Question

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

• A. 0
• B. $\frac{1}{e}$
• C. $-e$
• D. $e$

Answer 1

Answer: (B)

$\frac{1}{e}$

$f\left(x\right)=x{e}^{-x}{f}^{\prime }\left(x\right)=e^{-x}-x{e}^{-x}$

At maxima $f^{\prime }\left(x\right)=0$

$\Rightarrow e^{-x}-x{e}^{-x}=0\Rightarrow e^{-x}(1-x)=0\Rightarrow x=1$

By second derivative test

$f^{\prime \prime }\left(x\right)=-e^{-x}-e^{-x}+x{e}^{-x}{f}^{\prime \prime }\left(1\right)=-e^{-1}-e^{-1}+e^{-1}=-e^{-1}{f}^{\prime \prime }\left(1\right)<0$

So $x=1$ is the point of maxima

$f\left(1\right)=\left(1\right)e^{-1}=\frac{1}{e}$

So, option B is correct.