# How do you use the second derivative test to find the local extrema for f(x)=e^(x^2)?

Apr 4, 2018

The function $f \left(x\right) = {e}^{{x}^{2}}$ has a local minimum in $x = 0$ and no local maximum.

#### Explanation:

Evaluate the first and second derivatives of the function:

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

$f ' ' \left(x\right) = 2 {e}^{{x}^{2}} + 4 {x}^{2} {e}^{{x}^{2}} = 2 \left(1 + 2 {x}^{2}\right) {e}^{{x}^{2}}$

Solving the equation:

$f ' \left(x\right) = 0$

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

as the exponential is never null we can see that the only critical point for the function is $x = 0$.

Then we see that;

$f ' ' \left(0\right) > 0$

so that the critical point is a local minimum.

graph{e^(x^2) [-2, 2, -1, 10]}