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MCV4U Rates of Change & Limits
Grade 12 Unit 1 — practice questions with full solutions

Average versus instantaneous rate of change, the difference quotient, and evaluating limits algebraically — the ideas every derivative rule is built on.

Free MCV4U rates of change & limits practice for Grade 12 students in Ontario. These 5 questions cover the same material as a typical unit test or exam review on this topic, and every one comes with a complete worked solution — not just an answer key. Useful whether you are preparing for a MCV4U unit test, catching up on a lesson, or reviewing before the final exam.
MCV4U · Grade 12 Rate of Change 4 core 1 challenge
1Core
For f(x) = x3 − 4x, find the average rate of change from x = 1 to x = 3.
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  1. Average rate of change is the slope of the secant: [f(3) − f(1)] ÷ (3 − 1).
  2. f(3) = 27 − 12 = 15.
  3. f(1) = 1 − 4 = −3.
  4. Slope = [15 − (−3)] ÷ 2 = 18 ÷ 2 = 9.
Final answer9
2Core
Evaluate: lim as x → 3 of (x2 − 9) ÷ (x − 3).
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  1. Substituting x = 3 gives 0 ÷ 0, so the expression must be simplified first.
  2. Factor the numerator as a difference of squares: x2 − 9 = (x − 3)(x + 3).
  3. Cancel the common factor (x − 3), which is valid because x approaches 3 without equalling 3.
  4. The limit is then lim (x + 3) = 3 + 3 = 6.
Final answer6
3Core
Use the limit definition (first principles) to find the instantaneous rate of change of f(x) = x2 − 5x at x = 3.
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  1. First principles: lim as h → 0 of [f(3 + h) − f(3)] ÷ h.
  2. f(3 + h) = (3 + h)2 − 5(3 + h) = 9 + 6h + h2 − 15 − 5h = h2 + h − 6.
  3. f(3) = 9 − 15 = −6.
  4. Difference quotient = (h2 + h − 6 + 6) ÷ h = (h2 + h) ÷ h = h + 1.
  5. Letting h → 0 gives 1.
Final answer1
4Core
A ball is thrown so its height is h(t) = −4.9t2 + 20t + 1, where h is in metres and t in seconds. Find its average velocity between t = 1 and t = 3, and explain what the sign of your answer means.
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  1. Average velocity is the average rate of change of height: [h(3) − h(1)] ÷ (3 − 1).
  2. h(1) = −4.9 + 20 + 1 = 16.1 m.
  3. h(3) = −4.9(9) + 60 + 1 = −44.1 + 61 = 16.9 m.
  4. Average velocity = (16.9 − 16.1) ÷ 2 = 0.8 ÷ 2 = 0.4 m/s.
  5. The value is positive but small: the ball rose, passed its peak and came back down, ending only slightly higher than it started.
Final answer0.4 m/s
5Challenge
Evaluate: lim as x → 0 of (√(x + 4) − 2) ÷ x.
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  1. Direct substitution gives 0 ÷ 0, so the expression needs rewriting.
  2. Multiply numerator and denominator by the conjugate √(x + 4) + 2.
  3. The numerator becomes (x + 4) − 4 = x.
  4. So the expression is x ÷ [x(√(x + 4) + 2)] = 1 ÷ (√(x + 4) + 2).
  5. Now substitute x = 0: 1 ÷ (√4 + 2) = 1 ÷ 4.
Final answer1⁄4

MCV4U Rates of Change & Limits — common questions

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

What is the difference between average and instantaneous rate of change?

Average rate of change is the slope of a secant line between two points. Instantaneous rate of change is the slope of the tangent at a single point, found by letting the interval shrink to zero — that limit is the derivative.

Do I need to know limits for MCV4U?

Yes. Limits are how the derivative is defined, and first-principles questions appear on most unit tests even after the shortcut rules are introduced.

More MCV4U practice — grade 12 calculus and vectors

MCV4U DerivativesMCV4U Applications of Derivatives & Curve SketchingMCV4U Derivatives of Exponential & Trigonometric FunctionsMCV4U Geometric VectorsAll MCV4U units
← All assignmentsUnit 2: Derivatives →
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