ONTO FUNCTION

Example 1 :

Check whether the following function is onto.

f : N → N defined by f(n) = n + 2

Solution :

Domain and co-domains are containing a set of all natural numbers.       

If x = 1, then f(1)  =  1 + 2  =  3.

If x = 2, then f(2)  =  2 + 2  =  4.

From this we come to know that every elements of codomain except 1 and 2 are having pre image with.

In order to prove the given function as onto, we must satisfy the condition.

Co-domain of the function = range 

Since the given question does not satisfy the above condition, it is not onto.

Example 2 :

Check whether the following function is onto.

f : R → R defined by f(n) = n²

Solution :

Domain = All real numbers.

Co-domain = All real numbers.

Since negative numbers and non perfect squares are not having preimage. It is not onto function.

Example 3 :

Check whether the following function are one-to-one.

f : R - {0} → R defined by f(x) = 1/x

Solution :

Domain = all real numbers except 0.

Co-domain = All real numbers including zero.

In co-domain all real numbers are having pre-image. But zero is not having preimage, it is not onto.

Example 4 :

f : [−2, 2] → B is given by f(x) = 2x³, then find B so that f is onto.

Solution :

Here domain is [-2, 2] and range is B.

B is [-16, 16].

If f : R → R is defined by

f(x) = 2x − 3

prove that f is a bijection and find its inverse.

When the function is one to one and onto, that can be considered as bijection.

To prove the given function as one to one, let us assume

f(x) = f(y)

2x - 3 = 2y - 3

2x = 2y

x = y

So, it is one to one function.

To prove the function is onto, we have to prove the codomain and range are equal.

Then let x = (y + 3)/2

Then f(x) = 2[(y + 3)/2] - 3

= y + 3 - 3

= y

f(x) = y

Then it is onto function.

Finding inverse function :

Let y = 2x - 3

Solve for x,

y + 3 = 2x

x = (y + 3)/2

Replace x by f^-1(x) and y by x.

f^-1(x) = (x + 3)/2

Example 5 :

State whether the following relations are functions or not. If it is a function check for one-to- oneness and ontoness. If it is not a function, state why?

(i) If A = {a, b, c} and f = {(a, c),(b, c),(c, b)}; (f : A → A).

(ii) If X = {x, y, z} and f = {(x, y),(x, z),(z, x)}; (f : X → X).

Solution :

i)

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

Here the out put c is having more than one preimage, then it is not a one to function. In codomain the output a does not has any preimage, then it is not onto function.

ii)  

f = {(x, y), (x, z), (z, x)

Here the input x is associated with more than one output, then it is not a function.

Example 6 :

If f : R → R is defined by f(x) = 3x − 5, prove that f is a bijection and find its inverse.

Solution :

Proving the function f(x) is one to one :

f(x) = f(y)

3x - 5 = 3y - 5

3x = 3y

x = y

Proving the function f(x) is onto :

To prove codomain and range are equal, 

Let x = (y + 5)/3

f(x) = 3[(y + 5)/3] - 5

= y + 5 - 5

f(x) = y

Then it is onto function.

Finding inverse :

Let y = 3x - 5

y + 5 = 3x

x = (1/3)(y + 5)

f^-1(x) = (1/3)(x + 5)

Example 7 :

The function f : [0, 2π] → [−1, 1] defined by f(x) = sin x is

(1) one-to-one        (2) onto

(3) bijection           (4) cannot be defined

Solution :

f(x) = sin x

Solving trigonometric function, for one out put we have have more than one inputs.

For example,

1/2 = sin x

x = sin^-1(1/2)

initial value, x = π/6

x = nπ ± (-1)ⁿ α

When n = 1

x = 1π ± (-1)¹ (π/6)

= 1π - 1(π/6)

x = 5π/6

For  x = π/6 and 5π/6, we get the value 1/2.

 So, it is not one to one function

It is onto function.

Example 8 :

If the function f : [−3, 3] → S defined by f(x) = x² is onto, then S is 

(1) [−9, 9]       (2) R     (3) [−3, 3]      (4) [0, 9]

Solution :

f(x) = x²

So, option (4) is the correct answer.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.