#### Answer

$0$ and $2$ are critical points. There is local minimum at $x=0$ and local maximum at $x=2$.

#### Work Step by Step

$f'(x) = x^2 · (−e^{−x}) + e^{−x} · 2x = e^{−x}(2x − x^2)$, which is zero for $x = 0$ and $x = 2$. $$f''(x) = e^{−x}(2 − 2x) + (2x − x^2)(−e^{−x}) = e^{−x}(2 − 4x + x^2).$$Note that $f''(0) = 2 > 0$, so there is a local minimum at $x = 0$. Also, $f''(2) = −2e−2 < 0$, so there is a local maximum at $x = 2$.