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 all contexts and data are invented. Notation: lim (h → 0) g(h) means “the limit as h approaches 0 of g(h)”. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
What is the average rate of change of f(x) = x³ − 2x over the interval [1, 3]?
- (A) 7
- (B) 10
- (C) 11
- (D) 22
Answer and explanation
Answer: (C). f(3) = 27 − 6 = 21 and f(1) = 1 − 2 = −1. Average rate = (f(3) − f(1))/(3 − 1) = (21 − (−1))/2 = 22/2 = 11.
- (A) divides one output by its input: f(3)/3 = 7. That is not a change in output over a change in input.
- (B) makes a sign slip with f(1), using +1 instead of −1: (21 − 1)/2 = 10.
- (D) finds the change in output, 22, but forgets to divide by the change in input, 2.
Question 2 (multiple choice · foundation)
A student wants the rate of change of a function f at x = 4. She writes the average rate over [4, 4 + h] as (f(4 + h) − f(4))/h and then puts h = 0. Why does this fail?
- (A) Putting h = 0 gives 0/0, which is undefined, because an average rate needs a nonzero change in x.
- (B) Putting h = 0 always gives 0, so the rate would be zero for every function.
- (C) The method only works when h is negative.
- (D) It fails only if f is not a polynomial.
Answer and explanation
Answer: (A). With h = 0 the top is f(4) − f(4) = 0 and the bottom is 0. Division by zero is undefined for every function f, because the average rate divides by the change in the input. The rate at x = 4 must come from what the quotient approaches as h gets close to 0, not from h = 0 itself.
- (B) treats 0/0 as if it were 0. It is not a number at all.
- (C) is wrong: intervals on either side of 4 (h positive or negative) are both useful, and neither allows h = 0.
- (D) is wrong: the problem is the zero denominator, which happens for every function, polynomials included.
Question 3 (multiple choice · core)
The table shows average rates of change of a function g over intervals that contain x = 2.
| Interval | [1.9, 2] | [1.99, 2] | [1.999, 2] | [2, 2.001] | [2, 2.01] | [2, 2.1] |
|---|---|---|---|---|---|---|
| Average rate | 10.41 | 10.9401 | 10.994001 | 11.006001 | 11.0601 | 11.61 |
Which value is the best estimate of the rate of change of g at x = 2?
- (A) 6
- (B) 10.41
- (C) 11
- (D) 11.61
Answer and explanation
Answer: (C). As the intervals shrink, the averages from the left rise towards 11 (10.41, 10.9401, 10.994001) and the averages from the right fall towards 11 (11.61, 11.0601, 11.006001). Both sides approach 11.
- (A) confuses the value of the function with its rate of change. (For the invented function behind this table, g(2) = 6.) Nothing in the table approaches 6.
- (B) and (D) are averages over the longest intervals in the table, so they are the least accurate estimates. Each also uses one side only.
Question 4 (multiple choice · core)
A weather balloon’s height is A(t) metres, t minutes after launch. Which expression gives the balloon’s vertical velocity at t = 2, in metres per minute?
- (A) A(2)/2
- (B) (A(4) − A(2))/(4 − 2)
- (C) lim (h → 0) (A(2 + h) − A(2))/h
- (D) (A(2 + h) − A(2))/h evaluated at h = 0
Answer and explanation
Answer: (C). The velocity at an instant is the value that average velocities over intervals [2, 2 + h] approach as h gets close to 0. That is exactly this limit.
- (A) divides the height by the time. At best (when A(0) = 0) this is the average velocity over [0, 2], not the velocity at t = 2.
- (B) is the average velocity over [2, 4]. It contains t = 2 and is an estimate, but not the exact rate at the instant.
- (D) gives (A(2) − A(2))/0 = 0/0, which is undefined.
Question 5 (table · core)
Rain fills a water tank. The volume W(t), in litres, is recorded t hours after the rain starts.
| t (hours) | 0 | 3 | 5 | 6 | 7 | 10 |
|---|---|---|---|---|---|---|
| W(t) (litres) | 120 | 165 | 200 | 222 | 246 | 330 |
(a) Find the average rate of change of W over [0, 10]. Include units. (b) Use the table to estimate the rate at which water is entering the tank at t = 6. Show the interval you use and say why you chose it. (c) Interpret your answer to (b) in context. (d) Explain why the table cannot give the rate at t = 6 using the value W(6) alone.
Worked solution and suggested Marlbridge rubric
(a) (W(10) − W(0))/(10 − 0) = (330 − 120)/10 = 210/10 = 21 litres per hour.
(b) Use the shortest interval in the table with t = 6 inside it: [5, 7]. (W(7) − W(5))/(7 − 5) = (246 − 200)/2 = 46/2 = 23 litres per hour. It is short, and it uses one data point on each side of t = 6. (One-sided estimates: [5, 6] gives 22 and [6, 7] gives 24 litres per hour; 23 lies between them.)
(c) At t = 6 hours, the volume of water in the tank is increasing at about 23 litres per hour.
(d) A rate needs a change in volume divided by a change in time. With only t = 6, the change in time is 0, so the quotient would be (222 − 222)/0 = 0/0, which is undefined. You need intervals that contain t = 6.
| Point | What earns it |
|---|---|
| 1 | 21, with units of litres per hour |
| 1 | An estimate from an interval containing t = 6, preferably [5, 7] (23), with the quotient shown; [5, 6] (22) or [6, 7] (24) also earn the point |
| 1 | Interpretation that names the time (t = 6 hours), the quantity (volume of water), “increasing” and the units |
| 1 | Explains that a single time gives a zero change in time, so the quotient is 0/0 (undefined) |
No point for (b) is earned by using [0, 10] or any interval that does not contain t = 6.
Question 6 (constructed response · core)
A particle moves along a straight line. Its position is s(t) = 2t² − t + 1 metres at time t seconds.
(a) Find the average velocity of the particle over [1, 3]. (b) Show that, for h ≠ 0, the average velocity over the interval from t = 1 to t = 1 + h is 3 + 2h. (c) Use (b) to find the velocity of the particle at t = 1. Explain your reasoning. (d) Explain why you may divide by h in (b), but may not set h = 0 in the original quotient.
Worked solution and suggested Marlbridge rubric
(a) s(3) = 18 − 3 + 1 = 16 and s(1) = 2 − 1 + 1 = 2. Average velocity = (16 − 2)/(3 − 1) = 14/2 = 7 m/s.
(b) s(1 + h) = 2(1 + 2h + h²) − (1 + h) + 1 = 2 + 3h + 2h². So s(1 + h) − s(1) = 3h + 2h², and (3h + 2h²)/h = h(3 + 2h)/h = 3 + 2h, for h ≠ 0.
(c) As the interval shrinks, h gets close to 0 (from either side), and 3 + 2h gets close to 3. For example, h = 0.1 gives 3.2 and h = −0.1 gives 2.8. So the velocity at t = 1 is lim (h → 0) (3 + 2h) = 3 m/s.
(d) Every interval used has h ≠ 0, so dividing by h is allowed for every quotient you actually calculate. Setting h = 0 in the original quotient (s(1 + h) − s(1))/h gives 0/0, because the interval would have length zero. The velocity at t = 1 comes from what the quotients approach, not from h = 0.
| Point | What earns it |
|---|---|
| 1 | Average velocity 7 m/s, with s(3) and s(1) correct |
| 1 | Correct expansion of s(1 + h) and simplification to 3 + 2h |
| 1 | Velocity at t = 1 is 3 m/s, justified by 3 + 2h approaching 3 as h approaches 0 (a limit statement or a clear verbal equivalent) |
| 1 | Explains that h ≠ 0 for every interval, and that h = 0 gives an undefined 0/0 |
Acceptable alternative for (c): a table of 3 + 2h for h = ±0.1, ±0.01, ±0.001 showing values approaching 3 from both sides, with a stated conclusion.
Question 7 (constructed response · stretch)
Let g(x) = 3 + |x − 2|.
(a) Find the average rate of change of g over [2, 2.1] and over [1.9, 2]. (b) Show that the average rate over [2, 2 + h] is 1 for every h > 0, and that the average rate over [2 + h, 2] is −1 for every h < 0. (c) A student uses the interval [1.9, 2.1], finds an average rate of 0, and concludes that “the rate of change of g at x = 2 is 0”. Explain why this conclusion is not justified. (d) Does g have a single rate of change at x = 2? Explain using your answers above, and describe what the graph of g looks like at x = 2.
Worked solution and suggested Marlbridge rubric
(a) g(2) = 3, g(2.1) = 3.1 and g(1.9) = 3.1. Over [2, 2.1]: (g(2.1) − g(2))/(2.1 − 2) = (3.1 − 3)/0.1 = 1. Over [1.9, 2]: (g(2) − g(1.9))/(2 − 1.9) = (3 − 3.1)/0.1 = −1.
(b) For h > 0, g(2 + h) = 3 + h, so (g(2 + h) − g(2))/h = h/h = 1. For h < 0, g(2 + h) = 3 + |h| = 3 − h, so the average over [2 + h, 2] is (g(2) − g(2 + h))/(2 − (2 + h)) = (3 − (3 − h))/(−h) = h/(−h) = −1.
(c) Over [1.9, 2.1], (g(2.1) − g(1.9))/0.2 = (3.1 − 3.1)/0.2 = 0. But this one interval hides what happens on each side: g falls at rate 1 to the left of 2 and rises at rate 1 to the right. The 0 is just the two sides cancelling. One average, however short the interval, does not show what the averages approach.
(d) No. As the intervals shrink, the averages from the right stay at 1 and the averages from the left stay at −1. They do not approach one common value, so there is no single rate of change at x = 2. The graph of g is a V shape: a line of slope −1 to the left of 2 and a line of slope 1 to the right, meeting at a sharp corner at (2, 3). No single tangent line fits there. (Unit 2 returns to this when it discusses where a derivative exists.)
| Point | What earns it |
|---|---|
| 1 | Both averages correct: 1 and −1 |
| 1 | General result for each side, including the correct treatment of |
| 1 | Explains that the symmetric average 0 comes from two different one-sided behaviours cancelling, so it does not show a rate at x = 2 |
| 1 | Concludes there is no single rate at x = 2 because the left and right averages approach different values (−1 and 1), and describes the graph as a V shape with a sharp corner at (2, 3) |
How did you do?
- Q1 or Q5(a) wrong: revisit “Average rate of change” in the study guide. The formula is change in output over change in input.
- Q2, Q4 or Q6(d) wrong: reread “Why one division cannot give the rate at an instant”.
- Q3 or Q6(c) wrong: reread “Shrinking the interval” and Worked example 1.
- Q5(b)–(c) wrong: redo Worked example 2: use the shortest interval containing the instant, and interpret with units and sign.
- Q7 wrong: see the misconception “Use only one side”: averages from both sides must agree.
Then tick off the topic checklist.