Inverse Functions
Understand inverse functions for Victorian Year 11 Mathematical Methods (VCAA) — the function that undoes another, swapping its inputs and its outputs.
You will learn to find the rule of an inverse, swap the domain and range, graph it as a reflection in the line y equals x, and restrict a domain so a function can be inverted.
Theory
In Year 11 Mathematical Methods (Unit 1), the inverse function \(f^{-1}\) undoes \(f\): it sends each output back to the input it came from. Its graph is the reflection of \(y=f(x)\) in the line \(y=x\), so the coordinates of every point swap. This page shows how to find an inverse rule (swap \(x\) and \(y\), then solve), how domain and range swap, why a function must be one-to-one to have an inverse, and how to evaluate \(f^{-1}\) at a value.
The inverse function \(f^{-1}\) of a one-to-one function \(f\) reverses the effect of \(f\). If \(f\) maps \(a\) to \(b\), then \(f^{-1}\) maps \(b\) back to \(a\); in symbols \(f(a)=b\iff f^{-1}(b)=a\).
Because the roles of input and output swap, the graph of \(y=f^{-1}(x)\) is the reflection of \(y=f(x)\) in the line \(y=x\). Every point \((a,b)\) on \(f\) becomes the point \((b,a)\) on \(f^{-1}\), so the domain and range swap:
An inverse function exists only when \(f\) is one-to-one (each output comes from exactly one input). A many-to-one function such as \(f(x)=x^2\) over all of \(\mathbb{R}\) must first be restricted to a one-to-one branch, for example \(x\ge 0\).
To find the inverse rule, start from \(y=f(x)\), swap \(x\) and \(y\), then solve for \(y\):
Points and the domain/range relationship reflect in \(y=x\):
How to find an inverse rule for \(y=f(x)\)
- Check one-to-one: confirm \(f\) is one-to-one (restrict the domain to a single branch first if it is not, e.g. \(x\ge 0\) for a parabola).
- Swap: write \(y=f(x)\), then interchange \(x\) and \(y\) to get \(x=f(y)\).
- Solve: make \(y\) the subject. Where a square root appears, choose the sign that matches the range of \(f^{-1}\) (which is the original domain of \(f\)).
- State the inverse: write \(f^{-1}(x)=\dots\); if the domain matters, record \(\text{domain}(f^{-1})=\text{range}(f)\).
To evaluate \(f^{-1}\) at a number
- Either substitute into the inverse rule, or find the input whose output is that number (solve \(f(x)=\text{value}\)), or reflect a known point \((a,b)\to(b,a)\).
Set \(y=f(x)\), then swap \(x\) and \(y\):
| \(y\) | \(=\) | \(3x-6\) |
| \(x\) | \(=\) | \(3y-6\) |
Make \(y\) the subject:
| \(x+6\) | \(=\) | \(3y\) |
| \(y\) | \(=\) | \(\dfrac{x+6}{3}\) |
Write in inverse notation:
| \(f^{-1}(x)\) | \(=\) | \(\dfrac{x+6}{3}\) |
Inverse rule: \(f^{-1}(x)=\dfrac{x+6}{3}\).
Set \(y=f(x)\) and swap \(x\) and \(y\):
| \(y\) | \(=\) | \(2x+3\) |
| \(x\) | \(=\) | \(2y+3\) |
Solve for \(y\) to get the inverse rule:
| \(x-3\) | \(=\) | \(2y\) |
| \(y\) | \(=\) | \(\dfrac{x-3}{2}\) |
| \(f^{-1}(x)\) | \(=\) | \(\dfrac{x-3}{2}\) |
Substitute \(x=11\):
| \(f^{-1}(11)\) | \(=\) | \(\dfrac{11-3}{2}\) |
| \(=\) | \(\dfrac{8}{2}\) | |
| \(=\) | \(4\) |
\(f^{-1}(11)=4\).
Key idea: reflecting in \(y=x\) swaps the domain and the range, so \(\text{domain}(f^{-1})=\text{range}(f)\) and \(\text{range}(f^{-1})=\text{domain}(f)\).
Apply the swap:
| \(\text{domain}(f^{-1})\) | \(=\) | \(\text{range}(f)=[5,29]\) |
| \(\text{range}(f^{-1})\) | \(=\) | \(\text{domain}(f)=[2,8]\) |
\(f^{-1}\) has domain \([5,29]\) and range \([2,8]\).
The restriction \(x\ge 0\) makes \(f\) one-to-one, so an inverse function exists.
Set \(y=f(x)\) and swap \(x\) and \(y\):
| \(y\) | \(=\) | \(x^2+1\) |
| \(x\) | \(=\) | \(y^2+1\) |
Solve for \(y\), taking the positive root (the domain of \(f\) is \(x\ge 0\)):
| \(y^2\) | \(=\) | \(x-1\) |
| \(y\) | \(=\) | \(\sqrt{x-1}\) |
Write in inverse notation:
| \(f^{-1}(x)\) | \(=\) | \(\sqrt{x-1}\) |
Inverse rule: \(f^{-1}(x)=\sqrt{x-1}\) (with domain \(x\ge 1\)).
Common pitfalls
Frequently asked questions
How do you find the inverse of a function?
Write \(y=f(x)\), swap \(x\) and \(y\) to get \(x=f(y)\), then solve for \(y\). The result is \(f^{-1}(x)\).
What does the graph of an inverse function look like?
It is the reflection of \(y=f(x)\) in the line \(y=x\). Every point \((a,b)\) on \(f\) becomes \((b,a)\) on \(f^{-1}\).
Does the domain and range swap for an inverse?
Yes. \(\text{domain}(f^{-1})=\text{range}(f)\) and \(\text{range}(f^{-1})=\text{domain}(f)\).
Why must a function be one-to-one to have an inverse?
If two inputs share an output, the reverse mapping is ambiguous. A many-to-one function like \(f(x)=x^2\) must be restricted (e.g. \(x\ge 0\)) first.
Is \(f^{-1}(x)\) the same as \(\dfrac{1}{f(x)}\)?
No. \(f^{-1}\) is the inverse function; \(\dfrac{1}{f(x)}\) is the reciprocal. They are usually completely different.
How do you evaluate \(fig(f^{-1}(5)ig)\)?
Use the identity \(f\big(f^{-1}(x)\big)=x\), so \(f\big(f^{-1}(5)\big)=5\) — no need to find the whole inverse rule.