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Module 1 · L7 of 15 ~40 min ⚡ +50 XP in Learn · +25 to complete

Composite Functions

In a supply chain, raw materials go to a factory, and the factory's output goes to a distribution centre. The final product depends on two connected processes, one after the other. Composite functions work exactly the same way: one function feeds its output directly into another.

Today's hook, Imagine two machines. Machine A doubles its input. Machine B adds 3 to its input. If you put a number into Machine A, then take the output and feed it into Machine B, what is the overall rule? Does it matter whether Machine A or Machine B goes first?
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Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.

Worksheets

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Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

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Recall, your gut answer first
+5 XP warm-up

Imagine two machines. Machine A doubles its input. Machine B adds 3 to its input. If you put a number into Machine A, then take the output and feed it into Machine B, what is the overall rule? Does it matter whether Machine A or Machine B goes first?

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Formula reference · this lesson
core notation

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

$g$  = inner function (applied first)
$f$  = outer function (applied second)
Domain: $x$ must be valid for $g$, and $g(x)$ must be valid for $f$

Key insight: Order matters. In almost all cases, $f(g(x)) \neq g(f(x))$.

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What you'll master
Know

Key facts

  • the notation $(f \circ g)(x) = f(g(x))$
  • how to evaluate composite functions numerically
  • the domain conditions for composite functions
Understand

Concepts

  • why composite functions model chained processes
  • why order matters in composition
  • how the domain of a composite is restricted by both inner and outer functions
Can do

Skills

  • evaluate composite functions for given inputs
  • find the algebraic expression for a composite function
  • determine the domain of a composite function
  • compare $f \circ g$ and $g \circ f$ and explain when they differ
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Key terms
Function
A relation where each input has exactly one output.
Domain
The set of all possible input values for a function.
Range
The set of all possible output values for a function.
Composite Function
A function formed when the output of one function becomes the input of another.
Inner Function
The function applied first in a composition; $g$ in $(f \circ g)(x)$.
Outer Function
The function applied second in a composition; $f$ in $(f \circ g)(x)$.
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What is a composite function?

Work through the core explanation before applying it.

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What is a composite function?
core concept · +3 XP at end

A composite function is formed when the output of one function becomes the input of another. We write this as $(f \circ g)(x)$, which means $f(g(x))$: first apply $g$, then apply $f$ to the result.

Supply chains are composite. A factory takes raw materials and produces components. A second factory takes those components and assembles the final product. The overall transformation from raw materials to finished product is a composite of two separate functions. You cannot understand the whole process by looking at either factory in isolation.

The notation can be confusing at first. Remember:

  • $(f \circ g)(x)$ is read "$f$ composed with $g$" or "$f$ of $g$ of $x$"
  • The function on the right ($g$) is applied first
  • The function on the left ($f$) is applied second

How to evaluate a composite function

  1. Start with the inner function. Evaluate $g(x)$ for the given input.
  2. Take that result and use it as the input for the outer function $f$.
  3. Simplify.

How to find the algebraic expression

To find $(f \circ g)(x)$ algebraically, substitute the entire expression for $g(x)$ into every $x$ in $f(x)$. Use brackets to avoid errors.

Domain of a composite function

The domain of $f \circ g$ has two requirements:

  1. $x$ must be in the domain of $g$
  2. $g(x)$ must be in the domain of $f$

Both conditions must be satisfied. If either fails, the composite is undefined at that point.

$(f \circ g)(x) = f(g(x))$, $g$ is the inner function (applied first), $f$ is the outer function (applied second); Evaluation: Step 1: evaluate $g(x)$; Step 2: use that result as input for $f$; Step 3: simplify

Pause, copy the composite notation $(f \circ g)(x) = f(g(x))$ and the three-step evaluation procedure (evaluate inner $g$ first, feed result into outer $f$, simplify) into your book.

Did you get this? True or false: in $(f \circ g)(x)$, the function $f$ is applied first.

Quick check: If $f(x) = 2x$ and $g(x) = x + 3$, what is $(f \circ g)(x)$?

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Order matters
core concept

We just saw that $f(g(x))$ means: apply $g$ first, then feed the result into $f$. That raises a question: what if we reverse the order, does $f(g(x))$ always equal $g(f(x))$? This card answers it → composition is not commutative: swapping the order almost always gives a different function.

In most cases, $f(g(x)) \neq g(f(x))$. Composition is not commutative. The order in which you apply the functions usually changes the result.

For example, if $f(x) = x + 1$ and $g(x) = x^2$:

  • $(f \circ g)(x) = f(g(x)) = f(x^2) = x^2 + 1$
  • $(g \circ f)(x) = g(f(x)) = g(x + 1) = (x + 1)^2 = x^2 + 2x + 1$

These are clearly different expressions. In real-world terms, it matters whether you double a price and then add tax, or add tax and then double the price.

The inner function runs first, then the outer function acts on its output.

$f(g(x))$ is generally NOT equal to $g(f(x))$; Composition is not commutative, order always matters

Pause, copy the non-commutativity rule ($f(g(x)) \neq g(f(x))$ in general) with a worked counter-example showing the two different results into your book.

Odd one out: Which statement about composite functions is FALSE?

Worked examples · reveal as you go

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Work examples end to end

Follow the reasoning through complete worked solutions.

Worked example 1 · evaluating a composite function +5 XP on full reveal

If $f(x) = 2x + 1$ and $g(x) = x^2 - 3$, find $(f \circ g)(2)$ and $(g \circ f)(2)$.

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$g(2) = (2)^2 - 3 = 4 - 3 = 1$
Find $(f \circ g)(2)$: evaluate the inner function first
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$f(1) = 2(1) + 1 = 3$
Use $g(2) = 1$ as the input for $f$: $(f \circ g)(2) = \mathbf{3}$
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$f(2) = 2(2) + 1 = 5$
Find $(g \circ f)(2)$: evaluate the inner function $f$ first
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$g(5) = (5)^2 - 3 = 25 - 3 = 22$ ✓
$(g \circ f)(2) = \mathbf{22}$, notice how the order gave a different result
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Worked example 2 · finding the algebraic expression +5 XP on full reveal

If $f(x) = 3x - 2$ and $g(x) = x + 4$, find $(f \circ g)(x)$ and $(g \circ f)(x)$.

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$f(g(x)) = f(x + 4) = 3(x + 4) - 2$
Substitute $g(x) = x + 4$ into $f$, use brackets!
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$= 3x + 12 - 2 = 3x + 10$
Expand and simplify: $(f \circ g)(x) = \mathbf{3x + 10}$
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$g(f(x)) = g(3x - 2) = (3x - 2) + 4$
Substitute $f(x) = 3x - 2$ into $g$
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$= 3x + 2$ ✓
$(g \circ f)(x) = \mathbf{3x + 2}$, different from $(f \circ g)(x)$
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Worked example 3 · domain of a composite function +5 XP on full reveal

Find the domain of $(f \circ g)(x)$ where $f(x) = \sqrt{x}$ and $g(x) = x - 5$.

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$(f \circ g)(x) = f(g(x)) = f(x - 5) = \sqrt{x - 5}$
Write the composite expression
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$g(x) = x - 5$ is defined for all real $x$
Check domain of the inner function $g$
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$f$ requires input $\geq 0$, so we need $x - 5 \geq 0 \Rightarrow x \geq 5$
Check domain of the outer function $f$ applied to $g(x)$
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Domain: $[5, \infty)$ ✓
Combine both conditions: $x \geq 5$
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Common mistakes · the 4 traps that cost marks

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Dodge the traps, then apply

Meet the mistakes that cost marks, then do it yourself.

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Applying the functions in the wrong order

$(f \circ g)(x)$ means $g$ first, then $f$. Many students do the opposite, especially when both functions are simple. This almost always leads to a wrong answer.

✓ Fix: Write "INNER = $g(x)$" and "OUTER = $f(\text{inner})$" before you substitute any numbers.

2

Assuming $f(g(x)) = g(f(x))$

Composition is not commutative. Unless there is a specific reason (like when $f$ and $g$ are inverses), the order matters.

✓ Fix: Treat every composition problem as order-dependent unless you have proven otherwise.

3

Forgetting brackets when substituting algebraic expressions

If $f(x) = x^2 + 1$ and you want $f(x + 3)$, writing $x + 3^2 + 1$ is wrong. The entire expression $x + 3$ must be squared: $(x + 3)^2 + 1$.

✓ Fix: Always use brackets around the substituted expression before simplifying.

4

Ignoring the domain of the inner function

When finding the domain of a composite, students often check only the outer function's domain and forget that the input $x$ must first be valid for the inner function $g$.

✓ Fix: Domain of $f \circ g$ = domain of $g$ intersected with the set of $x$ where $g(x)$ is in the domain of $f$.

Activity 1, Evaluate the composites

Let $f(x) = 2x + 3$ and $g(x) = x^2 - 1$. Find each of the following.

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$(f \circ g)(1)$

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$(g \circ f)(1)$

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$(f \circ g)(x)$ in simplified form

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$(g \circ f)(x)$ in simplified form

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Fill the blanks: drag each token into the matching blank.

inner outer first second

In $(f \circ g)(x)$, the function $g$ is the ___ function and is applied ___. The function $f$ is the ___ function and is applied ___.

Quick-fire practice · 5 reps +2 XP per reveal

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Drill it, then lock it in

Run the quick drill and copy the summary into your book.

1

If $f(x) = 2x$ and $g(x) = x + 3$, find $(f \circ g)(4)$.

2

If $f(x) = x^2$ and $g(x) = x - 1$, find $(g \circ f)(3)$.

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Find $(f \circ g)(x)$ where $f(x) = 3x + 1$ and $g(x) = x^2$.

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Find the domain of $(f \circ g)(x)$ where $f(x) = \sqrt{x}$ and $g(x) = x + 2$.

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Is there a case where $(f \circ g)(x) = (g \circ f)(x)$? Give an example.

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Revisit your thinking

Earlier you were asked: If Machine A doubles its input and Machine B adds 3, what is the overall rule? Does it matter which machine goes first?

If A goes first and then B, the overall rule is $B(A(x)) = 2x + 3$. If B goes first and then A, the overall rule is $A(B(x)) = 2(x + 3) = 2x + 6$. These are different results, so yes, the order matters. This is the fundamental nature of composite functions: $(f \circ g)(x)$ is not the same as $(g \circ f)(x)$ unless the functions have a special relationship (like being inverses). In supply chains, manufacturing, and even cooking, the order in which you apply processes almost always changes the final outcome.

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Multiple choice

Answer the drill bank and rate your confidence.

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

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Short answer

Write full responses, then check them against the model answers.

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Short answer
ApplyBand 34 marks

Q8. Let $f(x) = 2x - 1$ and $g(x) = x^2 + 3$. (a) Find $(f \circ g)(x)$ in expanded form. (b) Find $(g \circ f)(x)$ in expanded form. (c) Show that $(f \circ g)(x) \neq (g \circ f)(x)$. (4 marks)

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ApplyBand 43 marks

Q9. Find the domain of $(f \circ g)(x)$ where $f(x) = \sqrt{x + 2}$ and $g(x) = x - 3$. Show the inequality you solve and write your final answer in interval notation. (3 marks)

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EvaluateBand 53 marks

Q10. A student claims that $(f \circ g)(x)$ and $(g \circ f)(x)$ are always different. Evaluate this claim by considering the functions $f(x) = x + 2$ and $g(x) = x - 2$. Calculate both composites and explain what this example tells you about the student's claim. (3 marks)

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Comprehensive answers (click to reveal)

Multiple choice, drill bank

MC answers and feedback are shown inline as you complete each question. Use the retry button to attempt a fresh set.

Composing $f \circ g$. $(f \circ g)(x) = f(g(x)) = f(x + 3) = 2(x + 3) = 2x + 6$.

Evaluating $g \circ f$. $(g \circ f)(2) = g(f(2)) = g(4) = 4 + 1 = 5$.

Order of composition. $(f \circ g)(x) = f(g(x))$.

Domain of a composite. $\sqrt{x - 4}$ requires $x \geq 4$.

Composing with a square. $(f \circ g)(x) = f(x^2) = 3x^2 - 1$.

Activity 1, Evaluate the composites model answers

1. $g(1) = 1^2 - 1 = 0$, so $(f \circ g)(1) = f(0) = 2(0) + 3 = 3$.

2. $f(1) = 2(1) + 3 = 5$, so $(g \circ f)(1) = g(5) = 5^2 - 1 = 24$.

3. $(f \circ g)(x) = f(x^2 - 1) = 2(x^2 - 1) + 3 = 2x^2 + 1$.

4. $(g \circ f)(x) = g(2x + 3) = (2x + 3)^2 - 1 = 4x^2 + 12x + 9 - 1 = 4x^2 + 12x + 8$.

Short answer model answers

Q8 (4 marks):

(a) $(f \circ g)(x) = f(x^2 + 3) = 2(x^2 + 3) - 1 = 2x^2 + 5$ [1]
(b) $(g \circ f)(x) = g(2x - 1) = (2x - 1)^2 + 3 = 4x^2 - 4x + 1 + 3 = 4x^2 - 4x + 4$ [1]
(c) $2x^2 + 5 \neq 4x^2 - 4x + 4$ for most values of $x$ [1]. Therefore $(f \circ g)(x) \neq (g \circ f)(x)$ [1].

Q9 (3 marks): $(f \circ g)(x) = f(x - 3) = \sqrt{(x - 3) + 2} = \sqrt{x - 1}$ [1]. We need $x - 1 \geq 0$ [1]. Domain: $[1, \infty)$ [1].

Q10 (3 marks): The student's claim is false [1]. $(f \circ g)(x) = f(x - 2) = (x - 2) + 2 = x$ [0.5]. $(g \circ f)(x) = g(x + 2) = (x + 2) - 2 = x$ [0.5]. In this case, both composites equal $x$ because $f$ and $g$ are inverses of each other [1]. This shows that while composition is not generally commutative, there are special cases where the two composites are the same.

Module overview · Maths Advanced · Module quiz