Posted by Dylan on .
Find constants a and b in the function f(x)=axe^(bX) such that f(1/8)=1 and the function has a local maximum at x=1/8.

Calculus 
Steve,
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) 
Calculus 
Anonymous,
Calculus  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) 
Calculus 
Anonymous,
you suck steve