What are composite functions?

The composite fg means "do g first, then f": fg(x) = f(g(x)). The order matters, and in general fg ≠ gf.

What are inverse functions?

The inverse f^-1 undoes f: f^-1(f(x)) = x. An inverse exists only if f is one-to-one. To find it:

Given f(x) = x^2 for x ≥ 0 and g(x) = x - 4 for x R, find gf(x) and fg(x). [3 marks]

The function f is defined by f(x) = 3 - 2x for x R. Find f^-1(x) and verify f^-1f(x) = x. [3 marks]

Explain the geometric relationship between the graphs of f and f^-1, and state where they can intersect. [2 marks]

SEAB Original-style practice: The functions f and g are defined by f(x) = 2x + 1 for x R and g(x) = 1x for x R, x ≠ 0. Find fg(x), state its domain, and find an expression for f^-1(x).

fg(x) = f(g(x)) = 2(1x) + 1 = 2x + 1. The domain of fg is the domain of g (the inner function), namely x R, x ≠ 0, since every output of g is a valid input to f. For the inverse of f: let y = 2x + 1, so x = y - 12. Hence f^-1(x) = x - 12 for x R. Markers reward correct composition order (apply g first), the domain taken from the inner function, and a correct rearrangement for the inverse.

SEAB Original-style practice: The function h is defined by h(x) = (x - 3)^2 + 1 for x ≥ 3. Find h^-1(x) and state its domain.

Let y = (x - 3)^2 + 1, so (x - 3)^2 = y - 1 and x - 3 = (y - 1). Because the domain is x ≥ 3 we have x - 3 ≥ 0, so we take the positive root: x = 3 + sqrt(y - 1). Hence h^-1(x) = 3 + sqrt(x - 1). The domain of h^-1 is the range of h. Since h has minimum 1 at x = 3 and increases, the range is [1, ∞), so the domain of h^-1 is x ≥ 1. Markers reward choosing the correct root using the domain, the inverse expression, and the domain of the inverse equal to the range of h.

SingaporeMathsSyllabus dot point

How do we combine functions and reverse them, and when does an inverse exist?

Form and find the domain of composite functions, determine when a composite is defined, find inverse functions and their domains, and use the graphical relationship between a function and its inverse

A focused answer to the H2 Mathematics outcome on composite and inverse functions. Forming composites and their domains, the condition for a composite to exist, finding inverses, and the reflection in y equals x.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

SEAB wants you to combine functions by composition, decide when a composite is defined and state its domain, find inverse functions and their domains, and use the fact that the graph of $\mathrm{f}^{-1}$ is the reflection of the graph of $\mathrm{f}$ in the line $y = x$ .

The answer

Composite functions

The composite $\mathrm{fg}$ means "do $\mathrm{g}$ first, then $\mathrm{f}$ ": $\mathrm{fg}(x) = \mathrm{f}(\mathrm{g}(x))$ . The order matters, and in general $\mathrm{fg} \neq \mathrm{gf}$ .

The composite $\mathrm{fg}$ is defined only if the range of $\mathrm{g}$ is contained in the domain of $\mathrm{f}$ , so that every output of $\mathrm{g}$ is a legal input for $\mathrm{f}$ . When it exists, the domain of $\mathrm{fg}$ is the domain of the inner function $\mathrm{g}$ .

When the composite fails

If some output of $\mathrm{g}$ lies outside the domain of $\mathrm{f}$ , the composite $\mathrm{fg}$ is not defined. For example, if $\mathrm{f}(x) = \sqrt{x}$ (domain $x \geq 0$ ) and $\mathrm{g}$ produces negative values, then $\mathrm{fg}$ cannot be formed for those inputs.

Inverse functions

The inverse $\mathrm{f}^{-1}$ undoes $\mathrm{f}$ : $\mathrm{f}^{-1}(\mathrm{f}(x)) = x$ . An inverse exists only if $\mathrm{f}$ is one-to-one. To find it:

Write $y = \mathrm{f}(x)$ .
Rearrange to make $x$ the subject.
Swap to get $\mathrm{f}^{-1}(x)$ .
The domain of $\mathrm{f}^{-1}$ is the range of $\mathrm{f}$ , and the range of $\mathrm{f}^{-1}$ is the domain of $\mathrm{f}$ .

The graphical relationship

The graph of $\mathrm{f}^{-1}$ is the reflection of the graph of $\mathrm{f}$ in the line $y = x$ . A point $(a, b)$ on $\mathrm{f}$ corresponds to $(b, a)$ on $\mathrm{f}^{-1}$ . Where the graphs of $\mathrm{f}$ and $\mathrm{f}^{-1}$ intersect, they meet on the line $y = x$ (for increasing functions).

Worked example

The function $\mathrm{f}$ is defined by $\mathrm{f}(x) = \dfrac{2x + 1}{x - 1}$ for $x \in \mathbb{R}$ , $x > 1$ . Find $\mathrm{f}^{-1}(x)$ and state its domain.

Step 1: Set y equal to the rule

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

Step 2: Make x the subject

Multiply through: $y(x - 1) = 2x + 1$ , so $yx - y = 2x + 1$ . Collect $x$ terms: $yx - 2x = 1 + y$ , giving $x(y - 2) = y + 1$ . Hence

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

Step 3: Write the inverse

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

Step 4: Find the domain of the inverse

The domain of $\mathrm{f}^{-1}$ is the range of $\mathrm{f}$ . As $x \to 1^+$ , $\mathrm{f}(x) \to +\infty$ ; as $x \to \infty$ , $\mathrm{f}(x) \to 2^+$ . On $x > 1$ the function is decreasing toward $2$ , so the range is $(2, \infty)$ . Hence the domain of $\mathrm{f}^{-1}$ is $x > 2$ .

Examples in context

Example 1. Checking a composite exists. With $\mathrm{f}(x) = \ln x$ (domain $x > 0$ ) and $\mathrm{g}(x) = x^2 + 1$ , the range of $\mathrm{g}$ is $[1, \infty) \subseteq (0, \infty)$ , so $\mathrm{fg}(x) = \ln(x^2 + 1)$ is defined for all real $x$ . Reversing, $\mathrm{gf}(x) = (\ln x)^2 + 1$ needs $x > 0$ .

Example 2. Inverse of an exponential model. A population $P(t) = 200 e^{0.1t}$ is one-to-one, so its inverse $t = 10 \ln\left(\dfrac{P}{200}\right)$ recovers the time from a measured population. The domain of the inverse is $P \geq 200$ , matching the range of the original model.

Try this

Q1. Given $\mathrm{f}(x) = x^2$ for $x \geq 0$ and $\mathrm{g}(x) = x - 4$ for $x \in \mathbb{R}$ , find $\mathrm{gf}(x)$ and $\mathrm{fg}(x)$ . [3 marks]

Cue. $\mathrm{gf}(x) = x^2 - 4$ ; $\mathrm{fg}(x) = (x - 4)^2$ but only valid where $x - 4 \geq 0$ , that is $x \geq 4$ .

Q2. The function $\mathrm{f}$ is defined by $\mathrm{f}(x) = 3 - 2x$ for $x \in \mathbb{R}$ . Find $\mathrm{f}^{-1}(x)$ and verify $\mathrm{f}^{-1}\mathrm{f}(x) = x$ . [3 marks]

Cue. $\mathrm{f}^{-1}(x) = \dfrac{3 - x}{2}$ ; substituting $\mathrm{f}(x)$ gives $\dfrac{3 - (3 - 2x)}{2} = x$ .

Q3. Explain the geometric relationship between the graphs of $\mathrm{f}$ and $\mathrm{f}^{-1}$ , and state where they can intersect. [2 marks]

Cue. $\mathrm{f}^{-1}$ is the reflection of $\mathrm{f}$ in $y = x$ ; for increasing functions any intersection lies on $y = x$ .

Exam-style practice questions

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

Original5 marksThe functions

\mathrm{f}

and

\mathrm{g}

are defined by

\mathrm{f}(x) = 2x + 1

for

x \in \mathbb{R}

and

\mathrm{g}(x) = \dfrac{1}{x}

for

x \in \mathbb{R}

x \neq 0

. Find

\mathrm{fg}(x)

, state its domain, and find an expression for

\mathrm{f}^{-1}(x)

Show worked answer →

$\mathrm{fg}(x) = \mathrm{f}(\mathrm{g}(x)) = 2\left(\dfrac{1}{x}\right) + 1 = \dfrac{2}{x} + 1$ .

The domain of $\mathrm{fg}$ is the domain of $\mathrm{g}$ (the inner function), namely $x \in \mathbb{R}$ , $x \neq 0$ , since every output of $\mathrm{g}$ is a valid input to $\mathrm{f}$ .

For the inverse of $\mathrm{f}$ : let $y = 2x + 1$ , so $x = \dfrac{y - 1}{2}$ . Hence $\mathrm{f}^{-1}(x) = \dfrac{x - 1}{2}$ for $x \in \mathbb{R}$ .

Markers reward correct composition order (apply $\mathrm{g}$ first), the domain taken from the inner function, and a correct rearrangement for the inverse.

Original4 marksThe function

\mathrm{h}

is defined by

\mathrm{h}(x) = (x - 3)^2 + 1

for

x \geq 3

. Find

\mathrm{h}^{-1}(x)

and state its domain.

Show worked answer →

Let $y = (x - 3)^2 + 1$ , so $(x - 3)^2 = y - 1$ and $x - 3 = \pm\sqrt{y - 1}$ .

Because the domain is $x \geq 3$ we have $x - 3 \geq 0$ , so we take the positive root: $x = 3 + \sqrt{y - 1}$ .

Hence $\mathrm{h}^{-1}(x) = 3 + \sqrt{x - 1}$ .

The domain of $\mathrm{h}^{-1}$ is the range of $\mathrm{h}$ . Since $\mathrm{h}$ has minimum $1$ at $x = 3$ and increases, the range is $[1, \infty)$ , so the domain of $\mathrm{h}^{-1}$ is $x \geq 1$ .

Markers reward choosing the correct root using the domain, the inverse expression, and the domain of the inverse equal to the range of $\mathrm{h}$ .