Composite Functions Revision Notes for VCE SSCE Mathematical Methods

Composite Functions

What is a composite function?

Think of a function like a machine that processes inputs to create outputs. The input values form the domain, and the output values form the range.

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The function machine analogy helps visualize how functions work: you feed in an input value, the machine processes it according to a rule, and out comes the output value. This visual model becomes especially useful when we start combining multiple functions together.

In many real-world situations, we need multiple operations to get from an input to a final output. When we apply one function and then apply another function to that result, we create a composite function.

For example, imagine you have:

Function $f(x) = 3x + 2$
Function $g(x) = x^2$

If you put an input through $f$ first and then put that output through $g$ , you're creating a composite function. Using input $3$ :

First, $f(3) = 3 \times 3 + 2 = 11$
Then, $g(11) = 11^2 = 121$

This creates a new function $h$ where $h(x) = (3x + 2)^2$ .

Understanding composite function notation

When we combine functions in this way, we write it as g ∘ f, which we read as "the composition of $f$ followed by $g$ " or simply " $g$ of $f$ ".

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Read the notation from right to left!

In $g \circ f$ , you apply $f$ first, then apply $g$ to the result. This is the opposite of the order you might expect from reading left to right, so it's a common source of confusion.

The rule for the composite function is written as:

g \circ f(x) = g(f(x))

This means: take the input $x$ , apply function $f$ to get $f(x)$ , then apply function $g$ to that result.

Conditions for composite functions to exist

Not all pairs of functions can be composed. For $g \circ f$ to exist, we need a crucial condition:

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The range of $f$ must be a subset of the domain of $g$

In mathematical notation: ran f ⊆ dom g

This makes sense when you think about it - the outputs from $f$ need to be valid inputs for $g$ . If $f$ produces an output that $g$ can't accept as input, the composition won't work.

When the composite function exists, its domain is:

\text{dom}(g \circ f) = \text{dom } f

The domain of the composite function is the same as the domain of the first function applied.

Finding composite functions

To find a composite function, follow these steps:

Check that the range of the first function fits within the domain of the second function
Substitute the entire rule of the first function into the second function
Simplify the resulting expression
Determine the domain and range of the composite function

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Worked Example: Finding and Comparing Composite Functions

Find both $f \circ g$ and $g \circ f$ , stating the domain and range of each, where:

$f: \mathbb{R} \to \mathbb{R}$ , $f(x) = 2x - 1$ and $g: \mathbb{R} \to \mathbb{R}$ , $g(x) = 3x^2$

Solution:

First, check if the compositions are defined by examining domains and ranges:

Function	Domain	Range
$g$	$\mathbb{R}$	$\mathbb{R}^+ \cup \{0\}$
$f$	$\mathbb{R}$	$\mathbb{R}$

Since $\text{ran } g \subseteq \text{dom } f$ and $\text{ran } f \subseteq \text{dom } g$ , both $f \circ g$ and $g \circ f$ are defined.

Finding $f \circ g$ :

f \circ g(x) = f(g(x))

\text{dom}(f \circ g) = \text{dom } g = \mathbb{R}

\text{ran}(f \circ g) = [-1, \infty)

Finding $g \circ f$ :

g \circ f(x) = g(f(x))

\text{dom}(g \circ f) = \text{dom } f = \mathbb{R}

\text{ran}(g \circ f) = [0, \infty)

Key observation: Notice that $f \circ g \neq g \circ f$ . The functions $6x^2 - 1$ and $12x^2 - 12x + 3$ are completely different! This shows that composite functions are not commutative - the order matters.

When composite functions aren't defined

Sometimes only one of the two possible compositions will be defined, or neither may be defined. Understanding why a composition fails to exist helps you recognize when function domains and ranges are incompatible.

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Worked Example: Identifying When Compositions Exist

For the functions $g(x) = 2x - 1$ , $x \in \mathbb{R}$ , and $f(x) = \sqrt{x}$ , $x \geq 0$ :

a) State which of $f \circ g$ and $g \circ f$ is defined.

b) For the composite function that is defined, state the domain and rule.

Solution:

a) Check the domains and ranges:

Function	Domain	Range
$g$	$\mathbb{R}$	$\mathbb{R}$
$f$	$\mathbb{R}^+ \cup \{0\}$	$\mathbb{R}^+ \cup \{0\}$

For $f \circ g$ : We need $\text{ran } g \subseteq \text{dom } f$

The range of $g$ is all real numbers, but the domain of $f$ only includes non-negative numbers. Since the range of $g$ is NOT a subset of the domain of $f$ , the composition $f \circ g$ is not defined.

For $g \circ f$ : We need $\text{ran } f \subseteq \text{dom } g$

The range of $f$ is $\mathbb{R}^+ \cup \{0\}$ , and the domain of $g$ is $\mathbb{R}$ . Since non-negative numbers are a subset of all real numbers, $g \circ f$ is defined.

b) Finding $g \circ f$ :

g \circ f(x) = g(f(x))

\text{dom}(g \circ f) = \text{dom } f = \mathbb{R}^+ \cup \{0\}

Function restrictions

Sometimes we need to restrict the domain of a function to make a composition work. This technique allows us to "fix" situations where ranges and domains don't naturally align.

lightbulbExample

Worked Example: Restricting Functions for Valid Composition

For the functions $f(x) = x^2 - 1$ , $x \in \mathbb{R}$ , and $g(x) = \sqrt{x}$ , $x \geq 0$ :

a) State why $g \circ f$ is not defined.

b) Define a restriction $f^*$ of $f$ such that $g \circ f^*$ is defined, and find $g \circ f^*$ .

Solution:

a) Check domains and ranges:

Function	Domain	Range
$f$	$\mathbb{R}$	$[-1, \infty)$
$g$	$\mathbb{R}^+ \cup \{0\}$	$\mathbb{R}^+ \cup \{0\}$

The range of $f$ includes negative values (from $-1$ upwards), but $g$ can only accept non-negative inputs. Since $\text{ran } f \not\subseteq \text{dom } g$ , the composition $g \circ f$ is not defined.

b) For $g \circ f^*$ to work, we need $\text{ran } f^* \subseteq \mathbb{R}^+ \cup \{0\}$ .

This means the range of $f^*$ must only include non-negative values. Looking at the graph of $y = x^2 - 1$ , the function equals zero when $x = \pm 1$ and is positive when $x < -1$ or $x > 1$ .

To ensure the range is non-negative, we restrict the domain to:

\{x : x \leq -1\} \cup \{x : x \geq 1\} = \mathbb{R} \setminus (-1, 1)

So we define $f^*$ by:

$f^*: \mathbb{R} \setminus (-1, 1) \to \mathbb{R}$ , $f^*(x) = x^2 - 1$

Then:

g \circ f^*(x) = g(f^*(x))

\text{dom}(g \circ f^*) = \text{dom } f^* = \mathbb{R} \setminus (-1, 1)

The composite function is: $g \circ f^*: \mathbb{R} \setminus (-1, 1) \to \mathbb{R}$ , $g \circ f^*(x) = \sqrt{x^2 - 1}$

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Key Points to Remember:

A composite function combines two functions: apply the first function, then apply the second function to that result
The notation $g \circ f$ means "apply $f$ first, then $g$ " - read from right to left
For $g \circ f$ to exist, the range of $f$ must be a subset of the domain of $g$ : ran f ⊆ dom g
In general, $f \circ g \neq g \circ f$ - order matters in function composition
The domain of $g \circ f$ equals the domain of $f$ : $\text{dom}(g \circ f) = \text{dom } f$
Sometimes you need to restrict a function's domain to make a composition possible

Composite Functions (VCE SSCE Mathematical Methods): Revision Notes