# absolute extrema

• Mar 4th 2009, 05:41 PM
calcbeg
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
= (1-x^2)(-2xe^x^2)

Now the critical point of (1-x^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
• Mar 4th 2009, 06:02 PM
skeeter
$\displaystyle f(x) = x^2 e^{-x^2}$

$\displaystyle f'(x) = x^2(-2xe^{-x^2}) + 2xe^{-x^2}$

$\displaystyle f'(x) = 2xe^{-x^2}(1 - x^2)$

$\displaystyle f'(x) = 0$ 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 $\displaystyle x$