Functions, domains, inverses and composition
Track the order of input–output rules and state where they are valid.
Section E2.13 only · Functions, inverses and composition
Extended only. This is the separate 9–1 qualification 0980. Cambridge intends Core for grades 4–1 and Extended for grades 9–5; the eligible award bands are grades 5–1 for the Core papers and grades 9–3 for the Extended papers.
Before you begin
- Substitute expressions and rearrange a formula.
- Understand squares, roots and denominator restrictions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For f(x) = 4x − 3, what is f(2)?
What you will learn
- Use function notation with a stated domain and range.
- Find an inverse by reversing a one-to-one rule.
- Form composite functions in the stated order.
A function assigns one output to each allowed input
For f(x) = 3x − 4, f(2) = 2. The notation f(2) means apply the rule to input two; it is not f multiplied by two. The domain is the permitted input set and the range is the output set actually reached. If the domain is 0 ≤ x ≤ 2, the range of this increasing rule is −4 ≤ f(x) ≤ 2.
An algebraic formula can impose restrictions: f(x) = 1/(x − 2) cannot accept x = 2, and f(x) = √x requires x ≥ 0 for real outputs. A domain may be narrower still when the problem states it. For f(x) = x² with −2 ≤ x ≤ 3, the range is 0 to 9; substituting only the two endpoints would miss the minimum at zero.
An inverse undoes a one-to-one rule
To invert f(x) = 3x − 4, write y = 3x − 4 and solve x = (y + 4)/3. Rename the inverse input to x, giving f⁻¹(x) = (x + 4)/3. The notation f⁻¹ does not mean 1/f. Check f⁻¹(f(x)) = x and f(f⁻¹(x)) = x on the appropriate domains.
The unrestricted square rule is not one-to-one because two and negative two both map to four. If its domain is x ≥ 0, its inverse is √x, and the inverse's domain is the original non-negative range. If the original domain were x ≤ 0, its inverse would instead use −√x. Stating the restriction prevents an inverse rule from returning two outputs for one input.
Composition applies the inside function first
In Cambridge notation gf(x) = g(f(x)): apply f first, then g. If f(x) = 2x + 1 and g(x) = x², gf(x) = (2x + 1)² while fg(x) = 2x² + 1. At x = 3, the first order gives 49 and the reverse gives 19. Composition usually depends on the order and is not multiplication of outputs.
Replace every occurrence of the outer function's input with the complete inner expression, using brackets. For f(x) = 3/(x + 1) and g(x) = x², fg(x) = 3/(x² + 1). First ensure an actual input is allowed by the inner rule and its output is allowed by the outer rule. This course does not require students to derive domains and ranges of composite functions as a separate assessed technique.
Pause and explain
In gf(x) = g(f(x)), which rule acts first?
Worked example
For f(x) = 2x + 1 and g(x) = x², find gf(x), fg(x) and f⁻¹(x).
Show the worked solution
- Apply f before g: gf(x) = (2x + 1)².
- Apply g before f: fg(x) = 2x² + 1.
- Rearrange y = 2x + 1 to obtain f⁻¹(x) = (x − 1)/2.
Answer gf(x) = (2x + 1)²; fg(x) = 2x² + 1; f⁻¹(x) = (x − 1)/2
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For f(x) = 5x + 2, find f(−1) and f⁻¹(x).
Give me a hint
Undo adding two, then multiplying by five.
Compare my reasoning
- f(−1) = 5(−1) + 2 = −3.
- From y = 5x + 2, rearrange x = (y − 2)/5.
- Thus f⁻¹(x) = (x − 2)/5; applying it to −3 returns −1.
f(−1) = −3; f⁻¹(x) = (x − 2)/5
Look for these in your work
- Substituted the signed input correctly.
- Checked the inverse using the original output.
For f(x) = x − 4 and g(x) = 3x², find gf(x) and fg(x), then evaluate both at x = 2.
Give me a hint
Replace the outer rule's input with the complete inner expression.
Compare my reasoning
- gf(x) = 3(x − 4)², so gf(2) = 12.
- fg(x) = 3x² − 4, so fg(2) = 8.
- The different results confirm that changing the order changes this composite.
gf(x) = 3(x − 4)²; fg(x) = 3x² − 4; gf(2) = 12, fg(2) = 8
Look for these in your work
- Kept the correct composition order.
- Bracketed the expression before squaring.
A sensor calibration uses f(x) = 2x + 6 on raw readings 0 ≤ x ≤ 10. State its calibrated range, inverse, and the raw reading corresponding to calibrated output 18.
Give me a hint
The increasing rule maps the domain endpoints to its range endpoints.
Compare my reasoning
- Outputs run from f(0) = 6 to f(10) = 26.
- The inverse is f⁻¹(y) = (y − 6)/2, defined for calibrated readings 6 ≤ y ≤ 26.
- For y = 18, the raw reading is (18 − 6)/2 = 6, within its allowed input interval.
Range [6, 26]; inverse (y − 6)/2; raw reading 6
Look for these in your work
- Kept the original domain and inverse domain distinct.
- Checked the recovered reading is allowed.
Common mistakes
- Reading f(x) as multiplication.
- Confusing an inverse function with a reciprocal.
- Applying composite functions in the opposite order.
What can you explain now?
For f(x) = x² on x ≥ 0, give its range and inverse.
Compare with the explanation
Range y ≥ 0; f⁻¹(x) = √x on x ≥ 0
The non-negative domain makes squaring one-to-one. Its non-negative outputs form the inverse's domain.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Build two simple functions and apply them in both orders to the same input. Find an inverse for one and state its permitted inputs.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.