(B)(B) Let . Then f ′ increases for 1 < x < 2, then begins to
decrease. In the figure above, the area below the x-axis, from 2 to 3, is
equal in magnitude to that above the x-axis, hence, .(D) P′(x) = 2g(x) · g′(x); P′(3) = 2g(3) · g′(3) = 2 · 2 · 3 = 12.(D) Note that H(3) = f −^1 (3) = 2. Therefore(C) Note that the domain of y is all x such that |x| 1 and that the graph
is symmetric to the origin. The area is given by(D) Since