IB · MATH AI SL

Mathematics: Applications & Interpretation SL

Introductory calculus — Topic 5

Name: ____________________Date: October 2, 2026

1.
Find the derivative of f(x) = 3x⁴ − 2x² + 5x, and hence find f'(2).
[3 marks]
Marking points
- Applies the power rule to each term to obtain f'(x) = 12x³ − 4x + 5.
- Substitutes x = 2.
- Obtains f'(2) = 93.
Examiner tip: Differentiate term by term, reducing each power by one and multiplying by the original exponent — a constant term always differentiates to zero.
2.
Marking analysis: A learner attempts the following task: “Find the derivative of f(x) = 3x⁴ − 2x² + 5x, and hence find f'(2).” Their response addresses only this point: “Applies the power rule to each term to obtain f'(x) = 12x³ − 4x + 5.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Marking points
- Recognises credit for the stated point: Applies the power rule to each term to obtain f'(x) = 12x³ − 4x + 5.
- Identifies the missing requirement: Substitutes x = 2.
- Identifies the missing requirement: Obtains f'(2) = 93.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
3.
The curve y = x² − 3x + 1 passes through the point where x = 4. Find the equation of the tangent to the curve at this point.
[4 marks]
Marking points
- Calculates y(4) = 4² − 3(4) + 1 = 5, giving the point (4, 5).
- Differentiates to obtain y' = 2x − 3, and evaluates the gradient at x = 4 as 5.
- Uses the point-gradient form y − 5 = 5(x − 4).
- Simplifies to y = 5x − 15.
Examiner tip: The gradient of the tangent at a point equals the value of the derivative at that point — always find both the y-coordinate and the derivative before forming the tangent equation.
4.
Marking analysis: A learner attempts the following task: “The curve y = x² − 3x + 1 passes through the point where x = 4. Find the equation of the tangent to the curve at this point.” Their response addresses only this point: “Calculates y(4) = 4² − 3(4) + 1 = 5, giving the point (4, 5).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Calculates y(4) = 4² − 3(4) + 1 = 5, giving the point (4, 5).
- Identifies the missing requirement: Differentiates to obtain y' = 2x − 3, and evaluates the gradient at x = 4 as 5.
- Identifies the missing requirement: Uses the point-gradient form y − 5 = 5(x − 4).
- Identifies the missing requirement: Simplifies to y = 5x − 15.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
5.
Find the stationary points of f(x) = x³ − 6x² + 9x + 2, and use the second derivative to determine whether each is a local maximum or a local minimum.
[6 marks]
Marking points
- Differentiates to obtain f'(x) = 3x² − 12x + 9.
- Sets f'(x) = 0 and simplifies to x² − 4x + 3 = 0.
- Factorises to obtain x = 1 and x = 3.
- Finds the second derivative f''(x) = 6x − 12.
- Evaluates f''(1) = −6 < 0, so (1, 6) is a local maximum.
- Evaluates f''(3) = 6 > 0, so (3, 2) is a local minimum.
Examiner tip: The second derivative test is quick: a negative value confirms a maximum, a positive value confirms a minimum, so there is no need to check gradient signs either side of the point.
6.
Marking analysis: A learner attempts the following task: “Find the stationary points of f(x) = x³ − 6x² + 9x + 2, and use the second derivative to determine whether each is a local maximum or a local minimum.” Their response addresses only this point: “Differentiates to obtain f'(x) = 3x² − 12x + 9.” Evaluate the response against the complete 6-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[6 marks]
Marking points
- Recognises credit for the stated point: Differentiates to obtain f'(x) = 3x² − 12x + 9.
- Identifies the missing requirement: Sets f'(x) = 0 and simplifies to x² − 4x + 3 = 0.
- Identifies the missing requirement: Factorises to obtain x = 1 and x = 3.
- Identifies the missing requirement: Finds the second derivative f''(x) = 6x − 12.
- Identifies the missing requirement: Evaluates f''(1) = −6 < 0, so (1, 6) is a local maximum.
- Identifies the missing requirement: Evaluates f''(3) = 6 > 0, so (3, 2) is a local minimum.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
7.
A company's profit, in thousands of dollars, from selling x units is modelled by P(x) = −2x² + 80x − 200. Find the number of units that maximises profit, and find the maximum profit.
[4 marks]
Marking points
- Differentiates to obtain P'(x) = −4x + 80.
- Sets P'(x) = 0 and solves to obtain x = 20.
- Confirms this is a maximum since P''(x) = −4 < 0.
- Substitutes x = 20 into P(x) to obtain a maximum profit of $600,000.
Examiner tip: For an optimisation problem, always confirm the stationary point is a maximum (not a minimum) using the second derivative before stating the final answer.
8.
Marking analysis: A learner attempts the following task: “A company's profit, in thousands of dollars, from selling x units is modelled by P(x) = −2x² + 80x − 200. Find the number of units that maximises profit, and find the maximum profit.” Their response addresses only this point: “Differentiates to obtain P'(x) = −4x + 80.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Differentiates to obtain P'(x) = −4x + 80.
- Identifies the missing requirement: Sets P'(x) = 0 and solves to obtain x = 20.
- Identifies the missing requirement: Confirms this is a maximum since P''(x) = −4 < 0.
- Identifies the missing requirement: Substitutes x = 20 into P(x) to obtain a maximum profit of $600,000.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
9.
The volume of water in a tank, in litres, is modelled by V(t) = 100 − 5t², where t is time in minutes. Find the rate of change of volume at t = 3 minutes, and interpret the sign of your answer.
[3 marks]
Marking points
- Differentiates to obtain V'(t) = −10t.
- Substitutes t = 3 to obtain V'(3) = −30.
- Interprets the negative sign as the volume decreasing at a rate of 30 litres per minute at t = 3.
Examiner tip: The sign of a rate of change always tells you the direction: negative means decreasing, positive means increasing — always state this in context for an 'interpret' command term.
10.
Marking analysis: A learner attempts the following task: “The volume of water in a tank, in litres, is modelled by V(t) = 100 − 5t², where t is time in minutes. Find the rate of change of volume at t = 3 minutes, and interpret the sign of your answer.” Their response addresses only this point: “Differentiates to obtain V'(t) = −10t.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Marking points
- Recognises credit for the stated point: Differentiates to obtain V'(t) = −10t.
- Identifies the missing requirement: Substitutes t = 3 to obtain V'(3) = −30.
- Identifies the missing requirement: Interprets the negative sign as the volume decreasing at a rate of 30 litres per minute at t = 3.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
11.
Use the chain rule to differentiate f(x) = (3x + 1)⁴, and find f'(1).
[3 marks]
Marking points
- Applies the chain rule: f'(x) = 4(3x + 1)³ × 3 = 12(3x + 1)³.
- Substitutes x = 1 to obtain f'(1) = 12(4)³.
- Obtains f'(1) = 768.
Examiner tip: The chain rule always multiplies the derivative of the outer function by the derivative of the inner function — here the inner function 3x + 1 contributes the extra factor of 3.
12.
Marking analysis: A learner attempts the following task: “Use the chain rule to differentiate f(x) = (3x + 1)⁴, and find f'(1).” Their response addresses only this point: “Applies the chain rule: f'(x) = 4(3x + 1)³ × 3 = 12(3x + 1)³.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Marking points
- Recognises credit for the stated point: Applies the chain rule: f'(x) = 4(3x + 1)³ × 3 = 12(3x + 1)³.
- Identifies the missing requirement: Substitutes x = 1 to obtain f'(1) = 12(4)³.
- Identifies the missing requirement: Obtains f'(1) = 768.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
13.
Use the product rule to differentiate f(x) = x²(2x + 3), and find f'(2).
[4 marks]
Marking points
- Applies the product rule with u = x², v = (2x + 3): f'(x) = 2x(2x + 3) + x²(2).
- Simplifies to f'(x) = 6x² + 6x.
- Substitutes x = 2.
- Obtains f'(2) = 36.
Examiner tip: You can check a product-rule result by expanding the original function first and differentiating term by term — here x²(2x+3) = 2x³ + 3x² gives the same derivative 6x² + 6x.
14.
Marking analysis: A learner attempts the following task: “Use the product rule to differentiate f(x) = x²(2x + 3), and find f'(2).” Their response addresses only this point: “Applies the product rule with u = x², v = (2x + 3): f'(x) = 2x(2x + 3) + x²(2).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Applies the product rule with u = x², v = (2x + 3): f'(x) = 2x(2x + 3) + x²(2).
- Identifies the missing requirement: Simplifies to f'(x) = 6x² + 6x.
- Identifies the missing requirement: Substitutes x = 2.
- Identifies the missing requirement: Obtains f'(2) = 36.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
15.
Use the quotient rule to differentiate f(x) = (2x + 1)/(x − 3), and find f'(5).
[4 marks]
Marking points
- Applies the quotient rule with u = 2x + 1, v = x − 3: f'(x) = [2(x−3) − (2x+1)(1)]/(x−3)².
- Simplifies the numerator to −7, giving f'(x) = −7/(x−3)².
- Substitutes x = 5.
- Obtains f'(5) = −1.75.
Examiner tip: In the quotient rule, the order of subtraction in the numerator matters: it is (u'v − uv'), not (uv' − u'v) — reversing it flips the sign of the whole answer.
16.
Marking analysis: A learner attempts the following task: “Use the quotient rule to differentiate f(x) = (2x + 1)/(x − 3), and find f'(5).” Their response addresses only this point: “Applies the quotient rule with u = 2x + 1, v = x − 3: f'(x) = [2(x−3) − (2x+1)(1)]/(x−3)².” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Applies the quotient rule with u = 2x + 1, v = x − 3: f'(x) = [2(x−3) − (2x+1)(1)]/(x−3)².
- Identifies the missing requirement: Simplifies the numerator to −7, giving f'(x) = −7/(x−3)².
- Identifies the missing requirement: Substitutes x = 5.
- Identifies the missing requirement: Obtains f'(5) = −1.75.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
17.
Evaluate the definite integral ∫₀³ (3x² − 4x + 2) dx.
[4 marks]
Marking points
- Finds the antiderivative x³ − 2x² + 2x.
- Evaluates at the upper limit: 3³ − 2(3²) + 2(3) = 15.
- Evaluates at the lower limit: 0³ − 2(0²) + 2(0) = 0.
- Subtracts to obtain a value of 15.
Examiner tip: Increase the power of each term by one and divide by the new power to integrate — then always subtract the lower-limit value from the upper-limit value.
18.
Marking analysis: A learner attempts the following task: “Evaluate the definite integral ∫₀³ (3x² − 4x + 2) dx.” Their response addresses only this point: “Finds the antiderivative x³ − 2x² + 2x.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Finds the antiderivative x³ − 2x² + 2x.
- Identifies the missing requirement: Evaluates at the upper limit: 3³ − 2(3²) + 2(3) = 15.
- Identifies the missing requirement: Evaluates at the lower limit: 0³ − 2(0²) + 2(0) = 0.
- Identifies the missing requirement: Subtracts to obtain a value of 15.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
19.
Find the area enclosed between the curve y = 4 − x² and the x-axis, between x = −2 and x = 2.
[4 marks]
Marking points
- Recognises the area is given by ∫₋₂² (4 − x²) dx.
- Finds the antiderivative 4x − x³/3.
- Evaluates the antiderivative at x = 2 and at x = −2.
- Obtains an area of 32/3 ≈ 10.7 square units.
Examiner tip: Since the curve lies entirely above the x-axis on this interval (4 − x² ≥ 0 for −2 ≤ x ≤ 2), the definite integral directly gives a positive area with no sign adjustment needed.
20.
Marking analysis: A learner attempts the following task: “Find the area enclosed between the curve y = 4 − x² and the x-axis, between x = −2 and x = 2.” Their response addresses only this point: “Recognises the area is given by ∫₋₂² (4 − x²) dx.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Recognises the area is given by ∫₋₂² (4 − x²) dx.
- Identifies the missing requirement: Finds the antiderivative 4x − x³/3.
- Identifies the missing requirement: Evaluates the antiderivative at x = 2 and at x = −2.
- Identifies the missing requirement: Obtains an area of 32/3 ≈ 10.7 square units.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
21.
A particle moves with velocity v(t) = 6t² − 4t m/s. Find the displacement of the particle between t = 1 and t = 3 seconds.
[4 marks]
Marking points
- Recognises that displacement is the definite integral of velocity: ∫₁³ (6t² − 4t) dt.
- Finds the antiderivative 2t³ − 2t².
- Evaluates at t = 3 and t = 1.
- Obtains a displacement of 36 m.
Examiner tip: Integrating velocity gives displacement, and integrating acceleration gives velocity — keep this chain in mind when deciding whether to differentiate or integrate in a kinematics problem.
22.
Marking analysis: A learner attempts the following task: “A particle moves with velocity v(t) = 6t² − 4t m/s. Find the displacement of the particle between t = 1 and t = 3 seconds.” Their response addresses only this point: “Recognises that displacement is the definite integral of velocity: ∫₁³ (6t² − 4t) dt.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Recognises that displacement is the definite integral of velocity: ∫₁³ (6t² − 4t) dt.
- Identifies the missing requirement: Finds the antiderivative 2t³ − 2t².
- Identifies the missing requirement: Evaluates at t = 3 and t = 1.
- Identifies the missing requirement: Obtains a displacement of 36 m.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
23.
The displacement of a particle is given by s(t) = t³ − 6t² + 9t. Find the velocity and acceleration of the particle at t = 2 seconds.
[4 marks]
Marking points
- Differentiates once to obtain velocity v(t) = 3t² − 12t + 9.
- Substitutes t = 2 to obtain v(2) = −3 m/s.
- Differentiates again to obtain acceleration a(t) = 6t − 12.
- Substitutes t = 2 to obtain a(2) = 0 m/s².
Examiner tip: The negative velocity shows the particle is momentarily moving backward at t = 2, while zero acceleration means velocity is neither increasing nor decreasing at that exact instant.
24.
Marking analysis: A learner attempts the following task: “The displacement of a particle is given by s(t) = t³ − 6t² + 9t. Find the velocity and acceleration of the particle at t = 2 seconds.” Their response addresses only this point: “Differentiates once to obtain velocity v(t) = 3t² − 12t + 9.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Differentiates once to obtain velocity v(t) = 3t² − 12t + 9.
- Identifies the missing requirement: Substitutes t = 2 to obtain v(2) = −3 m/s.
- Identifies the missing requirement: Differentiates again to obtain acceleration a(t) = 6t − 12.
- Identifies the missing requirement: Substitutes t = 2 to obtain a(2) = 0 m/s².
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
25.
The cost, in dollars, of producing x units is modelled by C(x) = 0.01x³ − 0.6x² + 13x + 100. Find the marginal cost when x = 10, and interpret its meaning.
[4 marks]
Marking points
- Differentiates to obtain C'(x) = 0.03x² − 1.2x + 13.
- Substitutes x = 10.
- Obtains C'(10) = 4.
- Interprets this as the approximate cost, $4, of producing one additional unit beyond the 10th.
Examiner tip: Marginal cost is the derivative of the cost function — it approximates the cost of producing one more unit at the current production level, not the total cost.
26.
Marking analysis: A learner attempts the following task: “The cost, in dollars, of producing x units is modelled by C(x) = 0.01x³ − 0.6x² + 13x + 100. Find the marginal cost when x = 10, and interpret its meaning.” Their response addresses only this point: “Differentiates to obtain C'(x) = 0.03x² − 1.2x + 13.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Differentiates to obtain C'(x) = 0.03x² − 1.2x + 13.
- Identifies the missing requirement: Substitutes x = 10.
- Identifies the missing requirement: Obtains C'(10) = 4.
- Identifies the missing requirement: Interprets this as the approximate cost, $4, of producing one additional unit beyond the 10th.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
27.
A farmer has 200 m of fencing to enclose a rectangular field. Let the width be x metres. (a) Express the area A of the field in terms of x. (b) Find the value of x that maximises the area, and find the maximum area.
[5 marks]
Marking points
- Uses the perimeter condition 2x + 2(length) = 200 to express the length as 100 − x.
- Expresses the area as A(x) = x(100 − x) = 100x − x².
- Differentiates and sets A'(x) = 100 − 2x = 0 to obtain x = 50.
- Confirms this is a maximum since A''(x) = −2 < 0.
- Substitutes x = 50 to obtain a maximum area of 2,500 m².
Examiner tip: When a fixed perimeter encloses a rectangle, the area is always maximised when the rectangle is a square — here x = 50 and the length 100 − 50 = 50 confirm this pattern.
28.
Marking analysis: A learner attempts the following task: “A farmer has 200 m of fencing to enclose a rectangular field. Let the width be x metres. (a) Express the area A of the field in terms of x. (b) Find the value of x that maximises the area, and find the maximum area.” Their response addresses only this point: “Uses the perimeter condition 2x + 2(length) = 200 to express the length as 100 − x.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]
Marking points
- Recognises credit for the stated point: Uses the perimeter condition 2x + 2(length) = 200 to express the length as 100 − x.
- Identifies the missing requirement: Expresses the area as A(x) = x(100 − x) = 100x − x².
- Identifies the missing requirement: Differentiates and sets A'(x) = 100 − 2x = 0 to obtain x = 50.
- Identifies the missing requirement: Confirms this is a maximum since A''(x) = −2 < 0.
- Identifies the missing requirement: Substitutes x = 50 to obtain a maximum area of 2,500 m².
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
29.
Find the inflection points of the curve f(x) = x⁴ − 8x², using the second derivative.
[5 marks]
Marking points
- Differentiates twice to obtain f''(x) = 12x² − 16.
- Sets f''(x) = 0 and solves to obtain x² = 4/3.
- Obtains x ≈ ±1.15.
- Substitutes each x-value into f(x) to obtain y ≈ −8.89 at both points, by symmetry of the even function.
- States the inflection points as approximately (1.15, −8.89) and (−1.15, −8.89).
Examiner tip: An inflection point occurs where the second derivative changes sign — solving f''(x) = 0 only locates candidates, but for a smooth polynomial like this one, a genuine sign change is guaranteed at each simple root.
30.
Marking analysis: A learner attempts the following task: “Find the inflection points of the curve f(x) = x⁴ − 8x², using the second derivative.” Their response addresses only this point: “Differentiates twice to obtain f''(x) = 12x² − 16.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]
Marking points
- Recognises credit for the stated point: Differentiates twice to obtain f''(x) = 12x² − 16.
- Identifies the missing requirement: Sets f''(x) = 0 and solves to obtain x² = 4/3.
- Identifies the missing requirement: Obtains x ≈ ±1.15.
- Identifies the missing requirement: Substitutes each x-value into f(x) to obtain y ≈ −8.89 at both points, by symmetry of the even function.
- Identifies the missing requirement: States the inflection points as approximately (1.15, −8.89) and (−1.15, −8.89).
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
31.
Find the average value of the function f(x) = x² + 1 over the interval [0, 4].
[4 marks]
Marking points
- Uses the average value formula: (1/(4−0)) ∫₀⁴ (x² + 1) dx.
- Finds the antiderivative x³/3 + x.
- Evaluates the integral as (64/3 + 4) − 0 = 76/3.
- Divides by 4 to obtain an average value of approximately 6.33.
Examiner tip: The average value of a function over an interval is the definite integral divided by the width of the interval — this is the continuous analogue of averaging a list of numbers.
32.
Marking analysis: A learner attempts the following task: “Find the average value of the function f(x) = x² + 1 over the interval [0, 4].” Their response addresses only this point: “Uses the average value formula: (1/(4−0)) ∫₀⁴ (x² + 1) dx.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Uses the average value formula: (1/(4−0)) ∫₀⁴ (x² + 1) dx.
- Identifies the missing requirement: Finds the antiderivative x³/3 + x.
- Identifies the missing requirement: Evaluates the integral as (64/3 + 4) − 0 = 76/3.
- Identifies the missing requirement: Divides by 4 to obtain an average value of approximately 6.33.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.