1. ## Critical points

Hello,

Can someone show me how to find the critical points for the function f(x)=x^2*e^(-5x^2)?

I appreciate your help! Thank you!

2. ## Re: Critical points

Find f'(x) using the product rule, then solve f'(x) = 0

3. ## Re: Critical points

Yes, I understand that I have to do that. But I'm getting confused when I'm differentiating. Could you help me with that part? Thank you!

4. ## Re: Critical points

$\displaystyle \displaystyle f'(x) = (x^2)'e^{-5x^2} +(e^{-5x^2})'x^2$

Recall if $\displaystyle \displaystyle y= e^{f(x)} \implies \frac{dy}{dx} = f'(x)e^{f(x)}$

5. ## Re: Critical points

Right, so this is what I get from using the product rule and chain rule:

f'(x)=[2x]e^(-5x^2)+x^2[e^(-5x^2)*(-10x)]

Then I factored it (hopefully correctly) and got:

2xe^(-5x^2)[1-5x^2*e^(-5x^2)]

I set that equal to 0. But since e can never equal zero, I threw out 2xe^(-5x^2). So I'm looking for (1-5x^2*e^(-5x^2))=0. This is where I got confused. I'm not quite sure how to solve for x. Any ideas?

6. ## Re: Critical points

True, that should've been $\displaystyle 0 = 2x- 10x^3 \implies x = 0,\frac{1}{\sqrt{5}},\frac{-1}{\sqrt{5}}$

7. ## Re: Critical points

Ok, so I did factor it wrong. The answer given to me on my worksheet is x=0, root(1/5), -root(1/5).

After factoring it correctly i got:

0=2xe^(-5x^2)[1-5x^2]
0=[1-5x^2]
x= +/- root(1/5) and x also equals 0 because of the 2xe^(-5x^2)

Thanks for the help!