f(x)=axe^(bx)
f(5/4)=1
then 1 = a(5/4) e^(5b/4)
f'(x) = ax(b)(e^(bx)) + a(e^(bx))
= a e^(bx) (bx + 1)
when x = 5/4, f'(5/4) = 0
a(e^(5b/4)) (5b/4+1) = 0
e^(5b/4) = 0 , no solution
or
5b/4 = -1
b = -4/5
in a(5/4) e^(5b/4)
5a/4 e^(5/4(-4/5) = 1
5a/4 e = 1
ae = 4/5
a = 4e/5
better check my arithmetic
Find values for a and b so that the function f(x)=axe^(bx) has the following properties:
f(5/4)=1
f has a local maximum when x = 5/4.
a =
b=
2 answers
Thanks so much for your help!!!