Mathematics Advanced • Year 11 • Module 1 • Lesson 3

Domain & Range

Build fluency in interval notation, the three classic domain restrictions, and finding ranges of basic functions.

Build · Skill Drill

1. Quick recall

Answer in the space provided. 1 mark each

Q1.1 Complete: the domain is the set of all allowed ______-values and the range is the set of all possible ______-values.

Q1.2 List the three classic domain restrictions covered in Year 11:

1) Denominator: ____________ 2) Radicand (under √): ____________ 3) Log argument: ____________

Q1.3 Why must we always use a round bracket next to ∞ or −∞? __________________________________________________

Stuck? Revisit lesson § Domain & Interval Notation.

2. Worked example — domain of f(x) = 5 / (x − 3)

Follow each line.

Problem. Find the domain of f(x) = 5 / (x − 3).

Step 1 — Identify the restriction.

The denominator cannot equal zero.

Step 2 — Solve for the excluded value.

x − 3 ≠ 0 ⇒ x ≠ 3

Step 3 — Write the domain as a union of intervals.

Domain = (−∞, 3) ∪ (3, ∞)

Reason: all reals except 3; round brackets at both ∞ and at the excluded 3.

Conclusion. Domain: (−∞, 3) ∪ (3, ∞).

3. Faded example — domain of a square-root function

Find the domain of f(x) = √(2x + 6). Fill in each blank. 3 marks

Step 1 — State the restriction. For a real-valued square root, the radicand must be __________ 0.

2x + 6 ____ 0

Step 2 — Solve the inequality:

2x ____ −6 ⇒ x ____ ____

Step 3 — Write in interval notation. Because the endpoint ______ included (radicand can equal 0), use a ______ bracket.

Domain = [ ____ , ____ )

Conclusion. Domain: ______________.

Stuck? Revisit lesson § Worked Example 2 — Square Root.

4. Graduated practice — find the domain (and range where stated)

Write your answers in interval notation. Use [ ] for included endpoints, ( ) for excluded.

Foundation — single restriction (4 questions)

Q	Function	Domain (interval notation)
4.1 1	f(x) = 3x + 1
4.2 1	f(x) = 1 / x
4.3 1	f(x) = √(x − 2)
4.4 1	f(x) = log_e(x)

Standard — typical HSC difficulty (6 questions)

State the restriction(s) you used, then the interval(s).

4.5 Find the domain of f(x) = 7 / (x² − 9). 2 marks

4.6 Find the domain of f(x) = √(6 − 2x). 2 marks

4.7 Find the domain and range of f(x) = x² − 4x + 5. 3 marks

4.8 Find the domain of f(x) = log_e(x + 2). 2 marks

4.9 Find the range of f(x) = |x − 1| + 3. 2 marks

4.10 Find the domain of f(x) = 1 / √(x − 4). (Two restrictions.) 2 marks

Extension — combined / harder (2 questions)

4.11 Find the domain of f(x) = √(x² − 9). Show how you handled the quadratic inequality. 3 marks

4.12 Find the domain and range of f(x) = √(9 − x²). Identify the geometric curve and use it to justify the range. 3 marks

Stuck on 4.11/4.12? Solve x² ≥ 9 by factorising, or recognise upper-semicircle.

5. Self-check the easy 3

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

For 4.1 I wrote the domain of any non-piecewise polynomial as (−∞, ∞), round brackets at both infinities.

For 4.2 I excluded only x = 0, not x < 0. The natural domain of 1/x is (−∞, 0) ∪ (0, ∞).

For 4.3 I used a [ at 2 because √0 = 0 is defined.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Definitions

Domain = allowed x-values. Range = possible y-values.

Q1.2 — Three classic restrictions

1) Denominator ≠ 0. 2) Radicand ≥ 0 (for √). 3) Log argument > 0.

Q1.3 — Round bracket at ∞

Infinity is not a real number, so it can never be "reached" or "included" as an endpoint. We always use a round bracket: (a, ∞), (−∞, b], etc.

Q3 — Faded square-root example

Radicand must be ≥ 0. 2x + 6 ≥ 0 ⇒ 2x ≥ −6 ⇒ x ≥ −3. Endpoint is included (radicand can equal 0), use a [ bracket. Domain = [−3, ∞).

Q4.1 — Linear f(x) = 3x + 1

Polynomial; no restrictions. Domain = (−∞, ∞).

Q4.2 — f(x) = 1/x

x ≠ 0. Domain = (−∞, 0) ∪ (0, ∞).

Q4.3 — f(x) = √(x − 2)

x − 2 ≥ 0 ⇒ x ≥ 2. Domain = [2, ∞).

Q4.4 — f(x) = ln(x)

Argument > 0 ⇒ x > 0. Domain = (0, ∞).

Q4.5 — f(x) = 7/(x² − 9)

x² − 9 ≠ 0 ⇒ (x − 3)(x + 3) ≠ 0 ⇒ x ≠ ±3. Domain = (−∞, −3) ∪ (−3, 3) ∪ (3, ∞).

Q4.6 — f(x) = √(6 − 2x)

6 − 2x ≥ 0 ⇒ −2x ≥ −6 ⇒ x ≤ 3 (sign flipped on division by −2). Domain = (−∞, 3].

Q4.7 — f(x) = x² − 4x + 5

Polynomial, domain = (−∞, ∞). Vertex at x = −b/(2a) = 4/2 = 2; f(2) = 4 − 8 + 5 = 1. Coefficient of x² is positive, parabola opens up, so range = [1, ∞).

Q4.8 — f(x) = ln(x + 2)

x + 2 > 0 ⇒ x > −2. Domain = (−2, ∞).

Q4.9 — Range of f(x) = |x − 1| + 3

|x − 1| ≥ 0 for all x, with minimum 0 at x = 1. So f ≥ 3, with minimum 3 attained. Range = [3, ∞).

Q4.10 — f(x) = 1 / √(x − 4)

Two restrictions: (a) radicand ≥ 0 ⇒ x ≥ 4; (b) denominator ≠ 0 ⇒ √(x − 4) ≠ 0 ⇒ x ≠ 4. Combined: x > 4. Domain = (4, ∞).

Q4.11 — f(x) = √(x² − 9)

x² − 9 ≥ 0 ⇒ (x − 3)(x + 3) ≥ 0. Sign analysis: positive when both factors share the same sign. x ≤ −3 or x ≥ 3. Domain = (−∞, −3] ∪ [3, ∞).

Q4.12 — f(x) = √(9 − x²)

9 − x² ≥ 0 ⇒ −3 ≤ x ≤ 3 ⇒ Domain = [−3, 3]. Geometrically, y = √(9 − x²) is the upper semicircle of x² + y² = 9 (only the y ≥ 0 half). The minimum y is 0 (at x = ±3) and the maximum is 3 (at x = 0). Range = [0, 3].