# Invertible Functions

Last Updated : 12 Jun, 2023

As the name suggests Invertible means inverse“, and Invertible function means the inverse of the function. Invertible functions, in the most general sense, are functions that “reverse” each other. For example, if f takes a to b, then the inverse, f-1, must take b to a.

## Invertible Function Definition

Inverse of a function is denoted by f -1

In other words, we can define it as, If f is a function the set of ordered pairs obtained by interchanging the first and second coordinates of each ordered pair in f is called the inverse of f. Let’s understand this with the help of an example.

Example:

function g = {(0, 1), (1, 2), (2,1)}, here we have to find the g-1

As we know that g-1 is formed by interchanging X and Y co-ordinates.

g = {(0, 1), (1, 2), (2, 1)}  -> interchange X and Y, we get

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

So this is the inverse of function g.

## Graph of Invertible Function

We can check for the function is invertible or not by plotting on the graph. We can plot the graph by using the given function and check for the invertibility of that function, whether the function is invertible or not.

Example: Find and plot the inverse of the function f(x) = 3x + 6.

Solution:

f(x) = 3x + 6

y = 3x + 6

Interchange x with y

x = 3y + 6

x – 6 = 3y

y = (x – 6) / 3

y = x/3 – 2

Thus, f -1(x) = x/3 – 2

Now let’s plot the graph for f-1(x). The inverse of a function having intercept and slope 3 and 1 / 3 respectively.

A function and its inverse will be symmetric around the line y = x. Then the function is said to be invertible. So let’s draw the line between both the function and the inverse of the function and check whether it separated symmetrically or not.

After drawing the straight line y = x, we observe that the straight line intersects the line of both of the functions symmetrically. So, the function f(x) is an invertible function and in this way, we can plot the graph for an inverse function and check the invertibility.

## Conditions for the Function to Be Invertible

Condition: To prove the function to be invertible, we need to prove that, the function is both One to One and Onto, i.e, Bijective.

### Explanation

We can say the function is One to One when every element of the domain has a single image with a codomain after mapping. We can say the function is Onto when the Range of the function should be equal to the codomain. When we prove that the given function is both One to One and Onto then we can say that the given function is invertible. Let’s see some examples to understand the condition properly.

Example 1: Let A: R – {3} and B: R – {1}. Consider the function f : A -> B defined by f(x) = (x – 2) / (x – 3). Show that function f(x) is invertible and hence find f-1.

Solution:

To show the function is invertible, we have to verify the condition of the function to be invertible as we discuss above. To show that the function is invertible we have to check first that the function is One to One or not so let’s check.

Let x, y âˆˆ A such that f(x) = f(y)

â‡’ (x – 2) / (x – 3) = (y – 2) / (y – 3)

â‡’ (x – 2) (y – 3) = (x – 3) (y – 2)

â‡’ xy – 3y – 2y + 6 = xy – 2x – 3y + 6

â‡’ -3x + 2y + 6 = xy – 2x – 3y + 6

â‡’ -3x + 2x = -3 + 2y

â‡’ -x = -y

â‡’ x = y

Since f(x) = f(y) â‡’ x = y, âˆ€x, y âˆˆ A, so function is One to One.

We have proved the function to be One to One. Now let’s check for Onto. To show that f(x) is onto, we show that range of f(x) = its codomain.

Let y = (x – 2) / (x – 3)

Put f(x) = y.

â‡’ xy – 3y = x – 2

â‡’ xy – x = 3y – 2

â‡’ x(y – 1) = 3y – 2

â‡’ x = (3y – 2) / (y -1)    —-(1)

Since x âˆˆ  R – {3}, âˆ€y R – {1}, so range of f is given as = R – {1}. Also codomain of f = R – {1}.

Therefore, Range = Codomain â‡’ f is Onto function

As both conditions are satisfied function is both One to One and Onto, Hence function f(x) is Invertible. Now as the question asked after proving function Invertible we have to find f -1

from eq(1) we get,

f -1(y) = (3y – 2) / (y – 1)

=> f-1(x) = (3x – 2) / (y – 1)

Example 2: Show that f: R – {0} -> R – {0} given by f(x) = 3 / x is invertible.

Solution:

To show the function f(x) = 3 / x is invertible.

We have to check first whether the function is One to One or not.

Let x1, x2 âˆˆ R – {0}, such that  f(x1) = f(x2). Then,

f(x1) = f(x2)

â‡’ 3 / x1 = 3 / x2
â‡’ x1 = x2

Thus, f(x1) = f(x2

â‡’ x1 =x2 âˆ€x, y âˆˆ  R – {0}

So, function f is One to One.

We have proved that the function is One to One, now let’s check whether the function is Onto or not.

Let y be an arbitrary element of  R – {0}.

Then for y in the codomain  R – {0},

There exist its pre-image in the domain  R – {0}. So f is Onto.

Since we proved the function both One to One and Onto, the function is Invertible.

Example 3: Consider f: R+ -> [4, âˆž] given by f(x) = x2 + 4. Show that f is invertible, where R+ is the set of all non-negative real numbers.

Solution:

To show that the function is invertible or not we have to prove that the function is both One to One and Onto i.e, Bijective

Let’s check for One to One.

Here function f: R+ -> [4, infinity)

Given by f(x) = x2 + 4,

Let x, y âˆˆ R+, such that f(x) = f(y)

â‡’ x2 + 4  = y2 + 4

â‡’ x2 = y2

â‡’ x = y [Since we have to take only +ve sign as x, y âˆˆ R+]

Therefore, f is One to One function.

Now, we have to check for Onto.

For y âˆˆ [4, infinity), let y = x2 + 4

â‡’ x2 = y – 4 â‰¥ 0

â‡’ x = âˆš(y – 4) â‰¥ 0 [we take only +ve sign, as x âˆˆ R+]

Therefore, for any y âˆˆ R+ (codomain), there exists

x = âˆš(y – 4) R+ (domain) such that

f(x) = f(âˆš(y-4)) = (âˆš(y – 4))2 + 4 = y – 4 + 4 = y

Therefore, f is Onto function.

Since function f(x) is both One to One and Onto, function f(x) is Invertible.

## How to find If a Function is Invertible?

As we had discussed above, the conditions for the function to be invertible are the same conditions we will check to determine whether the function is invertible or not. So let’s take some of the problems to understand properly how can we determine whether the function is invertible or not.

Example 1: If f is an invertible function, defined as f(x) = (3x -4) / 5, then write f-1(x).

Solution:

In the question given that f(x) = (3x – 4) / 5 is an invertible and we have to find the inverse of x. So, firstly we have to convert the equation in the terms of x. In the below figure, the last line we have found out the inverse of x and y. So, this is our required answer.

Given, f(x) (3x – 4) / 5 is an invertible function.

Let, y = (3x – 5) / 5
â‡’ 5y = 3x – 4
â‡’ 3x = 5y + 4
â‡’ x = (5y – 4) / 3

Therefore, f-1(y) = (5y – 4) / 3 or f-1(x) = (5x – 4) / 3

Example 2: f : R -> R defined by f(x) = 2x -1, find f-1(x)?

Solution:

As we done in the above question, the same we have to do in this question too. In the question we know that the function f(x) = 2x – 1 is invertible. Let us have y = 2x – 1, then to find its inverse only we have to interchange the variables.

Given,

f(x) = 2x -1 = y is an invertible function.

Let, y = 2x – 1

Inverse: x = 2y – 1

therefore, f-1(x) = (x + 1) / 2

Example 3: Show that the function f: R -> R, defined as f(x) = 4x – 7 is invertible or not, also find f-1.

Solution:

For function to be invertible it must follow two conditions,

• Function should be one-one
• Function should be onto

Given, f : R -> R such that f(x) = 4x – 7

For one to one:

Let x1 and x2 be any elements of R such that f(x1) = f(x2), Then

f(x1) = f(x2)
â‡’ 4x1 – 7 = 4x2 – 7
â‡’ 4x1 = 4x2
â‡’ x1 = x2
So, f is one to one

For onto:

Let y = f(x), y belongs to R. Then,
y = 4x – 7
â‡’ x = (y+7) / 4

From above it is seen that for every value of y, there exist it’s pre-image x.

So, f is onto

Thus, f is being One to One Onto, it is invertible.

## Inverse Trigonometric Functions

Inverse functions are of many types such as Inverse Trigonometric Functions, inverse log functions, inverse rational functions, inverse rational functions, etc. In the following table, there is the list of Inverse Trigonometric Functions with their Domain and Range.

## Finding Inverse Function Using Algebra

Example 1: Find the inverse of the function f(x) = (x + 1) / (2x – 1), where x â‰  1 / 2

Solution:

For finding the inverse function we have to apply very simple process, we  just put the function in equals to y.

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

â‡’ x +1 = 2xy – y

â‡’ x- 2xy = -y – 1

â‡’ x(1- 2y) = -y – 1

â‡’ x = (-y – 1) / (1 – 2y)

By taking negative sign common, we can write ,

x = (1 + y) / (2y – 1)

â‡’ f-1(x) = (1 + x / (2x – 1)

This is the required inverse of the function.

Example 2: Solve: f(x) = 2x  /  (x -1)

Solution:

As we done above, put the function equal to y, we get

2x / (x – 1) = y

â‡’ 2x = xy – y

â‡’ 2x – xy = -y

Taking x common from the left hand side

x (2 – y) = -y

â‡’ x = -y / (2 – y)

Taking y common from the denominator we get,

x = y / (y – 2)

â‡’ f -1(x) = x / (x – 2)

This is required inverse of the function.

Example 3: Find the inverse for the function f(x) = 2x2 – 7x +  8

Solution:

We follow the same procedure for solving this problem too,

put the function equal to y, we get

2x2 – 7x +  8 = y

Taking 2 common from left hand side

2 [ x2 – (7 / 2)x +  4] = y

To make the perfect square of (x – y)2

we have to divide and multiply by 2 with second term of the expression.

Adding and subtracting 49 / 16 after second term of the expression.

We get,

2[ x2 – 2. (7 / 2*2). x + 49 / 16 – 49 / 16 +4] = y

See carefully the underlined portion, it is the formula (x – y)2 = x2 – 2xy + y2

We can write this,

2(x – (7 / 4))2 – 49 / 8 + 8 = y

â‡’ 2(x – (7 / 4))2 = y – (15 / 8)

â‡’ (x – (7 / 4))2 = (y / 2) – (15 / 8)

Now, remove square root,

x – (7 / 4) = âˆš((y / 2) – (15 / 32))

â‡’ x = (7 / 4) + âˆš((y / 2) – (15 / 32))

â‡’ f-1(x) = (7 / 4) + âˆš((x / 2) – (15 / 32))

This is the required inverse of the function.

### Restricting Domains of Functions to Make Them Invertible

As the name of the heading suggests, here we will make the function that is not invertible to invertible by restricting or setting the domain such that our function should become an invertible function in the restricted domain. We know that the function is something that takes a set of number, take each of those numbers, and map them to another set of numbers. So if we start with a set of numbers.

The above table shows that we are trying different values in the domain and by seeing the graph we took the idea of the f(x) value. When x = 0 then our graph tells us that the value of f(x) is -8, in the same way for 2 and -2 we get -6 and -6 respectively. As we see in the above table on giving 2 and -2 we have the output -6 it is ok for the function, but it should not be a longer invertible function. So, in the graph, the function is defined as not invertible, why it should not be invertible?, because two of the values of x map the single value of f(x) as we saw in the above table. In order for the function to be invertible, you should find a function that maps the other way means you can find the inverse of that function, so let’s see

So if we find the inverse, and we give -8 the inverse is 0 it should be ok, but when we give -6 we find something interesting we are getting 2 or -2, it means that this function is no longer to be invertible, demonstrated in the below graph.

In the same way, if we check for 4 we are getting two values of x as shown in the above graph. Now, we have to restrict the domain so how that our function should become invertible. So, we can restrict the domain in two ways

• (0, âˆž)
• (-âˆž, 0)

Let’s try the first approach, if we restrict the domain from 0 to infinity then we have a graph like this

We have this graph and now when we check the graph for any value of y we are getting one value of x, in the same way, if we check for any positive integer of y we are getting only one value of x. Now, let’s try our second approach, in which we are restricting the domain from -infinity to 0. If we plot the graph our graph looks like this.

In this graph we are checking for y = 6 we are getting a single value of x. Now if we check for any value of y we are getting a single value of x. So in both of our approaches, our graph is giving a single value, which makes it invertible. So, our restricted domain to make the function invertible are

• (0, âˆž)
• (-âˆž, 0)

## FAQs on Invertible Functions

### Q1: Define Invertible Function.

Invertible functions are those function that has a unique inverse function, which can “reverse” the effect

### Q2: What is the Condition for a Function to be Invertible?

For a function to be  invertible, functions needs to be a bijective (one-one and onto function).

### Q3: How to Find the Inverse of a Function?

To find the inverse of a function switch the roles of dependent and independent variables and solve for original independent varible. The result of this process results in inverse of the function.

### Q4: What is the Notation used for the Inverse of a function?

The inverse of a function f(x) is denoted as f-1(x).

### Q5: Can all Functions be Inverted?

No, all functions can’t be inverted only functions which are bijective i.e., one-one and onto, can be inverted.

### Q6: What is the Domain of the Inverse of a Function?

Domain of the inverse of a function is the range of the original function.

### Q7: How to Determine if a Function is One-One?

To determine function is one-one, take two arbitrary values in it’s domaine and equate value of function at those values. If those two aribitrary values cames out to be same, then function is one-one.