Analyzing Functions Test Answer Key with Scoring – AP Calculus AB

Question 1 Absolute maximum (2 points)

f′(x) = 3x² − 12 = 0 gives x = ±2. Candidates: f(−3) = 9, f(−2) = 16, f(2) = −16, f(5) = 65. The maximum is 65. Answer (B).

Scoring: 2 points for the correct choice; 1 point for listing the candidates.

Question 2 Local maximum from a graph (2 points)

f′(x) is negative for x < −2, positive on (−2, 1) and negative for x > 1. The sign changes from + to − at x = 1. Answer (C).

Scoring: 2 points for the correct choice; 1 point for identifying where f′(x) changes sign.

Question 3 Mean Value Theorem application (3 points)

(a) f is a polynomial, so it is continuous on [0, 3] and differentiable on (0, 3).

(b) The average rate is (f(3) − f(0))/3 = (21 − 0)/3 = 7. Solve 3c² − 2 = 7, so c² = 3 and c = √3 ≈ 1.732 (the negative root is outside (0, 3)).

Scoring: Part (a): 1 point. Part (b): 1 point for the average rate, 1 point for c.

Question 4 Concavity analysis (3 points)

(a) g′ = 4x³ − 24x² + 36x, so g″ = 12x² − 48x + 36 = 12(x − 1)(x − 3).

(b) g″ > 0 for x < 1 and x > 3, and < 0 on (1, 3). So concave up on (−∞, 1) and (3, ∞), concave down on (1, 3).

(c) The sign of g″ changes at x = 1 and x = 3, so those are the inflection points, with g(1) = 11 and g(3) = 27.

Scoring: 1 point per part.

Question 5 Minimizing the surface of a box (3 points)

(a) The height is h = 1,728/s², so S = 2s² + 4sh = 2s² + 6,912/s.

(b) S′ = 4s − 6,912/s² = 0 gives s³ = 1,728, so s = 12 cm (and h = 12).

(c) S″ = 4 + 13,824/s³ > 0 for s > 0, so the graph is concave up and s = 12 gives a minimum.

Scoring: 1 point per part.

Question 6 First derivative test (3 points)

(a) f′(x) = 6x² − 18x + 12 = 6(x − 1)(x − 2).

(b) Signs: positive for x < 1, negative on (1, 2), positive for x > 2. So x = 1 is a local maximum and x = 2 a local minimum.

(c) f(1) = 4 and f(2) = 3. The local maximum value is 4 and the local minimum value is 3.

Scoring: 1 point per part.

Question 7 Analyzing a graph (2 points)

The graph rises between the minimum at x = −1 and the maximum at x = 1, so f is increasing on (−1, 1). It bends upward to the left of the origin and downward to the right, so f is concave up on (−∞, 0), which on the graph shown is the stretch from the left edge to the origin.

Scoring: 1 point for each interval.

Question 8 Extreme Value Theorem hypotheses (2 points)

The theorem needs continuity on the closed interval. g is not continuous at x = 1, where it has a vertical asymptote, so the theorem does not apply. Indeed, g(x) becomes arbitrarily large as x → 1⁺, so g has no absolute maximum on [0, 2].

Scoring: 1 point for the discontinuity, 1 point for the conclusion about the maximum.