Question

Let $f\left(x\right)=x{e}^{-x}$. The maximum value of the function in the interval $(0,\infty )$ is

• A. $e^{-1}$
• B. $e$
• C. $1-e^{-1}$
• D. $1+e^{-1}$

Answer 1

The correct option is A $e^{-1}$

$f\left(x\right)=x{e}^{-x},xϵ(0,\infty )$
$f^{\prime }\left(x\right)=-x{e}^{-x}+e^{-x}=(1-x)e^{-x}$
For critical points:
$f^{\prime }\left(x\right)=0$
$(1-x)e^{-x}=0$
$x=1$
$\Rightarrow x=1$ is a critical point.
$f^{\prime \prime }\left(x\right)=(x-2)e^{-x}$
$f^{\prime \prime }\left(1\right)=-e^{-1}<0$
$\Rightarrow f\left(x\right)$ has maxima at $x=1$
Maxima value $=f\left(1\right)=1.e^{-1}=\frac{1}{e}$.