We use two methods to find if function has inverse or not

  1. If function is one-one and onto, it is invertible.
  2. We find g, and check fog = I Y and gof = I X

 

We discussed how to check one-one and onto previously.

 

Let’s discuss the second method

 

We find g, and check fog = I Y and gof = I X

 

Steps are

Checking inverse of f : X → Y

Step 1 : Calculate g: Y → X

Step 2 : Prove gof = I X

Step 3 : Prove fog = I Y

 


Example

Let f : N → Y,

f (x) = 2x + 1, where, Y = {y ∈ N : y = 4x + 3 for some x ∈ N }.

Show that f is invertible

 

Checking by One-One and Onto Method

Checking one-one

f(x 1 ) = 2x 1 + 1

f(x 2 ) = 2x 2 + 1

 

Putting f(x 1 ) = f(x 2 )

2x 1 + 1 = 2x 2 + 1

  2x 1 = 2x 2

  x 1 = x 2

 

If f(x 1 ) = f(x 2 ) , then x 1 = x 2

∴  f is one-one

 

Checking onto

f(x) = 2x + 1

 

Let f(x) = y, where y ∈ Y

  y = 2x + 1

  y – 1 = 2x

  2x =  y – 1

  x = (y - 1)/2

 

For every y in Y = {y ∈ N : y = 2x + 1 for some x ∈ N }.

There is a value of x which is a natural number

Thus, f is onto

 

Since f is one-one and onto

f is invertible

 

Checking by fog = I Y and gof = I X  method

Checking inverse of f: X → Y

Step 1 : Calculate g: Y → X

Step 2 : Prove gof = I X

Step 3 : Prove fog = I Y

g is the inverse of f

 

Step 1

f(x) = 2x + 1

Let f(x) = y

  y = 2x + 1

  y – 1 = 2x

  2x =  y – 1

  x = (y - 1)/2

 

Let g(y) = (y - 1)/2

where g: Y → N

 

Step 2 :

gof = g(f(x))

        = g(2x + 1)

        = ((2x + 1) - 1)/2

= (2x + 1 - 1)/2

= 2x/2

= x

= I N

 

Step 3 :

fog = f(g(y))

        = f((y - 1)/2)

= 2 ((y - 1)/2) + 1

= y – 1 + 1

= y

= I Y

 

Since gof   = I N and fog = I Y ,

f is invertible

  1. Chapter 1 Class 12 Relation and Functions
  2. Concept wise

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.