Reverse an invertible function and exchange domain with range
Find an inverse when one exists · Read inverse values from a graph or table · Restrict a many-to-one rule before reversing it
Section Paper 4 · Historical Extended · calculator
Exam scope and assessment notes
Extended content; not a Core requirement. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Functions and domain/range
- Rearranging equations
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Find an inverse when one exists
- Read inverse values from a graph or table
- Restrict a many-to-one rule before reversing it
Undo the mapping
For a one-to-one function, f⁻¹(y) returns the original input whose output was y. Solve y = f(x) for x, then rename the input variable. f⁻¹ is not the reciprocal 1/f. On a table or graph, inverse pairs exchange coordinates: (2, 3) becomes (3, 2).
The inverse's domain is the original range; its range is the original domain. A horizontal-line test checks one-to-one behaviour on a graph: if a horizontal line meets the allowed graph more than once, reversing would give two outputs.
Choose a branch before taking a root
The rule x² on all real x has no inverse function because 2 and −2 both give 4. Restricting x ≥ 0 permits √y; restricting x ≤ 0 permits −√y. The restriction determines the sign.
For the displayed f(x) = (x − 1)² + 2, x ≥ 1, the inverse is 1 + √(y − 2), y ≥ 2. The two graphs reflect in y = x; the drawing's axis limit is not an upper domain limit.
Sources for this lesson
Pause and explain
If f(3) = 6 and f is invertible, what is f⁻¹(6)?
Worked example
Invert f(x) = (x − 1)² + 2 on x ≥ 1.
Show the worked solution
- y − 2 = (x − 1)². Since x − 1 ≥ 0, take x − 1 = √(y − 2).
- x = 1 + √(y − 2), requiring y ≥ 2.
- Check f(3) = 6, so the inverse maps 6 back to 3.
Answer f⁻¹(x) = 1 + √(x − 2), x ≥ 2; inverse range outputs ≥ 1.
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.
Invert f(x) = 3x − 2 on all real inputs, and find f⁻¹(10).
Give me a hint
Reverse subtracting 2 and multiplying by 3.
Compare my reasoning
- y = 3x − 2 gives x = (y + 2)/3.
- At inverse input 10, (10 + 2)/3 = 4.
f⁻¹(x) = (x + 2)/3; f⁻¹(10) = 4.
Look for these in your work
- Reverse operation order
- Check by applying f
An invertible table has 1 ↦ 4, 2 ↦ 7, 5 ↦ 9. Give the inverse domain, range and value at 7.
Give me a hint
Swap inputs and outputs in the table.
Compare my reasoning
- Inverse mappings are 4 ↦ 1, 7 ↦ 2, 9 ↦ 5.
- Inverse domain is {4, 7, 9}; inverse range is {1, 2, 5}.
Domain {4, 7, 9}; range {1, 2, 5}; inverse value at 7 is 2.
Look for these in your work
- Use only listed values
- Exchange the sets correctly
A model f(x) = x² is restricted to x ≤ 0. Find its inverse and explain why +√x is wrong.
Give me a hint
The original input must remain non-positive.
Compare my reasoning
- Outputs satisfy y ≥ 0. Solving x² = y on the allowed branch gives x = −√y.
- +√y returns a positive input for y > 0, outside the original domain.
f⁻¹(x) = −√x for x ≥ 0, with range ≤ 0.
Look for these in your work
- Choose the correct branch
- State both inverse domain and range
Common mistakes
- Inverse notation means undo, not reciprocal.
- The square-root branch must agree with the original restricted domain.
What can you explain now?
If f(3) = 6 and f is invertible, what is f⁻¹(6)?
Compare with the explanation
3.
The inverse undoes the original mapping.
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Make it stick
If f(3) = 6 and f is invertible, what is f⁻¹(6)?
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.