calculus
posted by casidhe on .
Consider the function f(x)=(x^3)(e^9x),
2 is less than or equal to x is less than or equal to 4
*The absolute maximum value is _____ and this occurs at x equals 4.
*The absolute minimum value is _____ and this occurs at x equals ______
I got the 4 but I must be doing something wrong with the other numbers because it is not working

let's take the derivative
f'(x) = 3x^2 (e^(9x) ) + x^3 (9) e^(9x)
= 3x^2 e^(9x) [ 1 + 3x]
= 0 for a max/min
x = 0 , x = 1/3
f(0) = 0
f(1/3) = (1/27) e^3 = 1/(27e^3) = appr .00184
f(2) = 8 e^18 = 8/e^18 = appr .0000000121
so the absolute minimum is 8/e^18 , when x = 2 
wrong is x=4