Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 6

Given: f(x)=x^5-5x+2
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)=5x^4-5
f''(x)=20x^3=0
x^3=0
x=0
The critical value for the second derivative is x=0.
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 0.
f''(-1)=-20 Since f''(-1)<0 the graph is concave down in the interval (-oo ,0).
Choose a value of x that is greater than 0.
f''(1)=20 Since f''(1)>0 the graph is concave up in the interval (0, oo).