`h(x) = x^5 - 5x + 2` Determine the open intervals on which the graph is concave upward or downward.
Find the critical values for x by setting the second derivative of the function equal to zero and solving for the x value(s).
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).`