Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 34

Given: f(x)=36x+3x^2-2x^3
Find the critical numbers by setting first derivative equal to zero and solving for the x value(s).
f'(x)=36+6x-6x^2=0
f'(x)=6+x-x^2=0
f'(x)=x^2-x-6=0
(x-3)(x+2)=0
The critical values are x=3 and x=-2
Part a)
If f'(x)>0 then the function is increasing on the interval.
If f'(x)<0 then the function is decreasing on the interval.
Select an x value in the interval (- oo ,-2).
Since f'(-3)<0 the function is decreasing in the interval (-oo ,-2).
Select an x value in the interval (-2,3).
Since f(0)>0 the function is increasing in the interval (-2, 3 ).
Select an x value in the interval (3,oo).
Since f(4)<0 the function is decreasing in the interval (3,oo ).
Part b)
Because the function changed direction from decreasing to increasing at x=-2 a local minimum exists. The local minimum occurs at the point (-2,-44).
Because the function changed direction from increasing to decreasing at x=3 a local maximum exists. The local maximum occurs at the point (3, 81).
Part c)
Find the critical values for the second derivative.
f''(x)=6-12x=0
6=12x
x=1/2
The critical value is x=1/2.
If f''(x)>0 then the graph is concave up in the interval.
If f"(x)>0 then the graph is concave down in the interval.
If f'(x)=0 then an inflection point will exist.
Select an x value in the interval (-oo , 1/2).
Since f"(0)>0 the graph is concave up in the interval (-oo ,1/2).
Select an x value in the interval (1/2, oo ).
Since f"(1)<0 the graph is concave down in the interval (1/2, oo ).
Since f"(1/2)=0 there is an inflection point at x=1/2. The inflection point occurs at the coordinate (1/2, 18.5).