If we combine two functions in such a way that the output of one function becomes the input to another function, then this is called as composite function.

Consider three sets X, Y and Z and let f: *X → Y* and g: Y → Z.

As per this, under f, an element x∈ X is mapped to an element y = f(x) ∈ Y. This in turn is mapped by g to an element z ∈ Z in such a manner that z = g(y) = g[(*f*(x)] .

The composite function is denoted as:

(gof)(x) = g(*f (X) )*

Similarly, (*f*og) (x) = *f* (g(x))

So, take f(x) as argument for the function g to find (gof) (x),

Consider f: A → B and g: B → C be two functions. In this case the composition of f and g, denoted by g o f, is defined as the function g o f: A → C given by g o f (x) = g (f (x)), ∀ x ∈ A.

(ii) If f: A → B and g: B → C are one-one, then g o f: A → C is also one-one

(iii) If f: A → B and g: B → C are onto, then g o f A → C is also onto. However, it is not necessary that the converse of above stated results (ii) and (iii) be true. Moreover, we have the following results in this direction.

(iv) If: A → B and g: B → C be the given functions such that g o f is one-one. Then f is one-one.

(v) If: A→ B and g: B → C be the given functions such that g o f is onto. Then g is onto.

**Example: Given the function f(x) = 3x + 5 and g(x) = **2×3** .Find ( gof)(x) and ( fog)(x).**

**Solution:** We know, **(gof) (x)** = g(f(x)) = g (3x+5) = 2(3x+5)^{3}

Using Binomial Expansion, we have

(gof)(x)=2[(3x)^{3}+3.(3x)^{2}.5+3.(3x)(5)** ^{2}**+(5)

**]**

^{3}⇒(gof)(x)=2[27x^{3}+135x^{2}+225x+125]

⇒(gof)(x)=54x^{3}+270x^{2}+450x+250

Now, **(fog)(x)** = f (g(x)) = f ( 2x** ^{3}**) = 3(2x

**)+5**

^{3}⇒(fog)(x)=6x** ^{3}**+5

**Example:** **Let f(x) = **x** ^{2} and g(x) = **√1–x

^{2}.Find (gof)(x) and ( fog)(x) .**Solution:** (gof)(x) = g( f(x)) = g(x2) = (√ 1–x^{2})^{2} =√ 1–x^{4}

(fog) (x) = f(g(x)) = f(√ 1–x2) = (√ 1–x^{2})^{2} = 1–x^{2}