Understanding Inverse Functions Revision Notes | GCSE AQA Higher Further Maths

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 $f^{-1}(x)$.

1. $y = 3x - 7$ — Replace $f(x)$ with $y$ so we can rearrange. [M1 — setting up the equation]
2. $x = 3y - 7$ — This is the core step: swapping the roles of input and output reverses the function.
3. $x + 7 = 3y$ — Add 7 to both sides to isolate the term with $y$. [M1 — correct rearrangement]
4. $y = \frac{x + 7}{3}$ — Dividing by 3 gives $y$ on its own.
5. $f^{-1}(x) = \frac{x + 7}{3}$ — Replace $y$ with $f^{-1}(x)$. [A1 — correct final answer]
6. $f(2) = 3(2) - 7 = 6 - 7 = -1$, 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.

1. $y = x^2$, where $x \geq 0$ — State the domain restriction — it is essential for the inverse to exist. [M1]
2. $x = y^2$, where $y \geq 0$ — The restriction on $x$ becomes a restriction on $y$ after swapping.
3. $y = \sqrt{x}$ — Taking the positive square root only, because $y \geq 0$. [M1 — correct root selection]
4. $f^{-1}(x) = \sqrt{x}$, 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.

1. $y = (x + 1)^2 - 4$, for $x \geq -1$ — State the domain restriction alongside the equation. [M1]
2. $x = (y + 1)^2 - 4$, for $y \geq -1$ — The restriction on $x$ transfers to $y$ after swapping.
3. $x + 4 = (y + 1)^2$ — Isolate the squared bracket. [M1 — correct rearrangement]
4. $\sqrt{x + 4} = y + 1$ — Since $y \geq -1$, we have $y + 1 \geq 0$, so the positive root is correct. Taking the negative root here would be wrong.
5. $y = \sqrt{x + 4} - 1$ — Rearranging to make $y$ the subject. [M1]
6. $f^{-1}(x) = \sqrt{x + 4} - 1$ 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]

Answer: $f^{-1}(x) = \sqrt{x + 4} - 1$, domain $x \geq -4$

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.