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, and every function named is continuous on the interval given. Notation: f′ is the first derivative of f and f″ the second. This set is for both Calculus AB and Calculus BC students.
Questions 3 and 6 use Figure 1.
Question 1 (multiple choice · foundation)
On an open interval, the graph of f′ lies above the x-axis and is falling. Which statement describes f on that interval?
- (A) f is increasing and concave down.
- (B) f is increasing and concave up.
- (C) f is decreasing and concave down.
- (D) f is decreasing and concave up.
Answer and explanation
Answer: (A). f′ above the axis means f′ > 0, so f is increasing. f′ falling means f′ is decreasing, so f is concave down. Example: a hill on the way up to its top.
- (B) gets the direction right but reads “f′ falling” as concave up. Falling f′ means the slopes of f are getting smaller.
- (C) uses “f′ is falling” to decide direction. Direction comes from the sign of f′, not from whether f′ rises or falls.
- (D) swaps both facts.
Question 2 (multiple choice · core)
A function f is twice differentiable for all x. f″(x) < 0 for x < 3 and f″(x) > 0 for x > 3. No other information is given. Which statement must be true?
- (A) f has a relative minimum at x = 3.
- (B) f′ has a relative minimum at x = 3.
- (C) f is increasing for x > 3.
- (D) f(3) = 0.
Answer and explanation
Answer: (B). f″ is the derivative of f′. It changes from negative to positive at x = 3, so f′ is decreasing and then increasing: f′ has a relative minimum at x = 3. (f also has an inflection point there.)
- (A) applies the first derivative test to the wrong function. A sign change in f″ is an extremum of f′, not of f. To find extrema of f you would need the sign of f′.
- (C) confuses “f″ > 0” with “f′ > 0”. f″ > 0 only says f is concave up; f could still be falling.
- (D) has no basis. Nothing about f″ gives a value of f.
Question 3 (multiple choice · core)
Use Figure 1. On which interval is f both decreasing and concave up?
- (A) (0, 2)
- (B) (2, 4)
- (C) (4, 6)
- (D) (6, 7.5)
Answer and explanation
Answer: (B). f is decreasing where f′ < 0, which is (0, 4) and (7.5, 8). f is concave up where f′ is increasing, which is (2, 6). The overlap is (2, 4).
- (A) f′ is negative there, so f is decreasing, but f′ is falling, so f is concave down.
- (C) f′ is rising, so f is concave up, but f′ is positive, so f is increasing.
- (D) f′ is positive and falling, so f is increasing and concave down.
Question 4 (multiple choice · core)
The second derivative of a function g is g″(x) = x(x − 5)². At which x-values does the graph of g have an inflection point?
- (A) x = 0 only
- (B) x = 5 only
- (C) x = 0 and x = 5
- (D) x = 5/3 only
Answer and explanation
Answer: (A). Check the sign of g″ around each zero. g″(−1) = −36 < 0, g″(1) = 16 > 0 and g″(6) = 6 > 0. g″ changes sign at x = 0 only. At x = 5 the squared factor keeps g″ positive on both sides, so concavity does not change.
- (B) picks the zero where g″ does not change sign.
- (C) treats every zero of g″ as an inflection point. A zero is only a candidate; g″ must change sign.
- (D) is where g″ has a turning point (its own derivative is (x − 5)(3x − 5)). Turning points of g″ are not inflection points of g; that rule belongs one level up, to the turning points of g′.
Question 5 (constructed response · core)
The graphs of a function f, its derivative f′ and its second derivative f″ are called K, L and M, in some order. You are told:
- Curve K has horizontal tangents at x = −1 and x = 3, and nowhere else.
- Curve L is a rising straight line that crosses the x-axis at x = 1.
- Curve M is U-shaped. It crosses the x-axis at x = −1 and x = 3, and its lowest point is at x = 1.
(a) Decide which curve is f, which is f′ and which is f″. Explain how you decided. (b) Classify each critical point of f, with a reason. (c) Give the x-coordinate of the inflection point of f, with a reason.
Worked solution and suggested Marlbridge rubric
(a) M crosses the axis at x = −1 and x = 3, exactly where K has horizontal tangents, so M is K′. L crosses the axis at x = 1, exactly where M has its lowest point, so L is M′. The signs agree: M falls then rises around x = 1, and L goes from negative to positive there. The chain is K → M → L, so K is f, M is f′ and L is f″.
(b) M (that is, f′) is U-shaped and crosses at −1 and 3, so f′ > 0 for x < −1, f′ < 0 for −1 < x < 3 and f′ > 0 for x > 3.
- x = −1: f′ changes from positive to negative, so relative maximum.
- x = 3: f′ changes from negative to positive, so relative minimum.
(c) x = 1, because f″ (curve L) changes sign there, from negative to positive. Equivalently, f′ (curve M) has its minimum at x = 1.
| Point | What earns it |
|---|---|
| 1 | Matches zeros of M with horizontal tangents of K (M = f′ of K) |
| 1 | Matches the zero of L with the turning point of M, giving K = f, M = f′, L = f″ |
| 1 | Relative max at x = −1 and relative min at x = 3, each justified by the sign change of f′ |
| 1 | Inflection point at x = 1, justified by f″ changing sign or f′ having a relative extremum |
A consistent example (not needed for the marks): f(x) = x³/3 − x² − 3x, with f′(x) = (x + 1)(x − 3) and f″(x) = 2x − 2.
Question 6 (constructed response · core)
Use Figure 1. The function f is continuous on [0, 8].
(a) Find the x-values in the open interval (0, 8) at which f has a relative minimum or a relative maximum. Justify each. (b) Find the open intervals on which f is concave down. Justify. (c) Find the x-coordinates of the inflection points of f. Justify. (d) Find f″(5) and f″(7). (e) f″(4) does not exist. A student says this means f has an inflection point at x = 4. Is the student right? Explain.
Worked solution and suggested Marlbridge rubric
(a) f′ changes from negative to positive at x = 4, so f has a relative minimum at x = 4. f′ changes from positive to negative at x = 7.5, so f has a relative maximum at x = 7.5. (At x = 2, f′ = −2 is not zero; that is where f is decreasing most steeply, not an extremum.)
(b) f′ is decreasing on (0, 2) and on (6, 8). So f is concave down on (0, 2) and (6, 8), because f′ is decreasing there.
(c) f′ changes from decreasing to increasing at x = 2 and from increasing to decreasing at x = 6. So f has inflection points at x = 2 and x = 6.
(d) f″ is the slope of f′. On (4, 6), f′ is the segment from (4, 0) to (6, 3), with slope 3/2. So f″(5) = 3/2. On (6, 8), the segment from (6, 3) to (8, −1) has slope −4/2. So f″(7) = −2.
(e) No. An inflection point needs a change in concavity. f′ is increasing on (2, 4) and also on (4, 6), so f is concave up on both sides of x = 4. f″ fails to exist at x = 4 because the semicircle meets the segment at a sharp point, but concavity does not change.
| Point | What earns it |
|---|---|
| 1 | Relative minimum at x = 4 and relative maximum at x = 7.5 |
| 1 | Justifies both using the sign change of f′ |
| 1 | Concave down on (0, 2) and (6, 8), because f′ is decreasing |
| 1 | Inflection points at x = 2 and x = 6, because f′ changes between increasing and decreasing |
| 1 | f″(5) = 3/2 and f″(7) = −2 |
| 1 | Rejects x = 4: f′ is increasing on both sides, so concavity does not change |
Using f″ language is also acceptable, for example “f″ < 0 on (0, 2) because the graph of f′ has negative slope there”.
Question 7 (constructed response · stretch)
A function f is twice differentiable on the open interval (0, 6). You know only these facts:
- f′(1) = 0 and f′(5) = 0
- f″(x) < 0 for 0 < x < 3, and f″(x) > 0 for 3 < x < 6
(a) Show that f is decreasing on the interval (1, 5). (b) Classify x = 1 and x = 5 as relative maxima or minima of f. Justify. (c) At which x-value in (0, 6) does f′ take its least value? Justify. (d) Does the graph of f have an inflection point at x = 3? Justify.
Worked solution and suggested Marlbridge rubric
(a) f″ < 0 on (0, 3), so f′ is decreasing there. Since f′(1) = 0, f′(x) < 0 for 1 < x ≤ 3. f″ > 0 on (3, 6), so f′ is increasing there. Since f′(5) = 0, f′(x) < 0 for 3 ≤ x < 5. So f′ < 0 on the whole of (1, 5), and f is decreasing on (1, 5).
(b) On (0, 1), f′ is decreasing towards f′(1) = 0, so f′ > 0 there. On (5, 6), f′ is increasing from f′(5) = 0, so f′ > 0 there.
- x = 1: f′ changes from positive to negative, so relative maximum.
- x = 5: f′ changes from negative to positive, so relative minimum.
(c) f′ decreases on (0, 3) and increases on (3, 6), so f′ takes its least value at x = 3. (This is the point where f is falling most steeply.)
(d) Yes. f is twice differentiable, so f is continuous at x = 3, and f″ changes from negative to positive there. The concavity of f changes from down to up.
| Point | What earns it |
|---|---|
| 1 | Uses f″ < 0 on (0, 3) and f′(1) = 0 to get f′ < 0 on (1, 3] |
| 1 | Uses f″ > 0 on (3, 6) and f′(5) = 0 to get f′ < 0 on [3, 5), concluding f is decreasing |
| 1 | Relative max at x = 1 and relative min at x = 5, with sign changes of f′ |
| 1 | f′ is least at x = 3, because f′ changes from decreasing to increasing there |
| 1 | Inflection point at x = 3, because f″ changes sign (with f continuous) |
Acceptable alternative for (b): the second derivative test. f′(1) = 0 with f″(1) < 0 gives a relative maximum; f′(5) = 0 with f″(5) > 0 gives a relative minimum. A consistent example is f′(x) = (x − 1)(x − 5), with f″(x) = 2x − 6.
How did you do?
- Q1 or Q3 wrong: revisit “Reading a graph of f′” and the table in “One story told by three graphs” in the study guide.
- Q2 or Q4 wrong: reread “Reading a graph of f″” and Worked example 3. A sign change of f″ is an inflection point of f and an extremum of f′.
- Q5 wrong: redo Worked example 2 (matching graphs).
- Q6 or Q7 wrong: redo Worked example 1, and check each reason names f′ or f″ and says how it changes.
Then tick off the topic checklist.