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\begin{document}


\begin{center}\section*{\textcolor{red}{Correction des exercices d'Ã©tudes de variations}}\end{center}


Ã‰tudions les variations des fonctions suivantes :



\begin{multicols}{2}
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\begin{enumerate}[1)]
\item $f(x)=-x^3+3x$\\
$f'(x)=-3x^2+3=3\left(1-x^2\right)=\boxed{\textcolor{red}{3(1+x)(1-x)}}$.\\
$f'(x)$ a pour racines -1 et 1 et $f'(x)&lt;0$ Ã&nbsp; l'extÃ©rieur de l'intervalle formÃ© par les racines.\\
\textbf{\textcolor{blue}{Tableau de variation :}}
\begin{center}
\begin{variations}
xz&amp;\mI&amp;&amp;-1&amp;&amp;1&amp;&amp;\pI\\
\hline
f'(x)&amp;&amp;-&amp;\z&amp;+&amp;\z&amp;-&amp;\\
\hline
\m{f(x)}&amp;\h{\pI}&amp;\d&amp;-2&amp;\c&amp;\h{2}&amp;\d&amp;\mI&amp;\\
\hline
\end{variations}
\end{center}

\bigskip

\item $f(x)=x^3-x^2-x+1$\\
$\boxed{\textcolor{red}{f'(x)=3x^2-2x-1}}$.\\
$\Delta=4+12=16&gt;0$.\\
Deux racines : $x_2=-\dfrac{1}{3}$ et $x_2=1$.\\
$f'(x)$ a pour racines $-\dfrac{1}{3}$ et 1 et $f'(x)&gt;0$ Ã&nbsp; l'extÃ©rieur de l'intervalle formÃ© par les racines.\\
\textbf{\textcolor{blue}{Tableau de variation :}}
\begin{center}
\begin{variations}
x&amp;\mI&amp;&amp;-\dfrac{1}{3}&amp;&amp;1&amp;&amp;\pI\\
\hline
f'(x)&amp;&amp;+&amp;\z&amp;-&amp;\z&amp;+&amp;\\
\hline
\m{f(x)}&amp;\mI&amp;\c&amp;\h{\dfrac{32}{27}}&amp;\d&amp;0&amp;\c&amp;\h{\pI}\\
\hline
\end{variations}
\end{center}

\bigskip

\item $x^4-8x^2 +8$.\\
$f'(x)=4x^3-16x=4x\left(x^2-4\right)=\boxed{\textcolor{red}{4x(x+2)(x-2)}}$
\begin{center}
\begin{variations}
x&amp;\mI&amp;&amp;-2&amp;&amp;0&amp;&amp;2&amp;&amp;\pI\\
\hline
x&amp;&amp;-&amp;\l&amp;-&amp;\z&amp;+&amp;\l&amp;+&amp;\\
\hline
x+2&amp;&amp;-&amp;\z&amp;+&amp;\l&amp;+&amp;\l&amp;+&amp;\\
\hline
x-2&amp;&amp;-&amp;\l&amp;-&amp;\l&amp;-&amp;\z&amp;+&amp;\\
\hline
f'(x)&amp;&amp;-&amp;\z&amp;+&amp;\z&amp;-&amp;\z&amp;+&amp;\\
\hline
\m{f(x)}&amp;\h{\pI}&amp;\d&amp;-8&amp;\c&amp;\h{8}&amp;\d&amp;-8&amp;\c&amp;\h{\pI}\\
\hline
\end{variations}
\end{center}

\bigskip

\item $f(x)=\dfrac{-4x}{x^2+1}=-4\times \dfrac{x}{x^2+1}$.\\
$f'(x)=-4\times \dfrac{x^2+1-2x^2}{\left(x^2+1\right)^2}=-4\times \dfrac{1-x^2}{\left(x^2+1\right)^2}\\
=\boxed{\textcolor{red}{\dfrac{4(x+1)(x-1)}{\left(x^2+1\right)^2}}}$.\\
\textbf{\textcolor{blue}{Tableau de variation :}}
\begin{center}
\begin{variations}
x&amp;\mI&amp;&amp;-1&amp;&amp;1&amp;&amp;\pI\\
\hline
f'(x)&amp;&amp;+&amp;\z&amp;-&amp;\z&amp;+&amp;\\
\hline
\m{f(x)}&amp;0&amp;\c&amp;\h{2}&amp;\d&amp;-2&amp;\c&amp;\h{0}\\
\hline
\end{variations}
\end{center}

\bigskip

\item $f(x)=\dfrac{x^2-x-2}{(x-1)^2}$ $(x\neq 1)$\\
$f'(x)=\dfrac{(2x-1)\left(x-1\right)^2-2(x-1)\left(x^2-x-2\right)}{\left((x-1)\right)^4}\\
=\dfrac{(x-1)\left[(2x-1)(x-1)-2\left(x^2-x-2\right)\right]}{\left(x-1\right)^4}\\
=\dfrac{(x-1)\left(2x^2-2x-x+1-2x^2+2x+4\right)}{(x-1)^4}=\boxed{\textcolor{red}{\dfrac{-x+5}{(x-1)^3}}}$\\
\textbf{\textcolor{blue}{Tableau de variation :}}
\begin{center}
\begin{variations}
x&amp;\mI&amp;&amp;&amp;1&amp;&amp;&amp;5&amp;&amp;\pI\\
\hline
-x+5&amp;&amp;+&amp;&amp;\bb&amp;&amp;+&amp;\z&amp;-&amp;\\
\hline
(x-1)^3&amp;&amp;-&amp;&amp;\bb&amp;&amp;+&amp;\l&amp;+&amp;\\
\hline
f'(x)&amp;&amp;-&amp;&amp;\bb&amp;&amp;+&amp;\z&amp;-&amp;\\
\hline
\m{f(x)}&amp;\h{1}&amp;\d&amp;\mI&amp;\bb&amp;\mI&amp;\c&amp;\h{\dfrac{9}{8}}&amp;\d&amp;1\\
\hline
\end{variations}
\end{center}

\bigskip

\item $f(x)=x-1+\dfrac{4}{x-2}$ ($x\neq 2$)\\
$f'(x)=1+4\times \dfrac{-1}{(x-2)^2}=1-\dfrac{4}{(x-2)^2}=\dfrac{(x-2)^2-4}{(x-2)^2}=\dfrac{(x-2)^2-2^2}{(x-2)^2}
=\boxed{\textcolor{red}{\dfrac{x(x-4)}{(x-2)^2}}}$.\\`
$f'(x)$ est du signe du numÃ©rateur.\\
\textbf{\textcolor{blue}{Tableau de variation :}}
\begin{center}
\begin{variations}
x&amp;\mI&amp;&amp;0&amp;&amp;&amp;2&amp;&amp;&amp;4&amp;&amp;\pI\\
\hline
f'(x)&amp;&amp;+&amp;\z&amp;-&amp;&amp;\bb&amp;&amp;-&amp;\z&amp;+&amp;\\
\hline
\m{f(x)}&amp;\mI&amp;\c&amp;\h{-3}&amp;\d&amp;\mI&amp;\bb&amp;\h{\pI}&amp;\d&amp;5&amp;\c&amp;\h{\pI}\\
\hline
\end{variations}
\end{center}
\end{enumerate}


\end{multicols}

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