\documentclass[12pt]{article}
\usepackage{ammaths}
\usepackage{wasysym}
\usetikz




\begin{document}

\module{Functions and calculators}{01}{AP07}{1 period}


\prereq{None.}

\object{\begin{itemize}
        \item Discover how to use a calculator to obtain the table of values of a function.
        \item Discover how to use a calculator to graph a function.
        \end{itemize}}

\mater{\begin{itemize}
	\item Calculator.
	\item Slideshow.
	\item Texas and Casio Emulator.
	\end{itemize}}

\modpart{Instruction}{10 mins}

Using a slideshow or a calculator emulator, the teacher explains how to obtain the table of values of a function and how
to graph it. Students are handed out a paper with the explanation for their own calculator.


\modpart{Practice}{Remaining time}

Working alone with their calculator, students have to fill out an exercise sheet.



\pagebreak

\moddocdis{Functions and calculators}{01}{AP07}{TI}

Let $f$ be the function define over $\R$ by $f(x) = x^2 + 1$.\\

\sectionblack{Table of values}
To get the table of values of $f$ for $x$ from $-5$ to $5$ with step $1$ by $1$, follow the steps below.
\begin{center}
\fbox{\parbox{14cm}{
\begin{itemize}
\itemb Press \touche{Y=} and input the expression of $f$.
\itemb Go to Tblset with \touche{2$^\textrm{nd}$}\touche{WINDOW} and input the first value and the step
($\Delta$ Tbl) of the table.
\itemb Show the table with \touche{2$^\textrm{nd}$}\touche{GRAPH}.
\end{itemize}}
}
\end{center}


You should get the following table.
\begin{center}\begin{tabular}{|*{12}{c|}}\hline
$x$ & $-5$ & $-4$ & $-3$ & $-2$ & $-1$ & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ \\\hline
$f(x)$ & $26$ & $17$ & $10$ & $5$ & $2$ & $1$ & $2$ & $5$ & $10$ & $17$ & $26$ \\\hline
\end{tabular}\end{center}

\medskip

\sectionblack{Graph of a function}

The most difficult part of displaying a graph is to choose the right view window. To do so, you can use the Zoom
options provided by your calculator. Another method is to look up in the table of values the minimum and maximum for
the $x$ values and for the $y$ values, and set up the window accordingly. This is the method that is described below.

\begin{center}
\fbox{\parbox{14cm}{
\begin{itemize}
	\itemb Press \touche{Y=} and input the expression of $f$ (done for the table of values).
	\itemb Press \touche{WINDOW} and set the appropriate values for X$_\textrm{min}$,
X$_\textrm{max}$,Y$_\textrm{min}$,Y$_\textrm{max}$.
	\itemb Show the graph with \touche{GRAPH}.
\end{itemize}}
}
\end{center}

Check that you get this graph.
\begin{center}\psset{xunit=0.5cm,yunit=0.125cm,algebraic=true}
\begin{pspicture}(-5,-1)(5,28)
\psplot{-5}{5}{x*x+1}
\psline{->}(-5,0)(5,0)
\psline{->}(0,-1)(0,28)
\rput(-5,4){$-5$}\rput(5,4){$5$}\rput(0.5,-1.5){$0$}\rput(0.5,28){$26$}
\end{pspicture}\end{center}


\pagebreak

\moddocdis{Functions and calculators}{01}{AP07}{Casio}

Let $f$ be the function define over $\R$ by $f(x) = x^2 + 1$.\\

\sectionblack{Table of values}
To get the table of values of $f$ for $x$ from $-5$ to $5$ with step $1$ by $1$, follow the steps below.
\begin{center}
\fbox{\parbox{14cm}{
\begin{itemize}
\itemb Choose the \includegraphics[scale=0.3]{Euro.s01AP07.menutable.eps} menu and input the expression of $f$.
\itemb Go to the range page with \touche{F5} and input the first value, the last value and the pitch.
\itemb Exit and show the table with \touche{Exit}\touche{F6}
\end{itemize}}
}
\end{center}


Check that you get this table.
\begin{center}\begin{tabular}{|*{12}{c|}}\hline
$x$ & $-5$ & $-4$ & $-3$ & $-2$ & $-1$ & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ \\\hline
$f(x)$ & $26$ & $17$ & $10$ & $5$ & $2$ & $1$ & $2$ & $5$ & $10$ & $17$ & $26$ \\\hline
\end{tabular}\end{center}

\medskip

\sectionblack{Graph of a function}

The most difficult part of displaying a graph is to choose the right view window. To do so, you can use the Zoom
options provided by your calculator. Another method is to look up in the table of values the minimum and maximum for
the $x$ values and for the $y$ values, and set up the window accordingly. This is the method that is described below.

\begin{center}
\fbox{\parbox{14cm}{
\begin{itemize}
\itemb Choose the \includegraphics[scale=0.3]{Euro.s01AP07.menugraph.eps} menu and input the expression of $f$.
\itemb Set the table with \touche{V-Window} and set the value for X$_\textrm{min}$,
X$_\textrm{max}$,Y$_\textrm{min}$,Y$_\textrm{max}$ according to the table of values.
\itemb Draw the graph with \touche{F4}
\end{itemize}}
}
\end{center}


Check that you get this graph.
\begin{center}\psset{xunit=0.5cm,yunit=0.125cm,algebraic=true}
\begin{pspicture}(-5,-1)(5,28)
\psplot{-5}{5}{x*x+1}
\psline{->}(-5,0)(5,0)
\psline{->}(0,-1)(0,28)
\rput(-5,4){$-5$}\rput(5,4){$5$}\rput(0.5,-1.5){$0$}\rput(0.5,28){$26$}
\end{pspicture}\end{center}



\pagebreak

\moddocdis{Functions and calculators}{01}{AP07}{Exercises}

\exo{
\begin{enumerate}
\item Let $f_1$ be the function defined over $\R$ by $f_1:x\mapsto -x^2+3x+1$.
  \begin{enumerate}
  \item	Fill out the table below :
  \begin{center}\large
    \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
    \hline
    $x$ & $-4$ & $-3$ & $-2$ & $-1$ & $0$  & $1$ & $2$ & $3$ & $4$\\
    \hline
    $f_1(x)$ & & & &\rule{19pt}{0pt} &\rule{19pt}{0pt} &\rule{19pt}{0pt} &\rule{19pt}{0pt} &\rule{19pt}{0pt}
    &\rule{19pt}{0pt} \\
    \hline
    \end{tabular}
    \end{center}
  \item Deduce from the table of values the best parameters for the view window :
  \begin{center}\large
    \begin{tabular}{|c|p{1cm}|c|p{1cm}|c|p{1cm}|c|p{1cm}|}
    \hline
    Xmin &  & Xmax &  & Ymin &  & Ymax & \\
    \hline
    \end{tabular}
  \end{center}
  \item Draw a quick sketch of the graph in the space below :
  \begin{center}
  \fbox{\parbox{10cm}{\rule{0pt}{4cm}}}
  \end{center}
  \end{enumerate}
\item Let $f_2$ be the function defined over $\R$ by $f_2:x\mapsto x^3-15$.
  \begin{enumerate}
  \item	Fill out the table below :
  \begin{center}\large
    \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|}
    \hline
    $x$ & $-5$ & $-4.5$ &  &  &  &  &  &  &  & & \\
    \hline
    $f_2(x)$ & & & \rule{19pt}{0pt} &\rule{19pt}{0pt} &\rule{19pt}{0pt} &\rule{19pt}{0pt} &\rule{19pt}{0pt}
&\rule{19pt}{0pt}
    &\rule{19pt}{0pt}  &\rule{19pt}{0pt} &\rule{19pt}{0pt}\\
    \hline
    \hline
    $x$ &  & &  &  &  &  &  &  &  & & $5$\\
    \hline
    $f_2(x)$ &  & &  &  &  &  &  &  &  & &\\
    \hline
    \end{tabular}
    \end{center}
  \item Deduce from the table of values the best parameters for the view window :
  \begin{center}\large
    \begin{tabular}{|c|p{1cm}|c|p{1cm}|c|p{1cm}|c|p{1cm}|}
    \hline
    Xmin &  & Xmax &  & Ymin &  & Ymax & \\
    \hline
    \end{tabular}
  \end{center}
  \item Draw a quick sketch of the graph in the space below :
  \begin{center}
  \fbox{\parbox{10cm}{\rule{0pt}{4cm}}}
  \end{center}
  \end{enumerate}
\end{enumerate}}


\pagebreak

\exo{Let $g$ be the function defined by $g(x)=x^2-x-1$. 
\begin{enumerate}
\item \begin{enumerate}
    \item Fill out the following table of value :
      \begin{center}\begin{tabular}{|*{9}{c|}}\hline
      $x$ & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$\\\hline
      $g(x)$ & \rule{19pt}{0pt} & \rule{19pt}{0pt} & \rule{19pt}{0pt} & \rule{19pt}{0pt} 
	     & \rule{19pt}{0pt} & \rule{19pt}{0pt} & \rule{19pt}{0pt} & \rule{19pt}{0pt}\\\hline
      \end{tabular}\end{center}
    \item Between what values is the solution of the equation $g(x)=0$ ? \smallskip \\
    \hspace*{-0.9cm}\fbox{\parbox{15cm}{\rule{0pt}{1cm}}}
    \end{enumerate}
\item \begin{enumerate}
      \item Build the table of value of $g$ between 1 and 2 with a pitch of $0.1$. Copy it below. 
      \begin{center}\begin{tabular}{|*{12}{c|}}\hline
      $x$ &  &  &  &  &  &  &  &  &  & & \\\hline
      $g(x)$ & \rule{19pt}{0pt} & \rule{19pt}{0pt} & \rule{19pt}{0pt} & \rule{19pt}{0pt} & \rule{19pt}{0pt} 
	     & \rule{19pt}{0pt} & \rule{19pt}{0pt} & \rule{19pt}{0pt} 
	     & \rule{19pt}{0pt} & \rule{19pt}{0pt} & \rule{19pt}{0pt}\\\hline
      \end{tabular}\end{center}
      \item Deduce an approximation at $1$ DP of the solution of the equation $g(x)=0$. \smallskip \\
      \hspace*{-0.9cm}\fbox{\parbox{15cm}{\rule{0pt}{1cm}}}
      \end{enumerate}
\item Use the same process to give an approximate value at 4 DP of the solution of the equation $g(x)=0$. \smallskip \\
\fbox{\parbox{15cm}{\rule{0pt}{1cm}}}
\end{enumerate}}

\exo{For each function, give the values that define the best fit window and do a quick sketch of the graph.

\begin{multicols}{2}
$f_3(x)=x^5+32$ over $[-2;2]$\\ \smallskip\\
{\large
\begin{tabular}{|c|p{1cm}|c|p{1cm}|}
\hline
Xmin &  & Ymin &  \\
\hline
Xmax &  & Ymax &  \\
\hline
\end{tabular}}
\bigskip
\columnbreak

\fbox{\parbox{7cm}{\rule{0pt}{4cm}}}
\end{multicols}

\begin{multicols}{2}
$f_4(x)=\cfrac{1}{1+x^2}$ over $[-4;4]$\\ \smallskip\\
{\large
\begin{tabular}{|c|p{1cm}|c|p{1cm}|}
\hline
Xmin &  & Ymin &  \\
\hline
Xmax &  & Ymax &  \\
\hline
\end{tabular}}
\bigskip
\columnbreak

\fbox{\parbox{7cm}{\rule{0pt}{4cm}}}
\end{multicols}

\begin{multicols}{2}
$f_5(x)=(4x+1)^3(4x-9)+10$ over $[-2;3]$\\ \smallskip\\
{\large
\begin{tabular}{|c|p{1cm}|c|p{1cm}|}
\hline
Xmin &  & Ymin &  \\
\hline
Xmax &  & Ymax &  \\
\hline
\end{tabular}}
\bigskip
\columnbreak

\fbox{\parbox{7cm}{\rule{0pt}{4cm}}}
\end{multicols}

\begin{multicols}{2}
$f_6(x)=\cfrac{x-5}{100}$ over $[0;5]$\\ \smallskip\\
{\large
\begin{tabular}{|c|p{1cm}|c|p{1cm}|}
\hline
Xmin &  & Ymin &  \\
\hline
Xmax &  & Ymax &  \\
\hline
\end{tabular}}
\bigskip
\columnbreak

\fbox{\parbox{7cm}{\rule{0pt}{4cm}}}
\end{multicols}


}


\exo{Let $f$ and $g$ be two functions defined over $\R$ by $f(x) =\cfrac{x^3-2x}{x^2+1}$ et $g(x) = 0.5x$. The aim of
this exercise is to get an approximation of the coordinates of the intersection points of the graphs of $f$ and $g$.
\begin{enumerate}
\item On your calculator, display the graphs of the two functions over the interval $[-5;5]$. Draw a quick graph of the
screen below. \smallskip \\
\fbox{\parbox{15cm}{\rule{0pt}{4cm}}}
\item Use \touche{TRACE} to read the coordinates of the intersection points of the graph of $f$ and $g$. \smallskip \\
\fbox{\parbox{15cm}{\rule{0pt}{1cm}}}
\item Solve the equation $f(x) = g(x)$ to check the result. \smallskip \\
\fbox{\parbox{15cm}{\rule{0pt}{4cm}}}
\end{enumerate}
}

\pagebreak


\exo{Let $g$ be the function defined by  $g : x  \longmapsto
5x^3-0.1x+0.005.$
\begin{enumerate}
\item Draw with your calculator the graph ${\cal C}_g$ of this function in a standard window
(the minimum on each axis must be $-10$ and the maximum $10$).
\item How many times does ${\cal C}_g$ seem to pass through the $x$-axis ?
What does it mean about the equation $g(x)=0$ ? \smallskip \\
\fbox{\parbox{15cm}{\rule{0pt}{1cm}}}
\item Give approximate values of the solutions to the previous equation. For each of these solutions check that
$f(x)=0$. \smallskip \\
\fbox{\parbox{15cm}{\rule{0pt}{3cm}}}
\item Set the minimum and maximum on each axis respectively to $-1$ and $1$.
\item How many times does ${\cal C}_g$ seem to pass through the $x$-axis ?
What does it mean about the equation $g(x)=0$ ? \smallskip \\
\fbox{\parbox{15cm}{\rule{0pt}{1cm}}}
\item Give approximate values of the solutions to the previous equation. For each of these solutions check that
$f(x)=0$. \smallskip \\
\fbox{\parbox{15cm}{\rule{0pt}{3cm}}}
\item Set the $x$-axis from $-0.2$ to $0.2$ and the $y$-axis from $-0.1$ to $0.1$.
\item How many times does ${\cal C}_g$ seem to pass through the $x$-axis ?
What does it mean about the equation $g(x)=0$ ? \smallskip \\
\fbox{\parbox{15cm}{\rule{0pt}{1cm}}}
\item Give approximate values of the solutions to the previous equation. For each of these solutions check that
$f(x)=0$. \smallskip \\
\fbox{\parbox{15cm}{\rule{0pt}{3cm}}}
\end{enumerate}
}


 
\pagebreak

\moddocdis{Functions and calculators}{01}{AP07}{Solutions}

\exo{
\begin{enumerate}
\item Let $f_1$ be the function defined over $\R$ by $f_1:x\mapsto -x^2+3x+1$.
  \begin{enumerate}
  \item	Fill out the table below :
  \begin{center}\large
    \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
    \hline
    $x$ & $-4$ & $-3$ & $-2$ & $-1$ & $0$  & $1$ & $2$ & $3$ & $4$\\
    \hline
    $f_1(x)$ &$-27$&$-17$&$-9$&$-3$&$1$&$3$&$
    3$&$1$&$-3$\\
    \hline
    \end{tabular}
    \end{center}
  \item Deduce from the table of values the best parameters for the view window :
  \begin{center}\large
    \begin{tabular}{|c|p{1cm}|c|p{1cm}|c|p{1cm}|c|p{1cm}|}
    \hline
    Xmin & $-4$ & Xmax & $4$ & Ymin & $-27$ & Ymax &$3$ \\
    \hline
    \end{tabular}
  \end{center}
  \item Draw a quick sketch of the graph in the space below :
  \begin{center}
  \fbox{\parbox{10cm}{
  $$\psset{xunit=1cm,yunit=0.1cm,algebraic=true,arrowsize=5pt}
    \def\xmin{-4}\def\ymin{-27}\def\xmax{4}\def\ymax{3}
    \begin{pspicture*}(\xmin,\ymin)(\xmax,\ymax)
    \psline{->}(\xmin,0)(\xmax,0)\psline{->}(0,\ymin)(0,\ymax)
    \psplot[linecolor=rouge]{\xmin}{\xmax}{-x*x+3*x+1}
    \end{pspicture*}$$
  }}
  \end{center}
  \end{enumerate}
\item Let $f_2$ be the function defined over $\R$ by $f_2:x\mapsto x^3-15$.
  \begin{enumerate}
  \item	Fill out the table below :
  \end{enumerate}
  \begin{center}
    \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|}
    \hline
    $x$ & $-5$ & $-4.5$ &$-4$  & $-3.5$ & $-3$ &$-2.5$  &$-2$ & $-1.5$ &$-1$  &$-0.5$ & $0$ \\
    \hline
    $f_2(x)$ & $-140$&$106.1$&$-79$&$57.9$&$-42$&$30.6$&$-23$&$18.4$&$-16$&$15.1$&$-15$\\
    \hline
    \hline
    $x$ & $0.5$ & $1$ &$1.5$  & $2$ & $2.5$ &$3$  &$3.5$ & $4$ &$4.5$  & & $5$ \\
    \hline
    $f_2(x)$ & $-14.9$& $-14$&$-11.6$&$-7$&$0.6$&$12$&$27.9$&$49$&$76.1$& & $110$\\
    \hline
    \end{tabular}
    \end{center}
   \begin{enumerate}\setcounter{enumi}{1} 
  \item Deduce from the table of values the best parameters for the view window :
  \begin{center}\large
    \begin{tabular}{|c|p{1cm}|c|p{1cm}|c|p{1cm}|c|p{1cm}|}
    \hline
    Xmin & $-5$ & Xmax & $5$ & Ymin & $-140$ & Ymax &$110$ \\
    \hline
    \end{tabular}
  \end{center}
  \item Draw a quick sketch of the graph in the space below :
  \begin{center}
  \fbox{\parbox{10cm}{$$\psset{xunit=1cm,yunit=0.02cm,algebraic=true,arrowsize=5pt}
    \def\xmin{-5}\def\ymin{-140}\def\xmax{5}\def\ymax{110}
    \begin{pspicture*}(\xmin,\ymin)(\xmax,\ymax)
    \psline{->}(\xmin,0)(\xmax,0)\psline{->}(0,\ymin)(0,\ymax)
    \psplot[linecolor=rouge]{\xmin}{\xmax}{x*x*x-15}
    \end{pspicture*}$$}}
  \end{center}
  \end{enumerate}
\end{enumerate}}


\pagebreak

\exo{Let $g$ be the function defined by $g(x)=x^2-x-1$. 
\begin{enumerate}
\item \begin{enumerate}
    \item Fill out the following table of value :
      \begin{center}\begin{tabular}{|*{9}{c|}}\hline
      $x$ & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$\\\hline
      $g(x)$ & $-1$ & $-1$ & $1$ & $5$ & $11$ & $19$ & $29$ & $41$\\\hline
      \end{tabular}\end{center}
    \item Between what values is the solution of the equation $g(x)=0$ ? \smallskip \\
    \hspace*{-0.9cm}\fbox{\parbox{15cm}{between $1$ and $2$}}
    \end{enumerate}
\item \begin{enumerate}
      \item Build the table of value of $g$ between 1 and 2 with a pitch of $0.1$. Copy it below. 
      \begin{center}\begin{tabular}{|*{12}{c|}}\hline
      $x$ & $1$ & $1.1$ & $1.2$ & $1.3$ & $1.4$ & $1.5$ &$1.6$ & $1.7$ & $1.8$ &$1.9$ & $2$ \\\hline
      $g(x)$ & $-1$&$-0.89$&$-0.76$&$0.61$&$-0.44$&$-0.25$&$-0.4$&$0.19$&$0.44$&$0.71$&$1$\\\hline
      \end{tabular}\end{center}
      \item Deduce an approximation at $1$ DP of the solution of the equation $g(x)=0$. \smallskip \\
      \hspace*{-0.9cm}\fbox{\parbox{15cm}{between $1.6$ and $1.7$}}
      \end{enumerate}
\item Use the same process to give an approximate value at 4 DP of the solution of the equation $g(x)=0$. \smallskip \\
\fbox{\parbox{15cm}{$1.6180$}}
\end{enumerate}}

\exo{For each function, give the values that define the best fit window and do a quick sketch of the graph.

\begin{multicols}{2}
$f_3(x)=x^5+32$ over $[-2;2]$\\ \smallskip\\
{\large
\begin{tabular}{|c|p{1cm}|c|p{1cm}|}
\hline
Xmin & $-2$ & Ymin & $0$ \\
\hline
Xmax & $2$ & Ymax & $64$ \\
\hline
\end{tabular}}
\bigskip
\columnbreak

\fbox{\parbox{7cm}{
$$\psset{xunit=1.5cm,yunit=0.05cm,algebraic=true,arrowsize=5pt}
    \def\xmin{-2}\def\ymin{0}\def\xmax{2}\def\ymax{64}
    \begin{pspicture*}(\xmin,\ymin)(\xmax,\ymax)
    \psline{->}(\xmin,0)(\xmax,0)\psline{->}(0,\ymin)(0,\ymax)
    \psplot[linecolor=rouge]{\xmin}{\xmax}{x*x*x*x*x+32}
    \end{pspicture*}$$
}}
\end{multicols}

\begin{multicols}{2}
$f_4(x)=\cfrac{1}{1+x^2}$ over $[-4;4]$\\ \smallskip\\
{\large
\begin{tabular}{|c|p{1cm}|c|p{1cm}|}
\hline
Xmin & $-4$ & Ymin & $0$ \\
\hline
Xmax & $4$ & Ymax & $1$ \\
\hline
\end{tabular}}
\bigskip
\columnbreak

\fbox{\parbox{7cm}{
$$\psset{xunit=0.8cm,yunit=3cm,algebraic=true,arrowsize=5pt}
    \def\xmin{-4}\def\ymin{0}\def\xmax{4}\def\ymax{1}
    \begin{pspicture*}(\xmin,\ymin)(\xmax,\ymax)
    \psline{->}(\xmin,0)(\xmax,0)\psline{->}(0,\ymin)(0,\ymax)
    \psplot[linecolor=rouge]{\xmin}{\xmax}{1/(1+x*x)}
    \end{pspicture*}$$
}}
\end{multicols}

\begin{multicols}{2}
$f_5(x)=(4x+1)^3(4x-9)+10$ over $[-2;3]$\\ \smallskip\\
{\large
\begin{tabular}{|c|p{1cm}|c|p{1cm}|}
\hline
Xmin & $-2$ & Ymin & $-719$ \\
\hline
Xmax & $3$ & Ymax & $6601$ \\
\hline
\end{tabular}}
\bigskip
\columnbreak

\fbox{\parbox{7cm}{
$$\psset{xunit=1cm,yunit=0.0004cm,algebraic=true,arrowsize=5pt}
    \def\xmin{-2}\def\ymin{-719}\def\xmax{3}\def\ymax{6601}
    \begin{pspicture*}(\xmin,\ymin)(\xmax,\ymax)
    \psline{->}(\xmin,0)(\xmax,0)\psline{->}(0,\ymin)(0,\ymax)
    \psplot[linecolor=rouge]{\xmin}{\xmax}{(4*x+1)*(4*x+1)*(4*x+1)*(4*x-9)+10}
    \end{pspicture*}$$
}}
\end{multicols}

\begin{multicols}{2}
$f_6(x)=\cfrac{x-5}{100}$ over $[0;5]$\\ \smallskip\\
{\large
\begin{tabular}{|c|p{1cm}|c|p{1cm}|}
\hline
Xmin & $0$  & Ymin & $-0.05$ \\
\hline
Xmax & $5$ & Ymax & $0$ \\
\hline
\end{tabular}}
\bigskip
\columnbreak

\fbox{\parbox{7cm}{
$$\psset{xunit=1cm,yunit=80cm,algebraic=true,arrowsize=5pt}
    \def\xmin{0}\def\ymin{-0.05}\def\xmax{5}\def\ymax{0}
    \begin{pspicture*}(\xmin,\ymin)(\xmax,\ymax)
    \psline{->}(\xmin,0)(\xmax,0)\psline{->}(0,\ymin)(0,\ymax)
    \psplot[linecolor=rouge]{\xmin}{\xmax}{(x-5)*0.01}
    \end{pspicture*}$$
}}
\end{multicols}


}


\exo{Let $f$ and $g$ be two functions defined over $\R$ by $f(x) =\cfrac{x^3-2x}{x^2+1}$ et $g(x) = 0.5x$. The aim of
this exercise is to get an approximation of the coordinates of the intersection points of the graphs of $f$ and $g$.
\begin{enumerate}
\item On your calculator, display the graphs of the two functions over the interval $[-5;5]$. Draw a quick graph of the
screen below. \smallskip \\
\fbox{\parbox{15cm}{
$$\psset{xunit=1cm,yunit=0.5cm,algebraic=true,arrowsize=5pt}
    \def\xmin{-5}\def\ymin{-5}\def\xmax{5}\def\ymax{5}
    \begin{pspicture*}(\xmin,\ymin)(\xmax,\ymax)
    \psaxes[labels=all,labelsep=1pt, Dx=1,Dy=1]{->}(0,0)(\xmin,\ymin)(\xmax,\ymax)
    \psgrid[gridlabels=0pt,gridwidth=.3pt, gridcolor=gray, subgridwidth=.3pt, subgridcolor=gray, subgriddiv=1](\xmin,\ymin)(\xmax,\ymax)
    \psplot[linecolor=rouge]{\xmin}{\xmax}{(x^3-2*x)/(x^2+1)}
    \psplot[linecolor=bleu]{\xmin}{\xmax}{0.5*x}
    \end{pspicture*}$$
}}
\item Use \touche{TRACE} to read the coordinates of the intersection points of the graph of $f$ and $g$. \smallskip \\
\fbox{\parbox{15cm}{The coordinates of the intersection points seem to be $$(-2,-1)\qquad (0,0)\qquad (2,1)$$}}
\item Solve the equation $f(x) = g(x)$ to check the result. \smallskip \\
\fbox{\parbox{15cm}{
\begin{eqnarray*}
f(x)&=&g(x)\\
 \cfrac{x^3-2x}{x^2+1}&=&0.5x \\
 x^3-2x&=&0.5x(x^2+1)\\
 x^3-0.5x^3-2x-0.5x&=&0\\
 x(0.5x^2-2.5)&=&0\\
 x=0 &OU& x^2-5=0\\
 x=0 &OU& x=\pm\sqrt{5}\\
\end{eqnarray*}

}}
\end{enumerate}
}

\pagebreak


\exo{Let $g$ be the function defined by  $g : x  \longmapsto
5x^3-0.1x+0.005.$
\begin{enumerate}
\item Draw with your calculator the graph ${\cal C}_g$ of this function in a standard window
(the minimum on each axis must be $-10$ and the maximum $10$).
\item How many times does ${\cal C}_g$ seem to pass through the $x$-axis ?
What does it mean about the equation $g(x)=0$ ? \smallskip \\
\fbox{\parbox{15cm}{\rule{0pt}{1cm}}}
\item Give approximate values of the solutions to the previous equation. For each of these solutions check that
$f(x)=0$. \smallskip \\
\fbox{\parbox{15cm}{\rule{0pt}{3cm}}}
\item Set the minimum and maximum on each axis respectively to $-1$ and $1$.
\item How many times does ${\cal C}_g$ seem to pass through the $x$-axis ?
What does it mean about the equation $g(x)=0$ ? \smallskip \\
\fbox{\parbox{15cm}{\rule{0pt}{1cm}}}
\item Give approximate values of the solutions to the previous equation. For each of these solutions check that
$f(x)=0$. \smallskip \\
\fbox{\parbox{15cm}{\rule{0pt}{3cm}}}
\item Set the $x$-axis from $-0.2$ to $0.2$ and the $y$-axis from $-0.1$ to $0.1$.
\item How many times does ${\cal C}_g$ seem to pass through the $x$-axis ?
What does it mean about the equation $g(x)=0$ ? \smallskip \\
\fbox{\parbox{15cm}{\rule{0pt}{1cm}}}
\item Give approximate values of the solutions to the previous equation. For each of these solutions check that
$f(x)=0$. \smallskip \\
\fbox{\parbox{15cm}{\rule{0pt}{3cm}}}
\end{enumerate}
}


 



\end{document}
