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December 21, 2014

December 21, 2014

Posted by **Dylan** on Wednesday, March 6, 2013 at 4:09pm.

- Calculus -
**Steve**, Wednesday, March 6, 2013 at 5:00pmf = ax e^(bx)

f' = ae^(bx) + abx e^(bx)

= ae^(bx) (1+bx)

f'(1/8) = 0, so b = -8

f(1/8) = 1 = a(1/8)e^(-1), so a=8e

f(x) = (8e)x e^(-8x)

- Calculus -
**Anonymous**, Thursday, November 6, 2014 at 10:50amCalculus - Steve, Wednesday, March 6, 2013 at 5:00pm

f = ax e^(bx)

f' = ae^(bx) + abx e^(bx)

= ae^(bx) (1+bx)

f'(1/8) = 0, so b = -8

f(1/8) = 1 = a(1/8)e^(-1), so a=8e

f(x) = (8e)x e^(-8x)

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