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When can a function be reversed, and how do composition and inversion interact with domain and range?

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

Building composite functions and tracking their domains, the one-to-one condition for an inverse to exist, finding inverse functions algebraically, and the reflection relationship in y equals x.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

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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

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 $f \circ g \ne g \circ f$ .

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.

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.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20222 marksCalculator-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 20221 marksCalculator-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 20233 marksCalculator-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 $x \ne 1$ .