Composition of Functions

A function connects each input to exactly one output. When one function is applied first and the result is then used as the input of another function, the new function formed is called the composition of functions.

This topic is important because it helps in understanding multistep mathematical processes, identity functions, inverse functions, and several board-level problem patterns.

Let f: A → B and g: B → C be any two functions. Then, the composition of f and g, denoted by gof, is defined as a function gof: A → C given by

gof(x) = g[f(x)], ∀ x ∈ A

• Domain (gof) = Domain (f)
• g∘f(x) = g(f(x)) → first apply f, then g

1. Order matters

In general,

So, composition is not commutative in general.

2. Associative property

If three functions are composable, then

Thus, composition is associative.

3. Identity function

If \[I_A\] is the identity function on set A, then

whenever the compositions are defined.

[Composition works only when: Range of inner function fits domain of outer function.]

Let f : {2, 3, 4, 5} → {3, 4, 5, 9} and g : {3, 4, 5, 9} → {7, 11, 15} be functions defined as
f(2) = 3, f(3) = 4, f(4) = 5, f(5) = 5 and g(3) = 7, g(4) = 7, g(5) = 11, g(9) = 11 Find gof.

Solution
gof(2) = g(f(2)) = g(3) = 7
gof(3) = g(f(3)) = g(4) = 7
gof(4) = g(f(4)) = g(5) = 11
gof(5) = g(f(5)) = g(5) = 11

Let f, g and h be functions from R to R. Show that
(i) (f + g)oh = foh + goh
(ii) (f.g)oh = (foh) . (goh).

Solution:

(i) For all x ∈ R,

((f + g) ∘ h)(x) = (f + g)(h(x))
= f(h(x)) + g(h(x))
= (f ∘ h)(x) + (g ∘ h)(x)

∴ (f + g) ∘ h = f ∘ h + g ∘ h.

(ii) ((f · g) ∘ h)(x) = (f · g)(h(x)) = f(h(x)) · g(h(x))
= (f ∘ h)(x) · (g ∘ h)(x)= {(f ∘ h) · (g ∘ h)}(x), ∀ x ∈ R.

Hence,

(f · g) ∘ h = (f ∘ h) · (g ∘ h).

If f: R → R be the signum function defined as \[f(x)=
\begin{cases}
1, & x>0 \\
0, & x=0 \\
-1, & x<0 & 
\end{cases}\] and g : R → R be the greatest integer function defined as g(x) = [x], then determine whether fog and gof coincide on (0, 1].

Solution:
Clearly gof: R → R and fog: R → R.

Consider x = 1/2, which lies in (0, 1].

Now

(gof)(1/2) = g(f(1/2)) = g(1) = [1] = 1

and (fog)(1/2) = f(g(1/2)) = f([1/2]) = f(0) = 0

⇒ (gof)(1/2) ≠ (fog)(1/2)

⇒ gof ≠ fog on (0, 1]

If f(x) = \[\frac{x-1}{x+1}\], find (f ∘ f)(x).

Solution:

Given \[f(x)=\frac{x-1}{x+1}\] ⇒ D_f = R − {−1}

D_f∘f = { x : x ∈ D_f , f(x) ∈ D_f }

= { x : x ∈ R − {−1} , \[\frac{x-1}{x+1}\] ∈ D_f } \[=\left\{x:x\neq-1,\frac{x-1}{x+1}\neq-1\right\}\]

\[=\{x:x\neq-1,x\neq0\}\]

⇒ D_f∘f = R − {−1, 0}

∴ f ∘ f is defined with domain R − {−1, 0} and

(f ∘ f)(x) = f(f(x)) \[=f\left({\frac{x-1}{x+1}}\right)\] \[=\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}\]

\[=\frac{x-1-x-1}{x-1+x+1}=\frac{-2}{2x}=-\frac{1}{x}\]

• In \[g \circ f\], first apply f, then apply g.

• \[(g \circ f)(x) = g(f(x))\].

• Composition is defined only when the output of the first function is acceptable as input to the second function.

• In general, \[g \circ f \neq f \circ g\].

• Composition is associative whenever defined.

• Identity function leaves a function unchanged under composition.

• Misconception 1: g∘f and f∘g are always equal.
  Correction: They are usually different.

• Misconception 2: In g∘f, apply g first.
  Correction: Always apply f first, then g.

• Misconception 3: Composition is always possible for any two functions.
  Correction: It is possible only when the outputs of the first function fit into the domain of the second.

• Misconception 4: Composition and multiplication of functions are the same.
  Correction: Composition means one rule after another; multiplication of functions is a different operation.