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 exact answers. All contexts and data are invented. Notation: lim (h → 0) g(h) means “the limit as h approaches 0 of g(h)”; f′(x), dy/dx and y′ all denote the derivative. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
Let y = f(x). Which of the following does not denote the derivative of f?
- (A) dy/dx
- (B) f′(x)
- (C) y′
- (D) Δy/Δx
Answer and explanation
Answer: (D). Δy/Δx is the average rate of change over an interval, the difference quotient itself. The derivative is its limit as Δx → 0.
- (A), (B) and (C) are the three standard notations for the derivative of y = f(x). They mean the same thing, so none of them is the answer.
Question 2 (multiple choice · core)
Let f(x) = 4 − x². From the definition, f′(x) = −2x. Which is an equation of the line tangent to the graph of f at x = 1?
- (A) y = −2x + 5
- (B) y = −2x + 3
- (C) y − 3 = −2x(x − 1)
- (D) y = 2x + 1
Answer and explanation
Answer: (A). The point is (1, f(1)) = (1, 3). The slope is f′(1) = −2. So y − 3 = −2(x − 1), which rearranges to y = −2x + 5.
- (B) uses the correct slope but makes f(1) = 3 the y-intercept. The line must pass through (1, 3); this one passes through (0, 3).
- (C) uses the formula f′(x) = −2x as the slope instead of the number f′(1). The result is a parabola, not a line.
- (D) drops the minus sign in f′(1), using slope 2: y − 3 = 2(x − 1) gives y = 2x + 1.
Question 3 (multiple choice · core)
For a function g, lim (h → 0) (g(x + h) − g(x))/h = 3x² − 4x for all x. Which statement must be true?
- (A) g′(1) = −1
- (B) g(1) = −1
- (C) The derivative of g is 0 only at x = 0.
- (D) The line tangent to the graph of g at x = 1 is y = −x.
Answer and explanation
Answer: (A). The limit is the definition of g′(x), so g′(x) = 3x² − 4x and g′(1) = 3 − 4 = −1.
- (B) confuses g′(1) with g(1). The derivative gives no information about the height of g.
- (C) is false: 3x² − 4x = x(3x − 4) = 0 at x = 0 and at x = 4/3.
- (D) has the correct slope, −1, but assumes the line passes through the origin. Without g(1) you cannot write the tangent line.
Question 4 (multiple choice · foundation)
In a fictional shop, C(n) is the cost, in dollars, of n kilograms of rice. What are the units of C′(n)?
- (A) Dollars
- (B) Kilograms per dollar
- (C) Dollars per kilogram
- (D) Dollar-kilograms
Answer and explanation
Answer: (C). C′(n) is the limit of ΔC/Δn, a change in dollars divided by a change in kilograms. So its units are the units of C over the units of n.
- (A) is the unit of C(n) itself, not of its rate of change.
- (B) inverts the ratio; that would be the unit of the rate of change of n with respect to C.
- (D) multiplies the units instead of dividing.
Question 5 (constructed response · core)
Let f(x) = √(x + 3).
(a) Use the definition of the derivative to show that f′(x) = 1/(2√(x + 3)). (b) For which values of x does your formula for f′(x) exist? (c) Write an equation of the line tangent to the graph of f at x = 1. (d) A student writes the tangent line at x = 1 as y − 2 = (1/(2√(x + 3)))(x − 1). Explain why this is not a correct answer.
Worked solution and suggested Marlbridge rubric
(a) f′(x) = lim (h → 0) (√(x + h + 3) − √(x + 3))/h. Substitution gives 0/0. Multiply top and bottom by the conjugate √(x + h + 3) + √(x + 3). The top becomes (x + h + 3) − (x + 3) = h. So, for h ≠ 0,
(f(x + h) − f(x))/h = h / (h(√(x + h + 3) + √(x + 3))) = 1/(√(x + h + 3) + √(x + 3)).
As h → 0, the bottom approaches 2√(x + 3). So f′(x) = 1/(2√(x + 3)).
(b) The formula needs √(x + 3) to be defined and nonzero, so x > −3. (f is defined at x = −3, but f′ is not.)
(c) f(1) = √4 = 2 and f′(1) = 1/(2 × 2) = 1/4. Tangent line: y − 2 = (1/4)(x − 1), or y = x/4 + 7/4.
(d) The student used the formula f′(x) as the slope instead of the number f′(1) = 1/4. The slope of a line must be a constant; with x still inside the “slope”, the equation is not a straight line at all. Substituting x = 1 into f′ first gives the correct line from (c).
| Point | What earns it |
|---|---|
| 1 | Correct difference quotient and multiplication by the conjugate |
| 1 | Simplifies the top to h, cancels h (h ≠ 0) and takes the limit to reach 1/(2√(x + 3)) |
| 1 | Domain of f′: x > −3, with the reason that the denominator is 0 at x = −3 |
| 1 | Tangent line with point (1, 2) and slope 1/4 |
| 1 | Explains that the slope must be the number f′(1) = 1/4, not the expression f′(x) |
Question 6 (constructed response · core)
Let f(x) = x³.
(a) Use the definition of the derivative to find f′(x). (b) Find all points on the graph of f where the tangent line is parallel to the line y = 12x − 5. (c) Write an equation of the tangent line at each point you found in (b). Are they the same line? (d) Find f′(0) and describe the tangent line at the origin.
Worked solution and suggested Marlbridge rubric
(a) (x + h)³ − x³ = 3x²h + 3xh² + h³. Divide by h (h ≠ 0): 3x² + 3xh + h². Let h → 0: f′(x) = 3x².
(b) Parallel lines have equal slopes, so f′(x) = 12. Then 3x² = 12, x² = 4 and x = 2 or x = −2. The points are (2, 8) and (−2, −8).
(c) At (2, 8): y − 8 = 12(x − 2), so y = 12x − 16. At (−2, −8): y + 8 = 12(x + 2), so y = 12x + 16. They are different lines: parallel, with the same slope but different intercepts.
(d) f′(0) = 0. The tangent line at (0, 0) is horizontal: y = 0, the x-axis. (The curve crosses this tangent line at the origin, which shows a tangent line can cross the curve at the point of tangency.)
| Point | What earns it |
|---|---|
| 1 | Correct expansion of (x + h)³ − x³ and division by h |
| 1 | f′(x) = 3x² from the limit |
| 1 | Sets f′(x) = 12 and finds both x = 2 and x = −2 |
| 1 | Both tangent lines correct, with the conclusion that they are parallel but not the same |
| 1 | f′(0) = 0 and the tangent line y = 0 |
Question 7 (constructed response · stretch)
In a fictional model, a toy rocket’s height above the ground is y = 40t − 5t² metres, t seconds after launch, for 0 ≤ t ≤ 8.
(a) Use the definition of the derivative to find dy/dt. (b) Find dy/dt at t = 3 and at t = 5. Interpret both values, with units. (c) Find the time when dy/dt = 0 and the height at that time. What is the rocket doing at that instant? (d) A student writes: “dy/dt at t = 3 is 40(3) − 5(3)² = 75 m/s.” Explain the student’s error.
Worked solution and suggested Marlbridge rubric
(a) y(t + h) − y(t) = 40h − 5(2th + h²) = 40h − 10th − 5h². Divide by h (h ≠ 0): 40 − 10t − 5h. Let h → 0: dy/dt = 40 − 10t metres per second.
(b) At t = 3: dy/dt = 40 − 30 = 10 m/s. Three seconds after launch, the rocket is rising at 10 metres per second. At t = 5: dy/dt = 40 − 50 = −10 m/s. Five seconds after launch, the rocket is falling at 10 metres per second.
(c) 40 − 10t = 0 gives t = 4. The height is 40(4) − 5(16) = 80 m. At that instant the rocket is neither rising nor falling: its vertical velocity is 0.
(d) The student substituted t = 3 into y, not into dy/dt. The value 75 is the height y at t = 3, in metres, not a rate in metres per second. The correct value is dy/dt at t = 3 = 10 m/s. In notation, y(3) = 75 but y′(3) = 10.
| Point | What earns it |
|---|---|
| 1 | Correct difference quotient simplified to 40 − 10t − 5h (h ≠ 0) |
| 1 | dy/dt = 40 − 10t from the limit |
| 1 | Both values, 10 and −10, with units m/s and the correct meaning of each sign |
| 1 | t = 4 and height 80 m, with the rocket momentarily neither rising nor falling |
| 1 | Identifies 75 as y(3), a height, and gives y′(3) = 10 |
How did you do?
- Q1 or Q4 wrong: reread “Derivative notation” and the units paragraph under “Four representations” in the study guide.
- Q2 or Q6 wrong: redo Worked example 3 and the tangent line box: point (a, f(a)), slope the number f′(a).
- Q3 or Q7(d) wrong: review the difference between f(a) and f′(a) in “Common misconceptions”.
- Q5, Q6(a) or Q7(a) wrong: redo Worked examples 1 and 2, writing f(x + h) in full before subtracting.
Then tick off the topic checklist.