# What are the absolute extrema of f(x)=x / e^(x^2) in[1,oo]?

Aug 15, 2017

$\left(1 , \frac{1}{e}\right)$ is an absolute maximum in the given domain
There is no minimum

#### Explanation:

The derivative is given by

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

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

Critical values will occur when the derivative equals $0$ or is undefined. The derivative will never be undefined (because ${e}^{{x}^{2}}$ and $x$ are continuous functions and ${e}^{{x}^{2}} \ne 0$ for any value of $x$.

So if $f ' \left(x\right) = 0$:

$0 = {e}^{{x}^{2}} - 2 {x}^{2} {e}^{{x}^{2}}$

$0 = {e}^{{x}^{2}} \left(1 - 2 {x}^{2}\right)$

As mentioned above ${e}^{{x}^{2}}$ will never equal $0$, so our only two critical numbers will occur at the solution of

$0 = 1 - 2 {x}^{2}$

$2 {x}^{2} = 1$

${x}^{2} = \frac{1}{2}$

$x = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}}$

But neither of these lie in our given domain. Therefore, $x = 1$ is going to be a maximum (because $f \left(x\right)$ converges to $0$ as x->+oo).

There will be no minimum

Hopefully this helps!