Mathematics Advanced • Year 11 • Module 1 • Lesson 2

Function Notation & Evaluation

Build fluency with f(x) notation, careful bracket substitution, and the difference quotient.

Build · Skill Drill

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) = __________________________

Stuck? Revisit lesson § Formula Reference and § Algebraic Inputs.

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 ______________________.

Stuck? Revisit lesson § Worked Example 3, The Difference Quotient.

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
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

Stuck on 4.12? Substitute the rule into f(x + h), subtract f(x), divide by h, and watch the c-terms cancel.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 I wrote f(3) = 4(3) + 7 with brackets, then simplified to 19.

For 4.2 I wrote g(−3) = (−3)² − 5 = 9 − 5 = 4, using brackets so the square doesn't drop the sign.

For 4.4 I checked C(10) = 5 + 1.8(10) = 23 and read it as "$23 for a 10 km trip".

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers, Do not peek before attempting

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. ▮