Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 16

Given: f(x)=-x^3+6x^2-5
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
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 up in the interval (-oo 2).
Choose a value for x that is greater than 2.
f''(3)=-6 Since f''(3)<0 the graph is concave down in the interval (2, oo).
Since the concavity changed directions and f''(2)=0 the inflection point occurs at x=2. The inflection point is (2, 11).