Composition of Functions

Introduction:

In algebra, the composition of functions refers to combining two functions such that the output of one function becomes the input of another function. If we have two functions, $f(x)$ and $g(x)$, the composition of $f$ and $g$ is written as:

\[(f ∘ g)(x) = f(g(x)) \]

This is read as "$f$ composed with $g$ of $x$" or "$f$ of $g$ of $x$." In composition, the inner function g is applied first, and then the result is used as input into the outer function $ f$. This means you first apply $g(x)$, then take the result and apply $f$ to it.

Definition:

Let 𝑓 and 𝑔 be two functions.
\[ g : 𝐴 → 𝐵 \] means function $g$ takes input from set 𝐴 and gives output in set 𝐵.
\[ f : 𝐵 → 𝐶 \] means function $f$ takes input from set 𝐵 and gives output in set 𝐶.
Then the composition of functions 𝑓 ∘ 𝑔 is a function from
\[ 𝐴 → 𝐶 \] defined by: \[ ( 𝑓 ∘𝑔 ) ( 𝑥) = 𝑓 ( 𝑔 ) ( 𝑥 ) \] for all 𝑥 ∈𝐴

Example 1: Basic Polynomial Functions
Let

\[f(x) = 2x + 3\]
\[ g(x) = x^2 \]

\[ (f ∘ g)(x) = f(g(x)) = f(x^2) = 2x^2 + 3 \]
\[ (g ∘ f)(x) = g(f(x)) = g(2x + 3) = (2x + 3)^2 \]

Note: In general, \[(f ∘ g)(x) ≠ (g ∘ f)(x)\]

Example 2: Real-World Interpretation
Let

$f(t) = t + 5$; adds 5 to the input.
$ g(x) = 3t $; triples the input

find $ (f ∘ g)(x) $ and $ (g ∘ f)(x)$ \[ (f ∘ g)(x) = f(g(x)) = f(3x) = 3x + 35\] \[ (g ∘ f)(x) = g(f(x)) = g(x + 5) = 3(x+5) = 3x + 15 \] Interpretation

In 𝑓 ∘ 𝑔, you triple first, then add 5.
In 𝑔∘𝑓, you add 5 first, then triple the result.

Example 3: Value of composition
Let

$f(x) = x^2 + 1$;
$ g(x) = 2x $;

When Are Compositions Equal?

While it's generally true that: \[(f ∘ g)(x) ≠ (g ∘ f)(x)\] there are special cases where \[(f ∘ g)(x) = (g ∘ f)(x)\]

Let’s explore some examples where they are equal

Example 1: Identity Function

\[f(x) = x \]
\[g(x) = x^2 + 3x\]
\[(f ∘ g)(x) = f(g(x))= f(x^2 + 3x) = x^2 + 3x\]
\[(g ∘ f)(x) = g(f(x)) = g(x) = x^2 + 3x\]

Note: Why? Because the identity function does nothing—it returns the input as-is. So composition with the identity function always preserves the other function.

Example 2: Same Functions

\[f(x) = g(x) = 2x\]
\[(f ∘ g)(x) = f(g(x)) = f(2x)= 4x\]
\[(g ∘ f)(x) = g(f(x)) = (2x) = 4x\]

Note: Why? Because when 𝑓 = 𝑔, the order of composition doesn’t matter—you’re applying the same function twice.

Example 3: Simple Linear Functions

\[f(x) = x + 1, g(x) = x + 2\]
\[(f ∘ g)(x) = f(g(x)) = f(x+2)= x +1 + 2 = x + 3\]
\[(g ∘ f)(x) = g(f(x)) = g(x + 1) = x + 2 + 1 = x + 3\]

Why? Because addition is commutative: \[(x +1) + 2 = (x + 2) + 1 \]

Example 4: When f(x) and g(x) are inverse functions of each other

\[(f ∘ g)(x) = f(g(x)) = x \] \[(g ∘ f)(x) = g(f(x)) = x \]

Why Does This Happen?
By definition of inverse functions, if f(x) is the inverse of g(x), then: \[f(g(x)) = x\] \[g(f(x)) = x \] This means both compositions lead to the identity function, which always returns the input $x$.

Examples of Inverse Functions
Example 1
Let $$ f(x) = 2x $$ $$ g(x) = x / 2 $$ Check the compositions: $$ f(g(x)) = 2 * (x / 2) = x $$ $$ g(f(x)) = (2x) / 2 = x $$ Both compositions return $x$.
Example 2:
Let $$ f(x) = x + 3 $$ $$ g(x) = x - 3 $$ Check the compositions: $$ f(g(x)) = (x - 3) + 3 = x $$ $$ g(f(x)) = (x + 3) - 3 = x $$ ✅ Again, both compositions return $x$.
Example 3:
Let $$ f(x) = 1/x $$ $$ g(x) = 1/x $$ (Note: f(x) is its own inverse here.)
Check the compositions: $$ f(g(x)) = f(1/x) = 1 / (1/x) = x $$ $$ g(f(x)) = g(1/x) = 1 / (1/x) = x $$ ✅ Both return $x$.

Summary:

Composition of functions involves plugging one function into another.
Not commutative: $𝑓∘𝑔 ≠ 𝑔∘𝑓 $ generally.
Evaluate inside out: compute $g(x)$ first, then $f(g(x))$.
Useful in real-world models where multiple processes happen sequentially.

General Conditions When \[(f ∘ g)(x) = (g ∘ f)(x) \]

f = g: If both functions are the same, composition in any order gives the same result.
f or g is the identity function: 𝑓(𝑥) = 𝑥 or 𝑔(𝑥) = 𝑥, then 𝑓∘𝑔=𝑔∘𝑓=𝑓=𝑔.
f and g commute: If 𝑓(𝑔(𝑥))=𝑔(𝑓(𝑥)) for all 𝑥, then 𝑓∘𝑔=𝑔∘𝑓. This is rare and typically happens with linear functions of the form 𝑓(𝑥)=𝑎𝑥+𝑏, 𝑔(𝑥)=𝑐𝑥+𝑑 only when 𝑎=𝑐.
Functions act on disjoint variables: In more advanced math, if functions affect different parts of a variable (e.g., in multivariable functions), they might commute.

General Condition for Equality

For linear functions f(x) = ax + b, g(x) = cx + d, the compositions are equal if:

\[d(a - 1) = b(c - 1)\]

Proof:
let,
\[f(x) = ax +b\] and \[(x) = cx +d\] To find when \[(f ∘ g)(x) = (g ∘ f)(x) \] Compute: \[f(g(x))=f(cx+d)=a(cx+d)+b=acx+ad+b \] Compute: \[ g(f(x))=g(ax+b)=c(ax+b)+d=acx+bc+d \] Set equal: \[ acx+ad+b=acx+bc+d \] Equating constants: \[ad+b=bc+d\] Solve for this: \[ad−d=bc−b⇒d(a−1)=b(c−1)\] This equation must be true for 𝑓∘𝑔=𝑔∘𝑓 for linear functions.