**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(sinx) = (sinx)² = sin²x

And, (fog)(x) = f(g(x)) = f(x²) = sinx².

**Example: **Let f(x) \(=\left\{ \begin{align}& 1+x,\,\,\,\,\,0\le x\le2 \\& 3-x,\,\,\,\,\,2<x\le 3 \\\end{align} \right.\) find fof (x).

**Solution: **We have, fof (x) = f(f(x))

x \(=\left\{ \begin{align}& f\left( 1+x \right),\,\,\,\,\,0\le x\le 2 \\& f\left( 3-x\right),\,\,\,\,2<x\le 3 \\\end{align} \right.\)

\(=\left\{ \begin{align}& f\left( 1+x \right),\,\,\,\,\,0\le x<1 \\& f\left( 1+x \right),\,\,\,\,\,1<x\le 2 \\& f\left( 3-x \right),\,\,\,\,2<x\le 3 \\\end{align} \right.\).

\(=\left\{ \begin{align}& 1+\left( 1+x \right),\,\,\,\,\,0\le x<1 \\ & 3-\left( 1+x \right),\,\,\,\,\,1<x\le 2 \\& 1+\left( 3-x \right),\,\,\,\,2<x\le 3 \\\end{align} \right.\).

\(=\left\{ \begin{align}& 2+x,\,\,\,\,\,0\le x\le 1 \\& 2-x,\,\,\,\,\,1<x\le 2 \\& 4-x,\,\,\,\,2<x\le 3 \\\end{align} \right.\).

**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_{B}of = f i.e., the composition of any function with the identity function is the function itself.

**Inverse of an 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

⇒ x = (y – 3)^{⅓}

⇒ f¯¹ (y) = (y – 3)^{⅓}

Thus, f¯¹ : R → R is given by f¯¹ (x) = (x – 3)^{⅓} 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 → C are two bijections, then gof : A → C is a bijection and (gof)¯¹ = f¯¹og¯¹

**Property 5: **Let f : A → B and g : A → B 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 \(f\left( x \right)={{\log }_{a}}\left( x+\sqrt{{{x}^{2}}+1} \right)\), a > 0, a ≠ 1 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.

⇒ \(x+\sqrt{{{x}^{2}}+1}\ne y+\sqrt{{{y}^{2}}+1}\).

⇒ \({{\log }_{a}}\left( x+\sqrt{{{x}^{2}}+1} \right)\ne {{\log }_{a}}\left( y+\sqrt{{{y}^{2}}+1} \right)\).

⇒ f (x) ≠ f (y)

⇒ f is an injection.

**f is a surjection: **Let f (x) = y Then, \({{\log }_{a}}\left( x+\sqrt{{{x}^{2}}+1} \right)=y\).

\(x+\sqrt{{{x}^{2}}+1}={{a}^{y}}\).

\(-x+\sqrt{{{x}^{2}}+1}={{a}^{-y}}\).

∴ 2x = (a^{y} – a^{-y})

\(x=\left( \frac{{{a}^{y}}-{{a}^{-y}}}{2} \right)\).

Thus, for every y ϵ R, there exists \(x=\frac{{{a}^{y}}-{{a}^{-y}}}{2}\in R\) 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

⇒ \({{\log }_{a}}\left( x+\sqrt{{{x}^{2}}+1} \right)=y\).

⇒ \(x+\sqrt{{{x}^{2}}+1}={{a}^{y}}\).

⇒ \(-x+\sqrt{{{x}^{2}}+1}={{a}^{-y}}\).

∴ 2x = (a^{y} – a^{-y})

⇒ x = ½ (a^{y} – a^{-y})

⇒ f ¯¹ = ½ (a^{y} – a^{-y})

Hence, f ¯¹ (x) = ½ (a^{x} – a^{-x})