Calculus of a Single Variable, Chapter 3, 3.1, Section 3.1, Problem 22

Given: f(x)=2x^3-6x,[0, 3]
Find the critical values for x by setting the derivative equal to zero and solving for the x value(s).
f'(x)=6x^2-6=0
6(x^2-1)=0
6(x+1)(x-1)=0
x=-1, x=1
The critical values for x are x=1 and x=-1. Plug in the critical value(s) and the endpoints of the interval into f(x). Because x=-1 is not in the interval [0, 3], it is not necessary to plug in the x=-1
f(x)=2x^3-6x
f(0)=2(0)^3-6(0)=0
f(1)=2(1)^3-6(1)=-4
f(3)=2(3)^2-6(3)=36
Examine the f(x) values to determine the absolute extrema.
The absolute minimum value is the point (1, -4).
The absolute maximum value is the point (3, 36).