mathcalc
posted by mkcolema on .
f(x)=(5x4)e^2x
How do you find the one critical number??

Differentiate and put equal to zero.
f'[x] = 5e^(2x) + 2e^(2x)(4+5x)
f'[0] = 3
So the xcoordinate is 3
I'd guess that that is what the question wants.
Hope that helps 
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