# Composite Functions
Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/composite-functions/
Fetched from algebrica.org post 15487; source modified 2025-12-06T17:56:46.
## What are composite functions
When we talk about composite functions, we refer to the process of applying one [function](../functions.md) to the result of another. In other words, given two functions \\( f(x) \\) and \\( g(x) \\), a composite function is formed by evaluating \\( g \\) at the output of \\( f \\). This is denoted by:
\\[
g \\circ f= g(f(x))
\\]
This means that we first apply \\( f \\) to the input \\( x \\), and then apply \\( g \\) to the result.
##### This diagram illustrates the concept of a composite function: the input \\( x \\) from set \\( A \\) is first mapped to \\( f(x) \\) in set \\( B \\), and then \\( f(x) \\) is mapped to \\( g(f(x)) \\) in set \\( C \\), resulting in the composition \\( g \\circ f \\).
* * *
More formally, let two functions \\( f(x) \\) and \\( g(x) \\) be given such that:
- \\(f \\colon A \\rightarrow B \\)
- \\(g \\colon B \\rightarrow C \\)
- \\(f(A) \\subseteq B\\)
The composite function is defined as follows:
\\[
g \\circ f \\colon x \\in A \\rightarrow g(f(x)) \\in C
\\]
This means that the function \\( g \\circ f \\) maps each element \\( x \\) in the [domain](../determining-the-domain-of-a-function.md) \\( A \\) to the value \\( g(f(x)) \\), provided that the image of \\( f \\) is contained in the domain of \\( g \\).
## Example
Consider the following functions:
- \\( f(x) = 2x + 3 \\)
- \\( g(x) = x^2 \\)
* * *
We wanto to define the composite function \\(g \\circ f = g(f(x)) \\). Let’s start by evaluating \\( f(x) \\):
\\[
f(x) = 2x + 3
\\]
Now plug that into \\( g(x) \\):
\\[
g(f(x)) = g(2x + 3) = (2x + 3)^2
\\]
Therefore, the composite function is:
\\[
g \\circ f = (2x + 3)^2
\\]
## Composition with the inverse function
If a function \\( f \\) is composed with its [inverse](../inverse-function.md) \\( f^{-1} \\), the result is the identity function, which maps each element of a set to itself:
\\[
f(f^{-1}(x)) = f^{-1}(f(x)) = x
\\]
##### This operation is valid only if the function \\( f \\) is invertible, meaning that it is both one-to-one (injective) and onto (surjective) over its domain.
When the composition between two functions is well-defined, that is, when the output of the first function lies within the domain of the second, we can write:
\\[
g(f(x)) \\equiv g \\circ f \\quad \\text{and} \\quad f(g(x)) \\equiv f \\circ g
\\]
This notation highlights that function composition is **not commutative**: in general, the order in which functions are composed affects the outcome, and the following holds:
\\[
g \\circ f \\neq f \\circ g
\\]
## Example
Let’s demonstrate with a simple example that function composition is not a commutative operation, that is, in general, \\( g \\circ f \\neq f \\circ g \\). Consider the two functions:
- \\( f(x) = e^x \\)
- \\( g(x) = x + 1 \\)
* * *
Compute \\( f \\circ g \\):
\\[
f \\circ g = f(g(x)) = f(x + 1) = e^{x + 1}
\\]
Compute \\( g \\circ f \\):
\\[
g \\circ f = g(f(x)) = g(e^x) = e^x + 1
\\]
We have:
- \\( (f \\circ g)(x) = e^{x + 1} = e \\cdot e^x \\)
- \\( (g \\circ f)(x) = e^x + 1 \\)
These expressions are not equal and this proves that function composition is not commutative.
