# Inverse Function Let A = {2, 3, 4}, B = {a, b, c} and the function f : A → B is defined by f = {(2, a), (3, b), (4, c)}. Here,

2 A, f(2) = a B

3 A, f(3) = b B

4 A, f(4) = c B

Also domain of f = {2, 3, 4} and range of f = {a, b, c}.

Now we define a relation g : B → A so that g = {(a, 2), (b, 3), (c, 4)}. Here,

a B, g(a) = 2

b B, g(b) = 3

c B, g(c) = 4

So, g is also one-to-one onto function.

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The function g is defined from B to A is said to be the inverse of f is denoted by f-1. But for every function, there may not exist its inverse. Let us study the following examples.

1. f = {(2, 3), (3, 3), (4, 5)}. Here, f is “many to one function.” Here a relation from B to A is not a function as 3 B corresponds with more than one element of A. So, f-1 does not exist.

2. f = {(2, 3), (4, 8), (5, 2)}. Here, f is a one-to-one into function. Here, a relation from B to A is not a function because 9 B does not correspond with any element of A. So, f-1 does not exist.

Hence, from the above examples, it is clear that to define an inverse function f must be a one-to-one onto function.

## Definition of Inverse Function:

Let f : A → B be a one-to-one onto function. Then a function from B to A which associates each element of B with a unique element of A is called the inverse function of f. The inverse function from B to A is denoted by f-1 : B → A.

### Worked Out Examples:

Example 1: Let A = {a, b, c} and f : A → A is defined as f = {(a, c), (b, a), (c, b)}, find f-1.

Solution:

Here,

f = {(a, c), (b, a), (c, b)}

Clearly, the function f is one-to-one onto, hence f-1 exists and

f-1 = {(c, a), (a, b), (b, c)}.

Example 2: Let f : R → R be defined by f(x) = 2x – 3, find the formula that defines f-1.

Solution:

Here,

f : R → R and x is a variable, f is a one-to-one function, f-1 exists.

y = f(x) = 2x – 3,

To find the inverse function we interchange the role of x and y,

i.e.  x = 2y – 3

or,  2y = x + 3

or,  y = (x + 3)/2

Hence, inverse function y = f-1(x) = (x + 3)/2