The undo button
Every function does something to its input. The inverse function does the opposite, so that it sends each output back to the input it came from. If \(f\) turns \(5\) into \(13\), then the inverse turns \(13\) back into \(5\). We write the inverse of \(f\) as \(f^{-1}(x)\), read as "\(f\) inverse of \(x\)".
- Inverse function
- The function that reverses another function. If \(f\) maps an input to an output, then \(f^{-1}\) maps that output back to the original input. The small \(-1\) is a label, not a power.
Take care with the notation: \(f^{-1}(x)\) does not mean \(\dfrac{1}{f(x)}\). The \(-1\) here is a name that means "the inverse", not the reciprocal.
Where this is assessed
Understanding what an inverse does is Criterion A (knowing and understanding). Writing your rearrangement steps clearly is Criterion C (communicating), and using an inverse to undo a real calculation reaches into Criterion D (applying maths in context).
Reversing the operations
The quickest way to picture the inverse is to list what the function does, step by step, then undo each step in the opposite order. This is exactly how you retrace your steps to get home: last thing done, first thing undone.
Consider \(f(x) = 2x + 3\). Reading it as a set of instructions, it says: start with \(x\), multiply by 2, then add 3. To reverse, undo those in the opposite order: first subtract 3, then divide by 2. Written as a rule,
Test it. We said \(f(5) = 2(5) + 3 = 13\). Feed \(13\) into the inverse: \(f^{-1}(13) = \dfrac{13 - 3}{2} = \dfrac{10}{2} = 5\). It sends \(13\) straight back to \(5\), just as an undo button should.
Opposite operation, opposite order
Addition undoes subtraction, multiplication undoes division, and the order flips: the operation done last is undone first. Getting the order wrong is the most common slip, so always undo from the outside in.
Swapping \(x\) and \(y\)
The second method is more mechanical and never lets you down. Write the function as \(y = \ldots\), swap every \(x\) with \(y\) and every \(y\) with \(x\), then rearrange to make \(y\) the subject. The result is \(f^{-1}(x)\).
Using \(f(x) = 2x + 3\) again: write \(y = 2x + 3\), swap to get \(x = 2y + 3\), then solve for \(y\). Subtract \(3\): \(x - 3 = 2y\). Divide by \(2\): \(y = \dfrac{x-3}{2}\). This is the same \(f^{-1}(x) = \dfrac{x-3}{2}\) we found by reversing operations, which is a good sign both methods agree.
Find the inverse of \(f(x) = 3x - 4\) by swapping \(x\) and \(y\).
Check yourself
Try these before opening the answers. Show the working, not just the final value.
1. Find the inverse of \(f(x) = x + 7\). +
The function adds \(7\), so the inverse subtracts \(7\). Formally: \(y = x + 7\), swap to \(x = y + 7\), then \(y = x - 7\). So \(f^{-1}(x) = x - 7\).
2. Find the inverse of \(f(x) = 5x\). +
The function multiplies by \(5\), so the inverse divides by \(5\). Formally: \(y = 5x\), swap to \(x = 5y\), then \(y = \dfrac{x}{5}\). So \(f^{-1}(x) = \dfrac{x}{5}\).
3. Find the inverse of \(f(x) = \dfrac{x - 1}{4}\). +
Write \(y = \dfrac{x-1}{4}\), then swap: \(x = \dfrac{y-1}{4}\). Multiply both sides by \(4\): \(4x = y - 1\). Add \(1\): \(y = 4x + 1\). So \(f^{-1}(x) = 4x + 1\). Check: \(f(9) = \dfrac{9-1}{4} = 2\), and \(f^{-1}(2) = 4(2) + 1 = 9\).
Part of the Standard Maths library. Spotted an error or want a topic added? That feedback makes the notes better.