Question

for the function $f\left(x\right)=x^2{e}^{-x},$, the maximum occurs when $x$ is equal to

• A. $2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

The correct option is A $2$

$f\left(x\right)=x^2{e}^{-x}$
$f^{\prime }\left(x\right)=-x^2.e^{-x}+e^{-x}.2x$
For maximam or minima, $f^{\prime }\left(x\right)=0$
$\Rightarrow x=0,2$
$f^{\prime \prime }\left(x\right)=x^2{e}^{-x}-2x{e}^{-x}+2e^{-x}-2x{e}^{-x}$
$=e^{-x}[x^2-2x+3-2x]$
$=e^x[x^2-4x+2]$
$f^{\prime \prime }\left(0\right)=2>0\Rightarrow minima$
$f^{\prime \prime }\left(2\right)=e^{-2}[+4-8+2]=-2e^{-2}<0$
$\Rightarrow$ Maxima at $x=2$