Calculus (pleas help!)
posted by Fernando .
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.
a=
b=
Please help me with this one...

Calculus (pleas help!) 
Reiny
you know that f(1/8) = 1, that is, ...
1 = a(1/8)e^(b/8)
f'(x) = ax(b)(e^(bx) + a(e^(bx))
= e^(bx)(abx + a)
when x = 1/8 this is zero
0 = e^(b/8)(ab/8 + a)
ab/8 + a = 0 or e^(b/8) = 0, the last part has no solution
so ab/8 = a
ab = 8a
b = 8
then in original, x = 1/8, b = 8
1 = (a/8)e^(8/8)
8 = a(e^1)
a = 8e
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