(D) Since ab > 0, a and b have the same sign; therefore f ′′(x) = 12ax^2 +
2 b never equals 0. The curve has one horizontal tangent at x = 0.
(C) Since the first derivative is positive, the function must be increasing.
However, the negative second derivative indicates that the rate of
increase is slowing down, as seen in table C.
(B) Since , therefore, at t = 1, . Also, x = 3 and y = 2.
(A) Let , and find the slope of the tangent line at (64, 4). Since
, . If we move one unit to the left of 64, the tangent
line will drop approximately unit.
(D)
(E) ekh e k · 0 + ke k · 0 (h − 0) = 1 + kh
(E) Since the curve has a positive y-intercept, e > 0. Note that f ′(x) = 2cx
+ d and f ′′(x) = 2c. Since the curve is concave down, f ′′(x) < 0, implying
that c < 0. Since the curve is decreasing at x = 0, f ′(0) must be negative,
implying, since f ′(0) = d, that d < 0. Therefore c < 0, d < 0, and e > 0.
(A) . Solving the equation of the line for y yields , so to
be parallel the slope of the tangent must also be . If , then 2x +
1 = 9.
