Functions: notation, inverse and composite [Higher tier]
A function is a rule that assigns each input exactly one output. In GCSE we use the notation f(x) (read "f of x"). Higher-tier candidates need to evaluate, compose and invert these.
Notation and evaluation
If f(x) = 2x + 3, then f(5) = 2(5) + 3 = 13. The letter inside the brackets just tells you what to substitute. f(a + 1) = 2(a + 1) + 3 = 2a + 5.
Composite functions: fg(x)
fg(x) means "do g first, then apply f to the result". Read it right to left.
If f(x) = 2x + 3 and g(x) = x², then:
fg(x) = f(g(x)) = f(x²) = 2x² + 3gf(x) = g(f(x)) = g(2x + 3) = (2x + 3)²
These are different — composition is not commutative.
Inverse function: f⁻¹(x)
The inverse "undoes" f. To find it:
- Write
y = f(x). - Swap x and y.
- Solve for y. That y is
f⁻¹(x).
Example: f(x) = 3x - 4. Write y = 3x - 4; swap → x = 3y - 4; solve → y = (x + 4)/3. So f⁻¹(x) = (x + 4)/3.
Check: f(f⁻¹(x)) = 3 × (x + 4)/3 - 4 = x + 4 - 4 = x. ✓ Always verify.
When does an inverse exist?
For a function to have an inverse it must be one-to-one — each output corresponds to exactly one input. f(x) = x² is not one-to-one over all reals (4 has square-root inputs +2 and -2), so it has no inverse unless its domain is restricted (e.g. x ≥ 0).
✦Worked example— Worked example — chained composite
f(x) = x + 2, g(x) = 3x. Find fg(4).
Method: g(4) = 12; f(12) = 14. So fg(4) = 14.
Or symbolically: fg(x) = f(3x) = 3x + 2; substitute 4: 14.
✦Worked example— Worked example — inverse of a composite
If f(x) = 5x and g(x) = x - 7, find (fg)⁻¹(x).
fg(x) = 5(x - 7) = 5x - 35. To invert: y = 5x - 35; swap; solve.
x = 5y - 35 → y = (x + 35)/5. So (fg)⁻¹(x) = (x + 35)/5.
Useful identity: (fg)⁻¹(x) = g⁻¹f⁻¹(x) — the inverses compose in the opposite order. (Like undressing: socks then shoes off.)
⚠Common mistakes— Common mistakes (examiner traps)
- Composing left to right.
fg(x)means do g first, then f. Right-to-left. - Forgetting to swap when finding the inverse — students often "rearrange y = ..." without exchanging variables.
- Treating
f⁻¹as1/f. It is not a reciprocal — it is the inverse function.f⁻¹(2)is the input that gives output 2, not1/f(2). - Squaring without keeping ± when inverting an x² function. State the domain restriction.
- Confusing
fg(x)withf(x) × g(x).fgis composition (a function), not a product.
➜Try this— Quick check
f(x) = 4 - x. Find f⁻¹(x).
Swap and solve: x = 4 - y ⇒ y = 4 - x. So f⁻¹(x) = 4 - x — this function is its own inverse.
AI-generated · claude-opus-4-7 · v3-deep-algebra