EST. 2024 · LONDON·MMXXVI SPECIFICATION
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Inverse Functions –

GCSEMathsAI Team·7 min read·23 May 2026

Inverse functions reverse the effect of the original function. If f takes 3 to 11, then the inverse function f⁻¹ takes 11 back to 3. This is a Higher-tier topic that appears on AQA, Edexcel and OCR papers, typically worth three to four marks.

What Is an Inverse Function?

The inverse of a function f, written f⁻¹(x), undoes what f does. If f(a) = b, then f⁻¹(b) = a.

To find the inverse algebraically, you swap x and y in the equation y = f(x), then rearrange to make y the subject. The result is f⁻¹(x).

Key Formulas

If f(a) = b, then f⁻¹(b) = a
To find f⁻¹(x): write y = f(x), swap x and y, rearrange for y

Graphical interpretation

The graph of f⁻¹(x) is the reflection of the graph of f(x) in the line y = x. This means if the point (3, 7) lies on y = f(x), then (7, 3) lies on y = f⁻¹(x).

Step-by-Step Method

  1. Write y = f(x) — replace f(x) with y.
  2. Swap x and y in the equation.
  3. Rearrange the new equation to make y the subject.
  4. Write f⁻¹(x) = the expression you found for y.

Worked Example 1 — Foundation Level

Question: f(x) = 3x + 7. Find f⁻¹(x).

Working:

Step 1: Write y = 3x + 7.

Step 2: Swap x and y: x = 3y + 7.

Step 3: Rearrange for y: x − 7 = 3y → y = (x − 7)/3.

Answer: f⁻¹(x) = (x − 7)/3

Check: f(2) = 6 + 7 = 13. f⁻¹(13) = (13 − 7)/3 = 6/3 = 2 ✓

Worked Example 2 — Higher Level

Question: f(x) = (4x − 1) / 5. Find f⁻¹(x).

Working:

Step 1: Write y = (4x − 1) / 5.

Step 2: Swap x and y: x = (4y − 1) / 5.

Step 3: Rearrange: 5x = 4y − 1 → 5x + 1 = 4y → y = (5x + 1)/4.

Answer: f⁻¹(x) = (5x + 1)/4

Check: f(4) = (16 − 1)/5 = 3. f⁻¹(3) = (15 + 1)/4 = 16/4 = 4 ✓

Worked Example 3 — Exam Style

Question: f(x) = 2x³ + 5. Find f⁻¹(x) and evaluate f⁻¹(59).

Working:

Step 1: Write y = 2x³ + 5.

Step 2: Swap: x = 2y³ + 5.

Step 3: Rearrange: x − 5 = 2y³ → y³ = (x − 5)/2 → y = ∛((x − 5)/2).

So f⁻¹(x) = ∛((x − 5)/2).

Evaluate: f⁻¹(59) = ∛((59 − 5)/2) = ∛(54/2) = ∛27 = 3.

Answer: f⁻¹(x) = ∛((x − 5)/2); f⁻¹(59) = 3

Common Mistakes

  • Confusing f⁻¹(x) with 1/f(x). The notation f⁻¹(x) means the inverse function, not the reciprocal. f⁻¹(x) undoes f, whereas 1/f(x) divides 1 by the output.
  • Forgetting to swap x and y. Simply rearranging y = 2x + 3 to get x = (y − 3)/2 is not enough. You must swap the variables first, then rearrange for y.
  • Making rearrangement errors with fractions. When the function involves a fraction, multiply both sides first to clear the denominator before isolating y.

Exam Tips

  • Always verify your inverse by checking that f(f⁻¹(x)) = x. Pick a simple number and confirm that applying f then f⁻¹ gets you back to where you started.
  • If a question asks you to sketch f⁻¹(x), reflect the graph of f(x) in the line y = x. Key points swap their coordinates.
  • Show each algebraic step clearly — write the swap step explicitly so the examiner can see your method.

Practice Questions

Q1 (Higher): f(x) = 5x − 4. Find f⁻¹(x).

Answer: y = 5x − 4. Swap: x = 5y − 4. Rearrange: x + 4 = 5y, y = (x + 4)/5. So f⁻¹(x) = (x + 4)/5.

Q2 (Higher): f(x) = (x + 3)/2. Find f⁻¹(x) and evaluate f⁻¹(7).

Answer: y = (x + 3)/2. Swap: x = (y + 3)/2. Rearrange: 2x = y + 3, y = 2x − 3. So f⁻¹(x) = 2x − 3. f⁻¹(7) = 14 − 3 = 11.

Q3 (Higher): g(x) = x² − 1 for x ≥ 0. Find g⁻¹(x).

Answer: y = x² − 1. Swap: x = y² − 1. Rearrange: y² = x + 1, y = sqrt(x + 1) (positive root only since x ≥ 0). So g⁻¹(x) = sqrt(x + 1).

Practise inverse functions questions with instant feedback — completely free on GCSEMathsAI.

Summary

  • The inverse function f⁻¹(x) reverses the effect of f(x): if f(a) = b, then f⁻¹(b) = a.
  • To find f⁻¹(x): write y = f(x), swap x and y, then rearrange for y.
  • The graph of f⁻¹(x) is the reflection of f(x) in the line y = x.
  • f⁻¹(x) is not the same as 1/f(x) — it is the inverse function, not the reciprocal.
  • Always check your answer by verifying that f(f⁻¹(x)) gives x.

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§Academic References

Further reading from leading academic institutions — free and open-access.

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Function notation, composition and inverses — Cambridge.

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Function notation, composite and inverse functions.

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Cambridge problem-solving with ratio and proportion.

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RatioCorbett Maths

Simplifying, sharing in a ratio, and proportion problems.

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