Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 17

Given: f(x)=(1/2)x^4+2x^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).
f'(x)=2x^3+6x^2
f''(x)=6x^2+12x=0
6x(x+2)=0
x=0,x=-2
The critical value for the second derivative is x=0 and 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''(-3)=18 Since f''(-3)>0 the graph is concave up in the interval (-oo,-2 ).
Choose a value for x that is between -2 and 0.
f''(-1)=-6 Since f''(-1)<0 the graph is concave down in the interval (-2, 0).
Choose a value for x that is greater than 0.
f''(1)=18 Since f''(1)>0 the graph is concave up in the interval (0, oo).
Because the direction of concavity changes twice and because f''(-2)=0 and
f"(0)=0 there will be an inflection point at x=-2 and x=0.
The inflection points are (-2, -8) and (0, 0).