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, every function is differentiable where it is used, and every context is fictional. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
A differentiable function f has f(4) = 7 and f′(4) = −2. What is the estimate of f(4.3) given by the tangent line to the graph of f at x = 4?
- (A) 6.4
- (B) 7.6
- (C) −0.6
- (D) −1.6
Answer and explanation
Answer: (A). L(x) = 7 − 2(x − 4). So f(4.3) ≈ L(4.3) = 7 − 2(0.3) = 7 − 0.6 = 6.4.
- (B) adds 0.6 instead of subtracting it: it ignores the negative slope.
- (C) is only the estimated change, f′(4)(0.3). It forgets to add the starting value f(4) = 7.
- (D) uses x instead of (x − 4): 7 − 2(4.3) = −1.6.
Question 2 (multiple choice · core)
Let g(x) = √(1 + 3x). Which of the following is the linearization of g at x = 1?
- (A) L(x) = 2 + (3/4)(x − 1)
- (B) L(x) = 2 + (1/4)(x − 1)
- (C) L(x) = 2 + (3/4)x
- (D) L(x) = (3/4)(x − 1)
Answer and explanation
Answer: (A). g(1) = √4 = 2. By the chain rule, g′(x) = 3/(2√(1 + 3x)), so g′(1) = 3/(2 × 2) = 3/4. Then L(x) = g(1) + g′(1)(x − 1) = 2 + (3/4)(x − 1).
- (B) forgets the factor 3 from the chain rule: 1/(2√4) = 1/4.
- (C) uses x in place of (x − 1), so the line does not pass through (1, 2).
- (D) leaves out g(1) = 2, so the line passes through (1, 0) instead of (1, 2).
Question 3 (multiple choice · core)
A function f has f(2) = 10, f′(2) = 3 and f″(x) > 0 for all x. The tangent line at x = 2 is used to estimate f(2.5). Which statement is true?
- (A) The estimate is 11.5, and it is an overestimate.
- (B) The estimate is 11.5, and it is an underestimate.
- (C) The estimate is 11.5, and it is exact because f is increasing.
- (D) The estimate is 8.5, and it is an underestimate.
Answer and explanation
Answer: (B). L(2.5) = 10 + 3(0.5) = 11.5. Because f″ > 0, the graph bends upward and the tangent line lies below it, so 11.5 is less than f(2.5): an underestimate.
- (A) swaps the rule. Bending upward puts the line below the curve.
- (C) confuses increasing with straight. An increasing function can still bend, so the line and curve differ away from x = 2.
- (D) subtracts the change instead of adding it, even though the slope is positive.
Question 4 (multiple choice · core)
T(t) is the temperature, in °C, of a loaf of bread t minutes after it comes out of an oven. At t = 10, T(10) = 78 and T′(10) = −2.4. Using the tangent line at t = 10, what is the best estimate of the temperature at t = 12.5?
- (A) 72 °C
- (B) 75.6 °C
- (C) 84 °C
- (D) 48 °C
Answer and explanation
Answer: (A). The step is 12.5 − 10 = 2.5 minutes. T(12.5) ≈ 78 + (−2.4)(2.5) = 78 − 6 = 72 °C.
- (B) uses a step of 1 minute instead of 2.5: 78 − 2.4.
- (C) adds 6 instead of subtracting it. The bread is cooling, so the temperature should fall.
- (D) multiplies the rate by 12.5 instead of the step 2.5: 78 − 2.4 × 12.5 = 48.
Question 5 (constructed response · core)
Let f(x) = 1/x.
(a) Write the linearization of f at x = 5. (b) Use it to estimate 1/4.9, as a decimal. (c) Is your estimate an overestimate or an underestimate? Justify your answer using f″. (d) The actual value of 1/4.9 is 0.20408 to five decimal places. Is this consistent with your answer to (c)?
Worked solution and suggested Marlbridge rubric
(a) f(5) = 1/5 = 0.2. f′(x) = −1/x², so f′(5) = −1/25 = −0.04. L(x) = 0.2 − 0.04(x − 5).
(b) The step is 4.9 − 5 = −0.1. L(4.9) = 0.2 − 0.04(−0.1) = 0.2 + 0.004 = 0.204.
(c) f″(x) = 2/x³, which is positive for x > 0, including every x between 4.9 and 5. So the graph bends upward near x = 5 and the tangent line lies below it. The estimate 0.204 is an underestimate.
(d) 0.204 < 0.20408, so the estimate is slightly too low, which is consistent with (c). The error is about 0.00008.
| Point | What earns it |
|---|---|
| 1 | Correct f(5) and f′(5) = −1/25 in the linearization |
| 1 | L(4.9) = 0.204, with the negative step handled correctly |
| 1 | Underestimate, justified by f″(x) = 2/x³ > 0 near x = 5 (or “the graph bends upward, so the line is below it”) |
| 1 | Correct comparison in (d) |
Question 6 (constructed response · core)
H(t) is the height, in metres, of a young tree t years after it was planted. At t = 9 the tree is 8 m tall, and for t > 0 its growth rate is H′(t) = 1.2/√t metres per year.
(a) Find H′(9) and explain its meaning in context. (b) Use the tangent line at t = 9 to estimate the height of the tree at t = 9.5. (c) Is the estimate in (b) too high or too low? Justify your answer. (d) A student uses the same line to estimate the height at t = 15 and gets 10.4 m. Explain why this estimate is less reliable than the one in (b), and say whether it is too high or too low.
Worked solution and suggested Marlbridge rubric
(a) H′(9) = 1.2/√9 = 1.2/3 = 0.4 m per year. At 9 years after planting, the tree’s height is increasing at 0.4 metres per year.
(b) L(t) = 8 + 0.4(t − 9). H(9.5) ≈ 8 + 0.4(0.5) = 8.2 m. About nine and a half years after planting, the tree is approximately 8.2 m tall.
(c) H″(t) = −0.6/t^(3/2), which is negative for all t > 0. The growth rate is decreasing, so the graph of H bends downward and the tangent line lies above it. The estimate of 8.2 m is too high (an overestimate).
(d) 15 is 6 years from 9, while 9.5 is only half a year away. Over 6 years the growth rate falls a lot (from 0.4 m per year to 1.2/√15, about 0.31 m per year), but the line assumes it stays at 0.4. So the error is much bigger. Since H″ < 0 on the whole interval, 10.4 m is still an overestimate, and by more than in (b).
| Point | What earns it |
|---|---|
| 1 | H′(9) = 0.4 with units and a correct interpretation (rate of change of height at t = 9) |
| 1 | Estimate 8.2 m from a correct linearization |
| 1 | Overestimate, justified by H″ < 0 (or by H′ decreasing) |
| 1 | Explains that the step is much larger, so the rate changes more over the interval, and states that 10.4 m is an overestimate |
Question 7 (constructed response · stretch)
Let f(x) = x³ − 3x + 1.
(a) Write the linearization L(x) of f at x = 0, and use it to estimate f(0.2) and f(−0.2). (b) Find f″(x). Use it to decide whether each estimate in (a) is an overestimate or an underestimate, and explain why the two answers differ. (c) The equation f(x) = 0 has a solution between 0 and 1. Solve L(x) = 0 to estimate it. Then evaluate f at your estimate and comment on the result.
Worked solution and suggested Marlbridge rubric
(a) f(0) = 1. f′(x) = 3x² − 3, so f′(0) = −3. L(x) = 1 − 3x. Then f(0.2) ≈ 1 − 0.6 = 0.4 and f(−0.2) ≈ 1 + 0.6 = 1.6.
(b) f″(x) = 6x. For 0 < x ≤ 0.2, f″(x) > 0: the graph bends upward, the tangent line is below it, so 0.4 is an underestimate. For −0.2 ≤ x < 0, f″(x) < 0: the graph bends downward, the line is above it, so 1.6 is an overestimate. The answers differ because f″ changes sign at x = 0, the point of tangency. (Check: f(0.2) = 0.408 and f(−0.2) = 1.592.)
(c) 1 − 3x = 0 gives x = 1/3. Then f(1/3) = 1/27 − 1 + 1 = 1/27, about 0.037. This is close to 0 but not equal to it, so x = 1/3 is an approximate solution, not an exact one. The value is positive and f is decreasing there (f′(1/3) = −8/3 < 0), so the actual solution is a little more than 1/3.
| Point | What earns it |
|---|---|
| 1 | L(x) = 1 − 3x with both estimates, 0.4 and 1.6 |
| 1 | f″(x) = 6x, with underestimate for x = 0.2 and overestimate for x = −0.2, each justified by the sign of f″ on that side |
| 1 | Explains that f″ changes sign at the point of tangency, so the sides behave differently |
| 1 | x = 1/3 from L(x) = 0, with f(1/3) = 1/27 and a comment that it is close to, but not exactly, a solution |
How did you do?
- Q1, Q2 or Q5(a)–(b) wrong: revisit “The linearization formula” and Worked example 1 in the study guide.
- Q3, Q5(c) or Q7(b) wrong: reread “Overestimate or underestimate?” and Figure 1.
- Q4 or Q6 wrong: redo Worked example 2 (context) and Worked example 3.
- Q6(d) or Q7(c) wrong: look again at the error table in Worked example 1: estimates get worse further from a.
Then tick off the topic checklist.