Find the absolute maximum and absolute minimum values of f on the given interval. (If an answer does not exist, enter DNE.)
f(x) = x^3 − 6x^2 + 9x + 8
on
[−1, 7]

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.

2^y = ln 5, or   y(ln 2) = ln 5.   solving for y,   y = (ln 5) / (ln 2).   y=5

step-by-step explanation: