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, (fog) (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[27x3+135x2+225x+125]
⇒(gof)(x)=54x3+270x2+450x+250
Now, (fog)(x) = f (g(x)) = f ( 2x3) = 3(2x3)+5
⇒(fog)(x)=6x3+5
Example: Let f(x) = x2 and g(x) = √1–x2.Find (gof)(x) and ( fog)(x) .
Solution: (gof)(x) = g( f(x)) = g(x2) = (√ 1–x2)2 =√ 1–x4
(fog) (x) = f(g(x)) = f(√ 1–x2) = (√ 1–x2)2 = 1–x2
