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NotesMath AA SLTopic 2.2
Unit 2 · Functions · Topic 2.2

IB Math AA SL — Functions, domain & range

Topic 2.2 of IB Mathematics: Analysis and Approaches covers Functions, domain & range, which is part of Unit 2: Functions. Students explore key concepts including Function notation, Domain & range, Inverse as reflection. A strong understanding of functions, domain & range is essential for IB Math AA SL exams and builds the foundation for connected topics across the syllabus.

Exam technique guidePractice questions

Key concepts in Functions, domain & range

Key Idea: This topic is the language of every function question: f(x) names the output, the domain is what you may put in and the range is what comes out, and f⁻¹ undoes f. It's a Paper 1, by-hand skill that underpins almost everything in the functions unit.

🔧 Function notation: f(x)

y=f(x)y = f(x)y=f(x)
xxx
the input — the number you put in
f(x)f(x)f(x)
"f of x" (NOT f × x) — the one output the rule gives
f(a)f(a)f(a)
the output when x = a — substitute a for every x
Important: When you substitute, wrap the value in brackets so signs and powers behave: (−3)² = 9, never −9. The same care applies when the input is a whole expression, e.g. f(2a) = 3(2a) − 5.

🚪 Domain & range

Domain = inputs x, read off the x-axis (left↔right spread). Range = outputs y, read off the y-axis (down↕up spread) — a quadratic turns at its vertex, aˣ stays positive. √0 = 0 is allowed (≥), but a root in a denominator must be > 0. If nothing is banned, the domain is all reals (x ∈ ℝ).

🔄 Inverse: reflect in y = x

Key Idea: The inverse reverses f: f(a) = b ⟺ f⁻¹(b) = a, so its graph is f reflected in the line y = x (each point (a, b) → (b, a)). Find it: (1) write y = f(x), (2) swap x and y, (3) solve for y. Domain and range trade places. Note f⁻¹(x) ≠ 1/f(x) — the −1 is a label, not a power.

✏️ IB-style worked examples

IB-style question — evaluate, then solve

For f(x) = 3x − 5, find f(−2), then solve f(x) = 7.

Step by step:

  1. Evaluate: replace x with −2 in brackets.

    f(−2)=3(−2)−5=−11f(-2) = 3(-2) - 5 = -11f(−2)=3(−2)−5=−11
  2. Solve: set the rule equal to 7.

    3x−5=73x - 5 = 73x−5=7
  3. Rearrange for x.

    3x=12  ⇒  x=43x = 12 \;\Rightarrow\; x = 43x=12⇒x=4
Final answer:

f(−2) = −11, and f(x) = 7 when x = 4.

IB-style question — domain and range

Find the domain of g(x) = √(x − 4), and state the range of h(x) = (x − 1)² + 2.

Step by step:

  1. Domain: what's under the root can't be negative.

    x−4≥0  ⇒  x≥4x - 4 \geq 0 \;\Rightarrow\; x \geq 4x−4≥0⇒x≥4
  2. Range: vertex form a(x − h)² + k has vertex (1, 2), opening up.

    (x−1)2≥0(x-1)^2 \geq 0(x−1)2≥0
  3. So the smallest output is k = 2.

    h(x)≥2h(x) \geq 2h(x)≥2
Final answer:

Domain of g: x ≥ 4. Range of h: y ≥ 2.

IB-style question — find the inverse

Find the inverse of f(x) = 2x + 3.

Step by step:

  1. Write y = f(x).

    y=2x+3y = 2x + 3y=2x+3
  2. Swap x and y.

    x=2y+3x = 2y + 3x=2y+3
  3. Solve for y.

    y=x−32y = \frac{x - 3}{2}y=2x−3​
Final answer:

f⁻¹(x) = (x − 3)/2. Check: f(1) = 5, and f⁻¹(5) = 1. ✓

Important: When you solve f(x) = k for a quadratic (or other curve), two inputs can give the same output — x² = 9 gives x = ±3, not just 3. Write every solution. And before declaring x ∈ ℝ, always check the two bans: denominator ≠ 0 and nothing negative under an even root.

Tap each card to reveal the answer.

Exam Tips

  • Substitute in brackets so signs and powers behave: (−3)² = 9, never −9.
  • Evaluate goes input → output; solving f(x) = k goes backwards — and may have more than one answer.
  • Check the two domain bans before x ∈ ℝ: denominator ≠ 0, and nothing negative under an even root (√0 is OK, ≥).
  • Read domain off the x-axis (left↔right), range off the y-axis (down↕up); a quadratic's range starts at its vertex.
  • Inverse: swap x and y then solve; its graph is f reflected in y = x, and domain/range swap.

What you'll learn in Topic 2.2

  • 2.2.1 Function notation
  • 2.2.2 Domain & range
  • 2.2.3 Inverse as reflection
Suggested study order: Read the notes for each sub-topic below → test yourself with flashcards → attempt practice questions → review exam technique.

Study resources — 2.2 Functions, domain & range

2.2.1

Function notation

Notes
2.2.2

Domain & range

Notes
2.2.3

Inverse as reflection

Notes

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Topic 2.2 Functions, domain & range forms a core part of Unit 2: Functions in IB Math AA SL. Mastering these concepts will strengthen your understanding of connected topics across the syllabus and prepare you for exam questions that require analysis, evaluation, and real-world application.

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