The maximum value

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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