Composite functions 

Composite functions: Let A, B, C are three sets .
Let f: A -> B, g : B -> C be two functions. Here we have taken the domain of g to be the
co-domain of f.
g o f : A -> C as
g o f (a) = g [f (a) ] for a ∈ A
Since f(a) ∈ B
g [f(a) ] ∈ C
The function g o f so obtained is called the composition of f and g.

Examples on composite functions

Example 1: A = {1, 2, 3}, B = {4, 5}, C = {5, 6}
Let f : A -> B, g: B -> C be defined by f(1) = 4, f(2) = 5, f(3) = 4, g(4) = 5, g(5) = 6. Find g o f : A -> C
Solution:We have,
f(1) = 4 and g (4) = 5
So g [f(1) ] = g o f (1) = 5
f(2) = 5 and g(f) = 6
∴ g [f(2)] = g o f (2) = 6
f(3) = 4 and g(4) = 5
So g [f(3)] = g o f (4) = 5
So, g o f = {(1, 5), (2, 6), (3, 5)}

Example 2: Let f:R -> R be defined by f(x) = 2x – 3 and g:R -> R be defined by g(x) = (x + 3)/ 2. Show that f o g = g o f
Solution : f o g is a composite-function.
f o g means g(x) function is in f(x) function.
f o g = f[g(x)]
f(x) = 2x -3 and g(x)= (x+3)/2
∴ f[g(x)] =[2(x+3)/2] - 3
= x + 3 - 3
f o g = x -----(1)
g o f means f(x) function is in g(x) function.
g o f = g[f(x)]
f(x) = 2x -3 and g(x) = (x+3)/2
g[f(x)] = [(2x-3) + 3]/2
= (2x -3 + 3)/2
= 2x/2
g o f = x ----(2)
So from equation (1) and (2)
f o g = g o f

Example 3: Let f, g: R -> R be defined respectively, by f(x) = x² + 3x + 1 , g(x) = 2x -3. Find f o g (2).
solution : f o g means g(x) function is in f(x) function.
This means put x = 2x -3 in f(x) function.
f[g(x)] = (2x - 3)² + 3 (2x -3)+ 1
= (2x)² - 2(2x)(3) + 3² + 6x - 9 + 1
= 4x² - 12x + 9 + 6x - 9 +1
f[g(x)] = f o g (x)= 4x² -6x + 1
Now we have to find f o g(2), so put x = 2 in f o g
f o g (2)= 4(2)² - 6(2) +1
= 4 x 4 - 12 + 1
= 16 -12 +1
∴ f o g(2) = 5


11th grade math

From Composite functions to Home page