## Calculus (3rd Edition)

$x=0, -2$, $f(x)$ has a local minimum at $x=0$ and local maximum at $x=-2$.
To find the critical point, we put $f'(x)=0$, so we have $$f(x)=x^2 e^{x}\Longrightarrow f'(x)=2xe^{x}+x^2e^{x}=(x^2+2x)e^{x}=0,$$ then $x(x+2)=0$ and hence $x=0$ or $x=-2$. So the critical points are $x=0$ and $x=-2$.. Moreover, we have $$f''(x)=(2x+2 )e^{x}+(x^2+2x)e^{x}$$ then $$f''(0)=2>0, \quad f''(-2)=-2e^{-2}<0.$$ Then $f(x)$ has a local minimum at $x=0$ and local maximum at $x=-2$.