Question

f(x)=(5x-4)e^2x
How do you find the one critical number??

Answer 1

Differentiate and put equal to zero.
f'[x] = 5e^(2x) + 2e^(2x)(-4+5x)
f'[0] = -3
So the x-coordinate is -3

I'd guess that that is what the question wants.

Hope that helps

Answer 2

I made a slight error. i don't want f'[0], i want f'[x] = 0 to find x = something.

so we get 5e^(2x) + 2e^(2x)(-4+5x) = 0
divide across by e^(2x)
we get 5 + 2(-4+5x) = 0
5 - 8 + 10x =0
10x = 3
x = 3/10
Sorry about the mistake in my first post.

hope that helps