COMPOSITION OF FUNCTIONS

Definition:

Let f: A → B and g: B → C be two functions. Then a function gof: A → C defined by (gof) (x) = g (f(x)), for all x ϵ A

Is called the composition of f and g

Note 1:

It is evident from the definition that gof is defined only if for each x ϵ A, f(x) is an element of g so that we can take its g-image. Hence, for the composition gof to exist, the range of f must be a subset of the domain of g.

Note 2:

It should be noted that gof exists if the range of f is a subset of domain of g similarly, fog exists if range of g is a subset of domain of f.

Example:

Let f: R →R; f(x) = sin x and g: R→R; g(x) = x² find fog and gof.

Solution:

Clearly, fog and gof both exist.

Now, (gof) (x) = g (f(x)) = g (sin x) = (sin x) ² = sin²x

And, (fog) (x) = f (g(x)) = f(x²) = sin x²

Example:

Let  find fof(x).

Solution:

We have,

fof (x) = f (f(x))

Properties of Composition of Functions:

Property 1:

The composition of functions is not commutative i.e. fog ≠ gof.

Property 2:

The composition of functions is associative i.e., if f, g, h are three functions such that (fog)oh and fo(goh) exist, then (fog)oh = fo(goh)

Property 3:

The composition of two bijections is a bijection i.e., if f and g are two bijections, then gof is also a bijection.

Property 4:

Let f: A→B Then foI_A = I_Bof =f i.e., the composition of any function with the identity function is the function itself.

Inverse of a Function:

If f: A→B  is a bijection, we can define a new function from B to A which associates each element y ϵ B to its pre-image f^-1 (y) ϵ A. such a function is known as the inverse of function f and is denoted by
f^-1.

Example:

If f: R→R is a bijection given by f(x) = x³ + 3 find f^-1(x).

Solution:

Let f(x) =y Then,

   

      f (x) = y

     => x³ + 3 = y

Thus, f^-1: R →R is given by  for all x ϵ R.

Properties of Inverse of a Function:

Property 1:

The inverse of a bijection is unique.

Property 2:

The inverse of a bijection is also a bijection.

Property 3:

If f: A→B is a bijection and g: B→A is the inverse of f, then fog = I_B and gof = I_A, Where I_A and I_B are the identity functions on the sets A and B respectively.

Property 4:

If f: A→B and g: B→A are two bijections, then gof: A → C is a bijection and (gof)^-1 = f^-1og^-1.

Property 5:

Let f: A→B and g: B→A be two functions such that gof = I_A and fog = I_B Then, f and g are bijections and g = f ^‑1.

Example:

Show that the function f: R→R given by  is invertible and find its inverse.

Solution:

In order to prove that f(x) is invertible, it is sufficient to show that f is a bijection.

f is an injection:

Let x, y be any two distinct real numbers. Then, x ≠ y

=>f (x) ≠ g(x)

=> f is an injection.

f is a surjection:

Let f(x) = y. Then,

Thus, for every y ϵ R, there exists  such that f(x) = y. Therefore, f is a surjection.

Hence f is a bijection. Consequently, it is invertible.

In order to find f^-1 let

f (x) = y