These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Assumptions for every question: no calculator, exact answers, and concavity stated on open intervals. The context in Question 3 is invented. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
On which of these intervals is the graph of f(x) = x³ + 3x² − 5 concave down on the whole interval?
- (A) (0, ∞)
- (B) (−∞, −1)
- (C) (−2, 0)
- (D) (−1, ∞)
Answer and explanation
Answer: (B). f′(x) = 3x² + 6x and f″(x) = 6x + 6 = 6(x + 1). f″ < 0 when x < −1, so the graph is concave down on (−∞, −1).
- (A) is an interval where f′ = 3x(x + 2) > 0, so f is increasing there. That is not concavity: in fact f″ > 0 there, so the graph is concave up.
- (C) is where f′ < 0, so f is decreasing there. Again, the sign of f′ does not decide concavity: the graph is concave down only on (−2, −1) and concave up on (−1, 0).
- (D) is where f″ > 0: concave up.
Question 2 (multiple choice · core)
Let f(x) = x⁵ − 5x⁴. At which x-values does the graph of f have a point of inflection?
- (A) x = 3 only
- (B) x = 0 and x = 3
- (C) x = 0 and x = 4
- (D) x = 4 only
Answer and explanation
Answer: (A). f″(x) = 20x³ − 60x² = 20x²(x − 3). The candidates are x = 0 and x = 3. The factor x² is never negative, so the sign of f″ is the sign of (x − 3): negative for x < 0, negative for 0 < x < 3, positive for x > 3. The sign changes only at x = 3.
- (B) treats x = 0 as an inflection point because f″(0) = 0. But f″ is negative on both sides of 0, so the graph is concave down on both sides.
- (C) and (D) use the zeros of f′(x) = 5x³(x − 4). Those are critical points, not inflection points.
Question 3 (multiple choice · foundation)
The volume of water V(t) in an invented storage tank, t hours after a pump is switched on, satisfies V′(t) > 0 and V″(t) < 0 for 0 < t < 6. Which statement is true for 0 < t < 6?
- (A) The volume is increasing at an increasing rate.
- (B) The volume is increasing at a decreasing rate.
- (C) The volume is decreasing at a decreasing rate.
- (D) The volume is decreasing at an increasing rate.
Answer and explanation
Answer: (B). V′ > 0 means the volume is increasing. V″ < 0 means V′, the rate of increase, is itself decreasing. So the tank is filling, but more slowly as time passes. The graph of V is increasing and concave down.
- (A) would need V″ > 0.
- (C) and (D) would need V′ < 0. They read the sign of V″ as the direction of V.
Question 4 (multiple choice · core)
The derivative of a function f is f′(x) = (x − 1)²(x + 2). At which x-values does the graph of f have a point of inflection?
- (A) x = −2 and x = 1
- (B) x = −1 and x = 1
- (C) x = −1 only
- (D) x = 1 only
Answer and explanation
Answer: (B). Product rule: f″(x) = 2(x − 1)(x + 2) + (x − 1)² = (x − 1)(2x + 4 + x − 1) = (x − 1)(3x + 3) = 3(x − 1)(x + 1). f″ is positive for x < −1, negative for −1 < x < 1 and positive for x > 1. It changes sign at both x = −1 and x = 1.
- (A) gives the zeros of f′, which are critical points.
- (C) rejects x = 1 because f′(1) = 0. A critical point can also be an inflection point: here f′ has a local minimum of 0 at x = 1.
- (D) misses x = −1. It is not a zero of f′, so it only appears once you find and factor f″.
Question 5 (graph · core)
The function f is continuous on [0, 8]. The graph of its derivative f′ consists of three line segments, joining the points (0, −2), (3, 4), (5, 0) and (8, 3).
(a) Find the open intervals on which the graph of f is concave up and concave down. Give a reason. (b) Find the x-coordinates of all points of inflection of f. Give a reason. (c) A student says: “x = 1 is a point of inflection because f′(1) = 0.” Explain the error. (d) Find f″(4).
Worked solution and suggested Marlbridge rubric
(a) f′ is increasing on (0, 3) and (5, 8), so the graph of f is concave up there. f′ is decreasing on (3, 5), so the graph of f is concave down there.
(b) Concavity changes at x = 3 (up to down) and at x = 5 (down to up), and f is continuous there. These are the points of inflection. (f″ does not exist at the corners x = 3 and x = 5, but that does not matter: the concavity still changes.)
(c) f′(1) = 0, and f′ changes from negative to positive there, so f has a local minimum at x = 1. That is a statement about f′ crossing the axis. Concavity depends on whether f′ is increasing or decreasing, and f′ is increasing on both sides of x = 1, so the concavity does not change there.
(d) f″(4) is the slope of the f′ graph at x = 4, which is on the segment from (3, 4) to (5, 0). Slope = (0 − 4)/(5 − 3) = −2.
| Point | What earns it |
|---|---|
| 1 | Concave up on (0, 3) and (5, 8), concave down on (3, 5), with a reason based on f′ increasing or decreasing |
| 1 | Inflection points at x = 3 and x = 5, with a reason (concavity or the sign of f″ changes) |
| 1 | Explains that f′ = 0 at x = 1 signals a possible extremum, and that f′ is increasing on both sides, so no change of concavity |
| 1 | f″(4) = −2 from the slope of the segment |
Note: x = 5 is also a point where f′ = 0, but f′ ≥ 0 on both sides, so f has no local extremum there. It is an inflection point only.
Question 6 (constructed response · core)
Let h(x) = x² + 8/x, for x ≠ 0.
(a) Show that h″(x) = 2(x³ + 8)/x³. (b) Find the open intervals on which the graph of h is concave up and concave down. (c) Find all points of inflection of the graph of h. Explain why there is no point of inflection at x = 0.
Worked solution and suggested Marlbridge rubric
(a) h′(x) = 2x − 8/x². h″(x) = 2 + 16/x³ = (2x³ + 16)/x³ = 2(x³ + 8)/x³.
(b) h″ is 0 when x³ = −8, so x = −2. h″ does not exist at x = 0. Check signs:
| Interval | x³ + 8 | x³ | h″ | Concavity |
|---|---|---|---|---|
| (−∞, −2) | − | − | + | concave up |
| (−2, 0) | + | − | − | concave down |
| (0, ∞) | + | + | + | concave up |
(c) At x = −2, h″ changes sign and h(−2) = 4 + 8/(−2) = 0. The point of inflection is (−2, 0). Concavity also changes at x = 0, but h(0) is not defined, so there is no point on the graph at x = 0 and therefore no point of inflection.
| Point | What earns it |
|---|---|
| 1 | Correct h′ and h″ in the given form |
| 1 | Candidates x = −2 (h″ = 0) and x = 0 (h″ undefined) and a correct sign analysis |
| 1 | Correct intervals: up on (−∞, −2) and (0, ∞), down on (−2, 0) |
| 1 | Inflection point (−2, 0) with a sign-change reason, and x = 0 rejected because h(0) is not defined |
Question 7 (constructed response · stretch)
Let f(x) = x³ + ax² + bx, where a and b are constants.
(a) The graph of f has a point of inflection at (1, −3). Find a and b. (b) Using your values, confirm that the concavity of f really changes at x = 1. (c) Let k(x) = (x − 1)⁴ − 3. Show that k″(1) = 0 and that k(1) = −3, but that (1, −3) is not a point of inflection of k.
Worked solution and suggested Marlbridge rubric
(a) f″(x) = 6x + 2a. For an inflection point at x = 1 we need f″(1) = 0 (f″ is a polynomial, so it exists everywhere): 6 + 2a = 0, so a = −3. The point (1, −3) is on the graph: f(1) = 1 + a + b = 1 − 3 + b = −3, so b = −1.
(b) f″(x) = 6x − 6 = 6(x − 1). It is negative for x < 1 and positive for x > 1. The sign changes, so the graph changes from concave down to concave up at x = 1. And f(1) = 1 − 3 − 1 = −3, as required.
(c) k′(x) = 4(x − 1)³ and k″(x) = 12(x − 1)². So k″(1) = 0 and k(1) = 0 − 3 = −3. But 12(x − 1)² > 0 for every x ≠ 1, so the graph is concave up on both sides of x = 1. The concavity does not change, so (1, −3) is not a point of inflection (it is in fact the minimum point of k).
| Point | What earns it |
|---|---|
| 1 | Uses f″(1) = 0 to find a = −3 |
| 1 | Uses f(1) = −3 to find b = −1 |
| 1 | Shows f″ = 6(x − 1) changes sign at x = 1 |
| 1 | For k: k″(1) = 0, k(1) = −3, and k″ > 0 on both sides, so no change of concavity |
How did you do?
- Q1 or Q2 wrong: revisit “The second derivative decides concavity” and Worked example 1 in the study guide.
- Q4 or Q5 wrong: redo Worked example 2 (concavity from a graph of f′).
- Q6 or Q7 wrong: reread “Points of inflection”, especially the table with x⁴, ∛x and 1/x.
- Q3 wrong: see “Concavity in context”.
Then tick off the topic checklist.