\documentclass[mathserif, t]{beamer}
%\documentclass[mathserif, handout, t]{beamer}

\input{/home/sarah/Dropbox/Travail_S/Divers/Preambule/Preambule_Cours}

\def\mySample{100}
\Teachtrue

\begin{document}

\titreCours{10}{Inéquations et tableaux de signes}{}{}

%%%%%%%%%%%%
% Evaluation graphique, images et zéros
%%%w
%
%
\begin{frame}{1. Notation en intervalle}

\def\a{2}
\def\b{5}
\def\max{11}
\def\min{-11}

\underline{Exemple 1.2} Dessiner les solutions des ces inéquations et les écrire sous forme d'intervalle.
\begin{multicols}{2}
\begin{itemize}
\item $x \ro{>} 2 \hunc{3}{\Rightarrow S=\ro{]}2,\infty[}$
\begin{tikzpicture}[scale=0.2]
\scriptsize
\draw[->] (-7,0) node[left]{$-\infty$} -- coordinate (x axis mid) (7,0) node[right]{$+\infty$};
\draw (\a,5pt) -- (\a,-5pt) node[anchor=south, yshift=0.05cm] {\a};			
\hunc{2}{\draw[dblue, very thick, ->] (\a, 0) -- (7, 0);
\draw[dred] (\a,0) circle (20pt);}
\end{tikzpicture}

\item<4-> $x \bl{\geq} 2 \hunc{6}{\Rightarrow S=\bl{[} 2,\infty[}$
\begin{tikzpicture}[scale=0.2]
\scriptsize
\draw[->] (-7,0) node[left]{$-\infty$} -- coordinate (x axis mid) (7,0) node[right]{$+\infty$};
\draw (\a,5pt) -- (\a,-5pt) node[anchor=south, yshift=0.05cm] {\a};			
\hunc{5}{\draw[dblue, very thick, ->] (\a, 0) -- (7, 0);
\fill[dblue] (\a,0) circle (20pt);}
\end{tikzpicture}

\item<7-> $x \ro{<} 2 \hunc{9}{\Rightarrow S=]-\infty, 2 \ro{[}}$
\begin{tikzpicture}[scale=0.2]
\scriptsize
\draw[->] (-7,0) node[left]{$-\infty$} -- coordinate (x axis mid) (7,0) node[right]{$+\infty$};
\draw (\a,5pt) -- (\a,-5pt) node[anchor=south, yshift=0.05cm] {\a};			
\hunc{8}{\draw[dblue, very thick] (-7, 0) -- (\a, 0);
\draw[dred] (\a,0) circle (20pt);}
\end{tikzpicture}

\item<10-> $x \bl{\leq} 2 \hunc{12}{\Rightarrow S=]-\infty, 2\bl{]}}$
\begin{tikzpicture}[scale=0.2]
\scriptsize
\draw[->] (-7,0) node[left]{$-\infty$} -- coordinate (x axis mid) (7,0) node[right]{$+\infty$};
\draw (\a,5pt) -- (\a,-5pt) node[anchor=south, yshift=0.05cm] {\a};			
\hunc{11}{\draw[dblue, very thick] (-7, 0) -- (\a, 0);
\fill[dblue] (\a,0) circle (20pt);}
\end{tikzpicture}
\end{itemize}
\end{multicols}

\vspace{0.5cm}
%
\uncover<13->{\underline{Exemple 1.3} Ecrire sous forme d'intervalle l'expression \only<13-15>{$$-2 < x \leq 5$$}\hunc{16}{\only<16->{$$-2 \,\ro{<}\, x \,\bl{\leq}\, 5$$}}.}
%
\hunc{14}{
\def\a{-2}
\def\b{5}
\begin{center}
\begin{tikzpicture}[scale=0.25]
\draw[->] (-11,0)node[left]{$-\infty$} -- coordinate (x axis mid) (11,0) node[right]{$+\infty$};
\foreach \x in {-10, -5, ..., 10}
     		\draw (\x,1pt) -- (\x,-3pt)
			node[anchor=north] {\x};
		
\draw (\a,5pt) -- (\a,-5pt) node[anchor=south, yshift=0.2cm] {\a};			
\draw (\b,5pt) -- (\b,-5pt) node[anchor=south, yshift=0.2cm] {\b};			

\uncover<15->{	
\draw[dblue, very thick] (\a, 0) -- (\b, 0);}

\uncover<17->{\draw[dred] (\a,0) circle (20pt);}
\uncover<18->{\fill[dblue] (\b,0) circle (20pt);}

\end{tikzpicture}\end{center}}
%
\hunc{19}{On a donc $S= \uncover<11->{\,\ro{]}\, -2; 5 \,\bl{]}}$.}

\end{frame}

\begin{frame}

\underline{Exemple 1.4} Noter sous forme d'intervalle l'ensemble illustré ci-dessous:

\def\a{1.9}
\def\b{7.2}
\def\max{11}
\def\min{-11}
\def\c{-7.7}
\def\d{-2.4}
\begin{center}
\begin{tikzpicture}[scale=0.4]
\draw[->] (-11,0)node[left]{$-\infty$} -- coordinate (x axis mid) (11,0) node[right]{$+\infty$};
\foreach \x in {-10, -5, ..., 10}
     		\draw (\x,1pt) -- (\x,-3pt)
			node[anchor=north] {\x};
		
\draw (\a,5pt) -- (\a,-5pt) node[anchor=south, yshift=0.2cm] {\a};			
\draw (\b,5pt) -- (\b,-5pt) node[anchor=south, yshift=0.2cm] {\b};			
\draw (\c,5pt) -- (\c,-5pt) node[anchor=south, yshift=0.2cm] {\c};			
\draw (\d,5pt) -- (\d,-5pt) node[anchor=south, yshift=0.2cm] {\d};			

\draw[dblue, very thick] (\a, 0) -- (\b, 0);
\draw[dred] (\a,0) circle (10pt);
\fill[dblue] (\b,0) circle (10pt);

\draw[dblue, very thick] (\c, 0) -- (\d, 0);
\fill[dblue] (\c,0) circle (10pt);
\draw[dred] (\d,0) circle (10pt);

\end{tikzpicture}
\end{center}

\hunc{2}{
\uncover<2->{On utilise le symbole $\ve{\cup}$ ("union") pour dénoter \ve{la réunion de deux intervalles}:}
\uncover<3->{$$S=\bl{[} \c; \d \ro{[} \,\ve{\cup}\, \ro{]} \a; \b \bl{]}$$}}

\ifTeach\else\vspace{2cm}\fi

\uncover<4->{\underline{Remarque 1.1} On peut noter les intervalles sous la forme ensembliste:
\begin{enumerate}
\item $\bl{[}-2; 3\bl{]} \hunc{5}{= \{ x \in \mathbb{R} | -2 \bl{\leq} x \bl{\leq} 3 \}}$
\item<6-> $\ro{]}-2; 3\bl{]} \hunc{7}{= \{ x \in \mathbb{R} | -2 \ro{<} x \bl{\leq} 3 \}}$
\item<8-> $\bl{[}-2; 3\ro{[} \hunc{9}{= \{ x \in \mathbb{R} | -2 \bl{\leq} x \ro{<} 3 \}}$
\item<10-> $\ro{]}-2; 3\ro{[} \hunc{11}{= \{ x \in \mathbb{R} | -2 \ro{<} x \ro{<} 3 \}}$
\end{enumerate}
} 


\end{frame}

\begin{frame} 

\underline{Exemple 1.5} Résoudre l'inéquation $x -3 > 4x -2$. Indiquer la solution sous forme d'intervalle. 

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Your Answer
 %Tableaux de signes 
%%%%%%

\begin{frame}{2. Tableaux de signes}

\uncover<2->{
\underline{Exemple 2.1} Soit $f$ une fonction dont le graphe est donné ci-dessous. Etudier le signe de cette fonction. 
\begin{center}
\begin{tikzpicture}[scale=0.4]
\scriptsize
\draw[gray, very thin] (-7.9,-5.9) grid (7.9,5.9);
\draw[-latex, thick] (-8,0) -- (8,0) node[right]{$x$};
\draw[-latex, thick] (0,-6) -- (0,6) node[above]{$y$};
\hunc{3}{\draw[thick, dred] (0,-6)node[right]{-} --  (0,0);}
\hunc{2}{\draw[thick, dblue] (0,0) -- (0,6) node[right]{+} ;}

\draw (1, -0.1) -- (1, 0.1) node[below]{1};
\draw (-2, -0.1) -- (-2, 0.1) node[below]{-2};
\draw (3, -0.1) -- (3, 0.1) node[below]{3};
\draw (0.1,1) -- (-0.1,1) node[left]{1};
\clip (-8,-6) rectangle (8,6);
\draw[very thick, samples=\mySample, domain=-8:8] plot(\x,{0.5*(\x-1)*(\x -3)*(\x+2)});
\hunc{4}{\draw[dred, very thick, samples=\mySample, domain=-8:-2] plot(\x,{0.5*(\x-1)*(\x -3)*(\x+2)});}
\hunc{5}{\node[dred, left] at (-3,-3) {$f(x)<0$};}
\hunc{6}{\fill[dgreen] (-2,0) circle (5pt) node[above left] {\only<7->{$f(x)=0$}};}
\hunc{8}{\draw[dblue, very thick, samples=\mySample, domain=-2:1] plot(\x,{0.5*(\x-1)*(\x -3)*(\x+2)});}
\hunc{9}{\node[dblue, left] at (-2,3) {$f(x)>0$};}
\hunc{10}{\fill[dgreen] (1,0) circle (5pt) node[above right] {\only<11->{$f(x)=0$}};}
\hunc{12}{\draw[dred, very thick, samples=\mySample, domain=1:3] plot(\x,{0.5*(\x-1)*(\x -3)*(\x+2)});}
\hunc{13}{\node[dred] at (2,-3) {$f(x)<0$};}
\hunc{14}{\fill[dgreen] (3,0) circle (5pt) node[below right] {\only<15->{$f(x)=0$}};}
\hunc{16}{\draw[dblue, very thick, samples=\mySample, domain=3:8] plot(\x,{0.5*(\x-1)*(\x -3)*(\x+2)});}
\hunc{17}{\node[dblue, right] at (4,3) {$f(x)>0$};}
\end{tikzpicture}
\end{center}}

\hunc{18}{
On construit un \bl{tableau de signes} pour résumer la situation

\small
\hunc{19}{
$$\begin{array}{ c | l c c c c c c c r}
x & -\infty & &\hunc{20}{\ve{-2}} & & \hunc{21}{\ve{1}} & & \hunc{22}{\ve{3}} & & +\infty \\
\hline
\hunc{15}{f(x) & & \hunc{23}{\ro{-}} & \hunc{24}{\ve{0}} & \hunc{25}{\bl{+}} & \hunc{26}{\ve{0}} & \hunc{27}{\ro{-}} & \hunc{28}{\ve{0}} & \hunc{29}{\bl{+} }& }\\
\end{array}$$
}}

\end{frame}

\begin{frame}

\underline{Exercice 2.1} Esquisser le graphe des fonctions suivantes et compléter leur tableau de signes. \\[0.2cm]

\begin{minipage}{0.49\textwidth}
a) $f(x)= 2x -3$
\myTikz{0.4}{-5}{5}{-5}{5}{ 
\hunc{2}{\fill (0,-3) circle (4pt) node[left]{-3};}
\hunc{3}{\draw (0,-3) -- node[below]{1} (1, -3);}
\hunc{4}{\draw (1,-3) -- node[right]{2} (1, -1);}
\hunc{5}{\draw[thick, samples=2, domain=\xmin:\xmax] plot(\x, {2*\x-3});}
\hunc{6}{\fill[dgreen] (1.5,0) circle (4pt) node[below]{$\frac{3}{2}$};}
\hunc{7}{\draw[thick, samples=2, domain=\xmin:1.5, dred] plot(\x, {2*\x-3});}
\hunc{8}{\draw[thick, samples=2, domain=1.5:\xmax, dblue] plot(\x, {2*\x-3});}
}

\scriptsize
\begin{tikzpicture}[scale=0.4]
\tkzTabInit[lgt=2.5,deltacl=1, nocadre]{$x$/1,$2x-3$/1}{-$\infty$, $\hunc{9}{\frac{3}{2}}$, +$\infty$}
\hunc{10}{\tkzTabLine{, - , 0, +,};}
\end{tikzpicture}

\normalsize
\hunc{20}{\centering Pente positive ($m=2$)}
\end{minipage}\begin{minipage}{0.49\textwidth}
\uncover<10->{b) $f(x)= 3 - x$
\myTikz{0.4}{-5}{5}{-5}{5}{ 
\hunc{11}{\fill (0,3) circle (4pt) node[left]{3};}
\hunc{12}{\draw (0,3) -- node[above]{1} (1, 3);}
\hunc{13}{\draw (1,3) -- node[right]{1} (1, 2);}
\hunc{14}{\draw[thick, samples=2, domain=\xmin:\xmax] plot(\x, {-\x+3});}
\hunc{15}{\fill[dgreen] (3,0) circle (4pt) node[below]{$3$};}
\hunc{16}{\draw[thick, samples=2, domain=\xmin:3, dblue] plot(\x, {-\x+3});}
\hunc{17}{\draw[thick, samples=2, domain=3:\xmax, dred] plot(\x, {-\x+3});}
}

\scriptsize
\begin{tikzpicture}[scale=0.4]
\tkzTabInit[lgt=2.5, deltacl=1, nocadre]{$x$/1,$3-x$/1}{-$\infty$, $\hunc{18}{3}$, +$\infty$}
\hunc{19}{\tkzTabLine{, + , 0, -,};}
\end{tikzpicture}}

\normalsize
\hunc{21}{\centering Pente négative ($m=-1$)}
\end{minipage}

\end{frame}

\begin{frame}

\underline{Règle 2.1} Le signe d'une fonction affine $f(x)=mx+h$ est déterminée par sa \bl{pente} et son \ve{zéro}:\\[0.2cm]

\begin{minipage}{0.49\textwidth}
\uncover<2->{
\begin{center}
\textbf{Pente positive $(m>0)$}

\begin{tikzpicture}[scale=0.4]
\tkzTabInit[lgt=4, deltacl=1, nocadre]{$x$/1,$mx+h$/1}{-$\infty$, \ve{zéro} , +$\infty$}
\tkzTabLine{, - , 0, +,};
\end{tikzpicture}
\end{center}}
\end{minipage}
\begin{minipage}{0.49\textwidth}
\uncover<3->{
\begin{center}
\textbf{Pente négative $(m<0)$}

\begin{tikzpicture}[scale=0.4]
\tkzTabInit[lgt=4, deltacl=1, nocadre]{$x$/1,$mx+h$/1}{-$\infty$, \ve{zéro} , +$\infty$}
\tkzTabLine{, + , 0, -,};
\end{tikzpicture}
\end{center}}
\end{minipage}

\vspace{0.5cm}
\uncover<4->{
\underline{Exemple 2.2} Donner le tableau de signes de la fonction donnée par $f(x)= 2x -1$.

\hunc{2}{
\vfill
On commence par calculer le zéro de la fonction:
\begin{equa}
2x -1 &=& 0  & \hunc{3}{+1}\\
\hunc{4}{\LRq 2x&=& 1  & \hunc{5}{\div 2}}\\
\hunc{6}{\LRq x &=& \frac{1}{2} \uzero \hunc{7}{\qqR S=\{\frac{1}{2}\}}}
\end{equa}
\hunc{8}{La pente est \hunc{9}{positive ($m=2>0$)}}

\vfill
\hunc{10}{
\centering
\begin{tikzpicture}[scale=0.5]
\tkzTabInit[lgt=2.5, deltacl=1, nocadre]{$x$/1,$2x-1$/1}{-$\infty$, $\hunc{11}{\frac{1}{2}}$, +$\infty$}
\hunc{12}{\tkzTabLine{, - , 0, +,};}
\end{tikzpicture}}}}

\end{frame}

\begin{frame}

\underline{Exemple 2.3} Donner le tableau de signes de la fonction donnée par $f(x)= x^2 -3x - 4$ et esquisser le graphe de $f$.

\end{frame}


\begin{frame}

\underline{Exemple 2.4} Donner le tableau de signes de la fonction donnée par $f(x)= 4x -x^2$ et esquisser le graphe de $f$.


\end{frame}

\begin{frame}{3. Résolution d'inéquations}

\underline{Exemple 3.1} Soit $f(x)=x^2-3x-4$ la fonction dont le graphe est représenté. Résoudre graphiquement l'inéquation $\bl{f(x) >0}$. \\
\begin{minipage}{0.44\textwidth}

\begin{tikzpicture}[scale=0.27]

\scriptsize

\coordinate (O) at (0,0);

\def\xmax{10};\def\xmin{-6};\def\ymax{8};\def\ymin{-8};
\def\xqmax{9};\def\xqmin{-5};\def\yqmax{7};\def\yqmin{-7};

% quadrillage

\foreach \x in {\xqmin, ..., \xqmax}{
\draw[black!30, very thin] (\x, \ymin) -- (\x, \ymax);}

\foreach \y in {\yqmin, ..., \yqmax}{
\draw[black!30,very thin]  (\xmin , \y) -- (\xmax, \y );}

% origine
\node[below left] at (O) {O};
\node[left] at (0,1) {1};
\node[below] at (1,0) {1};
\draw[thick] (-0.1, 1) -- (0.1, 1); 
\draw[thick] (1,-0.1) -- (1, 0.1); 

% axes
\draw[->] (0:\xmin) -- (0:\xmax) node[right]{$x$};
\draw[->] (90:\ymin) -- (90:\ymax) node[above]{$y$};

%\fill (1.5, -6.25) circle (0.15cm) node[below]{$S(\frac{3}{2}, -\frac{25}{4})$};

\clip (\xmin,\ymin) rectangle (\xmax,\ymax);
\draw[thick, samples=\mySample, domain=\xmin:\xmax] plot(\x, {(\x)^2 -3*\x - 4});

\hunc{2}{
\uncover<3->{
\fill[dblue, opacity=0.2] (\xmin,0) -- (\xmax,0) -- (\xmax, \ymax) -- (\xmin,\ymax);
\node[dblue] at (7,4){$f(x)>0$};
}

\uncover<4->{
\fill[dred, opacity=0.2] (\xmin,0) -- (\xmax,0) -- (\xmax, \ymin) -- (\xmin,\ymin);
\node[dred] at (7,-4){$f(x)<0$};
}

\uncover<5->{
\draw[thick, samples=\mySample, domain=\xmin:-1, dblue] plot(\x, {(\x)^2 -3*\x - 4});
\draw[thick, samples=\mySample, domain=4:\xmax, dblue] plot(\x, {(\x)^2 -3*\x - 4});
\draw[thick, samples=\mySample, domain=-1:4, dred] plot(\x, {(\x)^2 -3*\x - 4});
}

\uncover<6->{
\fill[dgreen] (-1,0) circle (5pt) node[below left]{$Z_1(-1,0)$};
\fill[dgreen] (4,0) circle (5pt) node[below right]{$Z_2(4,0)$};
}

\uncover<7->{
\draw[very thick, dgreen] (\xmin,0) -- (-1,0);
\draw[very thick, dgreen] (4,0) -- (\xmax,0);
}}

\end{tikzpicture}

\end{minipage}\begin{minipage}{0.54\textwidth}

\hunc{2}{

\uncover<8->{On observe que $\bl{f(x)>0}$ lorsque $\bl{x<\ve{-1}}$ \textbf{ou} lorsque $\bl{x>\ve{4}}$.\uncover<9->{ On notera cette solution:
$$\bl{S = ]-\infty; \ve{-1}[ \,\cup\, ]\ve{4};\infty[ }$$}}}

\end{minipage}

\vspace{0.2cm}
\hunc{3}{
\uncover<10->{On peut résumer la situation à travers un \textbf{tableau de signes}:} \\[0.1cm]

\uncover<11->{
\begin{center}
\begin{array}{| c | l l c c c r r |}
\hline
x & \uncover<12->{-\infty} & \hspace{0.5cm} & \uncover<13->{\ve{-1}} & \hspace{0.5cm} & \uncover<14->{\ve{4}} & \hspace{0.5cm} & \uncover<12->{\infty}\\
\hline
f(x) &  & \uncover<17->{\bl{\bf+}} & \uncover<15->{\ve{0}} & \uncover<18->{\ro{-}} & \uncover<16->{\ve{0}} & \uncover<19->{\bl{+}} &\\
\hline
\end{array}
\end{center}
}}


\end{frame}

\begin{frame}


\underline{Exemple 3.2} Résoudre l'inéquation $\bl{-5(x+2)(3-x) \geq 0}$. \\[0.2cm]

\hunc{2}{
\uncover<2->{On cherche les zéros de la fonction $\bl{f(x)= -5(x+2)(3-x)}$: \uncover<3->{$S=\{-2, 3\}$}

\uncover<4->{
On peut ensuite construire le tableau de signes
\begin{center}
\begin{array}{| c | l l c c c r r |}
\hline
x & -\infty & \hspace{0.5cm} & \ve{-2} & \hspace{0.5cm} & \ve{3} & \hspace{0.5cm} & \infty\\
\hline
\uncover<5->{-5} & & \uncover<6->{\ro{-}} & & \uncover<6->{\ro{-}} &  & \uncover<6->{\ro{-}} &\\
\uncover<7->{x+ 2} & & \uncover<9->{\ro{-}} & \uncover<8->{\ve{0}} & \uncover<10->{\bl{+}} & & \uncover<10->{\bl{+}} &\\
\uncover<11->{3 - x} & & \uncover<13->{\bl{+}} & & \uncover<13->{\bl{+}}  & \uncover<12->{\ve{0}} & \uncover<14->{\ro{-}} &\\
\hline
\uncover<15->{f(x)} & & \uncover<17->{\bl{+}} & \uncover<16->{\ve{0}} & \uncover<18->{\ro{-}} & \uncover<16->{\ve{0}} & \uncover<19->{\bl{+}} &\\
\hline
\end{array}
\end{center}
}

}

\uncover<20->{La solution de l'inéquation $f(x) \geq 0$ est donc: }
\uncover<21->{$$\bl{S=]-\infty; -2] \cup [3 ; \infty [ }$$}}

\end{frame}

\begin{frame}


\underline{Exemple 3.3} Résoudre l'inéquation $\bl{-x^2 + 5x - 6 \geq 0}$. \\[0.2cm]

\hunc{2}{
\uncover<2->{On commence par factoriser:
$$-x^2 + 5x - 6 = - (x^2 -5x + 6) = \ve{-1}\cdot(x - \ve{2})(x - \ve{3}) $$
 

On peut ensuite construire le tableau de signes
\begin{center}
\begin{array}{| c | l l c c c r r |}
\hline
x & -\infty & \hspace{0.5cm} & \ve{2} & \hspace{0.5cm} & \ve{3} & \hspace{0.5cm} & \infty\\
\hline
-1 & & \ro{-} & & \ro{-} & & \ro{-} & \\
x-2 & & \ro{-} & 0 & \bl{+} & & \bl{+} & \\
x-3 & & \ro{-} &  & \ro{-} & 0 & \bl{+} & \\
\hline
f(x) & & \ro{-} & \ve{0} & \bl{+} & \ve{0} & \ro{-} &\\
\hline
\end{array}
\end{center}

On peut donc observer que $\bl{f(x) \geq 0}$ lorsque $\bl{\ve{2} \leq x \leq \ve{3}}$, donc la solution de l'inéquation est $\bl{S=[\ve{2};\ve{3}]}$.

}}


\end{frame}

\begin{frame}{4. Fractions rationnelles}

\underline{Exemple 4.1} Construire le tableau de signes de la fonction $f(x)= \frac{2(x+5)(x+1)(x-3)}{(x+2)(x-1)^2}$ dont le graphe est donné ci-dessous. 

\begin{minipage}{0.48\textwidth}

\begin{tikzpicture}[scale=0.8,y=0.25cm, x=0.25cm]

\scriptsize

\coordinate (O) at (0,0);

\def\lim{11}\def\qlim{10}
\def\xmax{\lim}\def\xmin{-\lim}\def\ymax{\lim}\def\ymin{-\lim}
\def\xqmax{\qlim}\def\xqmin{-\qlim}\def\yqmax{\qlim}\def\yqmin{-\qlim}

% quadrillage

\foreach \x in {\xqmin, ..., \xqmax}{
\draw[black!30, very thin] (\x, \ymin) -- (\x, \ymax);}

\foreach \y in {\yqmin, ..., \yqmax}{
\draw[black!30,very thin]  (\xmin , \y) -- (\xmax, \y );}

% origine
%\node[below left] at (O) {O};
\node[left] at (0,1) {1};
\node[below] at (1,0) {1};
\draw[thick] (-0.1, 1) -- (0.1, 1); 
\draw[thick] (1,-0.1) -- (1, 0.1); 

% axes
\draw[->] (0:\xmin) -- (0:\xmax) node[right]{$x$};
\draw[->] (90:\ymin) -- (90:\ymax) node[above]{$y$};



\clip (\xmin, \ymin) rectangle (\xmax,\ymax);

\draw[very thick, domain=\xmin:-2.1, samples=\mySample] plot(\x,{2*(\x-3)*(\x+1)/((\x-1)^2*(\x+2))*(\x+5)}); 
\draw[very thick, domain=-1.95:0.9, samples=\mySample] plot(\x,{2*(\x-3)*(\x+1)/((\x-1)^2*(\x+2))*(\x+5)}); 
\draw[very thick, domain=1.1:\xmax, samples=\mySample] plot(\x,{2*(\x-3)*(\x+1)/((\x-1)^2*(\x+2))*(\x+5)}); 


\hunc{2}{
\uncover<3->{
\draw[violet] (-5,0.5) -- (-5,-0.5) node[below]{$-5$};
\draw[violet] (-1,0.5) -- (-1,-0.5) node[below]{$-1$};
\draw[violet] (3,0.5) -- (3,-0.5) node[below]{$3$};
}

\uncover<7->{
\draw[dorange, very thick] (-2,\ymin) -- (-2,\ymax) node[below left]{$x=-2$};
\draw[dorange, very thick] (1,\ymax) node[below right]{$x=1$} -- (1,\ymin);
\uncover<8->{
\node[below right, dorange, fill=white] at (1.2,\ymax-1.5) {asymptotes};}
}}


\end{tikzpicture}
\end{minipage}\begin{minipage}{0.5\textwidth}

\hunc{2}{
\uncover<2->{Les zéros sont \vi{-5}, \vi{-1} et \vi{3}.} \\

\uncover<4->{Le domaine de définition est donné par \uncover<5->{$D =\mathbb{R} \backslash \{ \ora{-2}, \ora{1}\}$. \uncover<6->{ On appelle ces valeurs
\ora{valeurs interdites}.}\\

}}

}

\end{minipage}

\hunc{2}{
\vspace{0.2cm}
\uncover<9->{
\begin{tikzpicture}[scale=0.45]
\small
\tkzTabInit{$x$/1,$f(x)$/1}{$-\infty$,$\vi{-5}$,$\ora{-2}$,$\vi{-1}$,$\ora{1}$,$\vi{3}$,$\infty$}
\uncover<10->{\tkzTabLine{, + , z, -, d, + , z, -, d , - , z, +}}
\end{tikzpicture}}

\uncover<11->{ On indique les valeurs interdites par des doubles barres. }
}


\end{frame}

\begin{frame}

\underline{Exemple 4.2} Résoudre l'inéquation $\dfrac{2}{x-1} + \dfrac{5-3x}{x-x^2} \leq 0$. 

\end{frame}

\end{document}
