How to Calculate Inverse Functions (IGCSE Mathematics)

Knowing how to calculate inverse functions is a core Edexcel IGCSE Maths skill: an inverse, written f⁻¹(x), reverses what a function does, sending each output back to its input. This page shows the reliable method (swap x and y, then make y the subject), how to handle the domain and range, and how to check your inverse by composing it with the original. Worked examples and the auto-marked practice rooms below give instant feedback.

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Prior Knowledge This topic builds on Composite Functions and Changing the Subject of a Formula. Make sure you are comfortable with both before continuing.

⇩ Jump to Practice Questions ⇩

What is an inverse function?

An inverse function reverses what the original function does. If $f$ maps an input $x$ to an output $y$, then $f^{-1}$ maps $y$ back to $x$. In other words, applying $f$ then $f^{-1}$ (or vice versa) returns the original value.

Notation: the inverse of $f(x)$ is written $f^{-1}(x)$. Note that $f^{-1}(x)$ does not mean $\dfrac{1}{f(x)}$.

$Diagram showing f(x) mapping x to y, and the inverse f^{-1}(x) mapping y back to x$

Worked example showing the swap-and-solve method for finding an inverse function

How to Find an Inverse Function

Write the function as $y = f(x)$.
Swap $x$ and $y$ to get $x = f(y)$.
Rearrange to make $y$ the subject.
Rename $y$ as $f^{-1}(x)$.
Check by composition: $f\!\left(f^{-1}(x)\right) = x$.

Core Ideas

Inverse undoes the function Applying $f$ then $f^{-1}$ returns $x$: $f\!\left(f^{-1}(x)\right) = x$.

Swap, then solve Start with $y = f(x)$, swap $x$ and $y$, then rearrange for $y$.

Domains and ranges swap The domain of $f$ becomes the range of $f^{-1}$, and vice versa.

Worked Examples

💡 Example 1: Linear (Room 1)

Find $f^{-1}(x)$ where $f(x) = 4x + 7$.

Write $y = f(x)$$y = 4x + 7$

Swap $x$ and $y$$x = 4y + 7$

Subtract 7 from both sides$x - 7 = 4y$

Divide both sides by 4$y = \dfrac{x-7}{4}$

Rename $y$ as $f^{-1}(x)$$f^{-1}(x) = \dfrac{x-7}{4}$

💡 Example 2: Reciprocal (Room 2)

Find $f^{-1}(x)$ where $f(x) = 5 + \dfrac{6}{x}$.

Write $y = f(x)$$y = 5 + \dfrac{6}{x}$

Swap $x$ and $y$$x = 5 + \dfrac{6}{y}$

Subtract 5 from both sides$x - 5 = \dfrac{6}{y}$

Multiply by $y$, then divide by $(x-5)$$y = \dfrac{6}{x-5}$

Rename $y$ as $f^{-1}(x)$$f^{-1}(x) = \dfrac{6}{x-5}$

💡 Example 3: Rational function (Room 3)

Find $f^{-1}(x)$ where $f(x) = \dfrac{3x+2}{x-1}$.

Write $y = f(x)$$y = \dfrac{3x+2}{x-1}$

Swap $x$ and $y$$x = \dfrac{3y+2}{y-1}$

Multiply both sides by $(y-1)$$x(y-1) = 3y + 2$

Collect the $y$ terms$y(x-3) = x + 2$

Divide by $(x-3)$ and rename$f^{-1}(x) = \dfrac{x+2}{x-3}$

💡 Example 4: Square root (Room 4)

Find $f^{-1}(x)$ where $f(x) = \sqrt{x+5}$.

Write $y = f(x)$$y = \sqrt{x+5}$

Swap $x$ and $y$$x = \sqrt{y+5}$

Square both sides$x^2 = y + 5$

Subtract 5 to make $y$ the subject$y = x^2 - 5$

Rename (domain $x \geq 0$, the range of $f$)$f^{-1}(x) = x^2 - 5$

💡 Example 5: Quadratic with a stated domain (Room 4)

Find $f^{-1}(x)$ where $f(x) = (x+3)^2$, $x \geqslant -3$.

Write $y = f(x)$$y = (x+3)^2$

Swap $x$ and $y$$x = (y+3)^2$

Square root, and a square root has two signs$\pm\sqrt{x} = y + 3$

The stated domain says $y \geqslant -3$, so $y+3$ cannot be negative: take the $+$ root$\sqrt{x} = y + 3$

Subtract 3 to make $y$ the subject$y = \sqrt{x} - 3$

Rename $y$ as $f^{-1}(x)$$f^{-1}(x) = \sqrt{x} - 3$

A quadratic sends two inputs to the same output, so it has no inverse until a domain is stated. That printed domain is not decoration: it is what tells you which sign of the root to keep, and an answer that leaves the $\pm$ in loses the accuracy mark.

🔑 Key Points

Swap $x$ and $y$, then rearrange to find the inverse.
Check: $f\!\left(f^{-1}(x)\right) = x$.
The graph of $f^{-1}$ is a reflection of $f$ in the line $y = x$.
Domains and ranges swap between $f$ and $f^{-1}$.
A quadratic only has an inverse once a domain is stated, and that domain chooses the sign of the square root.

⚠️ Common Pitfalls

Rearranging without swapping $x$ and $y$ first.
Confusing $f^{-1}(x)$ with $\dfrac{1}{f(x)}$: they are different things.
Forgetting domain restrictions when inverting quadratics (two branches exist).
Leaving $\pm$ in the final answer when the question has printed a domain: the method marks still score, the accuracy mark does not.

⇩ Jump to Practice Questions ⇩

Ready to practise? Work through all four rooms below.

Next Topic: Line Meets Parabola →

Inverse Functions: Practice Room

Rooms 1-4: find $f^{-1}(x)$ and enter the expression only, not f⁻¹(x) = (any correct equivalent form is accepted), e.g. (x-3)/2, 6/(x+1), x^2-4, sqrt(x)+3. Room 5 evaluates the inverse at a value and Room 6 solves $f(x)=f^{-1}(x)$: type a number (two, where a card gives two boxes). Room 7 mixes every type.

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How to calculate inverse functions: Step by step video guide

This YouTube video provides a comprehensive explanation of inverse functions, along with visual examples and step-by-step solutions. It covers key concepts related to inverse functions and offers additional insights to enhance your understanding. Take the time to watch the video, and you’ll gain a solid foundation in solving inverse functions.