\chapter{Functions}   
   
\Citation{Still glides the Stream, and shall for ever glide;\\   
The Form remains, the Function never dies.\\   
\auteur{William}{Wordsworth}{1770}{1850}{English poet}{The River Duddon}{1820}   
}   
   
\newcommand{\domain}{\ensuremath{\EuScript D}\xspace}   
   
\section{Definitions}   
   
\begin{Def}   
  A \nuvo{function} \pron{f£v£nk£sn} from \domain to \reel is a rule that maps each real of a certain   
  subset \domain of \reel onto one and only one real.   
\end{Def}   
   
Let $f$ be a function from \domain to \reel. Let $x$ be a real belonging to \domain. The unique real associated to $x$
by $f$ is denoted by $f(x)$ `$f$ of $x$' and called the \nuvo{image} \pron{£h£im£i£d} of $x$ under $f$.\index{$f(x)$} In
such a case, $x$ is called the \emph{independant variable} \pron{£b£ind£i£hpend£ent
  £hv£Pr£Qbl}.\index{variable!independant} 
The set \domain is called the \nuvo{domain} \pron{d£w£hm£qn} of $f$.  If
$y=f(x)$, then $x$ is the \nuvo{pre-image} \pron{pr£I£h£im£i£d} or \nuvo{counterimage} \pron{£bk£Wnt£e£hr£im£i£d} of $y$
under $f$ and $y$ is called the \emph{dependant} \pron{d£i£hpend£ent} \emph{variable}.\index{variable!dependant}
   
To denote a function, we write $f: \domain \dans \reel$; $x \donne f(x)$ `$f$ maps \domain to \reel   
and $x$ onto $f(x)$'. For example, let $f$ be the function $f: \interfer{0}{10} \dans \reel$; $x   
\donne x^2+4$. The image of $2$ under $f$ is $f(2) = 2^2+4 =8$. So $8$ is the image of $2$ under $f$   
and $2$ is a pre-image of $8$ under $f$.  We can also say that $8$ is the \emph{value} of $f$ for   
the \nuvo{argument} \pron{£h£Agj£um£ent} $2$.\index{value!of $f$}   
   
\section{Graphs of functions}   
   
\begin{Def}   
  Let $f$ be a function from \domain to \reel. The \nuvo{graph} of $f$ in the coordinate system   
  \repere is the set, denoted \graff, of points $M$ the coordinates of which are \couple{x}{f(x)}   
  with $x \in \domain$. The curve \graff is said to have equation $y=f(x)$.   
\end{Def}   % correction 2003-06-16
   
\begin{prop}   
  Let $M:\couple{x}{y}$ be a point in the plane. Let $f$ be a function from \domain to \reel and   
  \graff its graph.   
     
  $$M \in \graff \Iff x\in\domain \textbf{ and } y=f(x).$$   
\end{prop}   
   
\section{Increasing, decreasing functions}   
   
In this section the domain of the function $f$ is an \textbf{interval} $I$.   
   
\begin{Def}   
  Let $f : I \to \reel$, $x \donne f(x)$. $f$ is increasing \pron{£in£hkr£Is£i£n} on $I$ iff for all pair   
  \couple{a}{b} of numbers belonging to $I$ $a < b$ implies $f(a) < f(b)$.   
\end{Def}   
   
So an increasing function is `order-preserving': the images $f(a)$ and $f(b)$   % correction 2003-06-16
are in the same order as the arguments $a$ and $b$.   
   
\begin{exemple}   
\item Let $f : \reel \to \reel$, $x \donne 4x+5$. Let $a$ and $b$ be two real numbers.   
$a < b$ implies $4a < 4b$ which in turn implies $4a+5 < 4b+5$ \idest\  $f(a) < f(b)$.   
So $f$ is increasing on \reel.   
   
\item Let $f : \reel \to \reel$, $x \donne x^2 - 6x$.   
First we have $f(1) = -5$ and $f(2) = -8$ so $1<2$ and $f(1) > f(2)$.   
That proves that $f$ is not increasing on \reel.   
   
We have $x^2-6x = (x-3)^2-9$. So if $3 \leq a < b$ then $0 \leq a-3 < b-3$ and, as   
the square of two positive numbers are ordered as the numbers, $(a-3)^2 < (b-3)^2$   
which implies $(a-3)^2 - 9< (b-3)^2 - 9$ \idest\  $f(a) < f(b)$. We have proven   
that $f$ is increasing on \intferouv{3}{\plusinf}.   
\end{exemple}   
   
\begin{Def}   
  Let $f : I \to \reel$, $x \donne f(x)$. $f$ is decreasing \pron{d£I£hkr£Is£i£n} on $I$ iff for all pair   
  \couple{a}{b} of numbers belonging to $I$ $a < b$ implies $f(a) > f(b)$.   
\end{Def}   
   
So a decreasing function is `order-reversing': the images $f(a)$ and $f(b)$   
are in the reverse order as the arguments $a$ and $b$.   
   
\begin{exemple}   
\item Let $f : \reel \to \reel$, $x \donne -3x+8$. Let $a$ and $b$ be two real numbers.   
$a < b$ implies $-3a > -3b$ which in turn implies $-3a+8 > -3b+8$ \idest\  $f(a) > f(b)$.   
So $f$ is decreasing on \reel.   
   
\item Let $f : \reel \to \reel$, $x \donne x^2 - 6x$.   
   
We have proven that $f$ is increasing on \intferouv{3}{\plusinf}.   
   
Let us cast a glance at the behaviour of $f$ on \intouvfer{\moinsinf}{3}.   
$a \in \intouvfer{\moinsinf}{3}$ iff $a \leq 3$. Let us consider two numbers $a$ and $b$   
in \intouvfer{\moinsinf}{3} such that $a < b$. Then $a < b \leq 3$ so   
$a-3 < b-3 \leq 0$ and $(a-3)^2 > (b-3)^2$   
which implies $(a-3)^2 - 9 > (b-3)^2 - 9$ \idest\  $f(a) > f(b)$.   
So $f$ is decreasing on \intouvfer{\moinsinf}{3}.   
\end{exemple}   
   
   
\begin{Def}   
Let $f : I \to \reel$, $x \donne f(x)$.   
\begin{enumerate}   
\item $f$ is non-increasing on $I$ iff,   
for all pair \couple{a}{b} of numbers belonging to $I$, $a < b$ implies $f(a) \geq f(b)$.   
\item $f$ is non-decreasing on $I$ iff,   
for all pair \couple{a}{b} of numbers belonging to $I$, $a < b$ implies $f(a) \leq f(b)$.   
\end{enumerate}   
\end{Def}   
   
\section{Maxima, minima, turning points}   
   
In this section too the domain of the function $f$ is an \textbf{interval} $I$.   
   
\begin{Def}   
  $f$ has a \nuvo{maximum} \pron{£hm£aks£im£em} at $a$ on $I$ iff, for all $x \in I$, $f(x) \leq f(a)$.   
  Then $f(a)$ is the maximum of $f$ on $I$.   
     
  $f$ has a \nuvo{minimum} \pron{£hm£in£im£em} at $a$ on $I$ iff, for all $x \in I$, $f(x) \geq f(a)$.   
  Then $f(a)$ is the minimum of $f$ on $I$.   
\end{Def}   
   
\begin{rem}   
  The word `maximum' \resp{`minimum'} has two plurals: `maximums' \resp{`minimums'} and `maxima'   
  \pron{£hm£aks£im£e} \resp{`minima'}.   
\end{rem}   
   
Maxima and minima are also called \emph{turning points}.\index{point!turning}   
   
\begin{fgre}
  \includegraphics{figures/func01}
\end{fgre}
   
\section{Linear functions}   
   
\begin{Def}   
  Let $a$ and $b$ be two reals. The function $f: \reel \dans \reel$; $x \donne f(x)$ is   
  \nuvo{linear} \pron{£hl£in£Q£Z} iff $f(x) = ax+b$.   
\end{Def}   
\begin{rem}   
If $f$ is linear with $f(x) = ax+b$ and if $a=0$ then the function is \nuvo{constant} \pron{£hk£onst£ent} 
\idest\ all the reals have the same image $b$.   
\end{rem}   
\begin{prop}   
If $f$ is a non-constant linear function, then the increment of $y$ is proportional to the increment   
of $x$. % correction du 17/11/2000   
\end{prop}   
\begin{dem}   
  Let $f$ be a non-constant linear function such that $f(x) = ax+b$. Let $x$ and $x'$ be two reals.   
  The increment of the independent variable is $x'-x$. The increment of the dependent variable is   
  $f(x')-f(x)$.   
   
We have $f(x')-f(x)= \parent{ax'+b}-\parent{ax+b} = ax'+b-ax-b=ax'-ax=a\parent{x'-x}$.   
\end{dem}   
\begin{prop}[converse of the preceding]   
Let $f$ be a function from \reel to \reel. If the increment of $y$ is proportional to the increment of $x$,   
then $f$ is a linear function.   
\end{prop}   
\begin{prop}   
The graph of a linear function is a line.   
\end{prop}   
\begin{prop}[Behaviour of a linear function]   
Let $a$ and $b$ be two reals. Let $f$ be defined by $f: \reel \dans \reel$; $x \donne f(x)= ax+b$.   
\begin{itemize}   
\item If $a=0$ then $f$ is constant on \reel.   
\item If $a>0$ then $f$ is increasing on \reel.   
\item If $a<0$ then $f$ is decreasing on \reel.   
\end{itemize}   
\end{prop}   
   
\begin{rem}[a]   
  In French the linear functions are called \textbf{`affine'}. For a function $f$ to be   
  \textbf{`linéaire'}, $f$ has to map $x$ onto $ax$ with $a$ a constant real. Such a function has a   
  graph which is a line passing through $O$ the origin of the plane.   
     
  In English too it is possible to use the words `affine' \pron{£e£hf£pn} and `linear' with the   
  same meanings as in French but it is usually so just in University texts.   
\end{rem}\index{affine}\index{lineaire@linéaire}   
   
\section{The `square' function}   
   
Let $s$ be defined by $s: \reel \dans \reel$; $x \donne x^2$.\index{square!function}   
   
\begin{prop}   
The function $s$ is decreasing on \intouvfer{\moinsinf}{0} and increasing on \intferouv{0}{\plusinf}.   
\end{prop}   
\begin{dem}   
Let $a$ and $b$ be two reals.   
   
First let $a$ and $b$ be such that $0\leq a < b$. Let us determine the sign of $s(b)-s(a)$.   
$$   
s(b)-s(a)=b^2 - a^2= \parent{b-a}\parent{b+a}   
$$   
Now, as $a<b$, $b-a > 0$ and, as $0\leq a$ and $0 < b$, $b+a > 0$ therefore $s(b)-s(a) > 0$   
\idest\  $s(a) < s(b)$.   
   
So, $s$ is increasing on \intferouv{0}{\plusinf}.   
   
Secondly let $a$ and $b$ be such that $a < b \leq 0$. Then, as $a<b$, $b-a > 0$ and, as $a < 0$ and $b   
\leq 0$, $b+a < 0$ therefore $s(b)-s(a) < 0$ \idest\  $s(a) > s(b)$.   
   
So, $s$ is decreasing on \intouvfer{\moinsinf}{0}. \Qed   
\end{dem}   
   
We can sum up the behaviour of the `square' function with the following table:   
   
\begin{fgre}
  \IG{func02} % tableau de variation de x^2   
\end{fgre}


\newpage   
   
As consequences of the preceding proposition, we can state:   
   
\begin{prop}   
\begin{itemize}   
\item The `square' function has a minimum at $0$. That minimum is $0$.   
\item If $a$ and $b$ are negative reals then $a<b$ is equivalent to $a^2 > b^2$.   
\item If $a$ and $b$ are positive reals then $a<b$ is equivalent to $a^2 < b^2$.   
\end{itemize}   
\end{prop}   
   
The graph \graf{s} of the function $s$ is a \nuvo{parabola} \pron{p£e£hr£ab£el£e} with \nuvo{vertex} at   
$O$ and the $y$-axis as   
line of symmetry.\index{symmetry!line} So \graf{s} is invariant under the reflection in the $y$-axis.%   
\index{vertex!parabola}   
   
\begin{fgre}
  \IG{func03} % parabole y=x^2   
\end{fgre}
   
\begin{prop}   
Let $a$ be a positive real ($a >0$). Then:   
\begin{itemize}   
\item $\dps x^2 = a \Iff x=\sqrt{a} \text{ or } x=-\sqrt{a}$ \idest\  the equation $\dps x^2 = a$ has   
  exactly two solutions which are $\dps \sqrt{a}$ and $\dps -\sqrt{a}$;   
\item $\dps x^2 \leq a \Iff x \in \interfer{-\sqrt{a}}{\sqrt{a}}$;   
\item $\dps x^2 \geq a \Iff x \in \intouvfer{\moinsinf}{-\sqrt{a}}\cup\intferouv{\sqrt{a}}{\plusinf}$.   
\end{itemize}   
\end{prop}   
   
\begin{fgre}
  \IG{func04} % parabole y=x^2 et résolution graphique équation et inéquations   
\end{fgre}

\section{The `inverse' function}   
   
Let $i$ be defined by $i: \reeletoile \dans \reel$; $x \donne \dfrac{1}{x}$.\index{inverse!function}   
   
\begin{prop}   
The function $i$ is decreasing on \interouv{\moinsinf}{0} and on \interouv{0}{\plusinf}.   
\end{prop}   
\begin{dem}   
Let $a$ and $b$ be two reals not equal to $0$. We have   
$$   
i(b)-i(a)= \dfrac{1}{b}-\dfrac{1}{a}=\dfrac{a-b}{ab}   
$$   
now, if $a<b$ then $a-b <0$, therefore the sign of $i(b)-i(a)$ is minus times the sign of $ab$.   
   
If, on one hand, $a<b<0$ then $ab >0$ and $i(b)-i(a)>0$, and if, on the other hand, $0<a<b$ then   
$ab>0$ again and $i(b)-i(a)>0$. Therefore whether $a$ and $b$ are both negative or positive, if   
$a<b$ then $i(a) > i(b)$ \idest\  $i$ is decreasing on both \interouv{\moinsinf}{0} and   
\interouv{0}{\plusinf}.\Qed   
\end{dem}   
\enlargethispage{-\baselineskip}   
   
We can sum up the behaviour of the `inverse' function with the following table:   
   
\begin{fgre}
  \IG{func05} % tableau de variation de 1/x   
\end{fgre}
   
\begin{prop}   
\begin{itemize}   
\item If $a$ and $b$ are negative ($a<0$ and $b<0$) then $a\leq b \Iff \dfrac{1}{a}\geq\dfrac{1}{b}$.   
\item If $a$ and $b$ are positive ($a<0$ and $b<0$) then $a\leq b \Iff \dfrac{1}{a}\geq\dfrac{1}{b}$.   
\end{itemize}   
\end{prop}   
   
The graph \graf{i} of the function $i$ is an \nuvo{hyperbola} \pron{h£p£hp£Eb£el£e} with the $x$-axis and   
the $y$-axis as `axes' or `asymptotes'. \graf{i} is invariant under the reflection in $0$.   
   
\begin{fgre}
  \IG{func06} % courbe y=1/x   
\end{fgre}
   
\section{The trigonometric functions}   
   
Among the \nuvo{trigonometric} \pron{£btr£ig£en£e£hmetr£ik} \emph{functions} or \nuvo{circular} \pron{£hs£Ekj£ul£e£Z}   
\emph{functions} are the \nuvo{sine} \pron{s£pn} and the \nuvo{cosine} \pron{£hk£ws£pn}.   % correction 2003-06-16
   
\subsection{Definitions}   
\setlength{\templong}{.495\linewidth-\tabcolsep}   
\noindent\begin{tabular}{@{}b{1.05\templong} b{0.95\templong}@{}}   
Let \repere be a coordinate system in the plane. Let $I$ and $J$ be points in the plane such that $\dps\vecteur{OI}=\vecti$   
and $\dps\vecteur{OJ}=\vectj$. Let the triangle $IOJ$ be positive.   
Let \graf{c} be a directed circle with center $O$ and radius $1$ and \graf{d} be the tangent to \graf{c} at $I$.   
Let $A$ be a point on \graf{d} such that $IA=1$ and $A$ and $J$ are on the same side of $(OI)$.   
Let $x$ be a real. There exists one and only one point $M$ on \graf{d} such that $x$ is the abscissa   
of $M$ relatively to \couple{I}{A}. If we wind \graf{d} round \graf{c} as shown on the figure, $M$    
coincides with a point   
$m$ on \graf{c}.   
Then   
&   
\begin{Sbox}  
  \IG{func07}
\end{Sbox}
\parbox[t]{0.999\templong}{\framebox[6.2cm]{\TheSbox}}  
\end{tabular}    
   
\begin{Def}   
\begin{itemize}   
\item The cosine of $x$, denoted by $\cos x$, is the abscissa of $m$ in \repere.   
\item The sine of $x$, denoted by $\sin x$, is the ordinate of $m$ in \repere.   
\end{itemize}   
\end{Def}   
   
The function \emph{sine} is the function which associates $\sin x$ to any real $x$.    
The function \emph{cosine} is the function which associates $\cos x$ to any real $x$.    
   
\subsection{Graphs}   
   
\begin{fgre}
  \IG{func08}  % sinus   
\end{fgre}
   
\begin{rem}   
  We observe, and we could prove, that the graph of sine is symmetrical about $O$ the origin of the   
  coordinate system.  That means that, for all real $x$, $\sin(-x)= -\sin x$. The function sine is   
  said to be \nuvo{odd}.   
\end{rem}   
   
\begin{fgre}
  \IG{func09}
\end{fgre}
   
\begin{rem}   
We observe, and we could prove, that the graph of cosine is symmetrical about the $y$-axis.   
That means that, for all real $x$, $\cos(-x)= \cos x$. The function cosine is said to be \nuvo{even}.   
\end{rem}   
   
\begin{fgre}
  \IG{func10}  % les deux    
\end{fgre}
   
\begin{rem}   
  We observe, and we could prove, that the graph of cosine is the image of the graph of sine under   
  the translation by vector $\dfrac{§p}{2}\vecti$.   
     
  For all $x$ real, $\sin\parent{x+2§p}=\sin x$ and $\cos\parent{x+2§p}=\cos x$. The functions sine   
  and cosine are thus said to be \nuvo{periodic} \pron{£bp£Qr£i£h£od£ik} with \nuvo{period} \pron{£hp£Qr£Qd}   
  $2§p$.   
\end{rem}   
   
\subsection{The radian}   
   
The \nuvo{radian} \pron{£hr£qd£Qn} is a unit of measurement of angles. Its abbreviation is `rad'. An angle between two
radii\,\footnote{\emph{plural of} radius \pron{£hr£qd£Qs}.} \pron{£hr£qd£i£b£p} that cut off on the circumference of a
circle an arc equal in length to the radius has a measure of 1~rad. A right angle measures \Dfrac{§p}{2}~rad.
   
\subsection{Some remarkable values}   
   
$$   
{\renewcommand{\arraystretch}{1.2}   
\begin{array}{|c|c|c|c|c|c|}\hline   
\makebox[1.4cm]{}&\makebox[1.4cm]{}&\makebox[1.4cm]{}&\makebox[1.4cm]{}&\makebox[1.4cm]{}&\makebox[1.4cm]{}\\   
x \text{ (rad) }& 0&\dfrac{§p}{6}&\dfrac{§p}{4}&\dfrac{§p}{3}&\dfrac{§p}{2}\\   
& & & & &\\ \hline   
& & & & &\\   
\sin x& 0&\dfrac{1}{2}&\dfrac{\sqrt{2}}{2}&\dfrac{\sqrt{3}}{2}&1\\   
& & & & &\\ \hline   
& & & & &\\   
\cos x& 1&\dfrac{\sqrt{3}}{2}&\dfrac{\sqrt{2}}{2}&\dfrac{1}{2}&0\\   
& & & & &\\ \hline   
\end{array}}   
$$   

\vspace*{\stretch{1}}

\begin{fgre}
  \IG{trigo1}  %cercle trigo et définition des sinus et cosinus. 2002-06-12
\end{fgre}

\vspace*{\stretch{4}}

\endinput   
   
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% correction pour v0.0 2000-11-17
% correction pour v1.1 2001-02-12
% correction pour v1.2 2002-03-11
% correction pour v1.3 2003-06-16
% correction pour v1.4 2003-07-04

