Solution

Given,

$f(x)=|x+1|e^{-x^2}$

For $x<-1$,

$f(x)=-(x+1)e^{-x^2}$

Differentiate:

$f'(x)=e^{-x^2}(2x^2+2x-1)$

For critical points,

$2x^2+2x-1=0$

Using quadratic formula,

$x=\frac{-2\pm\sqrt{4+8}}{4}$

$x=\frac{-2\pm 2\sqrt{3}}{4}$

$x=\frac{-1\pm\sqrt{3}}{2}$

For $x<-1$, the valid critical point is

$x=\frac{-1-\sqrt{3}}{2}$

This lies in the interval $(-2,-1)$.

At this point, $f$ has a point of maxima.