Question

The least and greatest values of $f\left(x\right)=x{e}^{-x}$ on $[0,\infty )$ are

• A. $0,\frac{1}{e}$
• B. $\frac{1}{e},\frac{2}{e^2}$
• C. $\frac{2}{e^2},\frac{3}{e^3}$
• D. $\frac{1}{e},\frac{2}{e}$

Answer 1

The correct option is A $0,\frac{1}{e}$

$f^{\prime }\left(x\right)=e^{-x}-x{e}^{-x}$
$=e^{-x}(1-x)$
$f^{\prime }\left(x\right)>0$ If $0<x<1$
$f^{\prime }\left(x\right)<0$ If $x>1$
$f\left(x\right)$ is maximum at $x=1$
$f\left(1\right)=\frac{1}{e}$
$f\left(x\right)$ is minimum at $x=\infty$
$f\left(\infty \right)=0$