) $$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psline[linewidth=1.4pt,linecolor=brown](-5,3)(-3,-3)(3,3)(5,-3) \psdots[dotscale=1,dotstyle=*](-5,3)(-3,-3)(3,3)(5,-3) $$ \meinecaption{1}{C} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psline[linewidth=1.4pt,linecolor=brown](-5,-4)(-2,-3)(2,3)(4,5) \psdots[dotscale=1,dotstyle=*](-5,-4)(-2,-3)(2,3)(4,5) $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \vspace{.5cm} \psovalbox{A}A\hfill\psovalbox{B} B \hfill\psovalbox{C} C\hfill\psovalbox{D} D \hfill\psovalbox{E} none of these \item Find the exact value of $(f\circ f)(2)$.\\ \psovalbox{A} $ 4 $ \hfill \psovalbox{B} $ \dfrac{14}{3} $ \hfill \psovalbox{C} $ 8 $ \hfill \psovalbox{D} $ 3$ \hfill \psovalbox{E} none of these \item Which one could not possibly be a possible value for $\underbrace{(f\circ \cdots \circ f)}_{n \ \mathrm{compositions}}(a)$, where $n$ is a positive integer and $a\in [-5;5]$?.\\ \psovalbox{A} $ 0 $ \hfill \psovalbox{B} $ -5 $ \hfill \psovalbox{C} $ 5 $ \hfill \psovalbox{D} $ 6$ \hfill \psovalbox{E} none of these \item \label{pro:piecewise_linear_end1} Which one most resembles the graph of $f^{-1}$? \\ \vspace{2cm} \begin{figure}[h] \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psline[linewidth=1.4pt,linecolor=brown](-5,-5)(-4,-2)(4,2)(5,5) \psdots[dotscale=1,dotstyle=*](-5,-5)(-4,-2)(4,2)(5,5) $$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psline[linewidth=1.4pt,linecolor=brown](-5,-5)(-3,-2)(3,2)(5,5) \psdots[dotscale=1,dotstyle=*](-5,-5)(-3,-2)(3,2)(5,5)$$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psline[linewidth=1.4pt,linecolor=brown](-5,-5)(-2,-4)(2,4)(5,5) \psdots[dotscale=1,dotstyle=*](-5,-5)(-2,-4)(2,4)(5,5) $$ \meinecaption{1}{C} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psline[linewidth=1.4pt,linecolor=brown](-4,-5)(-2,-3)(2,3)(4,5) \psdots[dotscale=1,dotstyle=*](-4,-5)(-2,-3)(2,3)(4,5) $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \vspace{.5cm} \psovalbox{A}A\hfill\psovalbox{B} B \hfill\psovalbox{C} C\hfill\psovalbox{D} D \hfill\psovalbox{E} none of these \\ \item Let $f(x)=x-2$ and $g(x)=2x+1$. Find $(f^{-1}\circ g^{-1})(x).$\\ \pscirclebox{A} $\dfrac{x+1}{2}$ \hfill \pscirclebox{B} $\dfrac{x+3}{2}$ \hfill \pscirclebox{C} $2x-3$\hfill \pscirclebox{D} $2x-1$ \hfill \pscirclebox{E} none of these \clearpage \item Which of the following graphs represents an invertible function?\\ \vspace{2cm} \begin{figure}[!hptb] \begin{minipage}{4cm} \psset{unit=.75pc}\centering \psaxes[linewidth=2pt,labels=none,ticks=none,arrows={->}](0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown,algebraic,arrows={<->}]{0}{6}{x^2-6*x+5} \meinecaption{2}{A} \end{minipage} \hfill \begin{minipage}{4cm} \psset{unit=.75pc}\centering \psaxes[linewidth=2pt,labels=none,ticks=none,arrows={->}](0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown,algebraic]{-5}{5}{x+2*sin(x)} \meinecaption{2}{B} \end{minipage} \hfill \begin{minipage}{4cm} \psset{unit=.75pc}\centering \psaxes[linewidth=2pt,labels=none,ticks=none,arrows={->}](0,0)(-6,-6)(6,6) \parametricplot[algebraic,linewidth=2pt,linecolor=brown,plotpoints=5001]{-30}{10}{(2^(t/5))*cos(t)|(2^(t/5))*sin(t)} \meinecaption{2}{C} \end{minipage} \hfill \begin{minipage}{4cm} \psset{unit=.75pc}\centering \psaxes[linewidth=2pt,labels=none,ticks=none,arrows={->}](0,0)(-6,-6)(6,6) \psplot[linewidth=2pt,linecolor=brown,algebraic,arrows={<->}]{-5}{3}{2^x-2} \meinecaption{2}{D} \end{minipage} \end{figure} \psovalbox{A} A\hfill \psovalbox{B} B\hfill \psovalbox{C} C\hfill \psovalbox{D} D\hfill \psovalbox{E} none of these \item Let $f(x) = \left(\dfrac{x}{3}-1\right)^3+2$. Then $f^{-1}(x)=$\\ \psovalbox{A} $3\sqrt[3]{x+2}-3$ \hfill \psovalbox{B} $3\sqrt[3]{x-2}-3$ \hfill \psovalbox{C} $3\sqrt[3]{x-2}+3$ \hfill \psovalbox{D} $3\sqrt[3]{x+2}+3$ \hfill \psovalbox{E} none of these \item Let $f(x)=\dfrac{2x}{x+1}$. Find $f^{-1}(x)$. \vfill \psovalbox{A} $\dfrac{x+1}{2x}$\hfill \psovalbox{B} $\dfrac{x}{x-2}$ \hfill \psovalbox{C} $\dfrac{x-2}{x}$\hfill \psovalbox{D} $\dfrac{x}{2-x}$ \hfill \psovalbox{E} none of these \item Let $f(x)= (x+1)^5-2$. Find $f^{-1}(x)$. \\ \psovalbox{A} $\sqrt[5]{x+1}-2$\hfill \psovalbox{B} $\sqrt[5]{x-2}+1$ \hfill \psovalbox{C} $\dfrac{1}{(x+1)^5-2}$ \hfill \psovalbox{D} $\sqrt[5]{x+2}-1$\hfill \psovalbox{E} none of these % \item Let $f(x)= -\dfrac{x}{2}+1$. Find $f^{-1}(x)$. \\ \psovalbox{A} $\dfrac{2}{x}-1$\hfill \psovalbox{B} $-2x-1$ \hfill \psovalbox{C} $2x-1$ \hfill \psovalbox{D} $-2x+2$\hfill \psovalbox{E} none of these \item Let $f(x)=\dfrac{x}{x-1}$ and $g(x) = 1-x$. Determine $(g\circ f)^{-1}(x)$. \\ \psovalbox{A} $\dfrac{x-1}{x}$\hfill \psovalbox{B} $\dfrac{1-x}{x}$\hfill \psovalbox{C} $\dfrac{1}{x-1}$\hfill \psovalbox{D} $\dfrac{1}{1-x}$\hfill \psovalbox{E} none of these \item Let $f(x)=\dfrac{x+1}{x}$. Determine $f^{-1}(x)$. \\ \vfill \psovalbox{A} $f^{-1}(x)=\dfrac{x}{x-1}$\vfill \psovalbox{B} $f^{-1}(x)=\dfrac{1}{x+1}$ \vfill \psovalbox{C} $f^{-1}(x)=\dfrac{1}{x-1}$ \vfill \psovalbox{D} $f^{-1}(x)=\dfrac{x}{x+1}$ \vfill \psovalbox{E} none of these \subsection{Polynomial Functions} \item Let $p$ be a polynomial of degree $3$ with roots at $x=1$, $x=-1$, and $x=2$. If $p(0)=4$, find $p(4)$. \\ \psovalbox{A} $0$\hfill \psovalbox{B} $4$\hfill \psovalbox{C} $30$\hfill \psovalbox{D} $60$\hfill \psovalbox{E} none of these \item A polynomial of degree $3$ satisfies $p(0)=0$, $p(1)=0$, $p(2)=0$, and $p(3)=-6$. What is $p(4)$?\\ \psovalbox{A} $0$\hfill \psovalbox{B} $1$ \hfill \psovalbox{C} $-24$ \hfill \psovalbox{D} $24$ \hfill \psovalbox{E} none of these \item Factor the polynomial $x^3-x^2-4x+4$.\\ \psovalbox{A} $(x+1)(x-2)(x+2)$\vfill \psovalbox{B} $(x-1)(x+1)(x-4)$ \vfill \psovalbox{C} $(x-1)(x-2)(x+2)$ \vfill \psovalbox{D} $(x-1)(x+1)(x+4)$ \vfill \psovalbox{E} none of these \item Determine the value of the parameter $a$ so that the polynomial $x^3+2x^2+ax-10$ be divisible by $x-2$.\\ \psovalbox{A} $a=3$\hfill \psovalbox{B} $a=-3$ \hfill \psovalbox{C} $a=-2$ \hfill \psovalbox{D} $a=-1$ \hfill \psovalbox{E} none of these \item A polynomial leaves remainder $-1$ when divided by $x-2$ and remainder $2$ when divided by $x+1$. What is its remainder when divided by $x^2-x-2$? \\ \psovalbox{A} $x-1$\hfill \psovalbox{B} $2x-1$ \hfill \psovalbox{C} $-x-1$ \hfill \psovalbox{D} $-x+1$ \hfill \psovalbox{E} none of these \item[] Questions \ref{pro:polyna} through \ref{pro:polynb} refer to the polynomial $p$ in figure \ref{fig:polyn}. The polynomial has degree $5$. You may assume that the points marked with dots have integer coordinates. \vspace{3cm} \begin{figure}[!hptb] \centering \psset{unit=1pc} \psgrid[subgriddiv=0,gridlabels=.5](-7,-7)(7,7) \psplot[algebraic,linecolor=blue,linewidth=2pt]{-4.35}{3.57}{(x-3)*(x+2)*(x+4)*(x-1)^2/24} \psdots[dotscale=1.5](0,-1)(3,0)(-2,0)(1,0)(2,-1)(-3,4)(-1,-2) \meinecaption{3}{Problems \ref{pro:polyna} through \ref{pro:polynb}.}\label{fig:polyn} \end{figure} \vspace{1cm} \item \label{pro:polyna} Determine the value of $p(0)$.\\ \psovalbox{A} $0$\hfill \psovalbox{B} $-1$\hfill \psovalbox{C} $4$\hfill \psovalbox{D} $-2$\hfill \psovalbox{E} none of these \item Determine the value of $p(-3)$.\\ \psovalbox{A} $0$\hfill \psovalbox{B} $-1$\hfill \psovalbox{C} $4$\hfill \psovalbox{D} $-2$\hfill \psovalbox{E} none of these \item Determine $p(x)$.\\ \psovalbox{A} $\dfrac{(x-3)(x+2)(x+4)(x-1)^2}{24}$\vfill \psovalbox{B} $(x-3)(x+2)(x+4)(x-1)^2$\vfill \psovalbox{C} $\dfrac{(x-3)(x+2)(x+4)(x-1)}{24}$\vfill \psovalbox{D} $(x-3)(x+2)(x+4)(x-1)$\vfill \psovalbox{E} none of these \item \label{pro:polynb} Determine the value of $(p\circ p)(-3)$.\\ \psovalbox{A} $4$\hfill \psovalbox{B} $18$\hfill \psovalbox{C} $20$\hfill \psovalbox{D} $24$\hfill \psovalbox{E} none of these \item The polynomial $p$ whose graph is shewn below has degree $4$. You may assume that the points marked below with a dot through which the polynomial passes have have integer coordinates. Find its equation. \\ \vspace{1.5cm} \vfill \begin{figure}[h] $$\psset{unit=1pc} \psgrid[subgriddiv=0,linewidth=.5pt, gridlabels=5pt](-5,-5)(5,5) \psplot[linewidth=1.5pt, linecolor=brown, algebraic=true, arrows={<->}]{-3.703115981}{3.626962318}{x*(x+2)^2*(x-3)/18} \psdots[dotscale=1.4,dotstyle=*](1,-1)(3,0)(-2,0)(0,0)(-3,1) $$ \label{fig:quintic-f06} \end{figure} \vspace{1.5cm} \vfill \psovalbox{A} $p(x)=x(x+2)^2(x-3)$\vfill \psovalbox{B} $p(x)=-\dfrac{x(x+2)^2(x-3)}{18}$\vfill \psovalbox{C} $p(x)=\dfrac{x(x+2)^2(x-3)}{12}$ \vfill \psovalbox{D} $p(x)=\dfrac{x(x+2)^2(x-3)}{18}$ \vfill \psovalbox{E} none of these \vfill \item[] Problems \ref{pro:quartic-s07a} through \ref{pro:quartic-s07b} refer to the polynomial in figure \ref{fig:quartic-f06}, which has degree $4$. You may assume that the points marked below with a dot through which the polynomial passes have have integer coordinates. \\ \vspace{2.5cm} \vfill \begin{figure}[h] $$\psset{unit=.8pc} \psgrid[subgriddiv=0,linewidth=.5pt, gridlabels=5pt](-7,-7)(7,7) \psplot[linewidth=1.5pt, linecolor=brown, algebraic=true, arrows={<->}]{-3.3}{2.5}{x*(x+2)^2*(x-2)/3} \psdots[dotscale=1.4,dotstyle=*](0,0)(2,0)(-2,0)(-1,1) $$ \meinecaption{2}{Problems \ref{pro:quartic-s07a}through \ref{pro:quartic-s07b}.}\label{fig:quartic-f06} \end{figure} \vspace{1.5cm} \item \label{pro:quartic-s07a}Determine $p(-1)$. \\ \psovalbox{A} $1$\hfill \psovalbox{B} $-1$\hfill \psovalbox{C} $3$ \hfill \psovalbox{D} $-3$ \hfill \psovalbox{E} none of these \item $p(x)=$ \\ \psovalbox{A} $x(x+2)^2(x-2)$\vfill \psovalbox{B}$\dfrac{x(x-2)^2(x+2)}{3}$ \vfill \psovalbox{C} $\dfrac{x(x+2)^2(x-2)}{3}$ \vfill \psovalbox{D} $x(x+2)(x-2)^2$ \vfill \psovalbox{E} none of these \item \label{pro:quartic-s07b} Determine $(p\circ p)(-1)$.\\ \vfill \psovalbox{A} $1$\hfill \psovalbox{B} $3$\hfill \psovalbox{C} $-3$\hfill \psovalbox{D} $-1$\hfill \psovalbox{E} none of these \vfill \subsection{Rational Functions} \item Which graph most resembles the curve $y=\dfrac{1}{x-1}+2$?\\ \vspace{1cm}\vfill \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{1.166666667}{6}{1/(x-1)} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-6}{.8333333333}{1/(x-1)} $$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{.25}{6}{1/x+2} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-6}{-.125}{1/x+2} $$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc}\psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{1.25}{6}{1/(x-1)+2} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-6}{.875}{1/(x-1)+2} $$ \meinecaption{1}{C} \end{minipage} \hfill \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-.75}{6}{1/(x+1)+2} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-6}{-1.125}{1/(x+1)+2} $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \vfill \psovalbox{A} \hfill \psovalbox{B} \hfill \psovalbox{C} \hfill \psovalbox{D} \hfill \psovalbox{E} none of these \item Which graph most resembles the curve $y=\absval{\dfrac{1}{x-1}+2}$?\\ \vspace{1cm}\vfill \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{1.166666667}{6}{abs(1/(x-1))} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-6}{.8333333333}{abs(1/(x-1))} $$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{.25}{6}{abs(1/x+2)} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-6}{-.125}{abs(1/x+2)} $$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc}\psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{1.25}{6}{abs(1/(x-1)+2)} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-6}{.875}{abs(1/(x-1)+2)} $$ \meinecaption{1}{C} \end{minipage} \hfill \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-.75}{6}{abs(1/(x+1)+2)} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-6}{-1.125}{abs(1/(x+1)+2)} $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \vfill \psovalbox{A} \hfill \psovalbox{B} \hfill \psovalbox{C} \hfill \psovalbox{D} \hfill \psovalbox{E} none of these \item Which graph most resembles the curve $y=\dfrac{1}{\absval{x}-1}+2$?\\ \vspace{1cm}\vfill \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{1.166666667}{6}{1/(abs(x)-1)} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-1.166666667}{-6}{1/(abs(x)-1)} $$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{.25}{6}{1/abs(x)+2} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-.25}{-6}{1/abs(x)+2} $$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc}\psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{1.25}{6}{1/(abs(x)-1)+2} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-.875}{.875}{1/(abs(x)-1)+2} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-1.25}{-6}{1/(abs(x)-1)+2} $$ \meinecaption{1}{C} \end{minipage} \hfill \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={->},linecolor=brown]{-.75}{6}{1/(abs(x)+1)+2} \psplot[algebraic=true,linewidth=1.5pt, arrows={->},linecolor=brown]{.75}{-6}{1/(abs(x)+1)+2} $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \vfill \psovalbox{A} \hfill \psovalbox{B} \hfill \psovalbox{C} \hfill \psovalbox{D} \hfill \psovalbox{E} none of these \clearpage \item[] {\bf Situation:} Problems \ref{pro:rat-a1} through \ref{pro:rat-b1} refer to the rational function $f$, with $f(x) =\dfrac{x^2+x}{x^2+x-2}$. \item \label{pro:rat-a1} As $x\rightarrow +\infty$, $y\rightarrow $\\ \psovalbox{A} $+\dfrac{1}{2}$\hfill \psovalbox{B} $-\dfrac{1}{2}$\hfill \psovalbox{C} $0$ \hfill \psovalbox{D} $1$\hfill \psovalbox{E} none of these \item The $y$-intercept of $f$ is located at \\ \psovalbox{A} $(0,-1)$\hfill \psovalbox{B} $(0,\frac{1}{2})$\hfill \psovalbox{C} $(0,1)$ \hfill \psovalbox{D} $(0,0)$\hfill \psovalbox{E} none of these \item Which of the following is true?\\ \psovalbox{A} $f$ has zeroes at $x=0$ and $x=-1$, and poles at $x=1$ and $x=-2$. \vfill \psovalbox{B} $f$ has zeroes at $x=0$ and $x=1$, and poles at $x=1$ and $x=2$.\vfill \psovalbox{C} $f$ has zeroes at $x=0$ and $x=-1$, and poles at $x=-1$ and $x=2$.\vfill \psovalbox{D} $f$ has no zeroes and no poles \vfill \psovalbox{E} none of these \item Which of the following is the sign diagram for $f$?\\ \renewcommand{\arraystretch}{2.1} \psset{unit=.85pc} \def\plus{\psline(-.2,0)(.2,0)\psline(0,-.2)(0,.2)} \def\minus{\psline(-.2,0)(.2,0)} {\footnotesize \psovalbox{A} $\begin{array}{|c|c|c|c|c|}\hline ]-\infty;-2[ & ]-2;-1[ & ]-1;0[ & ]0;1[ & ]1;+\infty[ \\ \hline \plus & \minus & \plus & \minus & \minus \\ \hline \end{array}$ \vfill \psovalbox{B} $\begin{array}{|c|c|c|c|c|}\hline ]-\infty;-2[ & ]-2;-1[ & ]-1;0[ & ]0;1[ & ]1;+\infty[ \\ \hline \minus & \minus & \plus & \plus & \plus \\ \hline \end{array}$ \vfill \psovalbox{C} $\begin{array}{|c|c|c|c|c|}\hline ]-\infty;-2[ & ]-2;-1[ & ]-1;0[ & ]0;1[ & ]1;+\infty[ \\ \hline \plus & \plus & \minus & \plus & \minus \\ \hline \end{array}$ \vfill \psovalbox{D} $\begin{array}{|c|c|c|c|c|}\hline ]-\infty;-2[ & ]-2;-1[ & ]-1;0[ & ]0;1[ & ]1;+\infty[ \\ \hline \plus & \minus & \plus & \minus & \plus \\ \hline \end{array}$ \vfill \psovalbox{E} none of these } \item \label{pro:rat-b1} The graph of $y=f(x)$ most resembles\\ \vspace{2cm} \begin{figure}[h] \def\graph{\psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-11}{-2.065247584}{(x^2+x)/(x^2+x-2)} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-1.943375673}{.9433756730}{(x^2+x)/(x^2+x-2)} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{1.065247584}{11}{(x^2+x)/(x^2+x-2)}} \begin{minipage}{4cm} $$\psset{unit=.4pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-12,-12)(12,12) \rput{180}(0,0){\graph} $$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.4pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-12,-12)(12,12) \graph$$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.4pc}\psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-12,-12)(12,12) \rput(-2,0){\graph} $$ \meinecaption{1}{C} \end{minipage} \hfill \hfill \begin{minipage}{4cm} $$\psset{unit=.4pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-12,-12)(12,12) \rput(2,0){\graph}$$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \vfill \psovalbox{A} \hfill \psovalbox{B} \hfill \psovalbox{C} \hfill \psovalbox{D} \hfill \psovalbox{E} none of these \vfill \clearpage \item[] {\bf Situation:} Problems \ref{pro:rat-a} through \ref{pro:rat-b2} refer to the rational function $f$, with $f(x) =\dfrac{(x+1)^2(x-2)}{(x-1)(x+2)^2}$. \item \label{pro:rat-a2} As $x\rightarrow +\infty$, $y\rightarrow $\\ \psovalbox{A} $+\dfrac{1}{2}$\hfill \psovalbox{B} $-\dfrac{1}{2}$\hfill \psovalbox{C} $0$ \hfill \psovalbox{D} $1$\hfill \psovalbox{E} none of these \item The $y$-intercept of $f$ is located at \\ \psovalbox{A} $(0,-1)$\hfill \psovalbox{B} $(0,1)$\hfill \psovalbox{C} $(0,-\frac{1}{2})$ \hfill \psovalbox{D} $(0,\frac{1}{2})$\hfill \psovalbox{E} none of these \item Which of the following is true?\\ \psovalbox{A} $f$ has zeroes at $x=-1$ and $x=2$, and poles at $x=1$ and $x=-2$. \vfill \psovalbox{B} $f$ has zeroes at $x=1$ and $x=-2$, and poles at $x=-1$ and $x=2$.\vfill \psovalbox{C} $f$ has zeroes at $x=1$ and $x=2$, and poles at $x=-1$ and $x=-2$.\vfill \psovalbox{D} $f$ has no zeroes and no poles \vfill \psovalbox{E} none of these \item Which of the following is the sign diagram for $f$?\\ \renewcommand{\arraystretch}{2.1} \psset{unit=.85pc} \def\plus{\psline(-.2,0)(.2,0)\psline(0,-.2)(0,.2)} \def\minus{\psline(-.2,0)(.2,0)} {\footnotesize \psovalbox{A} $\begin{array}{|c|c|c|c|c|}\hline ]-\infty;-2[ & ]-2;-1[ & ]-1;1[ & ]1;2[ & ]2;+\infty[ \\ \hline \plus & \minus & \plus & \minus & \minus \\ \hline \end{array}$ \vfill \psovalbox{B} $\begin{array}{|c|c|c|c|c|}\hline ]-\infty;-2[ & ]-2;-1[ & ]-1;1[ & ]1;2[ & ]2;+\infty[ \\ \hline \minus & \minus & \plus & \plus & \plus \\ \hline \end{array}$ \vfill \psovalbox{C} $\begin{array}{|c|c|c|c|c|}\hline ]-\infty;-2[ & ]-2;-1[ & ]-1;1[ & ]1;2[ & ]2;+\infty[ \\ \hline \plus & \plus & \minus & \plus & \minus \\ \hline \end{array}$ \vfill \psovalbox{D} $\begin{array}{|c|c|c|c|c|}\hline ]-\infty;-2[ & ]-2;-1[ & ]-1;1[ & ]1;2[ & ]2;+\infty[ \\ \hline \plus & \plus & \plus & \minus & \plus \\ \hline \end{array}$ \vfill \psovalbox{E} none of these } \item \label{pro:rat-b2} The graph of $y=f(x)$ most resembles\\ \vspace{2cm} \begin{figure}[h] \def\graph{\psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-11}{-2.519009939}{(x+1)^2*(x-2)/((x+2)^2*(x-1))} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-1.739497661}{.9585075992}{(x+1)^2*(x-2)/((x+2)^2*(x-1))} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{1.039319160}{11}{(x+1)^2*(x-2)/((x+2)^2*(x-1))}} \begin{minipage}{4cm} $$\psset{unit=.4pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-12,-12)(12,12) \rput(0,-2){\graph} $$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.4pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-12,-12)(12,12) \graph$$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.4pc}\psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-12,-12)(12,12) \rput(-2,0){\graph} $$ \meinecaption{1}{C} \end{minipage} \hfill \hfill \begin{minipage}{4cm} $$\psset{unit=.4pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-12,-12)(12,12) \rput(2,0){\graph}$$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \vfill \psovalbox{A} \hfill \psovalbox{B} \hfill \psovalbox{C} \hfill \psovalbox{D} \hfill \psovalbox{E} none of these \vfill \clearpage \item[] {\bf Situation:} Problems \ref{pro:rat-a3} through \ref{pro:rat-b3} refer to the rational function $f$ whose graph appears in figure \ref{fig:rat3}. The function $f$ is of the form $$ f(x)=K\dfrac{(x-a)(x-b)^2}{(x-c)^4}, $$where $K, a, b, c$ are real constants that you must find. It is known that $f(x)\to +\infty$ as $x\to 1$. \vspace{3cm} \begin{figure}[h] $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-12,-12)(12,12) \def\fx{20*(x+1)*(x-2)^2/(x-1)^4} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-10}{-.5534304004}{\fx} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{1.756351042}{10}{\fx} \psgrid[subgriddiv=0,linewidth=.8pt, gridlabels=.5](-12,-12)(12,12) \psdots[dotscale=1.5](3,5)(-1,0)(2,0) $$ \meinecaption{3}{Problems \ref{pro:rat-a} through \ref{pro:rat-b}. }\label{fig:rat} \end{figure} \item \label{pro:rat-a3} Which of the following is true?\\\ \psovalbox{A} $a=1$, $b=-1$, $c=2$\\ \psovalbox{B} $a=-1$, $b=2$, $c=1$\\\ \psovalbox{C} $a=-1$, $b=1$, $c=2$\\\ \psovalbox{D} $a=2$, $b=-1$, $c=1$\\ \psovalbox{E} none of these \vfill \item What is the value of $K$? \\ \vfill \psovalbox{A} $10$ \hfill \psovalbox{B} $20$ \hfill \psovalbox{C} $-20$ \hfill \psovalbox{D} $1$ \hfill \psovalbox{E} none of these \vfill \item \label{pro:rat-b3} As $x\to +\infty$, $f(x)\to$\\ \vfill \psovalbox{A} $0$ \hfill \psovalbox{B} $1$\hfill \psovalbox{C} $+\infty$ \hfill \psovalbox{D} $-\infty$ \hfill \psovalbox{E} none of these \vfill \clearpage \item[] {\bf Situation:} Problems \ref{pro:rat-a4} through \ref{pro:rat-b4} refer to the rational function $f$, with $f(x) =\dfrac{x^3}{x^2-4}$. \item \label{pro:rat-a4} As $x\rightarrow +\infty$, $y\rightarrow $\\ \psovalbox{A} $+\infty$\hfill \psovalbox{B} $-\infty$\hfill \psovalbox{C} $0$ \hfill \psovalbox{D} $1$\hfill \psovalbox{E} none of these \item As $x\rightarrow -\infty$, $y\rightarrow $\\ \psovalbox{A} $+\infty$\hfill \psovalbox{B} $-\infty$\hfill \psovalbox{C} $0$ \hfill \psovalbox{D} $1$\hfill \psovalbox{E} none of these \item Where are the poles of $f$?\\ \psovalbox{A} $x=2$ and $x=-2$\hfill \psovalbox{B} $x=-1$ and $x=-2$\hfill \psovalbox{C} $x=0$ and $x=2$ \hfill \psovalbox{D} $x=0$ and $x=-2$\hfill \psovalbox{E} none of these \item Which of the following is true?\\ \psovalbox{A} $x=0$ is the only zero of $f$ \vfill \psovalbox{B} $x=-2$ and $x=+2$ are the only zeroes of $f$ \vfill \psovalbox{C} $x=0$, $x=2$, and $x=-2$ are all zeroes of $f$ \vfill \psovalbox{D} $f$ has no zeroes \vfill \psovalbox{E} none of these \item \label{pro:rat-b4} The graph of $y=f(x)$ most resembles\\ \vspace{1cm}\vfill \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.4pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-12,-12)(12,12) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-11.64610031}{-2.214807191}{-(x^3)/(x^2-4)} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-1.860907499}{1.860907499}{(x^3)/(x^2-4)} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{2.214807191}{11.64610031}{-(x^3)/(x^2-4)} $$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.4pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-12,-12)(12,12) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-11.64610031}{-2.214807191}{(x^3)/(x^2-4)} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-1.860907499}{1.860907499}{-(x^3)/(x^2-4)} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{2.214807191}{11.64610031}{(x^3)/(x^2-4)} $$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.4pc}\psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-12,-12)(12,12) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-11.64610031}{-2.214807191}{-(x^3)/(x^2-4)} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-1.860907499}{1.860907499}{-(x^3)/(x^2-4)} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{2.214807191}{11.64610031}{-(x^3)/(x^2-4)} $$ \meinecaption{1}{C} \end{minipage} \hfill \hfill \begin{minipage}{4cm} $$\psset{unit=.4pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-12,-12)(12,12) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-11.64610031}{-2.214807191}{(x^3)/(x^2-4)} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-1.860907499}{1.860907499}{(x^3)/(x^2-4)} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{2.214807191}{11.64610031}{(x^3)/(x^2-4)} $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \vfill \psovalbox{A} \hfill \psovalbox{B} \hfill \psovalbox{C} \hfill \psovalbox{D} \hfill \psovalbox{E} none of these \clearpage \item[] {\bf Situation:} Problems \ref{pro:rat-a5} through \ref{pro:rat-b5} refer to the rational function $f$, with $f(x) =\dfrac{(x-1)(x+2)}{(x+1)(x-2)}$. \item \label{pro:rat-a5} Which of the following is a horizontal asymptote for $f$?\\ \psovalbox{A} $y=-1$\hfill \psovalbox{B} $y=1$\hfill \psovalbox{C} $y=0$ \hfill \psovalbox{D} $y=2$\hfill \psovalbox{E} none of these \item Where are the poles of $f$?\\ \psovalbox{A} $x=1$ and $x=-2$\hfill \psovalbox{B} $x=-1$ and $x=-2$\hfill \psovalbox{C} $x=-1$ and $x=2$ \hfill \psovalbox{D} $x=1$ and $x=2$\hfill \psovalbox{E} none of these \item Where are the zeroes of $f$?\\ \psovalbox{A} $x=1$ and $x=-2$\hfill \psovalbox{B} $x=-1$ and $x=-2$\hfill \psovalbox{C} $x=-1$ and $x=2$ \hfill \psovalbox{D} $x=1$ and $x=2$\hfill \psovalbox{E} none of these \item What is the $y$-intercept of $f$?\\ \psovalbox{A} $(0,1)$ \hfill \psovalbox{B} $(0,2)$ \hfill \psovalbox{C} $(0,-1)$ \hfill \psovalbox{D} $(0,-2)$ \hfill \psovalbox{E} none of these \item \label{pro:rat-b5} The graph of $y=f(x)$ most resembles\\ \vspace{1cm}\vfill \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-6 }{-1.101469847}{-(x-1)*(x+2)/((x+1)*(x-2))} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-.8779733838 }{1.815755561}{-(x-1)*(x+2)/((x+1)*(x-2))} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{2.277973384}{6}{-(x-1)*(x+2)/((x+1)*(x-2))} $$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-6 }{-1.101469847}{-(x-1)*(x+2)/((x+1)*(x-2))} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-.8779733838 }{1.815755561}{(x-1)*(x+2)/((x+1)*(x-2))} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{2.277973384}{6}{(x-1)*(x+2)/((x+1)*(x-2))}$$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc}\psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-6 }{-1.101469847}{(x-1)*(x+2)/((x+1)*(x-2))} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-.8779733838 }{1.815755561}{(x-1)*(x+2)/((x+1)*(x-2))} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{2.277973384}{6}{-(x-1)*(x+2)/((x+1)*(x-2))} $$ \meinecaption{1}{C} \end{minipage} \hfill \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-6 }{-1.101469847}{(x-1)*(x+2)/((x+1)*(x-2))} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{-.8779733838 }{1.815755561}{(x-1)*(x+2)/((x+1)*(x-2))} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{2.277973384}{6}{(x-1)*(x+2)/((x+1)*(x-2))} $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \vfill \psovalbox{A} \hfill \psovalbox{B} \hfill \psovalbox{C} \hfill \psovalbox{D} \hfill \psovalbox{E} none of these \subsection{Algebraic Functions} \item \label{pro:root} The graph in figure \ref{fig:root} below belongs to a curve with equation of the form $y = A\sqrt{x+3}-2$. Find $A$.\\ \vspace{3cm} \begin{figure}[h] \psset{unit=1pc}\centering \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \psplot[algebraic,linewidth=2pt,linecolor=blue, arrows={*->}]{-3}{5.5}{2*sqrt(x+3)-2}\psdots[dotscale=1.5](-3,-2)(1,2)(-2,0) \meinecaption{2.5}{Problem \ref{pro:root}.}\label{fig:root} \end{figure} \vspace{1cm} \pscirclebox{A} $A=\dfrac{1}{2}$ \hfill \pscirclebox{B} $A=1$ \hfill \pscirclebox{C} $A=-2$\hfill \pscirclebox{D} $A=2$ \hfill \pscirclebox{E} none of these \item Which one of the following graphs best represents the curve $y = -\sqrt{-x}$? \vfill\vspace{1.5cm} \vfill \begin{figure}[hptb] \begin{minipage}{4cm} \centering \psset{unit=1pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none,ticks=none](0,0)(-5,-5)(5,5) \psplot[algebraic=true,linecolor=brown,linewidth=2pt]{-5.7}{0}{-sqrt(-x)}\meinecaption{2}{A} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=1pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none,ticks=none](0,0)(-5,-5)(5,5) \psplot[algebraic=true,linecolor=brown,linewidth=2pt]{-5}{0}{sqrt(-x)} \meinecaption{2}{B} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=1pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none,ticks=none](0,0)(-5,-5)(5,5) \psplot[algebraic=true,linecolor=brown,linewidth=2pt]{0}{5}{-sqrt(x)} \meinecaption{2}{C} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=1pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none,ticks=none](0,0)(-5,-5)(5,5) \psplot[algebraic=true,linecolor=brown,linewidth=2pt]{0}{-2.236}{-x^2} \meinecaption{2}{D} \end{minipage} \end{figure} \vfill \vspace{1cm} \pscirclebox{A} A \hfill \pscirclebox{B} B \hfill \pscirclebox{C} C\hfill \pscirclebox{D} D \hfill \pscirclebox{E} none of these \item Which graph most resembles the curve $y=-\sqrt{x-1}$?\\ \vspace{1cm}\vfill \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-5.5,-5.5)(5.5,5.5) \psplot[algebraic=true,linewidth=1.5pt, arrows={*->},linecolor=brown]{1}{6}{-sqrt(x-1)} $$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-5.5,-5.5)(5.5,5.5) \psplot[algebraic=true,linewidth=1.5pt, arrows={*->},linecolor=brown]{-1}{6}{sqrt(x+1)} $$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc}\psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-5.5,-5.5)(5.5,5.5) \psplot[algebraic=true,linewidth=1.5pt, arrows={*->},linecolor=brown]{0}{6}{1-sqrt(x)} $$ \meinecaption{1}{C} \end{minipage} \hfill \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-5.5,-5.5)(5.5,5.5) \psplot[algebraic=true,linewidth=1.5pt, arrows={*->},linecolor=brown]{0}{6}{-sqrt(x)-1} $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \vfill \psovalbox{A} \hfill \psovalbox{B} \hfill \psovalbox{C} \hfill \psovalbox{D} \hfill \psovalbox{E} none of these \item Which graph most resembles the curve $y=\sqrt{1-x}$?\\ \vspace{1cm}\vfill \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={*->},linecolor=brown]{1}{6}{sqrt(x-1)} $$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<-*},linecolor=brown]{-6}{1}{sqrt(1-x)} $$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc}\psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<-*},linecolor=brown]{-6}{0}{sqrt(-x)+1} $$ \meinecaption{1}{C} \end{minipage} \hfill \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={*->},linecolor=brown]{0}{6}{1-sqrt(x)} $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \vfill \psovalbox{A} \hfill \psovalbox{B} \hfill \psovalbox{C} \hfill \psovalbox{D} \hfill \psovalbox{E} none of these \item[] {\bf Situation:} Problems \ref{pro:algebraic-sum1-06a} through \ref{pro:algebraic-sum1-06b} refer to the assignment rule given by $a(x) = \sqrt{\dfrac{x+1}{x-1}}$. \item \label{pro:algebraic-sum1-06a} What is the domain of definition of $a$?\\ \psovalbox{A} $\co{-1;1}$ \hfill \psovalbox{B} $\cc{-1;1}$ \hfill \psovalbox{C} $\oc{-\infty;-1} \cup \co{1;+\infty}$\hfill \psovalbox{D} $\oc{-\infty;-1} \cup \oo{1;+\infty}$ \hfill \psovalbox{E} none of these \item What is $a(2)$?\\ \psovalbox{A} $\sqrt{3}$ \hfill \psovalbox{B} $\dfrac{1}{\sqrt{3}}$ \hfill \psovalbox{C} $\sqrt{2}$\hfill \psovalbox{D} undefined \hfill \psovalbox{E} none of these \item $a^{-1}(x)=$ \\ \psovalbox{A} $\dfrac{1-x^2}{1+x^2}$ \hfill \psovalbox{B} $\left(\dfrac{1+x}{1-x}\right)^2$ \hfill \psovalbox{C} $\dfrac{1+x^2}{1-x^2}$ \hfill \psovalbox{D} $\dfrac{1+x^2}{x^2-1}$ \hfill \psovalbox{E} none of these \item \label{pro:algebraic-sum1-06b} The graph of $a$ most resembles\\ \vspace*{2cm} \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<-*},linecolor=brown]{-7}{-1}{sqrt((x+1)/(x-1))} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{1.041666667}{7}{-sqrt((x+1)/(x-1))} $$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<-*},linecolor=brown]{-7}{-1}{sqrt((x+1)/(x-1))} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{1.041666667}{7}{sqrt((x+1)/(x-1))}$$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc}\psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<-*},linecolor=brown]{-7}{-1}{-sqrt((x+1)/(x-1))} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{1.041666667}{7}{-sqrt((x+1)/(x-1))} $$ \meinecaption{1}{C} \end{minipage} \hfill \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-7,-7)(7,7) \psplot[algebraic=true,linewidth=1.5pt, arrows={<-*},linecolor=brown]{-7}{-1}{-sqrt((x+1)/(x-1))} \psplot[algebraic=true,linewidth=1.5pt, arrows={<->},linecolor=brown]{1.041666667}{7}{sqrt((x+1)/(x-1))} $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \vfill \psovalbox{A} \hfill \psovalbox{B} \hfill \psovalbox{C} \hfill \psovalbox{D} \hfill \psovalbox{E} none of these \clearpage \subsection{Conics} \item \label{pro:ellipse}Find the equation of the ellipse in figure \ref{fig:ellipse}. \\ \vfill \vspace{2cm} \vfill \begin{figure}[h] $$\psset{unit=.7pc} \renewcommand{\pshlabel}[1]{{\tiny #1}} \renewcommand{\psvlabel}[1]{{\tiny #1}} \psgrid[subgriddiv=0,linewidth=.8pt, gridlabels=.5](-8,-8)(8,8) \psline[linewidth=1.3pt](0,-8)(0,8) \psline[linewidth=1.3pt](-8,0)(8,0) \psellipse[linewidth=2pt,linecolor=brown](2,3)(1,4)\psdots[dotscale=1,dotstyle=*](2,3)(3,3)(1,3)(2,7)(2,-1) $$ \meinecaption{1.5}{Problem \ref{pro:ellipse}.} \label{fig:ellipse} \end{figure} \vfill \psovalbox{A} $(x-2)^2 + \dfrac{(y-3)^2}{16}=1$\vfill \psovalbox{B} $(x+2)^2 + \dfrac{(y+3)^2}{16}=1$ \vfill \psovalbox{C} $(x-2)^2 + \dfrac{(y-3)^2}{4}=1$ \vfill \psovalbox{D} $(x+2)^2 + \dfrac{(y+3)^2}{4}=1$ \vfill \psovalbox{E} none of these \vfill \item \label{pro:hyperbola}Find the equation of the hyperbola in figure \ref{fig:hyperbola}. \\ \vfill \vspace{2cm} \vfill \begin{figure}[h] $$\psset{unit=.7pc} \renewcommand{\pshlabel}[1]{{\tiny #1}} \renewcommand{\psvlabel}[1]{{\tiny #1}} \psgrid[subgriddiv=0,linewidth=.8pt, gridlabels=.5](-8,-8)(8,8) \psline[linewidth=1.3pt](0,-8)(0,8) \psline[linewidth=1.3pt](-8,0)(8,0) \psplotImp[linewidth=2pt,linecolor=brown, algebraic](-4,-6)(6,8){ (x-1)^2-(y+1)^2-1}\psdots[dotscale=1,dotstyle=*](1,-1)(0,-1)(2,-1) $$ \meinecaption{1.5}{Problem \ref{pro:hyperbola}.} \label{fig:hyperbola} \end{figure} \vfill \psovalbox{A} $(x-1)^2 - (y-1)^2=1$\vfill \psovalbox{B} $(x-1)^2 - (y+1)^2=1$ \vfill \psovalbox{C} $(y-1)^2 - (x-1)^2=1$ \vfill \psovalbox{D} $(y+1)^2 - (x-1)^2=1$ \vfill \psovalbox{E} none of these \vfill \subsection{Geometric Series} \item Find the sum of the terms of the infinite geometric progression $$ 1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}+\cdots .$$ \vfill \psovalbox{A} $\dfrac{4}{3}$ \hfill \psovalbox{B} $\dfrac{3}{4}$ \hfill \psovalbox{C} $\dfrac{1}{4}$ \hfill \psovalbox{D} $\dfrac{1}{3}$ \hfill \psovalbox{E} none of these \subsection{Exponential Functions} \item Which of the following best resembles the graph of the curve $y=2^{-|x|}$? \\ \vspace{2cm} \begin{figure}[h] \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, ticks=none, linewidth=.7pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psplot[linecolor=black,linewidth=2pt, arrows={<->}]{-2.5}{2.5}{2^(-x)}$$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none,ticks=none, linewidth=.7pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psplot[linecolor=black,linewidth=2pt, arrows={<->}]{-4.5}{4.5}{2^(-abs(x))}$$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, ticks=none, linewidth=.7pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psplot[linecolor=black,linewidth=2pt, arrows={<->}]{-2.5}{2.5}{2^(abs(x))} $$ \meinecaption{1}{C} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, ticks=none, linewidth=.7pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psplot[linecolor=black,linewidth=2pt, arrows={<->}]{-2.5}{2.5}{2^(x)} $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \vspace{.5cm} \psovalbox{A} A\hfill \psovalbox{B} B \hfill \psovalbox{C} C\hfill \psovalbox{D} D \hfill \psovalbox{E} none of these \item If $3^{x^2} = 81 $, then \vfill \psovalbox{A} $x\in \{-4,4\}$ \hfill \psovalbox{B} $x\in \{-9,9\}$ \hfill \psovalbox{C} $x\in \{-2,2\}$ \hfill \psovalbox{D} $x\in\{-3,-3\}$ \hfill \psovalbox{E} none of these \item If the number $5^{2000}$ is written out (in decimal notation), how many digits does it have? \vfill \psovalbox{A} $1397$ \hfill \psovalbox{B} $1398$ \hfill \psovalbox{C} $1396$ \hfill \psovalbox{D} $2000$ \hfill \psovalbox{E} none of these \vfill \subsection{Logarithmic Functions} \item Which of the following best resembles the graph of the curve $y=\log_{1/2}x$? \\ \vspace{2cm} \begin{figure}[h] \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, ticks=none, linewidth=.7pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psplot[linecolor=black,linewidth=2pt, arrows={<->}]{.01}{5}{(ln(x))/(ln(1/2))}$$ \meinecaption{1}{A} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none,ticks=none, linewidth=.7pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psplot[linecolor=black,linewidth=2pt, arrows={<->}]{.01}{5}{(ln(x))/(ln(2))}$$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, ticks=none, linewidth=.7pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psplot[linecolor=black,linewidth=2pt, arrows={->}]{.01}{2.5}{2^x} $$ \meinecaption{1}{C} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{algebraic=true, unit=.7pc} \psaxes[labels=none, ticks=none, linewidth=.7pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5) \psplot[linecolor=black,linewidth=2pt, arrows={->}]{.01}{5}{2^(-x)} $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \vspace{.5cm} \psovalbox{A} A\hfill \psovalbox{B} B \hfill \psovalbox{C} C\hfill \psovalbox{D} D \hfill \psovalbox{E} none of these \item Find the smallest integer $n$ for which the inequality $2^n > 4n^2+n$ will be true. \\ \psovalbox{A} $n=4$\hfill \psovalbox{B} $n=7$ \hfill \psovalbox{C} $n=8$ \hfill \psovalbox{D} $n=9$ \hfill \psovalbox{E} none of these \item Solve the equation $9^x+3^x-6=0$. \\ \psovalbox{A} $x\in\{1,\log_3 2\}$\hfill \psovalbox{B} $x\in\{\log_3 2\}$ only \hfill \psovalbox{C} $x\in\{1\}$ only \hfill \psovalbox{D} $x\in\{\log_2 3,\log_3 2\}$ \hfill \psovalbox{E} none of these \item Find the exact value of $\log_{3\sqrt{3}} 729$. \\ \psovalbox{A} $ \dfrac{1}{9} $\hfill \psovalbox{B} $\dfrac{1}{4}$ \hfill \psovalbox{C} $9$\hfill \psovalbox{D} $4$ \hfill \psovalbox{E} none of these \item Let $a$ and $b$ be consecutive integers such that $a<\log_5 100 1$, $t>0$, $s>0$ and that $$\log_a t^3 = p, \qquad \log _{\sqrt{a}}s^2 = q,$$ find $\log_a st$ in terms of $p$ and $q$. \\ \psovalbox{A} $\dfrac{p}{3} + \dfrac{q}{2}$\hfill \psovalbox{B} $\dfrac{p}{3} + \dfrac{q}{4}$ \hfill \psovalbox{C} $3p+4q$ \hfill \psovalbox{D} $\dfrac{p}{3} + q$ \hfill \psovalbox{E} none of these \item Given that $a>1$, $s>1$, $t>1$, and that $$\log_a \sqrt{t} = p, \qquad \log _s a^2 = 2p^2,$$ find $\log_st$ in terms of $p$. \\ \psovalbox{A} $p^3$\hfill \psovalbox{B} $\dfrac{2}{p^3}$ \hfill \psovalbox{C} $2p^3$ \hfill \psovalbox{D} $\dfrac{p^2}{2}$ \hfill \psovalbox{E} none of these \item What is the domain of definition of $$ x\mapsto \log_x (1-x^2) ?$$ \\ \psovalbox{A} $[-1;1]$\hfill \psovalbox{B} $]0;1]$ \hfill \psovalbox{C} $]0;1[$ \hfill \psovalbox{D} $]-1;1[$ \hfill \psovalbox{E} none of these \subsection{Goniometric Functions} \item How many solutions does $1-\cos 2x = \dfrac{1}{2}$ have in the closed interval $[-\frac{\pi}{2};\pi]$?\\ \vfill \psovalbox{A} $0$\hfill \psovalbox{B} $1$ \hfill \psovalbox{C} $2$ \hfill \psovalbox{D} $3$ \hfill \psovalbox{E} none of these \item How many of the following assertions are true for all real numbers $x$?\\ $$I: \csc^2x+\sec^2x=1; \qquad II: |\csc x| \geq 1; \qquad III: |\arcsin x| \leq 1; \qquad IV: \sin (2\pi +x) = \sin x $$ \vfill \psovalbox{A} none\hfill \psovalbox{B} exactly one\hfill \psovalbox{C} exactly two \hfill \psovalbox{D} exactly three \hfill \psovalbox{E} all four \item Which of the following is a solution to the equation \hspace{1cm} $\cos (2x-1) = \dfrac{1}{2}$? \\ \vfill \psovalbox{A} $\dfrac{\pi}{6}+\dfrac{1}{2}$ \hfill \psovalbox{B} $\dfrac{\pi}{3}+\dfrac{1}{2}$\hfill \psovalbox{C} $\dfrac{\pi}{6}-\dfrac{1}{2}$ \hfill \psovalbox{D} $\dfrac{\pi}{3}-\dfrac{1}{2}$ \hfill \psovalbox{E} none of these \item If $\tan \theta = \dfrac{1}{4}$ and $\winding{\theta}$ is in the third quadrant, find $\sin \theta$. \vfill \psovalbox{A} $\dfrac{-\sqrt{17}}{4}$\hfill \psovalbox{B} $-\dfrac{4}{\sqrt{17}}$ \hfill \psovalbox{C} $-\dfrac{1}{\sqrt{17}}$ \hfill \psovalbox{D} $\dfrac{1}{\sqrt{17}}$ \hfill \psovalbox{E} none of these \item Find $\arcsin (\sin 10)$. \vfill \psovalbox{A} $10$\hfill \psovalbox{B} $10-3\pi$ \hfill \psovalbox{C} $3\pi-10$ \hfill \psovalbox{D} $10-\dfrac{7\pi}{2}$ \hfill \psovalbox{E} none of these \item Find $\sin (\arcsin 4)$. \vfill \psovalbox{A} $4$\hfill \psovalbox{B} $\sqrt{15}$ \hfill \psovalbox{C} $\sqrt{17}$ \hfill \psovalbox{D} $4-\pi$ \hfill \psovalbox{E} not a real number \item $\sec^2x+\csc^2x =$\vfill \psovalbox{A} $(\sec^2x)(\csc^2x)$\hfill \psovalbox{B} $(\sec x)(\csc x)$\hfill \psovalbox{C}$\sec x + \csc x$\hfill \psovalbox{D} $\tan^2x+\cot^2x$\hfill \psovalbox{E} none of these \vfill\clearpage \item[] {\bf Situation:} Let $\sin x = \frac{1}{3}$ and $\sin y=\frac{1}{4}$ where $x$ and $y$ are acute angles. Problems \ref{pro:sum_angles_begin} through \ref{pro:sum_angles_end} refer to this situation. \item \label{pro:sum_angles_begin} Find $\cos x$.\vfill \psovalbox{A} $\dfrac{2}{3}$\hfill \psovalbox{B} $\dfrac{2\sqrt{2}}{3}$ \hfill \psovalbox{C} $-\dfrac{2}{3}$ \hfill \psovalbox{D} $-\dfrac{2\sqrt{2}}{3}$ \hfill \psovalbox{E} none of these \hfill \item Find $\cos 2x$.\vfill \psovalbox{A} $\dfrac{2}{3}$\hfill \psovalbox{B} $\dfrac{4\sqrt{2}}{3}$ \hfill \psovalbox{C} $\dfrac{7}{9}$ \hfill \psovalbox{D} $\dfrac{\sqrt{2}}{3}$ \hfill \psovalbox{E} none of these \hfill \item Find $|\cos \frac{x}{2}|$.\vfill \psovalbox{A} $\dfrac{1}{3}$\hfill \psovalbox{B} $\frac{1}{2}\sqrt{\frac{1}{2}-\frac{\sqrt{3}}{3}}$ \hfill \psovalbox{C} $\sqrt{\dfrac{17}{18}}$ \hfill \psovalbox{D} $\frac{1}{2}\sqrt{\frac{1}{2}+\frac{\sqrt{3}}{3}}$ \hfill \psovalbox{E} none of these \hfill \item Find $\cos y$.\vfill \psovalbox{A} $\dfrac{3}{4}$\hfill \psovalbox{B} $\dfrac{\sqrt{15}}{4}$ \hfill \psovalbox{C} $-\dfrac{3}{4}$ \hfill \psovalbox{D} $-\dfrac{\sqrt{15}}{4}$ \hfill \psovalbox{E} none of these \hfill \item Find $\sin (x+y)$.\vfill \psovalbox{A} $\dfrac{7}{12}$\hfill \psovalbox{B} $\dfrac{1}{12}$ \hfill \psovalbox{C} $\dfrac{2\sqrt{2}}{9}+\dfrac{\sqrt{15}}{16}$ \hfill \psovalbox{D} $\dfrac{\sqrt{15}+2\sqrt{2}}{12}$ \hfill \psovalbox{E} none of these \hfill \item \label{pro:sum_angles_end} Find $\cos (x+y)$.\vfill \psovalbox{A} $\dfrac{\sqrt{30}}{6}+\dfrac{1}{12}$\hfill \psovalbox{B} $\dfrac{\sqrt{30}}{12}-\dfrac{1}{12}$ \hfill \psovalbox{C} $\dfrac{\sqrt{30}}{12}+\dfrac{1}{12}$ \hfill \psovalbox{D} $\dfrac{\sqrt{30}}{6}-\dfrac{1}{12}$ \hfill \psovalbox{E} none of these \hfill \item Which of the following is a real number solution to $2^{\cos x} =3$? \vfill \psovalbox{A} $\arccos \left(\dfrac{\ln 2}{\ln 3}\right)$ \hfill \psovalbox{B} $\arccos \left(\ln \dfrac{3}{2}\right)$\hfill \psovalbox{C} $\arccos \left(\dfrac{\ln 3}{\ln 2}\right)$ \hfill \psovalbox{D} $\arccos \left(\ln 6\right)$ \hfill \psovalbox{E} there are no real solutions \item $(\cos 2x)(\cos \frac{x}{2}) =$ \vfill \psovalbox{A} $\frac{1}{2}\sin \frac{5}{2}x - \frac{1}{2}\sin \frac{3}{2}x$ \vfill \psovalbox{B} $\frac{1}{2}\sin \frac{5}{2}x + \frac{1}{2}\sin \frac{3}{2}x$\vfill \psovalbox{C} $\frac{1}{2}\cos \frac{5}{2}x + \frac{1}{2}\cos \frac{3}{2}x$\vfill \psovalbox{D} $\frac{1}{2}\cos \frac{5}{2}x - \frac{1}{2}\cos \frac{3}{2}x$\vfill \psovalbox{E} none of these \item It is known that $\cos \frac{2\pi}{5} = \dfrac{\sqrt{5}-1}{2}$. Find $\cos \frac{\pi}{5}$. \vfill \psovalbox{A} $\dfrac{\sqrt{5}-1}{4}$\hfill \psovalbox{B} $\dfrac{\sqrt{5}+1}{2}$ \hfill \psovalbox{C} $-\dfrac{\sqrt{1+\sqrt{5}}}{2}$ \hfill \psovalbox{D} $\dfrac{\sqrt{1+\sqrt{5}}}{2}$ \hfill \psovalbox{E} none of these \item It is known that $\cos \frac{2\pi}{5} = \dfrac{\sqrt{5}-1}{2}$. Find $\cos \frac{4\pi}{5}$. \\ \psovalbox{A} $2-\sqrt{5}$\hfill \psovalbox{B} $\sqrt{5}-2$ \hfill \psovalbox{C} $3-\sqrt{5}$ \hfill \psovalbox{D} $\dfrac{3-\sqrt{5}}{2}$ \hfill \psovalbox{E} none of these \vfill \item Find the smallest positive solution to the equation $\cos x^2 = 0$.\\ \psovalbox{A} $0$ \hfill \psovalbox{B} $\dfrac{\sqrt{2\pi}}{2}$ \hfill \psovalbox{C} $\dfrac{\sqrt{\pi}}{2}$ \hfill \psovalbox{D} $\dfrac{\pi}{2}$ \hfill \psovalbox{E} none of these \item $\cos\frac{223\pi}{6} =$ \\ \psovalbox{A} $\frac{1}{2}$\hfill \psovalbox{B} $-\frac{1}{2}$\hfill \psovalbox{C} $-\frac{\sqrt{3}}{2}$\hfill \psovalbox{D} $\frac{\sqrt{3}}{2}$\hfill \psovalbox{E} none of these \hfill \item If $2\cos^2x+\cos x-1=0$ and $x\in[0;\pi]$ then \\ \vfill \psovalbox{A} $x\in\left\{\dfrac{\pi}{3},\pi\right\}$ \hfill \psovalbox{B} $x\in\left\{\dfrac{\pi}{2},\pi\right\}$ \hfill \psovalbox{C} $x\in\left\{\dfrac{\pi}{3},\dfrac{\pi}{4}\right\}$ \hfill \psovalbox{D} $x\in\left\{\dfrac{\pi}{3},\dfrac{\pi}{6}\right\}$\hfill \psovalbox{E} none of these \item If $2\sin^2x-\cos x-1=0$ and $x\in[0;\pi]$ then \\ \vfill \psovalbox{A} $x\in\left\{\dfrac{\pi}{3},\pi\right\}$ \hfill \psovalbox{B} $x\in\left\{\dfrac{\pi}{2},\pi\right\}$ \hfill \psovalbox{C} $x\in\left\{\dfrac{\pi}{3},\dfrac{\pi}{4}\right\}$ \hfill \psovalbox{D} $x\in\left\{\dfrac{\pi}{3},\dfrac{\pi}{6}\right\}$\hfill \psovalbox{E} none of these \subsection{Trigonometry} \item[] \noindent {\bf Situation:} Questions \ref{pro:m162_s06_trigs1} through \ref{pro:m162_s06_trigs2} refer to the following. Assume that $\alpha$ and $\beta$ are acute angles. Assume also that $\tan \alpha = \dfrac{1}{3}$ and that $\sec \beta =3$. \item \label{pro:m162_s06_trigs1} Find $\sin \alpha$. \\ \vfill \psovalbox{A} $\dfrac{1}{4}$ \hfill \psovalbox{B} $\dfrac{3\sqrt{10}}{10}$ \hfill \psovalbox{C} $\dfrac{\sqrt{10}}{30}$ \hfill \psovalbox{D} $\dfrac{\sqrt{10}}{10}$ \hfill \psovalbox{E} none of these \item Find $\sin \beta$. \\ \vfill \psovalbox{A} $\dfrac{1}{3}$ \hfill \psovalbox{B} $\dfrac{\sqrt{10}}{3}$ \hfill \psovalbox{C} $\dfrac{2\sqrt{2}}{3}$ \hfill \psovalbox{D} $\dfrac{\sqrt{2}}{3}$ \hfill \psovalbox{E} none of these \item Find $\cos\alpha$. \\ \vfill \psovalbox{A} $\dfrac{1}{4}$ \hfill \psovalbox{B} $\dfrac{3\sqrt{10}}{10}$ \hfill \psovalbox{C} $\dfrac{\sqrt{10}}{30}$ \hfill \psovalbox{D} $\dfrac{\sqrt{10}}{10}$ \hfill \psovalbox{E} none of these \item Find $\cos\beta$. \\ \vfill \psovalbox{A} $\dfrac{1}{3}$ \hfill \psovalbox{B} $\dfrac{\sqrt{10}}{3}$ \hfill \psovalbox{C} $\dfrac{2\sqrt{2}}{3}$ \hfill \psovalbox{D} $\dfrac{\sqrt{2}}{3}$ \hfill \psovalbox{E} none of these \item \label{pro:m162_s06_trigs2} Find $\cos (\alpha + \beta)$. \\ \vfill \psovalbox{A} $\dfrac{\sqrt{10}}{10}-\dfrac{2\sqrt{5}}{15}$ \hfill \psovalbox{B} $\dfrac{\sqrt{10}}{10}+\dfrac{2\sqrt{5}}{15}$ \hfill \psovalbox{C} $\dfrac{2\sqrt{5}}{5}-\dfrac{\sqrt{10}}{30}$ \hfill \psovalbox{D} $\dfrac{2\sqrt{5}}{5}+\dfrac{\sqrt{10}}{30}$ \hfill \psovalbox{E} none of these \vfill \clearpage \item[] \noindent {\bf Situation:} Questions \ref{pro:m162_s06_triags1} through \ref{pro:m162_s06_triags2} refer to the following. $\triangle ABC$ is right-angled at $A$, $a = 4$ and $\sec B = 4$. Assume standard labelling. \item \label{pro:m162_s06_triags1} Find $\sin C$. \\ \vfill \psovalbox{A} $\dfrac{1}{4}$ \hfill \psovalbox{B} $\dfrac{3\sqrt{15}}{15}$ \hfill \psovalbox{C} $\dfrac{\sqrt{15}}{4}$ \hfill \psovalbox{D} $\dfrac{4\sqrt{15}}{15}$ \hfill \psovalbox{E} none of these \item Find $\angle C$, in radians. \\ \vfill \psovalbox{A} $\arcsin \dfrac{1}{4}$ \hfill \psovalbox{B} $\arccos \dfrac{1}{4}$ \hfill \psovalbox{C} $\arcsin \dfrac{\sqrt{15}}{4}$ \hfill \psovalbox{D} $\arccos \dfrac{4}{\sqrt{15}}$ \hfill \psovalbox{E} none of these \item Find $b$. \\ \vfill \psovalbox{A} $1$ \hfill \psovalbox{B} $\sqrt{15}$ \hfill \psovalbox{C} $4$ \hfill \psovalbox{D} $16$ \hfill \psovalbox{E} none of these \item Find $R$, the radius of the circumscribed circle to $\triangle ABC$. \\ \vfill \psovalbox{A} $2$ \hfill \psovalbox{B} $\dfrac{\sqrt{15}}{2}$ \hfill \psovalbox{C} $2\sqrt{15}$ \hfill \psovalbox{D} $\sqrt{15}$ \hfill \psovalbox{E} none of these \item Find the area of $\triangle ABC$. \\ \vfill \psovalbox{A} $2$ \hfill \psovalbox{B} $\dfrac{\sqrt{15}}{2}$ \hfill \psovalbox{C} $2\sqrt{15}$ \hfill \psovalbox{D} $\sqrt{15}$ \hfill \psovalbox{E} none of these \item \label{pro:m162_s06_triags2} Find $r$, the radius of the inscribed circle to $\triangle ABC$. \\ \vfill \psovalbox{A} $\dfrac{\sqrt{15}}{2\sqrt{15}+10}$ \hfill \psovalbox{B} $\dfrac{\sqrt{15}}{\sqrt{15}+5}$ \hfill \psovalbox{C} $\dfrac{\sqrt{15}+5}{\sqrt{15}}$ \hfill \psovalbox{D} $2$ \hfill \psovalbox{E} none of these \vfill \clearpage \item[] \noindent {\bf Situation:} Questions \ref{pro:m162_s06_hexagon1} through \ref{pro:m162_s06_hexagon2} refer to the following. In figure \ref{fig:m162_s06_hexagon}, a {\bf regular hexagon} is inscribed in a circle of radius $1$. \vspace{2cm} \begin{figure}[h] \centering\psset{unit=4pc}\pscircle[fillstyle=solid,fillcolor=black](0,0){1} \rput(0,0){\PstHexagon[fillstyle=solid,fillcolor=white]} \pstGeonode[PosAngle={90,-90}](1;120){A}(1;240){B}\pstLineAB[linewidth=2pt]{A}{B} \meinecaption{2}{Problems \ref{pro:m162_s06_hexagon1} through \ref{pro:m162_s06_hexagon2}. } \label{fig:m162_s06_hexagon} \end{figure} \item \label{pro:m162_s06_hexagon1} Find the area of the hexagon. \\ \vfill \psovalbox{A} $\dfrac{3}{2}$ \hfill \psovalbox{B} $3\sqrt{2}$ \hfill \psovalbox{C} $\dfrac{3\sqrt{3}}{4}$ \hfill \psovalbox{D} $\dfrac{3\sqrt{3}}{2}$ \hfill \psovalbox{E} none of these \item Find the perimeter of the hexagon. \\ \vfill \psovalbox{A} $6$ \hfill \psovalbox{B} $6\sqrt{3}$ \hfill \psovalbox{C} $3\sqrt{3}$ \hfill \psovalbox{D} $3$ \hfill \psovalbox{E} none of these \item Find the length of the line segment $AB$. \\ \vfill \psovalbox{A} $2$ \hfill \psovalbox{B} $\sqrt{5}$ \hfill \psovalbox{C} $3\sqrt{3}$ \hfill \psovalbox{D} $\sqrt{3}$ \hfill \psovalbox{E} none of these \item \label{pro:m162_s06_hexagon2} Find the shaded area outside the hexagon but inside the circle. \\ \vfill \psovalbox{A} $\pi-\dfrac{3}{2}$ \hfill \psovalbox{B} $\pi-3\sqrt{2}$ \hfill \psovalbox{C} $\pi-\dfrac{3\sqrt{3}}{4}$ \hfill \psovalbox{D} $\pi-\dfrac{3\sqrt{3}}{2}$ \hfill \psovalbox{E} none of these \end{enumerate} \clearpage \section{Old Exam Match Questions} \begin{enumerate} \item[] Match the equation with the appropriate graph. Observe that there are fewer graphs than equations, hence, some blank spaces will remain blank. \begin{multicols}{3} \item $x-y^2=3$, \rule{2cm}{2pt} \item $x^2-y^2=9$, \rule{2cm}{2pt} \item $\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$, \rule{2cm}{2pt} \item $y^2-x^2=9$, \rule{2cm}{2pt} \item $x^2+y=3$, \rule{2cm}{2pt} \item $x+y^2=3$, \rule{2cm}{2pt} \item $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$, \rule{2cm}{2pt} \item $x^2+y^2=9$, \rule{2cm}{2pt} \item $y-x^2=3$, \rule{2cm}{2pt} \item $x+y=3$, \rule{2cm}{2pt} \end{multicols} \vfill \vspace{2cm} \begin{figure}[hptb] \begin{minipage}{4cm} \centering \psset{unit=.5pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none](0,0)(-10,-10)(10,10) \pscircle[linecolor=brown,linewidth=2pt](0,0){3} \meinecaption{2}{Allan} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=.5pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none](0,0)(-10,-10)(10,10) \parametricplot[algebraic,linecolor=brown,linewidth=2pt]{-2.5}{2.5}{3-t^2|t} \meinecaption{2}{Bob} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=.5pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none](0,0)(-10,-10)(10,10) \psplot[algebraic=true,linecolor=brown,linewidth=2pt]{-2.5}{2.5}{3-x^2} \meinecaption{2}{Carmen} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=.5pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none](0,0)(-10,-10)(10,10) \psellipse[linecolor=brown,linewidth=2pt](0,0)(3,2) \meinecaption{2}{Donald} \end{minipage} \end{figure} \vfill\vspace{2cm} \begin{figure}[hptb] \begin{minipage}{4cm} \centering \psset{unit=.5pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none](0,0)(-10,-10)(10,10) \psplotImp[linewidth=2pt,linecolor=brown, algebraic](-8,-8)(8,8){ y^2-x^2-9} \meinecaption{2}{Edgard} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=.5pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none](0,0)(-10,-10)(10,10) \psplotImp[linewidth=2pt,linecolor=brown, algebraic](-8,-8)(8,8){ x^2-y^2-9} \meinecaption{2}{Frances} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=.5pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none](0,0)(-10,-10)(10,10) \psellipse[linecolor=brown,linewidth=2pt](0,0)(2,3) \meinecaption{2}{Gertrude} \end{minipage}\hfill \begin{minipage}{4cm} \centering \psset{unit=.5pc} \psaxes[arrows={->},linewidth=1.7pt,labels=none](0,0)(-10,-10)(10,10) \parametricplot[algebraic,linecolor=brown,linewidth=2pt]{-2.5}{2.5}{3+t^2|t} \meinecaption{2}{Harry} \end{minipage} \end{figure} \vfill \clearpage Figure \ref{fig:transformations} shows a functional curve $y = f(x)$. You are to match the letters of figures \ref{fig:transformations1begin} to \ref{fig:transformations1end} with the equations on $\alpha$ through $\mu$ below. Some figures may not match with any equation, or viceversa. \vfill \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt]{-2.5}{1.8}{x 2 add x -1 add x mul mul}$$\meinecaption{.8}{$y = f(x)$} \label{fig:transformations} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt]{-1.8}{2.5}{x -1 mul 2 add x -1 mul -1 add x -1 mul mul mul} $$ \meinecaption{.8}{A} \label{fig:transformations1begin} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt]{-1.8}{2.5}{x -1 mul 2 add x -1 mul -1 add x -1 mul mul mul abs} $$ \meinecaption{.8}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt]{-2.5}{2.5}{x abs -1 mul 2 add x abs -1 mul -1 add x abs -1 mul mul mul} $$ \meinecaption{.8}{C} \end{minipage} \hfill \end{figure} \vfill \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt]{-2.5}{2.5}{x abs -1 mul 2 add x abs -1 mul -1 add x abs -1 mul mul mul abs} $$ \meinecaption{.8}{D} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt]{-5}{3.6}{x 2 div 2 add x 2 div -1 add x 2 div mul mul} $$ \meinecaption{.8}{E} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt]{-1.8}{1.8}{x abs 2 add x abs -1 add x abs mul mul} $$ \meinecaption{.8}{F} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt]{-1.8}{1.8}{x abs 2 add x abs -1 add x abs mul mul abs -1 mul} $$ \meinecaption{.8}{G} \end{minipage} \hfill \end{figure} \vfill \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt]{-2.5}{1.8}{x 2 add x -1 add x mul mul abs -1 mul}$$ \meinecaption{.8}{H} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt]{-2.5}{1.8}{x 2 add x -1 add x mul mul 2 div} $$ \meinecaption{.8}{I} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \rput(1,1){\psplot[linecolor=brown,linewidth=2pt]{-2.5}{1.8}{x 2 add x -1 add x mul mul}} $$ \meinecaption{.8}{J} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \rput(-1,2){\psplot[linecolor=brown,linewidth=2pt]{-2.5}{1.8}{x 2 add x -1 add x mul mul}}$$ \meinecaption{.8}{K} \label{fig:transformations1end} \end{minipage} \hfill \end{figure} \vfill \bigskip \noindent \begin{tabular}{lll} $\alpha$. $y = f(-x) = \rule{2cm}{.2pt}$ & $\beta$. $y = -f(-x) = \rule{2cm}{.2pt}$ & $\gamma$. $y = f(-|x|) = \rule{2cm}{.2pt}$ \\ $\delta$. $y = f(x+1)+2 = \rule{2cm}{.2pt}$ & $\epsilon$.$y = |f(-|x|)| = \rule{2cm}{.2pt}$ & $\zeta$. $y = -|f(|x|)| = \rule{2cm}{.2pt}$ \\ $\eta$. $y = |f(-x)| = \rule{2cm}{.2pt}$ & $\theta$.$y = |f(-|x|/2)| = \rule{2cm}{.2pt}$ & $\iota$. $y = f(x/2) = \rule{2cm}{.2pt}$ \\ $\kappa$. $y = -|f(x)| = \rule{2cm}{.2pt}$ & $\lambda$. $y = \frac{1}{2}f(x) = \rule{2cm}{.2pt}$ & $\mu$. $y = f(x-1)+1 = \rule{2cm}{.2pt}$ \\ \end{tabular} \clearpage \item[] You are to match the letters of figures \ref{fig:curvesbegin} to \ref{fig:curvesend} with the equations on 13 through 24 below. Some figures may not match with any equation, or viceversa. (0.5 mark each)\vfill \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt,algebraic=true, arrows={<-*}]{-5}{0}{sqrt(-x)}$$ \meinecaption{1}{A} \label{fig:curvesbegin} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt,algebraic=true, arrows={<->}]{-4}{5}{abs(x-1)-1} $$ \meinecaption{1}{B} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt,algebraic=true, arrows={<->}]{-1.5}{3.5}{(x-1)^2-1} $$ \meinecaption{1}{C} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt,algebraic=true, arrows={<->}]{-1.5}{3.5}{abs((x-1)^2-1)} $$ \meinecaption{1}{D} \end{minipage} \hfill \end{figure} \vfill \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt,algebraic=true, arrows={<->}]{-3.5}{3.5}{(abs(x)-1)^2-1} $$ \meinecaption{1}{E} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt,algebraic=true, arrows={<-*}]{-5}{0}{-sqrt(-x)+1} $$ \meinecaption{1}{F} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt,algebraic=true]{.2}{5}{abs(1/x)-1} \psplot[linecolor=brown,linewidth=2pt,algebraic=true]{-.2}{-5}{abs(1/x)-1} $$ \meinecaption{1}{G} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt,algebraic=true]{-.2}{-5}{abs(1/x-1)} \psplot[linecolor=brown,linewidth=2pt,algebraic=true]{.2}{5}{abs(1/x-1)} $$ \meinecaption{1}{H} \end{minipage} \hfill \end{figure} \vfill \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt,algebraic=true]{1.17}{5}{1/(abs(x)-1)}\psplot[linecolor=brown,linewidth=2pt,algebraic=true]{-1.17}{-5}{1/(abs(x)-1)}$$ \meinecaption{1}{I} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt]{-2.5}{1.8}{x 2 add x -1 add x mul mul 2 div} $$ \meinecaption{1}{J} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt, algebraic=true, arrows={*-*}]{-3}{3}{2-sqrt(9-x^2)} $$ \meinecaption{1}{K} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=.7pc} \psaxes[labels=none, tickstyle=bottom, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(7,7.5) \psplot[linecolor=brown,linewidth=2pt,algebraic=true, arrows={*-*}]{-2}{2}{1+sqrt(4-x^2)}$$ \meinecaption{1}{L} \label{fig:curvesend} \end{minipage} \hfill \end{figure} \vfill \item[] \begin{tabular}{lll} 13. $y = (x-1)^2-1 = \rule{2cm}{.2pt}$ & 14. $y = (|x|-1)^2-1 = \rule{2cm}{.2pt}$ & 15. $y = \sqrt{-x} = \rule{2cm}{.2pt}$ \\ 16. $y = |x-1|-1 = \rule{2cm}{.2pt}$ & 17.$y = |(x-1)^2-1| = \rule{2cm}{.2pt}$ & 18. $y = 2-\sqrt{9-x^2} = \rule{2cm}{.2pt}$ \\ 19. $y = 1+\sqrt{4-x^2} = \rule{2cm}{.2pt}$ & 20.$y = |x^2-1| = \rule{2cm}{.2pt}$ & 21. $y = 1-\sqrt{-x} = \rule{2cm}{.2pt}$ \\ 22. $y = \dfrac{1}{|x|}-1 =\rule{2cm}{.2pt}$ & 23. $y =\Big|\dfrac{1}{x}-1\Big| = \rule{2cm}{.2pt}$ & 24. $y = \dfrac{1}{|x|-1} = \rule{2cm}{.2pt}$ \\ \end{tabular} \clearpage \end{enumerate} \section{Essay Questions} \begin{enumerate} \item Find the solution set to the inequality $$\dfrac{(x-1)(x+2)}{(x-3)} \geq 0, $$ and write the answer in interval notation. \vfill \item For the points $P(-1, 2)$ and $Q(2, 3)$, find: \begin{enumerate} \item the distance between $P$ and $Q$, \item the midpoint of the line segment joining $P$ and $Q$, \item if $P$ and $Q$ are the endpoints of a diameter of a circle, find the equation of the circle. \end{enumerate} \vfill \item Show that if the graph of a curve has $x$-axis symmetry and $y$-axis symmetry then it must also have symmetry about the origin. \vfill \item \label{pro:domain_range}Consider the graph of the curve $y = f(x)$ in figure \ref{fig:domain_range}. You may assume that the domain of $f$ can be written in the form $\co{a;b} \cup \oc{b;c}$, where $a, b, c$ are integers, and that its range can be written in the form $\cc{u;v}$, with $u$ and $v$ integers. Find $a, b, c, u$ and $v$. \vspace{2cm} \begin{figure}[h] $$\psset{unit=1pc} \psaxes[labels=none,linewidth=2pt]{<->}(0,0)(-7,-7)(7,7) \psplot[linecolor=black,linewidth=2pt,algebraic=true, arrows={*-o}]{-4}{-2}{-2*x-4}\psplot[linecolor=black,linewidth=2pt,algebraic=true, arrows={o-*}]{-2}{2}{x^2-4}\psplot[linecolor=black,linewidth=2pt,algebraic=true, arrows={o-*}]{2}{4}{1.5*x-2}$$\meinecaption{2}{Problem \ref{pro:domain_range}.} \label{fig:domain_range} \end{figure} \vspace{1cm} \ \item If the points $(1, 3)$, $(-1,2)$, $(2,t)$ all lie on the same line, find the value of $t$. \vfill \item An apartment building has $30$ units. If all the units are inhabited, the rent for each unit is $\$ 700$ per unit. For every empty unit, management increases the rent of the remaining tenants by $\$ 25$. What will be the profit $P(x)$ that management gains when $x$ units are empty? What is the maximum profit? \vfill \item Draw a rough sketch of the graph of $y =x-\floor{x}$, where $\floor{x}$ is the the floor of $x$, that is, the greatest integer less than or equal to $x$. \item Sketch the graphs of the curves in the order given. Explain, by which transformations (shifts, compressions, elongations, squaring, reflections, etc.) how one graph is obtained from the preceding one. \begin{enumerate} \item $y = x-1$ \item $y = (x-1)^2$ \item $y = x^2-2x$ \item $y = |x^2-2x|$ \item $y = \dfrac{1}{|x^2-2x|}$ \item $y = -\dfrac{1}{|x^2-2x|}$ \item $y = \dfrac{1}{x^2-2|x|}$ \end{enumerate} \item The polynomial $$p(x) = x^4-4x^3+4x^2-1 $$ has a local maximum at $(1,0)$ and local minima at $(0,-1)$ and $(2,-1)$. \begin{enumerate} \item Factor the polynomial completely and sketch its graph. \item Determine how many real zeros the polynomial $q(x) = p(x) + c$ has for each constant $c$. \end{enumerate} \item The rational function $q$ in figure \ref{fig:rational1} has only two simple poles and satisfies $q(x)\rightarrow 1$ as $x\rightarrow \pm \infty$. You may assume that the poles and zeroes of $q$ are located at integer points. Problems \ref{pro:rational_begin} to \ref{pro:rational_end} refer to it. \vspace{3cm} \begin{figure}[h] $$\psset{unit=1pc} \renewcommand{\pshlabel}[1]{{\tiny #1}} \renewcommand{\psvlabel}[1]{{\tiny #1}} %\psgrid[subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-8,-8)(8,8) %\psline[linewidth=1.3pt](0,-8)(0,8)\psline[linewidth=1.3pt](-8, 0)(8, 0) \psaxes[linewidth=1pt,labels=none](0,0)(-8,-8)(8,8) \psline[linewidth=.7pt, linestyle=dotted](1, -8)(1, 8) \psline[linewidth=.7pt, linestyle=dotted](-2, -8)(-2, 8) \psplot[linewidth=1.5pt, linecolor=black,algebraic=true,arrows={<->}]{-7}{-2.230138587}{((x+1)*(x-2))/((x-1)*(x+2))} \psplot[linewidth=1.5pt, linecolor=black,algebraic=true,arrows={<->}]{-1.838087489}{.8968052533}{((x+1)*(x-2))/((x-1)*(x+2))} \psplot[linewidth=1.5pt, linecolor=black,algebraic=true,arrows={<->}]{1.088087489}{6}{((x+1)*(x-2))/((x-1)*(x+2))} \psdots[dotstyle=*,dotscale=1](-1,0)(2,0)(0,1) $$\meinecaption{3}{Problems \ref{pro:rational_begin} to \ref{pro:rational_end}.} \label{fig:rational1} \end{figure} \begin{enumerate} \item \label{pro:rational_begin} Find $q(0)$. \item Find $q(x)$ for arbitrary $x$. \item Find $q(-3)$. \item \label{pro:rational_end} Find $\lim _{x\rightarrow -2+} q(x)$. \end{enumerate} \item Find the solution to the absolute value inequality $$ |x^2-2x-1| \leq 1, $$ and express your answer in interval notation. \item Find all values of $x$ for which the point $(x, x+1)$ is at distance $2$ from $(-2,1)$. \item Determine any intercepts with the axes and any symmetries of the curve $$y^2=|x^3+1|. $$\ \item Let $f(x) = x^2$. Find $$\dfrac{f(x+h)-f(x-h)}{h}. $$ \item {\bf Situation: } Questions \ref{pro:f05_line_begin1} to \ref{pro:f05_line_end1} refer to the straight line $L_u$ given by the equation $$L_u: \ \ (u -2)y = (2u +4)x + 2u, $$where $u$ is a real parameter. \begin{enumerate} \item \label{pro:f05_line_begin1} For which value of $u$ is $L_u$ a horizontal line? \item For which value of $u$ is $L_u$ a vertical line? \item For which value of $u$ is $L_u$ parallel to the line $y=-2x+1$? \item For which value of $u$ is $L_u$ perpendicular to the line $y=-2x+1$? \item \label{pro:f05_line_end1} Is there a point which is on every line $L_u$ regardless the value of $u$? If so, find it. If not, prove that there is no such point. \end{enumerate} \item The polynomial $p$ in figure \ref{fig:cubic1} has degree $3$. You may assume that all its roots are integers. Problems \ref{pro:cubic_begin} to \ref{pro:cubic_end} refer to it. \vspace{2cm} \begin{figure}[ht] $$\psset{unit=.7pc} \renewcommand{\pshlabel}[1]{{\tiny #1}} \renewcommand{\psvlabel}[1]{{\tiny #1}} \psgrid[subgriddiv=0,linewidth=.5pt, gridlabels=0](0,0)(-8,-8)(8,8) \psline[linewidth=1.3pt](0,-8)(0,8)\psline[linewidth=1.3pt](-8, 0 )(8, 0) \psplot[linewidth=2pt, linecolor=black, arrows={<->}]{-6}{6}{ x 4 add x -3 add mul x mul -0.05 mul} \psdots[dotscale=1,dotstyle=*](3,0)(-4,0)(0,0)(-2,-1) $$\meinecaption{3}{Problems \ref{pro:cubic_begin} to \ref{pro:cubic_end}.} \label{fig:cubic1} \end{figure} \begin{enumerate} \item \label{pro:cubic_begin} Find $p(-2)$, assuming it is an integer. \item \label{pro:cubic_end} Find a formula for $p(x)$. \end{enumerate} \item A rectangular box with a square base of length $x$ and height $h$ is to have a volume of $20$ ft$^3$. The cost of the material for the top and bottom of the box is 20 cents per square foot. Also, the cost of the material for the sides is 8 cents per square foot. Express the cost of the box in terms of \begin{enumerate} \item the variables $x$ and $h$; \item the variable $x$ only; and \item the variable $h$ only. \end{enumerate} \item Sketch the graph of the curve $y=\sqrt{\dfrac{1-x}{x+1}}$ and label the axis intercepts and asymptotes. \item Find all the rational roots of $x^5+4x^4+3x^3-x^2-4x-3 = 0$. \item Given $f(x) = \dfrac{1}{x+1}$, graph \begin{enumerate} \item $y = |f(x)|$, \item $y = f(|x|)$, \item $y = |f(|x|)$, \item $y = f(-|x|)$. \end{enumerate} \item Graph $ y = (x-1)^{2/3}+2$ noting any intercepts with the axes. \item[] {\bf Problems \ref{pro:abs-val-begin} through \ref{pro:abs-val-end} refer to the curve with equation $y = |x+2|+|x-3|$.} \item \label{pro:abs-val-begin} Write the equation $ y = |x+2|+|x-3| $ without absolute values if $x\leq -2$. \item Write the equation $ y = |x+2|+|x-3| $ without absolute values if $-2\leq x\leq 3$. \item Write the equation $ y = |x+2|+|x-3| $ without absolute values if $x\geq 3$. \item Solve the equation $|x+2|+|x-3|= 7$. \item Solve the equation $|x+2|+|x-3|= 4$. \item Graph the curve $y = |x+2|+|x-3|$ on the axes below. Use a ruler or the edge of your ID card to draw the straight lines. \item Graph the curve $y = 4$ on the axes below. \item \label{pro:abs-val-end} Graph the curve $y = 7$ on the axes above. \vfill \begin{center}\psset{unit=1.5pc}\psaxes[linewidth=2pt,arrows={->},labelFontSize=\tiny](0,0)(-10,-10)(11,11)\end{center} \vfill \clearpage \item[] Questions \ref{pro:circle-line1} through \ref{pro:circle-line2} refer to the circle ${\cal C}$ having centre at $O(1, 2)$ and passing through the point $A(5,5)$, as shewn in figure \ref{fig:circle-line} below. \vspace{4cm} \begin{figure}[hptb] \centering \psset{unit=.6pc} \psaxes[arrows={->},linewidth=1.7pt,labelFontSize=\tiny](0,0)(-10,-10)(12,12) \pstGeonode[PosAngle={0,0}](1,2){O}(5,5){A} \pstGeonode[PointName=none,PointSymbol=none](8.75,0){B} \pstTranslation[PointName=none,PointSymbol=none]{A}{B}{B}[C] \pstTranslation[PointName=none,PointSymbol=none]{B}{A}{A}[C'] \pstCircleOA[linecolor=brown,linewidth=2pt]{O}{A} \pstLineAB[arrows={<->},linewidth=2pt,linecolor=blue]{C}{C'}\vspace{1cm} \meinecaption{2}{Problems \ref{pro:circle-line1} through \ref{pro:circle-line2} .} \label{fig:circle-line} \end{figure} \vspace{1cm} \item \label{pro:circle-line1} Find the equation of the circle ${\cal C}$. \item If the point $(2, a)$ is on the circle ${\cal C}$, find all the possible values of $a$. \item \label{pro:circle-line2} Find the equation of the line $L$ that is tangent to the circle ${\cal C}$ at $A$. (Hint: A tangent to a circle at a point is perpendicular to the radius passing through that point.) \item[] Problems \ref{pro:inversea} through \ref{pro:inverseb} refer to the graph of a function $f$ is given in figure \ref{fig:inversea}. \vspace{3cm} \begin{figure}[!hptb] \begin{minipage}{7cm} \centering \psset{unit=1pc} \psgrid[subgriddiv=0,gridlabels=.5](-7,-7)(7,7) \psline[showpoints=true,linewidth=2pt,linecolor=blue](-5,-5)(-3,-3)(0,-2)(2,3)(5,4) \meinecaption{3}{Problems \ref{pro:inversea} through \ref{pro:inverseb}.}\label{fig:inversea} \end{minipage} \hfill \begin{minipage}{7cm} \centering \psset{unit=1pc} \psgrid[subgriddiv=0,gridlabels=.5](-7,-7)(7,7) \meinecaption{3}{Problems \ref{pro:inversea} through \ref{pro:inverseb}.}\label{fig:inverseb} \end{minipage} \end{figure} \vspace{1cm} \item \label{pro:inversea} Give a brief explanation as to why $f$ is invertible. \item \label{pro:inversea} Determine $\dom{f}$. \item Determine $\im{f}$. \item Draw the graph of $f^{-1}$ in figure \ref{fig:inverseb}. \item Determine $f(-5)$. \item Determine $f^{-1}(3)$. \item \label{pro:inverseb} Determine $f^{-1}(4)$. \item[] Figure \ref{fig:s08_fun5} has the graph of a curve $y=f(x)$. Draw each of the required curves {\em very carefully}. \vspace{3cm} \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \pstGeonode[PointName=none](-4,1){A}(-3,1){B}(-2,-1){C}(-1,-1){C'}(0,0){D}(1,1){E}(2,0){F}(3,2){G}(4,0){H}(5,-2){I} \psline[linewidth=2pt,linecolor=blue](A)(B)(C)(C')(D)(E)(F)(G)(H)(I) \psdots(A)(B)(C)(C')(D)(E)(F)(G)(H)(I) $$ \meinecaption{2}{$y=f(x)$.} \label{fig:s08_fun5} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) $$ \meinecaption{2}{\mbox{$y=f(x)+1$}.} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) $$ \meinecaption{2}{\mbox{$y=|f(x)+1|$}.} \end{minipage} \end{figure} \vspace{3cm} \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) $$ \meinecaption{2}{\mbox{$y=f(-|x|)$}.} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) $$ \meinecaption{2}{$y=|f(-|x|)|$.} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) $$ \meinecaption{2}{$y=-f(-x)$.} \end{minipage} \end{figure} \clearpage \item Figure \ref{fig:s08_fun8} has the graph of a curve $y=f(x)$, which is composed of lines and a pair of semicircles. Draw each of the required curves {\em very carefully}. Use a ruler or the edge of your id card in order to draw the lines. Shapes with incorrect coordinate points will not be given credit. \vspace{3cm} \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) \pstGeonode[PointName=none,PointSymbol=none](-4,0){A}(-2,0){B}(0,0){C}(1,0){D}(2,0){E}(3,0){F}(4,0){G} \pstArcOAB[linewidth=2pt,linecolor=blue]{B}{A}{C} \pstArcOAB[linewidth=2pt,linecolor=blue]{F}{G}{E} \psline[linewidth=2pt,linecolor=blue](C)(E) \psdots[linewidth=1.5pt](C)(D)(E)(G)(3,1)(-2,-2)(A) $$ \meinecaption{2}{$y=f(x)$.} \label{fig:s08_fun8} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) $$ \meinecaption{2}{\mbox{$y=f(-x)$}.} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) $$ \meinecaption{2}{\mbox{$y=-f(x)$}.} \end{minipage} \end{figure} \vspace{3cm} \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) $$ \meinecaption{2}{\mbox{$y=|f(x)|$}.} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) $$ \meinecaption{2}{$y=f(-|x|)$.} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) $$ \meinecaption{2}{$y=f(|x|)$.} \end{minipage} \end{figure} \clearpage \item Use the following set of axes to draw the following curves in succession. Note all intercepts. \vspace{3cm} \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) $$ \meinecaption{2}{$y=x-2$.} \label{fig:s08_fun8} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) $$ \meinecaption{2}{\mbox{$y=|x-2|$}.} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) $$ \meinecaption{2}{\mbox{$y=|x|-2$}.} \end{minipage} \end{figure} \vspace{3cm} \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) $$ \meinecaption{2}{\mbox{$y=||x|-2|$}.} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) $$ \meinecaption{2}{\mbox{$y=|-|x|-2|$}.} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=1pc} \psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[ Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0) \psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0) \psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5) $$ \meinecaption{2}{\mbox{$|y|=x-2$}.} \end{minipage} \end{figure} \item[] {\bf Situation:} $\triangle ABC$ is right-angled at $A$, and $AB = 2$ and $\tan \angle B = \frac{1}{2}$. Problems \ref{pro:right_trianglebegin} through \ref{pro:right_triangleend} refer to this situation. \item \label{pro:right_trianglebegin} Find $AC$. \item Find $BC$. \item Find $\sin \angle B$. \item \label{pro:right_triangleend} Find $\tan \angle C$. \item Using the standard labels for a $\triangle ABC$, prove that $\dfrac{a-b}{a+b}=\dfrac{\sin A - \sin B}{\sin A + \sin B}$. \item A triangle has sides measuring $2, 3, 4$. Find the cosine of the angle opposite the side measuring $3$. \item Find the area of a triangle whose sides measure $2,3,4$. Find the radius of its circumcircle. \item If in a $\triangle ABC$, $a=5$, $b=4$, and $\cos (A-B) = \dfrac{31}{32}$, prove that $\cos C = \dfrac{1}{8}$ and that $c=6$. \item A triangle with vertices $A, B, C$ on a circle of radius $R$, has the side opposite to vertex $A$ of length $12$, and the angle at $A = \frac{\pi}{4}$. Find diameter of the circle. \item $\triangle ABC$ has sides of length $a, b, c$, and circumradius $R=4$. Given that the triangle has area $5$, find the product $abc$. \item Find, approximately, the area of a triangle having two sides measuring $1$ and $2$ respectively, and angle between these sides measuring $35^\circ$. What is the measure of the third side? \item Find the area and the perimeter of a regular octagon inscribed in a circle of radius $2$. \item Two buildings on opposite sides of a street are $45$ m apart. From the top of the taller building, which is $218$ m high, the angle of depression to the top of the shorter building is $13.75^\circ $. Find the height of the shorter building. \item A ship travels for $3$ hours at $18$ mph in a direction N$28^\circ$E. From its current direction, the ship then turns through an angle of $95^\circ$ to the right and continues traveling at $18$ mph. How long will it take before the ship reaches a point directly east of its starting point? \item Let $\tan x + \cot x = a$. Find $\tan^3x+\cot^3x$ as a polynomial in $a$. \item If $\cos \dfrac{\pi}{7} = a$, find the exact value of $\cos \dfrac{\pi}{14}$ and $\cos \dfrac{2\pi}{7}$ in terms of $a$. \item Given that $\csc x =-4$, and $\winding{x}$ lies in quadrant III, find the remaining trigonometric functions. \item Graph the curve $y = 2-\cos \dfrac{x}{2}$. \item Graph the curve $y = \Big|2-\cos \dfrac{x}{2}\Big|$. \item Find the smallest positive solution, if any, to the equation $3^{\cos 3x} = 2$. Approximate this solution to two decimal places. \item Find all the solutions lying in $[0;2\pi]$ of the following equations: \begin{enumerate}\item $2\sin^2x+\cos x -1 = 0$ \item $\sin 2x = \cos x$ \item $\sin 2x = \sin x$ \item $\tan x + \cot x = 2\csc 2x$ \end{enumerate} \item Find the exact value of $\sin \dfrac{88\pi}{3}$. \item Find the exact value of $\tan \left(\arcsin \dfrac{1}{3}\right)$. \item Is $\sin (\arcsin 30)$ a real number? \item Find the exact value of $ \arcsin (\sin 30)$. \item Find the exact value of $ \arcsin (\cos 30)$. \item If $x$ and $y$ are acute angles and $\sin \dfrac{x}{2} = \dfrac{1}{3}$ and $\cos y = \dfrac{3}{4}$, find the exact value of $\tan (x-y)$. \item Find the exact value of the product $$ P = \cos \frac{\pi}{7}\cdot\cos \frac{2\pi}{7}\cdot\cos \frac{4\pi}{7}.$$ \item How many digits does $5^{2000}3^{1000}$ have? \item What is $5^{2000}3^{1000}$ approximately? \item Let $a>1$, $x>1$, $y>1$. If $\log_a x^3=N$ and $\log_{a^{1/3}}y^4= M$, find $\log_{a^2} xy$ in terms of $N$ and $M$. Also, find $\log _x y$. \item Graph $y = 3^{-x}-2$. \item Graph $y = 3^{-|x|}-2$. \item Graph $y = |3^{-x}-2|$. \item Graph $y = \ln (x+1)$. \item Graph $y = \ln (|x|+1)$. \item Graph $y = |\ln (x+1)|$. \item Graph $y = |\ln |(x+1)||$. \item Solve the equation $3^{x} + \dfrac{1}{3^x} = 12$. \item The expression $$ (\log _2 3)\cdot (\log _3 4)\cdot (\log _4 5) \cdots (\log _{511} 512)$$ is an integer. Find it. \item The expression $$\log (\tan 1^\circ) + \log (\tan 2^\circ) + \log (\tan 3^\circ) + \cdots + \log (\tan 89^\circ) $$is an integer. Find it. \item Prove that the equation $$\cos \left(\left(\dfrac{3}{2}\right)^x-1\right) = \dfrac{1}{2}, $$ has only $ 4$ solutions lying in the interval $[0;2\pi ]$. \item Prove that the equation $$\cos (\log_3x-2) = \dfrac{1}{2}, $$ has only $2 $ solutions lying in the interval $[0;2\pi ]$. \end{enumerate} \chapter{Maple} The purpose of these labs is to familiarise you with the basic operations and commands of Maple. The commands used here can run on any version of Maple (at least V through X). \section{Basic Arithmetic Commands} Maple uses the basic commands found in most calculators: $+$ for addition, $-$ for subtraction, $*$ for multiplication, $/$ for division, and $\wedge$ for exponentiation. Maple also has other useful commands like {\tt expand} and {\tt simplify}. Be careful with capitalisation, as Maple distinguishes between capital and lower case letters. For example, to expand the algebraic expression $(\sqrt{8}-2^{1/2})^2$, type the following, pressing \fbox{ENTER} after the semicolon: \bigskip \begin{mapleinput} \mapleinline{active}{1d}{expand((sqrt(8)-2^(1/2))^2);}{} \end{mapleinput} \bigskip If you desire a decimal approximation of the above, either put a decimal point after the numbers, or use the command {\tt evalf}. Notice that on the first one Maple is dealing with two approximations and hence, it outputs an error! \bigskip \begin{mapleinput} \mapleinline{active}{1d}{(sqrt(8.)-2.^(1/2))^2;}{} \mapleinline{active}{1d}{evalf((sqrt(8)-2^(1/2))^2);}{} \end{mapleinput} \bigskip Now, prove that $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=a^3+b^3+c^3-3abc$ by expanding the expression on the sinistral side: \bigskip \begin{mapleinput} \mapleinline{active}{1d}{expand((a+b+c)*(a^2+b^2+c^2-a*b-b*c-c*a));}{} \end{mapleinput} \bigskip To simplify $a(a-1)+(a-2)(a^2+a+2)$ use the command {\tt simplify}: \bigskip \begin{mapleinput} \mapleinline{active}{1d}{simplify(a*(a-1)+(a-2)*(a^2+a+2));}{} \end{mapleinput} \bigskip The absolute value of a real quantity is found using the function {\em absolute value}: {\tt abs()} \bigskip \begin{mapleinput} \mapleinline{active}{1d}{abs(-5);}{} \end{mapleinput} \begin{exa}Factor $ x^{10}-x^8-2x^7-x^6-x^4+x^2+2x+1$ using Maple. \end{exa} \begin{solu} The required command line is \begin{maplegroup} \begin{mapleinput} \mapleinline{active}{1d}{factor(x^10-x^8-2*x^7-x^6-x^4+x^2+2*x+1);}{} \end{mapleinput} \mapleresult \begin{maplelatex} \mapleinline{inert}{2d}{}{\[( x-1)( x+1)(x^2-x+1)(x^2-x-1)(x^2+x+1)^2\]} \end{maplelatex} \end{maplegroup} \end{solu} \begin{exa} Obtain the partial fraction expansion of $\dfrac{x}{x^3+1}$ using Maple. \end{exa} \begin{solu} The required command line is \begin{maplegroup} \begin{mapleinput} \mapleinline{active}{1d}{convert(x/(x^3-1), parfrac,x);}{} \end{mapleinput} \mapleresult \begin{maplelatex} \mapleinline{inert}{2d}{}{\[\dfrac{1}{3}\cdot \dfrac{1}{x-1}-\dfrac{1}{3}\cdot \dfrac{x-1}{x^2+x+1}\]} \end{maplelatex} \end{maplegroup} \end{solu} \begin{exa} Reduce the fraction $\dfrac{x-1}{x^4-1}$. \end{exa} \begin{solu} The required command line is \begin{maplegroup} \begin{mapleinput} \mapleinline{active}{1d}{simplify((x-1)/(x^4-1));}{} \end{mapleinput} \mapleresult \begin{maplelatex} \mapleinline{inert}{2d}{}{\[\dfrac{1}{x^3+x^2+x+1}\]} \end{maplelatex} \end{maplegroup} \end{solu} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{pro}Write the exact command line to compute $(8^2 - 67)^{(8 - (3)(2))}$ and give the output (result) obtained. \begin{answer} The command line follows: \begin{mapleinput} \mapleinline{active}{1d}{(8^2 - 67)^(8 - (3)*(2));}{} \end{mapleinput} \mapleresult \begin{maplelatex} \mapleinline{inert}{2d}{}{\[9\]} \end{maplelatex} \end{answer} \end{pro} \begin{pro} Expand using Maple: $$ (a+b+c)^3-3(a+b)(b+c)(c+a). $$ \begin{answer} The required command line follows. \begin{mapleinput} \mapleinline{active}{1d}{expand((a+b+c)^3-3*(a+b)*(b+c)*(c+a));}{} \end{mapleinput} \mapleresult \begin{maplelatex} \mapleinline{inert}{2d}{}{\[a^3+b^3+c^3\]} \end{maplelatex} \end{answer} \end{pro} \begin{pro} Use Maple to verify that $$(x + y)^5 - x^5 - y^5 = 5xy(x + y)(x^2 + xy + y^2)$$ and $$(x + a)^7 - x^7 - a^7 = 7xa(x + a)(x^2 + xa + a^2)^2.$$ \begin{answer} The required command lines are \begin{maplegroup} \begin{mapleinput} \mapleinline{active}{1d}{factor((x + y)^5 - x^5 - y^5);}{} \end{mapleinput} \mapleresult \begin{maplelatex} \mapleinline{inert}{2d}{}{\[5xy(x+y)(y^2+xy+x^2)\]} \end{maplelatex} \begin{mapleinput} \mapleinline{active}{1d}{factor((x + y)^7 - x^7 - y^7);}{} \end{mapleinput} \mapleresult \begin{maplelatex} \mapleinline{inert}{2d}{}{\[7xy(x+y)(y^2+xy+x^2)^2\]} \end{maplelatex} \end{maplegroup} \end{answer} \end{pro} \begin{pro} Write Maple code to verify that a product of sums of squares can be written as a sum of squares, that is, verify that $$(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2.$$ \begin{answer} Here is one possible answer \begin{maplegroup} \begin{mapleinput} \mapleinline{active}{1d}{is((a^2 + b^2)*(c^2+ d^2)= (a*c + b*d)^2 + (a*d - b*c)^2);}{} \end{mapleinput} \mapleresult \begin{maplelatex} \mapleinline{inert}{2d}{}{\[true\]} \end{maplelatex} \end{maplegroup} \end{answer} \end{pro} \section{Solving Equations and Inequalities} Maple can be used to solve equations and inequalities. Type the following, pressing \fbox{ENTER} after the semicolon: \bigskip \begin{mapleinput} \mapleinline{active}{1d}{solve(x^2-3*x+2=0, x);}{} \mapleinline{active}{1d}{solve(abs(x-1)+abs(x+2)=4, x);}{} \mapleinline{active}{1d}{solve(abs(x-1)+abs(x+2)=3, x);}{} \mapleinline{active}{1d}{solve((x+1)/(x*(x-1))>=0, x);}{} \end{mapleinput} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{pro} Write the exact command line to find the solutions of the equation $x^2 + |x - 1| = 5$ and name all the solutions. \begin{answer} The command line follows: \begin{mapleinput} \mapleinline{active}{1d}{solve(x^2+abs(x-1)=5, x);}{} \end{mapleinput} \mapleresult \begin{maplelatex} \mapleinline{inert}{2d}{}{\[2, \dfrac{1}{2}-\dfrac{1}{2}\sqrt{17}\]} \end{maplelatex} \end{answer} \end{pro} \section{Maple Plotting Commands} Although there is no direct Maple command\footnote{To my knowledge,that is.} to express $|2x|+|x+2|$ without absolute values, we can graph the curve $y = |2x|+|x+2|$. To do so, we must first load the plot library. \bigskip \begin{mapleinput} \mapleinline{active}{1d}{with(plots);}{} \mapleinline{active}{1d}{plot(abs(2*x)+abs(x+2), x=-10..10,y=-10..10);}{} \end{mapleinput} \bigskip Plot now the curve $y = |x+1|+|x-1|$. \bigskip \begin{mapleinput} \mapleinline{active}{1d}{plot(abs(x+1)+abs(x-1), x=-10..10,y=-10..10);}{} \end{mapleinput} \bigskip Notice the difference between the above curve and $y = |x+1|-|x-1|$, and $y = |x-1|-|x+1|$. \bigskip \begin{mapleinput} \mapleinline{active}{1d}{plot(abs(x+1)-abs(x-1), x=-10..10,y=-10..10);}{} \mapleinline{active}{1d}{plot(abs(x-1)-abs(x+1), x=-10..10,y=-10..10);}{} \end{mapleinput} \bigskip The equation of a circle with centre at $(-1, 2)$ and radius $5$ is $(x+1)^2+(y-2)^2=25$. To graph it, type the following commands: \bigskip \begin{mapleinput} \mapleinline{active}{1d}{implicitplot((x+1)^2+(y-2)^2=25, x=-7..5, y=-4..8);}{} \end{mapleinput} What is the main difference between the commands {\tt plot} and {\tt implicitplot}? Since we haven't discussed functions yet, let us just say that you use {\tt plot} when you can solve uniquely for $y$. In this case, $y$ does not appear in the equation of the command (other than for stipulating ranges). If you can't solve uniquely for $y$, then use {\tt implicitplot}. \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{pro}\noindent \begin{enumerate} \item Write the exact command line to find the solution set to the inequality $\dfrac{x^3-x}{x-2} \geq -1$. \item Write the exact command line to graph the circle $(x+1)^2+(y-2)^2=4$. Pick as small ranges as possible for $x$ and $y$ that shew the whole graph. \end{enumerate} \end{pro} \section{Assignment Rules in Maple} Let us define the assignment rule $f:\reals\rightarrow \reals$, $f(x)=x^3- x^2+1$ and evaluate at a few points. \bigskip \begin{mapleinput} \mapleinline{active}{1d}{f:=x->x^3-x^2+1;}{} \mapleinline{active}{1d}{f(1);}{} \mapleinline{active}{1d}{f(-2);}{} \end{mapleinput} We now wish to plot $f$. For this type: \begin{mapleinput} \mapleinline{active}{1d}{with(plots);}{} \mapleinline{active}{1d}{plot(f(x), x=-10..10,y=-10..10);}{} \end{mapleinput} \bigskip Let us plot various transformations of $f$. \bigskip \begin{mapleinput} \mapleinline{active}{1d}{plot(f(-x), x=-10..10,y=-10..10);}{} \mapleinline{active}{1d}{plot(-f(x), x=-10..10,y=-10..10);}{} \mapleinline{active}{1d}{plot(-f(-x), x=-10..10,y=-10..10);}{} \mapleinline{active}{1d}{plot(f(abs(x)), x=-10..10,y=-10..10);}{} \mapleinline{active}{1d}{plot(f(-abs(x)), x=-10..10,y=-10..10);}{} \mapleinline{active}{1d}{plot(abs(f(x)), x=-10..10,y=-10..10);}{} \mapleinline{active}{1d}{plot(abs(f(abs(x))), x=-10..10,y=-10..10);}{} \mapleinline{active}{1d}{plot(abs(f(-abs(x))), x=-10..10,y=-10..10);}{} \mapleinline{active}{1d}{plot(f(x+5), x=-10..10,y=-10..10);}{} \mapleinline{active}{1d}{plot(f(x-5), x=-10..10,y=-10..10);}{} \mapleinline{active}{1d}{plot(f(x)+5, x=-10..10,y=-10..10);}{} \mapleinline{active}{1d}{plot(f(x)-5, x=-10..10,y=-10..10);}{} \end{mapleinput} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{pro} You are to match the following transformations with the corresponding equation, given the original curve $y=f(x)$. \begin{tabular}{lll} \rule{1cm}{2pt} $y = f(-x)$ & \hspace{2in} & A. moves the original graph five units down \\ \rule{1cm}{2pt} $y = -f(-x)$ & \hspace{2in} & B. moves the original graph five units up\\ \rule{1cm}{2pt} $y = f(x-5)$ & \hspace{2in} & C. moves the original graph five units left \\ \rule{1cm}{2pt} $y = f(x+5)$ & \hspace{2in} & D. moves the original graph five units right \\ \rule{1cm}{2pt} $y = f(x)+5$ & \hspace{2in} & E. a reflexion about the $x$-axis of the original graph \\ \rule{1cm}{2pt} $y = f(x)-5$ & \hspace{2in} & G. a reflexion about the $y$-axis of the original graph \\ \rule{1cm}{2pt} $y = -f(x)$ & \hspace{2in} & H. a reflexion about $(0,0)$ of the original graph \\ \rule{1cm}{2pt} $y = f(-|x|)$ & \hspace{2in} & I. every $y$-coordinate of the original graph becomes positive\\ \rule{1cm}{2pt} $y = |f(x)|$ & \hspace{2in} & J. recognises only $x<0$ of the original graph, and it's an even graph. \\ \rule{1cm}{2pt} $y = f(|x|)$ & \hspace{2in} & K. recognises only $x>0$ of the original graph, and it's an even graph. \\ \end{tabular} \end{pro} \section{Polynomials Splitting in the Real Numbers} We will use Maple in order to graph polynomials all whose zeroes are real numbers. \begin{mapleinput} \mapleinline{active}{1d}{with(plots);}{} \vfill \mapleinline{active}{1d}{a:=x->x*(x-1)*(x+1);}{} \vfill \mapleinline{active}{1d}{plot(a(x), x=-5..5, y=-5..5);}{} \vfill \mapleinline{active}{1d}{b:=x->x^2*(x-1)*(x+1);}{} \vfill \mapleinline{active}{1d}{plot(b(x), x=-5..5, y=-5..5);}{} \vfill \mapleinline{active}{1d}{c:=x->x*(x-1)^2*(x+1);}{} \vfill \mapleinline{active}{1d}{plot(c(x), x=-5..5, y=-5..5);}{} \vfill \mapleinline{active}{1d}{d:=x->x*(x-1)^2*(x+1)^2;}{} \vfill \mapleinline{active}{1d}{plot(d(x), x=-5..5, y=-5..5);}{} \vfill \mapleinline{active}{1d}{f:=x->x^3*(x-1)^3*(x+1)^3;}{} \vfill \mapleinline{active}{1d}{plot(f(x), x=-5..5, y=-5..5);}{} \vfill \mapleinline{active}{1d}{g:=x->((x-1)^2+1)*(x^2+1);}{} \vfill \mapleinline{active}{1d}{plot(g(x), x=-5..5, y=-5..5);}{} \vfill \mapleinline{active}{1d}{a(2); b(2); c(2); d(2); f(2); g(2);}{} \vfill \mapleinline{active}{1d}{a(-2); b(-2); c(-2); d(-2); f(-2); g(-2);}{} \vfill \end{mapleinput} \subsection*{Homework}\addcontentsline{toc}{subsection}{Homework} \begin{pro} Answer the following questions. In all items, $p:\reals\rightarrow\reals$ refers to a polynomial all whose zeroes are real numbers. \begin{enumerate} \item If the multiplicity of a zero of $p$ is one, then does the graph of $p$ cross the $x$-axis, or is it tangent to it? \item If the multiplicity of a zero of $p$ is even, then does the graph of $p$ cross the $x$-axis, or is it tangent to it? \item If the multiplicity of a zero of $p$ is odd and at least three, then does the graph of $p$ cross the $x$-axis, or is it tangent to it? \item Does the graph of $x\mapsto g(x)$ above cross the $x$-axis? Does $g$ have any real zeroes? \end{enumerate} \end{pro} \section{Sets, Lists, and Arrays} Maple has a rich variety of data structures, among them sets, lists, and arrays. Roughly speaking, a {\em set} corresponds to a set in combinatorics: the order of the elements is irrelevant, and repetitions are not taken into account. Sets are defined by using curly braces $\{\quad \}$. In a {\em list}, the order of the elements is important and repetitions are taken into account. Lists are defined by using square brackets $[\quad ]$. Arrays are a generalisations of matrices. They can be modified and are declared with the command \verb+array()+. We will first consider sets and set operations. In order to facilitate our presentation, we will give names to the various objects we will define. In order to attach a name, we need the assignment operator \verb+:=+, where there is no space between the colon and the equal sign. Maple is able to perform set operations with the commands \verb+union+, \verb+intersect+, and \verb+minus+. To check whether two sets are equal we may use the command \verb+evalb()+ (evaluate boolean). \begin{exa} Consider the sets $$ A=\{1,2,3,a,b,c,d\}, \quad B=\{3,4,5,a,b,e,f\}. $$ Use Maple to obtain $$ A\cup B, \quad A \cap B,\quad A \setminus B, $$and to verify that $$(A\setminus B)\cup (B\setminus A) = (A\cup B)\setminus (A\cap B).$$ \end{exa} \begin{solu} We first define the sets and then perform the desired operations. The following command lines accomplish what is required. \begin{maplegroup} \begin{mapleinput} \mapleinline{active}{1d}{A:=\{1,2,3,a,b,c,d\};}{} \end{mapleinput} \mapleresult \begin{maplelatex} \mapleinline{inert}{2d}{}{\[A:=\left\{ 1,2,3,a,b,c,d \right\}\] } \end{maplelatex} \begin{mapleinput} \mapleinline{active}{1d}{B:=\{3,4,5,a,b,e,f\};}{} \end{mapleinput} \mapleresult \begin{maplelatex} \mapleinline{inert}{2d}{}{\[B:=\{3,4,5,a,b,e,f\}\]} \end{maplelatex} \begin{mapleinput} \mapleinline{active}{1d}{A union B}{} \end{mapleinput} \mapleresult \begin{maplelatex} \mapleinline{inert}{2d}{}{\[\{1, 2, 3, 4, 5, f, a, b, c, d, e\}\]} \end{maplelatex} \begin{mapleinput} \mapleinline{active}{1d}{A intersect B;}{} \end{mapleinput} \mapleresult \begin{maplelatex} \mapleinline{inert}{2d}{}{\[\{3, a, b\}\]} \end{maplelatex} \begin{mapleinput} \mapleinline{active}{1d}{A minus B;}{} \end{mapleinput} \mapleresult \begin{maplelatex} \mapleinline{inert}{2d}{}{\[\{1, 2, c, d\}\]} \end{maplelatex} \begin{mapleinput} \mapleinline{active}{1d}{evalb((A union B) minus (A intersect B)=(A minus B) union (B minus A));}{} \end{mapleinput} \mapleresult \begin{maplelatex} \mapleinline{inert}{2d}{}{\[true\]} \end{maplelatex} \end{maplegroup} \end{solu} \chapter{Some Answers and Solutions} \Closesolutionfile{ans} \section*{Answers}\addcontentsline{toc}{section}{Answers} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\tiny \input{ansPreCalcI1} \end{multicols} \begin{thebibliography}{mmm.m} \bibitem[Chen]{Chen} William CHEN, \href{http://www.maths.mq.edu.au/~wchen/lnemfolder/lnem.html}{http://www.maths.mq.edu.au/~wchen/lnemfolder/lnem.html}, Elementary Mathematics Notes. \bibitem[Eves]{Eves} Howard EVES, ``An Introduction to the History of Mathematics'', 3rd ed., 1969. \bibitem[HaKn]{HaKn} H.S. 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Jerome KEISLER, ``Elementary Calculus, An Infinitesimal Approach''. \bibitem[Mit]{Mit} Dragoslav MITRINOVIC, ``Analytic Inequalities'', Springer. \bibitem[Penn]{Penn} Philip PENNANCE, \href{http://math.uprrp.edu/profs/pennance/courses/3041.php}{http://math.uprrp.edu/profs/pennance/courses/3041.php}, Matem\'{a}tica I para maestros. \bibitem[Toom]{Toom} Andrei TOOM, \href{http://www.de.ufpe.br/~toom/articles/index.htm}{http://www.de.ufpe.br/~toom/articles/index.htm}, Teaching Articles. \bibitem[Woo]{Woo} Mark R. WOODARD, \href{http://math.furman.edu/~mwoodard/data.html}{http://math.furman.edu/~mwoodard/data.html}, Quotes from the Mathematical Quotations Server. \bibitem[Apo]{Apo} Apostol, T. M., {\bf Calculus}, Vol 1 \& 2, 2nd ed., Waltham: Xerox, 1967. \bibitem[Art]{Art} Artino, R., Gaglione, A., and Shell, N., {\bf The Contest Problem Book IV}, Mathematical Association of America. \bibitem[Har]{Har}Hardy, G. H., {\bf Pure Mathematics}, 10th ed., New York: Cambridge University Press, 1952. \bibitem[HarWri]{HarWri}Hardy, G. H.; Wright, E. M. {\bf An Introduction to the Theory of Numbers}, Oxford Science Publications. \bibitem[Kla]{Kla} Klambauer, Gabriel, {\bf Aspects of Calculus}, New York: Springer-Verlag, 1986. \bibitem[Gel1]{Gel1} Gelfand, I.M., Glagoleva, E.G., Kirillov, A.A. {\bf The Method of Coordinates}, Mineola, New York: Dover Publications, Inc., 2002. \bibitem[Gel2]{Gel2} Gelfand, I.M., Glagoleva, E.G., Shnol, E.E. {\bf Functions and Graphs }, Mineola, New York: Dover Publications, Inc., 2002. \bibitem[Lan]{Lan} Landau, E., {\bf Differential and Integral Calculus}, New York, Chelsea Publishing Company, 1950. \bibitem[Olm]{Olm} Olmstead, J. M. H., {\bf Calculus with Analytic Geometry}, Vol 1 \& 2, New York: Appleton-Century-Crofts, 1966. \bibitem[Par]{Par}Parsonson, S. L., {\bf Pure Mathematics}, Vol 1 \& 2, New York: Cambridge University Press, 1970. \bibitem[Spi]{Spi} Spivak, Michael {\bf Calculus}, 3rd ed., Houston, Texas: Publish or Perish, Inc., 1994. \end{thebibliography} \lhead{}\chead{}\rhead{} {\tiny\chapter*{\rlap{GNU Free Documentation License}} \phantomsection % so hyperref creates bookmarks \addcontentsline{toc}{chapter}{GNU Free Documentation License} %\label{label_fdl} \begin{center} Version 1.2, November 2002 Copyright \copyright{} 2000,2001,2002 Free Software Foundation, Inc. \bigskip 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA \bigskip Everyone is permitted to copy and distribute verbatim copies of this license document, but changing it is not allowed. \end{center} \begin{center} {\bf\large Preamble} \end{center} The purpose of this License is to make a manual, textbook, or other functional and useful document ``free'' in the sense of freedom: to assure everyone the effective freedom to copy and redistribute it, with or without modifying it, either commercially or noncommercially. Secondarily, this License preserves for the author and publisher a way to get credit for their work, while not being considered responsible for modifications made by others. This License is a kind of ``copyleft'', which means that derivative works of the document must themselves be free in the same sense. It complements the GNU General Public License, which is a copyleft license designed for free software. We have designed this License in order to use it for manuals for free software, because free software needs free documentation: a free program should come with manuals providing the same freedoms that the software does. But this License is not limited to software manuals; it can be used for any textual work, regardless of subject matter or whether it is published as a printed book. We recommend this License principally for works whose purpose is instruction or reference. \begin{center} {\Large\bf 1. APPLICABILITY AND DEFINITIONS\par} \phantomsection \addcontentsline{toc}{section}{1. APPLICABILITY AND DEFINITIONS} \end{center} This License applies to any manual or other work, in any medium, that contains a notice placed by the copyright holder saying it can be distributed under the terms of this License. Such a notice grants a world-wide, royalty-free license, unlimited in duration, to use that work under the conditions stated herein. The ``\textbf{Document}'', below, refers to any such manual or work. Any member of the public is a licensee, and is addressed as ``\textbf{you}''. You accept the license if you copy, modify or distribute the work in a way requiring permission under copyright law. A ``\textbf{Modified Version}'' of the Document means any work containing the Document or a portion of it, either copied verbatim, or with modifications and/or translated into another language. A ``\textbf{Secondary Section}'' is a named appendix or a front-matter section of the Document that deals exclusively with the relationship of the publishers or authors of the Document to the Document's overall subject (or to related matters) and contains nothing that could fall directly within that overall subject. (Thus, if the Document is in part a textbook of mathematics, a Secondary Section may not explain any mathematics.) The relationship could be a matter of historical connection with the subject or with related matters, or of legal, commercial, philosophical, ethical or political position regarding them. The ``\textbf{Invariant Sections}'' are certain Secondary Sections whose titles are designated, as being those of Invariant Sections, in the notice that says that the Document is released under this License. If a section does not fit the above definition of Secondary then it is not allowed to be designated as Invariant. The Document may contain zero Invariant Sections. 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The ``\textbf{Title Page}'' means, for a printed book, the title page itself, plus such following pages as are needed to hold, legibly, the material this License requires to appear in the title page. For works in formats which do not have any title page as such, ``Title Page'' means the text near the most prominent appearance of the work's title, preceding the beginning of the body of the text. A section ``\textbf{Entitled XYZ}'' means a named subunit of the Document whose title either is precisely XYZ or contains XYZ in parentheses following text that translates XYZ in another language. (Here XYZ stands for a specific section name mentioned below, such as ``\textbf{Acknowledgements}'', ``\textbf{Dedications}'', ``\textbf{Endorsements}'', or ``\textbf{History}''.) To ``\textbf{Preserve the Title}'' of such a section when you modify the Document means that it remains a section ``Entitled XYZ'' according to this definition. The Document may include Warranty Disclaimers next to the notice which states that this License applies to the Document. These Warranty Disclaimers are considered to be included by reference in this License, but only as regards disclaiming warranties: any other implication that these Warranty Disclaimers may have is void and has no effect on the meaning of this License. \begin{center} {\Large\bf 2. VERBATIM COPYING\par} \phantomsection \addcontentsline{toc}{section}{2. VERBATIM COPYING} \end{center} You may copy and distribute the Document in any medium, either commercially or noncommercially, provided that this License, the copyright notices, and the license notice saying this License applies to the Document are reproduced in all copies, and that you add no other conditions whatsoever to those of this License. You may not use technical measures to obstruct or control the reading or further copying of the copies you make or distribute. However, you may accept compensation in exchange for copies. If you distribute a large enough number of copies you must also follow the conditions in section~3. You may also lend copies, under the same conditions stated above, and you may publicly display copies. \begin{center} {\Large\bf 3. COPYING IN QUANTITY\par} \phantomsection \addcontentsline{toc}{section}{3. COPYING IN QUANTITY} \end{center} If you publish printed copies (or copies in media that commonly have printed covers) of the Document, numbering more than 100, and the Document's license notice requires Cover Texts, you must enclose the copies in covers that carry, clearly and legibly, all these Cover Texts: Front-Cover Texts on the front cover, and Back-Cover Texts on the back cover. Both covers must also clearly and legibly identify you as the publisher of these copies. The front cover must present the full title with all words of the title equally prominent and visible. You may add other material on the covers in addition. Copying with changes limited to the covers, as long as they preserve the title of the Document and satisfy these conditions, can be treated as verbatim copying in other respects. If the required texts for either cover are too voluminous to fit legibly, you should put the first ones listed (as many as fit reasonably) on the actual cover, and continue the rest onto adjacent pages. If you publish or distribute Opaque copies of the Document numbering more than 100, you must either include a machine-readable Transparent copy along with each Opaque copy, or state in or with each Opaque copy a computer-network location from which the general network-using public has access to download using public-standard network protocols a complete Transparent copy of the Document, free of added material. If you use the latter option, you must take reasonably prudent steps, when you begin distribution of Opaque copies in quantity, to ensure that this Transparent copy will remain thus accessible at the stated location until at least one year after the last time you distribute an Opaque copy (directly or through your agents or retailers) of that edition to the public. It is requested, but not required, that you contact the authors of the Document well before redistributing any large number of copies, to give them a chance to provide you with an updated version of the Document. \begin{center} {\Large\bf 4. MODIFICATIONS\par} \phantomsection \addcontentsline{toc}{section}{4. MODIFICATIONS} \end{center} You may copy and distribute a Modified Version of the Document under the conditions of sections 2 and 3 above, provided that you release the Modified Version under precisely this License, with the Modified Version filling the role of the Document, thus licensing distribution and modification of the Modified Version to whoever possesses a copy of it. In addition, you must do these things in the Modified Version: \begin{itemize} \item[A.] Use in the Title Page (and on the covers, if any) a title distinct from that of the Document, and from those of previous versions (which should, if there were any, be listed in the History section of the Document). You may use the same title as a previous version if the original publisher of that version gives permission. \item[B.] List on the Title Page, as authors, one or more persons or entities responsible for authorship of the modifications in the Modified Version, together with at least five of the principal authors of the Document (all of its principal authors, if it has fewer than five), unless they release you from this requirement. \item[C.] State on the Title page the name of the publisher of the Modified Version, as the publisher. \item[D.] Preserve all the copyright notices of the Document. \item[E.] Add an appropriate copyright notice for your modifications adjacent to the other copyright notices. \item[F.] Include, immediately after the copyright notices, a license notice giving the public permission to use the Modified Version under the terms of this License, in the form shown in the Addendum below. \item[G.] Preserve in that license notice the full lists of Invariant Sections and required Cover Texts given in the Document's license notice. \item[H.] Include an unaltered copy of this License. \item[I.] Preserve the section Entitled ``History'', Preserve its Title, and add to it an item stating at least the title, year, new authors, and publisher of the Modified Version as given on the Title Page. If there is no section Entitled ``History'' in the Document, create one stating the title, year, authors, and publisher of the Document as given on its Title Page, then add an item describing the Modified Version as stated in the previous sentence. \item[J.] Preserve the network location, if any, given in the Document for public access to a Transparent copy of the Document, and likewise the network locations given in the Document for previous versions it was based on. These may be placed in the ``History'' section. You may omit a network location for a work that was published at least four years before the Document itself, or if the original publisher of the version it refers to gives permission. \item[K.] For any section Entitled ``Acknowledgements'' or ``Dedications'', Preserve the Title of the section, and preserve in the section all the substance and tone of each of the contributor acknowledgements and/or dedications given therein. \item[L.] Preserve all the Invariant Sections of the Document, unaltered in their text and in their titles. Section numbers or the equivalent are not considered part of the section titles. \item[M.] Delete any section Entitled ``Endorsements''. Such a section may not be included in the Modified Version. \item[N.] Do not retitle any existing section to be Entitled ``Endorsements'' or to conflict in title with any Invariant Section. \item[O.] Preserve any Warranty Disclaimers. \end{itemize} If the Modified Version includes new front-matter sections or appendices that qualify as Secondary Sections and contain no material copied from the Document, you may at your option designate some or all of these sections as invariant. To do this, add their titles to the list of Invariant Sections in the Modified Version's license notice. These titles must be distinct from any other section titles. You may add a section Entitled ``Endorsements'', provided it contains nothing but endorsements of your Modified Version by various parties--for example, statements of peer review or that the text has been approved by an organization as the authoritative definition of a standard. You may add a passage of up to five words as a Front-Cover Text, and a passage of up to 25 words as a Back-Cover Text, to the end of the list of Cover Texts in the Modified Version. Only one passage of Front-Cover Text and one of Back-Cover Text may be added by (or through arrangements made by) any one entity. If the Document already includes a cover text for the same cover, previously added by you or by arrangement made by the same entity you are acting on behalf of, you may not add another; but you may replace the old one, on explicit permission from the previous publisher that added the old one. The author(s) and publisher(s) of the Document do not by this License give permission to use their names for publicity for or to assert or imply endorsement of any Modified Version. \begin{center} {\Large\bf 5. COMBINING DOCUMENTS\par} \phantomsection \addcontentsline{toc}{section}{5. COMBINING DOCUMENTS} \end{center} You may combine the Document with other documents released under this License, under the terms defined in section~4 above for modified versions, provided that you include in the combination all of the Invariant Sections of all of the original documents, unmodified, and list them all as Invariant Sections of your combined work in its license notice, and that you preserve all their Warranty Disclaimers. The combined work need only contain one copy of this License, and multiple identical Invariant Sections may be replaced with a single copy. If there are multiple Invariant Sections with the same name but different contents, make the title of each such section unique by adding at the end of it, in parentheses, the name of the original author or publisher of that section if known, or else a unique number. Make the same adjustment to the section titles in the list of Invariant Sections in the license notice of the combined work. In the combination, you must combine any sections Entitled ``History'' in the various original documents, forming one section Entitled ``History''; likewise combine any sections Entitled ``Acknowledgements'', and any sections Entitled ``Dedications''. You must delete all sections Entitled ``Endorsements''. \begin{center} {\Large\bf 6. COLLECTIONS OF DOCUMENTS\par} \phantomsection \addcontentsline{toc}{section}{6. COLLECTIONS OF DOCUMENTS} \end{center} You may make a collection consisting of the Document and other documents released under this License, and replace the individual copies of this License in the various documents with a single copy that is included in the collection, provided that you follow the rules of this License for verbatim copying of each of the documents in all other respects. You may extract a single document from such a collection, and distribute it individually under this License, provided you insert a copy of this License into the extracted document, and follow this License in all other respects regarding verbatim copying of that document. \begin{center} {\Large\bf 7. AGGREGATION WITH INDEPENDENT WORKS\par} \phantomsection \addcontentsline{toc}{section}{7. AGGREGATION WITH INDEPENDENT WORKS} \end{center} A compilation of the Document or its derivatives with other separate and independent documents or works, in or on a volume of a storage or distribution medium, is called an ``aggregate'' if the copyright resulting from the compilation is not used to limit the legal rights of the compilation's users beyond what the individual works permit. When the Document is included in an aggregate, this License does not apply to the other works in the aggregate which are not themselves derivative works of the Document. If the Cover Text requirement of section~3 is applicable to these copies of the Document, then if the Document is less than one half of the entire aggregate, the Document's Cover Texts may be placed on covers that bracket the Document within the aggregate, or the electronic equivalent of covers if the Document is in electronic form. Otherwise they must appear on printed covers that bracket the whole aggregate. \begin{center} {\Large\bf 8. TRANSLATION\par} \phantomsection \addcontentsline{toc}{section}{8. TRANSLATION} \end{center} Translation is considered a kind of modification, so you may distribute translations of the Document under the terms of section~4. Replacing Invariant Sections with translations requires special permission from their copyright holders, but you may include translations of some or all Invariant Sections in addition to the original versions of these Invariant Sections. You may include a translation of this License, and all the license notices in the Document, and any Warranty Disclaimers, provided that you also include the original English version of this License and the original versions of those notices and disclaimers. In case of a disagreement between the translation and the original version of this License or a notice or disclaimer, the original version will prevail. If a section in the Document is Entitled ``Acknowledgements'', ``Dedications'', or ``History'', the requirement (section~4) to Preserve its Title (section~1) will typically require changing the actual title. \begin{center} {\Large\bf 9. TERMINATION\par} \phantomsection \addcontentsline{toc}{section}{9. TERMINATION} \end{center} You may not copy, modify, sublicense, or distribute the Document except as expressly provided for under this License. Any other attempt to copy, modify, sublicense or distribute the Document is void, and will automatically terminate your rights under this License. However, parties who have received copies, or rights, from you under this License will not have their licenses terminated so long as such parties remain in full compliance. \begin{center} {\Large\bf 10. FUTURE REVISIONS OF THIS LICENSE\par} \phantomsection \addcontentsline{toc}{section}{10. FUTURE REVISIONS OF THIS LICENSE} \end{center} The Free Software Foundation may publish new, revised versions of the GNU Free Documentation License from time to time. Such new versions will be similar in spirit to the present version, but may differ in detail to address new problems or concerns. See http://www.gnu.org/copyleft/. Each version of the License is given a distinguishing version number. If the Document specifies that a particular numbered version of this License ``or any later version'' applies to it, you have the option of following the terms and conditions either of that specified version or of any later version that has been published (not as a draft) by the Free Software Foundation. If the Document does not specify a version number of this License, you may choose any version ever published (not as a draft) by the Free Software Foundation. } \renewcommand{\chaptername}{Index} \printindex \end{document} \endinput