Answer:
absolute max is 120 and absolute min is -8
Step-by-step explanation:
Find critical numbers
f'(x) = 3x^2 - 12x + 9 = 0
= 3(x^2 - 4x + 3) = 0
3(x-3)(x-1) = 0
(x-3) = 0 or (x-1)=0
x = 1,3
Test them!
x<1 Sign of f' on this interval is positive
1<x<3 Sign of f' on this interval is negative
x>3 Sign of f' on this interval is positive
f(x) changes from positive to negative at x = 1 which means there is a relative maximum here.
f(x) changes from negative to positive at x = 3 which means there is a relative minimum here.
Test the endpoints to find the absolute max and min.
f(-1) = -8
f(1) = 12
f(3) = 8
f(7) = 120
The absolute maximum value of f is 120 and the absolute minimum value of f is -8.
