Calculus of a Single Variable, Chapter 3, 3.3, Section 3.3, Problem 22

Given: f(x)=x^3-6x^2+15
Find the critical values for x by setting the first derivative of the function equal to zero and solving for the x value(s).
f'(x)=3x^2-12x=0
3x(x-4)=0
x=0,x=4
The critical value for the first derivative are x=0 and x=4.
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 0.
f'(-1)=15 Since f'(-1)>0 the graph of the function is increasing in the interval (-oo,0).
Choose a value for x that is between 0 and 4.
f'(1)=-9 Since f'(1)<0 the graph of the function is decreasing in the interval
(0, 4).
Choose a value for x that is greater than 4.
f'(5)=15 Since f'(5)>0 the graph of the function is increasing in the interval
(4, oo).
Because the function changed direction from increasing to decreasing there will be a relative maximum at x=0. The relative maximum occurs at (0, 15).
Also, because the function changed direction from decreasing to increasing there will be a relative minimum at x=4. The relative minimum occurs at
(4, -17).