Find an inverse rule and exchange domain with range
Reverse a one-one function algebraically, keeping excluded values and checking the two compositions.
Section 1.2 · Functions
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Written practice is self-checked; show your reasoning and domain restrictions.
Before you begin
- Rearrange linear equations, including ones with fractional expressions.
- Distinguish a function's actual range from its domain.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What must be true for f to have an inverse function on its full stated domain?
What you will learn
- Find inverses of simple one-one linear and rational functions.
- State the inverse's domain and range and distinguish f⁻¹ from a reciprocal.
Reverse the mapping, rather than reciprocating the output
An inverse reverses the input-output pairs of a one-one function. If f sends a to b, then f⁻¹ sends b back to a. Write y = f(x), solve for x in terms of y, then rename the inverse input as x. The inverse's domain is f's range, and its range is f's domain.
For f(x) = 4x − 7 on all real x, y = 4x − 7 gives x = (y + 7)/4, so f⁻¹(x) = (x + 7)/4 on all real x. The notation f⁻¹(x) does not mean 1/f(x). Indeed f⁻¹(1) = 2, whereas 1/f(1) = −1/3.
Keep rational exclusions through the rearrangement
For f(x) = (3x + 2)/(x − 1), x = 1 is forbidden. Rearranging y(x − 1) = 3x + 2 gives x(y − 3) = y + 2. Thus x = (y + 2)/(y − 3), with y ≠ 3. Every such y is attained by an allowed x; y = 3 would require 3x + 2 = 3x − 3, which is impossible. The function is one-one because this rearrangement gives exactly one x for each output.
Therefore f⁻¹(x) = (x + 2)/(x − 3), with inverse domain x ≠ 3 and inverse range y ≠ 1. Check f⁻¹(f(x)) = x for x in f's domain, and f(f⁻¹(x)) = x for x in the inverse domain. These identities do not license evaluation at a forbidden denominator value.
Pause and explain
If f has domain x ≠ 1 and range y ≠ 3, what is the domain of f⁻¹?
Worked example
Find the inverse of f(x) = (3x + 2)/(x − 1), x ≠ 1, and state its domain and range.
Show the worked solution
- Write y(x − 1) = 3x + 2 and collect the x terms: x(y − 3) = y + 2.
- Solve x = (y + 2)/(y − 3); y = 3 is excluded and every other y is attained.
- Rename the input to get f⁻¹(x) = (x + 2)/(x − 3); exchange the original range and domain.
Answer f⁻¹(x) = (x + 2)/(x − 3); domain x ≠ 3; range y ≠ 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.
Find the inverse of f(x) = 5 − 2x on all real x and check it at the output of x = 3.
Give me a hint
Solve y = 5 − 2x for x before renaming the input.
Compare my reasoning
- Rearrange to 2x = 5 − y, hence x = (5 − y)/2.
- Thus f⁻¹(x) = (5 − x)/2, with all real inputs and outputs.
- Since f(3) = −1, f⁻¹(−1) = (5 + 1)/2 = 3, reversing the mapping.
f⁻¹(x) = (5 − x)/2; f⁻¹(f(3)) = 3.
Look for these in your work
- Divide the whole rearranged numerator by 2.
- Check that the inverse returns the original input.
For f(x) = 2/(x + 3), x ≠ −3, find the inverse and its domain and range.
Give me a hint
The original output cannot be zero; rearrange y(x + 3) = 2.
Compare my reasoning
- Solve for the original input: x = 2/y − 3, with y ≠ 0.
- Every non-zero y gives an allowed original x, so the original range is y ≠ 0.
- The inverse is f⁻¹(x) = 2/x − 3, with domain x ≠ 0 and range y ≠ −3.
f⁻¹(x) = 2/x − 3; domain x ≠ 0; range y ≠ −3.
Look for these in your work
- Retain the excluded inverse input zero.
- Exchange the original range and domain correctly.
A supplied temperature conversion is F = 1.8C + 32 for −10 ≤ C ≤ 40. Find the inverse conversion and its allowed input interval.
Give me a hint
Convert both endpoints before exchanging the input and output intervals.
Compare my reasoning
- The positive slope gives F(−10) = 14 and F(40) = 104, so the original range is 14 ≤ F ≤ 104.
- Rearrange the conversion to C = (F − 32)/1.8.
- The inverse accepts 14 ≤ F ≤ 104 and returns −10 ≤ C ≤ 40; for F = 68 it returns C = 20.
C = (F − 32)/1.8 for 14 ≤ F ≤ 104; output interval −10 ≤ C ≤ 40.
Look for these in your work
- Find the range from both original endpoints.
- Limit inverse inputs to outputs attained by the stated model.
Common mistakes
- f⁻¹ is an inverse mapping, not 1/f.
- Exchange domain and range along with the algebraic rule.
- Retain forbidden values when multiplying by or dividing by an expression.
What can you explain now?
Find f⁻¹(x) when f(x) = 4x − 7 for real x.
Compare with the explanation
f⁻¹(x) = (x + 7)/4, with all real inputs and outputs.
Solve y = 4x − 7 for x, then rename the inverse input; the non-zero slope ensures one-one behaviour.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain the difference between f⁻¹(x) and 1/f(x), then find an inverse rule and describe what happens to its input and output restrictions.
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.