# Inverse Functions

> Undo a function. One-to-one functions, finding f⁻¹(x), reflecting in y = x, and restricting the domain.

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/inverse-functions/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The secret code
Leo and Ava pass coded notes in class. Leo's rule is "double the number, then add 3". So the letter number 5 is sent as 13. To read the note, Ava needs the rule that **undoes** Leo's: "take away 3, then halve". From 13 she gets back to 5.

Ava's rule is the **inverse** of Leo's rule. Any function that can be undone has an inverse, and its graph is a mirror image of the original.

**In short:** An inverse function $f^{-1}$ undoes $f$: if $f(a) = b$, then $f^{-1}(b) = a$. Only one-to-one functions have inverses, found by swapping $x$ and $y$ and making $y$ the subject, and the graph of $f^{-1}$ is the reflection of $f$ in the line $y = x$.

## Starting point: what you already know

- A **function** gives exactly one output for each input. The **vertical line test**: if any vertical line cuts the graph more than once, it is not a function.
- The **domain** is the set of allowed inputs. The **range** is the set of outputs.
- Solving for a variable: from $y = 2x + 3$ you can get $x = \dfrac{y - 3}{2}$.
- From Mathematics Advanced: $y = e^x$ and $y = \ln x$ undo each other, so $\ln(e^x) = x$.

## Undoing a function

The **inverse function** $f^{-1}$ reverses what $f$ does. If $f$ sends 5 to 13, then $f^{-1}$ sends 13 back to 5.

$$f(a) = b \quad\Longleftrightarrow\quad f^{-1}(b) = a$$

### Watch out: The −1 is not a power
$f^{-1}(x)$ means the inverse function. It does **not** mean $\dfrac{1}{f(x)}$.

### Which functions can be undone?

Think of $f(x) = x^2$. Both 3 and $-3$ give 9. If you are told the answer is 9, you cannot tell which input it came from. So $x^2$ cannot be undone on all real numbers.

A function is **one-to-one** if every output comes from exactly one input. Only one-to-one functions have an inverse.

**Horizontal line test:** if every horizontal line cuts the graph at most once, the function is one-to-one.

### Finding the inverse

1. Write $y = f(x)$.
2. Swap $x$ and $y$.
3. Make $y$ the subject. That $y$ is $f^{-1}(x)$.

### Worked example: Leo's code
$f(x) = 2x + 3$. Write $y = 2x + 3$. Swap: $x = 2y + 3$. Solve: $y = \dfrac{x - 3}{2}$.

So $f^{-1}(x) = \dfrac{x - 3}{2}$. Check: $f^{-1}(13) = \dfrac{10}{2} = 5$.

## The mirror in y = x

Swapping $x$ and $y$ turns the point $(a, b)$ into $(b, a)$. On a graph, that is a reflection in the line $y = x$. So:

**The graph of $y = f^{-1}(x)$ is the reflection of $y = f(x)$ in the line $y = x$.**

*Diagram: Graphs of y equals e to the x and y equals ln x, mirror images in the dashed line y equals x. The point 0, 1 on the first matches the point 1, 0 on the second.*
*y = eˣ and y = ln x are mirror images in the line y = x.*

Because the coordinates swap, the domain and range swap too:

- domain of $f^{-1}$ = range of $f$
- range of $f^{-1}$ = domain of $f$

### Key idea: The two checks
For every $x$ in the right domain,
$$f\big(f^{-1}(x)\big) = x \quad\text{and}\quad f^{-1}\big(f(x)\big) = x$$
Doing a function and then undoing it gets you back to where you started.

**Mind map**

```text
Inverse functions
  Meaning
    Undoes f
    f(a) = b means f⁻¹(b) = a
  When it exists
    One-to-one only
    Horizontal line test
  Finding it
    Swap x and y
    Make y the subject
  Graph
    Reflect in y = x
    Domain and range swap
  Restrict domain
    Keep one branch
```

### Year 11: Restricting the domain
$f(x) = x^2$ is not one-to-one, but if we only allow $x \ge 0$, it is. Then $y = x^2$, swap: $x = y^2$, so $y = \sqrt{x}$. We take the positive root because the range of $f^{-1}$ must be $y \ge 0$.

So with domain $x \ge 0$, $f^{-1}(x) = \sqrt{x}$ with domain $x \ge 0$.

If instead we chose $x \le 0$, the inverse would be $f^{-1}(x) = -\sqrt{x}$.

**Restrict at the turning point.** For a parabola, split it at the vertex: $f(x) = x^2 - 6x + 2 = (x - 3)^2 - 7$ is one-to-one for $x \geq 3$, and then $f^{-1}(x) = 3 + \sqrt{x + 7}$ for $x \geq -7$. For a cubic, the largest interval is between its turning points, found where $f'(x) = 0$.

**Fractions.** $f(x) = \dfrac{x + 3}{x - 1}$: swap to $x = \dfrac{y + 3}{y - 1}$, then $xy - x = y + 3$, so $y(x - 1) = x + 3$ and $f^{-1}(x) = \dfrac{x + 3}{x - 1}$. This one is its **own inverse**, so its graph is symmetrical in $y = x$. Any $f(x) = \dfrac{ax + b}{cx - a}$ is its own inverse.

**Exponentials.** $f(x) = e^{3x} - 2$: swap to $x = e^{3y} - 2$, so $3y = \ln(x + 2)$ and $f^{-1}(x) = \frac{1}{3}\ln(x + 2)$ for $x > -2$.

**Harder ones.** For $f(x) = e^x + e^{-x}$ with $x \geq 0$, swap and multiply by $e^y$: $e^{2y} - xe^y + 1 = 0$, a quadratic in $e^y$. Then $e^y = \dfrac{x + \sqrt{x^2 - 4}}{2}$, taking the root that keeps $y \geq 0$.

### Year 11: Where f meets its inverse
For an **increasing** function, $f$ and $f^{-1}$ can only meet on the line $y = x$. So solve $f(x) = x$, which is much easier than solving $f(x) = f^{-1}(x)$.

**Example.** Where does $f(x) = x^2$ for $x \ge 0$ meet its inverse? Solve $x^2 = x$: $x = 0$ or $x = 1$. The points are $(0, 0)$ and $(1, 1)$.

Careful: a decreasing function can meet its inverse off the line $y = x$ as well.

**Gradient of the inverse.** Reflecting in $y = x$ swaps rise and run, so if $f$ has gradient $m$ at $(a, b)$, then $f^{-1}$ has gradient $\dfrac{1}{m}$ at $(b, a)$. For $f(x) = x^3 + 2x$, $f(1) = 3$ and $f'(1) = 5$, so the gradient of $y = f^{-1}(x)$ at $x = 3$ is $\dfrac{1}{5}$.

### Year 12: Inverse trigonometric functions
In Year 12, you will restrict the domains of $\sin x$, $\cos x$ and $\tan x$ to make them one-to-one. Their inverses $\sin^{-1} x$, $\cos^{-1} x$ and $\tan^{-1} x$ use every idea in this chapter, and you will also differentiate them.

## Quick summary

- $f^{-1}$ undoes $f$: if $f(a) = b$ then $f^{-1}(b) = a$.
- Only one-to-one functions have inverses. Use the horizontal line test.
- To find $f^{-1}$, swap $x$ and $y$, then make $y$ the subject.
- The graph of $f^{-1}$ is the reflection of $f$ in $y = x$. Domain and range swap.
- Restrict the domain to make a function one-to-one.

## Quiz

Try these questions from the chapter quiz.

1. If $f(2) = 7$, then $f^{-1}(7)$ is:
   - A. 2
   - B. 7
   - C. $\tfrac{1}{7}$
   - D. $-2$
   - Answer: A. 2. The inverse sends the output back to the input.

2. Which test checks whether a function has an inverse?
   - A. the vertical line test
   - B. the horizontal line test
   - C. the zero test
   - D. the diagonal test
   - Answer: B. the horizontal line test. If a horizontal line cuts twice, two inputs share an output, so it cannot be undone.

3. Find $f^{-1}(x)$ for $f(x) = 3x - 6$.
   - A. $3x + 6$
   - B. $\dfrac{x + 6}{3}$
   - C. $\dfrac{x - 6}{3}$
   - D. $\dfrac{1}{3x - 6}$
   - Answer: B. $\dfrac{x + 6}{3}$. Swap: $x = 3y - 6$, so $y = \dfrac{x + 6}{3}$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/inverse-functions/#quiz