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MCV4U Applications of Derivatives & Curve Sketching
Grade 12 Unit 3 — practice questions with full solutions

Critical points, increasing and decreasing intervals, concavity, optimization and related rates — turning derivatives into information about shape and motion.

MCV4U · Grade 12 Derivatives and their Applications 4 core 1 challenge
1Core
For f(x) = x3 − 3x2 − 9x + 5, find the critical points and classify each as a local maximum or local minimum.
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  1. Critical points occur where f′(x) = 0.
  2. f′(x) = 3x2 − 6x − 9 = 3(x2 − 2x − 3) = 3(x − 3)(x + 1).
  3. So x = −1 and x = 3.
  4. Use the second derivative test: f″(x) = 6x − 6.
  5. f″(−1) = −12 < 0, so x = −1 is a local maximum; f(−1) = −1 − 3 + 9 + 5 = 10.
  6. f″(3) = 12 > 0, so x = 3 is a local minimum; f(3) = 27 − 27 − 27 + 5 = −22.
Final answerLocal maximum (−1, 10); local minimum (3, −22)
2Core
For f(x) = x3 − 6x2 + 9x, state the intervals where the function is increasing and decreasing, and find the point of inflection.
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  1. f′(x) = 3x2 − 12x + 9 = 3(x − 1)(x − 3).
  2. f′ is positive when both factors share a sign: x < 1 or x > 3, so the function increases there.
  3. f′ is negative between the roots, so the function decreases for 1 < x < 3.
  4. f″(x) = 6x − 12, which is zero at x = 2 and changes sign there.
  5. f(2) = 8 − 24 + 18 = 2, so the inflection point is (2, 2).
Final answerIncreasing x < 1 and x > 3; decreasing 1 < x < 3; inflection at (2, 2)
3Core
A farmer has 240 m of fencing to enclose a rectangular field along a straight river, with no fence needed on the river side. What dimensions give the maximum area, and what is that area?
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  1. Let x be each of the two sides perpendicular to the river and y the side parallel to it.
  2. Fencing constraint: 2x + y = 240, so y = 240 − 2x.
  3. Area: A(x) = x(240 − 2x) = 240x − 2x2.
  4. A′(x) = 240 − 4x, which is zero at x = 60.
  5. A″(x) = −4 < 0, confirming a maximum.
  6. Then y = 240 − 120 = 120, and A = 60 × 120 = 7200 m2.
Final answer60 m × 120 m, area 7200 m2
4Core
Air is pumped into a spherical balloon at 100 cm3/s. How fast is the radius increasing at the moment the radius is 5 cm? Leave your answer in terms of π and also give a decimal.
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  1. Volume of a sphere: V = 4⁄3πr3.
  2. Differentiate both sides with respect to time: dV⁄dt = 4πr2(dr⁄dt).
  3. Substitute dV⁄dt = 100 and r = 5: 100 = 4π(25)(dr⁄dt) = 100π(dr⁄dt).
  4. So dr⁄dt = 1 ÷ π ≈ 0.32 cm/s.
Final answerdr⁄dt = 1⁄π ≈ 0.32 cm/s
5Challenge
An object moves along a line with position s(t) = t3 − 6t2 + 9t for t ≥ 0, in metres and seconds. Find (a) when the object is at rest, (b) when it is moving backward, and (c) its acceleration at t = 2.
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  1. Velocity is the derivative of position: v(t) = 3t2 − 12t + 9 = 3(t − 1)(t − 3).
  2. (a) At rest means v = 0, so t = 1 s and t = 3 s.
  3. (b) Moving backward means v < 0, which happens between the roots: 1 < t < 3.
  4. (c) Acceleration is the derivative of velocity: a(t) = 6t − 12.
  5. a(2) = 12 − 12 = 0 m/s2 — the instant velocity is most negative.
Final answer(a) t = 1 s and t = 3 s   (b) 1 < t < 3   (c) 0 m/s2

MCV4U Applications of Derivatives & Curve Sketching — common questions

Short answers to the things students ask most about this unit.

How do I tell a local maximum from a local minimum?

Use the second derivative test: at a critical point, a negative second derivative means the curve is concave down and you have a local maximum; a positive one means concave up and a local minimum.

What is the usual method for an optimization question?

Write the quantity to be optimized as a function of one variable using the constraint, differentiate, set the derivative to zero, then confirm it is a maximum or minimum before answering in context.

More MCV4U practice — grade 12 calculus and vectors

MCV4U Rates of Change & LimitsMCV4U DerivativesMCV4U Derivatives of Exponential & Trigonometric FunctionsMCV4U Geometric VectorsAll MCV4U units
← Unit 2: DerivativesUnit 4: Derivatives of Exponential & Trigonometric Functions →
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