Composition and domain restrictions | Year 12

01 / LEARN

The idea, step by step

Order matters

In f composed with g, g acts first. Substitute its complete output into the rule for f, including brackets. The notation f(g(x)) is not a product and composition generally does not commute.

A composite exists only when x belongs to the inner domain and its resulting value belongs to the outer domain. Check both conditions before simplifying.

Simplification does not restore inputs

For a root, the radicand must be nonnegative; for a real logarithm, its argument must be positive; for a denominator, its value must be nonzero. Carry these restrictions through a simplification even if the final expression looks less restricted.

Checking a proposed domain by trying an endpoint often exposes a lost exclusion. The domain is part of the function, not an optional annotation.

A common trap

Using only the domain of the simplified expression.

Instead: Intersect the inner domain with the condition that its output is accepted by the outer function.

02 / WORKED EXAMPLES

See the thinking

Example 1

Root after a line

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

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

Example 2

A hidden restriction survives

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

The inner square root must be defined.
After substitution, $g (f (x)) = x - 4$ , but the original inner-domain restriction remains.

x \geq 0

Check your understanding

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

Show answer

x=4.

It would send zero into the reciprocal.

03 / WORKSHEET

Your turn

6 questions · 20 marks · Work at your own pace.

Worksheet PDF 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 $f (x) = 1 / (x - 2)$ 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 $g (x) = 1 / x$ . 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 $f (x) = 1 / (x - 2)$ and $g (x) = x^{2}$ . Find the domain of $f \circ g$ .
3 marks
$f (g (x)) = 1 / (x^{2} - 2)$ .
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 $g (x) = 1 / x$ . Compare $f \circ g$ and $g \circ f$ .
4 marks
Substitute into the outer rule in each order.
The expressions and excluded inputs differ.
$1 / x + 1$ with $x \neq 0$ ; $1 / (x + 1)$ 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.

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