Source Latex
de la correction du devoir
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pdfauthor={Yoann Morel},
pdfsubject={Correction du devoir de mathématiques 1ère STI2D: dérivée},
pdftitle={Corrigé du devoir de mathématiques},
pdfkeywords={dérivée, dérivation des fonctions, tangente, correction, devoir corrigé, Mathématiques, 1ère STI2D, première STI2D}
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\lfoot{Y. Morel - \url{https://xymaths.fr/Lycee/1STI/}}
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\begin{document}
\ct{\bf\LARGE{Correction du devoir de math\'ematiques}}
\bgex
\bgmp{9.6cm}
On donne ci-contre une partie de la courbe $\mathcal{C}_f$
repr\'esentant une fonction $f$ d\'efinie et d\'erivable sur~$\R$.
\vspd
\bgen
\item Voir graphique.
\item Par lecture graphique:
\bgen[a.]
\item $f(-1)=-1$, $f(0)=1$, $f(1)=0$, $f'(-1)\simeq 4,5$,
$f'(1)\simeq -1,5$, $f'(2)=0$
\vspd
\item
\[\begin{tabular}{|c|ccccccc|}\hline
$x$ & $-\infty$ && $0$ && $2$ && $+\infty$ \\\hline
&&&$1$&&&&\\
$f(x)$ && \LARGE{$\nearrow$} && \LARGE{$\searrow$}
&&\LARGE{$\nearrow$} &\\
&&&&&$-1$&&\\\hline
\end{tabular}\]
\enen
\enen
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0.5 x mul x mul x mul
-1.5 x mul x mul add
1 add
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\enmp
\enex
\bgex
a)\ $f'(x)=5x^4+3$ \hfill
b)\ $g'(x)=24x^7-\dfrac{1}{x^2}$ \hfill
c)\ $h'(x)=-\dfrac{3}{(3x-2)^2}$ \hfill
d)\ $k'(x)=\dfrac{13}{(2x+3)^2}$ \hfill
\enex
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\bgex
\bgen
\item $f'(x)=3x^2-8x+2$.
\item $T_1: y=f'(1)(x-1)+f(1)$,
avec $f'(1)=-3$ et $f(1)=-2$,
d'o\`u, \ul{$T_1: y=-3(x-1)-2=-3x+1$}.
\vspd
$T_0: y=f'(0)(x-0)+f(0)$,
avec $f'(0)=2$ et $f(0)=-1$,
d'o\`u, \ul{$T_0: y=2(x-0)-1=2x-1$}.
\enen
\enex
\enmp
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3.
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\enmp
\bgex
Soit la fonction $f$ d\'efinie sur $\R\setminus\la -\dfrac14\ra$ par
l'expression
$f(x)=\dfrac{x^2+3}{4x+1}$.
$f=\dfrac{u}{v}$ avec
$\la\bgar{ll}
u(x)&=x^2+3 \\
v(x)&=4x+1
\enar\right.$
soit
$\la\bgar{ll}
u'(x)&=2x \\
v'(x)&=4
\enar\right.$
\vsp
On a donc,
$f'=\dfrac{u'v-uv'}{v^2}$,
soit
$f'(x)=\dfrac{2x(4x+1)-(x^2+3)\tm4}{(4x+1)^2}
=\dfrac{4x^2+2x-12}{(4x+1)^2}
$
\vsp
Le trin\^ome du num\'erateur a pour discriminant:
$\Delta=2^2+4\tm4\tm(-12)=196=14^2>0$, et admet donc deux racines
$x_1=\dfrac{-2-14}{2\tm4}=-2$
et
$x_1=\dfrac{-2+14}{2\tm4}=\dfrac32$ .
\[\begin{tabular}{|c|ccccccccc|}\hline
\rule[-0.3cm]{0.cm}{0.9cm}
$x$ & $-\infty$ && $-2$ && $-\dfrac14$ && $\dfrac32$ && $+\infty$ \\\hline
$4x^2+2x-12$ && $+$ &\zb&$-$&$|$ &$-$&\zb&$+$&\\\hline
$(4x+1)^2$ && $+$ &$|$&$+$&\zb &$+$&$|$&$+$&\\\hline
$f'(x)$ && $+$ &\zb&$-$&\db &$-$&\zb&$+$&\\\hline
&&&$-1$&&&&&&\\
$f(x)$ && \LARGE{$\nearrow$} && \LARGE{$\searrow$}
&\psline(0,-.8)(0,.8)\psline(0.08,-.8)(0.08,.8)&\LARGE{$\searrow$} && \LARGE{$\nearrow$}&\\
&&&&&&&$\dfrac34$&&\\\hline
\end{tabular}
\bgar{ll}
\bullet\
f(-2)=\dfrac{(-2)^2+3}{4\tm(-2)+1}=-1\\[0.5cm]
\bullet\
f\lp\dfrac32\rp=\dfrac{\lp\dfrac32\rp^2+3}{4\tm\lp\dfrac32\rp+1}
=\dfrac{\dfrac{21}{4}}{7}
=\dfrac34\\
\enar
\]
\enex
\label{LastPage}
\end{document}
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