Question

The maximum value of f(x) = x e^−x is _______________.

Answer 1

The given function is f(x) = xe^−x.

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

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=x\times e^{-x}\times \left(-1\right)+e^{-x}\times 1$

$\Rightarrow f\text{'}\left(x\right)=e^{-x}\left(-x+1\right)$

For maxima or minima,

$f\text{'}\left(x\right)=0$

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

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

$\Rightarrow x=1$

Now,

$f\text{'}\text{'}\left(x\right)=e^{-x}\times \left(-1\right)+\left(-x+1\right)\times e^{-x}\times \left(-1\right)$

$\Rightarrow f\text{'}\text{'}\left(x\right)=e^{-x}\left(x-2\right)$

At x = 1, we have

$f\text{'}\text{'}\left(1\right)=e^{-1}\left(1-2\right)=-\frac{1}{e}<0$

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

∴ Maximum value of f(x) = f(1) = 1 × e⁻¹ = $\frac{1}{e}$

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


The maximum value of f(x) = xe^−x is $\underline{\frac{1}{e}}$.