Mathematics Advanced • Year 11 • Module 3 • Lesson 1
Average & Instantaneous Rates of Change
Build procedural fluency in the difference quotient — calculating average rates from tables, equations and shrinking intervals.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete the formula for the average rate of change of a function f over the interval [a, b]:
Average rate = ( ____________ − ____________ ) ÷ ( ____________ − ____________ )
Q1.2 A straight line cutting a curve at two distinct points is called a ____________. A straight line touching the curve at a single point is called a ____________.
Q1.3 True or false (circle one):
(a) Average rate of change equals the gradient of a secant. T / F
(b) Instantaneous rate of change equals the gradient of a tangent. T / F
(c) You can compute an instantaneous rate using just one (x, y) value. T / F
2. Worked example — average rate of f(x) = x² on [1, 3]
Follow each line of algebra. Every step has a reason on the right.
Problem. Find the average rate of change of f(x) = x² over the interval [1, 3].
Step 1 — Evaluate the function at each endpoint.
f(1) = (1)² = 1
f(3) = (3)² = 9
Reason: we need f(b) and f(a) before applying the difference quotient.
Step 2 — Substitute into the difference quotient.
Average rate = ( f(3) − f(1) ) ÷ ( 3 − 1 )
= ( 9 − 1 ) ÷ ( 3 − 1 ) = 8 ÷ 2
Reason: keep the same order in numerator and denominator (b first, a second).
Step 3 — Simplify.
Average rate = 4
Reason: y changes by 8 while x changes by 2 — the secant has gradient 4.
Conclusion. The average rate of change of f(x) = x² over [1, 3] is 4 units of y per unit of x.
3. Faded example — fill in the missing steps
Find the average rate of change of f(x) = 2x² + 1 over the interval [−1, 2]. Fill in each blank. 3 marks
Step 1 — Evaluate at the endpoints:
f(−1) = 2(−1)² + 1 = ____________
f(2) = 2(2)² + 1 = ____________
Step 2 — Apply the difference quotient:
Average rate = ( ____________ − ____________ ) ÷ ( 2 − (−1) )
= ____________ ÷ ____________
Conclusion. Average rate of change = ____________ .
4. Graduated practice — calculate the average rate of change
Use the difference quotient f(b) − f(a) over b − a. Show one line of substitution and one line of arithmetic for each.
Foundation — clean numbers (4 questions)
| Q | Function | Interval | Average rate |
|---|---|---|---|
| 4.1 1 | f(x) = 3x + 2 | [0, 4] | |
| 4.2 1 | f(x) = x² | [2, 5] | |
| 4.3 1 | f(x) = 5x − 7 | [1, 6] | |
| 4.4 1 | f(x) = x² | [−3, 3] |
Standard — typical HSC difficulty (6 questions)
Show f(a), f(b) and the substitution into the difference quotient.
4.5 The temperature of a chemical solution at time t hours is T(t) = 15 + 7t − 0.5t². Find the average rate of change of T over [0, 4]. 2 marks
4.6 A runner's position is s(t) = t² metres at time t seconds. Find the average speed over [0, 3]. 2 marks
4.7 f(x) = x³ on [1, 2]. 2 marks
4.8 Distance d(t) for a runner, in metres at time t seconds, is given by the table:
t = 0, 1, 2, 3 → d = 0, 3, 8, 15
Find the average speed over (a) [0, 1], (b) [0, 2], (c) [0, 3]. 2 marks
4.9 f(x) = 1/x (x ≠ 0) on [1, 4]. 2 marks
4.10 Population P(t) = 200 + 30t − t² (thousands of bacteria after t hours). Find the average growth rate over [2, 6]. 2 marks
Extension — shrinking intervals (2 questions)
4.11 For s(t) = t², compute the average speed over each of these intervals and explain what value the speeds appear to converge to as the interval shrinks. 3 marks
[2, 2.5], [2, 2.1], [2, 2.01], [2, 2.001]
4.12 For f(x) = x², simplify ( f(2 + h) − f(2) ) ÷ h as far as possible without letting h = 0. State what happens to the expression as h is made very small, and what number the instantaneous rate at x = 2 must therefore equal. 3 marks
5. Self-check the easy 3
Tick the first three once you have checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Average rate formula
Average rate = ( f(b) − f(a) ) ÷ ( b − a ). Same order top and bottom.
Q1.2 — Secant vs tangent
Two points: secant. Single point: tangent.
Q1.3 — True/false
(a) T. (b) T. (c) F — a rate requires change, and change requires two values.
Q3 — Faded example f(x) = 2x² + 1 on [−1, 2]
f(−1) = 2(1) + 1 = 3; f(2) = 2(4) + 1 = 9.
Average rate = ( 9 − 3 ) ÷ ( 2 − (−1) ) = 6 ÷ 3 = 2.
Q4.1 — f(x) = 3x + 2 on [0, 4]
f(0) = 2; f(4) = 14. Average rate = (14 − 2) / (4 − 0) = 12 / 4 = 3. (For a linear function the average rate is always the gradient — here, 3.)
Q4.2 — f(x) = x² on [2, 5]
f(2) = 4; f(5) = 25. Average rate = (25 − 4) / (5 − 2) = 21 / 3 = 7.
Q4.3 — f(x) = 5x − 7 on [1, 6]
f(1) = −2; f(6) = 23. Average rate = (23 − (−2)) / (6 − 1) = 25 / 5 = 5. (Again, equals the slope.)
Q4.4 — f(x) = x² on [−3, 3]
f(−3) = 9; f(3) = 9. Average rate = (9 − 9) / (3 − (−3)) = 0 / 6 = 0. (The endpoints are at the same height, so the secant is horizontal.)
Q4.5 — Temperature T(t) = 15 + 7t − 0.5t² on [0, 4]
T(0) = 15; T(4) = 15 + 28 − 8 = 35. Average rate = (35 − 15) / 4 = 20 / 4 = 5 °C/h.
Q4.6 — Runner s(t) = t² on [0, 3]
s(0) = 0; s(3) = 9. Average speed = (9 − 0) / (3 − 0) = 3 m/s.
Q4.7 — f(x) = x³ on [1, 2]
f(1) = 1; f(2) = 8. Average rate = (8 − 1) / (2 − 1) = 7.
Q4.8 — Runner table
(a) [0, 1]: (3 − 0)/1 = 3 m/s. (b) [0, 2]: (8 − 0)/2 = 4 m/s. (c) [0, 3]: (15 − 0)/3 = 5 m/s. (The runner is accelerating — average speed over longer intervals is larger.)
Q4.9 — f(x) = 1/x on [1, 4]
f(1) = 1; f(4) = 1/4. Average rate = (1/4 − 1) / (4 − 1) = (−3/4) / 3 = −1/4. (Negative: f is decreasing on this interval.)
Q4.10 — Population P(t) = 200 + 30t − t² on [2, 6]
P(2) = 200 + 60 − 4 = 256 (thousand). P(6) = 200 + 180 − 36 = 344 (thousand). Average rate = (344 − 256) / (6 − 2) = 88 / 4 = 22 thousand bacteria per hour.
Q4.11 — Shrinking intervals for s(t) = t² at t = 2
[2, 2.5]: (6.25 − 4)/0.5 = 4.5. [2, 2.1]: (4.41 − 4)/0.1 = 4.1. [2, 2.01]: (4.0401 − 4)/0.01 = 4.01. [2, 2.001]: (4.004001 − 4)/0.001 = 4.001. The average speeds converge to 4 m/s — the instantaneous speed at t = 2.
Q4.12 — Algebraic shrinking for f(x) = x² at x = 2
f(2 + h) − f(2) = (2 + h)² − 4 = 4 + 4h + h² − 4 = 4h + h².
Divide by h (valid for h ≠ 0): (4h + h²) / h = 4 + h.
As h is made very small (h → 0), the expression 4 + h approaches 4. Hence the instantaneous rate of change of x² at x = 2 must be 4. (This is the same value the numerical table in 4.11 converged to — and it foreshadows the formal definition of the derivative.)