Question:
The maximum value of $f(x)=x e^{-x}$ is___________
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,
$f^{\prime}(x)=0$
$\Rightarrow e^{-x}(-x+1)=0$
$\Rightarrow-x+1=0$ $\left(e^{-x}>0 \forall x \in \mathrm{R}\right)$
$\Rightarrow x=1$
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