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 unless stated, and all contexts and data are fictional. MVT means the Mean Value Theorem; in your own written answers, name it in full. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
Let f(x) = √(2x + 1). What value of c satisfies the conclusion of the Mean Value Theorem for f on [0, 4]?
- (A) 0
- (B) 1/2
- (C) 3/2
- (D) 2
Answer and explanation
Answer: (C). f is continuous on [0, 4] and differentiable on (0, 4) (2x + 1 > 0 there). f(0) = 1 and f(4) = √9 = 3, so the average rate is (3 − 1)/4 = 1/2. By the chain rule, f′(x) = 1/√(2x + 1). Solve 1/√(2x + 1) = 1/2: √(2x + 1) = 2, so 2x + 1 = 4 and x = 3/2, which lies in (0, 4).
- (A) comes from forgetting the chain rule: 1/(2√(2x + 1)) = 1/2 gives x = 0, which is an endpoint anyway, so it could never be the answer.
- (B) is the average rate itself, not the value of x where the derivative equals it.
- (D) is the midpoint of [0, 4]. f′(2) = 1/√5, which is not 1/2.
Question 2 (multiple choice · core)
For which function are both conditions of the Mean Value Theorem satisfied on [−1, 2]?
- (A) f(x) = |x|
- (B) f(x) = x^(1/3), the cube root of x
- (C) f(x) = (x + 2)/(x − 1)
- (D) f(x) = e^(−x) + x²
Answer and explanation
Answer: (D). e^(−x) and x² are continuous and differentiable for every real x, so their sum is continuous on [−1, 2] and differentiable on (−1, 2), with f′(x) = −e^(−x) + 2x.
- (A) is continuous, but has a corner at x = 0, which is inside (−1, 2). It is not differentiable there.
- (B) is continuous, but has a vertical tangent at x = 0: the derivative (1/3)x^(−2/3) is undefined there.
- (C) is undefined at x = 1, which is inside the interval, so it is not even continuous on [−1, 2].
Question 3 (multiple choice · core)
A function f is differentiable for all real x. Some values are shown.
| x | 1 | 3 | 6 | 8 |
|---|---|---|---|---|
| f(x) | 5 | 9 | 3 | 7 |
Which statement must be true?
- (A) f′(c) = 2 for some c in (3, 6)
- (B) f′(c) = −2 for some c in (3, 6)
- (C) f(c) = 0 for some c in (1, 8)
- (D) f′(c) = 2/7 for some c in (6, 8)
Answer and explanation
Answer: (B). f is differentiable, so it is continuous on [3, 6] and differentiable on (3, 6). The average rate is (3 − 9)/(6 − 3) = −6/3 = −2. By the Mean Value Theorem, f′(c) = −2 for some c in (3, 6).
- (A) uses the wrong sub-interval. The average rate is 2 on [1, 3] and on [6, 8], so f′ = 2 is guaranteed in (1, 3) and in (6, 8), not in (3, 6).
- (C) needs a sign change for the Intermediate Value Theorem. Every table value is positive, so nothing guarantees a zero.
- (D) 2/7 is the average rate on [1, 8], which guarantees a c somewhere in (1, 8), not in a particular part of it. On [6, 8] the average rate is (7 − 3)/2 = 2. f could have gradient exactly 2 all the way from x = 6 to x = 8, so nothing guarantees 2/7 in (6, 8).
Question 4 (multiple choice · core)
Let f(x) = |x − 2| on [0, 5]. The average rate of change of f on [0, 5] is 1/5. Which statement is true?
- (A) By the Mean Value Theorem, f′(c) = 1/5 for some c in (0, 5).
- (B) No c in (0, 5) has f′(c) = 1/5. This does not contradict the Mean Value Theorem, because f is not differentiable at x = 2.
- (C) c = 5/2 works, because it is the midpoint of the interval.
- (D) The Mean Value Theorem does not apply because f is not continuous on [0, 5].
Answer and explanation
Answer: (B). f(0) = 2 and f(5) = 3, so the average rate is 1/5. But f′(x) = −1 for x < 2 and f′(x) = 1 for x > 2, and f′(2) does not exist (corner). So f′ never equals 1/5. The differentiability condition fails at x = 2, inside (0, 5), so the theorem made no promise.
- (A) applies the theorem without checking the conditions.
- (C) f′(5/2) = 1, not 1/5. The midpoint is not special.
- (D) f is continuous everywhere. The failing condition is differentiability, not continuity.
Question 5 (constructed response · core)
Let g(x) = x + 4/x.
(a) Explain why the Mean Value Theorem applies to g on [1, 4], and find every value of c it guarantees. (b) The average rate of change of g on [−1, 4] is 2. Show that no c in (−1, 4) has g′(c) = 2, and explain why this does not contradict the Mean Value Theorem.
Worked solution and suggested Marlbridge rubric
(a) g is a sum of a polynomial and a rational function whose only problem point is x = 0, which is not in [1, 4]. So g is continuous on [1, 4] and differentiable on (1, 4).
g(1) = 1 + 4 = 5 and g(4) = 4 + 1 = 5. Average rate: (5 − 5)/3 = 0.
g′(x) = 1 − 4/x². Solve 1 − 4/x² = 0: x² = 4, x = ±2. Only c = 2 lies in (1, 4).
(b) g(−1) = −1 − 4 = −5 and g(4) = 5, so the average rate is (5 − (−5))/5 = 2. Solve 1 − 4/x² = 2: −4/x² = 1, so x² = −4. There is no real solution, so no c in (−1, 4) works.
There is no contradiction: g is undefined at x = 0, which lies in [−1, 4]. So g is not continuous on [−1, 4], and the theorem does not apply.
| Point | What earns it |
|---|---|
| 1 | States that g is continuous on [1, 4] and differentiable on (1, 4), with a reason (only problem point is x = 0, outside the interval) |
| 1 | Average rate 0 and g′(x) = 1 − 4/x², solved to give c = 2 with −2 rejected |
| 1 | Shows g′(x) = 2 has no real solution |
| 1 | Explains that the continuity condition fails at x = 0 inside [−1, 4], so there is no contradiction |
Acceptable alternative for (b): note that g′(x) = 1 − 4/x² < 1 for every x ≠ 0, so g′ can never equal 2.
Question 6 (constructed response · core)
Let f(x) = ax² + 1 for x ≤ 2, and f(x) = 4x + b for x > 2, where a and b are constants.
(a) Find a and b so that f is differentiable for all x. (b) With these values, explain why the Mean Value Theorem applies to f on [0, 4]. (c) Find every value of c in (0, 4) that satisfies the conclusion.
Worked solution and suggested Marlbridge rubric
(a) Differentiable at 2 needs matching derivatives: 2a(2) = 4, so a = 1. Differentiable also needs continuity at 2: a(2²) + 1 = 4(2) + b, so 5 = 8 + b and b = −3.
(b) Each piece is a polynomial, so f is differentiable on each side of 2. With a = 1 and b = −3, both pieces give f(2) = 5 and both give derivative 4 at x = 2, so f is differentiable at 2 as well. Hence f is differentiable on [0, 4], and therefore continuous there too.
(c) f(0) = 0 + 1 = 1. f(4) = 16 − 3 = 13. Average rate: (13 − 1)/4 = 3.
On (0, 2), f′(x) = 2x. Solve 2x = 3: x = 3/2, which lies in (0, 2). ✓ At x = 2, f′(2) = 4 ≠ 3. On (2, 4), f′(x) = 4 ≠ 3.
So the only value is c = 3/2.
| Point | What earns it |
|---|---|
| 1 | a = 1 from matching derivatives at x = 2 |
| 1 | b = −3 from continuity at x = 2 |
| 1 | Explains that f is differentiable (hence continuous) on [0, 4], referring to both pieces and the join |
| 1 | Average rate 3 and c = 3/2, having checked the other piece and the join |
Question 7 (constructed response · stretch)
A fictional cyclist rides along a straight road. Her distance from the start is d(t) km, t minutes after starting. d is twice differentiable. Some values are shown.
| t (minutes) | 0 | 10 | 25 | 30 | 45 |
|---|---|---|---|---|---|
| d(t) (km) | 0 | 3.5 | 9.5 | 11.0 | 18.5 |
(a) Explain why there must be a time t in (10, 25) at which the cyclist’s velocity is 0.4 km per minute. (b) Explain why there must be a time t in (10, 45) at which her velocity is 0.45 km per minute. (c) A classmate says: “Her average velocity over [0, 45] is about 0.41 km per minute, so her velocity was never more than 0.5 km per minute.” Is this correct? Explain.
Worked solution and suggested Marlbridge rubric
(a) d is differentiable, so it is continuous on [10, 25] and differentiable on (10, 25). Average rate: (9.5 − 3.5)/(25 − 10) = 6/15 = 0.4 km per minute. By the Mean Value Theorem, d′(t₁) = 0.4 for some t₁ in (10, 25). Since d′ is velocity, her velocity was 0.4 km per minute (24 km/h) at that time.
(b) On [30, 45]: (18.5 − 11.0)/15 = 7.5/15 = 0.5. By the Mean Value Theorem, v(t₂) = d′(t₂) = 0.5 for some t₂ in (30, 45).
d is twice differentiable, so v = d′ is differentiable and therefore continuous. On [t₁, t₂], v is continuous, v(t₁) = 0.4 and v(t₂) = 0.5, and 0.4 < 0.45 < 0.5. By the Intermediate Value Theorem, v(t) = 0.45 for some t in (t₁, t₂), which lies inside (10, 45).
(c) Not correct: the conclusion does not follow. The Mean Value Theorem gives values the velocity must take; an average rate is never an upper bound. Part (b) already shows the velocity reaches 0.5, well above the 0.41 average, and nothing in the table stops it from being higher than 0.5 at other times.
| Point | What earns it |
|---|---|
| 1 | (a) Conditions stated (differentiable, so continuous), average rate 0.4 shown, Mean Value Theorem named |
| 1 | (b) Average rate 0.5 on [30, 45] and the Mean Value Theorem gives v = 0.5 at some t₂ in (30, 45) |
| 1 | (b) States that v is continuous because d is twice differentiable, and applies the Intermediate Value Theorem between 0.4 and 0.5 |
| 1 | (c) Explains that average rates give values the velocity reaches, not a maximum |
Note for (b): you need one guaranteed velocity below 0.45 and one above it. The values 0.4 (from [10, 25]) and 0.5 (from [30, 45]) do this and keep the time inside (10, 45). The value 0.35 from [0, 10] would also sit below 0.45, but it could place the time before t = 10.
How did you do?
- Q1, Q5(a) or Q6(c) wrong: redo Worked example 1 in the study guide and check every solution against the open interval.
- Q2, Q4 or Q5(b) wrong: reread “The theorem, part by part” and Worked example 3 (when a condition fails).
- Q3 or Q7 wrong: redo Worked example 2 (tables) and remember that any sub-interval may be used.
- Q6(a) wrong: review differentiability of piecewise functions (Topic 2.4).
Then tick off the topic checklist.