calculus help!!!?

Question

Consider the function f(x) = x^2 e^(8 x). Find the first derivative of f(x). Enter exp(8 x) when you write the exponent.
f′(x)=.
For the function f(x) there are two critical numbers, which correspond to a local maximum and a local minimum.
The x-coordinate of the LMAX is .
The x-coordinate of the LMIN is .

Answer 1

f(x) = x^2 e^(8 x)
f'(x) = 8x^2 e^(8 x) + 2x e^(8 x) = 0 for man & min
2x(4x + 1) = 0
x = 0, -1/4
f(0) = 0
f(-1/4) = (1/16)e^2

Answer 2

f'(x) = 8(x^2)(e^8x) + 2(e^8x)(x)
when f'(x) = 0, x = 0, x = -4
I don't have my calculator with me, but you should be able to see the max and min just fine by pluggin f(x) into the graph.