Calculus of a Single Variable, Chapter 3, 3.3, Section 3.3, Problem 26

Given: f(x)=x^4-32x+4
Find the critical numbers by setting the first derivative equal to zero and solving for the x value(s).
f'(x)=4x^3-32=0
4x^3=32
x^3=8
x=2
The critical value is x=2.
If f'(x)>0 the function is increasing in the interval.
If f'(x)<0 the function is decreasing in the interval.
Choose a value for x that is less than 2.
f'(0)=-32 Since f'(0)<0 the function is decreasing on the interval (-oo,2).
Choose a value for x that is greater than 2.
f'(3)=76 Since f'(3)>0 the function is increasing on the interval (2, oo).
Since the function changed directions from decreasing to increasing a relative minimum exists. The relative minimum occurs at the point (2, -44).