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 ln means the natural logarithm. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
The derivative of a function f is f′(x) = (x + 1)(x − 3)². Which statement is true?
- (A) f has a relative minimum at x = −1 and no other relative extrema.
- (B) f has a relative maximum at x = −1 and a relative minimum at x = 3.
- (C) f has a relative minimum at x = −1 and a relative maximum at x = 3.
- (D) f has a relative maximum at x = −1 and no other relative extrema.
Answer and explanation
Answer: (A). Critical points: x = −1 and x = 3. Test values: f′(−2) = (−1)(25) = −25, f′(0) = (1)(9) = 9, f′(4) = (5)(1) = 5. f′ changes from negative to positive at x = −1, so f has a relative minimum there. f′ is positive on both sides of x = 3, so f has neither at x = 3.
- (B) reverses the sign change at x = −1 and assumes the sign also changes at x = 3.
- (C) gets x = −1 right but assumes every zero of f′ is an extremum. The squared factor (x − 3)² does not change sign.
- (D) reverses the order of the signs at x = −1: f′ goes from − to +, which is a minimum.
Question 2 (multiple choice · core)
Let f(x) = x³ − 12x + 5. What is the relative maximum value of f?
- (A) −2
- (B) 21
- (C) −11
- (D) 2
Answer and explanation
Answer: (B). f′(x) = 3x² − 12 = 3(x − 2)(x + 2), so the critical points are x = −2 and x = 2. f′(−3) = 15 > 0, f′(0) = −12 < 0, f′(3) = 15 > 0. f′ changes from positive to negative at x = −2, so the relative maximum is at x = −2. Its value is f(−2) = −8 + 24 + 5 = 21.
- (A) is the location of the maximum, not its value.
- (C) is f(2), the relative minimum value.
- (D) is the location of the relative minimum.
Question 3 (multiple choice · core)
Let g(x) = x ln x for x > 0. Which statement is true?
- (A) g has a relative minimum at x = 1/e.
- (B) g has a relative maximum at x = 1/e.
- (C) g has a relative minimum at x = 1.
- (D) g has no relative extrema.
Answer and explanation
Answer: (A). By the product rule, g′(x) = ln x + x · (1/x) = ln x + 1. g′(x) = 0 when ln x = −1, so x = 1/e. Test values: g′(e⁻²) = −2 + 1 = −1 < 0 and g′(1) = 0 + 1 = 1 > 0. g′ changes from negative to positive at x = 1/e, so g has a relative minimum there (value g(1/e) = −1/e).
- (B) reverses the direction of the sign change.
- (C) forgets the “+ 1” from the product rule and solves ln x = 0.
- (D) would only be true if g′ kept one sign. But g′ = ln x + 1 is negative for 0 < x < 1/e and positive for x > 1/e.
Question 4 (multiple choice · core)
A differentiable function f has f′(2) = 0. Which statement is a correct justification that f has a relative maximum at x = 2?
- (A) f′(2) = 0.
- (B) f′(x) changes from positive to negative at x = 2.
- (C) The graph of f′ has a maximum at x = 2.
- (D) f(2) is greater than f(0).
Answer and explanation
Answer: (B). This is exactly the condition in the First Derivative Test, and f is continuous at 2 because it is differentiable.
- (A) only shows that x = 2 is a critical point. It could be a maximum, a minimum or neither.
- (C) is about the graph of f′, not f. In fact, if f′ has a maximum value of 0 at x = 2, then f′ ≤ 0 on both sides, so f′ cannot change from positive to negative at x = 2. Usually f is decreasing through x = 2 and has neither a maximum nor a minimum there.
- (D) compares f(2) with one other value. A relative maximum needs f(2) ≥ f(x) for all x near 2.
Question 5 (graph of f′ · core)
The function f is continuous on −4 ≤ x ≤ 5. The graph of its derivative f′ is shown below. It is made of straight segments joining (−4, 2), (−2, −2), (0, 0), (1, −2), (3, 2) and (5, 1).
(a) Find the x-coordinate of each relative maximum of f on −4 < x < 5. Justify your answer. (b) Find the x-coordinate of each relative minimum of f on −4 < x < 5. Justify your answer. (c) Does f have a relative extremum at x = 0? Justify your answer. (d) The graph of f′ has a lowest point at x = −2. A student says f has a relative minimum at x = −2. Is the student right? Explain.
Worked solution and suggested Marlbridge rubric
First locate the zeros of f′. The segment from (−4, 2) to (−2, −2) has slope −2 and crosses the axis at x = −3. The graph touches the axis at (0, 0). The segment from (1, −2) to (3, 2) has slope 2 and crosses at x = 2. So the critical points in −4 < x < 5 are x = −3, 0 and 2.
(a) f has a relative maximum at x = −3, because f′ changes from positive to negative at x = −3.
(b) f has a relative minimum at x = 2, because f′ changes from negative to positive at x = 2.
(c) No. f′(0) = 0, but f′(x) < 0 on both sides of x = 0 (on (−3, 0) and on (0, 2)). f′ does not change sign, so f has neither a relative maximum nor a relative minimum at x = 0.
(d) No. f′(−2) = −2, which is not 0, and f′ exists there, so x = −2 is not a critical point of f. f′ < 0 on both sides of x = −2, so f is decreasing through x = −2. The lowest point of the graph of f′ is where f is decreasing most steeply, not a minimum of f.
| Point | What earns it |
|---|---|
| 1 | Zeros of f′ at x = −3, 0 and 2 |
| 1 | (a) x = −3, with “f′ changes from positive to negative” as the reason |
| 1 | (b) x = 2, with “f′ changes from negative to positive” as the reason |
| 1 | (c) Neither at x = 0, because f′ does not change sign (negative on both sides) |
| 1 | (d) Not a minimum: f′(−2) ≠ 0 (or f′ < 0 on both sides), so f is decreasing there |
Question 6 (constructed response · core)
A function p is defined by
- p(x) = x² + 1 for x < 1
- p(x) = 1 − x for x ≥ 1
(a) Find p′(x) for x < 1 and for x > 1. Show that p′ changes from positive to negative at x = 1. (b) Is p continuous at x = 1? Show your reasoning. (c) A student says: “p′ changes from positive to negative at x = 1, so by the First Derivative Test p has a relative maximum at x = 1.” Explain why the test cannot be used here, and decide whether p has a relative maximum at x = 1. (d) Find any other relative extremum of p, and justify it.
Worked solution and suggested Marlbridge rubric
(a) For x < 1, p′(x) = 2x. For x > 1, p′(x) = −1. On 0 < x < 1, p′(x) = 2x > 0. For x > 1, p′(x) = −1 < 0. So p′ changes from positive to negative at x = 1.
(b) No. As x → 1 from the left, p(x) = x² + 1 → 2. But p(1) = 1 − 1 = 0. The left-hand limit (2) is not equal to p(1) (0), so p is not continuous at x = 1.
(c) The First Derivative Test needs p to be continuous at x = 1, and it is not. In fact p(1) = 0 is not a relative maximum: for x just less than 1, p(x) is close to 2, which is greater than p(1). (It is not a relative minimum either, because for x just greater than 1, p(x) = 1 − x < 0 = p(1).) So p has neither at x = 1.
(d) For x < 1, p′(x) = 2x = 0 at x = 0. p′(−1/2) = −1 < 0 and p′(1/2) = 1 > 0, so p′ changes from negative to positive at x = 0, and p is continuous there. p has a relative minimum at x = 0, with value p(0) = 1.
| Point | What earns it |
|---|---|
| 1 | p′(x) = 2x for x < 1 and p′(x) = −1 for x > 1, with the sign change shown |
| 1 | Not continuous at 1: left-hand limit 2 ≠ p(1) = 0 |
| 1 | States that the test requires continuity at x = 1 |
| 1 | Concludes no relative maximum at x = 1, using a nearby value greater than p(1) |
| 1 | Relative minimum at x = 0 (value 1), justified by p′ changing from negative to positive |
Question 7 (constructed response · stretch)
Let r(x) = (x² − 3)eˣ.
(a) Show that r′(x) = (x + 3)(x − 1)eˣ. (b) Find the x-coordinates of all relative extrema of r and classify each one. Justify your answers. (c) Find the exact value of each relative extremum. (d) A student writes: “r′(−3) = 0 and r′(1) = 0, so r has a relative maximum at one of them and a relative minimum at the other.” Explain why this is not a valid justification.
Worked solution and suggested Marlbridge rubric
(a) Product rule: r′(x) = 2x · eˣ + (x² − 3)eˣ = (x² + 2x − 3)eˣ = (x + 3)(x − 1)eˣ.
(b) eˣ > 0 for all x, so the sign of r′ is the sign of (x + 3)(x − 1). Critical points: x = −3 and x = 1.
| Interval | Test value | (x + 3)(x − 1) | Sign of r′ |
|---|---|---|---|
| (−∞, −3) | x = −4 | (−1)(−5) = 5 | + |
| (−3, 1) | x = 0 | (3)(−1) = −3 | − |
| (1, ∞) | x = 2 | (5)(1) = 5 | + |
r has a relative maximum at x = −3, because r′ changes from positive to negative there. r has a relative minimum at x = 1, because r′ changes from negative to positive there.
(c) r(−3) = (9 − 3)e⁻³ = 6e⁻³. r(1) = (1 − 3)e = −2e.
(d) r′(c) = 0 only shows that c is a critical point. A critical point need not be an extremum at all (f′ may keep its sign), and even if both are extrema, r′ = 0 does not say which is which. The justification must give the change of sign of r′ at each point.
| Point | What earns it |
|---|---|
| 1 | Correct product rule and factorisation to (x + 3)(x − 1)eˣ |
| 1 | Uses eˣ > 0 (or test values) to find the sign of r′ on each interval |
| 1 | Relative maximum at x = −3 and relative minimum at x = 1, each justified by the sign change of r′ |
| 1 | Values 6e⁻³ and −2e |
| 1 | Explains that r′ = 0 identifies candidates only; the change of sign decides |
Decimal values (about 0.299 and −5.44) are acceptable as a check, but exact answers are expected.
How did you do?
- Q1 or Q3 wrong: redo Worked example 1 in the study guide, especially the “neither” case.
- Q2 or Q7(c) wrong: reread “Mixing up location and value” in the misconceptions list.
- Q4 or Q7(d) wrong: reread “Writing a justification”.
- Q5 wrong: use the table “Reading extrema from a graph of f′”.
- Q6 wrong: reread “When the test does not apply”.
Then tick off the topic checklist.