Find the critical points of x^4.e^(-x^2)

[InfZero]

This is just a small mathematics tutorial question I'm having trouble with.

Critical points are where the f'(x) = 0, so I differentiate and I'm left with :

f'(x)=4x^3.-2x.e^(-x^2)

Where any of these terms is equal to 0, f'(x) is equal to 0.

Obviously, x=0 is a critical point but the question hints at there being 3 critical points. So, I need to figure out where exactly x^-x^2 is equal to 0, but I thought it was impossible for e^(anything) to equal 0!

So, unless I've missed something extremely obvious, I need some help.

 

[skeeter]

\(\displaystyle \L f(x) = x^4 e^{-x^2}\)

use the product rule to find f'(x) ...

\(\displaystyle \L f'(x) = x^4(-2x e^{-x^2}) + 4x^3 e^{-x^2}\)

factor out the common factors from each term ...

\(\displaystyle \L f'(x) = 2x^3 e^{-x^2}(2 - x^2)\)

now set f'(x) = 0 and solve for x

 

[InfZero]

Oh, I completely forgot about the product rule. Agh.