To find an inverse function, write y = f(x), swap x and y, then rearrange to make y the subject. Notes, worked examples and 13 questions answered.
Updated 14 Mar 2026 · Revise Right Team
Understanding Inverse Functions is part of Inverse functions in GCSE AQA Higher Further Maths.
4 min
Reading time
24
Lessons in this topic
GCSE AQA Higher
Pathway
Reviewed by Revise Right
Content review
What this lesson covers
To find an inverse function, write y = f(x), swap x and y, then rearrange to make y the subject. Notes, worked examples and 13 questions answered.
Define an inverse function and explain what it does to the output of the original function
Use the notation $f^{-1}$ correctly to represent the inverse of a function $f$
Find the inverse of a given function algebraically by swapping $x$ and $y$ and rearranging
Explain why a function must be one-one for its inverse to exist as a function
Key ideas to keep in view
Use this lesson to stay inside Inverse functions while connecting the detail back to the wider Algebra topic.
Inverse function: A function that reverses the effect of the original function. If $f$ maps $x$ to $y$, then $f^{-1}$ maps $y$ back to $x$. Written as $f^{-1}(x)$.
One-one function: A function in which every output value is produced by exactly one input value. Required for an inverse function to exist. Also written as one-to-one.
Domain: The set of all permitted input values for a function. Restricting the domain can make a many-one function become one-one, allowing an inverse to exist.
Range: The set of all output values a function can produce. The range of $f$ becomes the domain of $f^{-1}$, and vice versa.
Revision notes
Read the core explanations from Understanding Inverse Functions before testing yourself from memory.
1. What Is an Inverse Function?
You have already worked with functions as machines: you put a number in, follow a rule, and get a number out. An inverse function runs that machine in reverse — it takes the output and works back to the original input.
Think of it like a journey. If $f$ takes you from London to Edinburgh, then $f^{-1}$ takes you from Edinburgh back to London. The route is the same road; you are just travelling in the opposite direction.
Notice that $f(4) = 2(4) + 3 = 11$, and $f^{-1}(11) = \frac{11-3}{2} = \frac{8}{2} = 4$. The inverse undoes the function exactly.
2. Finding the Inverse Algebraically
There is a reliable algebraic method for finding $f^{-1}(x)$. The key move is to write $y = f(x)$, swap $x$ and $y$, and then rearrange to make $y$ the subject. That rearranged expression is $f^{-1}(x)$.
Worked example: Finding the inverse of $f(x) = 3x - 7$
Given $f(x) = 3x - 7$, find $f^{-1}(x)$.
$y = 3x - 7$ — Replace $f(x)$ with $y$ so we can rearrange. [M1 — setting up the equation]
$x = 3y - 7$ — This is the core step: swapping the roles of input and output reverses the function.
$x + 7 = 3y$ — Add 7 to both sides to isolate the term with $y$. [M1 — correct rearrangement]
$y = \frac{x + 7}{3}$ — Dividing by 3 gives $y$ on its own.
$f^{-1}(x) = \frac{x + 7}{3}$ — Replace $y$ with $f^{-1}(x)$. [A1 — correct final answer]
$f(2) = 3(2) - 7 = 6 - 7 = -1$, then $f^{-1}(-1) = \frac{-1 + 7}{3} = \frac{6}{3} = 2$ ✓ — Checking that $f^{-1}(f(2)) = 2$ confirms the inverse is correct.
Answer: $f^{-1}(x) = \frac{x + 7}{3}$
The verification step in the worked example above is worth doing in any exam question where you have time. If $f^{-1}(f(a)) = a$ for a chosen value $a$, your inverse is correct.
3. Why the Function Must Be One-One
Here is where things get interesting. Not every function has an inverse that is itself a function. The problem arises when two different inputs produce the same output.
Consider $f(x) = x^2$ with no domain restriction. Both $f(3) = 9$ and $f(-3) = 9$. If you try to reverse this, the output 9 would need to map back to both 3 and $-3$ simultaneously — but a function can only give one output per input. The inverse would fail to be a function.
The graph of $y = x^2$ is a U-shaped parabola symmetric about the $y$-axis. A horizontal line drawn at any positive $y$-value crosses the curve at two points — one on each side of the axis. This is the graphical test for a many-one function: if any horizontal line crosses the graph more than once, the function is not one-one and has no inverse function over that full domain.
4. Restricting the Domain to Make a Function One-One
The fix is elegant: restrict the domain so that only one branch of the function is used. For $f(x) = x^2$, choosing the domain $x \geq 0$ keeps only the right-hand branch of the parabola. Every output value now comes from exactly one input, so the function is one-one on this restricted domain.
With domain $x \geq 0$, the inverse is $f^{-1}(x) = \sqrt{x}$, defined for $x \geq 0$. The range of $f$ (which is $y \geq 0$) becomes the domain of $f^{-1}$, and the domain of $f$ (which is $x \geq 0$) becomes the range of $f^{-1}$. Domain and range always swap between a function and its inverse.
The reflection in $y = x$ is a key visual property. Every point $(a, b)$ on the graph of $f$ corresponds to the point $(b, a)$ on the graph of $f^{-1}$. This is a direct consequence of swapping $x$ and $y$ in the algebraic method.
Worked example: Finding the inverse of $f(x) = x^2$ on a restricted domain
The function $f$ is defined by $f(x) = x^2$ for $x \geq 0$. Find $f^{-1}(x)$ and state its domain.
$y = x^2$, where $x \geq 0$ — State the domain restriction — it is essential for the inverse to exist. [M1]
$x = y^2$, where $y \geq 0$ — The restriction on $x$ becomes a restriction on $y$ after swapping.
$y = \sqrt{x}$ — Taking the positive square root only, because $y \geq 0$. [M1 — correct root selection]
$f^{-1}(x) = \sqrt{x}$, domain $x \geq 0$ — The domain of $f^{-1}$ equals the range of $f$. Since $f(x) = x^2$ for $x \geq 0$ gives outputs $y \geq 0$, the domain of $f^{-1}$ is $x \geq 0$. [A1]
Answer: $f^{-1}(x) = \sqrt{x}$, domain $x \geq 0$
6. Exam Technique and Common Mistakes
Inverse function questions are reliable mark-earners once you know the method — but there are several specific traps that appear repeatedly in exam papers. The guidance below addresses each one directly.
Worked example: Exam-style question: inverse with a restricted domain
The function $f$ is defined by $f(x) = (x + 1)^2 - 4$ for $x \geq -1$. Find $f^{-1}(x)$ and state its domain.
$y = (x + 1)^2 - 4$, for $x \geq -1$ — State the domain restriction alongside the equation. [M1]
$x = (y + 1)^2 - 4$, for $y \geq -1$ — The restriction on $x$ transfers to $y$ after swapping.
$\sqrt{x + 4} = y + 1$ — Since $y \geq -1$, we have $y + 1 \geq 0$, so the positive root is correct. Taking the negative root here would be wrong.
$y = \sqrt{x + 4} - 1$ — Rearranging to make $y$ the subject. [M1]
$f^{-1}(x) = \sqrt{x + 4} - 1$
Domain: find the range of $f$ for $x \geq -1$.
At $x = -1$: $f(-1) = (0)^2 - 4 = -4$. As $x$ increases, $f(x)$ increases.
So range of $f$ is $y \geq -4$, meaning domain of $f^{-1}$ is $x \geq -4$. — Always state the domain of $f^{-1}$. [A1 for correct inverse; A1 for correct domain]
Every spec requirement for inverse functions connects back to one central idea: reversibility. A function can be reversed if and only if it is one-one, the notation $f^{-1}$ captures that reversal precisely, and restricting the domain is the tool that makes a non-one-one function reversible over a chosen interval.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-03-14.
To find an inverse function, write y = f(x), swap x and y, then rearrange to make y the subject. For f(x) = 3x − 7, x = 3y − 7 gives x + 7 = 3y, so f⁻¹(x) = (x + 7)/3. Check: f(2) = −1 and f⁻¹(−1) = 2.
What is an inverse function?
An inverse function, written f⁻¹, runs a function in reverse: it takes the output and works back to the original input. The superscript −1 means 'inverse', not 'reciprocal', so f⁻¹(x) is not the same as 1/f(x).
What is the horizontal line test?
The horizontal line test is the graphical test for a many-one function: if any horizontal line crosses the graph more than once, the function is not one-one and has no inverse function over that full domain. For y = x², a horizontal line at any positive y-value crosses the curve twice.
What is a one-one function?
A function is one-one (one-to-one) if no two different inputs produce the same output: if f(a) = f(b), then a = b. A function must be one-one for its inverse to exist as a function; if any horizontal line crosses the graph more than once, it is not one-one.
Why does f(x) = x² not have an inverse?
f(x) = x² with no domain restriction has no inverse function because f(3) = 9 and f(−3) = 9, so 9 would map back to two inputs. Restricting the domain to x ≥ 0 makes it one-one, giving f⁻¹(x) = √x with domain x ≥ 0.
How are the graphs of a function and its inverse related?
The graph of f⁻¹ is the reflection of the graph of f in the line y = x, so every point (a, b) on f corresponds to the point (b, a) on f⁻¹. Domain and range always swap: the range of f becomes the domain of f⁻¹.
Is f⁻¹(x) the same as 1/f(x)?
The inverse f⁻¹(x) is not the same as 1/f(x): the superscript −1 here means 'inverse', not 'reciprocal'. An inverse function takes the output of f and works back to the original input, so if f(4) = 11, then f⁻¹(11) = 4.
How do you check an inverse function is correct?
An inverse function is checked by choosing a value a and confirming f⁻¹(f(a)) = a. For f(x) = 3x − 7, f(2) = −1 and f⁻¹(−1) = (−1 + 7)/3 = 2, so f⁻¹(x) = (x + 7)/3 is correct.
How are the domain and range of an inverse function found?
The domain and range swap between a function and its inverse: the range of f becomes the domain of f⁻¹, and the domain of f becomes the range of f⁻¹. For f(x) = x² with x ≥ 0, the range is y ≥ 0, so f⁻¹(x) = √x has domain x ≥ 0.
What is a many-one function?
A many-one function is one where two different inputs produce the same output, such as f(x) = x² with f(3) = 9 and f(−3) = 9. Its inverse would fail to be a function, because the output 9 would need to map back to both 3 and −3.
Why do you take the positive square root when finding an inverse?
The positive square root is taken when the domain restriction makes the bracket non-negative. For f(x) = (x + 1)² − 4 with x ≥ −1, swapping gives x + 4 = (y + 1)² with y + 1 ≥ 0, so y = √(x + 4) − 1; the negative root would be wrong.
How do you find the inverse of (x + 1)² − 4?
For f(x) = (x + 1)² − 4 with x ≥ −1, write y = (x + 1)² − 4, swap x and y, add 4 and take the positive square root to get f⁻¹(x) = √(x + 4) − 1. Since f(−1) = −4 and f increases from there, the domain of f⁻¹ is x ≥ −4.
How do you make a function one-one?
A function is made one-one by restricting its domain so that only one branch is used. For f(x) = x², choosing the domain x ≥ 0 keeps only the right-hand branch of the parabola, so every output comes from exactly one input and the inverse f⁻¹(x) = √x exists.