Answer
At $x=0$, f has a relative maximum.
Work Step by Step
Critical points of $f$ are points that are either stationary or singular
(for which either $f'(c)=0$ or $f'(c)$ is undefined, but $f(c)$ is)
Given: $f(x)=e^{-x^{2}}$
$f'(x)=-2xe^{-x^{2}}, \qquad $
$f''(x)=(-2)e^{-x^{2}}-2x(-2xe^{-x^{2}})=e^{-x^{2}}(4x^{2}-2)$
Stationary points:
$f'(x)=0$
$-2xe^{-x^{2}}=0\qquad$... $e^{-x^{2}}$ is never zero, so
$x=0$
Second-derivative test:
$f''(0)=e^{0}(4(0)^{2}-2) =-2$, which is negative, so
at $x=0$, f has a relative maximum.
