# Calculus (pleas help!)

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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=

• Calculus (pleas help!) -

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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