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