Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 15

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