Composition of Functions: Definition, Illustration, Examples, and Solutions

Composition of Functions - Definition, Illustration, Example, Solution

Introduction to 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)) is formed in two steps.

1. Feed an input (say x) to f;
2. Feed the output f(x) to g to get g(f (x)) and call it gf(x).

Fig. 1.41

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. 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.

Fig. 1.42

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

Example 1.21

Represent the function f(x) = √(2x² − 5x + 3) as a composition of two functions.

Solution

We set f₁(x) = 2x² − 5x + 3 and f₂(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 ⇒ 2k = −2 ⇒ k = −1

Example 1.23

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

Solution

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

= 2(2k − 1) − 1 = 4k − 3

Thus, f o f (k) = 4k – 3

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

Therefore 4k - 3 = 5 ⇒ 4k = 8 ⇒ k = 2.

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 (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).

Fig. 1.43

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.