Posted by **mkcolema** on Friday, March 13, 2009 at 8:56pm.

f(x)=(5x-4)e^2x

How do you find the one critical number??

- math-calc -
**GanonTEK**, Friday, March 13, 2009 at 9:15pm
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

- math-calc -
**GanonTEK**, Friday, March 13, 2009 at 9:23pm
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

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