Calculus of a Single Variable, Chapter 3, 3.3, Section 3.3, Problem 25

Given: f(x)=(x^5-5x)/5=(1/5)x^5-x
Find the critical x values by setting the derivative equal to zero and solving for the x value(s).
f'(x)=x^4-1=0
(x^2+1)(x^2-1)=0
x=1, x=-1
The critical values for x are x=1 and x=-1.
If f'(x)>0 the function is increasing over an interval.
If f'(x)<0 the function is increasing over an interval.
Choose an x value less than -1.
f'(-2)=15 Since f'(-2)>0 the function is increasing in the interval (-oo, -1).
Choose an x value between -1 and 1.
f'(0)=-1 Since f'(0)<0 the function is decreasing in the interval (-1, 1).
Choose an x value greater than 1.
f'(2)=15 Since f'(2)>0 the function is increasing in the interval (1, oo).
Because the function changed direction from increasing to decreasing a relative maximum will exist at x=-1. The relative maximum occurs at the point
(-1, 4/5).
Because the function changed direction from decreasing to increasing a relative minimum will exist at x=1. The relative minimum occurs at the point (1, -4/5).