COMPOSITION OF THREE FUNCTIONS

Let A, B, C, D be four sets and let f : A--->B , g : B--->C and h : C--->D be three functions. 

Using composite functions f o g and g o h, we get two new functions like (f o g) o h and f o (g o h).

We observed that the composition of functions is not commutative. The natural question is about the associativity of the operation.

Composition of three functions is always associative. That is,

f o (g o h) = (f o g) o h

Consider the functions f(x), g(x) and h(x) as given below. Find (f o g) o h and f o (g o h) in each case and also show that (f o g) o h = f o (g o h).

Example 1 :

f(x) = x - 1 , g(x) = 3x + 1 and h(x) = x2

Solution :

f o (g o h) :

g o h = g[h(x)]

= g[x2]

= 3x2 + 1

f o (g o h) = f(3x2 + 1)

= 3x2 + 1 - 1

= 3x2 ----(1)

(f o g) o h :

f o g = f[g(x)]

= f[3x + 1]

= 3x + 1 - 1

= 3x

(f o g) o h = (f o g)[h(x)]

= (f o g)(x2)

= 3x2 ----(2)

From (1) and (2), 

f o (g o h) = (f o g) o h

Example 2 :

f(x) = x2, g(x) = 2x and h(x) = x + 4

Solution :

f o (g o h) :

g o h = g[h(x)]

= g[x + 4]

= 2(x + 4)

= 2x + 8

f o (g o h) = f(2x + 8)

= (2x + 8)2

= (2x)2 + 2(2x)(8) + 82

= 4x2 + 32x + 64 ----(1)

(f o g) o h :

f o g = f[g(x)]

= f[2x]

= (2x)2

= 4x2

(f o g) o h = (f o g)[h(x)]

= (f o g)(x + 4)

= 4(x + 4)2

= 4[x2 + 2(x)(4) + 42]

= 4[x2 + 8x + 16]

4x2 + 32x + 64 ----(2)

From (1) and (2), 

f o (g o h) = (f o g) o h

Example 3 :

f(x) = x - 4, g(x) = x2 and h(x) = 3x - 5

Solution :

f o (g o h) : 

g o h = g[h(x)]

= g[3x - 5]

= (3x - 5)2

f o (g o h) = f[(3x - 5)2]

(3x - 5)2 - 4 ----(1)

(f o g) o h :

f o g = f[g(x)]

= f[x2]

= x2 - 4

(f o g) o h = (f o g)[h(x)]

= (f o g)(3x - 5)

= (3x - 5)2 - 4 ----(2)

From (1) and (2), 

f o (g o h) = (f o g) o h

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