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)
- the input — the number you put in
- "f of x" (NOT f × x) — the one output the rule gives
- the output when x = a — substitute a for every x
| Going forwards — evaluate | Going backwards — solve f(x) = k |
|---|---|
| You're given the input x | You're given the output k |
| Replace every x with the value, in brackets, then simplify. | Set the rule equal to k and solve for x. |
| f(x) = 2x + 1 → f(3) = 2(3) + 1 = 7 | 2x + 1 = 9 → x = 4 |
| One input → one output. | May have more than one answer (e.g. x² = 9 → x = ±3). |
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
| Restriction | Rule | Example → domain |
|---|---|---|
| Division | Denominator ≠ 0 | $\frac{1}{x-3}$ → x ≠ 3 |
| Even root (√) | What's inside ≥ 0 | √(x − 2) → x ≥ 2 |
| Logarithm | Argument > 0 | ln(x) → x > 0 |
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:
Evaluate: replace x with −2 in brackets.
Solve: set the rule equal to 7.
Rearrange for x.
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:
Domain: what's under the root can't be negative.
Range: vertex form a(x − h)² + k has vertex (1, 2), opening up.
So the smallest output is k = 2.
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:
Write y = f(x).
Swap x and y.
Solve for y.
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.
For f(x) = x² − 4x, find f(−3) 21 — f(−3) = (−3)² − 4(−3) = 9 + 12. Brackets keep the square positive.
Solve f(x) = 6 for f(x) = x² − 3 x = ±3 — x² = 9, so both roots count.
What's the domain of 1/(x − 3)? All real x except x = 3 — the denominator can't be zero.
State the range of f(x) = 2ˣ y > 0 — an exponential is always positive, approaching but never reaching 0.
Given f(2) = 7, what is f⁻¹(7)? 2 — the inverse sends the output back to its input.
How do you find f⁻¹ algebraically? Write y = f(x), swap x and y, then solve for y.
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.