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 f⁻¹ means the inverse function (not 1/f). The heater in Question 5 is fictional. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
Let f(x) = x⁵ + 4x − 1, and let g be the inverse of f. What is g′(4)?
- (A) 1/1284
- (B) 1/81
- (C) 1/9
- (D) 9
Answer and explanation
Answer: (C). First find the matching input: f(1) = 1 + 4 − 1 = 4, so g(4) = 1. Then f′(x) = 5x⁴ + 4, so f′(1) = 9. The rule gives g′(4) = 1/f′(1) = 1/9.
- (A) is 1/f′(4) = 1/(5 · 256 + 4). It evaluates f′ at the output 4 instead of the matching input 1.
- (B) is 1/[f′(1)]² = 1/81. It squares the slope; the inverse rule needs only the plain reciprocal 1/f′(1).
- (D) is f′(1) itself. It forgets to take the reciprocal.
Question 2 (multiple choice · core)
The function p is differentiable and increasing, and q is the inverse of p. Selected values are shown.
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| p(x) | 2 | 4 | 7 | 11 |
| p′(x) | 3/2 | 5/2 | 7/2 | 5 |
What is q′(4)?
- (A) 1/11
- (B) 1/5
- (C) 2/5
- (D) 5/2
Answer and explanation
Answer: (C). Find 4 in the p(x) row: p(2) = 4, so q(4) = 2. Then q′(4) = 1/p′(2) = 1/(5/2) = 2/5.
- (A) is 1/p(4). It confuses the inverse function with the reciprocal of p.
- (B) is 1/p′(4). It reads the column x = 4 instead of the column where p(x) = 4.
- (D) is p′(2). It finds the right column but forgets to take the reciprocal.
Question 3 (multiple choice · core)
A function f is differentiable and decreasing. Its graph passes through (2, 5), and the tangent line to f at that point has slope −4. Which statement about the graph of f⁻¹ is true?
- (A) It passes through (5, 2) with slope −1/4 there.
- (B) It passes through (5, 2) with slope 1/4 there.
- (C) It passes through (5, 2) with slope −4 there.
- (D) It passes through (2, 5) with slope −1/4 there.
Answer and explanation
Answer: (A). Since f(2) = 5, the inverse has f⁻¹(5) = 2, so (5, 2) is on its graph. The slope there is 1/f′(2) = 1/(−4) = −1/4.
- (B) changes the sign. Reflection in y = x keeps the sign of a slope; a decreasing function has a decreasing inverse.
- (C) keeps the slope of f instead of taking the reciprocal.
- (D) has the right slope at the wrong point. The coordinates swap when you reflect.
Question 4 (multiple choice · core)
Let f(x) = x³ − 3x² + 3x + 2. Then f is increasing for all x, and it has an inverse g. Which statement about g at x = 3 is true?
- (A) g′(3) = 0
- (B) g′(3) = 1/12
- (C) g is not differentiable at x = 3; its graph has a vertical tangent there.
- (D) g is not defined at x = 3, because f is not one-to-one.
Answer and explanation
Answer: (C). Note that f(x) = (x − 1)³ + 3, so f(1) = 3 and g(3) = 1. But f′(x) = 3(x − 1)², so f′(1) = 0. The tangent to f at (1, 3) is horizontal, so the tangent to g at (3, 1) is vertical. The rule 1/f′(1) would need division by 0, so g′(3) does not exist.
- (A) confuses a horizontal tangent on f with a horizontal tangent on g. The reflection of a horizontal line is vertical.
- (B) is 1/f′(3) = 1/12. It evaluates f′ at 3, which is an output of f. The matching input is 1, where f′ is 0.
- (D) is false. A function can have f′ = 0 at a single point and still be increasing and one-to-one, as (x − 1)³ + 3 is.
Question 5 (constructed response · core)
A fictional laboratory heater warms a tank of water. The temperature, in degrees Celsius, t minutes after it is switched on is
T = f(t) = 15 + 4t + 0.1t², for 0 ≤ t ≤ 20.
Let g be the inverse of f, so g(T) is the time, in minutes, at which the temperature is T.
(a) Explain why g exists on the interval of temperatures 15 ≤ T ≤ 135. (b) Find g(65). (c) Find g′(65), and give its units. (d) Interpret g′(65) in the context of the heater. (e) Find g′(97.5). Explain why it is smaller than g′(65).
Worked solution and suggested Marlbridge rubric
(a) f′(t) = 4 + 0.2t, which is at least 4 for 0 ≤ t ≤ 20. So f is increasing, hence one-to-one, on this interval. Its outputs run from f(0) = 15 to f(20) = 135, so g exists for 15 ≤ T ≤ 135.
(b) Solve f(t) = 65: 0.1t² + 4t − 50 = 0, or t² + 40t − 500 = 0, so (t − 10)(t + 50) = 0. Only t = 10 is in the domain. So g(65) = 10.
(c) f′(10) = 4 + 2 = 6 °C per minute. So g′(65) = 1/f′(10) = 1/6 minute per °C.
(d) When the water is at 65 °C, the time needed for the temperature to rise by one more degree is about 1/6 minute (10 seconds).
(e) f(15) = 15 + 60 + 22.5 = 97.5, so g(97.5) = 15. f′(15) = 4 + 3 = 7, so g′(97.5) = 1/7 minute per °C (about 8.6 seconds). The water is warming faster at t = 15 than at t = 10 (7 °C per minute compared with 6), so each degree takes less time.
| Point | What earns it |
|---|---|
| 1 | Shows f′(t) > 0 on the interval (or another valid one-to-one argument) |
| 1 | g(65) = 10, rejecting t = −50 because it is outside the domain |
| 1 | g′(65) = 1/f′(10) = 1/6 |
| 1 | Correct units, minutes per °C, and an interpretation that names the temperature 65 °C |
| 1 | g′(97.5) = 1/7 with a reason that links the smaller value to the faster heating rate |
Acceptable check for (c): solving for t gives g(T) = −20 + 5√(10 + 0.4T). Differentiating gives g′(T) = 1/√(10 + 0.4T), and g′(65) = 1/√36 = 1/6.
Question 6 (constructed response · core)
Let h(x) = 2x + e^(x − 1), and let k be the inverse of h.
(a) Explain why k exists. (b) Find k(3) and k′(3). (c) Write an equation for the line tangent to the graph of k at x = 3. (d) Let P(x) = x · k(x). Find P′(3).
Worked solution and suggested Marlbridge rubric
(a) h′(x) = 2 + e^(x − 1). Since e^(x − 1) > 0, h′(x) > 2 for all x. So h is always increasing, which makes it one-to-one, and k exists.
(b) h(1) = 2 + e⁰ = 3, so k(3) = 1. Then h′(1) = 2 + 1 = 3, so k′(3) = 1/h′(1) = 1/3.
(c) The point is (3, 1) and the slope is 1/3: y = 1 + (1/3)(x − 3).
(d) By the product rule, P′(x) = k(x) + x · k′(x). So P′(3) = k(3) + 3k′(3) = 1 + 3(1/3) = 2.
| Point | What earns it |
|---|---|
| 1 | h′(x) = 2 + e^(x − 1) and the conclusion that h′ > 0, so h is one-to-one |
| 1 | k(3) = 1, found from h(1) = 3 |
| 1 | k′(3) = 1/h′(1) = 1/3, with h′ evaluated at 1, not 3 |
| 1 | Tangent line through (3, 1) with slope 1/3 |
| 1 | Product rule applied correctly, P′(3) = 2 |
A common error in (b) is 1/h′(3) = 1/(2 + e²). That uses the wrong input.
Question 7 (constructed response · stretch)
Let f(x) = x² + 2x for x ≥ −1, and let g be the inverse of f.
A student writes: “g′(8) = 1/f′(8) = 1/(2 · 8 + 2) = 1/18.”
(a) Explain the student’s error. (b) Find the correct value of g′(8). (c) Confirm your answer by finding a formula for g(x) and differentiating it. (d) Explain why the domain is restricted to x ≥ −1, and describe the graph of g at the point (−1, −1).
Worked solution and suggested Marlbridge rubric
(a) The student evaluated f′ at 8, which is an output of f (an input of g). The rule needs f′ at the matching input a, where f(a) = 8.
(b) Solve a² + 2a = 8: a² + 2a − 8 = 0, so (a + 4)(a − 2) = 0. Since a ≥ −1, a = 2, and g(8) = 2. f′(x) = 2x + 2, so f′(2) = 6. Therefore g′(8) = 1/6.
(c) Write y = x² + 2x = (x + 1)² − 1. Then (x + 1)² = y + 1, and x + 1 ≥ 0 on the domain, so x = −1 + √(y + 1). Swapping letters, g(x) = −1 + √(x + 1). Then g′(x) = 1/(2√(x + 1)), and g′(8) = 1/(2 · 3) = 1/6. The two methods agree.
(d) Without a restriction, f would not be one-to-one: for example f(0) = f(−2) = 0. The vertex is at x = −1, so x ≥ −1 keeps the right-hand half, where f is increasing. At x = −1, f(−1) = −1 and f′(−1) = 0, so the graph of f has a horizontal tangent at (−1, −1). The graph of g therefore has a vertical tangent at (−1, −1), and g is not differentiable there. This agrees with g′(x) = 1/(2√(x + 1)), which grows without bound as x → −1 from the right.
| Point | What earns it |
|---|---|
| 1 | Identifies that f′ was evaluated at the output 8 instead of the matching input |
| 1 | Finds a = 2, rejecting a = −4 because of the domain |
| 1 | g′(8) = 1/f′(2) = 1/6 |
| 1 | Correct inverse formula and derivative giving 1/6 |
| 1 | Explains the one-to-one reason for x ≥ −1 and states the vertical tangent at (−1, −1) because f′(−1) = 0 |
Acceptable alternative for (c): implicit differentiation of x = y² + 2y gives 1 = (2y + 2) dy/dx, and at y = 2 this gives dy/dx = 1/6.
How did you do?
- Q1, Q2 or Q7(a) wrong: revisit the three-step method in “Deriving the rule with the chain rule” and Worked example 2 in the study guide. Always find the matching input first.
- Q3 wrong: reread “The idea” and Figure 1. Coordinates swap; the sign of the slope stays.
- Q4 or Q7(d) wrong: see “When the inverse is not differentiable”.
- Q5 or Q6 wrong: redo Worked example 1, including the tangent line and the checks.
Then tick off the topic checklist.