Learn on PengiAoPS: Introduction to Algebra (AMC 8 & 10)Chapter 16: Functions

Lesson 16.3: Composition

In this Grade 4 AMC math lesson from AoPS: Introduction to Algebra, students learn how to perform function composition — connecting two functions so the output of one becomes the input of another — using notation such as f(g(x)) and the ∘ symbol. Students practice evaluating composed functions, working with iterated compositions like f(f(x)), and understanding why the order of composition matters. The lesson also covers the domain and range conditions required for composition and introduces the repeated application notation f^n(x).

Section 1

Composition of Functions

Property

The composition of functions f and g is written $f \circ g$ and is defined by

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

We read $f(g(x))$ as $f$ of $g$ of $x$ . In composition, the output of one function is the input of a second function.

Examples

For $f(x) = 5x + 1$ and $g(x) = x - 3$ , find $(f \circ g)(x)$ . We compute $f(g(x)) = f(x-3) = 5(x-3) + 1 = 5x - 15 + 1 = 5x - 14$ .

For $f(x) = 5x + 1$ and $g(x) = x - 3$ , find $(g \circ f)(x)$ . We compute $g(f(x)) = g(5x+1) = (5x+1) - 3 = 5x - 2$ . Note that $(f \circ g)(x) \neq (g \circ f)(x)$ .

Section 2

Domain Restrictions in Function Composition

Property

For the composition $f(g(x))$ to be valid, every output value in the range of $g(x)$ must be an input value in the domain of $f(x)$ . In other words: Range of $g \subseteq$ Domain of $f$ .

Examples

Section 3

Iterative Function Composition

Property

The notation $f^n(x)$ means applying function $f$ exactly $n$ times in succession:

f^1(x) = f(x)

f^2(x) = f(f(x))

f^3(x) = f(f(f(x)))

f^n(x) = f(f^{n-1}(x))

Examples

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Composition of Functions

Property

The composition of functions f and g is written $f \circ g$ and is defined by

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

We read $f(g(x))$ as $f$ of $g$ of $x$ . In composition, the output of one function is the input of a second function.

Examples

For $f(x) = 5x + 1$ and $g(x) = x - 3$ , find $(f \circ g)(x)$ . We compute $f(g(x)) = f(x-3) = 5(x-3) + 1 = 5x - 15 + 1 = 5x - 14$ .

For $f(x) = 5x + 1$ and $g(x) = x - 3$ , find $(g \circ f)(x)$ . We compute $g(f(x)) = g(5x+1) = (5x+1) - 3 = 5x - 2$ . Note that $(f \circ g)(x) \neq (g \circ f)(x)$ .

Section 2

Domain Restrictions in Function Composition

Property

For the composition $f(g(x))$ to be valid, every output value in the range of $g(x)$ must be an input value in the domain of $f(x)$ . In other words: Range of $g \subseteq$ Domain of $f$ .

Examples

Section 3

Iterative Function Composition

Property

The notation $f^n(x)$ means applying function $f$ exactly $n$ times in succession:

f^1(x) = f(x)

f^2(x) = f(f(x))

f^3(x) = f(f(f(x)))

f^n(x) = f(f^{n-1}(x))

Examples

Overview

Lesson overview

Review the lesson summary and core properties without leaving the current page.

Overview

Section 1

Composition of Functions

Property

The composition of functions f and g is written $f \circ g$ and is defined by

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

We read $f(g(x))$ as $f$ of $g$ of $x$ . In composition, the output of one function is the input of a second function.

Examples

For $f(x) = 5x + 1$ and $g(x) = x - 3$ , find $(f \circ g)(x)$ . We compute $f(g(x)) = f(x-3) = 5(x-3) + 1 = 5x - 15 + 1 = 5x - 14$ .

For $f(x) = 5x + 1$ and $g(x) = x - 3$ , find $(g \circ f)(x)$ . We compute $g(f(x)) = g(5x+1) = (5x+1) - 3 = 5x - 2$ . Note that $(f \circ g)(x) \neq (g \circ f)(x)$ .

Section 2

Domain Restrictions in Function Composition

Property

For the composition $f(g(x))$ to be valid, every output value in the range of $g(x)$ must be an input value in the domain of $f(x)$ . In other words: Range of $g \subseteq$ Domain of $f$ .

Examples

Section 3

Iterative Function Composition

Property

The notation $f^n(x)$ means applying function $f$ exactly $n$ times in succession:

f^1(x) = f(x)

f^2(x) = f(f(x))

f^3(x) = f(f(f(x)))

f^n(x) = f(f^{n-1}(x))