Functions: Composite & Inverse

🔢 Maths Years 10–11 Higher

Evaluate functions, build composite functions fg(x), and find inverses by rearranging.

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A function is a rule that takes a number in and gives exactly one number out. Write f(x) = 3x + 2 and it is a machine: feed it 5, it multiplies by 3 and adds 2, so f(5) = 17. In Extended IGCSE you do two new things with these machines — you chain them together (composite functions) and you run them backwards (inverse functions).

Why this matters: function notation is the language of every later topic — graphs, transformations, calculus and coding all lean on it. Cambridge examiners set at least one f(x) question every paper, and marks are lost mostly through getting the order of a composite wrong or forgetting to swap and rearrange for an inverse.

1. Evaluating and reading f(x)

3›g: x + 1gives 4›f: 2x + 3gives 11›11
fg(3): do g first (3 → 4), then f (4 → 11). fg(3) = 11.

2. Composite functions — inner one first

fg(x) means do g first, then feed the result into f. Read it right-to-left, like peeling from the inside out. The order matters: fg(x) is usually not the same as gf(x).

NotationMeaningExample (f: 2x+3, g: x+1)
fg(x)f of g(x) — g firstf(x+1) = 2(x+1)+3 = 2x+5
gf(x)g of f(x) — f firstg(2x+3) = (2x+3)+1 = 2x+4
ff(x)f applied twicef(2x+3) = 2(2x+3)+3 = 4x+9
Worked: f(x) = 2x + 3, g(x) = x + 1. Find fg(x). Replace every x in f with g(x): fg(x) = 2(x + 1) + 3 = 2x + 2 + 3 = 2x + 5. Check with x = 3: g(3) = 4, f(4) = 11, and 2(3) + 5 = 11 ✓.

3. Inverse functions — running the machine backwards

The inverse f⁻¹(x) undoes f: if f turns 6 into 15, then f⁻¹ turns 15 back into 6. To find it, follow three steps.

domain xrange f(x)123579
f: x → 2x + 3 maps 1,2,3 to 5,7,9. The inverse f⁻¹: x → (x − 3)/2 reverses every arrow.

ff⁻¹(x) = x: because an inverse undoes the function, applying one straight after the other lands you back where you started. This is a handy check.

Worked: f(x) = 2x + 3. Write y = 2x + 3. Swap: x = 2y + 3. Rearrange: x − 3 = 2y, so y = (x − 3)/2. Hence f⁻¹(x) = (x − 3)/2. Check: f⁻¹(5) = (5 − 3)/2 = 1, and indeed f(1) = 5 ✓.
Common exam mistake: writing fg(x) as "f times g". It is not multiplication — it is substitution, g first. Another slip: forgetting to swap x and y before rearranging for the inverse.

Exam technique: when the command word is Find fg(x), state the inner function's output explicitly before substituting — examiners award a method mark for showing g(x) fed into f. For an inverse, always show the swap line: it secures the method even if the algebra slips.

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