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 Apart from the stuff given above, if you need any other stuff in Math, please use our google custom search here.

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