
absolute extrema
Hi
My problem tonight is I don't have a firm enough grasp of e
The question is to find all local extreme values and determine if any are absolute
f(x) = x^2 e^x^2
So I think f'(x) = x^2 (2xe^x^2) + 2xe^x^2
= (1x^2)(2xe^x^2)
Now the critical point of (1x^2) is x= 1 or x=1 but I don't know what the critical point of the other part is
I am also not sure how to test for absoluteness and how do I know if it is a local min or max  how can I tell if the sign changes???
sorry lots of questions  but thanks for reading  hopefully you can help
thanks
a calculus beginner

$\displaystyle f(x) = x^2 e^{x^2}$
$\displaystyle f'(x) = x^2(2xe^{x^2}) + 2xe^{x^2}$
$\displaystyle f'(x) = 2xe^{x^2}(1  x^2)$
$\displaystyle f'(x) = 0$ at $\displaystyle x = 0$ , $\displaystyle x = \pm 1$
for $\displaystyle x < 1$ , $\displaystyle f'(x) > 0$ ... $\displaystyle f(x)$ is increasing
for $\displaystyle 1 < x < 0$ , $\displaystyle f'(x) < 0$ ... $\displaystyle f(x)$ is decreasing
for $\displaystyle 0 < x < 1$ , $\displaystyle f'(x) > 0$ ... $\displaystyle f(x)$ is increasing
for $\displaystyle x > 1$ , $\displaystyle f'(x) < 0$ ... $\displaystyle f(x)$ is decreasing
the above analysis indicates maximums at $\displaystyle x = 1$ and $\displaystyle x = 1$
since $\displaystyle f(x)$ is an even function, the absolute max will occur at both values ...
$\displaystyle f(1) = f(1) = \frac{1}{e} $
$\displaystyle f(0) = 0$ is an absolute minimum since $\displaystyle f(x) \geq 0$ for all $\displaystyle x$