Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 4

Given: h(x)=12x-x^3
Find the critical values for x by setting the second derivative of the function equal to zero and solving for the x value(s).
h'(x)=12-3x^2
h''(x)=-6x=0
x=0
The critical value for the second derivative is x=0.
If h''(x)>0, the curve is concave up in the interval.
If h''(x)<0, the curve is concave down in the interval.
Choose a value for x that is less than 0.
h''(-1)=6 Since h''(-1)>0 the graph is concave up in the interval (-oo,0).
Choose a value for x that is greater than 0.
h''(1)=-6 Since h''(1)<0 the graph is concave down in the interval (0, oo).
Because the function changed from concave up to concave down and h''(0)=0, there will be and inflection point and x=0
The inflection point is (0, 0).