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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