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## Types of Functions

### One-to-one function or one–one function

A function f: A→B where A,B are two non-empty sets is called a one-to-one function if no two distinct elements of A have the same image in B. That is, f: A→B is a one-to-one (or one–one) function if and only if (i f f).

x1, x2∈A and x1≠ x2 then f(x1) ≠f(x2)    or

if f(x1) = f(x2) x1= x2

### Example 1

f  = {(a,x), (b,y), (c,z)}

is a oneone function.

### Example 2

Let f: R→R be defined by f(x) = x2. Is f one–one?

Consider f(1) = (1)2

= 1

f(-1) = (-1)2 = 1

f(x1) = f(x2) = 1

But x1≠ x2

1 ≠-1

So f is not a one–one function.
A one–one function or one-to-one function is also called an injection.

### Onto function

A function f: A→B, A,B are non-empty sets, is called an onto function if f(A)=B. That is f is onto if every element of the
codomain B is the image of at least one element of the domain A.

f: A→B is onto if and only if (iff) for every y ∈B there exists at least one x∈ A such that f(x) = y.

### Example 3

Let f: {a, b, c} →{1,2,3} such that f(a) = 3, f(b) = 2, f(c) = 1.

f is an onto function, since each element of {1,2,3} is an image of an element of {a, b, c}.

### Example 4

Let f: {2, 4, 7} →{p, q, r}

such that f(2) = q, f(4) = r, f(7)= r

f is not onto as p is not the image of any element of {2, 4, 7}

### Example 5

Let   f: N→ {-1, 1} defined by

f(n) = 1 if f is odd,

= -1 if n is even.

f is onto.

### Example 6

Let f: R→R be defined by f(x) = 3x-5. Show that f is onto.

Solution:

Let y = f(x) = 3x - 5

y = 3x - 5

y + 5 = 3x

For every y ∈R there is an x ∈R

 such that

An onto function is also called a surjection.

### One–one and onto functions

A function f: A→B, A,B are non-empty, is called a one–one and onto function if it is both one–one and onto. This type of function is also called a bijection.

### Example 7

If f   = { (p,1), (q,2), (r,3) }

f is one–one and onto

Since f (p) = 1

f(q) = 2

f(r) = 3

For every element of {1,2,3} is the image of element of {p,q,r}.

So f is onto.

Also, f is one–one since two distinct elements of {p,q,r} have two distinct images in {1,2,3}.

### Example 8

Let f: R →R be defined by f(x) = 2x + 3

f is a one–one and onto function.

Consider      -1≠ 1

f(-1) ≠f(1)

as 2 (-1) +3 ≠2 * 1+3

-2+3 ≠2+3

1 ≠5

Let y = f(x) = 2x + 3

y - 3 = 2x

 So for every y ∈R there exists an x ∈R such that
So f is onto.

### Example 9

Let f: R→R be defined by   f (x) = x2 - 1

Consider    f (2) = 22 -1

= 4-1

= 3.

f(-2) = (-2)2-1

= 4-1

= 3

f(2) = f(-2)

But 2 ≠-2

f is not one–one.

So f is not a bijection.

### State if the following functions are One–One

a.

f1(x) = x2          f1: R→ R

b.

f2(x) = -3x        f2: R→ R

Solutions:

a.
f1 : R →R

f1(x) = x2

f1is not One–One since

f1(-2) = (-2)2

= 4

f1(2) = (2)2

=
4

f1(-2) = f1(2)

=
4

but -2 ≠2

b.
f2: R →R
f2(x) = -3x

f2(x1) ≠ f(x2)

-3x1 ≠-3x2

x1 ≠x2

So f2 is one–one.

2.

### State if the following functions are onto

a.

g1(x) = 2x3      g1: R→ R

b.

g3: Z→ Z defined by g3 (x) = x-1

Solutions:

a.

g1: R→ R

g1(x) = 2x3

Let y = g1 (x) = 2x3

y = 2x3

y/2 = x3

g1 is onto.

b.

g3: Z→ Z      g3(x) = x-1

g3is an onto function since g3 (Z) = Z.

Or

for every x ∈ Z there exists a y ∈ Z such that g(x) = y.

3.

### State if the following functions are bijections. (One–One and onto functions).

a.

f1 = {(a,-1), (b,-2), (c,-3), (d, -4)}

b.

f2 : R→ R defined by f2(x) = 4x-1

Solutions:

a.

f1 = {(a,-1), (b,-2), (c,-3), (d,(-4)}

f1 is One–One since

-1 ≠-2

f1(a) ≠f1(b)

a≠ b

f1 is onto as every element of {a,b,c,d} has an image in {-1,-2,-3,-4}.

f1 is a bijection.

b.

f2 : R→ R where f2(x) = 4x-1

f2 is One–One since

f2(x1) = f2(x2)

 On canceling like terms

x1 = x2

f2 is onto since for every element x∈R there exists an element y∈ R such that

f(x) = y.

f2 is a bijection.

4.

### Find the inverse functions (if they exist) of the following.

a.

f: R→ R f | x | = 2x2-1

b.
 f: R- {3} →R - {1} defined by f(x) =

Solutions:

a.

f: R→ R f (x) = 2x2-1
f is not One–One as

f(-1) = 2 - 1

= 1

f(1) = 2 - 1

= 1

So         f(-1) = f(1)

But -1 ≠1

f-1 does not exist.

b.

 f: R- {3} →R - {1} defined by f(x) =

f is One–One

f is onto

f-1 exists.

 let   y =f(x)= y = y(x-3) = (x + 3) yx-3y= x+3 yx - x = 3 + 3 x(y-1) = 3(y+1) x = x=f--1(y) = Or f--1(x) =