Question 1033806
{{{f(x) = xe^(-x^2)}}} ==> f'(x) = {{{(1-2x^2)e^(-x^2)}}}.

Let f'(x) be equal to 0.

==> {{{1-2x^2=0}}} ==> {{{x = sqrt(2)/2}}} or {{{x = -sqrt(2)/2}}}, the critical numbers of f(x).

As you mentioned {{{e^(-x^2)>0}}}, so no solution comes from this.

In ({{{-infinity}}}, {{{-sqrt(2)/2}}}), f'(x) < 0, so f(x) is decreasing there.
In ({{{-sqrt(2)/2}}},{{{sqrt(2)/2}}}), f'(x) > 0, so f(x) is increasing there.
In ({{{sqrt(2)/2}}},{{{infinity}}}), f'(x) < 0, so f(x) is decreasing there.

Therefore f(x) attains local min at {{{x = -sqrt(2)/2}}}, while it attains a local max at {{{x = sqrt(2)/2}}}.