The maximum value

Solution:

The given function is $f(x)=x e^{-x}$.

$f(x)=x e^{-x}$

Differentiating both sides with respect to x, we get

$f^{\prime}(x)=x \times e^{-x} \times(-1)+e^{-x} \times 1$

$\Rightarrow f^{\prime}(x)=e^{-x}(-x+1)$

For maxima or minima,

$f^{\prime}(x)=0$

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

$\Rightarrow-x+1=0$    $\left(e^{-x}>0 \forall x \in \mathrm{R}\right)$

$\Rightarrow x=1$

Now,

$f^{\prime \prime}(x)=e^{-x} \times(-1)+(-x+1) \times e^{-x} \times(-1)$

$\Rightarrow f^{\prime \prime}(x)=e^{-x}(x-2)$

At x = 1, we have

$f^{\prime \prime}(1)=e^{-1}(1-2)=-\frac{1}{e}<0$

So, $x=1$ is the point of local maximum of $f(x)$.

$\therefore$ Maximum value of $f(x)=f(1)=1 \times e^{-1}=\frac{1}{e}$

Thus, the maximum value of $f(x)=x e^{-x}$ is $\frac{1}{e}$.

The maximum value of $f(x)=x e^{-x}$ is $\frac{1}{e}$