Key Idea: Two ways to combine and reverse functions: a composite feeds one function's output into another, and an inverse undoes a function. Both are Paper 1, by-hand algebra — no GDC.
🔗 Composite: inside-out
- the inner function — do this one first
- the outer function — feed g's output into it
Important: In general f∘g ≠ g∘f — the inner function changes the answer. And when g(x) goes into a square or product, wrap it in brackets: (2x + 1)², not 2x + 1².
🔄 Inverse: swap and solve
The domain of f⁻¹ is the range of f, and they trade places — sometimes you must restrict the domain so f is one-to-one. Quick check: a correct inverse gives f(f⁻¹(x)) = x. Graphically, f⁻¹ is the reflection of f in y = x.
✏️ IB-style worked examples
IB-style question — form a composite expression
Let f(x) = 3x − 4 and g(x) = x². Find (f∘g)(x) and (g∘f)(x).
Step by step:
(f∘g)(x): put g(x) = x² into f.
(g∘f)(x): put f(x) = 3x − 4 into g — use brackets.
Final answer:
(f∘g)(x) = 3x² − 4 ≠ (g∘f)(x) = 9x² − 24x + 16 — order matters.
IB-style question — inverse of a rational function
Find the inverse of f(x) = (3x + 1)/(x − 2).
Step by step:
Set y = f(x), then swap x and y.
Multiply up and expand.
Gather y-terms and factor.
Divide to isolate y.
Final answer:
f⁻¹(x) = (2x + 1)/(x − 3).
IB-style question — inverse with a restricted domain
Let f(x) = x² for x ≥ 0. Find f⁻¹ and state its domain.
Step by step:
Swap and solve — take the positive root, since x ≥ 0.
Domain of f⁻¹ = range of f.
Final answer:
f⁻¹(x) = √x, with domain x ≥ 0.
Important: Don't swap the order. (f∘g)(x) = f(g(x)) does the inner function g first — read it right-to-left. Composing in the wrong order gives a different (wrong) answer.
Tap each card to reveal the answer.
Exam Tips
- (f∘g)(x) = f(g(x)) — do the inner function first; read it right-to-left.
- Evaluate inside-out for a number; substitute g(x) into f (in brackets) for the expression.
- f∘g ≠ g∘f in general — never swap the order.
- Inverse: write y =, swap x ↔ y, solve for y. For fractions, multiply up and gather y-terms.
- Domain and range swap; check with f(f⁻¹(x)) = x; f⁻¹ reflects f in y = x.