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Definition, Illustration, Example, Solution | Mathematics - Composition of Functions | 10th Mathematics : Relation and Function

Chapter: 10th Mathematics : Relation and Function

Composition of Functions

Let f : A → B and g : B → C be two functions (Fig). Then the composition of f and g denoted by g o f is defined as the function g o f (x ) = g( f (x )) for all x ∈ A .

Composition of Functions

When a car driver depresses the accelerator pedal, it controls the flow of fuel which in turn influences  the speed of the car. Likewise, the  composition  of two functions is a kind of ‘chain reaction’, where the functions act upon one after another (Fig.1.40).


We can explain this further with the concept that a function is a ‘process’. If f and g are two functions then the composition g(f (x)) (Fig.1.41) is formed in two steps.

(i) Feed an input (say x) to f ;

(ii) Feed the output f(x) to g to get g(f (x)) and call it gf(x).


 

Illustration

Consider the set A of all students, who appeared in class X of Board Examination. Each student appearing in the Board Examination is assigned a  roll number.  In order to have confidentiality, the Board arranges to deface the roll number of each student and assigns a code number to each roll number.

Let A be the set of all students appearing for the board exam. B N be the set all roll numbers and C N be the set of all code numbers (Fig.1.41). This gives rise to two functions f: A B and g : B C given by b = f (a) be the roll number assigned to student a, c = g(b) be the code number assigned to roll number b, where a A , b B and c C.

We can write c = g(b) = g(f (a)).

Thus, by the combination of these two functions, each student is eventually attached a code number. This idea leads to the following definition.

 

Definition

Let f : A B and g : B C be two functions (Fig.1.42). Then the composition of f and g denoted by g o  f  is defined as the function g o  f (x ) = g( f (x )) for all x A .


 

Example 1.20

Find f o g and g o f when f (x) = 2x + 1 and g(x) = x2 – 2

Solution

  f (x) = 2x + 1 , g(x) = x2 – 2

f o g(x) = f (g(x)) = f (x2 − 2) = 2(x2 − 2) + 1 = 2x2 – 3

g o f (x) = g(f (x)) = g(2x + 1) = (2x + 1)2 − 2 = 4x2 + 4x – 1

Thus f o g = 2x2 − 3, g o f = 4x2 + 4x − 1. From the above, we see that f o g g o f .

Note

Generally, f o gg o  f for any two functions f and g. So, composition of functions is not commutative.

 

Example 1.21

Represent the function f (x) =  as a composition of two functions.

Solution

We set f (x) = 2x2 − 5x + 3 and f (x) = √x

Then, 


 

Example 1.22

If f (x) = 3x − 2 , g(x) = 2x + k and if f o g = g o f , then find the value of k.

Solution

f (x) = 3x − 2 , g(x) = 2x + k

f o g(x) = f (g(x)) = f (2x + k) = 3(2x + k) − 2 = 6x + 3k – 2

Thus, f o g(x) = 6x + 3k – 2.

g o f (x) = g(3x − 2) = 2(3x − 2) + k

Thus, g o f (x) = 6x − 4 + k .

Given that f o g = g o f

Therefore, 6x + 3k − 2 = 6x − 4 + k

6x − 6x + 3k k = −4 + 2 k = −1

 

Example 1.23

Find k if f o f (k) = 5 where f (k) = 2k – 1.

Solution

o f (k) = f (f (k))

= 2(2k − 1) − 1 = 4k − 3

Thus, o f (k) = 4k – 3

But, it is given that f o f (k) = 5

Therefore 4k - 3 = 5 k = 2 .

 

Composition of three functions

Let  ABCbe  four  sets  and let : A B ,   g : B Cand   : C D be three functions (Fig.1.43). 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.

Note

Composition of three functions is always associative. That is, f o (g o h) = (f o g) o h

 

Example 1.24

If  f (x) = 2x + 3 , g(x) = 1 − 2x and h(x) = 3x . Prove that f o (g o h) = (f o g) o h

Solution

f (x) = 2x + 3 , g(x) = 1 − 2x , h(x) = 3x

Now, (f o g)(x) = f (g(x)) = f (1 − 2x) = 2(1 − 2x) + 3 = 5 − 4x

Then, (f o g) o h(x) = (f o g)(h(x)) = (f o g)(3x) = 5 − 4(3x) = 5 − 12x         ……… (1)

(g o h)(x) = g(h(x)) = g(3x) = 1 − 2(3x) = 1 − 6x

So, f o (g o h)(x) = f (1 − 6x) = 2(1 − 6x ) + 3= 5 − 12x               ………(2)

From (1) and (2), we get (f o g) o h = f o (g o h)

 

Example 1.25

Find x if gff(x) = fgg(x), given f (x) = 3x + 1 and g(x) = x + 3 .

Solution

gff(x) = g [f {f (x)}] (This means “g of f of f of x”)

= g [ f (3x +1)] = g [ 3(3x +1)+1] = g (9x + 4)

g (9x + 4) = [ (9x + 4) + 3] = 9x + 7

fgg(x) = f [g {g (x)}] (This means “f of g of g of x”)

= f [ g (x + 3)] = f [ (x + 3) + 3] = f (x + 6)

f (x + 6) = [ 3(x + 6) + 1 ] = 3x + 19

These two quantities being equal, we get 9x + 7 = 3x + 19. Solving this equation we obtain x = 2.


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