Answer
At $x=-\displaystyle \frac{\sqrt{2}}{2}$, f has a relative minimum.
At $x=\displaystyle \frac{\sqrt{2}}{2}$, 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: $f(x)=xe^{1-x^{2}}$
$f'(x)=(1)e^{1-x^{2}}+x(-2xe^{1-x^{2}})=e^{1-x^{2}}(1-2x^{2})$
$f''(x)=-2xe^{1-x^{2}}(1-2x^{2})+e^{1-x^{2}}(-4x)= e^{1-x^{2}}(4x^{2}-2x-4x)$
$= e^{1-x^{2}}(4x^{2}-6x)$
Stationary points:
$f'(x)=0$
$ e^{1-x^{2}}(1-2x^{2})=0\qquad$... $e^{1-x^{2}}$ is never zero, so
$1-2x^{2}=0$
$2x^{2}=1$
$x=\displaystyle \pm\frac{\sqrt{2}}{2}$
Second-derivative test:
$f''(-\displaystyle \frac{\sqrt{2}}{2})=e^{1-\frac{1}{2}}[4\cdot\frac{1}{2}-6(-\frac{\sqrt{2}}{2})]$, which is positive,
At $x=-\displaystyle \frac{\sqrt{2}}{2}$, f has a relative minimum.
$f''(\displaystyle \frac{\sqrt{2}}{2})=e^{1-\frac{1}{2}}[4\cdot(\frac{1}{2})-6(\frac{\sqrt{2}}{2})] =e^{1/2}(2-3\sqrt{2})$, which is negative,
At $x=\displaystyle \frac{\sqrt{2}}{2}$, f has a relative maximum.
