## Inverse of a Function

If f is a function, then the set of ordered pairs obtained by interchanging the first and second coordinates of each ordered pairs of f is called the inverse of f and is denoted by f -1 ( read as f – inverse ).

Let f -1 = { (y,x) | (x,y) ∈ f }

Consider these functions:

f = {(0,0), (1,1), (2,4), (3,9), . . . . }

g = {(0,1), (1,2), (2,3), . . . }

h = {(0,0), (1,-1), (2,-2), (3,-3), . . . }

p = {(0,1), (1,2), (2,2), . . . .}

The inverse of these functions are:

f -1 = {(0,0), (1,1), (4,2), (9,3), . . .}

g -1 = {(1,0), (2,1), (3,2), . . .}

h -1 = {(0,0), (-1,1), (-2,2), (-3,3),. . .}

p -1 = {(1,0), (2,1), (2,2), . . .}

We can see that f -1, g -1, h -1 are functions but p -1 is not a function because 2 has two images, 1 and 2. So we see that even if f: A→B is a function, this does not necessarily mean that f -1: A→ B is also a function.

Suppose f: A→B is one–one and onto, then for each b ∈B there exists a unique f -1(b) ∈A such that f -1(b) = a

OR

For every b ∈B there is only one element f -1(b) ∈A assigned by this correspondence f -1. So f -1 is a function from B to A and is denoted by

f -1: B→A

f -1 is called the inverse function of f. If f is one–one and onto, then f -1 is also a function.

Example 1:

Let f: R→R defined by f(x) = x3 where R is the set of real numbers.

f is one–one.

Since f(x1) ≠ f(x2)

⇒ x13 ≠ x23

On finding the cube roots x1≠ x2

f is onto.

Let y = f(x) = x3

⇒y = x3 ------------------- (1)

Taking cube roots ------------------(2)

So f is one–one and onto.

Therefore, f -1 exists.

f -1: R→R exists and is given by from (1) and (2).

Example 2:

Let f: R→R be defined by f(x) = 3x-1. Find f -1.

Solution:

f is one–one.

Let y = f(x) = 3x -1

y = 3x-1 So f is onto.

f is one–one and onto. So f -1 exists.

f -1: R →R is defined by f -1 (y) = x = for all y ∈ B.

f -1 is one–one and onto.

#### Try these questions:

1. Let f: R→R be defined by f(x) = 2x-7. Show that f has an inverse function f -1. Find the rule that defines f -1.
Solution:
f is one–one and onto. f -1 exists.
Let y = f(x) = 2x-7 and x = f -1 (y) 2. Let f be given by f(x) = x + 3 and f has the domain {x | 3 ≤ x ≤ 7}. Find the domain and range of f -1. Solution: Given f(x) = x+3 Domain f = {x | 3 ≤ x ≤ 7} Range f = { f(x) | 3+3 ≤ f(x) ≤ 7+3} = {f(x) | 6 ≤ x+3 ≤10} = {y | 6 ≤ y ≤ 10} Taking y = f(x) y = f(x) = x+3 y-3 = x ⇒ x = f -1(y) that is f -1 (y) = y - 3. The domain of f -1 = { y | 6 ≤ y ≤10} and range of f -1 = { x | 3 ≤ y ≤7}