Calculus of a Single Variable, Chapter 3, 3.3, Section 3.3, Problem 36

Given: f(x)=(x^2-2x+1)/(x+1)
Find the critical numbers by setting the first derivative equal to zero and solving for the x value(s).
f'(x)=[(x+1)(2x-2)-(x^2-2x+1)(1)]/(x+1)^2=0
2x^2-2x+2x-2-x^2+2x-1=0
x^2+2x-3=0
(x+3)(x-1)=0
x=-3,x=1
The critical numbers are x=-3 and x=1. Critical numbers also exist where f(x) is not defined. Therefore x=-1 is also a critical number.
If f'(x)>0 the function is increasing on the interval.
If f'(x)<0 the function is decreasing on the interval.
Choose an x value that is less than -3.
f'(-4)=.5556 Since f'(-4)>0 the function is increasing in the interval
(-oo,-3).
Choose an x value that is between -3 and -1.
f'(-2)=-3 Since f'(-2) >0 the function is decreasing in the interval (-3, -1).
Choose an x value that is between -1 and 1.
f'(0)=-3 Since f'(0)<0 the function is decreasing in the interval (-1, 1).
Choose an x value that is greater than 1.
f(2)=.5556 since f'(2)>0 the function is increasing in the interval (1, oo).
Since the function changed direction from increasing to decreasing there is a relative maximum at x=-3. The relative maximum is at the point (-3, -8).
Since the function changed direction form decreasing to increasing there is a relative minimum at x=1. The relative minimum is at the point (1, 0).