Bonjour,
Partie A:
[tex]f(x)= \dfrac{x^2}{2} -2x+6\\\\
1a)\ f( \sqrt{2} )= \dfrac{(\sqrt{2})^2}{2} -2*\sqrt{2}+6=1-2*\sqrt{2}+6=7-2*\sqrt{2}\\\\
1b)\ f(x)=6\\
\Rightarrow\ \dfrac{x^2}{2} -2x+6=6\\
\Rightarrow\ \dfrac{x^2}{2} -2x=0\\
\Rightarrow\ x(\dfrac{x}{2} -2)=0\\
\Rightarrow\ x=0\ or\ x=4\\\\
2)\\
f(x)= \dfrac{x^2}{2} -2x+6\\
=\dfrac{1}{2}(x^2-4x+12)\\
=\dfrac{1}{2}(x^2-2*2x+4+8)\\
=\dfrac{1}{2}(x-2)^2+4)\\\\
[/tex]
[tex]3) f(x)=4\\
\Rightarrow\ \dfrac{1}{2}(x-2)^2+4=4\\
\Rightarrow\ \dfrac{1}{2}(x-2)^2=0\\
\Rightarrow\ (x-2)^2=0\\\\
\boxed{x=2}\\\\
4)f(x)=12\\
\Rightarrow\ \dfrac{1}{2}(x-2)^2+4=12\\
\Rightarrow\ \dfrac{1}{2}(x-2)^2=8\\
\Rightarrow\ (x-2)^2=16\\
\Rightarrow\ (x-2)^2-4^2=0\\
\Rightarrow\ (x-2-4)(x-2+4)=0\\
\boxed{\Rightarrow\ x=6\ ou\ x=-2}\\
[/tex]
Partie B:
1) [tex]Aire\ AFD= \dfrac{4x}{2}=2x\\
Aire\ BEF= \dfrac{x*(3-x)}{2} \\
Aire\ ECD= \dfrac{3*(4-x) }{2} \\
2)\\
S(x)= \dfrac{12-3x+3x-x^2+4x}{2} =-\dfrac{x^2}{2}+2x+6\\\\
3)\\
A(x)=12-(-\dfrac{x^2}{2}+2x+6)\\
=\dfrac{x^2}{2}-2x+6)=f(x)\\\\
4)\\
donc\ x=2
[/tex]
