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MHF4U Rates of Change & Combining Functions
Grade 12 Unit 6 — practice questions with full solutions

Average and instantaneous rate of change, secants and tangents, and adding, multiplying and composing functions.

Free MHF4U rates of change & combining functions practice for Grade 12 students in Ontario. These 4 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 MHF4U unit test, catching up on a lesson, or reviewing before the final exam.
MHF4U · Grade 12 Characteristics of Functions 3 core 1 challenge
1Core
12345510152025xy(1, 1)(4, 16)y = x²secant
The graph shows y = x² with a secant drawn from (1, 1) to (4, 16). (a) Find the average rate of change over that interval. (b) Explain what the secant's slope represents. (c) Estimate the instantaneous rate of change at x = 1 using the interval from 1 to 1.01.
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  1. (a) Average rate of change = (16 − 1) ÷ (4 − 1) = 15 ÷ 3 = 5.
  2. (b) It is exactly the slope of the secant line joining the two points — the average steepness across the interval.
  3. (c) At x = 1.01: y = 1.0201. Rate = (1.0201 − 1) ÷ 0.01 = 2.01.
  4. As the interval shrinks toward zero this approaches 2, the slope of the tangent at x = 1.
Final answer(a) 5   (b) the slope of the secant line   (c) ≈ 2.01, approaching 2
2Core
For f(x) = 3x² − 2x, find the average rate of change from x = 2 to x = 5.
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  1. Evaluate both endpoints. f(2) = 3(4) − 4 = 8.
  2. f(5) = 3(25) − 10 = 65.
  3. Average rate of change = (65 − 8) ÷ (5 − 2).
  4. That is 57 ÷ 3 = 19.
Final answer19
3Core
Given f(x) = x² + 1 and g(x) = 2x − 3, find (a) (f + g)(x), (b) (fg)(2), (c) f(g(x)), and (d) g(f(x)).
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  1. (a) Add: (x² + 1) + (2x − 3) = x² + 2x − 2.
  2. (b) f(2) = 5 and g(2) = 1, so the product is 5 × 1 = 5.
  3. (c) Substitute g into f: (2x − 3)² + 1 = 4x² − 12x + 9 + 1 = 4x² − 12x + 10.
  4. (d) Substitute f into g: 2(x² + 1) − 3 = 2x² − 1.
  5. Note (c) and (d) differ — composition is not commutative.
Final answer(a) x² + 2x − 2   (b) 5   (c) 4x² − 12x + 10   (d) 2x² − 1
4Challenge
A population is modelled by P(t) = 800(1.05)t, where t is in years. Find the average rate of change from t = 0 to t = 10, to one decimal place, and explain why it understates the growth near t = 10.
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  1. P(0) = 800.
  2. P(10) = 800(1.05)¹⁰ ≈ 800 × 1.628895 ≈ 1303.1.
  3. Average rate = (1303.1 − 800) ÷ 10 ≈ 50.3 people per year.
  4. Exponential growth accelerates, so the curve is steeper later than earlier.
  5. The average spreads the total change evenly across all ten years, so it sits below the actual growth rate near t = 10.
Final answer≈ 50.3 per year; exponential growth accelerates, so later years exceed the average

MHF4U Rates of Change & Combining Functions — 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 the secant between two points. Instantaneous rate of change is the slope of the tangent at a single point, estimated by shrinking the interval.

Is f(g(x)) the same as g(f(x))?

Almost never. Composition is not commutative — the order changes which function is applied first, and usually gives a completely different result.

More MHF4U practice — grade 12 advanced functions

MHF4U Polynomial FunctionsMHF4U Polynomial Equations & InequalitiesMHF4U Rational FunctionsMHF4U Exponential & Logarithmic FunctionsAll MHF4U units
← Unit 5: Trigonometric FunctionsBack to all assignments →
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