## Calculus (3rd Edition)

$x=0$, local minima at $x=0$.
To find the critical point, we put $f'(x)=0$, so we have $$f(x)=e^{-x}+x\Longrightarrow f'(x)=-e^{-x}+1=0,$$ then $e^{-x}=1$ and hence $x=0$. So the critical point is $x=0$. Moreover, we have $$f''(x)=e^{-x}\Longrightarrow f''(0)=1>0$$ then $f(x)$ has local minima at $x=0$.