Composite and inverse functions (SACE Stage 2 Specialist Mathematics)

Syllabus dot point

“Form composite functions and find inverse functions, attending to domain and range and the condition for an inverse to exist.”

8 min readUpdated 2 Jun 2026

Quick answer

A composite $f \circ g$ means "do $g$ first, then $f$ ": $(f \circ g)(x) = f(g(x))$ , and it is only defined where $g(x)$ lands in the domain of $f$ . A function has an inverse exactly when it is one-to-one (passes the horizontal line test). To find $f^{-1}$ , swap $x$ and $y$ and solve for $y$ ; its graph is the reflection of $f$ in the line $y = x$ , with domain and range swapped.

Jump to a section

What this dot point is asking
Composite functions
When does an inverse exist?
Finding an inverse algebraically
Composing more than two functions
The graphical relationship
Exam-style questions

What this dot point is asking

You need to form composite functions with correct domains, determine when an inverse function exists, find inverse functions, and relate a function and its inverse graphically.

Composite functions

The composite $(f \circ g)(x) = f(g(x))$ applies $g$ first and feeds the result into $f$ . Order matters: in general $f \circ g \ne g \circ f$ .

Domain of a composite

$(f \circ g)(x)$ is defined only when $x$ is in the domain of $g$ and $g(x)$ lies in the domain of $f$ . Always check the inner output is allowed as an input to the outer function - the natural domain of the composite can be smaller than the domain of $g$ alone.

For example, if $f(x) = \sqrt{x}$ and $g(x) = x - 4$ , then $(f \circ g)(x) = \sqrt{x - 4}$ , defined only for $x \ge 4$ , even though $g$ itself is defined for all real $x$ .

When does an inverse exist?

An inverse function $f^{-1}$ reverses $f$ : if $f(a) = b$ then $f^{-1}(b) = a$ . For this reversal to be a function, each output must come from exactly one input.

One-to-one and the horizontal line test

A function is one-to-one (injective) if different inputs always give different outputs. Graphically, it passes the horizontal line test: no horizontal line meets the graph more than once. A function has an inverse function if and only if it is one-to-one. A many-to-one function (such as $f(x) = x^2$ on all of $\mathbb{R}$ ) has no inverse unless its domain is restricted.

Restricting the domain can create an inverse: $f(x) = x^2$ on $x \ge 0$ is one-to-one and has inverse $f^{-1}(x) = \sqrt{x}$ .

Finding an inverse algebraically

The mechanical recipe: write $y = f(x)$ , swap $x$ and $y$ , then solve for $y$ . The domain of $f^{-1}$ is the range of $f$ , and the range of $f^{-1}$ is the domain of $f$ .

Finding an inverse with domain and range

Find the inverse of $f(x) = \dfrac{2x + 1}{x - 3}$ , stating its domain

The domain of $f$ excludes $x = 3$ .

Step 1: Swap $x$ and $y$

The mechanical recipe for finding an inverse is to write $y = f(x)$ and then swap the roles of $x$ and $y$ . This reflects the graph in the line $y = x$ , which is exactly what inversion does geometrically.

Set $y = \dfrac{2x+1}{x-3}$ and swap variables:

x = \frac{2y + 1}{y - 3}.

Step 2: Solve for $y$ algebraically

Multiply both sides by $(y-3)$ to clear the denominator, then collect all $y$ terms on one side and factor.

x(y - 3) = 2y + 1 \implies xy - 3x = 2y + 1.

Collect $y$ terms:

xy - 2y = 3x + 1 \implies y(x - 2) = 3x + 1 \implies y = \frac{3x + 1}{x - 2}.

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

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

The domain of $f^{-1}$ is the range of $f$ , because the inverse swaps inputs and outputs. As $x \to \infty$ , $f(x) \to 2$ , and $y = 2$ is never attained (the horizontal asymptote is approached but not reached), so the range of $f$ is all reals except $2$ .

Hence the domain of $f^{-1}$ is $x \ne 2$ , consistent with the denominator $x - 2$ .

Final answer: $f^{-1}(x) = \dfrac{3x + 1}{x - 2}$ , domain $x \ne 2$ .

Composing more than two functions

Composition extends to three or more functions, applied from the inside out. For $h(x)=f(g(k(x)))$ you evaluate $k$ first, feed the result to $g$ , then to $f$ . Each layer narrows the domain: a value is allowed only if it survives every stage. This nesting is exactly the structure the chain rule differentiates, so a clear grasp of composition here pays off directly in the calculus topics. When asked to decompose a function - to write $\sqrt{3x+1}$ as $f(g(x))$ - identify the outer operation ( $\sqrt{\;}$ ) and the inner expression ( $3x+1$ ); the decomposition is rarely unique, but the natural one names the last operation performed as the outer function.

The graphical relationship

The graph of $f^{-1}$ is the reflection of the graph of $f$ in the line $y = x$ . Consequently a point $(a, b)$ on $f$ corresponds to $(b, a)$ on $f^{-1}$ , and any intersection of $f$ with its inverse lies on the line $y = x$ (for increasing functions). Composing a function with its inverse returns the input:

f^{-1}(f(x)) = x \quad \text{and} \quad f(f^{-1}(x)) = x,

each on the appropriate domain.

Verifying an inverse

To check your inverse is correct, compute $f(f^{-1}(x))$ ; it should simplify to $x$ . This catches algebra slips far more reliably than re-reading the rearrangement.

Common errors

Reversing the order of composition: $(f \circ g)(x) = f(g(x))$ applies $g$ first. Writing $g(f(x))$ instead is a frequent error.
Ignoring the composite's domain: The inner function's output must be a legal input for the outer function; the composite domain may be smaller than expected.
Inverting a many-to-one function: A function with no domain restriction may fail the horizontal line test and have no inverse. State and use a restricted domain when needed.
Forgetting to swap domain and range: The domain of $f^{-1}$ is the range of $f$ . Stating the inverse rule without its restricted domain is an incomplete answer.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2022-style2 marks

Calculator-assumed. The graph of

g(x) = \sin x + \cos x

is drawn for

-\pi \le x \le \pi

. Explain why

g(x)

is a function but does not have an inverse function.

Show worked answer →

$g(x)$ is a function because every $x$ in $-\pi \le x \le \pi$ maps to exactly one value of $g(x)$ ; the graph passes the vertical line test. [1 mark]

However, $g(x)$ is not one-to-one over this domain: the curve rises then falls (it has a maximum), so a horizontal line such as $y = 1$ cuts the graph more than once, meaning two different $x$ values give the same output. [1 mark]

For an inverse to exist the function must be one-to-one (pass the horizontal line test), so the domain would need restricting to an interval on which $g$ is strictly monotonic.

SACE 2022-style1 mark

Calculator-assumed. The rule

f(x) = \sin x + \cos x

is restricted to a domain on which it is one-to-one. Explain why this restricted function does have an inverse.

Show worked answer →

Since $\sin x + \cos x = \sqrt{2}\sin\!\left(x + \tfrac{\pi}{4}\right)$ , restricting the domain to an interval on which it is strictly monotonic makes the function one-to-one. [1 mark]

Because each output is now produced by exactly one input, the graph passes the horizontal line test, so the mapping can be reversed: every $y$ in the range corresponds to a single $x$ . A one-to-one function always has an inverse.

SACE 2023-style3 marks

Calculator-free. Given

f(x) = \dfrac{x + 2}{x - 1}

for

x \ne 1

, find

f^{-1}(x)

and state its domain.

Show worked answer →

Set $y = \dfrac{x + 2}{x - 1}$ and swap variables: $x = \dfrac{y + 2}{y - 1}$ .

Solve: $x(y - 1) = y + 2$ , so $xy - x = y + 2$ , giving $xy - y = x + 2$ , hence $y(x - 1) = x + 2$ and $y = \dfrac{x + 2}{x - 1}$ .

So $f^{-1}(x) = \dfrac{x + 2}{x - 1}$ (this function is self-inverse). The domain of $f^{-1}$ is the range of $f$ , which is $x \ne 1$ .

Marks: one for swapping and rearranging, one for solving for $y$ , one for stating the domain $x \ne 1$ .

Practise this

Sources & how we know this

Stage 2 Specialist Mathematics subject outline (2025 onwards) — SACE Board
Stage 2 Specialist Mathematics: external assessment — SACE Board

What this dot point is asking

Composite functions

When does an inverse exist?

Finding an inverse algebraically

Find the inverse of f(x)=2x+1x−3f(x) = \dfrac{2x + 1}{x - 3}f(x)=x−32x+1​, stating its domain

Step 1: Swap xxx and yyy

Step 2: Solve for yyy algebraically

Step 3: Determine the domain of f−1f^{-1}f−1 from the range of fff

Composing more than two functions

The graphical relationship

Exam-style questions

Practise this

Sources & how we know this

Find the inverse of $f(x) = \dfrac{2x + 1}{x - 3}$ , stating its domain

Step 1: Swap $x$ and $y$

Step 2: Solve for $y$ algebraically

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