Mathematics Advanced • Year 11 • Module 1 • Lesson 2
Function Notation & Evaluation
Build fluency with f(x) notation, careful bracket substitution, and the difference quotient.
1. Quick recall
Answer in the space provided. 1 mark each
Q1.1 The expression f(−2) is read aloud as ______________________________________ (not "f times negative two").
Q1.2 Write the difference quotient for a function f at x = a, step h:
[ __________________ − __________________ ] / __________
Q1.3 If f(x) = x², write f(x + h) fully expanded. f(x + h) = __________________________
2. Worked example — f(x) = 2x² − 3x + 1
Find f(2), f(−1) and f(a). Follow each line.
Problem. Evaluate f(2), f(−1), and f(a) for f(x) = 2x² − 3x + 1.
Step 1 — Substitute x = 2 with brackets.
f(2) = 2(2)² − 3(2) + 1 = 2(4) − 6 + 1 = 8 − 6 + 1 = 3
Reason: brackets protect (2)² so the exponent applies to the whole input.
Step 2 — Substitute x = −1 with brackets.
f(−1) = 2(−1)² − 3(−1) + 1 = 2(1) + 3 + 1 = 6
Reason: (−1)² = +1; without brackets you'd get −1²=−1 — wrong sign.
Step 3 — Substitute x = a (algebraic input).
f(a) = 2a² − 3a + 1
Reason: every x is replaced by a; no further simplification possible without a value.
Conclusion. f(2) = 3, f(−1) = 6, f(a) = 2a² − 3a + 1.
3. Faded example — difference quotient
For f(x) = x² + 2x, find the difference quotient [f(x + h) − f(x)] / h in simplest form. Fill in each blank. 4 marks
Step 1 — Expand f(x + h):
f(x + h) = (x + h)² + 2(x + h)
= ______________ + ______________ + ______________ + ______________ + 2h
Step 2 — Subtract f(x):
f(x + h) − f(x) = (the expansion above) − (x² + 2x)
= ______________ + ______________ + ______________
Step 3 — Divide by h and simplify:
[f(x + h) − f(x)] / h = ______________
Conclusion. The difference quotient is ______________________.
4. Graduated practice — evaluate the function
Substitute carefully, using brackets every time. Show the substitution line and the simplified result.
Foundation — single integer or simple algebraic input (4 questions)
| Q | Function | Find | Answer |
|---|---|---|---|
| 4.1 1 | f(x) = 4x + 7 | f(3) | |
| 4.2 1 | g(x) = x² − 5 | g(−3) | |
| 4.3 1 | h(t) = 2 − t/3 | h(6) | |
| 4.4 1 | C(d) = 5 + 1.8d | C(10) |
Standard — typical HSC difficulty (6 questions)
Show one line of substitution and one of simplification for each.
4.5 If f(x) = 3x² − 2x + 4, find f(2) and f(−1). 2 marks
4.6 If g(x) = (x − 1)/(x + 2), find g(3) and explain why g(−2) is not defined. 2 marks
4.7 If f(x) = 2x + 1, write expressions for f(a), f(a + h) and f(a) + h in simplest form. 2 marks
4.8 For the taxi function C(d) = 5 + 1.8d, find C(15), C(30), and interpret C(0) in one sentence. 2 marks
4.9 If f(x) = √(x − 4), find f(13). State a value of x for which f is not defined and why. 2 marks
4.10 If f(x) = x² − 4x, solve f(x) = 0. (Hint: factorise.) 2 marks
Extension — difference quotient and algebraic substitution (2 questions)
4.11 For f(x) = x² + 2x, simplify the difference quotient [f(x + h) − f(x)] / h. Then evaluate at x = 3, h = 0.01. 3 marks
4.12 Suppose f is any linear function f(x) = mx + c. Prove that the difference quotient [f(x + h) − f(x)] / h equals m, independent of x and h. 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Reading f(−2)
Read as "f of negative two" (the output of function f when the input is −2). Not "f times negative two".
Q1.2 — Difference quotient
[ f(a + h) − f(a) ] / h.
Q1.3 — f(x + h) for f(x) = x²
f(x + h) = (x + h)² = x² + 2xh + h². The middle term 2xh is the one most often missed.
Q3 — Difference quotient for f(x) = x² + 2x
Step 1: f(x + h) = (x + h)² + 2(x + h) = x² + 2xh + h² + 2x + 2h.
Step 2: f(x + h) − f(x) = 2xh + h² + 2h.
Step 3: Dividing by h: 2x + h + 2.
Difference quotient: 2x + h + 2.
Q4.1 — f(3) for f(x) = 4x + 7
f(3) = 4(3) + 7 = 19.
Q4.2 — g(−3) for g(x) = x² − 5
g(−3) = (−3)² − 5 = 9 − 5 = 4.
Q4.3 — h(6) for h(t) = 2 − t/3
h(6) = 2 − 6/3 = 2 − 2 = 0.
Q4.4 — C(10) for C(d) = 5 + 1.8d
C(10) = 5 + 1.8(10) = 5 + 18 = $23.
Q4.5 — f(2), f(−1) for f(x) = 3x² − 2x + 4
f(2) = 3(2)² − 2(2) + 4 = 12 − 4 + 4 = 12.
f(−1) = 3(−1)² − 2(−1) + 4 = 3 + 2 + 4 = 9.
Q4.6 — g(3) and g(−2) for g(x) = (x − 1)/(x + 2)
g(3) = (3 − 1)/(3 + 2) = 2/5. g(−2) is undefined because the denominator −2 + 2 = 0 (division by zero).
Q4.7 — f(a), f(a + h), f(a) + h for f(x) = 2x + 1
f(a) = 2a + 1. f(a + h) = 2(a + h) + 1 = 2a + 2h + 1. f(a) + h = (2a + 1) + h = 2a + 1 + h. Note: f(a + h) ≠ f(a) + h in general.
Q4.8 — Taxi function
C(15) = 5 + 1.8(15) = $32. C(30) = 5 + 1.8(30) = $59. C(0) = $5: the flag-fall paid before any distance is travelled.
Q4.9 — f(13) for f(x) = √(x − 4)
f(13) = √(13 − 4) = √9 = 3. Any x < 4 makes the radicand negative; f is undefined on (−∞, 4) over the reals.
Q4.10 — Solve f(x) = 0 for f(x) = x² − 4x
x² − 4x = x(x − 4) = 0 ⇒ x = 0 or x = 4.
Q4.11 — Difference quotient for f(x) = x² + 2x at x = 3, h = 0.01
From Q3 the simplified quotient is 2x + h + 2. At x = 3, h = 0.01: 2(3) + 0.01 + 2 = 8.01. (This estimates the slope of f near x = 3; in Year 12 it will become f ′(3) = 8.)
Q4.12 — Proof: difference quotient of linear f equals m
Let f(x) = mx + c.
f(x + h) = m(x + h) + c = mx + mh + c.
f(x + h) − f(x) = (mx + mh + c) − (mx + c) = mh.
[f(x + h) − f(x)] / h = mh / h = m (for h ≠ 0).
The result m is independent of x and h, confirming that linear functions have a constant rate of change equal to the slope m. ▮