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\title{{\Huge\cal The Elements of Infinitesimal Calculus}}
\author{ David A. Santos \\
\href{mailto:dsantos@ccp.edu}{dsantos@ccp.edu}}
\date{\today\ Version}
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\clearpage \section*{Preface} \markboth{}{}
\addcontentsline{toc}{chapter}{Preface} \markright{Preface}


These notes started the fall of 2004, when I taught Maths 165,
Differential Calculus, at Community College of Philadelphia.

\bigskip

The students at that course were Andrea BATEMAN, Kelly BLOCKER,
Alexandra LOUIS,  Cindy LY, Thoraya SABER, Stephanilee MAHONEY,
Brian McCLINTON, Jessica MENDEZ,  Labaron PALMER, Leonela TROKA, and
Samneak SAK. I would like to thank them for making me a better
teacher with their continuous input and questions.



I have profitted from conversations with Jos\'{e} Mason and Alain
Schremmer regarding approaches to teaching this course.

\bigskip


\hfill David A. Santos \\

\hfill Please send comments to
\href{mailto:dsantos@ccp.edu}{dsantos@ccp.edu}

\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


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This chapter introduces essential notation and terminology that will
be used throughout these notes. We will often use the symbol $\iff$
for ``if and only if'', and the symbol $\implies$, ``implies.'' The
symbol $\approx$ means {\em approximately}.

\section{Real Numbers}
In these notes we consider the following sets of numbers, assigning
to them special notation.

\begin{df}
The set $$\naturals=\{0,1,2,3,4,\ldots\}$$ is the set of {\em
natural numbers.}
\end{df}
Natural numbers allow us to count objects. The sum and product of
two natural numbers is also a natural number, and so we say that
natural numbers are {\em closed} under addition and subtraction. So,
for example, $1+1$ is a natural number, which we write as $1+1\in
\naturals$, read ``one times one belongs to the natural numbers.''
Similarly, $1\cdot 1 \in \naturals$. The natural numbers are not
closed under subtraction or division, since, for example, $1-2\nin
\naturals$, which we read ``one minus two is not in the natural
numbers.'' In order to have a set closed under subtraction, we must
adjoin the opposite of the natural numbers, creating thus the
following set.
\begin{df}
The set $$\integers=\{\ldots, -4,-3,-2,-1,0,1,2,3,4,\ldots\}$$ is
the set of {\em integers.}
\end{df}
The integers are not closed under division, since for example,
$1\div 2\nin \integers$. Starting from the integers, in order to
have a set closed under division, we must adjoin all the quotients
of integers, creating thus the following set.
\begin{df}
The set $$\rationals =\left\{\dfrac{a}{b}:a\in \integers, b\in
\integers, b\neq 0\right\}$$ is the set of {\em rational numbers.}
This is read as ``$\rationals$ is the set of all fractions $a$ over
$b$ such that $a$ is an integer, $b$ is an integer, and $b$ is
different from $0$.''
\end{df}
In other words, fractions, that is, rational numbers, are divisions
that we are too lazy to perform.

\bigskip

Notice that we do not allow division by $0$. What would happen if we
were not that lazy, and actually performed the implicit divisions in
the rational numbers? We would get objects like
$$\dfrac{2}{1}=2.0, \qquad \dfrac{1}{2}=0.5, \qquad \dfrac{1}{3}=0.33333\ldots = 0.\overline{3},  $$
where in this last division, the division goes on forever, but we
see that we repeatedly obtain $3$ in the quotient, that is, we get a
{\em periodic decimal.} It can be proved---but we will not do so in
these notes---that the set of rational numbers $\rationals$ is
precisely the set of numbers whose decimal expansion is either
finite, or a periodic decimal.

\bigskip

In $\rationals$ we have a very elegant system of numbers that is
closed under addition, subtraction, multiplication, and division
(except division by $0$). Do we need more numbers? What happens with
numbers like the {\em Champernowne-Mahler} constant
$$ 0.123456789101112131415161718192021\ldots, $$
which is the decimal number obtained by consecutively writing all
the natural number? This number is clearly not a periodic decimal,
and hence it is not rational.


To accommodate infinite non-periodic decimals, we must create the
following set.
\begin{df}
The set $\reals$ is the set of real numbers, that is, the set of all
numbers with either
\begin{enumerate}
\item a finite decimal expansion, or
\item an infinite periodic decimal expansion, or
\item an infinite non-periodic decimal.
\end{enumerate}
A real number which is not rational is called {\em irrational.}
\end{df}

We must remark that ``looking into the decimal expansion'' of a
number is not enough to {\em prove} that a number is irrational. For
example, it was known since the times of Pythagoras that the number
$\sqrt{2}$ is irrational. This guarantees that its decimal expansion
$$ \sqrt{2}=1.41421356\ldots $$ does not repeat. If we started,
however, with the number $1.41421356\ldots$ we would not know
whether it is rational or irrational, for, it may have a very long
decimal period, so long that our calculators and computers could not
store it. Again, although Archimedes suspected that $$ \pi =
3.14159265\ldots  $$was irrational, a proof of this was not obtained
until the eighteenth century by Lambert.

\subsection*{Homework}\addcontentsline{toc}{subsection}{Homework}
\begin{pro}
Give an example of a rational number between $\dfrac{1}{10}=0.1$ and
$\dfrac{1}{9}=0.\overline{1}$. Give an example of an irrational
number between $\dfrac{1}{10}=0.1$ and
$\dfrac{1}{9}=0.\overline{1}$.
\begin{answer}
There are infinitely many answers. A rational number between then is
$0.101$. An irrational number between then can be obtained by
letting $c=0.12345678910\ldots$ be the Champernowne-Mahler constant,
and then taking
$$ 0.1 + 0.1\cdot c = 0.101234567891011\ldots . $$
\end{answer}
\end{pro}

\section{Intervals}

\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.}



\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,arrowscale=1.5]{*-*}(-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,arrowscale=1.5]{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,arrowscale=1.5]{*-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,arrowscale=1.5]{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,arrowscale=1.5]{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,arrowscale=1.5]{*->}(-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,arrowscale=1.5]{<-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,arrowscale=1.5]{<-*}(-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,arrowscale=1.5]{<->}(-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}





\section{Inequalities}

\begin{rem}
\fcolorbox{brown}{yellow}{
    \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 newspeak terms as {\em non-negative} and {\em
    non-positive}.
\end{minipage}}
\end{rem}
The set of real numbers $\reals$  is  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
$,
$$\mathrm{if} \quad  x> y \quad \mathrm{and}\quad  z<0 \quad \mathrm{then}\quad xz<yz.$$
\end{axi}
\begin{rem}
$x<y$ means that $y>x$.  $x\leq y$ 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
$$2x<-10 \implies x < \dfrac{-10}{2} \implies x<-5. $$The solution
set is thus the interval $\oo{-\infty}{-5}$.
\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
$$-2x\leq -10 \implies x \geq \dfrac{-10}{-2} \implies x\geq -5. $$
The solution set is therefore $\co{-5}{+\infty}$.
\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
$$ \{x\in \reals : a_kx+b_k > 0\}\cup \{x\in \reals : a_kx+b_k = 0\}\cup\{x\in \reals : a_kx+b_k < 0\}.  $$
Here the sign $\cup$, read ``union'' means that elements of all the
sets involved are considered.\index{union} \index{$\cup$} 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}\label{exa:ineq-3}
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}

\subsection*{Homework}\addcontentsline{toc}{subsection}{Homework}
\begin{pro}
\noindent
\begin{enumerate}
\item Determine a sign diagram for the set
 $$\{x\in \reals \quad : \quad x(x+1)(x-1)<0\}.$$
 \item Using the sign diagram obtained, write the set $\{x\in \reals :
 x(x+1)(x-1)<0\}$ as a union of intervals.
 \item Write the set $\left\{x\in \reals : \dfrac{x(x-1)}{x+1}\geq
 0\right\}$ as a union of intervals.
 \item Write the set  $\left\{x\in \reals : \dfrac{x(x-1)}{x+1}\geq
 -1\right\}$ as an interval.
\end{enumerate}

\begin{answer}
\noindent
\begin{enumerate}
\item $\renewcommand{\arraystretch}{2.1}
\psset{unit=.85pc}
\def\plus{\psline(-.2,0)(.2,0)\psline(0,-.2)(0,.2)}
\def\minus{\psline(-.2,0)(.2,0)}\begin{array}{|c|c|c|c|}\hline ]-\infty;-1[ & ]-1;0[  & ]0;1[  & ]1;+\infty[ \\
\hline  \minus & \plus & \minus & \plus    \\ \hline
 \end{array}$
 \item $\oo{-\infty}{-1} \ \cup \ \oo{0}{1}$
 \item  $\oc{-1}{0}\ \cup \  \co{1}{+\infty}$
\item We have
$$ \dfrac{x(x-1)}{x+1}\geq -1 \iff \dfrac{x(x-1)}{x+1}+1\geq 0 \iff \dfrac{x(x-1)}{x+1}+\dfrac{x+1}{x+1}\geq 0 \iff \dfrac{x^2+1}{x+1}\geq 0.  $$
The numerator is always positive, since it is a square plus $1$. The
denominator is strictly positive for $x>-1$. Hence the solution set
is $\oo{-1}{+\infty}$.
\end{enumerate}
\end{answer}
\end{pro}

\section{Functions}
We will take a very narrow approach here to the definition of a
function, one that will be useful to our purposes.
\begin{df}
A {\em function} $f:\dom{f}\to \reals$, $x\mapsto f(x)$ from the set
$\dom{f}$ to the real numbers is the collection of the following
ingredients:
\begin{enumerate}
\item A collection of inputs $\dom{f}$ called the {\em domain of definition} of the
function, being the largest set of real numbers for which the output
of the function is a real number.
\item A name for the function, typically $f, g, h$, etc.
\item A name for a typical input, normally $x$.
\item An assignment rule or formula, that associates to every input
$x$ a unique output $f(x)$.
\end{enumerate}
\end{df}
We will give here some examples of functions.
\begin{exa}
Consider the function $$\fun{f}{x}{x-x^2}{\reals}{\reals} .
$$Observe that the domain of definition is $\reals$ since the
expression $x-x^2$ is a real number for every $x\in \reals$. We
have, for instance,
$$ f(0)=0-0^2=0, \qquad f(2)=2-2^2=-2, \qquad f(1+\sqrt{2})= 1+\sqrt{2}-(1+\sqrt{2})^2=1+\sqrt{2}-(1+2\sqrt{2}+2)=-2-\sqrt{2}.$$
\end{exa}
\begin{exa}
Consider the function $$\fun{f}{x}{\dfrac{1}{x-x^2}}{\reals
\setminus \{0,1\}}{\reals} .
$$Since the denominator vanishes when  $x-x^2=0 \implies x(1-x)=0 \implies x\in \{0,1\}$, the domain of definition is all real numbers excluding those two numbers,
which we write as $\reals \setminus \{0,1\}$, pronounced ``$\reals$
without the set containing $1$ and $2$.''  We have, for instance,
$$f\left(\dfrac{1}{2}\right) = \dfrac{1}{\dfrac{1}{2}-\dfrac{1}{4}}= 4.$$
\end{exa}
\begin{exa}
Consider the function
$$\fun{f}{x}{\sqrt{x^2+2x-35}}{\oc{-\infty}{-7}\cup \co{5}{+\infty}}{\reals} .
$$By example \ref{exa:ineq-3}, the quantity under the square root is strictly negative for
$x\in \oo{-7}{5}$, and hence, the formula will not give a real
number output. We must consider only those numbers for which the
quantity under the square root is positive and hence the domain of
definition of the function is $\oc{-\infty}{-7}\cup
\co{5}{+\infty}$.
\end{exa}


\begin{df}
Let $f$ and $g$ be two functions and let the point $x$ be in both of
their domains. Then $f+g$ is their sum, defined at each point $x$ by
\index{functions!sum of}
$$(f+g)(x) = f(x) + g(x).
$$The difference $f-g$ is defined by \index{functions!difference of}
$$(f-g)(x) = f(x) - g(x),  $$and their product $fg$ is defined by \index{functions!product of}$$(fg)(x) = f(x)\cdot g(x).  $$
Furthermore, if $g(x) \neq 0$, then their quotient is defined
as\index{functions!quotient of}
$$\left(\dfrac{f}{g}\right)(x) = \dfrac{f(x)}{g(x)}.  $$
The composition $f\circ g$ (``$f$ composed with $g$'') is defined at
the point $x$ by
$$(f\circ g)(x) = f(g(x)).
$$

\end{df}

\begin{exa}
Let $$\fun{f}{x}{\dfrac{1}{x}}{\reals\setminus\{0\}}{\reals}, \qquad
\fun{g}{x}{\sqrt{25-x^2}}{\cc{-5}{5}}{\reals}.
$$
Find
\begin{enumerate}
\item $(f+g)(2)$
\item $(fg)(2)$
\item $(f\circ g)(2)$
\item $(g\circ f)(2)$
\end{enumerate}
\end{exa}
\begin{solu}
We have
\begin{enumerate}
\item $(f+g)(2)=f(2)+g(2)=\dfrac{1}{2}+\sqrt{21}$
\item $(fg)(2)=\dfrac{1}{2}\cdot \sqrt{21}=\dfrac{\sqrt{21}}{2}$
\item $(f\circ g)(2)=f(g(2))=f(\sqrt{21})=\dfrac{1}{\sqrt{21}}$
\item $(g\circ f)(2)=g(f(2))=g\left(\dfrac{1}{2}\right)=\sqrt{25-\dfrac{1}{4}}=\dfrac{3\sqrt{11}}{2}$
\end{enumerate}
\end{solu}



We conclude this section with some special names that will be used
throughout these notes.

\begin{df}
A function $p:\reals \to \reals$ whose formula is of the form
$$ p(x) = a_nx^n+a_{n-1}x^{n-1}+\cdots + a_1x+a_0,  $$where the $a_k\in
\reals$ are constants, $a_n\neq 0$, is called a {\em polynomial
function} of degree $n$. \index{function!polynomial} If $n=0$ then
$$ p(x) = a$$for some real constant  numbers $a$, this is called a {\em constant function}. If $n=1$ then
$$ p(x) = a+bx $$for some real numbers $a, b$, with $b\neq 0$, this is called an {\em affine
function}.  If $n=2$ then
$$ p(x) = a+bx + cx^2 $$for some real numbers $a, b, c$, with $c\neq 0$,  this is called a {\em
quadratic function}.
\end{df}
\begin{df}
Let $p(x)$ and $q(x)$ be two polynomials with real coefficients, and
let $Z=\{x\in \reals: q(x)=0\}$. The function $r:\reals \setminus Z
\to \reals$ with formula $r(x)=\dfrac{p(x)}{q(x)}$ is called a {\em
rational function}.
\end{df}

\begin{df}
The function with assignment rule $x\mapsto f(x)$  is said to be
{\em algebraic} if $y=f(x)$ is a solution of an equation of the form
$$
p_n(x) y^n + \cdots + p_1(x) y + p_0(x) = 0,
$$
where the $p_0(x), p_1(x), \ldots, p_n(x)$ are polynomials in $x$. A
function that satisfies no such equation is said to be
\emph{transcendental}.
\end{df}

\begin{exa}
The function $$\fun{p}{x}{x^3-x+1}{\reals}{\reals}  $$is a
polynomial function of degree $3$.

\bigskip

The function $$\fun{r}{x}{\dfrac{x}{x^2+2x+2}}{\reals}{\reals}  $$is
a rational function.



\bigskip

The function
$$\fun{a}{x}{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\co{0}{+\infty}}{\reals}
$$is a algebraic function.

\end{exa}

\subsection*{Homework}\addcontentsline{toc}{subsection}{Homework}
\begin{pro}
Determine the domain of definition of the assignment rule $x\mapsto
\sqrt{x}+\sqrt{-x}$.
\begin{answer}
$\{0\}$
\end{answer}
\end{pro}
\begin{pro}
Determine the domain of definition of the assignment rule $x\mapsto
\dfrac{1}{\sqrt{x}+\sqrt{-x}}$.
\begin{answer}
$\varnothing$, the empty set.
\end{answer}
\end{pro}
\begin{pro}
Determine the domain of definition of the assignment rule $x\mapsto
\dfrac{x+1}{(x-1)\sqrt{x}}$.
\begin{answer}
$\oo{0}{1}\cup \oo{1}{+\infty}$
\end{answer}
\end{pro}
\begin{pro}
Consider the functions $$\fun{f}{x}{\sqrt{x^2+1}}{\reals}{\reals},
\qquad \fun{g}{x}{\sqrt{x^2-1}}{\oc{-\infty}{-1}\cup
\co{1}{+\infty}}{\reals}.
$$
\begin{enumerate}
\item Determine $(f+g)(2)$.
\item Determine $(fg)(2)$.
\item Determine $(f\circ g)(2)$.
\item Determine $(g\circ f)(2)$.
\end{enumerate}
\begin{answer}
\noindent
\begin{enumerate}
\item $(f+g)(2)=f(2)+g(2)=\sqrt{5}+\sqrt{3}$.
\item Determine $(fg)(2)=f(2)\cdot g(2)=\sqrt{5}\cdot \sqrt{3}=\sqrt{15}$.
\item Determine $(f\circ g)(2)=f(g(2))=f(\sqrt{3})=2$.
\item Determine $(g\circ f)(2)=g(f(2))=g(\sqrt{5})=2$.
\end{enumerate}

\end{answer}
\end{pro}

\section{Infinitesimals}
We now want to informally introduce the concepts of {\em nearness}
and {\em smallness}.



 What does it mean for one point to be
``near'' another point? We could argue that $1$ is near to $0$, but,
for some purposes, this distance could be too ``far.'' We could
certainly say that $0.5$ is closer to $0$ than $1$ is, but then
again, for some purposes, even this distance could be ``far.''
Mentioning a specific number ``near'' $0$, like $1$ or $0.5$ fails
in what we desire for ``nearness'' because mentioning a specific
point immediately gives a ``static'' quality to ``nearness'': once
you mention a specific point, you could mention infinitely many more
points which are closer than the point you mentioned. The points in
the sequence
$$ 0.1,\qquad 0.01,\qquad 0.001,\qquad 0.0001,\qquad \ldots
$$get closer and closer to $0$ with an arbitrary precision. Notice that this sequence approaches $0$ through values
$>0$. This arbitrary precision is what will be the gist of our
concept of ``nearness.'' ``Nearness'' is dynamic: it involves the
ability of getting closer to a point with any desired degree of
accuracy. It is not static.

\bigskip

Again, the points in the sequence $$-\dfrac{1}{2},\qquad
-\dfrac{1}{4},\qquad  -\dfrac{1}{8},\qquad  -\dfrac{1}{16},\qquad
\ldots $$ are arbitrarily close to $0$, but they ``approach''  $0$
from the left. Once again, the sequence
$$+\dfrac{1}{2},\qquad
-\dfrac{1}{3},\qquad  +\dfrac{1}{4},\qquad  -\dfrac{1}{5},\qquad
\ldots $$approaches $0$ from both above and below. After this long
preamble, we may formulate our first definition.
\begin{df}
The notation $x\rightarrow a$, read ``$x$ tends to $a$,'' means that
$x$ is very close, with an arbitrary degree of precision, to $a$.
Here $x$ can approach $a$ through values smaller or larger than $a$.
We write  $x\rightarrow a+$ (read ``$x$ tends to $a$ from the
right'') to mean that $x$ approaches $a$ through values larger than
$a$ and we write $x\rightarrow a-$ (read ``$x$ tends to $a$ from the
left'') we mean that $x$ approaches $a$ through values smaller than
$a$.
\end{df}

\begin{df}
Given a function $f$ and a point $a\in \reals$, we write $f(a+)$ for
the value that $f(x)$ attains when $x\to a+$. Similarly,  we write
$f(a-)$ for the value that $f(x)$ attains when $x\to a-$.
\end{df}
\begin{exa}
Given $f:\reals\to \reals$,
$$f(x)=\left\{\begin{array}{ll} \sqrt{4-x} & \mathrm{if}\ x < 4, \\
1  & \mathrm{if}\ x = 4, \\
3  & \mathrm{if}\ x > 4. \\

 \end{array}\right.  $$
Then $$ f(4-)=\sqrt{4-4}=0, \qquad f(4)=1, \qquad f(4+)=3. $$
Observe that also, for example,
 $$ f(0-)=2, \qquad f(0)=2, \qquad f(0+)=2, $$
 and
  $$ f(5-)=3, \qquad f(5)=3, \qquad f(5+)=3. $$

\end{exa}
\vspace{1cm}
\begin{figure}[h]
$$\psset{unit=1pc}\psline[linewidth=1.5pt]{<->}(-11, 0)(11,0)\uput[d](-11,0){-\infty}
\uput[d](11,0){+\infty}\uput[d](0,0){a}
\psline{<-}(.3,0)(1,1)\uput[ur](1,1){
a+}\psline{<-}(-.3,0)(-1,-1)\uput[dl](-1,-1){a-}\rput(0,0){|}
$$ \meinecaption{1}{A neighbourhood of $a$.}\label{fig:hyperrealline}
\end{figure}

\begin{df}
A {\em neighbourhood of a point $a$} is an interval containing $a$.
\end{df}
Notice that the definition of neighbourhood does not rule out the
possibility that $a$ may be an endpoint of the the interval. Our
interests will be mostly on arbitrarily small neighbourhoods of a
point. Schematically we have a diagram like figure
\ref{fig:hyperrealline}.

\begin{df}[Infinitesimal]
We say that a given formula $f(x)$ in the variable $x$ is {\em
infinitesimal} as $x\to 0$, written $f(x)=\soo{1}$ if $f(x)\to 0$ as
$x\to 0$. We also say that {\em $f$ is small oh of
$1$}.\index{infinitesimal}\index{small oh notation}
\end{df}
We make the following observations, which we will take as axioms.
\begin{axi}
If $x\to 0$ and if $n \geq 1$, $n \in \naturals$, then
$x^n=\soo{1}$. \label{axi:x^n->0}
\end{axi}

\begin{axi}[The Sum of Infinitesimals is an Infinitesimal]
If $x\to 0$ and if $a(x)=\soo{1}$ and $b(x) =\soo{1}$, then
$a(x)+b(x)=\soo{1}$. \label{axi:sum_of_infinitesimals}
\end{axi}
\begin{axi}[The Product of an Infinitesimal and a Constant is an Infinitesimal]
If $x\to 0$ and if $a(x)=\soo{1}$ and $\lambda\in \reals$, $\lambda
\neq 0$, then $\lambda a(x)=\soo{1}$. \label{axi:constant
and_infinitesimals}
\end{axi}
\begin{axi}[The Product of Infinitesimals is an Infinitesimal]
If $x\to 0$ and if $a(x)=\soo{1}$ and $b(x) =\soo{1}$, then
$a(x)b(x)=\soo{1}$. \label{axi:product_of_infinitesimals}
\end{axi}
\begin{exa}
Justify the assertion   that $f(x)=x^3-x=\soo{1}$ as $x\to 0$.
\end{exa}
\begin{solu} We have $x^3=\soo{1}$ and $x=\soo{1}$ as $x\to 0$ by
Axiom \ref{axi:x^n->0}. Also, $-x=\soo{1}$ by Axiom
\ref{axi:constant and_infinitesimals}. Finally,
$x^3-x=x^3+(-x)=\soo{1}$ by Axiom \ref{axi:sum_of_infinitesimals}.
\end{solu}

We would like to compare now how ``fast'' two infinitesimals
formulas tend toward zero. In order to motivate the following
definitions, let us consider a numerical example. Consider a very
small number, say,
$$ 0.0000000100023 = \dfrac{1}{10^8} +\dfrac{2}{10^{12}} +\dfrac{3}{10^{13}}. $$
In comparison to $ \dfrac{1}{10^8}$, the sum $\dfrac{2}{10^{12}}
+\dfrac{3}{10^{13}}$ is very small, practically negligible. In other
words, the number $ \dfrac{1}{10^8}$ overwhelms or dominates the sum
$\dfrac{2}{10^{12}} +\dfrac{3}{10^{13}}$.
\begin{df}\label{df:small_oh_x_small}
Let $m$ and $n$ be non-zero natural numbers with $m<n$. We say that
$x^m$ {\em dominates} $x^n$ as $x\to 0$. We also say that $x^n$ goes
{\em faster} to $0$ than $x^m$, or that $x^m$ goes {\em slower} to
$0$ than $x^n$. We write this as $x^n = \soo{x^m}$. In general, if
$\dfrac{f(x)}{g(x)}\to 0$, we write $f(x)=\soo{g(x)}$, and if
$a(x)-b(x)=\soo{c(x)}$, we write $a(x)=b(x) + \soo{c(x)}$. The term
$\soo{c(x)}$ is called the {\em error term}. \index{error
term}\end{df}
\begin{axi}Let $\lambda \in \reals$, $\lambda \neq 0$, and $f(x)=\soo{1}$
and $g(x)=\soo{1}$.  The $\soo{}$ symbol has the following
properties: as $x\to 0$,
\begin{eqnarray}
\lambda\soo{f(x)} & = & \soo{f(x)} \label{eq:kof=of}\\
\soo{\lambda f(x)} & = & \soo{f(x)} \label{eq:okf=of}\\
\soo{f(x)}\pm \soo{f(x)} &=& \soo{f(x)} \label{eq:of+of=of} \\
\soo{\soo{f(x)}} & = & \soo{f(x)}\label{eq:oof=of}\\
f(x)g(x) & = & \soo{f(x)}\label{eq:fg=of}\\
f(x)g(x) & = & \soo{g(x)}\label{eq:fg=og}\\
\soo{f(x)+\soo{f(x)}} & = & \soo{f(x)}\label{eq:o(f+of)=of}\\
\soo{f(x)^n} & = & \soo{f(x)^k}, \qquad n \geq 2, n\in \naturals, k=1, 2, \ldots , n-1\label{eq:o(f^n)=of^k}\\
\soo{f(x)}^n & = & \soo{(f(x))^n}, \qquad n \geq 1, n\in \naturals \label{eq:o(f)^n=of^n}\\
(f(x))^n\soo{f(x)} & = & \soo{(f(x))^{n+1}}, \qquad  n\in \naturals \label{eq:fo(f)=of^2}\\
\dfrac{\soo{(f(x))^n}}{f(x)} & = & \soo{(f(x))^{n-1}}, \qquad  n\geq
2, n\in \naturals \label{eq:o(f)/f=o1}
\end{eqnarray}
\end{axi}

\begin{exa}
As $\tends{x}{0}$ we have $\soo{x^3}+\soo{x^4} = \soo{x^3}$, since
the $x^3$ term dominates over the $x^4$ for  small $x$.
\end{exa}



\begin{df}
Let $n_1<n_2< \cdots <n_k$ be  natural numbers. The {\em dominant
term} as $x\to 0$ of the polynomial
$$p(x) = a_{n_1}x^{n_1}+a_{n_2}x^{n_2}+\cdots + a_{n_k}x^{n_k} $$
is $a_{n_1}x^{n_1}$. We write this as $p(x)\sim a_{n_1}x^{n_1}$,
read ``$p$ of $x$ is asymptotic to $a$ sub one times $x$ to the
power $n$ sub $1$.''
\end{df}

\begin{exa}
Given that $a(x) =2x^2+x^3+\soo{x^3}$ and $b(x)=x-4x^4+ \soo{x^4}$,
estimate the product $a(x)b(x)$ as $x\to 0$.
\end{exa}
\begin{solu}
We have
$$\begin{array}{lll}a(x)b(x) & = & \left(2x^2+x^3+\soo{x^3}\right)\left(x-4x^4+ \soo{x^4}\right) \\
& = &  2x^3-8x^6+\soo{2x^6}+
x^4-4x^7+\soo{x^7}+\soo{x^4}+\soo{-4x^7}+\soo{x^7}\\
& = & 2x^3+x^4+\soo{x^4},
\end{array}
$$since all other error terms are faster than $\soo{x^4}$.
\end{solu}



\begin{exa}
As $x\to 0$, $3x^2+2x+x \sim x$.
\end{exa}
\begin{axi}[Product of Asymptotic Estimates] If $p(x)$ and $q(x)$ are polynomials, with $p(x)\sim
ax^n$ and $q(x)\sim bx^m$ as $x\to 0$, then $p(x)q(x)\sim abx^{m+n}$
as $x\to 0$.
\end{axi}

\begin{exa}
As $x\to 0$, $x(2x+3x^2)(3+4x)(5+6x) \sim x\cdot 2x\cdot 3\cdot
5=20x^2$.
\end{exa}


\subsection*{Homework}\addcontentsline{toc}{subsection}{Homework}
\begin{pro}
Given $a(x)=2+x^2+\soo{x^2}$ and $b(x) = x+\soo{x}$,  as $x\to 0$,
estimate $a(x)b(x)$ as $x\to 0$.
\begin{answer}
$a(x)b(x) =2x+\soo{x}$ as $x\to 0$.
\end{answer}
\end{pro}



\section{Infinitely Large Quantities}


\begin{df}
We use the symbol $+\infty$ to denote a quantity that is larger than
any positive real number.
\end{df}
Thus if $P\in\reals$, no matter how large $P>0$ might be, we always
have $$0<P<+\infty .$$

\begin{df}
We use the symbol $-\infty$ to denote a quantity that is smaller
than any negative real number.
\end{df}
Thus if $N\in\reals$, $N<0$,  no matter how large $|N|$ might be, we
always have
$$-\infty<N<0.$$

\vspace{1cm}
\begin{figure}[h]
$$\psset{unit=.5}
\renewcommand{\pshlabel}[1]{{\tiny #1}}
\psaxes[yAxis=false](0, 0)(-7,0)(7, 0)
\psline[linestyle=dotted]{->}(7,0)(11,0)
\psline[linestyle=dotted]{->}(-7,0)(-11,0)
\uput[dr](11,0){+\infty}\uput[dl](-11,0){-\infty}
$$\meinecaption{1}{The Real Line.} \label{fig:the_real_line}
\end{figure}


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}. 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: $$\dfrac{\pm
\infty}{\pm \infty}, \ \ \ \ (+\infty) + (-\infty), \ \ \ 0(\pm
\infty).
$$

\begin{df}
The notation $x\to +\infty$, read ``$x$ tends to plus infinity''
means that increases without bound, eventually surpassing any
preassigned large positive constant. The notation $x\to -\infty$,
read ``$x$ tends to minus infinity'' means that decreases without
bound, eventually surpassing any preassigned large in magnitude
negative constant.
\end{df}



\begin{df}\label{df:small_oh_x_large}
Let $m$ and $n$ be non-zero natural numbers with $m<n$. We say that
$x^n$ {\em dominates} $x^m$ as $x\to \pm \infty$. We also say that
$x^n$ goes {\em faster} to $\pm \infty $ than $x^m$, or that $x^m$
goes {\em slower} to $\pm \infty$ than $x^n$. We write this as $x^m
= \soo{x^n}$. In general, if $\dfrac{f(x)}{g(x)}\to 0$, we write
$f(x)=\soo{g(x)}$ as $x\to \pm \infty$ and if
$a(x)-b(x)=\soo{c(x)}$, we write $a(x)=b(x) + \soo{c(x)}$. The term
$\soo{c(x)}$ is called the {\em error term}. \end{df}


The properties expounded in the preceding section for little oh as
$x\to 0$ also hold for little oh as $x\to \pm\infty$, except for
\ref{axi:x^n->0}, which must be replaced by the following.
\begin{axi}
If $x\to \pm \infty$ and if $n \geq 1$, $n \in \naturals$, then
$\dfrac{1}{x^n}=\soo{1}$. \label{axi:1/x^n->0}
\end{axi}

We now change the definition of dominant term.

\begin{df}
Let $n_1<n_2< \cdots <n_k$ be  natural numbers. The {\em dominant
term} as $x\to \pm \infty$ of the polynomial
$$p(x) = a_{n_1}x^{n_1}+a_{n_2}x^{n_2}+\cdots + a_{n_k}x^{n_k} $$
is $a_{n_k}x^{n_k}$. We write this as $p(x)\sim a_{n_k}x^{n_k}$,
read ``$p$ of $x$ is asymptotic to $a$ sub $k$ times $x$ to the
power $n$ sub $k$.''
\end{df}


\begin{exa}
As $\tends{x}{+\infty}$ we have $\soo{x^3}+\soo{x^4} = \soo{x^4}$,
since the $x^4$ term dominates over the $x^3$ for large $x$.
\end{exa}
\begin{exa}
Given that $a(x) =2x^3+x^2+\soo{x^2}$ and $b(x)=x^4-4x+ \soo{x}$,
estimate the product $a(x)b(x)$ as $x\to +\infty$.
\end{exa}
\begin{solu}
We have
$$\begin{array}{lll}a(x)b(x) & = & \left(2x^3+x^2+\soo{x^2}\right)\left(x^4-4x+ \soo{x}\right) \\
& = & 2x^7-8x^4+\soo{2x^4}+x^6-4x^3+\soo{x^3} + \soo{x^6} +
\soo{-4x^3}+\soo{x^3}\\
& = & 2x^7+x^6+\soo{x^6},
\end{array}
$$as $\tends{x}{+\infty}$ since all other error terms are weaker than $\soo{x^6}$.
\end{solu}



\begin{exa}
As $x\to +\infty$, $3x^2+2x+x \sim 3x^2$.
\end{exa}
\begin{axi}[Product of Asymptotic Estimates] If $p(x)$ and $q(x)$ are polynomials, with $p(x)\sim
ax^n$ and $q(x)\sim bx^m$ as $x\to 0$, then $p(x)q(x)\sim abx^{m+n}$
as $x\to \pm\infty$.
\end{axi}

\begin{exa}
As $x\to +\infty$, $x(2x+3x^2)(3+4x)(5+6x) \sim x\cdot 3x^2\cdot
4x\cdot 6x=72x^5$.
\end{exa}

\subsection*{Homework}\addcontentsline{toc}{subsection}{Homework}
\begin{pro}
Given $a(x)=x^3+2x^2+\soo{x^2}$ and $b(x)=x + \soo{1}$ as $x\to
+\infty$, estimate $a(x)b(x)$ as $x\to +\infty$.
\begin{answer}
$a(x)b(x) = x^4+2x^3+\soo{x^3}$ as $x\to +\infty$.
\end{answer}
\end{pro}


\section{Distance on the Plane and Lines} \label{sec:lines}
We now introduce some concepts from analytic geometry that will be
used throughout these notes.
\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 length of the line segment joining the points $(1,2)$ and
$(-3,4)$ on the plane is
$$\sqrt{(1-(-3))^2+(2-4)^2} =\sqrt{16+4}=2\sqrt{5}. $$
\end{exa}

\begin{df}\label{df:vert_hor_line}
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}
Definition \ref{df:vert_hor_line} is striking. It provides a link
between Algebra and Geometry, and viceversa, between Geometry and
Algebra. For example, if we are given the equation of a horizontal
line (Algebra), then we can graph it (Geometry). Conversely, if we
are given the picture of a horizontal line on the coordinate plane
(Geometry), we may find its equation (Algebra).

\vspace{2cm}
\begin{figure}[!hptb]
\begin{minipage}{.4\textwidth}
 \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}{.4\textwidth}
$$ \psset{unit=1pc}
\psaxes[labels=none,
ticks=none,linewidth=1.2pt]{<->}(0,0)(-7,-7)(7,7)
\psline[linewidth=2pt,linecolor=brown]{<->}(3,-6.5)(3,6.5)$$
\vspace{2cm}\meinecaption{.25}{A vertical line.}
\end{minipage} \end{figure}\vspace{3cm}\begin{figure}[h]
\begin{minipage}{.4\textwidth}
$$ \psset{unit=1pc}
\psaxes[labels=none,
ticks=none,linewidth=1.2pt]{<->}(0,0)(-7,-7)(7,7)
\psline[linewidth=2pt,linecolor=brown]{<->}(-6.5,-1)(6.5, -1)$$
\vspace{2cm}\meinecaption{.25}{A horizontal line.}
\end{minipage}
\hfill
\begin{minipage}{.4\textwidth}
$$
\psset{unit=4pc} \rput(-5, -2.5){\psline[linewidth=2pt]{*-*}(3,
1.5)(6,4.5) \psdots[dotstyle=*,dotscale=2](3, 1.5)(4,
2.5)(6,4.5)\uput{8pt}[l](3, 1.5){(x_1,y_1)} \uput{8pt}[l](4,
2.5){(x, y)} \uput{8pt}[l](6,4.5){(x_2,y_2)}
\psline[linestyle=dashed,linewidth=2pt](3,1.5)(6,1.5)(6,4.5)
\psline[linestyle=dashed, linewidth=2pt](4,2.5)(4,1.5)
\pcline[offset=10pt]{|-|}(6.5, 1.5)(6.5, 4.5) \lput*{:U}{y_2 - y_1}
\pcline[offset=10pt]{|-|}(4.5, 1.5)(4.5, 2.5)\lput*{:U}{y - y_1}
\pcline[offset=10pt]{|-|}(3, 1)(4, 1) \lput*{:U}{x - x_1}
\pcline[offset=10pt]{|-|}(3, .7)(6, .7)\lput*{:U}{x_2 - x_1}}
$$
\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}


\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{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}

\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}


\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}




\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{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 $u = -\frac{1}{4}$.
See figure \ref{fig:colli_pts}.\end{solu}

\vspace{2cm}

\begin{figure}[h]
 \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{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}


\subsection*{Homework}\addcontentsline{toc}{subsection}{Homework}
\begin{pro}
Find the distance of the line segments having $(1,-2)$ and $(-3,4)$
as endpoints.
\begin{answer}
This is $\sqrt{(1-(-3))^2+(-2-4)^2}=\sqrt{16+36}=2\sqrt{13}$.
\end{answer}
\end{pro}

\begin{pro}
Find the equation of the line passing through $(1,-2)$ and $(-3,4)$.
\begin{answer}
The slope of the line is
$\dfrac{-2-4}{1-(-3)}=\dfrac{-6}{4}=-\dfrac{3}{2}$.
 The equation of the line is of the form
$y=-\dfrac{3}{2}x+k$. Since $(1,-2)$ is on the line
$$ -2=-\dfrac{3}{2}(1)+k \implies k = -2+\dfrac{3}{2}=-\dfrac{1}{2}.
$$The equation sought is thus $y =-\dfrac{3}{2}x-\dfrac{1}{2}$.
\end{answer}
\end{pro}

\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}\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}
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}\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{figure}[h]
\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{2}{Problem
\ref{pro:coord_lines}.} \label{fig:coord_lines}
\end{figure}






\chapter{An Informal View of Graphs of Functions}\thispagestyle{empty}
\section{Graphs of Functions}
\begin{df}
The graph of a function $f$ is the set on the plane
$$\Gamma_f=\{(x,y)\in\reals^2:x\in\dom{f}, y=f(x)\}.$$
\end{df}
\begin{exa}
Consider the function $$ \fun{{\bf Id}}{x}{x}{\reals}{\reals},
$$called the {\em identity} function. By example \ref{ex:y=x}, we know that its graph is the straight
line that appears in figure \ref{fig:y=x}.
\end{exa}
\begin{exa}
Consider the function $$
\fun{f}{x}{-\dfrac{5}{3}x+5}{\reals}{\reals}
$$By example \ref{ex:line1}, we know that its graph is the straight
line that appears in figure \ref{fig:line1}.
\end{exa}

\vspace{1cm}
\begin{figure}[h]
\begin{minipage}{.3\textwidth}\centering
 \psgrid[gridwidth=1.2pt,
subgriddiv=5,linewidth=.8pt, gridlabels=0](-1,-1)(-1,-1)(1,1)
\psaxes[
Ox=-1,Oy=-1,labelFontSize=\tiny](-1,-1)(-1,-1)(1,1)\psline[linewidth=2pt,linecolor=brown]{->}(-1.5,0)(1.5,0)
\psline[linewidth=2pt,linecolor=brown]{->}(0,-1.5)(0,1.5)\psplot[linecolor=red,linewidth=2pt,algebraic=true,
plotpoints=1000]{-1}{1}{sqrt(1-x^2)} \meinecaption{2}{Example
\ref{pro:upper_semicircle}. $y=f(x)$} \label{fig:upper_semicircle}
\end{minipage}
\hfill
 \begin{minipage}{.3\textwidth}
$$ \psset{unit=.7pc}
\psaxes[labels=none, ](0,0)(-5,-5)(5,5)
\psline[linewidth=2pt,linecolor=brown]{<->}(-4,4)(-4,2)(3,2)(3,1)(-3,1)$$
\meinecaption{1}{Fails the vertical line test. Not a
function.}\label{fig:not_functions1}
\end{minipage}
\hfill \begin{minipage}{.3\textwidth}
$$\psset{unit=.7pc, algebraic=true}
\psaxes[labels=none,
](0,0)(-5,-5)(5,5)\psplot[linewidth=2pt,linecolor=brown]{-1}{2}{sqrt(x^3+1)}
\psplot[linewidth=2pt,linecolor=brown]{-1}{2}{-1*sqrt(x^3+1)}
$$\meinecaption{1}{Fails the vertical line test. Not a function.} \label{fig:not_functions2}
\end{minipage}\end{figure}

\begin{exa}\label{pro:upper_semicircle}
Demonstrate that the graph of the function
$$\fun{f}{x}{\sqrt{1-x^2}}{\cc{-1}{1}}{\reals}  $$is the upper
semicircle in figure \ref{pro:upper_semicircle}.
\end{exa}
\begin{solu}
If $y = f(x)=\sqrt{1-x^2}$, then $$x^2+y^2=1\implies
\sqrt{(x-0)^2+(y-0)^2}=1,$$which means that any point $(x, y)$ on
the graph of the function is at distance $1$ from the origin. Since
$y \geq 0$, the points satisfying these requirement form the
aforementioned semicircle.
\end{solu}

\begin{rem}

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}.

\end{rem}

Given a function $f$, it is generally difficult to know {\em a
priori} what its graph looks like. Most of the material on this
chapter will be dedicated to the investigation of tools for
sketching graphs of functions. In the subsections below we will
informally discuss some features of graphs of (algebraic) functions,
which will be elaborated later on.

\subsection{Symmetry}

\begin{df}
A function $f$ is {\em even} if for all $x$ in its domain, $-x$ is
also in its domain, and 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
$x$, $-x$ is also in its domain, and 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 function whose graph appears in figure \ref{fig:even} is even.
The function whose graph appears  in figure \ref{fig:odd} is odd.
\label{exa:even_odd}\end{exa}

\vspace{1cm}

\begin{figure}[!hptb]

\begin{minipage}{.4\textwidth}
 \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}{ The graph of an even function.} \label{fig:even}
\end{minipage}
\hfill
 \begin{minipage}{.4\textwidth}
 \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}{The graph of an odd function.} \label{fig:odd}
\end{minipage}
  \end{figure}


\subsection{Continuity}
 Heuristically
speaking, a {\em continuous function} at the point $x=a$ is one
whose graph has no ``breaks'' at the point $x=a$.

\bigskip
What could possibly go wrong and make a function discontinuos?
 Let us consider
the following three types of discontinuities. Let $f$ be a function
and $a\in \reals$. Assume that $f$ is defined in a neighbourhood of
$a$, but not precisely at $x=a$. Which value can we reasonably
assign to $f(a)$? Consider the situations depicted in figures
\ref{fig:cont1} through \ref{fig:cont3}. In figure  \ref{fig:cont1}
it seems reasonably to assign  $a(0)=0$. What value can we
reasonably assign in figure \ref{fig:cont2}? $b(0)=
\dfrac{-1+1}{2}=0$? In figure \ref{fig:cont3}, what value would it
be reasonable to assign? $c(0)=0$?, $c(0)=+\infty$?, $c(0)=-\infty$?
The situations presented here are typical, but not necessarily
exhaustive. Thus of the three discontinuous functions in figures
\ref{fig:cont1} through \ref{fig:cont3}, only that of figure
\ref{fig:cont1} has a chance of being continuous, and that happens
only if one defines $a(0)$ as $0$.


\subsection{Intercepts}
\begin{df}
Given a function $f$, the set
$$ \{(x,f(x))\in \reals ^2: x\in \dom{f}, f(x)=0\} $$ is called the set of
{\em $x$-intercepts (zeroes or roots)} of $f$. If $f$ is defined at
$0$, then the point $(0,f(0))$ is called the {\em $y$-intercept} of
$f$. In other words, the intercepts are the places where the graph
of a function touches the axes.
\end{df}
\begin{exa}
In figure \ref{fig:intercepts}, the points $A, B, C$ are the
$x$-intercepts of the function and the point $L$ is the
$y$-intercept.
\end{exa}


\vspace{2cm}
\begin{figure}[!hptb]
\begin{minipage}{.22\textwidth}
\centering\psset{unit=1pc}
\psaxes[labels=none,linewidth=2pt,arrows={->},
ticks=none](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}{$y=a(x)$.} \label{fig:cont1}
\end{minipage}\hfill
\begin{minipage}{.22\textwidth}
\centering\psset{unit=1pc}
\psaxes[labels=none,linewidth=2pt,arrows={->},
ticks=none](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}{$y=b(x)$.} \label{fig:cont2}
\end{minipage}
\hfill
\begin{minipage}{.22\textwidth}
\centering\psset{unit=1pc}
\psaxes[labels=none,linewidth=2pt,arrows={->},
ticks=none](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}{$y=c(x)$.} \label{fig:cont3}
\end{minipage}
\hfill
\begin{minipage}{.22\textwidth}
\centering \psset{unit=.8pc}
\psaxes[linewidth=2pt,arrows={->},labels=none,ticks=none](0,0)(-6,-6)(6,6)
\psplot[algebraic,linewidth=2pt,linecolor=brown,arrows={<->}]{-2.5}{4.3}{(x+2)^2*(x-2)^3*(x-4)/32}
\pstGeonode[PosAngle={-60,-90,-60,45}](-2,0){A}(2,0){B}(4,0){C}(0,4){L}
\meinecaption{2}{Intercepts}\label{fig:intercepts}
\end{minipage}

\end{figure}

\subsection{Asymptotes}

\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}



\vspace{2cm}
\begin{figure}[!hptb]
\begin{minipage}{.22\textwidth}
\centering\psset{unit=1pc}
\psaxes[labels=none,linewidth=2pt,arrows={->},
ticks=none](0,0)(-5,-5)(5,5)
\psplot[algebraic=true,linewidth=2pt,linecolor=brown,arrows={<->}]{-1.584962501}{4.5}{2+2^(-x)}
\psline[linewidth=2pt,linestyle=dotted,arrows=->](-5,1.8)(5,1.8)
\meinecaption{2}{$f(x)\to b$ as $x\to+\infty$.}
\label{fig:sin_tota1}
\end{minipage}\hfill
\begin{minipage}{.22\textwidth}
\centering\psset{unit=1pc}
\psaxes[labels=none,linewidth=2pt,arrows={->},
ticks=none](0,0)(-5,-5)(5,5)
\psplot[algebraic=true,linewidth=2pt,linecolor=brown,arrows={<->}]{-1.584962501}{4.5}{2-2^(-x)}
\psline[linewidth=2pt,linestyle=dotted,arrows=->](-5,2.2)(5,2.2)
\meinecaption{2}{$f(x)\to b$ as $x\to+\infty$.}
\label{fig:sin_tota2}
\end{minipage}\hfill
\begin{minipage}{.22\textwidth}
\centering\psset{unit=1pc}
\psaxes[labels=none,linewidth=2pt,arrows={->},
ticks=none](0,0)(-5,-5)(5,5)
\psplot[algebraic=true,linewidth=2pt,linecolor=brown,arrows={<->}]{-4.5}{1.584962501}{2+2^(x)}
\psline[linewidth=2pt,linestyle=dotted,arrows=->](-5,1.8)(5,1.8)
\meinecaption{2}{$f(x)\to b$ as $x\to-\infty$.}
\label{fig:sin_tota3}
\end{minipage}\hfill
\begin{minipage}{.22\textwidth}
\centering\psset{unit=1pc}
\psaxes[labels=none,linewidth=2pt,arrows={->},
ticks=none](0,0)(-5,-5)(5,5)
\psplot[algebraic=true,linewidth=2pt,linecolor=brown,arrows={<->}]{-4.5}{1.584962501}{2-2^(x)}
\psline[linewidth=2pt,linestyle=dotted,arrows=->](-5,2.2)(5,2.2)
\meinecaption{2}{$f(x)\to b$ as $x\to-\infty$.}
\label{fig:sin_tota4}
\end{minipage}
\end{figure}

Heuristically, the line $y=b$ is an asymptote of the function, if
eventually, the graph of the function gets closer and closer to it
without actually touching it. Figures \ref{fig:sin_tota1} through
\ref{fig:sin_tota4} shew different ways in which the line $y=b$ can
be the asymptote of a function.


\begin{exa}
Find the asymptotes, if any, of the function
$$\fun{f}{x}{\dfrac{(2x+1)^2(1-3x)^2}{x^4-1}}{\reals\setminus\{-1,1\}}{\reals}.
$$
\end{exa}
\begin{solu}
To do this, we find the dominant terms of the numerator and
denominator as $x\to \pm\infty$:
$$y= \dfrac{(2x+1)^2(1-3x)^2}{x^4-1}\sim \dfrac{(2x)^2(-3x)^2}{x^4}=36,  $$
which means that $y=36$ is its horizontal asymptote.
\end{solu}
\begin{exa}
Find the asymptotes, if any, of the function
$$\fun{g}{x}{\dfrac{(2x+1)^3(1-3x)^2}{x^4-1}}{\reals\setminus\{-1,1\}}{\reals}.
$$
\end{exa}
\begin{solu}
To do this, we find the dominant terms of the numerator and
denominator as $x\to \pm\infty$:
$$y= \dfrac{(2x+1)^3(1-3x)^2}{x^4-1}\sim \dfrac{(2x)^3(-3x)^2}{x^4}=72x,  $$
which tends to $\pm \infty$ as $x\pm \infty$, and hence does not
have a horizontal asymptote.\footnote{The expert will note that this
curve has a slanted asymptote, but we will discuss these later in
the text.}
\end{solu}
\begin{exa}
Find the asymptotes, if any, of the function
$$\fun{h}{x}{\dfrac{(2x+1)(1-3x)}{x^4-1}}{\reals\setminus\{-1,1\}}{\reals}.
$$
\end{exa}
\begin{solu}
To do this, we find the dominant terms of the numerator and
denominator as $x\to \pm\infty$:
$$y= \dfrac{(2x+1)(1-3x)}{x^4-1}\sim \dfrac{(2x)(-3x)}{x^4}=-\dfrac{6}{x^2}\to 0,  $$
as $x\pm \infty$, and hence $y=0$ is its horizontal asymptote.
\end{solu}




\subsection{Poles}

\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}
Find the poles, if any, of the function
$$\fun{h}{x}{\dfrac{(2x+1)(1-3x)}{x^4-1}}{\reals\setminus\{-1,1\}}{\reals}.
$$
\end{exa}
\begin{solu}
To do this, we find the real zeroes of the  denominator:
$$x^4-1 = 0\implies (x-1)(x+1)(x^2+1) \implies x\in \{-1,1\}, $$ from where the lines  $x=-1$
and $x=1$ are poles.
\end{solu}

\begin{exa}
Find the poles, if any, of the function
$$\fun{r}{x}{\dfrac{(2x+1)(1-3x)}{x^4+1}}{\reals\setminus\{-1,1\}}{\reals}.
$$
\end{exa}
\begin{solu}
To do this, we find the real zeroes of the  denominator. But since
$1+x^4=1+(x^2)^2$ is $1$ plus a real square, and real squares are
always positive, the denominator is never $0$, as it is, in fact,
always $\geq 1$. This means that $r$ does not have any poles.
\end{solu}











\vspace{2cm}
\begin{figure}[!hptb]
\begin{minipage}{.22\textwidth}
\centering\psset{unit=1pc}
\psaxes[labels=none,linewidth=2pt,arrows={->},
ticks=none](0,0)(-5,-5)(5,5)
\psplot[algebraic=true,linewidth=2pt,linecolor=brown,arrows={<->}]{-1.993262053}{4.5}{ln(x+2)}
\psline[linewidth=2pt,linestyle=dotted,arrows=->](-2.2,-5)(-2.2,5)\uput[d](-2.2,-5){$x=a$}
\meinecaption{3}{$f(x)\to -\infty$ as $x\to a+$.}
\label{fig:pollo_1}
\end{minipage}\hfill
\begin{minipage}{.22\textwidth}
\centering\psset{unit=1pc}
\psaxes[labels=none,linewidth=2pt,arrows={->},
ticks=none](0,0)(-5,-5)(5,5)
\psplot[algebraic=true,linewidth=2pt,linecolor=brown,arrows={<->}]{-1.993262053}{4.5}{-ln(x+2)}
\psline[linewidth=2pt,linestyle=dotted,arrows=->](-2.2,-5)(-2.2,5)\uput[d](-2.2,-5){$x=a$}
\meinecaption{3}{$f(x)\to +\infty$ as $x\to a+$.}
\label{fig:pollo_2}
\end{minipage}\hfill
\begin{minipage}{.22\textwidth}
\centering\psset{unit=1pc}
\psaxes[labels=none,linewidth=2pt,arrows={->},
ticks=none](0,0)(-5,-5)(5,5)
\psplot[algebraic=true,linewidth=2pt,linecolor=brown,arrows={<->}]{-4.5}{-2.006737947}{ln(-x-2)}
\psline[linewidth=2pt,linestyle=dotted,arrows=->](-1.8,-5)(-1.8,5)\uput[d](-1.8,-5){$x=a$}
\meinecaption{3}{$f(x)\to -\infty$ as $x\to a-$.}
\label{fig:pollo_3}
\end{minipage}\hfill
\begin{minipage}{.22\textwidth}
\centering\psset{unit=1pc}
\psaxes[labels=none,linewidth=2pt,arrows={->},
ticks=none](0,0)(-5,-5)(5,5)
\psplot[algebraic=true,linewidth=2pt,linecolor=brown,arrows={<->}]{-4.5}{-2.006737947}{-ln(-x-2)}
\psline[linewidth=2pt,linestyle=dotted,arrows=->](-1.8,-5)(-1.8,5)\uput[d](-1.8,-5){$x=a$}
\meinecaption{3}{$f(x)\to +\infty$ as $x\to a-$.}
\label{fig:pollo_4}
\end{minipage}
\end{figure}

\bigskip

Figures \ref{fig:pollo_1} through \ref{fig:pollo_4} exhibit the
various ways a function may have a pole.
\subsection{Monotonicity}

\begin{df}
A function $f$ is said to be {\em increasing} (respectively, {\em
strictly increasing}) if $a <b \implies f(a) \leq f(b)$
(respectively, $a <b \implies f(a) < f(b)$). A function $g$ is said
to be {\em decreasing} (respectively, {\em strictly decreasing}) if
$a <b \implies g(a) \leq g(b)$ (respectively, $a <b \implies g(a) <
g(b)$). A function is {\em monotonic} if it is either (strictly)
increasing or decreasing. By the {\em intervals of monotonicity of a
function} we mean the intervals where the function might be
(strictly) increasing or decreasing. \index{function!increasing}
\index{function!decreasing} \index{function!strictly
increasing}\index{function!strictly
decreasing}\index{function!monotonic}\index{monotonicity}
\end{df}
\begin{rem}
If the function $f$ is (strictly) increasing, its opposite $-f$ is
(strictly) decreasing, and viceversa.
\end{rem}
The following theorem is immediate.
\begin{thm}\label{thm:increasing_equivalence1}
A function $f$ is (strictly) increasing if for all $a<b$ for which
it is defined $$\dfrac{f(b)-f(a)}{b-a} \geq 0\qquad
\mathrm{(respectively}, \dfrac{f(b)-f(a)}{b-a} > 0).$$ Similarly, a
function $g$ is (strictly) decreasing if for all $a<b$ for which it
is defined
$$\dfrac{g(b)-g(a)}{b-a} \leq 0 \qquad \mathrm{(respectively},
\dfrac{g(b)-g(a)}{b-a} < 0).$$
\end{thm}

\begin{df}
If there is a point $a$ for which $f(x) \leq f(M)$ for all $x$ in a
neighbourhood centred at  $x=M$ then we say that $f$ has a {\em
local maximum} at $x=M$. Similarly, if there is a point $m$ for
which $f(x) \geq f(m)$ for all $x$ in a neighbourhood centred at
$x=m$ then we say that $f$ has a {\em local minimum} at $x=m$. The
maxima and the minima of a function are called its {\em extrema}.
\end{df}

\subsection{Convexity}
\begin{df}
A function $f: \dom{f} \rightarrow \reals$ is {\em (mid-point)
convex} in $\dom{f}$ if for all  $a, b, $ in $\dom{f}$ and
$$ f\left(\dfrac{a+b}{2}\right) \leq  \dfrac{f(a) + f(b)}{2}.$$ Similarly, a function $g: \dom{g} \rightarrow \reals$ is
{\em concave} in $\dom{g}$ if for all  $a, b, $ in $\dom{g}$
$$g\left(\dfrac{a+b}{2}\right) \geq  \dfrac{g(a) + g(b)}{2}.$$


By the {\em intervals of convexity (concavity) of a function} we
mean the intervals where the function is convex (concave). An {\em
inflexion point}\index{inflexion point} is a point where a graph
changes convexity.\footnote{American books, bowing to {\em
Newspeak}, have created such terms as ``concave up'' and concave
down.'' Their goal is to reduce vocabulary, thereby stultifying
students even more. We will not follow such practice here.}
\label{df:convexity_concavity}
\index{convexity}\index{concavity}\index{function!concave}\index{function!convex}
\end{df}
Consider figure \ref{fig:convex_curve1}, where $$A=(a,f(a)),\qquad
B=(b,f(b)),\qquad
M=\left(\dfrac{a+b}{2},\dfrac{f(a)+f(b)}{2}\right), \qquad
L=\left(\dfrac{a+b}{2},f\left(\dfrac{a+b}{2}\right)\right).$$ The
definition of convexity has the following geometric interpretation:
for any line segment joining two points on the curve, the midpoint
of the line segment lines on top of the curve, and hence, the line
segment itself lies on top of the curve. See figure
\ref{fig:convex_curve1}. Observe that if $f$ is convex, then $-f$ is
concave, and so in a concave curve, any line segment joining two
points on the curve, lies below the curve. See figure
\ref{fig:concave_curve2}.


\vspace{3cm}
\begin{figure}[h]
\begin{minipage}{.4\textwidth}
\centering\psset{unit=1pc}
\psaxes[linewidth=2pt,arrows=->,labels=none,ticks=none](0,0)(-6,-6)(6,6)
\pscurve[linewidth=2pt,linecolor=brown](-4,4)(-3,2)(1,-1)(4,0)(5,1)(6,2)
\pstGeonode[PosAngle={210,-10,-70}](-3,2){A}(5,1){B}(1,1.5){M}(1,-1){L}
\psline[linewidth=2pt,linecolor=orange](A)(B) \meinecaption{3}{A
convex curve.}\label{fig:convex_curve1}
\end{minipage}
\hfill
\begin{minipage}{.4\textwidth}
\centering\psset{unit=1pc}
\psaxes[linewidth=2pt,arrows=->,labels=none,ticks=none](0,0)(-6,-6)(6,6)
\pscurve[linewidth=2pt,linecolor=brown](-4,-4)(-3,-2)(1,1)(4,0)(5,-1)(6,-2)
\pstGeonode[PosAngle={210,-10,70}](-3,-2){A'}(5,-1){B'}(1,-1.5){M'}(1,1){L'}
\psline[linewidth=2pt,linecolor=orange](A')(B') \meinecaption{3}{A
concave curve.}\label{fig:concave_curve2}
\end{minipage}

\end{figure}

\subsection*{Homework}\addcontentsline{toc}{subsection}{Homework}

\begin{pro}\label{pro:curve0}
Draw the graph of a function $f$ meeting the following conditions:
\begin{enumerate}
\item $f$ is continuous everywhere;
\item $f$ has zeroes at $x=-3$, $x=1$, and $x=2$
\item $f(-\infty)=f(+\infty)=-\infty$
\item $f(-1)=2$
\item $f(5)=-16$
\end{enumerate}

\begin{answer}
There are infinitely many graphs satisfying the specified
conditions. Figure \ref{fig:curve0} shews one.\\

\vspace{1cm}
\begin{figure}[h]
\centering\psset{unit=1pc}
\psaxes[arrows=->,linewidth=2pt,labelFontSize=\tiny](0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red,arrows=<->,algebraic]{-3.705190256}{4.176575236}{-(1/18)*(x+3)*(x-1)*(x-2)^2}
 \meinecaption{3}{Problem
\ref{pro:curve0}.}\label{fig:curve0}
\end{figure}
\end{answer}

\end{pro}


\begin{pro}\label{pro:curve1}
Draw the graph of a function $f$ meeting the following conditions:
\begin{enumerate}
\item $f$ is everywhere continuous;
\item $f(-2)=f(0)=f(2)=0$;
\item $f(-\infty)=0$, $f(+\infty)=+\infty$
\item $f(-4)=-1$, $f(-1)=1$ , $f(1)=-1$, $f(4)=1$
\end{enumerate}
\begin{answer}
There are infinitely many graphs satisfying the specified
conditions. Figure \ref{fig:curve1} shews one.\\

\vspace{1cm}
\begin{figure}[h]
\centering\psset{unit=1pc}
\psaxes[arrows=->,linewidth=2pt,labelFontSize=\tiny](0,0)(-6,-6)(6,6)
\pscurve[linewidth=2pt,linecolor=red,arrows=<->](-5.5,-.2)(-5,-.25)(-4.5,-.3)(-4.5,-.5)(-4,-1)(-2,0)(-1,1)(0,0)(1,-1)(2,0)(4,1)(5,2)(5.5,3)
 \meinecaption{3}{Problem
\ref{pro:curve1}.}\label{fig:curve1}
\end{figure}

\end{answer}


\end{pro}
\begin{pro}\label{pro:curve2}
Draw the graph of a function $f$ meeting the following conditions:
\begin{enumerate}
\item $f$ is continuous in $\reals \setminus \{0\}$;
\item $f$ has a pole at $x=0$, with $f(0-)=-\infty$ and
$f(0+)=+\infty$;
\item $f(-\infty)=-\infty$, $f(+\infty)=+\infty$
\end{enumerate}
\begin{answer}
There are infinitely many graphs satisfying the specified
conditions. Figure \ref{fig:curve2} shews one.\\

\vspace{1cm}
\begin{figure}[h]
\centering\psset{unit=1pc}
\psaxes[arrows=->,linewidth=2pt,labelFontSize=\tiny](0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red,arrows=<->,algebraic]{.1715728752}{5.828427125}{x+1/x}
\psplot[linewidth=2pt,linecolor=red,arrows=<->,algebraic]{-.1715728752}{-5.828427125}{x+1/x}
 \meinecaption{3}{Problem
\ref{pro:curve2}.}\label{fig:curve2}
\end{figure}

\end{answer}


\end{pro}
\begin{pro}
 Consider the functions $f$ and $g$ in the figure below. You
may assume that they are composed solely of straight line segments.
\begin{enumerate}
\item Draw $f+g$.
\item Draw $f\circ g$.
\item Draw $2f + g$.
\end{enumerate}
\end{pro}

\vspace{2cm}
\begin{figure}[h]
\centering
\begin{minipage}{.4\textwidth}
$$\psset{unit=.8pc} \psaxes[linewidth=2pt,
labelFontSize=\tiny,linewidth=2pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5)\psline[linewidth=2pt,
linecolor=red](-5,-2)(-4,-2)(-1,2)(1,0)(4,3)(5,3)
\psdots[dotscale=1,dotstyle=*](-5,-2)(-4,-2)(-1,2)(1,0)(4,3)(5,3)$$
\meinecaption{3}{$f$}
\end{minipage}
\hfill
\begin{minipage}{.4\textwidth}
$$\psset{unit=.8pc} \psaxes[linewidth=2pt,
labelFontSize=\tiny,linewidth=2pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5)
\psline[linewidth=2pt,
linecolor=red](-5,-2)(-4,2)(-1,2)(1,-2)(4,-2)(5,2)
\psdots[dotscale=1,dotstyle=*](-5,-2)(-4,2)(-1,2)(1,-2)(4,-2)(5,2)
$$
\meinecaption{3}{$g$}
\end{minipage}
\end{figure}
\section{Transformations of Graphs of Functions}
\subsection{Translations}
\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}



\begin{exa}
Figures \ref{fig:translation2} through \ref{fig:translation4} shew
various translations of the graph of a function $f$.
\end{exa}



\vspace*{1cm}
\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}{.22\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)
\psaxes[linewidth=2pt,linecolor=brown,labels=none]{->}(0,0)(-5.5,-5.5)(5.5,5.5)\figurinetrans
\meinecaption{2}{$y=f(x)$. }\label{fig:translation1}
\end{minipage}
\hfill
\begin{minipage}{.22\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)
\psaxes[linewidth=2pt,linecolor=brown,labels=none]{->}(0,0)(-5.5,-5.5)(5.5,5.5)\rput(0,1){\figurinetrans}
\meinecaption{2}{\mbox{$y=f(x)+1$}. }\label{fig:translation2}
\end{minipage}
\hfill
\begin{minipage}{.22\textwidth}\centering\psset{unit=.6pc}
\psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt,
gridlabels=0](0,0)(-5,-5)(5,5) \psdots[dotstyle=*,dotscale=1](0,0)
\psaxes[
Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)\psaxes[linewidth=2pt,linecolor=brown,labels=none]{->}(0,0)(-5.5,-5.5)(5.5,5.5)
\rput(-1,0){\figurinetrans} \meinecaption{2}{\mbox{$y=f(x+1)$}.
}\label{fig:translation3}
\end{minipage}
\hfill
\begin{minipage}{.22\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)
\psaxes[linewidth=2pt,linecolor=brown,labels=none]{->}(0,0)(-5.5,-5.5)(5.5,5.5)
\rput(-1,1){\figurinetrans} \meinecaption{2}{\mbox{ $y=f(x+1)+1$}.
}\label{fig:translation4}
\end{minipage}
\end{figure}





\subsection{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 $f$.
Figure \ref{fig:reflexions2} shews the graph of its reflexion $a$
about the $x$-axis, and  figure \ref{fig:reflexions3} shews the
graph of its reflexion $b$ about the $y$-axis.
\end{exa}

\vspace*{1cm}
\begin{figure}[h]
\begin{minipage}{.22\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}{.22\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}{.22\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}{.22\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}




\subsection{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}





\vspace{2cm}
\begin{figure}[!hptb]
\begin{minipage}{4cm}
\centering\psset{unit=.6pc}
\psgrid[subgriddiv=0,griddots=5,gridlabels=0](-7,-7)(7,7) \psaxes[
Ox=-7,Oy=-7,labelFontSize=\tiny](-7,-7)(-7,-7)(7,7)\psaxes[linewidth=2pt,linecolor=brown,labels=none]{->}(0,0)(-7.5,-7.5)(7.5,7.5)
\pstGeonode[PointName=none](0,0){O}(2,4){A}(4,4){B}(-2,6){Y}(-4,-6){X}
\psline[linecolor=orange,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](-7,-7)(7,7)
 \psaxes[
Ox=-7,Oy=-7,labelFontSize=\tiny](-7,-7)(-7,-7)(7,7)\psaxes[linewidth=2pt,linecolor=brown,labels=none]{->}(0,0)(-7.5,-7.5)(7.5,7.5)
\pstGeonode[PointName=none](0,0){O}(2,2){A}(4,2){B}(-2,3){Y}(-4,-3){X}
\psline[linecolor=orange,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](-7,-7)(7,7)
 \psaxes[
Ox=-7,Oy=-7,labelFontSize=\tiny](-7,-7)(-7,-7)(7,7)\psaxes[linewidth=2pt,linecolor=brown,labels=none]{->}(0,0)(-7.5,-7.5)(7.5,7.5)
\pstGeonode[PointName=none](0,0){O}(1,4){A}(2,4){B}(-1,6){Y}(-2,-6){X}
\psline[linecolor=orange,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](-7,-7)(7,7)  \psaxes[
Ox=-7,Oy=-7,labelFontSize=\tiny](-7,-7)(-7,-7)(7,7)\psaxes[linewidth=2pt,linecolor=brown,labels=none]{->}(0,0)(-7.5,-7.5)(7.5,7.5)
\pstGeonode[PointName=none](0,0){O}(1,2){A}(2,2){B}(-1,3){Y}(-2,-3){X}
\psline[linecolor=orange,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}\label{exa:distorsiones} Consider the function
$f$ whose graph appears in figure \ref{fig:distorsiones1}.




In figure \ref{fig:distorsiones2}.
 we have the graph of the function $a$ with $a(x)=\dfrac{f(x)}{2}$.


In figure \ref{fig:distorsiones3} we have the graph of the function
$b$ with  $b(x)=f(2x)$.




In figure  \ref{fig:distorsiones4} we have the graph of the function
$c$ with $c(x)=\dfrac{f(2x)}{2}$.
\end{exa}

\subsection*{Homework}\addcontentsline{toc}{subsection}{Homework}


\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
$y=-f(1-x)$.

\vspace{2cm}
\begin{figure}[h]
\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{3}{Problem \ref{pro:reflexion}.}
\label{fig:reflexionA}
\end{figure}


\begin{answer}
Here is the graph of $x\mapsto f(x+1)$.\\ \vspace{3cm}

\begin{figure}[h]
$$\psset{unit=1pc}
\psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt,
gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[
Ox=-5,Oy=-5](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0)
\psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0)
\psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5)
\psline[linewidth=2pt,linecolor=blue](-5,2)(-3,-2)(-1,0)(0,2)(2,2)(4,-2)
\psdots(-1,0)(0,2)(2,2)(4,-2)(3,0)(-3,-2)(-5,2)(-4,0)
$$
\meinecaption{2}{$y=f(x+1)$.}
\end{figure}

Here is the graph of $x\mapsto f(-x+1)$.\\ \vspace{3cm}

\begin{figure}[h]
$$\psset{unit=1pc}
\psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt,
gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[
Ox=-5,Oy=-5](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0)
\psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0)
\psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5)
\psline[linewidth=2pt,linecolor=blue](5,2)(3,-2)(1,0)(0,2)(-2,2)(-4,-2)
\psdots(1,0)(0,2)(-2,2)(-4,-2)(-3,0)(3,-2)(5,2)(4,0)
$$
\meinecaption{2}{$y=f(1-x)$.}
\end{figure}


Here is the graph of $x\mapsto -f(-x+1)$.\\ \vspace{3cm}

\begin{figure}[h]
$$\psset{unit=1pc}
\psgrid[gridwidth=1.2pt, subgriddiv=0,linewidth=.8pt,
gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[
Ox=-5,Oy=-5](-5,-5)(-5,-5)(5,5)\psdots[dotstyle=*,dotscale=1](0,0)
\psline[linewidth=2pt,linecolor=brown]{<->}(-5.5,0)(5.5,0)
\psline[linewidth=2pt,linecolor=brown]{<->}(0,-5.5)(0,5.5)
\psline[linewidth=2pt,linecolor=blue](5,-2)(3,2)(1,0)(0,-2)(-2,-2)(-4,2)
\psdots(1,0)(0,-2)(-2,-2)(-4,2)(-3,0)(3,2)(5,-2)(4,0)
$$
\meinecaption{2}{$y=-f(1-x)$.}

\end{figure}

\end{answer}

\end{pro}


\chapter{Introduction to the Infinitesimal Calculus}\thispagestyle{empty}

\section{Continuity}
\begin{df}
A function $f$ is {\em continuous} at $x=a$ if as
$\tends{\varepsilon}{0}$, $$f(a+\varepsilon) =f(a)+\soo{1}.
$$
\end{df}
This means, essentially, that if $x\to a$, then $f(x)\to f(a)$.
\begin{exa}\label{exa:const_is_cont}Let $K\in \reals$. Demonstrate that the constant
function $$ \fun{f}{x}{K}{\reals}{\reals} $$ is continuous.
\end{exa}
\begin{solu}
Let $a\in \reals$ be fixed. Then
$$f(a+\varepsilon) = K = K+0 = K+\soo{1}=f(a)+\soo{1}, $$ proving
that a constant function is continuous.
\end{solu}

\begin{exa}\label{exa:id_is_cont} Demonstrate that the identity
function $$ \fun{{\bf Id}}{x}{x}{\reals}{\reals} $$ is continuous.
\end{exa}
\begin{solu}
Let $a\in \reals$ be fixed. Then
$${\bf Id}(a+\varepsilon) = a+\varepsilon . $$ Since
$\tends{\varepsilon}{0}$ means the same as $\varepsilon =\soo{1}$,
we have proved that
$${\bf Id}(a+\varepsilon) = a+\varepsilon = {\bf Id}(a)+\soo{1},
$$and so the identity function is continuous.
\end{solu}
\begin{exa}\label{exa:sq_is_cont} Demonstrate that the square
function $$ \fun{{\bf Sq}}{x}{x^2}{\reals}{\reals} $$ is continuous.
\end{exa}
\begin{solu}
Let $a\in \reals$ be fixed. Then
$${\bf Sq}(a+\varepsilon) = a^2+2a\varepsilon +\varepsilon ^2. $$
Since $a$ is fixed,  $\tends{\varepsilon}{0}$ implies that
$2a\varepsilon =\soo{1}$. Also, $\varepsilon ^2=\soo{1}$, and so we
have proved that
$${\bf Sq}(a+\varepsilon) = a^2+2a\varepsilon +\varepsilon^2= {\bf Sq}(a)+\soo{1} +\soo{1}={\bf Sq}(a)+\soo{1},
$$and so the square function is continuous.
\end{solu}
\begin{exa}
If  $$f(x) = \left\{\begin{array}{ll}\sqrt{x+2} + b & \mathrm{if} \ -2\leq x<0 \\
a + 2x & \mathrm{if} \ 0 \leq x<1 \\ 1-2x^2& \mathrm{if} \ x \geq 1
\end{array} \right.$$ is continuous in its domain of definition, determine the
value of $(a, b)$.
\end{exa}
\begin{solu}
 The conditions stipulate that $$ f(0-)=f(0+)\implies \sqrt{2}+b=a, \qquad f(1-)=f(1+)\implies a+2=-1,
$$whence $a=-3$ and $b=-3-\sqrt{2}$.
\end{solu}


\begin{thm}\label{thm:algebra_of_cont}
Let $f$ and $g$ be continuous at $x=a$. Then $f+g$ and $fg$ are also
continuous at $x=a$.
\end{thm}
\begin{pf}
We are given that as $\tends{\varepsilon}{0}$,
$$f(a+\varepsilon) =f(a)+\soo{1}, \qquad g(a+\varepsilon) =g(a)+\soo{1}.  $$
Then $$ (f+g)(a+\varepsilon)=f(a+\varepsilon)+g(a+\varepsilon)=
f(a)+\soo{1}+g(a)+\soo{1}=f(a)+g(a)+\soo{1}=(f+g)(a)+\soo{1},
$$proving that $f+g$ is continuous at $x=a$.

\bigskip

Similarly,
$$\begin{array}{lll}
(fg)(a+\varepsilon)& = & f(a+\varepsilon)\cdot g(a+\varepsilon)\\
& = &  (f(a)+\soo{1})\cdot
(g(a)+\soo{1})\\
& = & f(a)g(a)+\soo{f(a)}+\soo{g(a)}+\soo{1}\\
& = &(fg)(a)+\soo{1},
\end{array}$$since $f(a)$ and $g(a)$ are constants. This proves that  $fg$ is continuous at $x=a$.
\end{pf}

\begin{thm}[Law of Inertia]\label{thm:law_of_inertia}
If $g$ is continuous at $x=a$ and $g(a)\neq 0$, then there exists a
neighbourhood of $x=a$ where $g$ keeps the same sign as $g(a)$.
\end{thm}
\begin{pf}
We are given that $$ g(a+\varepsilon) = g(a)+\soo{1} .$$The quantity
$\soo{1}$ will eventually (as $\tends{\varepsilon}{0}$) be smaller
in absolute value than, say $\dfrac{|g(a)|}{2}$. But then this means
that $$g(a+\varepsilon)\approx g(a)+\dfrac{|g(a)|}{2} \implies
 g(a)-\dfrac{g(a)}{2} <g(a+\varepsilon)< g(a)+\dfrac{g(a)}{2}\implies \dfrac{g(a)}{2} <g(a+\varepsilon)< \dfrac{3g(a)}{2},$$and so
$g(a+\varepsilon)$ keeps the same sign as $g(a)$ for a sufficiently
small  neighbourhood of $a$.
 \end{pf}

\begin{lem}As $\varepsilon \to 0$, $$  \dfrac{1}{1+\soo{1}}=1+\soo{1}.
$$\label{lem:rec_1+soo}
\end{lem}
\begin{pf}
For if $|t|<1$  then$$ 1-t ^2 < 1 \implies (1+t)(1-t)<1 \implies 1-t
< \dfrac{1}{1+t}. $$ Now, if $0<t < 1$ then $1<1+t $ and so
$\dfrac{1}{1+t}<1$, obtaining
$$ 1-t <
\dfrac{1}{1+t}<1, \qquad 0<t < 1, $$ which gives the lemma in this
case.

\bigskip
Let us assume now that $-\dfrac{1}{2}<t <0$. We claim that
$$
\dfrac{1}{1+t}<1-t+2t^2. $$To prove this, observe that
$$-\dfrac{1}{2}<t\implies 1+2t>0\implies 1+t>-t>0,$$  and so
$$ \begin{array}{lll}0<t^2(1+2t) &  \implies &
1<1+t^2+2t^3\\
& \implies &  1<1-t^2+2t^2+2t^3 \\ & \implies &  1<(1+t)(1-t)+(1+t)2t^2\\
& \implies & \dfrac{1}{1+t}< 1-t+2t^2,
\end{array}$$as was to be shewn. Hence, for $-\dfrac{1}{2}<t <0$, we
have
$$1-t< \dfrac{1}{1+t}< 1-t+2t^2, $$
which gives the lemma in the remaining case.
\end{pf}

\begin{thm}\label{thm:reci_cont}
 If $g$ is continuous at $x=a$ and  $g(a)\neq 0$, then $\dfrac{1}{g}$ is also
continuous at $x=a$.
\end{thm}
\begin{pf}
By Theorem \ref{thm:law_of_inertia}, for sufficiently $\varepsilon$
close   $0$, $  g(a+\varepsilon)\neq 0$. Hence
$$\begin{array}{lll}
\left(\dfrac{1}{g}\right)(a+\varepsilon)& = &
\dfrac{1}{g(a+\varepsilon)}\\ & = & \dfrac{1}{g(a)+\soo{1}}\\
& = & \dfrac{1}{g(a)}\cdot \dfrac{1}{1+\soo{\dfrac{1}{g(a)}}}\\
& = & \dfrac{1}{g(a)}\cdot \dfrac{1}{1+\soo{1}}\\
& = & \dfrac{1}{g(a)}\cdot \left(1+\soo{1}\right)\\
& = & \dfrac{1}{g(a)}+\soo{\dfrac{1}{g(a)}}\\
& = & \dfrac{1}{g(a)}+\soo{1},\\
\end{array}$$by Lemma \ref{lem:rec_1+soo}.
 This proves that  $\dfrac{1}{g}$ is continuous at $x=a$.

\end{pf}
\begin{cor}
If the functions $f$ and $g$ are continuous at $x=a$ and if
$g(a)\neq 0$, then $\dfrac{f}{g}$ is continuous at $x=a$.
\end{cor}
\begin{pf}
Immediate from theorems \ref{thm:algebra_of_cont} and
\ref{thm:reci_cont}.
\end{pf}
\begin{cor}[Polynomials are continuous functions]
Let $p:\reals\to \reals$ be a polynomial with real coefficients.
Then $p$ is continuous at every $a\in
\reals$.\label{cor:poly_is_cunt}
\end{cor}
\begin{pf}
The proof is by induction on the degree on the degree of $p$. If
$\deg p =0$, then $p$ is a constant function, which is continuous by
example \ref{exa:const_is_cont}.


\bigskip

If $\deg p =1$ then $p(x)=ax+b$, for some constants $a, b$. Since
the identity function is continuous by example \ref{exa:id_is_cont},
the function $x\mapsto ax$ is continuous being the product of two
continuous functions $x\mapsto x$ and $x \mapsto a$. We then see
that  $p$ is the sum of the continuous functions $x\mapsto ax$ and
$x\mapsto b$.


\bigskip

Assume now that any polynomial of degree $n-1$ is continuous at
$x=a$. Then if
$$p(x) =a_nx^n+a_{n-1}x^{n-1}+\cdots
+a_1x+a_0=x(a_{n}x^{n-1}+a_{n-1}x^{n-2}+\cdots  +a_1)+a_0
:=xq(x)+a_0,
$$where $\deg q(x)=n-1$. By the induction hypothesis, $q(x)$ is
continuous at $x=a$. By example \ref{exa:id_is_cont}, the identity
functions is continuous, and so $x\mapsto xq(x)$ is the product of
two continuous functions, and hence continuous. Since the constant
function $x\mapsto a_0$ is continuous, $p$ is then the sum of two
continuous functions and hence continuous.

\end{pf}

\begin{cor}[Rational Functions are Continuous] Let $p(x)$ and $q(x)$
be two polynomials with real coefficients. Let $Z=\{x\in \reals:
q(x)=0\}$. Then the rational function
$$\fun{r}{x}{\dfrac{p(x)}{q(x)}}{\reals \setminus Z}{\reals}  $$
is continuous.\label{cor:ratio_is_cunt}
\end{cor}
\begin{pf}This is immediate from Corollary \ref{cor:poly_is_cunt}
and Theorem \ref{thm:reci_cont}.
\end{pf}


\vspace{3cm}
\begin{figure}[h]
\begin{minipage}{.4\textwidth}
\centering\psset{unit=1pc}
\psaxes[labels=none,ticks=none,linewidth=2pt,arrows=->](0,0)(-7,-7)(7,7)
\pscurve[linewidth=2pt,linecolor=orange,arrows=*-*](1,1)(2,2.1)(3,1.5)(4,.5)(5,5)
\psline[linestyle=dotted](1,1)(0,1)\uput[l](0,1){$f(a)$}
\psline[linestyle=dotted](5,5)(0,5)\uput[l](0,5){$f(b)$}
\psline[linestyle=dotted](2,2.1)(0,2.1)\uput[l](0,2.1){$t$}
\psline[linestyle=dotted](2,2.1)(2,0)\uput[d](2,0){$s$}
\psline[linestyle=dotted](1,1)(1,0)\uput[d](1,0){$a$}
\psline[linestyle=dotted](5,5)(5,0)\uput[d](5,0){$b$}
\meinecaption{3}{Intermediate Value
Theorem.}\label{fig:interm_val_thm1}
\end{minipage}
\hfill
\begin{minipage}{.4\textwidth}
\centering\psset{unit=1pc}
\psaxes[labels=none,ticks=none,linewidth=2pt,arrows=->](0,0)(-7,-7)(7,7)
\psplot[linewidth=2pt,linecolor=orange,arrows=*-*,algebraic,plotpoints=1001]{1}{5}{x
+ cos(sin(x))}
\psline[linestyle=dotted](1,1.6663667454)(0,1.6663667454)\uput[l](0,1.6663667454){$f(a)$}
\psline[linestyle=dotted](5,5.5744008792)(0,5.5744008792)\uput[l](0,5.5744008792){$f(b)$}
\psline[linestyle=dotted](3,3.9900590858)(0,3.9900590858)\uput[l](0,3.9900590858){$t$}
\psline[linestyle=dotted](3,3.9900590858)(3,0)\uput[d](3,0){$s$}
\psline[linestyle=dotted](1,1.6663667454)(1,0)\uput[d](1,0){$a$}
\psline[linestyle=dotted](5,5.5744008792)(5,0)\uput[d](5,0){$b$}
\meinecaption{3}{Intermediate Value Theorem.}
\label{fig:interm_val_thm2}
\end{minipage}
\end{figure}

\begin{thm}
Let $g$ be continuous at $x=a$ and $f$ continuous at $x=g(a)$. Then
$f\circ g$ is continuous at $x=a$.
\end{thm}
\begin{pf}We are given that as $\varepsilon \to 0$, then
$$f(g(a)+\varepsilon)=f(g(a))+\soo{1}, \qquad g(a+\varepsilon) = g(a)+\soo{1}.   $$
We must shew that
$$(f\circ g)(a+\varepsilon) = (f\circ g)(a)+\soo{1}.  $$
We have
$$(f\circ g)(a+\varepsilon) = f(g(a+\varepsilon))=f(g(a)+\soo{1})=f(g(a))+\soo{1}=(f\circ g)(a)+\soo{1},  $$
as was to be shewn.
\end{pf}

\begin{df}
Let $f:\dom{f}\to \reals$ and assume $\oo{a}{b}\subseteq \dom{f}$.
We say that $f$ is {\em continuous} in $\oo{a}{b}$ if $f$ is
continuous at every point $x\in\oo{a}{b}$.
\end{df}

\begin{df}
Let $f:\dom{f}\to \reals$ and assume $\cc{a}{b}\subseteq \dom{f}$.
We say that $f$ is continuous in $\cc{a}{b}$ if $f$ is continuous at
every point $x\in\oo{a}{b}$ and if $f(a+)=f(a)$ and $f(b-)=f(b)$.
\end{df}


\begin{thm}[Darboux's Intermediate Value
Theorem]\label{thm:intm_val} Let $f:\cc{a}{b}\to \reals$ be
continuous in $\cc{a}{b}$. Let $m=\min (f(a), f(b))$ and $M=\max
(f(a), f(b))$. Then, for every $t\in \cc{m}{M}$, there exists $s\in
\cc{a}{b}$ such that $f(s)=t$.
\end{thm}
\begin{pf}
We shall not give a rigorous proof of this assertion, but refer the
reader to figures  \ref{fig:interm_val_thm1} and
\ref{fig:interm_val_thm2}. Since there are no breaks in the graph,
the graph will not ``jump over'' the horizontal line $y=t$.
\end{pf}

\begin{cor} A continuous function defined on an
interval maps that interval into an interval.
\end{cor}
\begin{pf}
This follows at once from the Intermediate Value Theorem and the
definition of an interval.
\end{pf}
\begin{thm}[Bolzano's Theorem]If $f:\cc{u}{v}\rightarrow \reals$ is
continuous and $f(u)f(v)<0$, then there is a $w\in \oo{u}{v}$ such
that $f(w)=0$.
\end{thm}
\begin{pf}
This follows at once from the Intermediate Value Theorem by putting
$a=\min (f(u), f(v))<0$ and $b=\max (f(u), f(v))>0$ .
\end{pf}

\begin{cor}
Every polynomial of odd degree and real coefficients has at least
one real root.
\end{cor}
\begin{pf}
Since $p(-\infty)$ and $p(+\infty)$ have opposite sign, this follows
by the Intermediate Value Theorem.
\end{pf}
\begin{exa}
Demonstrate that the polynomial $p(x)=x^5-x-1$ has a root $\alpha $
in $\cc{1}{2}$. Further, find an interval of length $\dfrac{1}{10}$
or smaller that contains this root.
\end{exa}
\begin{solu}Observe that $p(1) = -1<0$ and $p(2) = 29>0$, and so, since the polynomial changes sign in $\cc{1}{2}$,  by
Bolzano's Theorem, the polynomial has a root $\alpha \in\cc{1}{2}$.
In fact, we find
$$p(1.1)\approx -0.5,\qquad  p(1.2) \approx 0.3,  $$and so
$\alpha \in\cc{1.1}{1.2}$, which is an interval of length
$\dfrac{1}{10}$ containing the zero. Still
$$ p(1.15) \approx -.14,\qquad  p(1.17) \approx 0.02,  $$meaning that  $\alpha \in\cc{1.15}{1.17}$.
This interval has length $1.17-1.15=0.02=\dfrac{1}{50}$, so we have
done better than the required $\dfrac{1}{10}$.

\end{solu}
\begin{thm}[Weierstrass Theorem]A continuous function $f:\cc{a}{b}\rightarrow
\reals$ attains a maximum and a minimum on
$\cc{a}{b}$.\label{thm:weierstrass-max-min}
\end{thm}
\begin{pf}
By Darboux's Theorem and by Bolzano's Theorem,
$f\left(\cc{a}{b}\right)$ is a closed finite interval. The minimum
sought is the left endpoint of this interval and the maximum sought
is the right endpoint of this interval.
\end{pf}

\subsection*{Homework}\addcontentsline{toc}{subsection}{Homework}
\begin{pro}Given that the equation $x^7-x^6-2x^4+x^3+1=0$ has
exactly three different real roots, find intervals, of length $1$ or
shorter, containing each root.
\begin{answer}
Let $\alpha< \beta<\gamma$ be the roots. Then
$$p(-1)=-4, \quad p(0) =1, \quad p(1) =0, \qquad p\left(\dfrac{3}{2}\right)=-\dfrac{7}{128}, \quad p(2) =41.  $$
Hence $\alpha \in \oo{-1}{0}$, $\beta =1$, $\gamma \in
\oo{\dfrac{3}{2}}{2}$.
\end{answer}
\end{pro}
\begin{pro}
Let $p(x), q(x)$ be polynomials with real coefficients such that $$
p(x^2+x+1) = p(x)q(x).$$Prove that $p$ must have even degree.
\begin{answer}
If $p$ had odd degree, then, by the Intermediate Value Theorem it
would have a real root. Let $\alpha$ be its largest real root. Then
$$
0=p(\alpha )q(\alpha )=p(\alpha ^2+\alpha+1)$$meaning that $\alpha
^2+\alpha+1>\alpha$ is a real root larger than the supposedly
largest real root $\alpha$, a contradiction.
\end{answer}
\end{pro}
\begin{pro}
A function $f$ defined over all real numbers is continuous and for
all real $x$ satisfies $$\left(f(x)\right) \cdot \left((f\circ
f)(x)\right) =1. $$Given that $f(1000)=999$, find $f(500)$.
\begin{answer}
Observe that $f(1000)f(f(1000)) = 1 \implies f(999) =
\dfrac{1}{999}$. So the range of $f$ include all numbers from
$\dfrac{1}{999}$ to $999$. By the intermediate value theorem, there
is a real number $a$ such that $f(a) = 500$. Thus
$$f(a)f(f(a)) = 1 \implies f(500) = \dfrac{1}{500}.  $$
\end{answer}
\end{pro}

\begin{pro}Suppose that $f:\cc{0}{1} \rightarrow \cc{0}{1}$ is continuous.
Prove that there is a number $c$ in $\cc{0}{1}$ such that $f(c) = 1
- c$.
\begin{answer}
If either $f(0)=1$ or $f(1)=0$, we are done. So assume that $0\geq
f(0)<1$ and $0<f(1)\leq 1$. Put $g(x)=f(x)+x-1$. Then $g(0) =
f(0)-1<0$ and $g(1)=f(1)>0$. By Bolzano's Theorem there is a $c\in
\oo{0}{1}$ such that $g(c)=0$, that is, $f(c)+c-1=0$, as required.
\end{answer}

 \end{pro}
\begin{pro}[Universal Chord Theorem]\label{pro:universal-chord-thm} Suppose that $f$ is a continuous
function of $\cc{0}{1}$ and that $f(0) = f(1)$. Let
$n$ be a strictly positive integer. Prove that there is some number $x \in \cc{0}{1}$ such that $f(x) = f(x + 1/n).$ \\

\begin{answer}
Consider $g(x) = f(x) - f(x + 1/n)$, which is clearly continuous. If
$g$ is never $0$ in $\cc{0}{1}$ then by the Law of Inertia $g$ must
be either strictly positive or strictly negative. But then
$$  0=f(0)-f(1)= \left(f(0)-f\left(\dfrac{1}{n}\right)\right) + \left(f\left(\dfrac{1}{n}\right)-f\left(\dfrac{2}{n}\right)\right)
+
\left(f\left(\dfrac{2}{n}\right)-f\left(\dfrac{3}{n}\right)\right)+
\cdots +
\left(f\left(\dfrac{n-1}{n}\right)-f\left(\dfrac{n}{n}\right)\right).
$$The sum of each parenthesis on the right is strictly positive or
strictly negative and hence never $0$, a contradiction.
\end{answer}


\end{pro}

\section{Graphical Differentiation}
In this section we will take a very informal approach to
differentiation. A more formal approach will be given in the next
section.

\begin{df}
Let $f$ be a continuous function and let $A(a, f(a))$ be a point on
the graph of the function. We say that $f$ is {\em smooth at $A$} if
upon imagining a particle travelling at some steady speed along the
curve, then the particle does not experience an abrupt change of
direction.
\end{df}

\vspace{2cm}
\begin{figure}[h]
\begin{minipage}{.22\textwidth}\centering
\psset{unit=1pc} \psplot[linewidth=2pt,algebraic]{-2}{2.2}{2^x}
\psdots[dotscale=1.1](0,1)\uput[d](0,1){$A$} \meinecaption{1}{Smooth
curve at $A$.}\label{fig:smooth_curve}
\end{minipage}
\hfill
\begin{minipage}{.22\textwidth}\centering
\psset{unit=1pc} \psplot[linewidth=2pt,algebraic]{-4}{4}{abs(x)}
\psdots[dotscale=1.1](0,0)\uput[d](0,0){$A$} \meinecaption{1}{Corner
at $A$.}\label{fig:corner_curve}
\end{minipage}
\hfill
\begin{minipage}{.22\textwidth}\centering
\psset{unit=1pc} \psplot[linewidth=2pt,algebraic]{0}{4}{x^(2/3)}
 \psplot[linewidth=2pt,algebraic]{0}{-4}{abs(x)^(2/3)}
\psdots[dotscale=1.1](0,0)\uput[d](0,0){$A$} \meinecaption{1}{Cusp
at $A$.}\label{fig:cusp_curve}
\end{minipage}
\hfill
\begin{minipage}{.22\textwidth}\centering
\psset{unit=1pc}\psplotTangent[linewidth=2pt,algebraic,linecolor=red,arrows=<->]{0}{4.2}{2^x}
 \psplot[linewidth=2pt,algebraic]{-4}{2.2}{2^x}
\psdots[dotscale=1.1](0,1)\uput[d](0,1){$A$}
\meinecaption{1}{Tangent line at
$x=a$.}\label{fig:tangent_line_at_a}
\end{minipage}

\end{figure}


\begin{exa}
Figure \ref{fig:smooth_curve} gives an example of a smooth curve.
The curve in figure \ref{fig:corner_curve} has a ``sharp corner'' at
$A$ and it is not smooth. The curve in figure \ref{fig:cusp_curve}
is a ``cusp.''
\end{exa}
\begin{df}If $f$ is an affine function, we define the {\em tangent
line} at any point $A(a, f(a))$ to be the line that is the graph of
$f$. Otherwise, if $f$ is a smooth function at the point $A(a,
f(a))$, then {\em tangent line at $A$} is the unique line with the
following properties:
\begin{enumerate}
\item for a sufficiently small neighbourhood of $A$, the line just
touches the curve at $A$.
\item for this sufficiently small neighbourhood of $A$, the portion
of the curve inside the neighbourhood is on only one side of the
line.
\end{enumerate}
The {\em derivative} of $f$ at $x=a$, denoted by $f'(a)$, is the
slope of the tangent line at $x=a$.\end{df}


\begin{exa}
Figure \ref{fig:tangent_line_at_a} gives an example of the tangent
line to a curve.
\end{exa}
\vspace{3cm}
\begin{figure}[h]
\begin{minipage}{.4\textwidth}\centering
\psset{unit=1.5pc} \psgrid[gridwidth=1.2pt,
subgriddiv=0,linewidth=.5pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[
Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)
 \psaxes[linewidth=2pt,arrows={->},labels=none](0,0)(-5.5,-5.5)(5.5,5.5)\psplot[linewidth=2pt,algebraic,linecolor=red]{-2.2}{2.2}{x^2}
\meinecaption{4}{Example
\ref{exa:graphical_differentiation3}.}\label{fig:graphical_differentiation3-1}
\end{minipage}
\hfill
\begin{minipage}{.4\textwidth}\centering
\psset{unit=1.5pc} \psgrid[gridwidth=1.2pt,
subgriddiv=0,linewidth=.5pt, gridlabels=0](0,0)(-5,-5)(5,5) \psaxes[
Ox=-5,Oy=-5,labelFontSize=\tiny](-5,-5)(-5,-5)(5,5)
 \psaxes[linewidth=2pt,arrows={->},labels=none](0,0)(-5.5,-5.5)(5.5,5.5)\psplot[linewidth=2pt,algebraic,linecolor=red]{-2.2}{2.2}{x^2}
 \psplotTangent[linewidth=1.2pt,algebraic,linecolor=blue,arrows=*-*]{1}{2.2}{x^2}
\psdots[dotscale=1.2](1,1) \meinecaption{4}{Example
\ref{exa:graphical_differentiation3}.}\label{fig:graphical_differentiation3-2}
\end{minipage}
\end{figure}



\begin{exa}\label{exa:graphical_differentiation3}
For the curve in figure, find, approximately, the value of $f'(1) $.
\end{exa}
\begin{solu}
We draw a line through the point $(1,f(1))$ that just grazes the
curve, as in figure \ref{fig:graphical_differentiation3-2}. We
compute the slope of this line:
$$ \dfrac{3-(-1)}{2-0}=2. $$Hence we find $f'(1)=2$.

\end{solu}


\begin{df}
If a function $f$ is differentiable at every point $x$ of its
domain, then the {\em derivative} function of $f$, denoted by $f'$,
is the function with assignment rule $x\mapsto f'(x)$.
\end{df}



Given the graph of a smooth curve, we can approximately obtain the
graph of its derivative by taking the following steps.
\begin{enumerate}
\item We  divide up the domain of $f$ into intervals of the same
length.
\item For each endpoint $x$ of an interval above, we look at the
point $(x,f(x))$ on the graph of $f$.
\item We place a ruler so that it is tangent to the curve at $(x,
f(x))$.
\item We find the slope of the ruler. Recall that any two points on
the tangent line (the ruler) can be used  to find the slope.
\item We tabulate the slopes obtained and we plot these values,
obtaining thereby an approximate graph of $f'$.
\end{enumerate}
To further aid our graphing of the derivative, we make the following
observations.

\begin{enumerate}
\item If the function increases, then the slope of the tangent is
positive.
\item If the function decreases, then the slope of the tangent is
negative.
\item If the curve is convex, then the slope of the tangent
increases.
\item If the curve is concave, then the slope of the tangent
decreases.
\end{enumerate}
See figures \ref{fig:increasing_convex} through
\ref{fig:decreasing_concave} for several examples.   \vspace{4cm}
\begin{figure}[h]
\begin{minipage}{.4\textwidth}\centering
\psset{unit=1.5pc, plotpoints=1001}
\psplot[linewidth=2pt,algebraic,linecolor=red]{-4}{2.5}{2^x}
\psplotTangent[linewidth=.3pt,algebraic,linecolor=blue,arrows=*-*]{-4}{2}{2^x}
\psplotTangent[linewidth=.3pt,algebraic,linecolor=magenta,arrows=*-*]{-2}{2}{2^x}
\psplotTangent[linewidth=.3pt,algebraic,linecolor=green,arrows=*-*]{0}{2}{2^x}
\psplotTangent[linewidth=.3pt,algebraic,linecolor=brown,arrows=*-*]{2}{2}{2^x}
\meinecaption{1}{Increasing and
convex.}\label{fig:increasing_convex}
\end{minipage}
\hfill
\begin{minipage}{.4\textwidth}\centering
\psset{unit=1.5pc, plotpoints=1001}
\psplot[linewidth=2pt,algebraic,linecolor=red]{4}{-2.5}{1/2^x}
\psplotTangent[linewidth=.3pt,algebraic,linecolor=blue,arrows=*-*]{4}{2}{1/2^x}
\psplotTangent[linewidth=.3pt,algebraic,linecolor=magenta,arrows=*-*]{2}{2}{1/2^x}
\psplotTangent[linewidth=.3pt,algebraic,linecolor=green,arrows=*-*]{0}{2}{1/2^x}
\psplotTangent[linewidth=.3pt,algebraic,linecolor=brown,arrows=*-*]{-2}{2}{1/2^x}
\meinecaption{1}{Decreasing and
convex.}\label{fig:decreasing_convex}
\end{minipage}
\end{figure}

\vspace{4cm}
\begin{figure}[h]
\begin{minipage}{.4\textwidth}\centering
\psset{unit=1.5pc, plotpoints=1001}
\rput(0,5){\psplot[linewidth=2pt,algebraic,linecolor=red]{4}{-2.5}{-1/2^x}
\psplotTangent[linewidth=.3pt,algebraic,linecolor=blue,arrows=*-*]{4}{2}{-1/2^x}
\psplotTangent[linewidth=.3pt,algebraic,linecolor=magenta,arrows=*-*]{2}{2}{-1/2^x}
\psplotTangent[linewidth=.3pt,algebraic,linecolor=green,arrows=*-*]{0}{2}{-1/2^x}
\psplotTangent[linewidth=.3pt,algebraic,linecolor=brown,arrows=*-*]{-2}{2}{-1/2^x}}
\meinecaption{1}{Increasing and
concave.}\label{fig:increasing_concave}
\end{minipage}
\hfill
\begin{minipage}{.4\textwidth}\centering
\psset{unit=1.5pc, plotpoints=1001}
\rput(0,5){\psplot[linewidth=2pt,algebraic,linecolor=red]{-4}{2.5}{-2^x}
\psplotTangent[linewidth=.3pt,algebraic,linecolor=blue,arrows=*-*]{-4}{2}{-2^x}
\psplotTangent[linewidth=.3pt,algebraic,linecolor=magenta,arrows=*-*]{2}{2}{-2^x}
\psplotTangent[linewidth=.3pt,algebraic,linecolor=green,arrows=*-*]{0}{2}{-2^x}
\psplotTangent[linewidth=.3pt,algebraic,linecolor=brown,arrows=*-*]{-2}{2}{-2^x}}
\meinecaption{1}{Decreasing and
concave.}\label{fig:decreasing_concave}
\end{minipage}


\end{figure}





\begin{exa}\label{exa:graphical_differentiation2}
Find an approximate graph for the derivative of $f$ given in figure
\ref{fig:graphical_differentiation2}.
\end{exa}
\begin{solu} Observe that from the remarks following figure
\ref{fig:decreasing_convex}, we expect $f'$ to be positive in
$[-1.4;-0.6]$, since $f$ increases there. We expect $f'$ to be $0$
at $x=-0.6$, since $f$ appears to have a (local) maximum there. We
expect $f'$ to be negative in $[-0.6;0.6]$ since $f$ decreases
there. We expect  $f'$ to be $0$ at $x=0.6$, since $f$ appears to
have a (local) minimum there. Finally we expect $f'$ to be positive
for $[0.6;1.4]$ since $f$ is increasing there.



In our case we obtain the following (approximate) values for
$f'(x)$.

\bigskip
$${\small
\begin{array}{|c|lllllllllllllll|}
\hline x &  -1.4 & -1.2 & -1 & -0.8 & -0.6 & -0.4 & -0.2 & 0 & 0.2 &
0.4 & 0.6 &  0.8 & 1 & 1.2 & 1.4 \\
\hline f'(x) &4.88  & 3.32&2 &0.92 & 0.08& -0.52&-0.88 & -1 &-0.88 &-0.52 & 0.08& 0.92&2 & 3.32 & 4.88 \\
\hline
\end{array}
}
$$

\bigskip
An approximate graph of the strong derivative  appears in figure
\ref{fig:graphical_differentiation2y'}.
\end{solu}


 \vspace{2cm}
\begin{figure}[h]
\centering
\begin{minipage}{.4\textwidth}$$\psset{unit=3pc}
\psgrid[subgriddiv=5,
gridlabels=0](-2,-2)(-2,-2)(2,2)\psplot[linecolor=red,linewidth=2pt,algebraic=true,
plotpoints=1000]{-1.414213562}{1.414213562}{x*(x-1)*(x+1)} \psaxes[
Ox=-2,Oy=-2](-2,-2)(-2,-2)(2,2)
$$\meinecaption{3}{Example \ref{exa:graphical_differentiation2}. $y=f(x)$}
\label{fig:graphical_differentiation2}\end{minipage}
\begin{minipage}{.4\textwidth}$$\psset{unit=3pc}
\psgrid[subgriddiv=5,
gridlabels=0](-2,-2)(-2,-2)(2,2)\psplot[linecolor=red,linewidth=2pt,algebraic=true,
plotpoints=1000]{-1}{1}{3*x^2-1} \psaxes[
Ox=-2,Oy=-2](-2,-2)(-2,-2)(2,2)
$$\meinecaption{3}{Example \ref{exa:graphical_differentiation2}. $y=f'(x)$}
\label{fig:graphical_differentiation2y'}\end{minipage}
\end{figure}




\begin{exa}\label{exa:graphical_differentiation1}
Figure \ref{fig:graphical_differentiation1y'} gives an approximate
graph of the derivative of the graph appearing in figure
\ref{fig:graphical_differentiation1}. We see that the curve is
increasing and so the derivative is positive. As the curve travels
the interval $\oo{-1}{0}$, the slope of the tangent is going from
almost horizontal (slope $0$) to almost vertical (infinite slope).
At $x=0$ the tangent is vertical, and so the derivative has a pole
there.  As the curve travels the interval $\oo{0}{+1}$,  the slope
of the tangent going from almost vertical to almost $0$.



\vspace{2cm}

\begin{figure}[h]
\centering
\begin{minipage}{.4\textwidth}$$\psset{unit=4pc}
\psgrid[subgriddiv=5,
gridlabels=0](-1,-1)(-1,-1)(1,1)\psplot[linecolor=red,linewidth=2pt,algebraic=true,
plotpoints=1000]{0}{1}{x^.3333333}
\psplot[linecolor=red,linewidth=2pt,algebraic=true,
plotpoints=1000]{-1}{0}{-1*abs(x)^.3333333}\psaxes[
Ox=-1,Oy=-1](-1,-1)(-1,-1)(1,1)
$$\meinecaption{2}{Example \ref{exa:graphical_differentiation1}. $y=f(x)$}\label{fig:graphical_differentiation1}\end{minipage}
\begin{minipage}{.4\textwidth}$$\psset{unit=4pc}
\psgrid[subgriddiv=5,
gridlabels=0](-1,-1)(-1,-1)(1,1)\psplot[linecolor=red,linewidth=2pt,algebraic=true,
plotpoints=1001]{.1924500897}{1}{.333333333*(1/x^.66666666)}
\psplot[linecolor=red,linewidth=2pt,algebraic=true,
plotpoints=1000]{-1}{-.1924500897}{.33333333333*(1/abs(x)^.6666666)}\psaxes[
 Ox=-1,Oy=-1](-1,-1)(-1,-1)(1,1)
$$\meinecaption{2}{Example \ref{exa:graphical_differentiation1}. $y=f'(x)$}\label{fig:graphical_differentiation1y'}\end{minipage}
\end{figure}


\bigskip

To obtain the graph of the derivative above, we have served
ourselves of the table below.
$$
\begin{array}{|c|rrrrrrrrrrr|}
\hline
x  &-1.0&-0.8&-0.6&-0.4&-0.2&0.0 &0.2&0.4&0.6&0.8&1.0\\
\hline f'(x) &0.3 &0.4 & 0.5 &0.6 &1.0 &+\infty &1.0&0.6&0.5&0.4&0.3
\\ \hline
\end{array}\\
$$

\end{exa}
\subsection*{Homework}\addcontentsline{toc}{subsection}{Homework}

\begin{pro}
Draw an example of a curve that is everywhere continuous, but fails
to be differentiable for $x\in \{-1,0,1\}$.
\begin{answer}
There are multiple answers here. One example appears below.
\vspace{2cm}
\begin{figure}[h]
\centering\psset{unit=1pc}
\psaxes[labelFontSize=\tiny,arrows=->,linewidth=2pt](0,0)(-5,-5)(5,5)
\psplot[linewidth=2pt,algebraic,arrows=<->,plotpoints=1001]{-5}{5}{abs(x+1)-2*abs(x)+abs(x-1)+2}
\end{figure}

\end{answer}


\end{pro}
\begin{pro}
Draw an example of a curve $f$ that is everywhere continuous,
differentiable in $\reals \setminus \{-2,2\}$, and such that $f'(x)
=-1$ for $x<-2$, $f'(x)=0$ for $-2<x<2$ and $f'(x)=1$ for $x>2$.
\begin{answer}
There are multiple answers here. One example appears below.
\vspace{2cm}
\begin{figure}[h]
\centering\psset{unit=1pc}
\psaxes[labelFontSize=\tiny,arrows=->,linewidth=2pt](0,0)(-5,-5)(5,5)
\psplot[linewidth=2pt,algebraic,arrows=<->]{-5}{5}{abs(x-2)+abs(x+2)-6}
\end{figure}

\end{answer}


\end{pro}

\begin{pro}
Draw an example of a curve $f$ that is continuous and differentiable
in $\reals \setminus \{0\}$, and such that $f(0+)=+\infty$,
$f'(0+)=+\infty$, $f(0-)=-\infty$, $f'(0-)=-\infty$,
$f'(-\infty)=0$, $f'(+\infty)=0$, $f(1)=1$ and $f(-1)=-1$.
\begin{answer}
There are multiple answers here. One example appears below.
\vspace{2cm}
\begin{figure}[h]
\centering\psset{unit=1pc}
\psaxes[labelFontSize=\tiny,arrows=->,linewidth=2pt](0,0)(-5,-5)(5,5)
\psplot[linewidth=2pt,algebraic,arrows=<->]{-5}{-.2}{1/x}
\psplot[linewidth=2pt,algebraic,arrows=<->]{.2}{5}{1/x}
\psdots(1,1)(-1,-1)
\end{figure}

\end{answer}


\end{pro}

\begin{pro}
Draw an example of an everywhere  differentiable curve $y=f(x)$ with
$$f(-2)=f(0)=f(2)=0, \qquad f(-1)=4, \quad f(1)=-2, \qquad f'(-1)=f'(0)=f'(1)=f'(2)=0, $$  $$ f(+\infty)=0, \quad f(-\infty)=-2,
\quad  f''(x)>0\ \mathrm{for}\ x\in\ \oo{-\infty}{-2}, \quad
  f''(x)<0\ \mathrm{for}\ x\in\ \oo{+\dfrac{5}{2}}{+\infty} $$
or explain why such a curve does not exist.
\begin{answer}
Here is an example of such a curve.\vspace{4cm}
$$\psset{unit=1pc}\psaxes[linewidth=2pt,labelFontSize=\tiny,arrows=->](0,0)(-9,-9)(9,9)
\pscurve[linewidth=2pt,arrows=<->](-8,-1.95)(-7,-1.9)(-6,-1.8)(-5,-1.7)(-4,-1.5)(-3,-1)(-2.5,-.5)(-2,0)(-1.5,1.5)(-1,4)(0,0)(.5,-1)(1,-2)(2,0)(3,-1.5)(4,-1)(5,-.5)(6,-.2)(7,-.1)(8,-.15)
\psline[linestyle=dotted](-8,-2)(8,-2)$$
\end{answer}
\end{pro}

\begin{pro}\label{pro:graphical_deriv1}
Use the graph in figure \ref{fig:graphical_deriv1} below to complete
the following table for a function $f$. Also, make a rough sketch of
$f'$. \vspace{2cm}

\begin{figure}[h]
\centering
\begin{minipage}{.4\textwidth}
\vfill
$$\begin{array}{|c|c|c|c|c|c|c|c|}\hline x & -1 & -0.4 & -0.2 & 0 &
0.2 & 0.4 & 1 \\
\hline f(x) & & & & & & & \\
\hline f'(x) & & & & & & & \\
\hline
\end{array}  $$
\vfill
\end{minipage}
\hfill
\begin{minipage}{.4\textwidth}
\centering\psset{unit=5pc} \psgrid[subgriddiv=5,
gridlabels=0](-1,-1)(-1,-1)(1,1)\psplot[linecolor=red,linewidth=2pt,algebraic,
plotpoints=1001]{-1}{1}{abs(x)}\psaxes[
 Ox=-1,Oy=-1](-1,-1)(-1,-1)(1,1) \meinecaption{3}{Problem
\ref{pro:graphical_deriv1}.}\label{fig:graphical_deriv1}

\end{minipage}
\end{figure}
\end{pro}
\begin{pro}\label{pro:graphical_deriv2}
Use the graph in figure \ref{fig:graphical_deriv2} below to complete
the following table for a function $f$. Also, make a rough sketch of
$f'$. \vspace{2cm}

\begin{figure}[h]
\begin{minipage}{.4\textwidth}
$$\begin{array}{|c|c|c|c|c|c|c|c|}\hline x & -1 & -0.4 & -0.2 & 0 &
0.2 & 0.4 & 1 \\
\hline f(x) & & & & & & & \\
\hline f'(x) & & & & & & & \\
\hline
\end{array}  $$
\end{minipage}
\hfill
\begin{minipage}{.4\textwidth}
\centering\psset{unit=5pc} \psgrid[subgriddiv=5,
gridlabels=0](-1,-1)(-1,-1)(1,1)\psplot[linecolor=red,linewidth=2pt,algebraic,
plotpoints=1001]{-1}{1}{x^3}\psaxes[
 Ox=-1,Oy=-1](-1,-1)(-1,-1)(1,1) \meinecaption{3}{Problem
\ref{pro:graphical_deriv2}.}\label{fig:graphical_deriv2}

\end{minipage}
\end{figure}
\end{pro}

\section{The Strong Derivative}
Given a finite number of points, we can find infinitely many curves
passing through them. See for example figure
\ref{fig:few_point_graph}, where we see three very different curves
$f, g, h$ each simultaneously passing through the points $A$, $B$,
$C$. Thus plotting a few points of the graph of a function can give
a misleading picture.



\vspace{1cm}
\begin{figure}[h]
$$ \psset{unit=1pc,algebraic=true}
\psline[linewidth=2pt,linecolor=red](-3,-3)(3,3)
\psplot[linewidth=2pt,linecolor=blue]{-1.5}{1.5}{x^3}
\psplot[linecolor=magenta,linewidth=2pt,algebraic=true,
plotpoints=1000]{0}{6}{x^.3333333}
\psplot[linecolor=magenta,linewidth=2pt,algebraic=true,
plotpoints=1000]{-6}{0}{-1*abs(x)^.3333333}\psdots[dotstyle=*,dotscale=1](0,0)(-1,-1)(1,1)
\uput[dr](1,1){A}\uput[dr](-1,-1){C}\uput[ul](0,0){B}
\uput[l](-3,-3){f}\uput[d](-1.5,-3.375){g}\uput[r](6,1.82){h}$$\meinecaption{1}{A
few points do not a graph determine.}\label{fig:few_point_graph}
\end{figure}

By the same token, given a formula, the plotting of a few points
does not give the salient features of a graph. For example, let us
say that we wanted to graph $y = 4x-x^3$.  In figures
\ref{fig:approx_by_lines_begin} through
\ref{fig:approx_by_lines_end} we have chosen a few selected points
on the curve and interpolated between them through lines. But
relying on this method does not give proof that the graph will not
have more turns or bends, say, or that it will grow indefinitely for
values of $x$ of large magnitude.





\vspace{1cm}

\begin{figure}[h]
\begin{minipage}{3cm}$$\psset{unit=.7pc, plotpoints=4, algebraic=true}
\psaxes[linewidth=1.5pt,labels=none](0,0)(-5,-5)(5,5)
\psplot[linewidth=2pt,linecolor=red]{-2.5}{2.5}{-1*(x+2)*((x-2))*x}
$$\meinecaption{1}{Four plot points.}\label{fig:approx_by_lines_begin}
\end{minipage}
\hfill
\begin{minipage}{3cm}$$\psset{unit=.7pc, plotpoints=7, algebraic=true}
\psaxes[linewidth=1.5pt,labels=none](0,0)(-5,-5)(5,5)
\psplot[linewidth=2pt,linecolor=red]{-2.5}{2.5}{-1*(x+2)*((x-2))*x}
$$\meinecaption{1}{Seven plot points.}
\end{minipage}\hfill
\begin{minipage}{3cm}$$\psset{unit=.7pc, plotpoints=10, algebraic=true}
\psaxes[linewidth=1.5pt,labels=none](0,0)(-5,-5)(5,5)
\psplot[linewidth=2pt,linecolor=red]{-2.5}{2.5}{-1*(x+2)*((x-2))*x}
$$\meinecaption{1}{Ten plot points.}
\end{minipage}\hfill
\begin{minipage}{3cm}$$\psset{unit=.7pc,plotpoints=1000, algebraic=true}
\psaxes[linewidth=1.5pt,labels=none](0,0)(-5,-5)(5,5)
\psplot[linewidth=2pt,linecolor=red]{-2.5}{2.5}{-1*(x+2)*((x-2))*x}
$$\meinecaption{1}{One thousand plot points.}\label{fig:approx_by_lines_end}
\end{minipage}
\end{figure}

But for all its faults, the progression of shapes in figures
\ref{fig:approx_by_lines_begin} through
\ref{fig:approx_by_lines_end} suggests that a ``reasonable'' graph
can  be approximated by a series of straight lines. By a
``reasonable'' graph we mean one that does not have many sharp
turns, does not oscillate wildly, does not have many jumps or many
asymptotes, and that it is mostly continuous and ``smooth.''
Admittedly, these concepts are vague, but we will gain more insight
into them as we progress.


 \vspace{2cm}
\begin{figure}[h]
\begin{minipage}{5cm}
\centering \psset{linewidth=2\pslinewidth,unit=1pc}
\psplot[linecolor=gray,algebraic]{-2.2}{2.2}{x^2}
\pstGeonode[PointName=none](0,0){O}(1,1){A}
\pstLineAB[nodesepA=-1cm,nodesepB=-1cm,arrows=<->]{O}{A}
\uput[dr](A){$(a,f(a))$}
 \meinecaption{1}{Left secant through $(a,f(a))$.}
\label{fig:left_secant}
\end{minipage} \hfill
\begin{minipage}{5cm}
\centering \psset{linewidth=2\pslinewidth,unit=1pc}
\psplot[linecolor=gray,algebraic]{-2.2}{2.2}{x^2}
\pstGeonode[PointName=none](1,1){A}(2,4){B}
\pstLineAB[nodesepA=-1cm,nodesepB=-1cm,arrows=<->]{A}{B}
 \uput[dr](A){$(a,f(a))$}
 \meinecaption{1}{Right secant through $(a,f(a))$.}
\label{fig:right_secant}
\end{minipage} \hfill
\begin{minipage}{5cm}
\centering \psset{linewidth=2\pslinewidth, unit=1pc}
\pstGeonode[PointName=none](1,1){A}
\psplot[linecolor=gray,algebraic]{-2.2}{2.2}{x^2}
\psplotTangent[linecolor=gray,algebraic,arrows=<->]{1}{4}{x^2}
 \uput[dr](A){$(a,f(a))$}
\meinecaption{1}{Line grazing $(a,f(a))$.} \label{fig:grazing}
\end{minipage}
\end{figure}


\bigskip

How do we choose the lines to approximate a given ``reasonable''
curve? Given a function $f$ consider the point $(a, f(a))$ on the
graph of the function. What happens around this point? If we
approached $a$ through values $x<a$ and joined the line with
endpoints $(x,f(x))$ and $(a,f(a))$, we would obtain a secant line
like that of figure \ref{fig:left_secant}. If we approached $a$
through  values $x>a$ and joined the line with endpoints $(a,f(a))$
and $(x,f(x))$, we would obtain a secant line like that of figure
\ref{fig:right_secant}. Eventually, on getting closer to $(a, f(a))$
we obtain a line just barely grazing the curve---that is,
``tangent'' to the curve---at the point $(a, f(a))$, as in figure
\ref{fig:grazing}.
\bigskip

In the simplest of cases, if our curve is the line  $L: y=mx+k$,
then in a neighbourhood of the point $x=a$ the tangent line to $L$
should be itself! Given now the formula for a function $f$ and a
point $(a, f(a))$ on the graph of $f$, how do we determine the
tangent line to $f$ at $(a, f(a))$?  Recall that if
$\tends{\epsilon}{0}$, then $a+\epsilon$ is in a neighbourhood of
$a$. The slope of the secant line joining $(a+\epsilon,
f(a+\epsilon))$ and $(a,f(a))$ is
\begin{equation} \dfrac{f(a+\epsilon)-f(a)}{a+\epsilon -\epsilon}= \dfrac{f(a+\epsilon)-f(a)}{\epsilon}\label{eq:slope_def_strong derivative } \end{equation}
We denote the value---if there is one---of \ref{eq:slope_def_strong
derivative }--- as $\tends{x}{\epsilon}$ by $f'(a)$. Hence for fixed
but small $\epsilon$ we have
$$\dfrac{f(a+\epsilon)-f(a)}{\epsilon} \approx f'(a) \implies f(a+\epsilon) \approx f(a) + f'(a)\epsilon .   $$
There is, generally, an error in taking the dextral side as an
approximation for the sinistral side on the above formula. We will
settle for having an error of the order of  $\soo{\varepsilon}$,
which will normally will be a good compromise for most of the
formul\ae\ we will encounter. This prepares the ground for our main
definition of this section.
\begin{df}Let $f$ be a function and let $a\in\dom{f}$.
When there is a number $f'(a)$ such that
$$f(a+\epsilon)=f(a)+f'(a)\epsilon+\soo{\varepsilon} \qquad
\mathrm{as}\qquad \tends{\epsilon}{0}  $$ then we say that the
function  $f$ has a {\em strong derivative } $f'(a)$ or that $f$ is
{\em strongly differentiable } at $x=a$. If we consider the set
$\{x\in\dom{f}: f'(x) \ \mathrm{exists}\}$ then we may form the
function $x\mapsto f'(x)$ with domain $\dom{f'} = \{x\in\dom{f}:
f'(x) \ \mathrm{exists}\}$. We call the function $f'$ the {\em
strong derivative} of $f$. We will also often use the  symbol
$\diff{f(x)}{x}$ to denote the function $x\mapsto
f'(x)$.\index{strong
derivative}\index{differentiable}\index{derivative}\index{strongly
differentiable}
\end{df}
\begin{df}
If $f'$ is itself differentiable, then the function $(f')' = f''$ is
the {\em second derivative} of $f$. It is also denoted by
$\dfrac{\mathrm{d}^2}{\mathrm{d}x^2}\ f(x)$. We similarly define the
third, fourth, etc., derivatives. It is customary to denote the
first three derivatives of a function with primes, as in $f', f'',
f'''$, and any higher derivative with either roman numbers or with
the order of the derivative enclosed in parenthesis, as in $f^{iv},
f^{v}$ or $f^{(4)}, f^{(5)}$, etc.
\end{df}

We will give but two very simple examples of the computation of
derivatives. These will be needed in this chapter. More substantial
examples will appear in subsequent chapters.



\begin{lem}\label{lem:D(k)=0}The strong derivative of a constant
function is the $0$ function. In symbols, if $f$ is a function with
assignment rule $f(x) = k$, constant, then for all $x$, $f'(x) = 0$.
\end{lem}
\begin{pf}
We have
$$f(x+\epsilon)=k= k+0\cdot\epsilon+0\epsilon^2=k+0\cdot\epsilon+\soo{\epsilon}, $$which proves the
assertion.
\end{pf}
\begin{rem}
For ellipsis we will write $(k)' = 0$ or $\diff{k}{x} = 0$.
\end{rem}

\begin{lem}\label{lem:D(x)=x} The strong derivative of the identity
function $x\mapsto x$ is the constant function $x \mapsto 1$.
\end{lem}
\begin{pf}
If $ f(x)=x$ we have
$$f(x+\epsilon)=x+\epsilon=x+1\cdot\epsilon+0\epsilon^2=x+1\cdot\epsilon+\soo{\epsilon},$$ and so
$f'(x) = 1$.
\end{pf}
\begin{rem}
For ellipsis we will write $(x)' = 1$ or $\diff{x}{x} = 1$.
\end{rem}

\section{Derivatives and Graphs}
In this section we prove the remarks preceding figures
\ref{fig:increasing_convex} through \ref{fig:decreasing_concave},
which will help us with the graphing in subsequent chapters.
\begin{thm}
If $f$ is strongly differentiable  at $x$ then $f$ is continuous at
$x$.
\end{thm}
\begin{pf}
We have $f(x+\epsilon) = f(x) + f'(x)\epsilon + \soo{\varepsilon}$.
If $\epsilon > 0$ and $\tends{\epsilon}{0}$ then $f(x+\epsilon) =
f(x+)$ and similarly if $\epsilon < 0$ and $\tends{\epsilon}{0}$
$f(x+\epsilon) = f(x-)$. Hence we have $f(x+)=f(x)=f(x-)$, and $f$
is continuous at $x$.
\end{pf}

\begin{thm}\label{thm:monotonicity_criteria} Let $f$ be
strongly differentiable  at $x$. If $f'(x)>0$ then  $f$ is
increasing in a neighbourhood of $x$, if $f'(x) <0$ then $f$ is
decreasing in a neighbourhood of $x$.
\end{thm}
\begin{pf}We have $f(x+\epsilon) - f(x) = f'(x)\epsilon +
\soo{\varepsilon}$. For $\epsilon$ very small, this means that
$$f(x+\epsilon) - f(x) \approx f'(x)\epsilon,$$ that is, the sign of
$f(x+\epsilon) - f(x)$ is the same as the sign of $f'(x)\epsilon$.
Thus if $\epsilon > 0$ and $f'(x) > 0$, then $f(x+\epsilon)> f(x)$,
that is, $f$ is increasing. If $\epsilon > 0$ and $f'(x) < 0$, then
$f(x+\epsilon)< f(x)$, that is, $f$ is decreasing. Similar
conclusions are reached when considering $\epsilon < 0$ and the
theorem is proved.
\end{pf}
\begin{df}
If $f$ is strongly differentiable  at $x$ and $f'(x) = 0$, then we
say that $x$ is a {\em stationary point of $f$.}
\end{df}
\begin{df}
If there is a point $a$ for which $f(x) \leq f(a)$ for all $x$ in a
neighbourhood centred at  $x=a$ then we say that $f$ has a {\em
local maximum} at $x=a$. Similarly, if there is a point $b$ for
which $f(x) \geq f(b)$ for all $x$ in a neighbourhood centred at
$x=b$ then we say that $f$ has a {\em local minimum} at $x=b$.
\end{df}
\begin{thm}\label{thm:local_extrema}
If $f$ is strongly differentiable  at $x=a$, $f'(a) = 0$, and $f'$
changes from $+$ to $-$ in a neighbourhood of $a$ then $x=a$ is a
local maximum. If $f$ is strongly differentiable  at $x=b$,
$f'(b)=0$, and $f'$ changes from $-$ to $+$ in a neighbourhood of
$b$ then $x=b$ is a local minimum.
\end{thm}
\begin{pf}
By Theorem \ref{thm:monotonicity_criteria}, when $f'$ changes from
$+$ to $-$, $f$ is increasing and then decreasing in a neighbourhood
of $x=a$. By Weierstrass's Theorem (Theorem
\ref{thm:weierstrass-max-min}), $f$ assumes a maximum on a closed
neighbourhood containing $a$. It cannot be to the left of $a$ since
the function is increasing there, and it cannot be to the right of
$a$ since the function is decreasing there. Hence the maximum must
be at $x=a$ and so $f(x)\leq f(a)$ for $x$ in a neighbourhood of
$a$. The result just obtained applied to $-f$ yields the second half
of the theorem.
\end{pf}

\begin{lem}\label{lem:convexity_criteria}
If $f'$ increases in a neighbourhood of $x$, then $f$ is convex in a
neighbourhood of $x$. Similarly, if $f'$ decreases in a
neighbourhood of $x$, then $f$ is concave in a neighbourhood of $x$.
\end{lem}
\begin{pf}
Let $\epsilon > 0$.  Then
$$f(x) = f(x-\epsilon+\epsilon) = f(x-\epsilon) + f'(x-\epsilon)\epsilon + \soo{\varepsilon},$$
$$f(x) = f(x+\epsilon-\epsilon) = f(x+\epsilon) - f'(x+\epsilon)\epsilon + \soo{\varepsilon}.$$
Adding, $$2f(x)   = f(x+\epsilon) + f(x-\epsilon) + \left(
f'(x-\epsilon) - f'(x+\epsilon)\right)\epsilon +
\soo{\varepsilon}.$$ For $\epsilon$ very small we then have
$$2f(x)  \approx  f(x+\epsilon) + f(x-\epsilon) + \left(
f'(x-\epsilon) - f'(x+\epsilon)\right)\epsilon .$$ If $f'$ is
increasing then $f'(x-\epsilon) - f'(x+\epsilon) < 0$. Since
$\epsilon > 0$ this implies that
$$2f(x)  <  f(x+\epsilon) + f(x-\epsilon), $$
which means that $f$ is convex in a neighbourhood of $x$. This
result now applied to $-f$ gives the second half of the theorem.
\end{pf}




\begin{thm}\label{thm:convexity_criteria} A twice strongly differentiable -function $f$ is convex in a neighbourhood of  $x=a$ if
$f''(a) > 0$. It is concave in a neighbourhood of  $x=b$ if $f''(b)
< 0$.
\end{thm}
\begin{pf}This follows from Lemma \ref{lem:convexity_criteria} and
Theorem \ref{thm:monotonicity_criteria}.
\end{pf}


\begin{exa}
\label{exa:graph_maxima_minima} The graph of the strong derivative
$f'$ of a function $f$ is given in figure
\ref{fig:graph_maxima_minima}. Then according to Theorem
\ref{thm:local_extrema} $f$ has a local minimum at $x = -2$ and
$x=2$, and a local maximum at $x=0$ and $x=4$.
\end{exa}



\begin{exa}\label{exa:curvaceous} Figure \ref{fig:curvaceous}
below shews the graph of a function $f$. At the point $A$, $f'>0$
and $f''<0$. At $B$, $f'=0$, $f''<0$. At $C$, $f'<0$, $f''<0$. At
$D$, $f'=0$ and $f''>0$. At $E$, $f'>0$, $f''>0$.
\end{exa}



\vspace{2cm}
\begin{figure}[h]
\begin{minipage}{.4\textwidth}\centering
$$\psset{unit=1pc}\psaxes[labelFontSize=\tiny](0,0)(-4,-3)(6,6)
\psdots[dotscale=1.5](-2,0)(4,0)(0,0)(2,0)
\psplot[linewidth=2pt,arrows=*-*,algebraic]{-2.5}{4.5}{-.1*(x+2)*(x-4)*(x-2)*x}
$$
 \meinecaption{2}{Example \ref{exa:graph_maxima_minima}.} \label{fig:graph_maxima_minima}
\end{minipage}
\hfill
\begin{minipage}{.4\textwidth}
$$\psset{algebraic=true, unit=1pc} \psaxes[labels=none, ticks = none, linewidth=1.5pt]{->}(0,0)(-6.5,-6.5)(6.5,6.5)
\psplot[linecolor=red,linewidth=1.4pt]{0}{6.4}{2+sin(x)}
\psdots[dotstyle=*,dotscale=1](.5,2.479425539)(1.570796327,3)(2.5,2.598472144)(4.712388981,1)(6,1.720584502)
\uput[d](.5,2.479425539){A} \uput[u](1.570796327,3){B}
\uput[d](2.5,2.598472144){C} \uput[u](4.712388981,1){D}
\uput[r](6,1.720584502){E}
 $$
\meinecaption{2}{Example \ref{exa:curvaceous}.}
\label{fig:curvaceous}
\end{minipage}
\end{figure}


\subsection*{Homework}\addcontentsline{toc}{subsection}{Homework}

\begin{pro}
Draw an example of a curve $p$ satisfying the following criteria:
\begin{enumerate}
\item $p$ is everywhere differentiable,
\item $p(-3)=p(0)=p(2)=0$,
\item $p(x)<0$ if $x\in \oo{-\infty}{-3}\cup \oo{0}{2}$,
\item $p(x)>0$ if $x\in \oo{-3}{0}\cup \oo{2}{+\infty}$,
\item $-2< p(x) < \dfrac{3}{2}$ for $x\in \cc{-3}{2}$,
\item $p$ has an inflexion point near $x=-\dfrac{1}{3}$,
\item $p(-\infty)=-\infty$, $p(+\infty)=+\infty$.
\end{enumerate}
\begin{answer}
Below is an example of such a curve. \vspace{2cm}
\begin{figure}[h]
\centering \psset{unit=1pc}
\psgrid[subgriddiv=0,gridlabels=0](0,0)(-5,-5)(5,5)
\psaxes[labels=none,ticks=none,arrows=->,linewidth=2pt](0,0)(-5.5,-5.5)(5.5,5.5)
\psaxes[Ox=-5,Oy=-5](-5,-5)(-5,-5)(5,5)
\psplot[linewidth=2pt,algebraic,arrows=<->]{-4.167279748}{3.386867835}{(x+3)*(x)*(x-2)/6}
\end{figure}
\end{answer}


\end{pro}

\begin{pro}
Draw an example of a curve $f$ satisfying the following criteria:
\begin{enumerate}
\item $f$ is everywhere differentiable,
\item $f'(x)>0$ for all $x\in \reals$,
\item $f(-\infty)=-1$, $f(+\infty)=1$,
\item $f''(x)>0$ for $x<0$, $f''(x)<0$ for $x>0$,
\item $f(0)=f''(0)=0$.
\end{enumerate}
\begin{answer}
Below is an example of such a curve. \vspace{2cm}
\begin{figure}[h]
\centering \psset{unit=1pc}
\psgrid[subgriddiv=0,gridlabels=0](0,0)(-5,-5)(5,5)
\psaxes[labels=none,ticks=none,arrows=->,linewidth=2pt](0,0)(-5.5,-5.5)(5.5,5.5)
\psaxes[Ox=-5,Oy=-5](-5,-5)(-5,-5)(5,5)
\psplot[linewidth=2pt,algebraic,arrows=<->]{-5}{5}{2*(2.718281828^x)/(2.718281828^x+1)-1}
\end{figure}
\end{answer}


\end{pro}

\begin{pro}\label{pro:max_min_conv1}
In  figure \ref{fig:max_min_conv1} below the graph of $f'$ is given
for a certain function $f:\cc{0}{6}\to \reals$. Find the
$x$-coordinate of the local maximum of $f$. How many inflexion
points does the graph of $f$ have?
\end{pro}
\begin{pro}
\label{pro:m103_s06_final_f'}The graph of the \emph{derivative} $f'$
of a certain function $f$ defined on~$[-3;2]$ is given in figure
\ref{fig:m103_s06_final_f'} below.

Find the $x$-coordinate of the local minimum  of $f$.
\end{pro}
\vspace{3cm}
\begin{figure}[h]
\begin{minipage}{.4\textwidth}
\psset{unit=1.2pc} \centering
\psgrid[subgriddiv=0,griddots=5,gridlabels=0](-6,-6)(6,6)
\psaxes[labelFontSize=\tiny]{->}(0,0)(-6.5,-6.5)(6.5,6.5)
\pscurve[linewidth=2pt](0,.5)(1,1)(3,-1)(4,0)(6,-2)
\psdots(0,.5)(1,1)(3,-1)(4,0)(6,-2) \meinecaption{3}{Problem
\ref{pro:max_min_conv1}.}\label{fig:max_min_conv1}
\end{minipage}
\hfill
\begin{minipage}{.4\textwidth}
$$
\psset{unit=1.2pc}
\psgrid[subgriddiv=0,griddots=5,gridlabels=0](-6,-6)(6,6)
\psaxes[labelFontSize=\tiny]{->}(0,0)(-6.5,-6.5)(6.5,6.5)
\psplot[linewidth=2pt,algebraic=true,
arrows={*-*}]{-3}{2}{-.4*(x+2)*(x+1)*(x-1)}
$$\meinecaption{3}{Problem
\ref{pro:m103_s06_final_f'}.} \label{fig:m103_s06_final_f'}
\end{minipage}
\end{figure}

\begin{pro}
Suppose $f:\reals \to \reals$ is everywhere differentiable and that
$f(+\infty)=K$, a constant. Must $f'(+\infty ) =0$?
\end{pro}

\section{Graphical Integration}
We now consider the following problem: Given a sufficiently ``decent
curve''\footnote{A curve that pays its taxes, observes the Torah,
and goes to synagogue every Shabath\ldots .}, what is the area
bounded by the curve and the $x$-axis in a given finite interval?

\vspace{3cm}
\begin{figure}[h]
\begin{minipage}{.3\textwidth}
$$\psset{unit=1pc}
\psaxes[linewidth=2pt, labels=none,
ticks=none,arrows=->](0,0)(-6,-6)(6,6)
\psplot[linecolor=red,linewidth=1.4pt]{-6}{4}{x  x 2 exp -1 12 div
mul x 3 exp -1 18 div mul add add}
\pscustom[fillstyle=solid,fillcolor=blue]{\psplot{1}{3}{x x 2 exp -1
12 div mul x 3 exp -1 18 div mul add
add}\psline(3,.75)(3,0)(1,0)(1,.8611111111) }\uput[d](1,0){a}
\uput[d](3,0){b}
$$\meinecaption{2}{$\dint _a ^b f \geq 0$} \label{fig:area_under_a_curve1}
\end{minipage}
\hfill
\begin{minipage}{.3\textwidth}
$$\psset{unit=1pc}
\psaxes[linewidth=2pt, labels=none,
ticks=none,arrows=->](0,0)(-6,-6)(6,6)
\psplot[linecolor=red,linewidth=1.4pt]{-6}{5}{x  x 2 exp -1 12 div
mul x 3 exp -1 18 div mul add add}
\pscustom[fillstyle=solid,fillcolor=green]{\psplot{-4}{-2}{x x 2 exp
-1 12 div mul x 3 exp -1 18 div mul add
add}\psline(-2,-1.888888889)(-2,0)(-4,0)(-4,-1.777777777)
}\uput[u](-4,0){c} \uput[u](-2,0){d}
$$\meinecaption{2}{$\dint _c ^d f \leq 0$} \label{fig:area_under_a_curve2}
\end{minipage}
\hfill
\begin{minipage}{.3\textwidth}
 $$\psset{unit=1pc} \psaxes[linewidth=2pt,labelFontSize=\tiny,arrows=->](0,0)(-6,-6)(6,6)\psline[linewidth=2pt,
linecolor=red]{-o}(-4,2)(-3,2)(-2,4)(0,-2)\psarc[linewidth=2pt,
linecolor=red](1,0){1}{0}{180}\psarc[linewidth=2pt,
linecolor=red](3,0){1}{180}{360}\psdots[dotscale=1.5, dotstyle=*,
linecolor=red](-4,2)(-3,2)(-2,4)(0,0)(2,0)(4,0)$$ \meinecaption{2}{
\ref{exa:integral1}} \label{fig:integral1}
\end{minipage}
\end{figure}
\begin{df}
Let $f$ be a function. We denote the signed area between $f$ and the
$x$-axis between $x = a$ and $x = b$ by $\dint _a ^b f$ or $\dint _a
^b f(x)\d{x}$. The symbol $\dint _a ^b f$ is read ``the (definite)
integral of $f$ from $a$ to $b$. See figures
\ref{fig:area_under_a_curve1} and \ref{fig:area_under_a_curve2}.
\end{df}
\begin{rem}
We will normally use the notation $\dint _a ^b f$ in the cases when
we are given a picture of $f$, and we will relegate the notation
$\dint _a ^b f(x)\d{x}$ when we are given a formula for $f$. Also,
there is nothing holy about which variable of integration to use.
Thus $\dint _a ^b f(x)\d{x}=\dint _a ^b f(t)\d{t}=\dint _a ^b
f(u)\d{u}$, etc.
\end{rem}
\begin{exa}\label{exa:integral1}
Figure \ref{fig:integral1} is composed of lines and semicircles. One
finds
\begin{enumerate}
\item $\dint _{-4} ^{-3} f = 2(1)$ since the figure between the
curve and the $x$-axis is a rectangle of base $1$ and height $2$.
\item $\dint _{-3} ^{-2} f = \left(\dfrac{2 + 4}{2}\right) 1 =
3$ since the figure between the curve and the $x$-axis is a
trapezoid with bases of length $2$ and $4$ and height $1$.
\item  $\dint _{-4} ^{-2} f = 2 + 3 = 5$ by adding the areas of the
rectangle and the trapezoid.
\item  $\dint _{-2} ^{0} f = 2$. The figure consists of two
triangles, one with height $4$ above the $x$-axis (and hence having
positive area, and another one with height $2$ below the $x$-axis
(thus having negative area). To find the area we enclose both
triangles in a rectangle $ABCD$ with vertices at $A=(-2,4)$, $B =
(0,4)$, $C=(0,-2)$, $D = (-2,-2)$. Put $E=(-2,0)$ and $F = (0,0)$.
Observe that the area of $ABCD$ is the area of rectangle $ABFE$
minus the area of rectangle $CDEF$ hence it is $2(4) - 2(2) = 4$.
The area of the triangles is half this, that is, $2$. Another
approach is the following. Let $a$ be the length of base of the
triangle with positive area and let $b$ the length of the base of
the triangle with negative area. Then $a+b = 2$. Also, by similar
triangles, $\dfrac{4}{a} = \dfrac{2}{b}$, and hence $a=2b$. Thus
$2b+b = 2\implies b = \dfrac{2}{3}$ and $a = \dfrac{4}{3}$. Whence
the area of the two triangles is $\frac{1}{2}(\frac{4}{3})(4)
-\frac{1}{2}(\frac{2}{3})(2) = 2$, as before.
\item  $\dint _{-1} ^{0} f= -\frac{5}{9}$. Observe that the area sought is composed
of a triangle of base $1 - \dfrac{2}{3} = \dfrac{1}{3}$ (with
positive area) and the triangle of the preceding part of base
$\dfrac{2}{3}$ and height $2$ (with negative area). Again, by
similar triangles, if $h$ is the height of the triangle of base
$\dfrac{1}{3}$ then $\dfrac{h}{\frac{1}{3}} =
\dfrac{2}{\frac{2}{3}}$, thus $h = 1$. Thus the area sought is
$\frac{1}{2}(\frac{1}{3})(1) - \frac{1}{2}(\frac{2}{3})(2) =
-\frac{1}{2}$.
\item  $\dint _{0} ^{2} f = \dfrac{\pi}{2}$  since the figure between the curve and the $x$-axis is a
semicircle of radius $1$.
\item  $\dint _{0} ^{1} f = \dfrac{\pi}{4}$  since the figure between the curve and the $x$-axis is a
quarter circle of radius $1$.
\item  $\dint _{0} ^{3} f = \dfrac{\pi}{2} - \dfrac{\pi}{4}=\dfrac{\pi}{4}$ by adding the (positive) area of the semicircle to the (negative) area of the
quarter circle.
\item  $\dint _{0} ^{4} f =0$ since the areas of the semicircles
cancel each other.
\end{enumerate}
\end{exa}


In the particular case when the function is a straight line, it is
easy to compute the integral.

\vspace{2cm}
\begin{figure}[h]
\centering \psset{unit=1pc}
\psaxes[linewidth=2pt,arrows=->,labelFontSize=\tiny](0,0)(-5,-5)(5,5)
\psplot[linewidth=2pt,linecolor=red,arrows=<->,algebraic]{-4.2}{4.2}{x/2+1}
\pscustom[fillcolor=red,fillstyle=solid]{\psline(1,0)(1,1)\psplot[algebraic]{1}{3}{x/2+1}\psline(3,2.5)(3,0)(1,0)}
\meinecaption{2}{Example
\ref{exa:area_under_line}.}\label{fig:area_under_line}
\end{figure}

\begin{exa}\label{exa:area_under_line}
Find the area of the plane region
$$ \{(x, y)\in \reals ^2: 1\leq x \leq 3, 0 \leq y \leq \dfrac{x}{2}+1\}. $$
\end{exa}
\begin{solu}
This is the area under the line $y=\dfrac{x}{2}+1$ and above the
$x$-axis, bounded by the vertical lines $x=1$ and $x=3$, as depicted
in figure \ref{fig:area_under_line}. Since this is a trapezoid with
height $2$ and parallel bases of length $\dfrac{3}{2}$ and
$\dfrac{5}{2}$, we obtain
$$ \dint _1 ^3 \left(\dfrac{x}{2}+1\right) \d{x} = \dfrac{2}{2}\left(\dfrac{3}{2}+\dfrac{5}{2}\right)=4. $$
\end{solu}

\subsection*{Homework}\addcontentsline{toc}{subsection}{Homework}
\begin{pro}
\label{pro:m103_s06_final_integral} The graph of a function $f$
defined on $[-3;4]$ is given in figure
\ref{fig:m103_s06_final_integral} below. It is composed of a
semicircle and lines.
\bigskip
\vspace{3cm}
\begin{figure}[h]
$$
\psset{unit=1.5pc}
\psgrid[subgriddiv=0,griddots=5,gridlabels=0](-5,-5)(5,5)
\psaxes[labelFontSize=\tiny]{->}(0,0)(-5.5,-5.5)(5.5,5.5)
\psarc[linewidth=1.5pt](-2,0){1}{0}{180}\psline[linewidth=1.5pt](-1,0)(0,2)(2,-2)(4,-2)
$$\meinecaption{3}{Problem
\ref{pro:m103_s06_final_integral}.}\label{fig:m103_s06_final_integral}
\end{figure}

If $g(x) = \dint_{-3}^{x} f(t) \d{t}$, find $g(2)$. Where does $g$
attain its absolute maximum value?

\end{pro}
\begin{pro}
Give a geometric argument to justify that $\dint _0 ^t
x\d{x}=\dfrac{t^2}{2}$.
\begin{answer}
The integral gives the area of the triangle with vertices at
$A(0,0)$, $B(t,0)$, $C(t,t)$. This is an isosceles right triangle of
area $\dfrac{t^2}{2}$.
\end{answer}
\end{pro}
\begin{pro}
Give a geometric argument to justify that $\dint _{-2} ^3
(2x+1)\d{x}=10$.
\begin{answer}Observe that $x\mapsto 2x+1$ is a straight line that
crosses the $x$-axis at $x=-\dfrac{1}{2}$. Since $\dint _{-2} ^3
(2x+1)\d{x}=\dint _{-2} ^{-1/2} (2x+1)\d{x}+\dint _{-1/2} ^3
(2x+1)\d{x}$, we decompose the region into two triangles: one with
negative area and the other with positive area. The triangle with
negative area has vertices $A(-2,0)$, $B(-2,-3)$,
$C(-\frac{1}{2},0)$ and area $-\dfrac{1}{2}\cdot \dfrac{3}{2}\cdot 3
= -\dfrac{9}{4}$. The triangle with positive area has vertices
 $C(-\frac{1}{2},0)$, $D(3,0)$, $B(3,7)$, and area $\dfrac{1}{2}\cdot
\dfrac{7}{2}\cdot 7 = \dfrac{49}{4}$. Thus the total signed area is
$-\dfrac{9}{4}+\dfrac{49}{4}=10$.
\end{answer}
\end{pro}
\begin{pro}
Give a geometric argument to justify  that $\dint _{-1} ^1
\sqrt{1-x^2}\d{x}=\dfrac{\pi}{2}$.
\begin{answer}
By example \ref{pro:upper_semicircle}, the graph of $f:\cc{-1}{1}\to
\reals$, $x\mapsto \sqrt{1-x^2}$ is an upper semicircle of radius
$1$. Hence, its area is half of that of a circle of radius $1$, that
is $\dfrac{\pi}{2}$.
\end{answer}
\end{pro}

\section{The Fundamental Theorem of Calculus}
In this section we seek to establish a link between differentiation
and integration.
\begin{df}
Let $f$ be a function. Then $t \mapsto \dint_a ^t f$ is the function
that for each value of  $t$ gives the area  between the graph of $f$
and the $x$-axis bounded by vertical lines at  $x=a$ and $x=t$. We
prefer to write, in brief,  $\dint _a f$ rather than $t \mapsto
\dint_a ^t f$.
\end{df}
From the additivity of area, one has
\begin{thm}[Chasles' Rule] \label{rul:chasles}$\dint _a ^b f + \dint _b ^c f = \dint _a ^c
f$.

\end{thm}
Letting $a = b = c$ in Chasles' Rule \ref{rul:chasles} one obtains
$$\dint _a ^a f + \dint _a ^a f = \dint _a ^a f\implies \dint _a ^a f =
\dint _a ^a f -\dint _a ^a f$$hence
\begin{thm}\label{rul:integral_of_a_point} $\dint _a ^a f = 0$.
\end{thm}
Letting $a=c$ and using Rule \ref{rul:integral_of_a_point} one
obtains
$$\dint _a ^b f + \dint _b ^a f = \dint _a ^a f\implies \dint _a ^a f =
\dint _a ^a f -\dint _b ^a f =  -\dint _b ^a f,$$whence
\begin{thm}\label{rul:interval_reversing} $\dint _a ^b f = -\dint _b ^a f$.
\end{thm}
Given constants $a$ and $b$ how are $\dint _a f$ and $\dint _b f$
related? We have $$\dint _a ^b f = \dint _a ^t + \dint _t ^b f =
\dint _a ^t - \dint _b ^t  f \implies \dint _a ^t = \dint _a ^b f +
\dint _b ^t f,
$$hence $\dint _a f$ and $\dint _b f$ differ by the constant $\dint _a ^b
f$.




 \vspace{2cm}
\begin{figure}[h]
\begin{minipage}{.45\textwidth}
\centering\psset{unit=2.5pc}
\begin{pspicture}(0,0)(5,5)
\psaxes[linewidth=2pt,labels=none,ticks=none,
arrows=->](0,0)(-1,-1)(5,5)
\psplot[algebraic,linewidth=2pt,linecolor=blue]{1.190905958}{5}{x/5-1/(x-1)+5}
\pstGeonode[PointName=none](2,4.4){A}(4,5.466666667){B}(2,0){A'}(4,0){B'}(1.190905958,0){X}
\pscustom[fillstyle=solid,fillcolor=yellow]{\psplot[algebraic,linewidth=2pt,linecolor=blue]{1.190905958}{2}{x/5-1/(x-1)+5}\psline[linewidth=2pt,linecolor=blue](A)(A')(X)}
\uput[d](A'){$t$} \uput[d](X){$a$}\uput[d](1.6,1.4){$\dint _a
^tf$}\uput[d](B'){$t+\varepsilon$}\end{pspicture}
\meinecaption{1}{Fundamental Theorem of
Calculus.}\label{fig:Barrow1}
\end{minipage}
\hfill
\begin{minipage}{.45\textwidth}
\centering
\begin{pspicture}(0,0)(5,5)
\psset{unit=2.5pc} \psaxes[linewidth=2pt,labels=none,ticks=none,
arrows=->](0,0)(-1,-1)(5,5)
\psplot[algebraic,linewidth=2pt,linecolor=blue]{1.190905958}{5}{x/5-1/(x-1)+5}
\pstGeonode[PointName=none](2,0){A'}(4,0){B'}(1.190905958,0){X}
\pstGeonode[PosAngle={0,135,90}](4,4.4){C}(2,4.4){A}(4,5.466666667){B}
\pscustom[fillstyle=solid,fillcolor=green]{\psline[linewidth=2pt,linecolor=blue](A')(A)\psplot[algebraic,linewidth=2pt,linecolor=blue]{2}{4}{x/5-1/(x-1)+5}\psline[linewidth=2pt,linecolor=blue](B)(B')(X)}
\uput[d](A'){$t$}\uput[d](X){$a$}
\uput[d](B'){$t+\varepsilon$}\uput[d](3,3){$\dint _t
^{t+\varepsilon}f$}
\pcline[offset=10pt]{|-|}(A')(A)\lput*{:U}{$f(t)$}
\pspolygon*[linewidth=2pt,linecolor=magenta](A)(B)(C)
\end{pspicture}\meinecaption{1}{Fundamental Theorem of
Calculus.}\label{fig:Barrow2}
\end{minipage}
\end{figure}


\bigskip

The area in diagram \ref{fig:Barrow2} below is given exactly by
$$\dint _t ^{t+\varepsilon} f=\dint _a ^{t+\varepsilon} f - \dint _a ^{t} f.   $$
If $\varepsilon$ is very small, the region is nearly trapezoidal,
composed of a rectangle of height $f(t)$ and width $\varepsilon$ and
$\triangle ABC$ of height $f(t+\varepsilon)-f(t)$ and width
$\varepsilon$. Hence, for $\varepsilon$ very small
$$ \dint _a ^{t+\varepsilon} f - \dint _a ^{t} f \approx f(t)\varepsilon + \dfrac{\varepsilon}{2}(f(t+\varepsilon)-f(t)). $$
If, besides, $f$ is continuous, then, as $\varepsilon\to 0$,
$f(t+\varepsilon)-f(t)=\soo{1}$. Whence we deduce,
$$ \dint _a ^{t+\varepsilon} f - \dint _a ^{t} f \approx f(t)\varepsilon + \dfrac{\varepsilon}{2}(f(t+\varepsilon)-f(t))= f(t)h+\soo{\varepsilon}. $$
If we put $I(t)=\dint _a ^{t} f$, what we are saying is that as
$\varepsilon\to 0$,
$$I(t+\varepsilon)\approx I(t)+f(t)\varepsilon+\soo{\varepsilon} \implies I'(t) \approx f(t).  $$
Upon replacing the approximation symbol with true equality we obtain
the following link between derivatives and integrals.

\bigskip

\begin{thm}[First Form of the Fundamental Theorem of Calculus]
If $f$ is continuous on $\cc{a}{b}$ and $x\in \cc{a}{b}$ then
$$I'(t)= \diff{\dint _a ^t f}{t}=f(t). $$\label{thm:fund_thm_calc_I}
\end{thm}


\begin{df}
A {\em primitive} (or {\em antiderivative}) of a function $f$ is a
function $F$ such that $F'=f$.
\end{df}
By the First Form of the Fundamental Theorem of Calculus,
$I(t)=\dint _a ^t f(x)\d{x}$ is a primitive of $f(x)$. If $c$ is a
constant and $F(t)=\dint _c ^t f(x)\d{x}$, then
$$I(t) - F(t) = \dint _a ^t f(x)\d{x}-\dint _c ^t f(x)\d{x} = \dint _a ^t f(x)\d{x}+\dint _t ^c f(x)\d{x} =\dint _a ^c f(x)\d{x},  $$
whence $I(t)$ and $F(t)$ differ by a constant. Since by Lemma
\ref{lem:D(k)=0}, the derivative of a constant function is $0$, we
have
$$(I(t)-F(t))'=0.  $$
Assuming for the moment that the derivative of a sum is the sum of
the derivatives (a fact that will be proved in Theorem
\ref{thm:sum_rule}), we have
$$I'(t)-F'(t)=(I(t)-F(t))'=0\implies I'(t)=F'(t),   $$
which means that any two primitives of $f$ only differ by a
constant. This means that we may write
$$F(t)= \dint _a ^t f(x)\d{x} +k,  $$
for some constant $k$. This in turn gives
$$F(b)-F(a)= \left(\dint _a ^b f(x)\d{x} +k\right)-\left(\dint _a ^a f(x)\d{x} +k\right)=\dint _a ^b f(x)\d{x},  $$
since by Theorem \ref{rul:integral_of_a_point}, $\dint _a ^a
f(x)\d{x}=0$. Furthermore, since
$$  \diff{F(x)}{x} =f(x),$$let us write then
\begin{equation}\d{F(x)} =f(x)\d{x}.  \label{eq:differential}\end{equation} We have then demonstrated
the following theorem.

\begin{thm}[Second  Form of the Fundamental Theorem of Calculus]
Let the  function $f$ be continuous on $\cc{a}{b}$ and let $F$ be a
primitive of $f$. Then
$$ \dint _a ^b f(x) \d{x}= \dint _a ^b  \d{F(x)}= F(x)\Big| _a ^b = F(b)-F(a). $$\label{thm:fund_thm_calc_II}
\end{thm}




\begin{exa}
Since $x\mapsto \dfrac{x+1}{x^8+1}$ is continuous everywhere, the
First Form of Fundamental Theorem of Calculus asserts that
$$  \diff{\dint _a ^x \dfrac{t+1}{t^8+1} \d{t}}{x}=\dfrac{x+1}{x^8+1}. $$
Notice that at this stage, we have no idea how to obtain $\dint _a
^x \dfrac{t+1}{t^8+1}\d{t}$.
\end{exa}



\vspace{2cm}
\begin{figure}[h]
\begin{minipage}{.3\textwidth}
\centering\psset{unit=1pc}\psaxes[labelFontSize=\tiny]{->}(0,0)(-5.5,-5.5)(5.5,5.5)
\psgrid[subgriddiv=0,griddots=4,gridlabels=0](-5,-5)(5,5)
\psplot[linecolor=blue,algebraic,linewidth=2pt]{-5}{5}{IfTE(x<-2,3+x,IfTE(x<2,1,3-x))}%
 \meinecaption{3}{$x\mapsto
f(x)$.}\label{fig:graphical_integration3-1}
\end{minipage}
\hfill
\begin{minipage}{.3\textwidth}
\centering\psset{unit=1pc}\psaxes[labelFontSize=\tiny]{->}(0,0)(-5.5,-5.5)(5.5,5.5)
\psgrid[subgriddiv=0,griddots=4,gridlabels=0](-5,-5)(5,5)
\psplot[linecolor=blue,algebraic,linewidth=2pt]{-5}{5}{IfTE(x<-2,3*x+(x^2)/2+5/2,IfTE(x<2,x+1/2,3*x-(x^2)/2-3/2))}%
\psdots(-5,0)(-4,-1.5)(-3,-2)(-2,-1.5)(-1,-.5)(0,.5)(1,1.5)(2,2.5)(3,3)(4,2.5)(5,1)
 \meinecaption{3}{$t\mapsto \dint _{-5} ^t f
$.}\label{fig:graphical_integration3-2}
\end{minipage}
\hfill
\begin{minipage}{.3\textwidth}
\centering\psset{unit=1pc}\psaxes[labelFontSize=\tiny]{->}(0,0)(-5.5,-5.5)(5.5,5.5)
\psgrid[subgriddiv=0,griddots=4,gridlabels=0](-5,-5)(5,5)
\psplot[linecolor=blue,algebraic,linewidth=2pt]{-5}{5}{IfTE(x<-2,3+x,IfTE(x<2,1,3-x))}%
\psdots(-5,-2)(-4,-1)(-3,0)(-2,1)(-1,1)(0,1)(1,1)(2,1)(3,0)(4,-1)(5,-2)
 \meinecaption{3}{$t\mapsto \diff{\dint _{-5} ^t
f}{t} $.}\label{fig:graphical_integration3-3}
\end{minipage}
\end{figure}

\bigskip

\begin{exa}\label{exa:graphical_integration3}
Given the curve $f:\cc{-5}{5}\to \reals$ in figure
\ref{fig:graphical_integration3-1} we have the graph of a function
$f$. We tabulate approximate values of $\dint _{-5}f$ below, and
plot these in the graph \ref{fig:graphical_integration3-2}.
$$ \begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|}\hline t & -5 & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline I(t)=\dint _{-5} ^tf &  0 & -\dfrac{3}{2} & -2&  -\dfrac{3}{2} &  -\dfrac{1}{2} & \dfrac{1}{2}  &  \dfrac{3}{2} & \dfrac{5}{2} & 3&\dfrac{5}{2}  & 1\\
\hline
 \end{array} $$
 We then tabulate approximate values of $I'(t)$, and plot them in
 the graph
 \ref{fig:graphical_integration3-3}.
$$ \begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|}\hline t & -5 & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline I'(t) &   -2&  -1 & 0&  1& 1& 1& 1& 1& 0& -1&  -2 \\
\hline
 \end{array} $$
 It seems that the graphs of $x\mapsto
 f(x)$ and $x \mapsto I'(x)$ resemble one another, giving empirical
 evidence of the First Form Fundamental Theorem of Calculus.
\end{exa}



\subsection*{Homework}\addcontentsline{toc}{subsection}{Homework}
\begin{pro}\label{pro:graphical_integration4-1}
For the function $f:\cc{-1}{1}\to \reals$ in figure
\ref{fig:graphical_integration4-1} below, graph $F: x\mapsto \dint
_{-1} ^x f$, $x\in \cc{-1}{1}$ and $F': x\mapsto \diff{\dint _{-1}
^x f}{x}$, $x\in \cc{-1}{1}$.
\end{pro}

\vspace{2cm}
\begin{figure}[h]
\begin{minipage}{.4\textwidth}
\centering\psset{unit=4pc}\psaxes[labels=none,linewidth=2pt,
ticks=none]{->}(0,0)(-1.1,-1.1)(1.1,1.1)
\psgrid[subgriddiv=5,griddots=4,gridlabels=0](-1,-1)(1,1)
\psplot[linecolor=blue,algebraic,linewidth=2pt]{-1}{1}{x}%
\psaxes[labelFontSize=\tiny,Ox=-1,Oy=-1,showorigin=false](-1,-1)(-1,-1)(1,1)
 \meinecaption{3}{Problem \ref{pro:graphical_integration4-1}.}\label{fig:graphical_integration4-1}
 \end{minipage}
 \hfill
 \begin{minipage}{.4\textwidth}
$$\psset{unit=1pc}  \psaxes[labelFontSize=\tiny,linewidth=2pt]{->}(0,0)(-6,-6)(6,6)\psline[linewidth=1.4pt,
linecolor=red](-5,-2)(-4,2)(-1,2)(1,-2)(4,-2)(5,2)
\psdots[dotscale=1,dotstyle=*](-5,-2)(-4,2)(-1,2)(1,-2)(4,-2)(5,2)$$
\meinecaption{2}{$f$ for problem
\ref{pro:f_difff_intf}.}\label{fig:f_difff_intf}

\end{minipage}
\end{figure}

\begin{pro}\label{pro:f_difff_intf}
Consider the function $f$ in  figure \ref{fig:f_difff_intf} below.
You may assume it is composed solely of straight line segments.
\begin{enumerate}
\item Draw $f'$, the derivative of $f$.
\item Complete the following table.
\renewcommand{\arraystretch}{2.5}
\begin{table}[h]
$$\begin{array}{|l|l|l|l|l|}
\hline \int _{-5} ^{-5} f = \hspace{1cm} & \int _{-5} ^{-4} f =
\hspace{1cm} & \int _{-5} ^{-3} f = \hspace{1cm} & \int _{-5} ^{-2}
f =
\hspace{1cm} & \int _{-5} ^{-1} f = \hspace{1cm} \\
\hline \int _{-5} ^{0} f = \hspace{1cm} & \int _{-5} ^{1} f =
\hspace{1cm} & \int _{-5} ^{2} f = \hspace{1cm} & \int _{-5} ^{3} f
=
\hspace{1cm} & \int _{-5} ^{4} f = \hspace{1cm} \\
\hline
\end{array}$$
\end{table}

\item \label{pro:table_for_integral_graph}Complete the following table.
\begin{table}[h]
$$\begin{array}{|l|l|l|l|l|}
\hline \int _{-5} ^{-5} f'  = \hspace{1cm} & \int _{-5} ^{-4} f'  =
\hspace{1cm} & \int _{-5} ^{-3} f'  = \hspace{1cm} & \int _{-5}
^{-2} f' =
\hspace{1cm} & \int _{-5} ^{-1} f'  = \hspace{1cm} \\
\hline \int _{-5} ^{0} f'  = \hspace{1cm} & \int _{-5} ^{1} f'  =
\hspace{1cm} & \int _{-5} ^{2} f'  = \hspace{1cm} & \int _{-5} ^{3}
f'  =
\hspace{1cm} & \int _{-5} ^{4} f'  = \hspace{1cm} \\
\hline
\end{array}$$
\end{table}

\item Using the table in part \ref{pro:table_for_integral_graph} or
otherwise, draw the graph of $\int _{-5} f'$.

\end{enumerate}
\end{pro}
\chapter{Polynomial Functions}\thispagestyle{empty}
In this chapter we study polynomials and their graphs. In order to
do the latter, we demonstrate the Power Rule, the Sum Rule, the
Product Rule, and the Chain Rule for derivatives. We also study the
integration of polynomial functions and some algebraic topics
related to the roots of a polynomial.
\section{Differentiating Power Functions}
By a {\em power function} we mean a function 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}

\bigskip
If $n$ is a positive integer, we are interested in how to graph
$x\mapsto x^n$. 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. We reproduce their graphs below in figures
\ref{fig:poly_y=1} through \ref{fig:poly_y=x} for easy reference.

\vspace{2cm}
\begin{figure}[h]
\begin{minipage}{.4\textwidth}
$$\psset{algebraic=true, unit=2pc} \psaxes[labels=none,
linewidth=1.5pt]{->}(0,0)(-2.5,-2.5)(2.5,2.5) \psline[linewidth=2pt,
linecolor=red]{<->}(-2.5,1)(2.5,1)
$$
\meinecaption{2}{$x\mapsto 1$.} \label{fig:poly_y=1}
\end{minipage}\hfill
\begin{minipage}{.4\textwidth}
$$\psset{algebraic=true, unit=2pc} \psaxes
[labels=none, linewidth=1.5pt]{->}(0,0)(-2.5,-2.5)(2.5,2.5)
\psline[linewidth=2pt, linecolor=red]{<->}(-2.5,-2.5)(2.5,2.5)
$$
\meinecaption{2}{$x\mapsto x$.} \label{fig:poly_y=x}
\end{minipage}
\end{figure}

\bigskip

By the groundwork from the preceding chapter, we know we can gather
information about the monotonicity and convexity of the function
$x\mapsto x^n$ by studying its first and second derivatives. For
that we first establish a series of lemmata.

\begin{lem}\label{exa:D(x^2)=2x} The strong derivative of the square function
$x\mapsto x^2$ is the function $x\mapsto 2x$.
\end{lem}
\begin{pf}
If $ f(x)=x^2$ we have
$$f(x+\epsilon) = (x+\epsilon)^2=x^2+2x\epsilon+\epsilon^2 =f(x)+2x\epsilon+\soo{\varepsilon},
$$proving the assertion. For ellipsis we will write
 $(x^2)' = 2x$ or $\diff{x^2}{x} = 2x$.
\end{pf}
\begin{lem}\label{exa:D(x^3)=3x^2} The strong derivative of the
cubic function $x\mapsto x^3$ is the function $x\mapsto 3x^2$.
\end{lem}
\begin{pf}If $ f(x)=x^3$ we have, using Lemma
\ref{exa:D(x^2)=2x},
$$\begin{array}{lll}(x+\epsilon)^3& = & (x+\epsilon)(x^2+2x\epsilon+\soo{\varepsilon})\\
& = & x^3 + 2x^2\epsilon +O(x\epsilon^2)+ x^2\epsilon +
2x\epsilon^2+ O(\epsilon ^3) \\
& = & x^3 + 3x^2\epsilon +\soo{\varepsilon},
\end{array}$$as $\tends{\epsilon}{0}$, and  so $(x^3)' = 3x^2$ (or $\diff{x^3}{x} = 3x^2$).
\end{pf}
We will now see that the pattern
$$\diff{x^0}{x} = 0, \qquad \diff{x^1}{x} = 1, \qquad \diff{x^2}{x} = 2x, \qquad \diff{x^3}{x} = 3x^2,   $$
is preserved for higher powers of the exponent. Arguing as in Lemma
\ref{exa:D(x^3)=3x^2}, we obtain the following theorem.
\begin{thm}[Power Rule] If $n$ is a positive integer, $\diff{x^n}{x} =
nx^{n-1}$. \label{thm:power_rule} \index{differentiation!power rule}
\end{thm}

\begin{pf} If the strong derivative  of $\diff{x^n}{x} = d_n(x)$, we have
$$\begin{array}{lll} (x+\epsilon)^{n+1}&=&(x+\epsilon)\left(x^n+d_n(x)\epsilon
+\soo{\varepsilon}\right)\\
&=&x^{n+1}+\left(xd_n(x)+x^n\right)\epsilon+\soo{\varepsilon}\\
\end{array}$$from where the strong derivative  of $x^{n+1}$ is $xd_n(x)+x^n$. Since $d_1(x) = 1$, we have by recurrence,
 $$\begin{array}{llllll} d_2(x) & = & xd_1(x)+x^1 & = x\cdot 1 + x & = & 2x\\
d_3(x) & = & xd_2(x)+x^2 & = x\cdot (2x) + x^2 & = & 3x^2\\
d_4(x) & = & xd_3(x)+x^3 & = x\cdot (3x^2) + x^3 & = & 4x^3\\
d_5(x) & = & xd_4(x)+x^4 & = x\cdot (4x^3) + x^4 & = & 5x^4\\
d_6(x) & = & xd_5(x)+x^5 & = x\cdot (5x^4) + x^5 & = & 6x^5\\
d_7(x) & = & xd_6(x)+x^6 & = x\cdot (6x^5) + x^6 & = & 7x^6,\\
 \end{array} $$
and so, by recursion, $d_n(x)=nx^{n-1}$.
\end{pf}
We may express the above result in the ``differential'' form
\begin{equation}
\d{x^n}=nx^{n-1}\d{x}.
\end{equation}

\begin{exa}
We have $$ \diff{x^3}{x} = 3x^2, \qquad \diff{x^7}{x} = 7x^6, \qquad
\diff{x^{1000}}{x} = 1000x^{999},
$$etc.

\end{exa}


\section{Graphs of Power Functions}
In this section we discuss how to graph functions with assignment
rules of the form $x\mapsto x^n$, where $n\in \naturals$. Observe
that the natural domain of definition for such rules is $\reals$.

\bigskip

We already know that the graphs for the cases $n=0$ and $n=1$ are
straight lines. Hence we will focus on the cases when $n \geq 2$.


 \vspace{2cm}
\begin{figure}[h]
\begin{minipage}{.3\textwidth}
$$\psset{algebraic=true, unit=2pc} \psaxes[labelFontSize=\tiny,linewidth=2pt]{->}(0,0)(-2.5,-2.5)(2.5,2.5) \psplot[linewidth=2pt,
linecolor=red]{-1.2}{1.2}{x^2}
$$
\meinecaption{2}{$y = x^2$.} \label{fig:xmapsx^2}
\end{minipage}\hfill
\begin{minipage}{.3\textwidth}
$$\psset{algebraic=true, unit=2pc} \psaxes
[labelFontSize=\tiny,linewidth=2pt]{->}(0,0)(-2.5,-2.5)(2.5,2.5)
\psplot[linewidth=2pt, linecolor=red]{-1.2}{1.2}{x^4}
$$
\meinecaption{2}{$y = x^4$.} \label{fig:xmapsx^4}
\end{minipage}\hfill
\begin{minipage}{.3\textwidth}
$$\psset{algebraic=true, unit=2pc} \psaxes
[labelFontSize=\tiny,linewidth=2pt]{->}(0,0)(-2.5,-2.5)(2.5,2.5)
\psplot[linewidth=2pt, linecolor=red]{-1.2}{1.2}{x^6}
$$
\meinecaption{2}{$y = x^6$.} \label{fig:xmapsx^6}
\end{minipage}\end{figure}




\bigskip

We will see in Theorem \ref{thm:graphs_power_functions} that for
even $n \geq 2$, the graphs of $x\mapsto x^n$ all resemble one
another, and look like those in figures \ref{fig:xmapsx^2} through
\ref{fig:xmapsx^6}. 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 will also  see that for odd $n \geq 3$, the graphs of $x\mapsto
x^n$ all resemble one another, and look like those in figures
\ref{fig:xmapsx^3} through \ref{fig:xmapsx^7}.


\vspace{2cm}
\begin{figure}[h]
\begin{minipage}{.3\textwidth}
$$\psset{algebraic=true, unit=2pc} \psaxes
[labelFontSize=\tiny,linewidth=2pt]{->}(0,0)(-2.5,-2.5)(2.5,2.5)
\psplot[linewidth=2pt, linecolor=red]{-1.2}{1.2}{x^3}
$$
\meinecaption{2}{$y = x^3$.} \label{fig:xmapsx^3}
\end{minipage}\hfill
\begin{minipage}{.3\textwidth}
$$\psset{algebraic=true, unit=2pc} \psaxes
[labelFontSize=\tiny,linewidth=2pt]{->}(0,0)(-2.5,-2.5)(2.5,2.5)
\psplot[linewidth=2pt, linecolor=red]{-1.2}{1.2}{x^5}
$$
\meinecaption{2}{$y = x^5$.} \label{fig:xmapsx^5}
\end{minipage}\hfill
\begin{minipage}{.3\textwidth}
$$\psset{algebraic=true, unit=2pc} \psaxes
[labelFontSize=\tiny,linewidth=2pt]{->}(0,0)(-2.5,-2.5)(2.5,2.5)
\psplot[linewidth=2pt, linecolor=red]{-1.2}{1.2}{x^7}
$$
\meinecaption{2}{$y = x^7$.} \label{fig:xmapsx^7}
\end{minipage}

\end{figure}


We now address the problem of how to graph $x\mapsto x^n$.
\begin{thm}\label{thm:graphs_power_functions}
Let $n \geq 2$ be an integer and $f(x) = x^n$. Then
\begin{itemize}
\item if $n$ is even, $f$ is convex, $f$ is decreasing for $x<0$,
and $f$ is increasing for $x>0$. Also, $f(-\infty) = f(+\infty) =
+\infty$.
\item 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{itemize}
\end{thm}
\begin{pf}If $n \geq 2$ is even, $n-1 \geq 1$ is odd, and $n-2$ is even. Now $f'(x) =
nx^{n-1}$ and $f''(x) = n(n-1)x^{n-2}$. Since $x^{n-2}>0$ for all
$x\neq 0$, $f''(x) > 0$ for $x\neq 0$ and so it is convex. Since $x
< 0 \implies x^{n-1}<0 \implies f'(x) <0$, $f$ is decreasing for
$x<0$. A similar argument shews that $f$ is increasing for $x> 0$.
It is clear that  $f(-\infty) = f(+\infty) = +\infty$.

\bigskip

If $n \geq 3$ is odd, $n-1 \geq 1$ is even, and $n-2$ is odd. Now
$f'(x) = nx^{n-1}$ and $f''(x) = n(n-1)x^{n-2}$. Since $x^{n-1}>0$
for all $x\neq 0$, $f'(x) > 0$ for $x\neq 0$ and so it is
increasing. Since $x < 0 \implies x^{n-2}<0 \implies f''(x) <0$, $f$
is concave for $x<0$. A similar argument shews that $f$ is convex
for $x> 0$. It is clear that $f(-\infty)= -\infty$ and $ f(+\infty)
= +\infty$.
\end{pf}
\subsection*{Homework}\addcontentsline{toc}{subsection}{Homework}
\begin{pro}
Graph the following transformations of power functions.
\begin{enumerate}
\item $x\mapsto (x+1)^2$
\item $x\mapsto x^2+1$
\item $x\mapsto (x+1)^2+1$
\item $x\mapsto -x^3$
\item $x\mapsto 1-x^3$
\item $x\mapsto 1-(x-1)^3$
\end{enumerate}
\end{pro}
\section{Integrals of Power Functions}
Let  $n\in \naturals$ and $$ F(x)=\dfrac{x^{n+1}}{n+1}. $$ Since $$
F'(x) = x^n,$$ $F$ is a primitive of $f:x\mapsto x^n$ and so by the
Second Form of the Fundamental Theorem of Calculus (Theorem
\ref{thm:fund_thm_calc_II}), we deduce the following result.
\begin{thm}\label{thm:int_power_func}
Let $n\in \naturals$. Then $$\dint _a ^bx^n\d{x}=\dint _a
^b\d{\dfrac{x^{n+1}}{n+1}}= \dfrac{x^{n+1}}{n+1}\Big|_a ^b =
\dfrac{b^{n+1}}{n+1}-\dfrac{a^{n+1}}{n+1}.$$
\end{thm}

Theorem \ref{thm:int_power_func} allows us to find areas of
``curvy'' regions.

\begin{exa}\label{exa:area_under_parabola}
Find the area of the plane region
$$ \{(x, y)\in \reals ^2: 1\leq x \leq 2, 0 \leq y \leq x^2\}. $$
\end{exa}
\begin{solu}
This is the area under the curve $y=x^2$ and above the $x$-axis,
bounded by the vertical lines $x=1$ and $x=2$, as depicted in figure
\ref{fig:area_under_parabola}. By Theorem \ref{thm:int_power_func},
this is
$$ \dint _1 ^2 x^2 \d{x} = \dfrac{x^3}{3}\Big| _1 ^2 = \dfrac{8}{3}-\dfrac{1}{3}=\dfrac{7}{3}. $$
\end{solu}
\vspace{2cm}
\begin{figure}[h]
\centering \psset{unit=1pc}
\psaxes[linewidth=2pt,arrows=->,labelFontSize=\tiny](0,0)(-5,-5)(5,5)
\pscustom[fillcolor=red,fillstyle=solid]{\psline(1,0)(1,1)\psplot[algebraic]{1}{2}{x^2}\psline(2,4)(2,0)(1,0)}
\psplot[linewidth=2pt,linecolor=red,arrows=<->,algebraic]{-2.2}{2.2}{x^2}
\meinecaption{2}{Example
\ref{exa:area_under_parabola}.}\label{fig:area_under_parabola}
\end{figure}
\subsection*{Homework}\addcontentsline{toc}{subsection}{Homework}
\begin{pro}
Let $a\in \reals$ be a constant and let $n\in \naturals$. Prove that
if $n$ is even then $$\dint _{-a} ^a x^n\d{x} =\dfrac{2}{n+1}a^{n+1}
$$ and that if $n$ is odd then $$\dint _{-a} ^a x^n\d{x} =0.
$$
\end{pro}

\section{Differentiating Polynomials}
In this section we prove some standard  differentiation rules.
\begin{thm}[Constant-Times-Function Rule]If $k\in\reals$ is a real
number constant, and $f$ is a strongly differentiable   function at
$x$, then $kf$ is strongly differentiable  at $x$ and $(kf)'(x) =
kf'(x)$. Alternatively, in differential notation,
$$ \d{kf(x)}=k\d{f(x)}.  $$


\label{thm:constant-times-function_rule}\index{differentiation!constant
times a function rule}
\end{thm}
\begin{pf}We have
$$\begin{array}{lll}
(kf)(x+\epsilon)& = & k(f(x+\epsilon)) \\
& = & k\left(f(x)+f'(x)\epsilon +
\soo{\varepsilon}\right) \\
& = & kf(x) +kf'(x)\epsilon + \soo{\varepsilon},
\end{array}$$from where the theorem follows.
\end{pf}



\begin{exa}
We have proved that $(x^2)' = 2x$. Hence $(-3x^2)' = -3(x^2)' =
-3(2x) = -6x$.
\end{exa}


\begin{thm}[Sum Rule]If $f, g$ are strongly differentiable   functions at $x$,
then $f+g$ is strongly differentiable  at $x$ and $$(f+g)'(x) =
f'(x)+g'(x).$$ Alternatively, in differential notation,
$$ \d{(f(x)+g(x))}=\d{f(x)}+\d{g(x)}.  $$\label{thm:sum_rule}\index{differentiation!sum rule}
\end{thm}
\begin{pf}We have
$$\begin{array}{lll}
f(x+\epsilon)+g(x+\epsilon)& = &(f(x)+f'(x)\epsilon +
\soo{\varepsilon})+(g(x)+g'(x)\epsilon + \soo{\varepsilon}) \\
& = & (f(x)+g(x)) +(f'(x)+g'(x))\epsilon + \soo{\varepsilon},
\end{array}$$from where the theorem follows.
\end{pf}

\begin{exa}
We have proved that $(x^2)' = 2x$ and that $(x)' = 1$. Hence $(x^2 +
x)' = (x^2)' + (x)'= 2x + 1$.
\end{exa}

\begin{exa} Let $f(x)=2x^3 -x^2 + 5x - 1$. Find $f'(1)$ and $f''(-1)$. \end{exa}
\begin{solu}We have $$f'(x)= 2(3x^2) -2x + 5 = 6x^2 -2x + 5,\qquad f''(x) =
2(6x) -2 = 12x -2.$$ Hence $f'(1) = 9$ and $f''(-1)= -14$.
\end{solu}


We now develop tools for differentiating more complex formul\ae.
\begin{thm}[Product Rule]If $f, g$ are are strongly differentiable   functions at $x$,
then $fg$ is strongly differentiable  at $x$ and then $(fg)'(x) =
f'(x)g(x) + f(x)g'(x)$. Alternatively, in differential notation,
$$ \d{f(x)g(x)}=f(x)\d{g(x)}+g(x)\d{f(x)}.  $$\label{thm:product_rule}

\end{thm}
\begin{pf}We have
$$\begin{array}{lll}f(x+\epsilon)g(x+\epsilon)& = &  \left(f(x)+f'(x)\epsilon + \soo{\varepsilon}\right)
\left(g(x)+g'(x)\epsilon + \soo{\varepsilon}\right) \\
& = & f(x)g(x)+\left(f'(x)g(x)+f(x)g'(x)\right)\epsilon
+\soo{\varepsilon}, \end{array}$$from where the theorem follows.
\end{pf}
\begin{rem}
It is {\bf not true} in general that $(fg)' = f'g'$.
\end{rem}
\begin{exa}
Let $f(x) = x^3$ and $g(x) = x^4$. Then from the Product Rule
$$ (x^7)' = (x^3\cdot x^4)' = x^4(x^3)' + x^3(x^4)' = x^4(3x^2) + x^4(4x^2) = 3x^6 + 4x^6 = 7x^6,  $$
which is what we expect from the Power Rule.
\end{exa}
By recurrence we can apply the product rule to more than two
functions.
\begin{exa}
$$\begin{array}{lll}(x(x+x^2)(1+x+x^2))' & = & (x)'(x+x^2)(1+x+x^2) + x(x+x^2)'(1+x+x^2) +
x(x+x^2)(1+x+x^2)'\\
& = & 1(x+x^2)(1+x+x^2)+ x(1+2x)(1+x+x^2) + x(x+x^2)(1+
2x)\\
& = & (x+2x^2+2x^3+x^4) +  (x+3x^2+3x^3+2x^4) + (x^2+3x^3+2x^4) \\
& = & 2x + 6x^2+8x^3 +5x^4.
\end{array}$$
\end{exa}


\begin{thm}[Chain Rule] If $g$ is strongly differentiable  at $x$ and  $f$ is  strongly differentiable  at $g(x)$,
then $f\circ g$ is strongly differentiable  at $x$ and $(f\circ
g)'(x)=f'\left(g(x)\right)g'(x)$.
 Alternatively, in differential notation,
$$ \d{(f(g(x)))}=f'(g(x))\d{g(x)}.  $$




\label{thm:chain_rule}
\end{thm}
\begin{pf}We have, putting $\epsilon_1 = g'(x)\epsilon +
\soo{\varepsilon}$,

$$\begin{array}{lll}f\left(g(x+\epsilon)\right)
& = & f(g(x)+g'(x)\epsilon + \soo{\varepsilon})
\\ & = &
f(g(x) + \epsilon_1)\\
& = & f(g(x)) + f'(g(x))\epsilon_1 + O(\epsilon_1 ^2) \\
& = & f(g(x)) + f'(g(x))(g'(x)\epsilon + \soo{\varepsilon}) + O((g'(x)\epsilon + \soo{\varepsilon}) ^2) \\
 & = & f\left(g(x)\right)+f'\left(g(x)\right)g'(x)\epsilon
+\soo{\varepsilon}\end{array}$$ and the theorem follows.
\end{pf}

\begin{exa}Consider $h(x) = (x+1)^2$. Then $h(x) = (f\circ g)(x)$ with
$f(x)=x^2$ and $g(x)=x+1$. Hence
$$((x+1)^2)' = (f\circ g)'(x) = f'(g(x))g'(x) =  2(x+1)^1(1) = 2x+2. $$
\end{exa}
\begin{exa}
$$((x^2 + x)^3)' = 3(x^2+x)^2(2x+1). $$
\end{exa}

\begin{exa}Using the Product Rule and the Chain Rule,
$$ (x(x+a)^2)' = (x+a)^2(x)' + x((x+a)^2)' = (x+a)^2 + 2x(x+a) = (x+a)(3x+a).  $$
\end{exa}
\begin{exa}Using the Product Rule and the Chain Rule,
$$ \begin{array}{lll}
(x(x+a)^2(x+b)^3)' & = & (x+a)^2(x+b)^3(x)' +
x(x+b)^3((x+a)^2)'+ x(x+a)^2((x+b)^3)' \\
& = & (x+a)^2(x+b)^3 + 2x(x+a)(x+b)^3 + 3x(x+a)^2(x+b)^2\\
& = & (x+a)(x+b)^2((x+a)(x+b) + 2x(x+b)+ 3x(x+a)) \\
& = & (x+a)(x+b)^2(6x^2+x(4a+3b)+ab).
\end{array}
$$
\end{exa}
\begin{exa}Let $f$ be strongly differentiable  with $f(4) = a$ and  $f'(4) =
b$. If $g(x) = x^2f(x^2)$, find $g'(2)$.
\end{exa}\begin{solu} Using both the Product Rule and the Chain Rule $$g'(x) = 2xf(x^2) + x^2f'(x^2)(2x) =
2xf(x^2) + 2x^3f'(x^2).$$ Hence $g'(2) = 2(2)f(4) + 2(8)f'(4) = 4a +
16b$.
\end{solu}

In some situations we may apply the Fundamental Theorem of Calculus.
\begin{exa}\label{exa:ftc_1}
Find $\diff{\dint _0 ^{x^2}f(t)\d{t}}{x}$.
\end{exa}
\begin{solu}
Put $F(x)=\dint _0 ^{x}f(t)\d{t}$. By Fundamental Theorem of
Calculus, $F'(x)=f(x)$. By the Chain Rule,
$$\diff{\dint _0
^{x^2}f(t)\d{t}}{x}=\diff{F(x^2)}{x}=2xF'(x^2)=2xf(x^2).$$
\end{solu}
\begin{exa}Let $g(x)=x^2\dint _0 ^{x^2}f(t)\d{t}$. Find
$g'(x)$.\label{exa:ftc_2}
\end{exa}
\begin{solu}
We use the product rule and example \ref{exa:ftc_1}
$$\begin{array}{lll} g'(x) & = &
2x\dint _0 ^{x^2}f(t)\d{t} + x^2\diff{\dint _0 ^{x^2}f(t)\d{t}}{x}\\
& = & 2x\dint _0 ^{x^2}f(t)\d{t} + 2x^3f(x^2).\\
\end{array}  $$
\end{solu}

\subsection*{Homework}\addcontentsline{toc}{subsection}{Homework}

\begin{pro}Find the following derivatives.
\begin{enumerate}
\item $f'(x) $ if $f(x)= x(x-1)(x+1)$
\item $g'(x) $ if $g(x)= (ax+(bx+(cx+1)^2)^3)^4$
\item $h'(x) $ if $h(x)= (x^3 + 2)^2(x^2 + 2x + 2)^3$.
\item $u'(x) $ if $u(x)= x(x-1)^2(x+2)^3$
\item $v'(x) $ if $v(x)= 1-(1-(1-(1-(1-x^2)^2)^2)^2)^2$
\end{enumerate}
\begin{answer}
\noindent
\begin{enumerate}
\item  By the Product Rule:
$$\begin{array}{lll} f'(x) & = & (x-1)(x+1) + x(x+1) + x(x-1) \\
& = & x^2-1 + x^2+x + x^2-x\\
& = & 3x^2-1  \end{array}$$
\item  Using the Chain Rule:
$$\begin{array}{lll} g'(x) & = & 4(ax + (bx + (cx + 1)^2)^3)^3(a + 3(bx + (cx + 1)^2)^2)(b + 2c(cx + 1)) \\
 \end{array}$$
\item  Using the Product Rule and the Chain Rule:
$$
\begin{array}{lll} h'(x) & = & 2(3x^2)(x^3+2)(x^2 + 2x + 2)^3 + (x^3+2)^23(2x+2)(x^2+2x +
2)^2\\ & = & 6x^2(x^3+2)(x^2 + 2x + 2)^3+6(x^3+2)(x+1)(x^2+2x+2)^2
\end{array}$$
\item  Using the Product Rule and the Chain Rule:
$$\begin{array}{lll} u'(x) & = & (x-1)^2(x+2)^3 + 2x(x-1)(x+2)^3 + 3x(x-1)^2(x+2)^2 \end{array} $$
\item Put $f(x) =1 -x^2$. Then $v(x)= f(f(f(f(f(x)))))$. Hence
$$\begin{array}{lll}v'(x) & =  &f'(x)\cdot f'(f(x))\cdot f'(f(f(x)))\cdot f'(f(f(f(x))))\cdot f'(f(f(f(f(x)))))\\
& = & -2x\cdot (-2(1-x^2))\cdot (-2(1-(1-x^2)^2))\cdot
(-2(1-(1-(1-x^2)^2)^2))\cdot (-2(1-(1-(1-(1-x^2)^2)^2)^2))\\
& = & -32x\cdot (1-x^2)\cdot (1-(1-x^2)^2)\cdot
(1-(1-(1-x^2)^2)^2)\cdot (1-(1-(1-(1-x^2)^2)^2)^2)\\
\end{array}$$
\end{enumerate}
\end{answer}
\end{pro}
\begin{pro}Find
$a'(x)$ if $a(x) = (1+x)(1+ x + x^2)(1+x + x^2+x^4)$\begin{answer}
Use the Product Rule: $$\begin{array}{lll}a'(x) & = & (1+ x +
x^2)(1+x + x^2+x^4) + (1 + 2x)(1+ x)(1+x + x^2+x^4) + (1+x)(1+ x +
x^2)(1 + 2x+4x^3) \\ &= &
           3 + 10x + 15x^2  + 15x^4  + 16x^3  + 12x^5  + 7x^6.
\end{array}$$
\end{answer}
\end{pro}
\begin{pro}
 Let $p$ be the polynomial of degree $100$
$$a_0x^{100}+a_1x^{99}+a_2x^{98}+\cdots + a_{99}x+a_{100} = (x-2)^{50}(2x-1)^{50}.  $$
 Find  $100a_0+99a_1+98a_2+\cdots + 2a_{98}+a_{99}$.
 \begin{answer}
We have $$100a_0x^{99}+99a_1x^{98}+98a_2x^{97}+\cdots +
2a_{98}x+a_{99} = p'(x) = 50(x-2)^{49}(2x-1)^{50} +
100(x-2)^{50}(2x-1)^{49}$$ and the desired quantity is
$$100a_0+99a_1+98a_2+\cdots + 2a_{98}+a_{99} = p'(1) = -50 + 100 = 50.$$
 \end{answer}
\end{pro}
\begin{pro}
 Let $f$ be an everywhere differentiable function and let $g = f\circ f$. It is known that
$$f(1) = 2;\quad f(2) = 3; \quad
f'(1) = 1; \quad f'(2) = 1, \quad f''(1) =-1,\quad  f''(2) = -1. $$
Find  $g'(1)$ and $g''(1)$.
\begin{answer}
We have $$g(x) = f(f(x)) \implies g'(x) = f'(x)f'(f(x)) \implies
g'(1) = f'(1)f'(f(1)) = 1\cdot f'(2) = 1.  $$ Also, $$g(x) = f(f(x))
\implies g'(x) = f'(x)f'(f(x)) \implies g''(x) = f''(x)f'(f(x)) +
(f'(x))^2f''(f(x)), $$which gives
$$  g''(1) = f''(1)f'(f(1)) + (f'(1))^2f''(f(1)) = -1f'(2) +
(1)^2f''(2) = -1(1) + 1(-1) = -2.
$$
\end{answer}
\end{pro}
\begin{pro}
Consider an {\em even} function $f:\reals\rightarrow \reals$ whose
derivative $f'$ exists everywhere and which satisfies $$f(1) =
2;\quad f(-2) = 1; \quad f'(-1) = 1; \quad f'(2) = 2, \quad f''(-1)
=0; \quad f''(2)=3.
$$
If $a(x) = xf(x^2)$, find $a''(-1)$.
\begin{answer}
Since $f$ is even, $f'$ is odd and $f''$ is even. Thus we have
$$f(1) = 2; \quad f(-1)=2;\quad f(-2) = 1; \quad f(2)=1 $$ $$f'(1) = -1\quad
f'(-1) = 1; \quad f'(2) = 2;\quad f'(-2) = -2;$$  $$f''(1) =0; \quad
f''(-1) =0; \quad f''(2)=3; \quad \quad f''(-2)=3.
$$

We have $a'(x) = f(x^2) + 2x^2f'(x^2)$ and $$a''(x) = 2xf'(x^2) +
4xf'(x^2) +4x^3f''(x^2), $$ whence
$$a''(-1) = -2f'(1) -4f'(1) -4f''(1) =-2(-1)-4(-1)=6.   $$
\end{answer}
\end{pro}
\begin{pro}
 Determine $\dfrac{\d{}}{\d{x}}\dint _x ^{x^2} \sqrt{1+t^2}\d{t}$.
 \begin{answer} Put $F(x)=\dint _a ^x\sqrt{1+t^2}\d{t}$, where $a$
 is a constant. Then
 $$ \dint _x ^{x^2} \sqrt{1+t^2}\d{t} = \dint _a ^{x^2} \sqrt{1+t^2}\d{t} -\dint _a ^{x} \sqrt{1+t^2}\d{t}=F(x^2)-F(x).$$
 By the Fundamental Theorem of Calculus, $F'(x) = \sqrt{1+x^2}$, and by  the Chain Rule,
$$\diff{F(x^2)-F(x)}{x}=2xF'(x^2)-F'(x)=2x\sqrt{1+x^4}-\sqrt{1+x^2}$$
 \end{answer}
\end{pro}

\section{Integrating Polynomials}
Each of the differentiation rules given in the preceding section has
a corresponding integration equivalent.

\bigskip

We recall here that the Second Form of the Fundamental Theorem of
Calculus asserts that if $f$ is continuous in $\cc{a}{b}$ and
$F'=f$, then
\begin{equation}
\dint _a ^b f(x)\d{x}=\dint _a ^b \d{F(x)}=
F(b)-F(a).\label{eq:fund_thm_calc}
\end{equation}
Most of the integration problems we will see in this course can be
solved by reducing them to the form in \ref{eq:fund_thm_calc}. The
trick is to recognise the integrand as the derivative of some
function, most times perhaps, as the result of a differentiation
through the Chain Rule.
\begin{exa}
Let $f:\cc{1}{2}\to \reals$ be such that $f'$ is  continuous and
suppose $f(2)=a$ and $f(5)=b$. Determine $\dint _1 ^2
xf'(x^2+1)\d{x}$.
\end{exa}
\begin{solu}
We recognise the integrand $xf'(x^2+1)$ as de derivative of
$\dfrac{f(x^2+1)}{2}$. Hence
$$ \dint _1 ^2
xf'(x^2+1)\d{x} = \dint _1 ^2
\d{\left(\dfrac{f(x^2+1)}{2}\right)}=\dfrac{f(x^2+1)}{2}\Big| _1
^2=\dfrac{f(5)}{2}-\dfrac{f(2)}{2}=\dfrac{b-a}{2}.
$$
\end{solu}
\begin{exa}We have
$$ \dint _0 ^1 (2x+1)(x^2+1)^{2008}\d{x}= \dint _0  ^1 \d{\left(\dfrac{(x^2+1)^{2009}}{2009}\right)}
=\dfrac{(x^2+1)^{2009}}{2009}\Big|_0 ^1 =
\dfrac{2^{2009}}{2009}-\dfrac{1^{2009}}{2009}=\dfrac{2^{2009}-1}{2009}.$$
\end{exa}
\begin{exa}We have
$$ \begin{array}{lll}\dint _0 ^t (2008x^{2007}+1004x^{1003}+502x^{501})(x^{2008}+x^{1004}+x^{502})^{10}\d{x}
& =  & \dint _0 ^t
\d{\left(\dfrac{(x^{2008}+x^{1004}+x^{502})^{11}}{11}\right)}\\
& = & \dfrac{(x^{2008}+x^{1004}+x^{502})^{11}}{11}\Big|_0 ^t\\
& = & \dfrac{(t^{2008}+t^{1004}+t^{502})^{11}}{11}.\end{array}$$
\end{exa}


\begin{thm}[Product of Function Times a Constant Rule]
Let $f$ be continuous on $\cc{a}{b}$ and let $k\in \reals$ be a
constant. Then $$\dint _a ^b (kf(x))\d{x}=k\dint _a ^b f(x)\d{x}.$$
\end{thm}
\begin{pf}
Let $\d{F(x)} =f(x)\d{x}$ and $\d{G(x)}=kf(x)\d{x}$. By Theorem
\ref{thm:constant-times-function_rule},
$$ \d{G(x)}=\d{(kF(x))}. $$Now, by
the Second Form of the Fundamental Theorem of Calculus
$$  \dint _a ^b kf(x)\d{x}= \dint _a ^b \d{G(x)} =\dint _a ^b \d{(kF(x))}=kF(b)-kF(a)=k(F(b)-F(a))=k\dint _a ^b f(x)\d{x}. $$


\end{pf}

\begin{thm}[Integral of a Sum Rule]
Let $f$  and $g$ be continuous on $\cc{a}{b}$  Then $$\dint _a ^b
(f(x)+g(x))\d{x}=\dint _a ^b f(x)\d{x}+\dint _a ^b g(x)\d{x}.$$
\label{thm:int_of_sum}
\end{thm}
\begin{pf}
Let $\d{F(x)} =f(x)\d{x}$ and $\d{G(x)}=g(x)\d{x}$. Now, by the
Second Form of the Fundamental Theorem of Calculus
$$  \begin{array}{lll}\dint _a ^b (f(x)+g(x))\d{x} & =  & \dint _a ^b \d{(F(x)+G(x))}\\
& = &  (F(x)+G(x))\Big| _a ^b\\ &  =  & (F(b)+G(b)) - (F(a)+G(a))\\
& = & F(b)-F(a) + G(b)-G(a)\\ &  =  & \dint _a ^b f(x)\d{x}+ \dint
_a ^b g(x)\d{x}.
\end{array}$$


\end{pf}

\vspace{2cm}


\begin{figure}[h]
\centering \psset{unit=4pc}
\psaxes[linewidth=2pt,arrows=->,labelFontSize=\tiny](0,0)(-1.5,-1.5)(1.5,1.5)
\psplot[linewidth=2pt,linecolor=red,arrows=<->,algebraic]{-1.27}{1.27}{x^2}
\psplot[linewidth=2pt,linecolor=red,arrows=<->,algebraic]{-1.25}{1.25}{x^3}
\pscustom[fillcolor=yellow,fillstyle=solid]{\psplot[algebraic]{0}{1}{x^3}\psplot[algebraic]{1}{0}{x^2}}
\meinecaption{3}{Example
\ref{exa:area_between_regions}.}\label{fig:area_between_regions}
\end{figure}


\begin{exa}\label{exa:area_between_regions}
Find the area of the plane region between the curves $y=x^2$ and
$y=x^3$.\end{exa}
\begin{solu}Observe that $x^2=x^3\implies x^2(1-x)=0\implies x\in
\{0,1\}$, and so the curves meet at $x=0$ and $x=1$. In the interval
$0 \leq x \leq 1$, $y=x^2$ lies atop of $y=x^3$, as  depicted in
figure \ref{fig:area_between_regions}. By Theorems
\ref{thm:int_power_func} and \ref{thm:int_of_sum}, this is
$$ \begin{array}{lll}
\dint _0 ^1 (x^2-x^3) \d{x} & = & \dint _0 ^1 x^2 \d{x} -  \dint _0
^1 x^3 \d{x}\\
& = & \dint _0 ^1  \d{\dfrac{x^3}{3}} -  \dint _0
^1  \d{\dfrac{x^4}{4}}\\ &  = & \dfrac{x^3}{3}\Big| _0 ^1- \dfrac{x^4}{4}\Big| _0 ^1\\
&  = &  \dfrac{1}{3}-\dfrac{1}{4}\\
& = & \dfrac{1}{12}.\\ \end{array} $$
\end{solu}

We will know demonstrate one of the most powerful integration
techniques, that of Integration by Parts.

\begin{thm}[Integration by Parts]
Let $f'$  and $g'$ be continuous on $\cc{a}{b}$  Then $$\dint _a ^b
f(x)\d{g(x)}=f(x)g(x)\Big| _a ^b -\dint _a ^b g(x)\d{f(x)}.$$
\end{thm}
\begin{pf}
 By the Product Rule for
derivatives,
$$\d{(f(x)g(x))}= f(x)\d{g(x)} + g(x)\d{f(x)}.
$$
Integrating on both sides and using the Sum Rule for Integrals
(Theorem \ref{thm:int_of_sum}),
$$\begin{array}{lll}\d{(f(x)g(x))}= f(x)\d{g(x)} + g(x)\d{f(x)}
& \implies & \dint _a ^b \d{(f(x)g(x))}=  \dint _a ^b f(x)\d{g(x)} +
 \dint _a ^b g(x)\d{f(x)}\\
& \implies & f(x)g(x)\Big| _a ^b=  \dint _a ^b f(x)\d{g(x)} +
 \dint _a ^b g(x)\d{f(x)},\\
 \end{array}
$$which readily gives the result.

\end{pf}
\begin{exa}
Let $f$ be a continuously differentiable function such that $f(0)
=0$, $f(1)=1$,  and $\dint _0 ^1 f(x)\d{x}=-\dfrac{1}{2}$. Determine
$\dint _0 ^1\ xf'(x) \d{x} $.
\end{exa}
\begin{solu}
Integrating by parts,
$$ \dint _0 ^1\ xf'(x) \d{x} = \dint _0 ^1\ x \d{f(x)} = xf(x)\Big|_0 ^1 - \dint _0 ^1 f(x)\d{x}= f(1) -\dint _0 ^1 f(x)\d{x}=
1-\left(-\dfrac{1}{2}\right)=\dfrac{3}{2}.$$
\end{solu}
\begin{thm}[Integration by Substitution Rule]
Let $f'$  and $g'$ be continuous on $\cc{a}{b}$  Then $$\dint _a ^b
f(g(x))\d{g(x)}=\dint _{g(a)} ^{g(b)} f(x)\d{x}.$$
\end{thm}
\begin{pf}
Let $P'(x) =f(g(x))g'(x)\d{x}=$ and $G'(x)=kf(x)$. Then
$$kF'(x)=G'(x)
$$
\end{pf}



\subsection*{Homework}\addcontentsline{toc}{subsection}{Homework}

\begin{pro}
Evaluate $\dint _0 ^a (x+1)^{100}\d{x}$, where $a$ is a real
constant.
\begin{answer}
We have
$$ \dint _{-1} ^a (x+1)^{100}\d{x}= \dint _{-1} ^a \d{\left(\dfrac{(x+1)^{101}}{101}\right)}=\dfrac{(x+1)^{101}}{101}\Big|_{-1} ^a =\dfrac{(a+1)^{101}}{101}.  $$
\end{answer}
\end{pro}
\begin{pro}Let $f:\cc{0}{4}\to\reals$ be continuous.
If $\dint _0 ^4 f(x)\d{x}=9$, find $\dint _0 ^2 xf(x^2)\d{x}$.
\begin{answer}
$36$
\end{answer}
\end{pro}

\begin{pro}
Consider a continuously twice differentiable function $f$ such that
$f(0)=a$, $f(1)=b$, $\dint _0 ^1 f(x)\d{x}=c$, $f'(0) =2b$, and
$f'(1)=2a$. Find $\dint _0 ^1 xf'(x)\d{x}$ and $\dint _0 ^1
xf''(x)\d{x}$.
\begin{answer}
We have, integrating by parts,
$$\begin{array}{lll}\dint _0 ^1 xf'(x) \d{x} & = & \dint _0 ^1  x\d{f(x)}\\
& = & xf(x)\Big|_0 ^1 - \dint _0 ^1 f(x)\d{x} \\
& = & 1f(1) -0f(0)-c \\
& = & b-c,
\end{array}
$$ and integrating by parts twice
$$\begin{array}{lll}\dint _0 ^1 xf''(x) \d{x} & = & \dint _0 ^1  x\d{f'(x)}\\
& = & xf'(x)\Big|_0 ^1 - \dint _0 ^1 f'(x)\d{x} \\
& = & 1f'(1) -0f'(0)-(f(1)-f(0)) \\
& = & 2a-(b-a).\\
& = & 3a-b.
\end{array}
$$
\end{answer}
\end{pro}




\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, $k = 0$, then we call the function $x\mapsto
mx$ a {\em linear function}.
\index{function!linear}\index{function!affine}\index{function!constant}
\end{df}
By virtue of Theorem \ref{thm:graph_mx+k}, the graph of the function
$x\mapsto mx + k$ is a straight line. Since the derivative of
$x\mapsto mx + k$ is $(mx+k)' = m$, we see that $x\mapsto mx + k$ is
strictly increasing if $m>0$ and strictly decreasing if $m<0$ in
view of Theorem \ref{thm:monotonicity_criteria}. If $m\neq 0$ then
$mx +k = 0\implies x=-\dfrac{k}{m}$, meaning that $x\mapsto mx+k$
has a unique zero (crosses the $x$-axis) at $x=-\dfrac{k}{m}$. This
information is summarised in the following tables.

\begin{table}[h]\centering
\begin{minipage}{8cm}
$$
\begin{array}{|c|ccccc|}
 \hline
  x& -\infty &  & -\dfrac{k}{m} &  & +\infty   \\
 \hline
  &  &  &  &  &    \\

  &  &   & & \nearrow &     \\

 f(x)=mx+k &  &  & 0 &  &    \\
     & & \nearrow & &  &    \\
 \hline
            \end{array}$$ \meinecaption{1}{$x\mapsto mx+k$, with $m>0$.}
            \end{minipage}
            \begin{minipage}{8cm}
$$
\begin{array}{|c|ccccc|}
 \hline
  x& -\infty &  & -\dfrac{k}{m} &  & +\infty   \\
 \hline
  &  &  &  &  &    \\

  &  & \searrow  & &  &     \\

 f(x)=mx+k &  &  & 0 &  &    \\
     & &  & & \searrow &    \\
 \hline
            \end{array}$$ \meinecaption{1}{$x\mapsto mx+k$, with $m<0$.}
            \end{minipage}
\end{table}


\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}.
\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 example
\ref{exa:square_function} and Theorems \ref{thm:translations} and
\ref{thm:distortions}, that the graph of $f$ is a parabola.

\bigskip

We have $f'(x) = 2ax+b$. Assume first that $a>0$. Then $$f'(x) > 0
\iff x>-\dfrac{b}{2a},\qquad  f'(x) < 0 \iff x<-\dfrac{b}{2a}.$$Thus
the function decreases for values $<-\dfrac{b}{2a}$ and increases
otherwise. Hence in view of Theorem \ref{thm:local_extrema}, it must
have a minimum at $x=-\dfrac{b}{2a}$. Since $f''(x) = 2a >0$, $f$ is
convex by virtue of Theorem \ref{thm:convexity_criteria}. The case
when $a<0$ can be similarly treated.
\end{pf}The information of Theorem
\ref{thm:behaviour_of_a_quadratic} is summarised in the following
tables.
\begin{table}[h]\centering
\begin{minipage}{8cm}
$$
\begin{array}{|c|ccccc|}
 \hline
  x& -\infty &  & -\dfrac{b}{2a} &  & +\infty   \\
 \hline
  &  &  &  &  &    \\

  &  & \searrow   & & \nearrow &     \\

 f(x)=ax^2+bx+c &  &  & 0 &  &    \\
 \hline
            \end{array}$$ \meinecaption{1}{$x\mapsto ax^2+bx+c$, with $a>0$.}
            \end{minipage}
            \begin{minipage}{8cm}
$$
\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}
\end{table}

\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.
\end{rem}

\vspace{2cm}
\begin{figure}[h]
\begin{minipage}{4.5cm}$$
\psset{unit=1pc} \psaxes[labels=none, ticks=none](0,0)(-4,-4)(4,4)
\rput(-1,1){\parabola[linewidth=2pt,linecolor=red]{<->}(2,4)(0,0)}
\rput{180}(1,-1){\parabola[linewidth=2pt,linecolor=blue]{<->}(2,4)(0,0)}$$
\meinecaption{1}{No real zeroes.} \label{fig:no_real_zero}
\end{minipage} \hfill
\begin{minipage}{4.5cm}$$
\psset{unit=1pc} \psaxes[labels=none, ticks=none](0,0)(-4,-4)(4,4)
\rput(-1,0){\parabola[linewidth=2pt,linecolor=red]{<->}(2,4)(0,0)}
\rput{180}(1,0){\parabola[linewidth=2pt,linecolor=blue]{<->}(2,4)(0,0)}$$
\meinecaption{1}{One real zero.}\label{fig:one_real_zero}
\end{minipage} \hfill
\begin{minipage}{4.5cm}$$
\psset{unit=1pc} \psaxes[labels=none, ticks=none](0,0)(-4,-4)(4,4)
\rput(-2,-2){\parabola[linewidth=2pt,linecolor=red]{<->}(2,4)(0,0)}
\rput{180}(2,2){\parabola[linewidth=2pt,linecolor=blue]{<->}(2,4)(0,0)}$$
\meinecaption{1}{Two real zeros.}\label{fig:two_real_zeros}
\end{minipage}

\end{figure}

\begin{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}

\begin{exa}\label{exa:parabola4}
Consider the quadratic function $f(x) = x^2 - 5x + 3$.
\begin{multicols}{2}
\begin{dingautolist}{202}
\item Find $f'(x)$. Solve $f'(x) = 0$ and hence find the vertex of $f$. Determine the
intervals of monotonicity of $f$.
\item Write this parabola in canonical form.
\item Determine $f''(x)$  and find the
convexity intervals of $f$.
\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{dingautolist}\end{multicols}
\end{exa}Solution:
\begin{dingautolist}{202}
\item We have $f'(x)=2x-5$. Now, $2x-5 = 0\implies x = \dfrac{5}{2}$. At $x=
\dfrac{5}{2}$ we have $f\left(\dfrac{5}{2}\right) = -\dfrac{13}{4}$,
whence the vertex is at $\left(\dfrac{5}{2}, -\dfrac{13}{4}\right)$.
Also,
$$f'(x) > 0 \implies 2x-5 > 0 \implies x > \dfrac{5}{2},$$and $f$
will be increasing for $x>\dfrac{5}{2}$. It will be decreasing for
$x < \dfrac{5}{2}$.
\item Completing squares $$ y = x^2-5x+3 = \left(x-\dfrac{5}{2}\right)^2  -\dfrac{13}{4}. $$
\item We have $f''(x) = (2x)' = 2$. Since $f''(x) = 2 > 0$ for all real values $x$, $f$ 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{dingautolist}
\vspace{2.5cm}
\begin{figure}[h]
\begin{minipage}{5cm}
$$\psset{algebraic=true, unit=.5pc} \psaxes
[labels=none, linewidth=1.5pt]{->}(0,0)(-10,-10)(10,10)
\psplot[linewidth=2pt, linecolor=red]{-1}{6}{x^2-5*x+3}$$
\meinecaption{2}{$y = x^2-5x + 3$} \label{fig:parabola4}
\end{minipage}
\hfill
\begin{minipage}{5cm}
$$\psset{algebraic=true, unit=.5pc} \psaxes
[labels=none, linewidth=1.5pt]{->}(0,0)(-10,-10)(10,10)
\psplot[linewidth=2pt, linecolor=red]{-1}{6}{abs(x^2-5*x+3)}$$
\meinecaption{2}{$y = |x^2-5x + 3|$} \label{fig:parabola4|f|}
\end{minipage}
\hfill
\begin{minipage}{5cm}
$$\psset{algebraic=true, unit=.5pc} \psaxes
[labels=none, linewidth=1.5pt]{->}(0,0)(-10,-10)(10,10)
\psplot[linewidth=2pt, linecolor=red]{-6}{6}{x^2-5*abs(x)+3}$$
\meinecaption{2}{$y = |x|^2-5|x| + 3$} \label{fig:parabola4f(|.|)}
\end{minipage}
\end{figure}


\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}Solution: 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 $\geq 0$.

\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}Solution:
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 $$ p(x) =
-\dfrac{2}{5}\left(x^2-2x-4\right).
$$
\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}
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 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}
\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 $C(x) = 1\equiv 1\cdot x^0$\footnote{The symbol $\equiv$ is read ``identically equal to'' and it means that both expressions
are always the same, regardless of the value of the input
parameter.}, 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  $$\deg p = \deg pq = \deg
p + 1 \implies 0 = 1\footnote{Much to the chagrin of our
Vice-President for Academic Affairs---who claims that $1=2$---it is
not true that $0=1$.},
$$nonsense. Thus the $0$-polynomial does not have any finite degree. We attach to it, by convention, degree
$-\infty$.
\end{exa}
\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{Taylor Polynomials}
In order to motivate the following theorem, let us consider the next
example.
\begin{exa} Write $x^2$ as a sum of powers of $x-1$.
\end{exa}
Solution: Observe that $x = x-1 + 1$ and use the identity
$(a+b)^2=a^2+2ab + b^2$ to obtain $$ x^2 = (x-1+1)^2 = (x-1)^2 +
2(x-1) + 1.
$$If such an identity is not known, one can proceed as follows,
giving an idea of a general procedure. Put
$$ x^2 = a + b(x-1) + c(x-1)^2, $$where we stop at the second power
since $x^2$ has degree $2$. Let $x=1$. Then $1 = a$. Differentiate
to obtain
$$2x = b + 2c(x-1).  $$Let again $x=1$. This gives $2=b$.
Differentiate a second time to obtain
$$ 2 = 2c, $$whence $c=1$. Hence we have $a = 1, b= 2, c= 1$ and so
$$ x^2 = a+b(x-1)+c(x-1)^2 = 1 + 2(x-1) + (x-1)^2,  $$as before.
\begin{thm}[Taylor
Polynomials]\label{thm:taylor_polynomials}
\index{polynomial!Taylor}Let $a\in \reals$. Then any polynomial
$p(x)$ of degree $n$ can be written as
$$ p(x) = b_0 + b_1(x-a) + b_2(x-a)^2 + \cdots + b_n(x-a)^n , $$for
some constants $b_k$.
\end{thm}
\begin{pf}First observe that we stop at $(x-a)^n$ since $p$ has
degree $n$. Differentiating $k$ times we obtain $$ p^{(k)}(x) =
b_kk!\qquad +\qquad  (k+1)(k)\cdots (2)b_{k+1}(x-a)\qquad +\qquad
\cdots \qquad + \qquad (n)(n-1)\cdots
(n-k+1)b_n(x-a)^{n-k}.\footnote{The symbol $k!$---read ``$k$
factorial''---is the product $1\cdot 2 \cdots k$. Thus for example
$5! = 1\cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120$. We define $0! = 1$.}
$$Letting $x=a$ we obtain   $$ b_k = \dfrac{p^{(k)}(a)}{k!},$$ proving
the theorem.
\end{pf}
\begin{df}
The expansion
\begin{equation}\label{eq:taylor_polynomial}
p(x) = p(a) + p'(a)(x-a) + \dfrac{p''(a)}{2!}(x-a)^2 + \cdots
+\dfrac{p^{(n)}(a)}{n!}(x-a)^n
\end{equation}is known as the {\em Taylor polynomial expansion} about $x=a$ of
$p$.
\end{df}
\begin{exa}
Find the Taylor polynomial expansion about $x=-2$ of $p(x) = x^3 +
2x + 1$.
\end{exa}Solution: We have
$$p'(x) = 3x^2+2, \qquad p''(x) = 6x, \qquad p'''(x) = 6.   $$
Hence $$p(-2) = -11,\qquad p'(-2) = 14, \qquad p''(-2) = -12, \qquad
p'''(-2) = 6,
$$ and
$$ x^3 +
2x + 1 = -11 + 14(x+2) + \dfrac{-12}{2}(x+2)^2 + \dfrac{6}{6}(x+2)^3
= -11 + 14(x+2) - 6(x+2)^2 + (x+2)^3.
$$

\subsection{Ruffini's Factor Theorem}
\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}The Taylor expansion of $p$ about $x=a$ is
$$p(x) = p(a) + (x-a)\left(p'(a) + \dfrac{p''(a)}{2!}(x-a) + \cdots
+\dfrac{p^{(n)}(a)}{n!}(x-a)^{n-1}\right),   $$ from where the
result quickly 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}
Solution: By Ruffini's Theorem \ref{thm:ruffini}, we must have
$$0 = t(-1) = (-1)^3 - 3a(-1)^2 + 2 \implies  a = \frac{1}{3}.$$
\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}
The number $a$ is a root of multiplicity $m$ if an only if $$p(a) =
p'(a) = p''(a) = \cdots = p^{(m-1)}(a) = 0, \qquad p^{(m)}(a) \neq
0.$$
\end{cor}
\begin{pf}
This follows immediately by considering the Taylor expansion of $p$
about $x=a$.
\end{pf}
\begin{exa}
Factor the polynomial $p(x) = x^5-5x^4+11x^3-13x^2+8x-2$ over
$\integers[x]$.
\end{exa}Solution: We see that $p(1) = 0, p'(1) = 0, p''(1) = 0, p'''(1) \neq
0$. Hence $(x-1)^3 = x^3-3x^2+3x-1$ divides $p$. By long division
$$ \polylongdiv{x^5-5x^4+11x^3-13x^2+8x-2}{x^3-3x^2+3x-1}  $$and so
$$x^5-5x^4+11x^3-13x^2+8x-2 = (x-1)^3(x^2-2x+2). $$Observe that
$x^2-2x+2$ does not factor over $\integers[x]$ and hence we are
finished.

\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}
Solution: 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
$$a(x) =  x^3 - 3x - 5x^2 + 15 = (x - 5)(x - \sqrt{3})(x + \sqrt{3}).$$
\begin{exa}
Factorise $b(x) = x^5 - x^4 - 4x  + 4$ over $\integers[x]$ and  over
$\reals[x]$.
\end{exa}
Solution: 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 $\reals$.
\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}

\section{Graphs of Polynomials} 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}
\begin{cor}
A polynomial of odd degree with real number coefficients always has
a real root.
\end{cor}
\begin{pf}
Since a polynomial of odd degree eventually changes sign, since it
is continuous, the corollary follows from Bolzano's Intermediate
Value Theorem \ref{thm:bolzano}.
\end{pf}


\begin{exa}\label{exa:polynom1}
Consider the polynomial $p(x) =x^3+4x^2+x-6$.
\begin{enumerate}
\item Prove that $p$ splits over $\integers$ and find its factorisation.
Also, determine its $y$-intercept.
\item Determine $p(-\infty)$ and $p(+\infty)$
\item Find $p'$ and determine the intervals of monotonicity of $p$.
\item Determine any local extrema of $p$.
\item Find $p''$ and determine the inflexion points of $p$ and its
convexity intervals.
\item Obtain an approximate graph of $p$.
\end{enumerate}
\end{exa}
Solution:
\begin{enumerate}
\item By Corollary \ref{cor:ruffini}, if there are integral roots of
$p$ they must divide $-6$. A quick inspection shews that $p(1)=0$
and so $x-1$ divides $p(x)$. By long division
$$\polylongdiv{x^3+4x^2+x-6}{x-1}  $$
whence $$p(x) = (x-1)(x^2+5x+6) = (x-1)(x+2)(x+3). $$This means that
$p$ crosses the $x$-axis at $x=-3$, $x=-2$, and $x=1$. Its
$y$-intercept is $(0,p(0)) = (0,-6)$.
\item Since the leading coefficient of $p$ is $1>0$ and since $p$
has odd degree, by Theorem \ref{thm:behaviour_polynomials_at_infty},
$p(x) \sim (x)(x)(x) = x^3$, as $\tends{x}{+\infty}$ and so
$p(-\infty) = -\infty$ and $p(+\infty) = +\infty$.
\item $p'(x) =3x^2+8x+1$, whose graph is a convex parabola. Using
the Quadratic Formula
$$ 3x^2+8x+1 = 0 \iff x = \dfrac{-4-\sqrt{13}}{3}\qquad  \mathrm{or}\qquad x = \dfrac{-4+\sqrt{13}}{3}  $$
and so $x\approx -2.54$ or $x\approx -0.13$. Since $p'$ is a convex
parabola this means that $$p'(x) > 0 \iff
x\in\oo{-\infty}{\dfrac{-4-\sqrt{13}}{3}}\cup
\oo{\dfrac{-4+\sqrt{13}}{3}}{+\infty},$$ and so $p$ is increasing
(approximately) in the intervals $\oo{-\infty}{-2.54}$ and
$\oo{-0.13}{+\infty}$.


\item Since at $x=-2.54$ $p'$ changes sign from $+$ to $-$, $p$ has
a local maximum there by virtue of Theorem \ref{thm:local_extrema},
which is $p(-2.54) \approx 0.88$. Also, $p'$ changes sign from $-$
to $+$ at $x=-0.13$  and so $p$ has a local minimum there, which is
$p(-0.13) \approx -6.06$.
\item We find $p''(x) = 6x+8$. Now, $p''(x) =0 \implies x = -\dfrac{4}{3} \approx
-1.33$ and $p(-1.33) \approx -2.61$. Hence $p$ changes convexity
(approximately) at $(-1.33,-2.61)$.
\item The graph of $p$ can be found in figure
\ref{fig:polynom1}. \end{enumerate}
\begin{exa}\label{exa:polynom2} Consider the polynomial
$p(x) =x^3+x+1$.
\begin{enumerate}
\item Prove that $p$ is strictly increasing.
\item Prove that $p$ has no positive roots.
\item Determine $p(-\infty)$ and $p(+\infty)$
\item Prove that $p$ has a unique real root and find an interval
$[a;b]$ of length $<\dfrac{1}{4}$ containing this root.
\item Find $p''$ and determine the inflexion points of $p$ and its
convexity intervals.
\item Obtain an approximate graph of $p$.
\end{enumerate}
\end{exa}
Solution:
\begin{enumerate}
\item We have $p'(x) = 3x^2+1\geq 1>0$ since $x^2$ is always
positive.\footnote{Another way of seeing that $3x^2+1>0$ always is
by checking its discriminant.} Since the derivative of $p$ is always
strictly positive, $p$ is always strictly increasing.
\item Since  $p$ is strictly increasing, $p(x) > p(0)=1$ for $x>0$.
Hence values $x>0$ can never make $p$ zero.
\item By Theorem \ref{thm:behaviour_polynomials_at_infty}, $p(-\infty) =
-\infty$ and $p(+\infty) = +\infty$.
\item Since $p$ changes sign, it must have a root. Since $p$ is
strictly increasing, it can cross the $x$-axis only once. Now,
observe that $$p(0) = 1, \qquad p(-1) = -1$$ so the root must lie in
$[-1;0]$. We bisect this interval and find   $p(-0.5) \approx
0.375$, so the root must lie in $[-1;-0.5]$. We again bisect this
interval and find that  $p(-0.75) \approx-0.171875$, so the root
must lie in $[-0.75;-0.5]$. Again, we bisect this interval and find
that $p(-0.625)\approx 0.13$, so the root must lie in
$[-0.75;-0.625]$. We now stop since we have reached an interval of
within the desired length.
\item $p''(x) = 6x$ and so $p$ is convex for $x>0$ and concave for
$x<0$.
\item An approximate graph is shewn in figure \ref{fig:polynom2}
\end{enumerate}
\vspace{2cm}
\begin{figure}[h]
\centering
\begin{minipage}{7cm}
$$\psset{unit=.8pc,algebraic=true} \psgrid[subgriddiv=3,
gridlabels=0,
gridwidth=1pt](-7,-7)(-7,-7)(7,7)\psplot[linecolor=red,linewidth=2pt,algebraic=true,
plotpoints=1000]{-3.7}{1.34}{(x-1)*(x+2)*(x+3)} \psaxes[
Ox=-7,Oy=-7](-7,-7)(-7,-7)(7,7)
 $$\meinecaption{2}{Example
 \ref{exa:polynom1}.}\label{fig:polynom1}
\end{minipage}
\begin{minipage}{7cm}
$$\psset{unit=.8pc,algebraic=true} \psgrid[subgriddiv=3,
gridlabels=0,
gridwidth=1pt](-7,-7)(-7,-7)(7,7)\psplot[linecolor=red,linewidth=2pt,algebraic=true,
plotpoints=1000]{-1.63}{1.34}{x^3+x+1} \psaxes[
Ox=-7,Oy=-7](-7,-7)(-7,-7)(7,7)
 $$\meinecaption{2}{Example
 \ref{exa:polynom2}.}\label{fig:polynom2}

\end{minipage}


\end{figure}
We now consider polynomials with real number coefficients and that
split in $\reals$. Such polynomials have the form
$$p(x) = a(x-r_1)^{m_1}(x-r_2)^{m_2} \cdots (x-r_k)^{m_k},
$$where $a\neq 0$ and the $r_i$ are real numbers and the $m_i \geq
1$ are integers. Graphing such polynomials will be achieved by
referring to the following theorem.
\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 Theorem \ref{thm:graphs_power_functions}.
\end{pf}
\begin{exa}
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}Solution: 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 $x=1$.
\bigskip

Now, by the Product Rule,
$$\begin{array}{lll} p'(x) & = &  x(x-1) + (x+2)(x-1) + (x+2)x \\
& = & 3x^2+2x-2. \end{array}$$Using the quadratic formula,
$$3x^2+2x-2=0\implies x=-\dfrac{1}{3}-\dfrac{\sqrt{7}}{3} \approx -1.22; \ \ \ x = -\dfrac{1}{3}+\dfrac{\sqrt{7}}{3}
\approx 0.55.$$From geometrical considerations, $x \approx -1.22$
will be the $x$-coordinate of a local maximum,  with $y$-coordinate
$p(-1.22)\approx 2.11$ and $x \approx 0.55$ will be the
$x$-coordinate of a local minimum,  with $y$-coordinate
$p(0.55)\approx -0.63$.

\bigskip

Also $$p''(x) = 6x + 2,$$so $p''(x) > 0$ for $x > -\dfrac{1}{3}$ and
$p''(x) < 0$ for $x< -\dfrac{1}{3}$. This means that $p$ is convex
for $x>-\dfrac{1}{3}$ and concave for $x<-\dfrac{1}{3}$. The graph
is shewn in figure \ref{y=(x+2)x(x-1)}.

\begin{exa}
Make a rough sketch of the graph of $y=(x+2)^3x^2(1-2x)$.
\end{exa}Solution: 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 $x= \frac{1}{2}$.

\bigskip


Now, $$\begin{array}{lll}p'(x) & = &  3(x+2)^2x^2(1-2x) +
2(x+2)^3x(1-2x) -2(x+2)^3x^2 \\
& = & x(x+2)^2(3x(1-2x) + 2(x+2)(1-2x) -2(x+2)x) \\
& = & -x(x+2)^2(12x^2 + 7x - 4), \end{array}$$ and $p'(x) = 0$ when
$x = 0, -2,$ $-\dfrac{7}{24}+\dfrac{\sqrt{241}}{24} \approx 0.36$,
$-\dfrac{7}{24}-\dfrac{\sqrt{241}}{24}\approx -0.94$. From
geometrical considerations, $x = 0$ and $x=-2$ are local minima,
both with $y$-coordinate $y=0$, and both $x\approx 0.36$ (with
corresponding $y = p(0.36) \approx 0.48$) and  $x \approx -0.94 $
(with corresponding $y$-coordinate $y =p(-0.94) \approx 3.03$) are
local maxima. The graph is shewn in figure \ref{y=(x+2)^3x^2(1-2x)}.


\begin{exa}
Make a rough sketch of the graph of $y=(x+2)^2x(1-x)^2$.
\end{exa}Solution: 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}.






\vspace{2cm}
\begin{figure}[h]
\begin{minipage}{4.5cm}
$$\psset{algebraic=true, unit=1pc} \psaxes
[labels=none]{->}(0,0)(-5,-5)(5,5) \psplot[linewidth=1.5pt,
linecolor=red]{-2.5}{1.7}{(x+2)*x*(x-1)}
$$
\meinecaption{2}{$y = (x+2)x(x-1)$.} \label{y=(x+2)x(x-1)}
\end{minipage}
\hfill
\begin{minipage}{4.5cm}
$$\psset{algebraic=true, unit=1pc} \psaxes
[labels=none]{->}(0,0)(-5,-5)(5,5) \psplot[linewidth=1.5pt,
linecolor=red]{-2.5}{.75}{(x+2)^3*x^2*(1-2*x)}
$$
\meinecaption{2}{$y =(x+2)^3x^2(1-2x)$.} \label{y=(x+2)^3x^2(1-2x)}
\end{minipage}
\hfill
\begin{minipage}{4.5cm}
$$\psset{algebraic=true, unit=1pc} \psaxes
[labels=none]{->}(0,0)(-5,-5)(5,5) \psplot[linewidth=1.5pt,
linecolor=red]{-2.45}{1.5}{(x+2)^2*x*(1-x)^2}
$$
\meinecaption{2}{$y = (x+2)^2x(1-x)^2$.} \label{y=(x+2)^2x(1-x)^2}
\end{minipage}
\end{figure}

 \chapter{Rational
Functions and Algebraic Functions}\thispagestyle{empty}
\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:D(x^-1)=-x^-2}The strong derivative of the
reciprocal function $x\mapsto \dfrac{1}{x}$ is the function
$x\mapsto -\dfrac{1}{x^2}$.
\end{thm}\begin{pf}
Put $f(x) = \dfrac{1}{x}$ and $g(x) = x$. Observe that $g'(x) = 1$
and $f(x)g(x) =1$. Hence by the product rule
$$0 = \diff{1}{x} = f'(x)g(x) + f(x)g'(x) = xf'(x) +  \dfrac{1}{x},$$
and solving for $f'(x)$ we obtain $f'(x) = -\dfrac{1}{x^2}$.

\bigskip

An alternate proof from the definition proceeds as follows. Let
$f(x) = \dfrac{1}{x}$ and $x\neq 0$. First observe the algebraic
identity
$$ \dfrac{1}{1+t} = 1-t + \dfrac{t^2}{1+t}. $$ Hence, if $x\neq 0$
is fixed,
$$\dfrac{1}{x+\epsilon} = \dfrac{1}{x}\cdot\dfrac{1}{1 + \epsilon/x} = \dfrac{1}{x}\left(1- \dfrac{\epsilon}{x} +
\dfrac{\epsilon^2}{x^2}\cdot\dfrac{1}{(1+\epsilon/x)} \right). $$
Now, since  $\tends{\epsilon}{0}$ we will have, eventually,
$|\epsilon| < \dfrac{|x|}{2}$. Hence $\dfrac{2}{3}
<\dfrac{1}{(1+\epsilon/x)} < 2$. This means that
$$\dfrac{\epsilon^2}{x^2}\cdot\dfrac{1}{(1+\epsilon/x)} =
\soo{\varepsilon},$$where the implied constant depends on (the fixed
value of) $x$, and so
$$ \dfrac{1}{x+\epsilon}= \dfrac{1}{x}\left(1- \dfrac{\epsilon}{x} + \soo{\varepsilon}\right)
= \dfrac{1}{x}- \dfrac{\epsilon}{x^2} + \soo{\varepsilon} ,$$from
where the assertion follows.
\end{pf}

\begin{thm}\label{thm:strong derivative _x^-n}
If $n>0$ is an integer and $x\neq 0$, $$
\left(\dfrac{1}{x^n}\right)' = \left(x^{-n}\right)' = -nx^{-n-1}. $$
\end{thm}
\begin{pf}
Let $f(x) = x^n$, $g(x) = \dfrac{1}{x}$, $h(x) = \dfrac{1}{x^n}$.
Then $h = f\circ g$. By the Chain Rule (Theorem
\ref{thm:chain_rule}) and Theorem  \ref{thm:D(x^-1)=-x^-2}, $$ h'(x)
= f'(g(x))g'(x) =
n\left(\dfrac{1}{x}\right)^{n-1}\left(-\dfrac{1}{x^2}\right) =
-\dfrac{n}{x^{n+1}} = -nx^{-n-1},
$$as it was to be demonstrated.
\end{pf}
\begin{rem}
Theorems \ref{thm:power_rule} and  \ref{thm:strong derivative _x^-n}
say that if $\alpha$ is an integer, then $(x^\alpha)' = \alpha
x^{\alpha -1}$.
\end{rem}
With the derivatives of reciprocal powers determined, we can now
address how to graph them.
\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{pf}
Let $h(x) = \dfrac{1}{x^n}$. By Theorem \ref{thm:strong derivative
_x^-n}, $h'(x) = -\dfrac{n}{x^{n+1}}$ and $h''(x) =
\dfrac{n(n+1)}{x^{n+2}}$. If $n$ is odd, then $n+1$ is even and
$n+2$ is odd. Hence  $h'(x) > 0$ for $x\neq 0$, proving that $h$ is
increasing and $h''(x)$ has the same sign as $x$, proving that $h$
is concave for $x<0$ and convex for $x>0$. A similar argument is
used for when $n$ is even, completing the proof.
\end{pf}
\vspace{1cm}
\begin{figure}[h]
\begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{-6}{-.166666667}{1/x}
\psplot[linewidth=2pt,linecolor=red]{.166666667}{6}{1/x}
$$\meinecaption{1}{$x\mapsto \dfrac{1}{x}$}
\label{fig:reciprocal_function}
\end{minipage}
\hfill
\begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{-6}{-.4082482905}{1/x^2}
\psplot[linewidth=2pt,linecolor=red]{.4082482905}{6}{1/x^2}
$$\meinecaption{1}{$x\mapsto \dfrac{1}{x^2}$}
\label{fig:1/x^2}
\end{minipage}
\hfill
\begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{-6}{-.5503212082}{1/x^3}
\psplot[linewidth=2pt,linecolor=red]{.5503212082}{6}{1/x^3}
$$\meinecaption{1}{$x\mapsto \dfrac{1}{x^3}$}
\label{fig:1/x^3}
\end{minipage}
\hfill
\begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{-6}{-.6389431042}{1/x^4}
\psplot[linewidth=2pt,linecolor=red]{.6389431042}{6}{1/x^4}
$$\meinecaption{1}{$x\mapsto \dfrac{1}{x^4}$}
\label{fig:1/x^4}
\end{minipage}
\hfill
\begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{-6}{-.6988271188}{1/x^5}
\psplot[linewidth=2pt,linecolor=red]{.6988271188}{6}{1/x^5}
$$\meinecaption{1}{$x\mapsto \dfrac{1}{x^5}$}
\label{fig:1/x^5}
\end{minipage}
\hfill
\begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{-6}{-.7418363756}{1/x^6}
\psplot[linewidth=2pt,linecolor=red]{.7418363756}{6}{1/x^6}
$$\meinecaption{1}{$x\mapsto \dfrac{1}{x^6}$}
\label{fig:1/x^6}
\end{minipage}
\hfill
\end{figure}
\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}
\begin{figure}[h]
\begin{minipage}{3.5cm}$$\psset{unit=.7pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \rput(1,-1){
\psline[linestyle=dashed](0,-6)(0,6)\psline[linestyle=dashed](-6,0)(6,0)
\psplot[linewidth=1.4pt,linecolor=red]{-6}{-.166666667}{1/x}
\psplot[linewidth=1.4pt,linecolor=red]{.166666667}{6}{1/x}}
$$\meinecaption{1}{$x\mapsto \dfrac{1}{x-1} -1$}
\label{fig:1/(x-1)-1}
\end{minipage}
\hfill
\begin{minipage}{3.5cm}$$\psset{unit=.7pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\rput(1,1){\psplot[linewidth=1.4pt,linecolor=red]{-6}{-.25}{1/abs(x)}}\rput(1,-1){
\psline[linestyle=dashed](0,1)(0,6)\psline[linestyle=dashed](-6,2)(6,2)
\psplot[linewidth=1.4pt,linecolor=red]{.166666667}{1}{1/x}}
\rput(1,1){\psplot[linewidth=1.4pt,linecolor=red]{1}{6}{-1/x}}
$$\meinecaption{1}{$x\mapsto \left|\dfrac{1}{x-1} -1\right|$}
\label{fig:1/|x-1|-1}
\end{minipage}
\hfill
\begin{minipage}{3.5cm}$$\psset{unit=.7pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6) \rput(1,-1){
\psline[linestyle=dashed](0,-6)(0,6)\psline[linestyle=dashed](-6,0)(6,0)
\psplot[linewidth=1.4pt,linecolor=red]{-1}{-.166666667}{1/x}
\psplot[linewidth=1.4pt,linecolor=red]{.166666667}{6}{1/x}}
\rput(-1,-1){
\psline[linestyle=dashed](0,-6)(0,6)\psline[linestyle=dashed](-6,0)(6,0)
\psplot[linewidth=1.4pt,linecolor=red]{-6}{-.166666667}{-1/x}
\psplot[linewidth=1.4pt,linecolor=red]{.166666667}{1}{-1/x}}
$$\meinecaption{1}{$x\mapsto \dfrac{1}{|x|-1}-1$}
\label{fig:1/(|x|-1)-1}
\end{minipage}
\hfill
\end{figure}
\begin{exa} Figures \ref{fig:1/(x-1)-1} through
\ref{fig:1/(|x|-1)-1} shew a few transformations of $x \mapsto
\dfrac{1}{x}$.
\end{exa}


\section{The Quotient Rule}
\begin{thm}\label{thm:deriv_1/g}
If $g$ is strongly differentiable  at $x$ and $g(x) \neq 0$ then $$
\left(\dfrac{1}{g}\right)' (x) =  -\dfrac{g'(x)}{(g(x))^2}.
$$
\end{thm}
\begin{pf}
Let $a(x) = \dfrac{1}{x}$. Then $\dfrac{1}{g} = a\circ g$. By the
Chain Rule  (Theorem \ref{thm:chain_rule}) and Theorem
\ref{exa:D(x^-1)=-x^-2}, $$ \left(\dfrac{1}{g}\right)' (x) =
a'(g(x))g'(x) = -\dfrac{g'(x)}{(g(x))^2},
$$as we needed to shew.
\end{pf}
\begin{cor}[Quotient Rule]If $f, g$ are strongly differentiable  at $x$ and if $g(x) \neq 0$,
then
$$ \left(\dfrac{f}{g}\right)'(x) =  \dfrac{f'(x)g(x) -f(x)g'(x)}{(g(x))^2}. $$
\end{cor}
\begin{pf}Using the Product Rule (Theorem \ref{thm:product_rule}) and Theorem \ref{thm:deriv_1/g},
$$\begin{array}{lll}  \left(\dfrac{f}{g}\right)'(x) & = & \left(f\cdot
\dfrac{1}{g}\right)'(x) \\ & = & f'(x)\dfrac{1}{g(x)} +
f(x)\left(\dfrac{1}{g}\right)'(x)\\
& = &  f'(x)\dfrac{1}{g(x)} + f(x)\left(-\dfrac{g'(x)}{(g(x))^2}\right)\\
& = & \dfrac{f'(x)g(x) -f(x)g'(x)}{(g(x))^2}, \end{array}$$ as
desired.
\end{pf}
\begin{exa} Find $b'(x)$ if $b(x) =
\dfrac{1+x+x^2}{(x-x^2)^2}$\end{exa} Solution: Use the Quotient Rule
and the Chain Rule: $$
\begin{array}{lll}b'(x) &=& \dfrac{(1 + 2x)(x-x^2)^2 -
(1+x+x^2)(2(1-2x)(x-x^2))}{(x-x^2)^4}\\ & = & \dfrac{(1 + 2x)(x-x^2)
- (1+x+x^2)(2(1-2x))}{(x-x^2)^3}\\ & = &
-\dfrac{-2+3x+3x^2+2x^3}{(x-x^2)^3}. \end{array}
$$






\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.
\begin{exa}Draw the curve
$x \mapsto \dfrac{x}{x^2 + 1}.$
\end{exa}Solution: Put $a(x) = \dfrac{x}{x^2 + 1}$. Observe that $a(-x) = -a(x)$, which means that $a$ is an odd function
and hence symmetric about the origin.  Also
$$a'(x) = \dfrac{(x^2+1) -x(2x)}{(x^2+1)^2} = \dfrac{1-x^2}{(x^2+1)^2} = \dfrac{(1-x)(1+x)}{(x^2+1)^2} .$$
Since $(x^2 + 1)^2$ is always positive, $a'$ changes sign when
$1-x^2 = (1-x)(1+x)$ changes sign. Hence $a'(x) \geq 0$ if $x\in
[-1;1]$ and $a'(x)< 0$ otherwise. This means that $a$ is increasing
for $x\in [-1;1]$ and decreasing otherwise. Moreover
$$a''(x) = \diff{\dfrac{1-x^2}{(x^2+1)^2}}{x} = \dfrac{-2x(x^2+1)^2 -2(2x)(1-x^2)(x^2+1)}{(x^2+1)^4} =
 \dfrac{2x(x^2-3)}{(x^2+1)^3} .$$Again $a''$ will change sign when
 $2x(x^2-3) = x(x-\sqrt{3})(x+\sqrt{3})$ changes sign. By means of a sign diagram
 we see that $a''(x) \geq 0$ for $x\in [-\sqrt{3};0] \cup
 [\sqrt{3};+\infty[$, and so $a$ is convex for $x\in [-\sqrt{3};0] \cup
 [\sqrt{3};+\infty[$ and concave otherwise. The graph is shewn in
 figure \ref{fig:x/(x^2+1)}.

\vspace{1cm}
\begin{figure}[h]
\begin{minipage}{4.5cm}$$\psset{unit=.7pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{-6}{6}{x/(x^2+1)}
$$\meinecaption{1}{$x\mapsto \dfrac{x}{x^2+1}$}
\label{fig:x/(x^2+1)}
\end{minipage}
\hfill
\begin{minipage}{4.5cm}$$\psset{unit=.7pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{-6}{6}{x^2/(x^2+1)}
$$\meinecaption{1}{$x\mapsto \dfrac{x^2}{x^2+1}$}
\label{fig:x^2/(x^2+1)}
\end{minipage}
\hfill
\begin{minipage}{4.5cm}$$\psset{unit=.7pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{-2.414213562}{-.4142135624}{x^2+1/x^2}
\psplot[linewidth=2pt,linecolor=red]{.4142135624}{2.414213562}{x^2+1/x^2}
$$\meinecaption{1}{$x\mapsto x^2+\dfrac{1}{x^2}$}
\label{fig:x^2+1/x^2}
\end{minipage}
\end{figure}

\begin{exa}Draw the curve
$x \mapsto \dfrac{x^2}{x^2 + 1}.$
\end{exa}Solution: Put $b(x) = \dfrac{x^2}{x^2 + 1}$. Observe that $b(-x) = b(x)$, which means that $b$ is an even function
and hence symmetric about the $y$-axis.  Also
$$b'(x) = \dfrac{(2x)(x^2+1) -2x(x^2)}{(x^2+1)^2} = \dfrac{2x}{(x^2+1)^2}.$$
Since $(x^2 + 1)^2$ is always positive, $b'$ changes sign when $x$
changes sign. Hence $b'(x) \geq 0$ if $x\geq 0$ and $b'(x)< 0$
otherwise. This means that $b$ is increasing for $x\geq 0$ and
decreasing otherwise. Moreover
$$b''(x) = \diff{\dfrac{2x}{(x^2+1)^2}}{x} = \dfrac{2(x^2+1)^2 -2(2x)(2x)(x^2+1)}{(x^2+1)^4} =
 \dfrac{2-6x^2}{(x^2+1)^3} .$$Again $b''$ will change sign when
 $2-6x^2 = 2(1-\sqrt{3}x)(1+\sqrt{3}x)$ changes sign. By means of a sign diagram
 we see that $b''(x) \geq 0$ for $x\in \cc{-\frac{1}{\sqrt{3}}}{\frac{1}{\sqrt{3}}}$, and so $b$ is convex for
 $x\in \cc{-\frac{1}{\sqrt{3}}}{ \frac{1}{\sqrt{3}}}$ and concave otherwise. The graph is shewn in
 figure \ref{fig:x^2/(x^2+1)}.
\begin{exa}Draw the curve
$x \mapsto x^2+\dfrac{1}{x^2}.$
\end{exa}Solution: Put $c(x) = x^2+\dfrac{1}{x^2}$. Observe that $c(-x) = c(x)$, which means that $c$ is an even function
and hence symmetric about the $y$-axis.  Also
$$c'(x) = 2x -\dfrac{2}{x^3} = \dfrac{2(x^4-1)}{x^3} =
\dfrac{2(x-1)(x+1)(x^2+1)}{x^3}.$$We make a sign diagram
investigating the sign changes of  $c'$ near $x=-1$, $x=0$, and
$x=1$. From this we gather that $c$ is increasing for $x\in
[-1;0[\cup [1;+\infty[$. Moreover
$$c''(x) = \diff{\left(2x-\dfrac{2}{x^3}\right)}{x} = 2+\dfrac{6}{x^4}.$$We see that  $c''$ is always positive and hence it is always convex. The graph is shewn in
 figure \ref{fig:x^2+1/x^2}.

\bigskip

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:graphs_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}Solution: 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 $y=1$. The graph of $r$
can be found in figure \ref{fig:(x-1)^2(x+2)/(x+1)(x-2)^2}.

\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}Solution:  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
$r(-\infty) = +\infty$. The graph of $r$ can be found in figure
\ref{fig:(x-3/4)^2(x+3/4)^2)/(x+1)(x-1)}.

\vspace{3cm}
\begin{figure}[h]
\centering
\begin{minipage}{7cm}$$\psset{unit=1pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{-6}{-1.083676189}{((x+2)*(x-1)^2)/((x+1)*(x-2)^2)}
\psplot[linewidth=2pt,linecolor=red]{-.9058688457}{1.655868846}{((x+2)*(x-1)^2)/((x+1)*(x-2)^2)}
\psplot[linewidth=2pt,linecolor=red]{3}{6}{((x+2)*(x-1)^2)/((x+1)*(x-2)^2)}
\psline[linestyle=dotted](-6,1)(6,1)
\psline[linestyle=dotted](-1,-6)(-1,6)
\psline[linestyle=dotted](2,-6)(2,6)
$$\meinecaption{2}{$x\mapsto \dfrac{(x-1)^2(x+2)}{(x+1)(x-2)^2}$}
\label{fig:(x-1)^2(x+2)/(x+1)(x-2)^2}
\end{minipage}
\hfill
\begin{minipage}{7cm}$$\psset{unit=2pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-3,-3)(3,3)
\psplot[linewidth=2pt,linecolor=red]{-1.740903741}{-1.046066042}{(((x-.75)^2)*((x+.75)^2))/((x+1)*(x-1))}
\psplot[linewidth=2pt,linecolor=red]{-.9746582010}{.9746582010}{(((x-.75)^2)*((x+.75)^2))/((x+1)*(x-1))}
\psplot[linewidth=2pt,linecolor=red]{1.046066042}{1.740903741}{(((x-.75)^2)*((x+.75)^2))/((x+1)*(x-1))}
\psline[linestyle=dotted](-3,1)(3,1)
\psline[linestyle=dotted](-1,-3)(-1,3)
\psline[linestyle=dotted](1,-3)(1,3)
$$\meinecaption{2}{$x\mapsto \dfrac{(x-3/4)^2(x+3/4)^2}{(x+1)(x-1)}$}
\label{fig:(x-3/4)^2(x+3/4)^2)/(x+1)(x-1)}
\end{minipage}

\end{figure}
\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:D(x^1/q)=x^(1/q-1)/q} Let $ q\neq 0$ be  an integer. The strong derivative of the
 function $x\mapsto x^{1/q}$ is the function $x\mapsto
\dfrac{1}{q}\cdot x^{1/q-1}$, whenever this last makes sense.
\end{thm}\begin{pf}
Put $f(x) = x^{1/q}$, assuming this quantity is real. Then $(f(x))^q
= x$. Differentiating both sides using the Chain Rule we have
$$q(f(x))^{q-1}f'(x) = \implies qx^{(q-1)/q}f'(x) = 1.
$$ Solving for $f'$ gives $$f'(x) = \dfrac{1}{q}\cdot x^{1/q-1},$$if this quantity is a real number, proving the result.
\end{pf}

\begin{rem}
Theorems \ref{thm:power_rule},  \ref{thm:strong derivative _x^-n},
and  \ref{thm:D(x^1/q)=x^(1/q-1)/q}, when combined with the Chain
Rule, say that if $\alpha$ is a rational number, then $(x^\alpha)' =
\alpha x^{\alpha -1}$.
\end{rem}
With the derivatives of rational powers determined, we can now
address how to graph them.
\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}
\begin{pf}
Let $h(x) = x^{1/q}$. By Theorem \ref{thm:D(x^1/q)=x^(1/q-1)/q},
$h'(x) = \dfrac{x^{(1-q)/q}}{q}$ and $h''(x) =
\dfrac{(1-q)x^{(2q-1)/q}}{q^2}$.

\bigskip
Assume first that $q$ is even. Then $x^{1/q}$ is not real for $x<0$
so we assume that $x>0$. The quantity $h'(x) =
\dfrac{x^{(1-q)/q}}{q}$ is $>0$ for $q\geq 2$ and negative for
$q\leq -2$. If $q\geq 2$ then $h''(x) =
\dfrac{(1-q)x^{(2q-1)/q}}{q^2} <0$ and if $q\leq -2$ then $h''(x)
>0$. Hence $h$ is increasing and concave for $q \geq 2$ and
decreasing and convex for $q \leq -2$

\bigskip

Assume now that $q$ is odd. Then $1-q$ is even and the sign of the
quantity $h'(x) = \dfrac{1}{q}\cdot \left(x^{1/q}\right)^{1-q}$ is
depends on the sign of $\dfrac{1}{q}$. Since $2q-1$ is odd, the sign
of $h''(x) = \dfrac{(1-q)x^{(2q-1)/q}}{q^2}$ is $\sgn{(1-q)(x)}$. We
have: if $q \geq 3$, $h'(x) > 0$, $h''(x)<0$ for $x>0$ and $h''(x)
>0$ for $x<0$. Hence for $q \geq 3$, $h$ is increasing and it is convex
for $x<0$ but concave for $x>0$. If $q\leq -3$ then $h$ is
decreasing and it is concave for $x<0$ and decreasing convex for
$x>0$.
\end{pf}
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}[h]
\begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{0}{6}{sqrt(x)}
$$\meinecaption{1}{$x\mapsto x^{1/2}$}
\label{fig:x^1/2}
\end{minipage}
\hfill
\begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{.02777777778}{6}{x^(-1/2)}
$$\meinecaption{1}{$x\mapsto x^{-1/2}$}
\label{fig:x^-1/2}
\end{minipage}
\hfill
\begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{0}{6}{x^(1/4)}
$$\meinecaption{1}{$x\mapsto x^{1/4}$}
\label{fig:x^1/4}
\end{minipage}
\hfill
\begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{.0007716049383}{6}{x^(-1/4)}
$$\meinecaption{1}{$x\mapsto x^{-1/4}$}
\label{fig:x^-1/4}
\end{minipage}
\hfill
\begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{0}{6}{x^(1/6)}
$$\meinecaption{1}{$x\mapsto x^{1/6}$}
\label{fig:x^1/6}
\end{minipage}
\hfill
\begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{.00002143347051}{6}{x^(-1/6)}
$$\meinecaption{1}{$x\mapsto x^{-1/6}$}
\label{fig:x^-1/6}
\end{minipage}
\hfill
\end{figure}
\vspace{1cm}
\begin{figure}[h]
\begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{0}{6}{x^(1/3)}
\psplot[linewidth=2pt,linecolor=red]{-6}{0}{-1*abs(x)^(1/3)}
$$\meinecaption{1}{$x\mapsto x^{1/3}$}
\label{fig:x^1/3}
\end{minipage}
\hfill
\begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{.004629629630}{6}{x^(-1/3)}
\psplot[linewidth=2pt,linecolor=red]{-.004629629630}{-6}{-1*abs(x)^(-1/3)}
$$\meinecaption{1}{$x\mapsto x^{-1/3}$}
\label{fig:x^-1/3}
\end{minipage}
\hfill
\begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{0}{6}{x^(1/5)}
\psplot[linewidth=2pt,linecolor=red]{0}{-6}{-1*abs(x)^(1/5)}
$$\meinecaption{1}{$x\mapsto x^{1/5}$}
\label{fig:x^1/5}
\end{minipage}
\hfill
\begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{.0001286008230}{6}{x^(-1/5)}
\psplot[linewidth=2pt,linecolor=red]{-.0001286008230}{-6}{-1*abs(x)^(-1/5)}
$$\meinecaption{1}{$x\mapsto x^{-1/5}$}
\label{fig:x^-1/5}
\end{minipage}
\hfill
\begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{0}{6}{x^(1/7)}
\psplot[linewidth=2pt,linecolor=red]{0}{-6}{-1*abs(x)^(1/7)}
$$\meinecaption{1}{$x\mapsto x^{1/7}$}
\label{fig:x^1/7}
\end{minipage}
\hfill
\begin{minipage}{2cm}$$\psset{unit=.5pc,algebraic=true} \psaxes[linewidth=1pt,
labels=none, ticks=none]{->}(0,0)(-6,-6)(6,6)
\psplot[linewidth=2pt,linecolor=red]{.000003572245085}{6}{x^(-1/7)}
\psplot[linewidth=2pt,linecolor=red]{-.000003572245085}{-6}{-1*abs(x)^(-1/7)}
$$\meinecaption{1}{$x\mapsto x^{-1/7}$}
\label{fig:x^-1/7}
\end{minipage}
\hfill
\end{figure}


\bigskip

We finish this section with an example.

\begin{exa}

Consider the function $y = f(x) = \sqrt{x-1}+\sqrt{2-x}$.
\begin{enumerate}
\item For which $x$ will the output of $f$ be a real number?

\item Find  $f'(x)$.\item  By examining $f'$, prove that  $f$ increases for $x< \dfrac{3}{2}$ and decreasing for $x>\dfrac{3}{2}$.

\item  Prove that for all $x$ in the domain of $f$ one has $f(x) \leq
\sqrt{2}$. \item Find $f''(x)$.
\item Determine in which intervals $f$ is convex or concave.

\item Graph $f$.\end{enumerate}
\end{exa}
Solution: \\
\begin{enumerate}\item  We need, simultaneously, $x\geq 1$ and $2 \geq x$. This
means that $x\in [1;2]$.
\item Via the Chain Rule: $$\diff{(x-1)^{1/2}+(2-x)^{1/2}}{x} =
\dfrac{1}{2}(x-1)^{-1/2}-\dfrac{1}{2}(2-x)^{-1/2} =
\dfrac{1}{2(x-1)^{1/2}}-\dfrac{1}{2(2-x)^{1/2}}$$
\item  $f$ has a stationary point when $f'(x) = 0$, that is, if
$$  \dfrac{1}{2(x-1)^{1/2}}=\dfrac{1}{2(2-x)^{1/2}} \implies x-1 = 2-x \implies
x=\dfrac{3}{2},
$$so $f'$ has only one zero in $[1;2]$. Since $f'$ is continuous in $]1;2[$ and has only one zero there, it must be negative in
a portion of the interval and positive in the other. Examining
values in $]1;\frac{3}{2}[$ we see that $f'$ is positive for there
and negative in $]\frac{3}{2}; 2[$.
\item  By the above, $x=\dfrac{3}{2}$ is a global
maximum in $[1;2]$, and hence
$$f(x) \leq f\left(\dfrac{3}{2}\right) = \sqrt{\dfrac{3}{2}-1} \sqrt{2-\dfrac{3}{2}} =  2\sqrt{\dfrac{1}{2}} = \sqrt{2}.   $$
\item  Via the Chain Rule: $$ f''(x) =
\diff{\dfrac{1}{2}(x-1)^{-1/2}-\dfrac{1}{2}(2-x)^{-1/2}}{x} =
-\dfrac{1}{4}(x-1)^{-3/2}-\dfrac{1}{4}(2-x)^{-3/2} =
-\dfrac{1}{4(x-1)^{3/2}}-\dfrac{1}{4(2-x)^{3/2}}. $$

\item  Observe that
$$f''(x) = -\dfrac{1}{4(x-1)^{3/2}}-\dfrac{1}{4(2-x)^{3/2}} = -\dfrac{1}{4}\left(\dfrac{1}{(x-1)^{3/2}}+\dfrac{1}{(2-x)^{3/2}}\right).  $$
Since the quantity in parenthesis is always positive, $f''$ is
always negative, and hence it is everywhere concave.

\item The graph appears in figure \ref{fig:sqrt{x-1}+sqrt{2-x}}.
\end{enumerate}
 \vspace{5cm}
\begin{figure}[h]
$$\psset{algebraic=true,unit=4pc}
\psgrid[subgriddiv=4,linewidth=1pt, gridlabels=0](0,0)(0,0)(3,3)
\psaxes(0,0)(0,0)(3,3)
\psplot[linecolor=red,linewidth=2pt]{1}{2}{sqrt(x-1)+sqrt(2-x)}
$$\meinecaption{1}{$y =\sqrt{x-1}+\sqrt{2-x}$}\label{fig:sqrt{x-1}+sqrt{2-x}}
\end{figure}

\chapter{Exponential and Logarithmic Functions}
\section{Exponential Functions}
\section{Differentiating Exponential Functions}
\section{Graphs of Exponential Functions}
\section{Integrating Exponential Functions}
\section{Logarithmic Functions}
\section{Algebraic Properties of Logarithms}
\section{Differentiating Logarithmic Functions Functions}
\section{Graphs of Exponential Functions}
\section{Integrating Exponential Functions}
\section{Exponential and Logarithmic Equations}

\chapter{Goniometric Functions}
\section{The Sine and Cosine Functions}
\section{Differentiation of the Sine and Cosine}

\section{Integration  of the Sine and Cosine}
\section{Graphs of the Sine and Cosine}
\section{Addition Formul\ae}
\section{Inverse Sine and Cosine}
\section{Equations in Sine and Cosine}
\section{Other Goniometric Functions}
\clearpage \Closesolutionfile{jose}
\appendix
\renewcommand{\chaptername}{Appendix}


\chapter{Answers to Selected Problems}
   \Readsolutionfile{jose}


\clearpage




\lhead{}\chead{}\rhead{}

{\tiny\chapter*{\rlap{GNU Free Documentation License}}
\phantomsection  % so hyperref creates bookmarks
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%\label{label_fdl}

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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. }












\printindex\end{document}
