Math
posted by Rebekah on .
Find the absolute extrema of
f(x)= (2x1)e^(x) on [0, 00)

f ' (x) = (2x1) e^(x) ( 1) + 2 e^(x)
= 0 for a max/min
e^(x) (2  (2x1)) = 0
e^(x) ( 3  2x) = 0
e^(x) = 0 > no solution
or
3  2x = 0
x = 3/2
then f(3/2) = 2e^(3/2) = 2/e^(3/2)