Calculus of a Single Variable, Chapter 3, 3.1, Section 3.1, Problem 26

Given: f(x)=(2x)/(x^2+1)[-2,2]
Find the critical values for x by setting the derivative equal to zero and solving for the x value(s).
f'(x)=[(x^2+1)(2)-(2x)(2x)]/(x^2+1)^2=0
2x^2+2-4x^2=0
-2x^2+2=0
-2(x^2-1)=0
x=1,x=-1
Plug in the critical x value(s) and the endpoints of the closed interval into the f(x) function.
f(x)=(2x)/(x^2+1)
f(-2)=-.8
f(-1)=-1
f(1)=1
f(2)=-.8
Examine the f(x) values to determine the absolute extrema.
The absolute maximum is the point (1, 1).
The absolute minimum is the point (-1, -1).