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, angles in radians, and every function in Questions 1–6 is twice differentiable on its domain unless the question says otherwise. Question 7 deliberately includes a point where the derivative does not exist. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
A function f has derivative f′(x) = x²(x − 4). Which statement is true?
- (A) f has a relative maximum at x = 0 and a relative minimum at x = 4.
- (B) f has a relative minimum at x = 4 and no relative extremum at x = 0.
- (C) f has a relative minimum at x = 0 and a relative maximum at x = 4.
- (D) f has a relative maximum at x = 4 and no relative extremum at x = 0.
Answer and explanation
Answer: (B). Sign chart for f′ = x²(x − 4): x² ≥ 0 always, so the sign comes from (x − 4). f′ < 0 for x < 0, f′ < 0 for 0 < x < 4, f′ > 0 for x > 4. At x = 4, f′ changes from − to +: relative minimum. At x = 0, f′ = 0 but stays negative on both sides: no extremum.
- (A) assumes every zero of f′ is an extremum and that they alternate.
- (C) also assumes alternation, and gets x = 4 the wrong way round.
- (D) reverses the sign change at x = 4.
Question 2 (multiple choice · foundation)
On the interval 1 < x < 4, f′(x) < 0 and f″(x) > 0. Which description of the graph of f on this interval is correct?
- (A) Falling, and flattening out as x increases
- (B) Falling, and getting steeper as x increases
- (C) Rising, and getting steeper as x increases
- (D) Rising, and flattening out as x increases
Answer and explanation
Answer: (A). f′ < 0 means f is decreasing (falling). f″ > 0 means f′ is increasing: the slopes are negative but moving towards 0, so the fall becomes less steep. The graph is concave up.
- (B) is the shape for f′ < 0 and f″ < 0 (concave down).
- (C) and (D) need f′ > 0. They take the sign of f″ as if it were the sign of f′.
Question 3 (multiple choice · core)
The graph of a function f has a relative maximum at (−1, 4) and a relative minimum at (3, −2). It is concave down for x < 1, concave up for x > 1 and has a point of inflection at (1, 1). Which describes the graph of f′?
- (A) It crosses the x-axis at x = −1 and x = 3 and has its lowest point at x = 1.
- (B) It crosses the x-axis at x = −1 and x = 3 and has its highest point at x = 1.
- (C) It crosses the x-axis only at x = 1.
- (D) It crosses the x-axis at x = −1, x = 1 and x = 3.
Answer and explanation
Answer: (A). The extrema of f are where f′ changes sign, so f′ crosses the axis at x = −1 and x = 3. Concave down for x < 1 means f′ is decreasing there; concave up for x > 1 means f′ is increasing. So f′ has its lowest point at x = 1.
- (B) reverses the link between concavity and the trend of f′.
- (C) puts a zero of f′ at the inflection point. At an inflection point f′ has a peak or trough, not a zero.
- (D) adds the inflection point as a zero, which is the same error as (C).
Question 4 (multiple choice · core)
The table gives values of f′ for a function f. You are also told that f″(x) > 0 for all x.
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| f′(x) | −3 | −1 | 0 | 2 | 5 |
Which statement must be true?
- (A) f has a relative minimum at x = 2, and the graph of f is concave up.
- (B) f has a relative maximum at x = 2.
- (C) The graph of f has a point of inflection at x = 2.
- (D) f is increasing on 0 < x < 4.
Answer and explanation
Answer: (A). f″ > 0 everywhere, so f′ is increasing everywhere and the graph of f is concave up. An increasing f′ with f′(2) = 0 is negative before x = 2 and positive after. So f′ changes from − to + and f has a relative minimum at x = 2.
- (B) reverses the sign change.
- (C) needs f″ to change sign, but f″ > 0 everywhere.
- (D) ignores the negative values f′(0) = −3 and f′(1) = −1: f is decreasing on 0 < x < 2.
Question 5 (constructed response · core)
A function f is continuous on −2 ≤ x ≤ 4 and twice differentiable on −2 < x < 4. Some values of f are f(−2) = 1, f(0) = 4, f(1) = 3, f(3) = −1 and f(4) = 0. The signs of f′ and f″ are:
| Interval or point | −2 < x < 0 | x = 0 | 0 < x < 1 | x = 1 | 1 < x < 3 | x = 3 | 3 < x < 4 |
|---|---|---|---|---|---|---|---|
| f′ | + | 0 | − | − | − | 0 | + |
| f″ | − | − | − | 0 | + | + | + |
(a) Find the x-coordinates of all relative extrema of f on −2 < x < 4. Classify each and justify. (b) Find the coordinates of the point of inflection. Justify. (c) On which interval is f decreasing and concave down? Describe what f′ is doing there. (d) Describe a sketch of f on −2 ≤ x ≤ 4, with labelled key points.
Worked solution and suggested Marlbridge rubric
(a) At x = 0, f′ changes from + to −: relative maximum at x = 0 (value 4). At x = 3, f′ changes from − to +: relative minimum at x = 3 (value −1). (The second derivative test also works: f′(0) = 0 with f″(0) < 0, and f′(3) = 0 with f″(3) > 0.)
(b) f″ changes from − to + at x = 1, so the graph has a point of inflection at (1, 3).
(c) On 0 < x < 1, f′ < 0 and f″ < 0. There f′ is negative and decreasing, so the slopes become more and more negative: the graph falls more and more steeply.
(d) From (−2, 1), rise as a cap to the maximum (0, 4). Fall more and more steeply to the inflection point (1, 3). Keep falling but flatten out (a cup) to the minimum (3, −1). Rise, getting steeper, to (4, 0).
| Point | What earns it |
|---|---|
| 1 | Relative maximum at x = 0 with the reason “f′ changes from positive to negative” (or f′(0) = 0 and f″(0) < 0) |
| 1 | Relative minimum at x = 3 with a correct reason |
| 1 | Inflection point (1, 3) because f″ changes sign there (not just “f″(1) = 0”) |
| 1 | Interval 0 < x < 1, with f′ negative and decreasing |
| 1 | Sketch or description passes through all five given points with the correct direction and bend in every interval |
Question 6 (graph · core)
The graph of f′, the derivative of a function f, is shown for 0 ≤ x ≤ 2π. It crosses the x-axis at x = π, has its highest point at (π/2, 2) and its lowest point at (3π/2, −2).
(a) On which interval is f increasing? Justify. (b) Find the x-coordinate of the relative extremum of f on 0 < x < 2π. Classify it and justify. (c) Find the x-coordinates of the points of inflection of the graph of f. Justify. (d) You are told f(0) = 1, f(π/2) = 3, f(π) = 5, f(3π/2) = 3 and f(2π) = 1. Describe the graph of f.
Worked solution and suggested Marlbridge rubric
(a) f′ > 0 (graph above the axis) for 0 < x < π, so f is increasing on 0 ≤ x ≤ π.
(b) At x = π, f′ changes from positive to negative, so f has a relative maximum there.
(c) f′ is increasing on 0 < x < π/2, decreasing on π/2 < x < 3π/2 and increasing on 3π/2 < x < 2π. A change from increasing to decreasing (or the reverse) in f′ is a change in the concavity of f. So the points of inflection are at x = π/2 and x = 3π/2.
(d) From (0, 1) the graph rises and steepens (concave up) to the inflection point (π/2, 3), where its slope is 2. It rises more slowly (concave down) to the maximum (π, 5), falls more and more steeply to the inflection point (3π/2, 3), with slope −2, then flattens out (concave up) as it falls to (2π, 1).
| Point | What earns it |
|---|---|
| 1 | (a): increasing on 0 ≤ x ≤ π because f′ > 0 there (open interval also acceptable) |
| 1 | (b): relative maximum at x = π because f′ changes from positive to negative |
| 1 | (c): x = π/2 and x = 3π/2 |
| 1 | (c): justification that f′ changes from increasing to decreasing (or decreasing to increasing) there, or that f′ has a relative extremum there |
| 1 | (d): correct order of shapes through all five points, with the cap around x = π and cups at both ends |
Question 7 (constructed response · stretch)
Let f(x) = 3x^(2/3) − x, defined for all real x.
(a) Find f′(x). Find all critical points, and state which one is a point where f′ does not exist. (b) Find the intervals on which f is increasing or decreasing, and classify each critical point. (c) Find f″(x). Explain why the graph of f has no point of inflection. (d) Describe the graph of f, including the points at x = −8, 0, 8 and 27 and the shape near x = 0.
Worked solution and suggested Marlbridge rubric
(a) f′(x) = 3 · (2/3)x^(−1/3) − 1 = 2x^(−1/3) − 1 = 2/x^(1/3) − 1. f′(x) = 0 when x^(1/3) = 2, so x = 8. f′(0) does not exist (division by 0), but f(0) = 0 is defined, so x = 0 is also a critical point.
(b) Test each interval:
- x < 0: x^(1/3) < 0, so 2/x^(1/3) < 0 and f′ < 0. For example f′(−1) = −3.
- 0 < x < 8: 0 < x^(1/3) < 2, so 2/x^(1/3) > 1 and f′ > 0. For example f′(1) = 1.
- x > 8: x^(1/3) > 2, so 2/x^(1/3) < 1 and f′ < 0. For example f′(27) = −1/3.
f is decreasing on x ≤ 0, increasing on 0 ≤ x ≤ 8 and decreasing on x ≥ 8. Relative minimum at x = 0 (f′ changes − to +). Relative maximum at x = 8 (f′ changes + to −).
(c) f″(x) = −(2/3)x^(−4/3) = −2/(3x^(4/3)). Since x^(4/3) = (x^(1/3))⁴ > 0 for every x ≠ 0, f″(x) < 0 for all x ≠ 0. The graph is concave down on both sides of x = 0, so the concavity never changes and there is no point of inflection.
(d) Key points: f(−8) = 12 + 8 = 20, f(0) = 0, f(8) = 12 − 8 = 4, f(27) = 27 − 27 = 0. The graph falls through (−8, 20) to a cusp at (0, 0), where f′ → −∞ from the left and +∞ from the right. It rises to the maximum (8, 4), then falls through (27, 0). It is concave down throughout.
| Point | What earns it |
|---|---|
| 1 | Correct f′, with critical points x = 8 (f′ = 0) and x = 0 (f′ undefined) |
| 1 | Correct sign of f′ on all three intervals |
| 1 | Relative minimum at x = 0 and relative maximum at x = 8, each with a sign-change reason |
| 1 | Correct f″ and the reason f″ < 0 for all x ≠ 0, so no change of concavity |
| 1 | Description with the four points, the cusp at the origin and concave down throughout |
Acceptable alternative for (b) at x = 8: the second derivative test (f′(8) = 0, f″(8) = −1/24 < 0). It cannot be used at x = 0, where f′ does not exist.
How did you do?
- Q1, Q2 or Q4 wrong: reread “What a sketch has to show”, the four basic shapes and Worked example 3 in the study guide.
- Q3 or Q6 wrong: revisit Worked example 2, Figure 2 and “Going the other way”. Remember: peaks of f′ are inflection points of f.
- Q5 or Q7 wrong: work through Worked example 1 again and build a full sign chart before you sketch.
Then tick off the topic checklist.