Question

The maximum value of $x{e}^{-x}$ is

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

Answer 1

The correct option is B

$\frac{1}{e}$

Explanation for the correct option:

Step 1: Find the critical points of the given function.

A function $f\left(x\right)=x{e}^{-x}$ is given.

Differentiate the given function with respect to $x$.

$$\begin{gathered} f\text{'}\left(x\right)=e^{-x}+(-x{e}^{-x}) \\ \Rightarrow f\text{'}\left(x\right)=e^{-x}-x{e}^{-x} \end{gathered}$$

Put $f\text{'}\left(x\right)$ equal to zero to find the critical points.

$$\begin{gathered} e^{-x}-x{e}^{-x}=0 \\ \Rightarrow e^{-x}\left(1-x\right)=0 \\ \Rightarrow 1-x=0 \\ \Rightarrow x=1 \end{gathered}$$

So, the critical point is $1$.

Step 2: Find the maximum value of the given function.

Since, the critical point is $1$

Evaluate $f\left(x\right)$ for $x=1$ as follows:

$$\begin{gathered} f\left(1\right)=1·e^{-1} \\ \Rightarrow f\left(1\right)=\frac{1}{e} \end{gathered}$$

So, the value of $f\left(x\right)$ for $x=1$ is $\frac{1}{e}$.

Therefore, the maximum value of the given function is $\frac{1}{e}$.

Hence, option $B$ is the correct answer.