(D)(C)(A) M′(1) = f ′(g(1)) · g′(1) = f ′(3)g′(1) = 4(−3).(B) [f(x^3 )]′ = f ′(x^3 ) · 3 x^2 , so P′(1) = f ′(1^3 ) · 3 · 12 = 2 · 3.(D) f(S(x)) = x implies that f ′(S(x)) · S′(x) = 1, so(E) Since g′(a) exists, g is differentiable and thus continuous; g′(a) > 0.(C) Near a vertical asymptote the slopes must approach ±∞.(A) There is only one horizontal tangent.(D) Use the symmetric difference quotient; then