Composite Functions
What are composite functions
When we talk about composite functions, we refer to the process of applying one function 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:
- $f : A \rightarrow B$
- $g : B \rightarrow C$
- $f ( A ) \subseteq B$
The composite function is defined as follows:
\[g \circ f : x \in A \rightarrow g ( f ( x ) ) \in C\]This means that the function $g \circ f$ maps each element $x$ in the domain $A$ to the value $g ( f ( x ) )$, provided that the image of $f$ is contained in the domain of $g$.
Example
Consider the following functions:
- $f ( x ) = 2 x + 3$
- $g ( x ) = x^{2}$
We wanto to define the composite function $g \circ f = g ( f ( x ) )$. Let’s start by evaluating $f ( x )$:
\[f ( x ) = 2 x + 3\]Now plug that into $g ( x )$:
\[g ( f ( x ) ) = g ( 2 x + 3 ) = ( 2 x + 3 )^{2}\]Therefore, the composite function is: \(g \circ f = ( 2 x + 3 )^{2}\)
Composition with the inverse function
If a function $f$ is composed with its inverse $f^{- 1}$, 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:
\[g ( f ( x ) ) \equiv g \circ f \text{and} f ( g ( x ) ) \equiv f \circ g\]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.