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) = x^{3} where R is the set of real numbers.

f is one–one.

Since f(x1) ≠ f(x2)

⇒ x1^{3} ≠ x2^{3}

On finding the cube roots x1≠ x2

f is onto.

Let y = f(x) = x^{3}

⇒y = x^{3} ------------------- (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.

- 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)

- 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}

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