Let f : A -> B and g : B -> C be two functions. Then the composition of f and g, denoted by $g \circ f$, is defined as the function $g \circ f$ : A -> C given by
$g \circ f=g(f(x))$ for all x ∈ A

Here's a step-by-step process to understand composition of functions

Start with the input x.

Apply the function f(x) to the input x, and obtain the result f(x).

Use the result f(x) as the input for the function g.

Apply the function g to f(x) to obtain the final output, g(f(x)).

$g \circ f$ is not necessary equal to $f \circ g$.

Domain of $g \circ f$ is the domain of f

if function f and g is one-one, the $g \circ f$ is one-one

if function f and g is onto, the $g \circ f$ is onto

Theorem Composition of three function

If f : X -> Y, g : Y -> Z and h : Z -> S are functions, then composite of these three function are
$h \circ (g \circ f)$ and $(h \circ g ) \circ f)$ Theorem
$h \circ (g \circ f) =(h \circ g ) \circ f$ Proof
$h \circ (g \circ f) (x) =h \circ (g(f(x))= h(g(f(x))$ for all x ∈ A
$(h \circ g ) \circ f (x) =h \circ g (f(x)= h(g(f(x))$ for all x ∈ A
Hence $h \circ (g \circ f) =(h \circ g ) \circ f$

Solved Example

Example 1
Consider f : N -> N, g : N -> N and h : N -> R defined as
$f(x) = 2x$,$g(y) = 3y + 4$ and $h(z) = sin z$, for x, y and z in N.
Show that $h \circ (g \circ f ) = (h \circ g) \circ f$. Solution
We have
$h \circ (g \circ f )= h(g(f (x))) = h(g(2x))= h(3(2x) + 4) = h(6x + 4) = sin (6x + 4)$ for x in N.
Also,
$(h \circ g) \circ f$ (x) = (hog) ( f (x)) = (hog) (2x) = h ( g (2x))
= h(3(2x) + 4) = h(6x + 4) = sin (6x + 4), for x in N.
This shows that $h \circ (g \circ f) =(h \circ g ) \circ f$

Example 2
Given $f(x) = x^2 + 1$ and $g(x) = 5x +3$, find
(a) (f o g)(x)
(b) (g o f)(x) Solution
(a) (f o g)(x)
$= f(5x+3)$
$= (5x+3)^2 + 1$
$= 25x^2 + 30x + 9 + 1$
$= 25x^2 + 30x + 10$
(b) (g o f)(x)
$= g(x^2 + 1)$
$= 5(x^2 + 1) + 3$
$= 5x^2 + 8$