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)
$= g(x^2 + 1)$
$= 5(x^2 + 1) + 3$
$= 5x^2 + 8$
