Calculus of a Single Variable, Chapter 3, 3.1, Section 3.1, Problem 21

Given: f(x)=x^3-(3/2)x^2,[-1,2]
First find the critical x value(s) of the function. To find the critical x value(s), set the derivative equal to zero and solve for the x value(s).
f'(x)=3x^2-3x=0
3x^2-3x=0
3x(x-1)=0
x=0,x=1
Plug the critical x value(s) and the endpoints of the closed interval into the original f(x) function.
f(x)=x^3-(3/2)x^2
f(-1)=(-1)^3-(3/2)(-1)^2=-5/2
f(0)=(0)^3-(3/2)(0)^2=0
f(1)=(1)^3-(3/2)(1)^2=-1/2
f(2)=(2)^3-(3/2)(2)^2=2
Examine the f(x) values.
The absolute maximum is at the point (2, 2).
The absolute minimum is at the point (-1,-5/2).