Calculus of a Single Variable, Chapter 3, 3.3, Section 3.3, Problem 23

Given: f(x)=(x-1)^2(x+3)
Find the critical values by setting the first derivative equal to zero and solving for the x value(s).
f'(x)=(x-1)^2(1)+(x+3)(2(x-1))=0
x^2-2x+1+(x+3)(2x-2)=0
x^2-2x+1+2x^2-2x+6x-6=0
3x^2+2x-5=0
(3x+5)(x-1)=0
x=-5/3, x=1
The critical numbers are x=-5/3 and x=1.
If f'(x)>0 the function increases in the interval.
If f'(x)<0 the function decreases in the interval.
Choose an x value less than -5/3.
f'(-2)=3 Since f'(-2)>0 the function increases in the interval (-oo,-5/3).
Choose an x value between -5/3 and 1.
f'(-1)=-4 Since f'(-1)<0 the function decreases in the interval (-5/3, 1).
Choose an x value greater than 1.
f'(2)=11 Since f'(2)>0 The function increases in the interval (1, oo).
Because the direction of the function changed from increasing to decreasing a relative maximum will exist at x=-5/3. The relative maximum is the point
(-5/3, 9.4815).
Because the direction of the function changed from decreasing to increasing a relative minimum will exist at x=1. The relative minimum is the point (1, 0).