Inverse Functions Revision Notes for VCE SSCE Mathematical Methods

Inverse Functions

What is an inverse function?

An inverse function reverses the effect of the original function. If function $f$ takes an input $x$ and produces output $y$ , then the inverse function $f^{-1}$ takes $y$ as input and produces $x$ as output.

However, not every function has an inverse. For an inverse function to exist, the original function must be one-to-one. This means that each output value corresponds to exactly one input value.

infoNote

A function is one-to-one if it passes the horizontal line test - any horizontal line intersects the graph at most once. This ensures that each output corresponds to exactly one input.

Definition: If $f$ is a one-to-one function, we can define its inverse $f^{-1}$ by:

f^{-1}(x) = y \text{ if } f(y) = x, \text{ for } x \in \text{ran } f \text{ and } y \in \text{dom } f

chatImportant

The notation $f^{-1}$ does not mean $\frac{1}{f(x)}$ . It represents the inverse function, not the reciprocal. These are two completely different concepts!

Geometric interpretation

The relationship between a function and its inverse has a beautiful geometric property. A point $(x, y)$ lies on the graph of $f^{-1}$ if and only if the point $(y, x)$ lies on the graph of $f$ . In other words, the coordinates are swapped.

This means that to obtain the graph of $f^{-1}$ from the graph of $f$ , you reflect the graph of $f$ across the line $y = x$ .

infoNote

Visual interpretation: Think of the line $y = x$ as a mirror. The graph of the inverse function is the mirror image of the original function across this diagonal line.

Domain and range relationships

The domain and range swap between a function and its inverse:

\text{dom } f^{-1} = \text{ran } f

\text{ran } f^{-1} = \text{dom } f

infoNote

This swap makes intuitive sense: the input values of the inverse are the output values of the original function, and vice versa. When finding an inverse, always identify the range of the original function to determine the domain of the inverse.

Composition properties

When you compose a function with its inverse, you get back the original input value:

f \circ f^{-1}(x) = x, \text{ for all } x \in \text{dom } f^{-1}

f^{-1} \circ f(x) = x, \text{ for all } x \in \text{dom } f

infoNote

This property confirms that the inverse function truly undoes what the original function does. It's like pressing "undo" - you get back to where you started. This composition property is often used as an alternative method for finding inverse functions.

Finding inverse functions algebraically

There are two main methods for finding inverse functions:

Method 1: Interchange and solve

Step 1: Write the function equation as $y = f(x)$

Step 2: Interchange $x$ and $y$ to get $x = f(y)$

Step 3: Solve this equation for $y$

Step 4: Replace $y$ with $f^{-1}(x)$

Step 5: Determine the domain of $f^{-1}$ (which equals the range of $f$ )

Method 2: Use composition

This method uses the property that $f(f^{-1}(x)) = x$ .

Step 1: Set up the equation $f(f^{-1}(x)) = x$

Step 2: Substitute the rule for $f$ and solve for $f^{-1}(x)$

Step 3: Determine the domain of $f^{-1}$

infoNote

Which method to choose? Method 1 (interchange and solve) is usually more straightforward and works well for most functions. Method 2 (composition) can be faster for simple functions and provides good practice with function composition.

Worked examples

lightbulbExample

Worked Example: Linear function

Find the inverse function $f^{-1}$ of the function $f(x) = 2x - 3$ .

Solution using Method 1:

The graph of $f$ has equation $y = 2x - 3$ .

Interchange $x$ and $y$ :

x = 2y - 3

x + 3 = 2y

y = \frac{1}{2}(x + 3)

Therefore $f^{-1}(x) = \frac{1}{2}(x + 3)$ and $\text{dom } f^{-1} = \text{ran } f = \mathbb{R}$ .

Solution using Method 2:

We require $f^{-1}$ such that $f(f^{-1}(x)) = x$

2f^{-1}(x) - 3 = x

f^{-1}(x) = \frac{1}{2}(x + 3)

Thus $f^{-1}(x) = \frac{1}{2}(x + 3)$ and $\text{dom } f^{-1} = \text{ran } f = \mathbb{R}$ .

lightbulbExample

Worked Example: Functions with restricted domains

Find the inverse of each function, stating the domain and range:

a) $f: [-2, 1] \to \mathbb{R}, f(x) = 2x + 3$

Solution:

$\text{ran } f^{-1} = \text{dom } f = [-2, 1]$

$\text{dom } f^{-1} = \text{ran } f = [-1, 5]$

Let $y = 2x + 3$ . Interchange $x$ and $y$ :

x = 2y + 3

x - 3 = 2y

y = \frac{x - 3}{2}

Therefore $f^{-1}: [-1, 5] \to \mathbb{R}, f^{-1}(x) = \frac{x - 3}{2}$

b) $g(x) = \frac{1}{5 - x}, x > 5$

Solution:

$\text{ran } g^{-1} = \text{dom } g = (5, \infty)$

$\text{dom } g^{-1} = \text{ran } g = (-\infty, 0)$

Let $y = \frac{1}{5 - x}$ . Interchange $x$ and $y$ :

x = \frac{1}{5 - y}

5 - y = \frac{1}{x}

y = 5 - \frac{1}{x}

Therefore $g^{-1}: (-\infty, 0) \to \mathbb{R}, g^{-1}(x) = 5 - \frac{1}{x}$

c) $h(x) = x^2 - 2, x \geq 1$

Solution:

$\text{ran } h^{-1} = \text{dom } h = [1, \infty)$

$\text{dom } h^{-1} = \text{ran } h = [-1, \infty)$

Let $y = x^2 - 2$ . Interchange $x$ and $y$ :

x = y^2 - 2

y^2 = x + 2

y = \pm\sqrt{x + 2}

Therefore $h^{-1}: [-1, \infty) \to \mathbb{R}, h^{-1}(x) = \sqrt{x + 2}$

The positive square root is taken because the range of $h^{-1}$ must equal the domain of $h$ , which is $[1, \infty)$ .

chatImportant

When dealing with functions involving squares or square roots, pay careful attention to domain restrictions. The domain restriction determines which sign (positive or negative) to choose when taking square roots.

Graphing inverse functions

Reflection property

The graph of an inverse function is obtained by reflecting the original function's graph across the line $y = x$ . This reflection transforms every point $(x, y)$ on the original graph to the point $(y, x)$ on the inverse graph.

infoNote

When graphing a function and its inverse together, always include the line $y = x$ to show the line of symmetry. This makes the reflection relationship visually clear.

Finding intersection points

The graphs of $y = f(x)$ and $y = f^{-1}(x)$ can intersect at points on the line $y = x$ (where $f(x) = x$ ) and potentially at other points as well.

To find intersections on the line $y = x$ , solve $f(x) = x$ .

To find all intersections, solve $f(x) = f^{-1}(x)$ .

lightbulbExample

Worked Example: Graphing a rational function and its inverse

Find the inverse of the function $f: \mathbb{R} \setminus \{0\} \to \mathbb{R}, f(x) = \frac{1}{x} + 3$ and sketch both functions, showing their points of intersection.

Solution:

Let $x \in \text{dom } f^{-1} = \text{ran } f$ . Then:

f(f^{-1}(x)) = x

\frac{1}{f^{-1}(x)} + 3 = x

\frac{1}{f^{-1}(x)} = x - 3

f^{-1}(x) = \frac{1}{x - 3}

The inverse function is $f^{-1}: \mathbb{R} \setminus \{3\} \to \mathbb{R}, f^{-1}(x) = \frac{1}{x - 3}$

To find intersections, solve $f(x) = f^{-1}(x)$ :

\frac{1}{x} + 3 = \frac{1}{x - 3}

3x^2 - 9x - 3 = 0

x^2 - 3x - 1 = 0

x = \frac{1}{2}(3 - \sqrt{13}) \text{ or } x = \frac{1}{2}(3 + \sqrt{13})

The points of intersection are $\left(\frac{1}{2}(3 - \sqrt{13}), \frac{1}{2}(3 - \sqrt{13})\right)$ and $\left(\frac{1}{2}(3 + \sqrt{13}), \frac{1}{2}(3 + \sqrt{13})\right)$ .

infoNote

Exam tip: You can often find intersection points more easily by solving $f(x) = x$ rather than $f(x) = f^{-1}(x)$ . Points on the line $y = x$ automatically satisfy both $f(x) = f^{-1}(x)$ and have equal coordinates.

lightbulbExample

Worked Example: Cube root function

Find the inverse of the function $f(x) = 3\sqrt{x + 2} + 4$ and sketch both functions.

Solution:

Consider $x = 3\sqrt{y + 2} + 4$ and solve for $y$ :

\frac{x - 4}{3} = \sqrt{y + 2}

y = \left(\frac{x - 4}{3}\right)^2 - 2

f^{-1}(x) = \left(\frac{x - 4}{3}\right)^2 - 2

The domain of $f^{-1}$ equals the range of $f$ :

f^{-1}: [4, \infty) \to \mathbb{R}, f^{-1}(x) = \left(\frac{x - 4}{3}\right)^2 - 2

Which can also be written as:

f^{-1}(x) = \frac{x^2}{9} - \frac{8x}{9} - \frac{2}{9}, \quad x \geq 4

Using technology

Modern calculators and software can help find and graph inverse functions.

Finding the inverse rule: Use the solve function to solve the equation $x = f(y)$ for $y$ .

Graphing the inverse: Many graphing calculators have a built-in feature to graph the inverse of a function by reflection.

lightbulbExample

Worked Example: Rational function in transformed form

Express $\frac{x + 4}{x + 1}$ in the form $\frac{a}{x + b} + c$ . Hence find the inverse of the function $f(x) = \frac{x + 4}{x + 1}$ and sketch both functions.

Solution:

\frac{x + 4}{x + 1} = \frac{3 + x + 1}{x + 1} = \frac{3}{x + 1} + \frac{x + 1}{x + 1} = \frac{3}{x + 1} + 1

Consider $x = \frac{3}{y + 1} + 1$ and solve for $y$ :

x - 1 = \frac{3}{y + 1}

y + 1 = \frac{3}{x - 1}

y = \frac{3}{x - 1} - 1

The range of $f$ is $\mathbb{R} \setminus \{1\}$ , so:

f^{-1}: \mathbb{R} \setminus \{1\} \to \mathbb{R}, f^{-1}(x) = \frac{3}{x - 1} - 1

The graphs meet where $\frac{3}{x + 1} + 1 = x$ (for $x \neq -1$ ), which gives $x = \pm 2$ .

The graphs intersect at (2, 2) and (-2, -2).

infoNote

For rational functions, expressing them in transformed form (like $\frac{a}{x + b} + c$ ) can make finding the inverse much easier. This form clearly shows the transformations applied to the basic reciprocal function.

Functions requiring domain restriction

Some functions are not one-to-one over their natural domain. To find an inverse, we must first restrict the domain to make the function one-to-one.

lightbulbExample

Worked Example: Reciprocal squared function

Let $f$ be the function $f(x) = \frac{1}{x^2}$ for $x \in \mathbb{R} \setminus \{0\}$ . Define a suitable restriction $g$ of $f$ such that $g^{-1}$ exists, and find $g^{-1}$ .

Solution:

The function $f$ is not one-to-one because, for example, $f(2) = f(-2) = \frac{1}{4}$ .

We can create one-to-one restrictions:

$f_1: (0, \infty) \to \mathbb{R}, f_1(x) = \frac{1}{x^2}$ with range $(0, \infty)$

$f_2: (-\infty, 0) \to \mathbb{R}, f_2(x) = \frac{1}{x^2}$ with range $(0, \infty)$

Let's find the inverse of $f_1$ :

f_1(f_1^{-1}(x)) = x

\frac{1}{(f_1^{-1}(x))^2} = x

f_1^{-1}(x) = \pm\frac{1}{\sqrt{x}}

Since $\text{ran } f_1^{-1} = \text{dom } f_1 = (0, \infty)$ , we take the positive square root:

f_1^{-1}(x) = \frac{1}{\sqrt{x}}

Therefore $f_1^{-1}: (0, \infty) \to \mathbb{R}, f_1^{-1}(x) = \frac{1}{\sqrt{x}}$

chatImportant

When restricting the domain of a function to create a one-to-one function, choose the restriction based on the context of the problem. Common choices include:

Restricting to positive values only $(0, \infty)$ or $[0, \infty)$
Restricting to negative values only $(-\infty, 0)$ or $(-\infty, 0]$
Choosing the principal branch for trigonometric functions

bookmarkSummary

Remember! Key Points About Inverse Functions:

A function has an inverse if and only if it is one-to-one (passes the horizontal line test)
The domain and range swap: $\text{dom } f^{-1} = \text{ran } f$ and $\text{ran } f^{-1} = \text{dom } f$
The graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y = x$
To find an inverse algebraically: interchange $x$ and $y$ , then solve for $y$
The notation $f^{-1}$ means inverse function, not reciprocal $\frac{1}{f(x)}$
When solving for inverses involving square roots, the domain restriction determines which sign to choose

Inverse Functions (VCE SSCE Mathematical Methods): Revision Notes