Composition and domain restrictions | Year 12

01 / LEARN

Understand the idea.

Order matters

In the composition $(f \circ g) (x) = f (g (x))$ , $g$ acts first. Treat its complete rule as the input of $f$ , using brackets where needed. Reversing the order usually changes both the rule and the allowed inputs.

Check two conditions: $x$ must belong to the domain of the inner function, and the resulting inner output must belong to the domain of the outer function. Neither condition can be replaced by simplifying the final rule.

Simplification does not restore inputs

For real-valued functions, a square root needs a nonnegative radicand, a logarithm needs a positive argument, and a denominator must be nonzero. Apply these restrictions at the stage where each operation occurs.

Simplification can hide an earlier restriction. If ${(\sqrt{x})}^{2}$ simplifies to $x$ , the original square root still required $x \geq 0$ . A function is its rule together with its domain, not the algebraic expression alone.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Root after a line

Let $f (x) = \sqrt{x - 1}$ and $g (x) = 2 x + 3$ . Find both compositions and their domains.

Worked solution

1
Substitute $g$ into $f$ .
$f (g (x)) = \sqrt{(2 x + 3) - 1} = \sqrt{2 x + 2} .$
The whole output of the linear function replaces the input of the square root.
2
Find the first composition domain.
$2 x + 2 \geq 0 \Rightarrow x \geq - 1.$
The inner linear function accepts every real input, and the outer radicand must be nonnegative.
3
Reverse the order.
$g (f (x)) = 2 \sqrt{x - 1} + 3.$
Now the root function acts first and its output enters the linear function.
4
Find the reverse composition domain.
$x - 1 \geq 0 \Rightarrow x \geq 1.$
The inner root restricts $x$ ; the outer linear function places no additional restriction.

Answer

$f (g (x)) = \sqrt{2 x + 2}$ , $x \geq - 1$ ; $g (f (x)) = 2 \sqrt{x - 1} + 3$ , $x \geq 1$ .

EXAMPLE 02

A hidden restriction survives

Let $f (x) = \sqrt{x}$ and $g (x) = x^{2} - 4$ . Find the domain of $g \circ f$ .

Worked solution

1
Apply the inner-domain restriction.
$f (x) = \sqrt{x} \Rightarrow x \geq 0.$
The real square root must exist before the outer function can act.
2
Compose and simplify on that domain.
$g (f (x)) = {(\sqrt{x})}^{2} - 4 = x - 4, x \geq 0.$
Squaring undoes the root only at inputs where the root was defined.
3
Check for further restrictions.
$dom (g \circ f) = [0, \infty) .$
The outer polynomial accepts every output of the root, so there is no additional exclusion.

Answer

$x \geq 0$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

For $f (x) = 1 / x$ and $g (x) = x - 4$ , which input is excluded from $f (g (x))$ ?

Show answer & reasoning

x=4.

The composition is $f (g (x)) = \frac{1}{x - 4}$ . At $x = 4$ , the inner function produces zero, which the reciprocal cannot accept; therefore $4$ is excluded.

Keep practising

6 worksheet questions · 20 marks

Open worksheet Worked solutions

Read the questions online

Let $f (x) = \sqrt{x - 1}$ and $g (x) = 2 x + 3$ . Find both compositions and their domains.
4 marks
Let $f (x) = \frac{1}{x - 2}$ and $g (x) = x^{2}$ . Find the domain of $f \circ g$ .
3 marks
For $f (x) = ln x$ and $g (x) = 3 - x$ , find $(f \circ g) (x)$ and its domain.
3 marks
Let $f (x) = \sqrt{x}$ and $g (x) = x^{2} - 4$ . Find the domain of $g \circ f$ .
3 marks
Let $f (x) = x + 1$ and $g (x) = \frac{1}{x}$ . Compare $f \circ g$ and $g \circ f$ .
4 marks
Let $f (x) = \sqrt{4 - x^{2}}$ and $g (x) = x - 1$ . Find the domain of $f \circ g$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Let $f (x) = \sqrt{x - 1}$ and $g (x) = 2 x + 3$ . Find both compositions and their domains.
4 marks
For $f$ after $g$ , require $2 x + 3 \geq 1$ .
For $g$ after $f$ , the inner root requires $x \geq 1$ ; $g$ accepts every real output.
$f (g (x)) = \sqrt{2 x + 2}$ , $x \geq - 1$ ; $g (f (x)) = 2 \sqrt{x - 1} + 3$ , $x \geq 1$ .
Let $f (x) = \frac{1}{x - 2}$ and $g (x) = x^{2}$ . Find the domain of $f \circ g$ .
3 marks
$f (g (x)) = \frac{1}{x^{2} - 2}$ .
Require $x^{2} - 2 \neq 0$ .
All real $x$ except $\pm \sqrt{2}$ .
For $f (x) = ln x$ and $g (x) = 3 - x$ , find $(f \circ g) (x)$ and its domain.
3 marks
The logarithm requires $3 - x > 0$ .
$ln (3 - x)$ , $x < 3$ .
Let $f (x) = \sqrt{x}$ and $g (x) = x^{2} - 4$ . Find the domain of $g \circ f$ .
3 marks
The inner square root must be defined.
After substitution, $g (f (x)) = x - 4$ , but the original inner-domain restriction remains.
$x \geq 0$ .
Let $f (x) = x + 1$ and $g (x) = \frac{1}{x}$ . Compare $f \circ g$ and $g \circ f$ .
4 marks
Substitute into the outer rule in each order.
The expressions and excluded inputs differ.
$\frac{1}{x} + 1$ with $x \neq 0$ ; $\frac{1}{x + 1}$ with $x \neq - 1$ .
Let $f (x) = \sqrt{4 - x^{2}}$ and $g (x) = x - 1$ . Find the domain of $f \circ g$ .
3 marks
Require $- 2 \leq x - 1 \leq 2$ .
Add 1 throughout.
$- 1 \leq x \leq 3$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist functions and sketching graphs. Calculus-based analysis assumes concurrent Methods differentiation; real-polynomial division detail is taught in the separate Real Polynomials chapter.

Official curriculum source ↗

In this lesson

Understand the idea.

Order matters

Simplification does not restore inputs

See the method in action.

Root after a line

Substitute g⁡ into f⁡.

Find the first composition domain.

Reverse the order.

Find the reverse composition domain.

A hidden restriction survives

Apply the inner-domain restriction.

Compose and simplify on that domain.

Check for further restrictions.

Try it yourself.

Check your understanding

Keep practising

Substitute $g$ into $f$ .