# Example 23

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 23 (Method 1) Let f : N → Y be a function defined as f (x) = 4x + 3, where, Y = {y ∈ N: y = 4x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse. Checking inverse Step 1 f(x) = 4x + 3 Let f(x) = y y = 4x + 3 y – 3 = 4x 4x = y – 3 x = 𝑦 − 34 Let g(y) = 𝑦 − 34 where g: Y → N Step 2: gof = g(f(x)) = g(4x + 3) = (4𝑥 + 3) − 34 = 4𝑥 + 3 − 34 = 4𝑥4 = x = IN Step 3: fog = f(g(y)) = f 𝑦 − 34 = 4 𝑦 − 34 + 3 = y – 3 + 3 = y + 0 = y = IY Since gof = IN and fog = IY, f is invertible & Inverse of f = g(y) = 𝒚 − 𝟑𝟒 Example 23 (Method 2) Let f : N → Y be a function defined as f (x) = 4x + 3, where, Y = {y ∈ N: y = 4x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse. f is invertible if f is one-one and onto Checking one-one f(x1) = 4x1 + 3 f(x2) = 4x2 + 3 Putting f(x1) = f(x2) 4x1 + 3 = 4x2 + 3 4x1 = 4x2 x1 = x2 If f(x1) = f(x2) , then x1 = x2 ∴ f is one-one Checking onto f(x) = 4x + 3 Let f(x) = y, where y ∈ Y y = 4x + 3 y – 3 = 4x 4x = y – 3 x = 𝑦 − 34 For every y in Y = {y ∈ N: y = 4x + 3 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 Finding inverse f(x) = 4x + 3 For finding inverse, we put f(x) = y and find x in terms of y We have done that while proving onto y = f(x) y = 4x + 3 ⇒ x = 𝑦 − 34 Let g(y) = 𝑦 − 34 where g: Y → N ∴ Inverse of f = g(y) = 𝒚 − 𝟑𝟒

Chapter 1 Class 12 Relation and Functions

Ex 1.2, 5
Important

Ex 1.2 , 10 Important

Example 23 Important You are here

Example 25 Important

Ex 1.3, 3 Important

Ex 1.3 , 6 Important

Ex 1.3 , 8 Important

Ex 1.3 , 9 Important

Ex 1.3 , 13 Important

Ex 1.3 , 14 Important

Ex 1.4, 11 Important

Misc 3 Important

Misc. 4 Important

Misc 14 Important

Misc 18 Important

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

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.