What is the Absolute Extrema of this function?

Question

I have the function f (x) = x^4/3 on the interval [-1,8]. Use the (EVT)

f(x)= x^4/3
f ' (x) = 4/3 x ^1/3

Then what do I do after that?

Answer 1

Set it equal to 0 to find the relative max's and min's.

4/3 x^(1/3) = 0
x = 0

Now test the relative max's and min's and the bounds to find the absolute max's and min's:
f(-1) = 1
f(0) = 0
f(8) = 16

Absolute max is 16 at x=8
Absolute min is 0 at x=0

Answer 2

Well this function is 0 at zero and goes to infinity as x -> infinity (+ or -). So absol. extrema on your interval is 8^(4/3)