Show that for each $a>0$ the function $e^{-ax}x^{a^2}$ has a maximum value, say $F(a)$, and that $F(x)$ has a minimum value $e^{-e/2}$

Question

Show that for each $a>0$ the function $e^{-ax}x^{a^2}$ has a maximum value say $F(a)$,and that $F(x)$ has a minimum value $e^{-e/2}$.

I differentiated the function $f(x)=e^{-ax}x^{a^2}$ to get critical point $x=a$, so the maximum value $F(a)=e^{-a^2}a^{a^2}$. Now I am confused what to take $F(x)$. I think $f(x)$ is same as $F(x)$ but $f(x)$ has no minimum value. I got stuck. Please help.

Answer 1

If you follow the same proceedure as you did for the first part of the problem, but for: F (x) = exp (-x^2)*x^(x^2)

You find another extremum, what does this tell you about the function F (x), and if you calculate the value of the function at this point, what would you end up with? (Hint, further local analysis can tell you the nature of the extremum)