IGCSE Maths · Algebra and Graphs · Extended only

Functions: Notation, Domain, Range, Inverses and Composites Extended only

This entire note is Extended-only, and it’s the final note in the Algebra and Graphs unit. A function is really just a rule for turning an input into an output — f(x) notation just makes that rule easy to name, evaluate, reverse, and chain together with other functions.

Why this matters Every formula you’ve used across this whole unit — substitution, rearranging, differentiating — has secretly been function work already. This note gives you the formal language and notation for it, plus two genuinely new skills: reversing a function (inverse) and chaining two functions together (composite).

1. Function Notation

f(x) means “the rule f, applied to x.” To evaluate f at a specific value, substitute that value in wherever x appears — exactly like substitution from earlier in the unit.

Worked example f(x) = 3x − 5. Find f(4) and f(−2).
f(4) = 3(4) − 5 = 7
f(−2) = 3(−2) − 5 = −11
Worked example — working backwards h(x) = 2x² + 3. Find the value(s) of x for which h(x) = 21.
2x² + 3 = 21 → 2x² = 18 → x² = 9 → x = 3 or x = −3
Try it — Function Notation
  1. f(x) = 4x + 1. Find f(3) and f(−5).
  2. g(x) = x² − 4. Find g(5) and g(−2).
  3. f(x) = 2x − 7. Find x when f(x) = 9.
Show answers
  1. f(3) = 13, f(−5) = −19
  2. g(5) = 21, g(−2) = 0
  3. 2x − 7 = 9 → x = 8

2. Domain and Range

The domain is the complete set of allowed inputs (x-values); the range is the resulting set of outputs (y-values, or f(x)-values). Most restrictions come from two sources: a denominator can never be zero, and a square root can never be negative.

Worked examples f(x) = 3x − 5. This is defined for every real number, so the domain is all real numbers, and so is the range.

h(x) = 2x² + 3. The domain is all real numbers, but since x² is never negative, h(x) is always at least 3.
Range: h(x) ⩾ 3

f(x) = 3/(x + 2). This is undefined when the denominator is 0.
Domain: x ≠ −2
Try it — Domain and Range
  1. State the domain of f(x) = 5/(x − 3).
  2. State the range of g(x) = x² + 1.
  3. State the domain of f(x) = 2x + 7.
Show answers
  1. x ≠ 3
  2. g(x) ⩾ 1
  3. All real numbers (no restriction on a simple linear function).

3. Inverse Functions

The inverse function f⁻¹(x) reverses f — if f turns a into b, f⁻¹ turns b back into a. Method: write y = f(x), swap x and y, then rearrange to make y the subject again (exactly like changing the subject of a formula).

Worked example Find the inverse of f(x) = 3x − 5.
Write as y = 3x − 5
Swap x and y: x = 3y − 5
Rearrange for y: x + 5 = 3y → y = (x + 5)/3
Answer: f⁻¹(x) = (x + 5)/3
Check: f⁻¹(f(4)) should return 4. f(4) = 7, and f⁻¹(7) = (7+5)/3 = 4 ✓
Worked example Find the inverse of g(x) = 3(x + 4)/5.
y = 3(x + 4)/5
Swap: x = 3(y + 4)/5
Rearrange: 5x = 3(y + 4) → 5x/3 = y + 4 → y = 5x/3 − 4
Answer: g⁻¹(x) = 5x/3 − 4
Try it — Inverse Functions
  1. Find the inverse of f(x) = 2x + 6.
  2. Find the inverse of f(x) = (x − 3)/4.
  3. Find the inverse of f(x) = 5 − 2x.
Show answers
  1. y=2x+6 → x=2y+6 → y=(x−6)/2. f⁻¹(x) = (x−6)/2
  2. y=(x−3)/4 → x=(y−3)/4 → 4x=y−3 → y=4x+3. f⁻¹(x) = 4x+3
  3. y=5−2x → x=5−2y → 2y=5−x → y=(5−x)/2. f⁻¹(x) = (5−x)/2

4. Composite Functions

A composite function chains two functions together: gf(x) means g(f(x)) — apply f first, then apply g to the result. Order matters: gf(x) and fg(x) are generally different.

gf(x) = g(f(x))
Worked example f(x) = 3x − 5, g(x) = 3(x + 4)/5. Find gf(x).
gf(x) = g(f(x)) — substitute f(x) = 3x − 5 into g in place of x:
g(3x − 5) = 3((3x − 5) + 4)/5 = 3(3x − 1)/5 = (9x − 3)/5
Worked example — the other order, for comparison Using the same f and g, find fg(x).
fg(x) = f(g(x)) — substitute g(x) = 3(x+4)/5 into f in place of x:
f(3(x+4)/5) = 3 × [3(x+4)/5] − 5 = 9(x+4)/5 − 5 = (9x + 36)/5 − 25/5 = (9x + 11)/5
Notice this is different from gf(x) = (9x − 3)/5 — confirming order matters.
Worked example — a quadratic composite f(x) = 3x − 5, h(x) = 2x² + 3. Find hf(x).
hf(x) = h(f(x)) = h(3x − 5) = 2(3x − 5)² + 3
Expand (3x−5)² = 9x² − 30x + 25
hf(x) = 2(9x² − 30x + 25) + 3 = 18x² − 60x + 50 + 3 = 18x² − 60x + 53
Try it — Composite Functions
  1. f(x) = 2x + 1, g(x) = x − 3. Find gf(x).
  2. f(x) = 2x + 1, g(x) = x − 3. Find fg(x), and confirm it’s different from gf(x).
  3. f(x) = x + 2, h(x) = x² − 1. Find hf(x).
Show answers
  1. gf(x) = g(2x+1) = (2x+1) − 3 = 2x − 2
  2. fg(x) = f(x−3) = 2(x−3)+1 = 2x−5. This differs from gf(x) = 2x−2.
  3. hf(x) = h(x+2) = (x+2)² − 1 = x²+4x+4−1 = x²+4x+3

5. Composite Functions with Algebraic Fractions

These combine composite functions with the algebraic fraction skills from earlier in the unit — substitute carefully, and simplify fully at the end.

Worked example f(x) = 3/(x + 2), g(x) = (3x + 5)². Find fg(x), giving your answer as a fraction in its simplest form.
fg(x) = f(g(x)) — substitute g(x) in place of x in f:
f((3x+5)²) = 3 / [(3x + 5)² + 2]
This is already in simplest form — the denominator doesn’t factor with the numerator.
Try it — Composite Functions with Fractions
  1. f(x) = 2/(x − 1), g(x) = x + 3. Find fg(x).
  2. f(x) = 4/x, g(x) = x² + 1. Find fg(x).
Show answers
  1. fg(x) = f(x+3) = 2/((x+3)−1) = 2/(x+2)
  2. fg(x) = f(x²+1) = 4/(x²+1)

6. Mapping Diagrams

A mapping diagram shows a function as arrows connecting each input directly to its output — useful for functions defined on a small, specific set of values rather than all real numbers.

Example — f(x) = x² for x ∈ {1, 2, 3} Domain Range 1 2 3 1 4 9 Each input (1, 2, 3) maps to exactly one output (1, 4, 9) via the rule x². This is what “function” formally means: every input has exactly one output.
Common mistakes to watch for
  • Confusing gf(x) with fg(x) — gf(x) means “apply f first, then g,” reading right to left, which trips people up constantly.
  • Forgetting to swap x and y when finding an inverse — that swap is the entire point of the method.
  • Not simplifying a composite function fully, especially after expanding a bracket like (3x−5)².
  • Assuming every function has an inverse that’s also a function — not required at this level, but worth knowing this isn’t always guaranteed in general.
  • Mixing up domain (inputs) and range (outputs) — domain restricts x, range restricts the result.

7. Quick-Fire Challenge Round

True or False?
  1. f(3) means “solve f(x) = 3.”
  2. gf(x) always equals fg(x).
  3. To find an inverse function, you swap x and y and rearrange.
  4. The domain of f(x) = 5/(x+1) excludes x = −1.
Show answers
  1. False — f(3) means “substitute 3 into f,” not “solve for x.”
  2. False — order generally matters; gf(x) and fg(x) are usually different.
  3. True.
  4. True — that value would make the denominator zero.
Spot the Error
  1. “f(x) = 2x − 3. To find f(5), I solved 2x − 3 = 5 to get x = 4.”
  2. “gf(x) means apply g first, then f.”
  3. “Inverse of f(x) = 4x: f⁻¹(x) = 4x.”
Show answers
  1. f(5) means substitute x=5 directly: f(5) = 2(5)−3 = 7, not solve an equation.
  2. gf(x) = g(f(x)) means apply f first, then g — the function closest to x is applied first, reading right to left.
  3. The inverse should undo the original: if f multiplies by 4, f⁻¹ should divide by 4: f⁻¹(x) = x/4, not 4x.

8. Mixed Practice — Bringing It All Together

  1. f(x) = 5x − 2. Find f(3).
  2. State the domain of f(x) = 7/(x − 5).
  3. Find the inverse of f(x) = 4x − 8.
  4. f(x) = x + 1, g(x) = 2x². Find gf(x).
  5. Using the same f and g, find fg(x).
Show answers
  1. 13
  2. x ≠ 5
  3. y=4x−8 → x=4y−8 → y=(x+8)/4. f⁻¹(x) = (x+8)/4
  4. gf(x) = g(x+1) = 2(x+1)²
  5. fg(x) = f(2x²) = 2x²+1

Practice at Home

A large set again, to close out the Algebra and Graphs unit.

Part A — Function notation
  1. f(x) = 6x − 1. Find f(2) and f(−3).
  2. g(x) = x² + 5. Find g(4) and g(−4).
  3. f(x) = 3x + 4. Find x when f(x) = 19.
Part B — Domain and range
  1. State the domain of f(x) = 8/(x + 6).
  2. State the range of g(x) = x² − 2.
Part C — Inverse functions
  1. Find the inverse of f(x) = 5x + 10.
  2. Find the inverse of f(x) = (x + 7)/2.
  3. Find the inverse of f(x) = 9 − 3x.
Part D — Composite functions
  1. f(x) = x − 4, g(x) = 3x. Find gf(x).
  2. Using the same f and g, find fg(x), and confirm it differs from gf(x).
  3. f(x) = 2x + 3, h(x) = x² − 4. Find hf(x), expanded fully.
Part E — Composite functions with fractions
  1. f(x) = 5/(x + 3), g(x) = 2x. Find fg(x).
  2. f(x) = 6/x, g(x) = x − 1. Find fg(x).
Part F — Applied / challenge
  1. f(x) = x − 1 and g(x) = x² + 2. Find gf(x) and fg(x), then find the value of x for which gf(x) = fg(x).
  2. A function converts temperature from Celsius to Fahrenheit: f(C) = (9/5)C + 32. Find f⁻¹, and explain what it would be used for.
  3. f(x) = x², defined only for x ⩾ 0. Explain why restricting the domain like this is necessary for f to have an inverse that is also a function.

Key Vocabulary Recap

Function — a rule where every input has exactly one output
Domain — the set of allowed inputs
Range — the set of resulting outputs
Inverse function, f⁻¹(x) — reverses the effect of f
Composite function, gf(x) — apply f first, then g, to the result
Mapping diagram — arrows connecting each input to its output

Syllabus Reference

This note covers Cambridge IGCSE Mathematics (0580) section E2.13 (Extended subject content) — Functions: understanding functions, domain and range, function notation, inverse functions, and composite functions. There is no Core equivalent for this section. Candidates are not expected to find the domain or range of a composite function.

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