Calculus of a Single Variable, Chapter 3, 3.3, Section 3.3, Problem 28

Given: f(x)=x^(2/3)-4
Find the critical value(s) by setting the first derivative equal to zero and solving for the x value(s).
f'(x)=(2/3)x^(-1/3)=0
2/(3x^(1/3))=0
2=0
Because 2=0 is not a true statement the first derivative will not produce a critical value.
If f'(x)>0 the function will increase in the interval.
If f'(x)<0 the function will decrease in the interval.
Notice f'(x)=undefined. This means the function is not differentiable at x=0.
Choose a value for x that is less than zero.
f'(-1)=-2/3 Since f'(-1)<0 the function is decreasing on the interval (-oo,0).
Choose a value for x that is greater than zero.
f'(1)=2/3 Since f'(1)>0 the function is increasing on the interval (0, oo).
Because the function changes direction from decreasing to increasing a relative minimum exists. The relative minimum exists at the point (0, -4) even though the derivative does not exist at that point.