Verified answer

D

You'd solve this by more or less trial and error but Ill start on the correct path

Graph III starts out with a very large negative slope that zeros out at about x=-4.5, then is positive until it zeros out again at x=-1, and then is negative for the rest of the visible graph

Because g' represents the value of the slope of the graph at a point, the g' graph will start out negative before hitting Y=0 at x=-4.5, going into positive Y values until it hits Y=0 at x=1, where it remains negative for the rest of the visible graph, this description matches graph I

By process of elimination II must be the g" graph

It's B.