section 3c inverse functions

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SECTION C Inverse Functions

By the end of this section you will be able to

understand what is meant by a bijective function show that a given function is bijective understand and find the inverse function see the relationship between bijective and inverse functions

You need to know your work on injections and surjections from section B thoroughly

to understand this section. It is a much more difficult section than the previous two

(3A and 3B).

C1 Bijective Functions

Guess what the term bijective function means?

A function which is both injective (one to one) and surjective (onto). A bijective

function is also called a bijection. Hence a bijection is a function from a set A (domain) to a setB (codomain) which is both injective and surjective. Therefore a

bijective function is both one to one and onto.

A B

Domain CodomainFig 16

Fig 16 shows a bijective function.

Example 17

Let be defined by:f ( ) 2 1f x x= + . Prove that f is a bijective function.

Solution

What do we need to prove?

We have to prove that the given function ( ) 2 1f x x= + is one to one and onto. First

we show that the function is one to one that is f is injective. We show

( ) ( ) givesf x f y x y= =

How?

Well

( )2 1f x x= + and

( )2 1f y y= + equating these we have


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2 1 2 1

2 2 which gives

x y

x y x

+ = +

y= =

Hence the given function f is one to one. Next we show f is onto (surjective).

How?

Let where( )y f x= y is in the codomain and then find an x in the domain.

[ ]

2 1

1Transposing

2

y x

yx

= +

=

Since (because it is in the codomain) which means that is a real number,y y

therefore1

2

yx

= is also a real number so we have found an x in the domain, .

Hence f is onto (surjective).

Since the given function ( ) 2 1f x x= + is both, one to one and onto, we conclude that

it is a bijective function.

One of the difficulties in this section is it is easy to get confused with all the notation

used to show a given function is bijective. For a function to be injective (one to one)

we consider elements x and y in the domain:

x y

f

( ) ( )f x f y=

This is an

INJECTION

Fig 17

To show surjection (onto) we consider to be in the codomain and then find any x in

the domain which corresponds to this via the functiony f .

x y

f This is a

SURJECTION

Fig 18

From the last section you might have noticed that any function can be made surjective

(onto) by changing the codomain of that function. Similarly we can change the

domain to make the function injective (one to one).

In the next example we use the method of completing the square to find an

appropriate codomain for the function to be surjective. If you have forgotten thismethod of completing the square or you have not covered it, then see Appendix B.


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Example 18

Let be defined by:f + ( ) 2 4 5f x x x= + . What is the largest codomain so

that f is surjective (onto)?

SolutionWe apply the technique of completing the square on ( ) 2 4 5f x x x= + .

( ) ( )

( ) [ ]

2 22 2

2

4 5 2 4 5 Because 2 4 4

2 1 Because 4 5 1

x x x x x x

x

+ = + = +

= + + =

Hence we have

( )

( )

2

2

4 5

2 1

f x x x

x

= +

= +

What do we know about the square part, ( ) ?2

2x

It is always positive or zero because squaring any real number does not give a

negative number. Hence

( )2

2x 0

Therefore . Hence( ) ( )2

2 1 0 1f x x= + + =1 ( ) 1f x . Therefore the largestcodomain is going to be the set of real numbers which are greater than or equal to 1.

How do we write this?

{ }and 1x x x . The codomain is the set B say where { }and 1 .B x x x=

Next we prove that the function given by the formula in Example 18 is bijective forthe appropriate domain and codomain.

Example 19

Let { }and 2A x x x= and { }and 1B x x x= . Let :f A B be

defined by ( ) 2 4 5f x x x= + . Prove that f is a bijective function.

Solution

We have to prove that the given function ( ) 2 4 5f x x x= + is one to one and onto.

First we show that the function is one to one that is f is injective. We need to show

( ) ( ) givesf x f y x y= =

How?

( ) ( )

[ ]

( ) ( ) ( ) ( ) ( )

( ) ( )

2 2

2 2

2 2

4 5 4 5

4 4 0 Rearranging

4 0 Remember

4 0

x x y y f x f y

x y x y

x y x y x y x y x y x y

x y x y

+ = + =

+ =

+ = = +

+ = [ ]

( )

[ ]

Factorizing

0 or 4 0

or 4 Solving for

x y x y

x y x y x

= + =

= =


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Consider the solution 4x y= . Remember is a member of the domain,y

{ }and 2A x x x= , which means it is a real number greater than or equal to2. If then2y = 4 2 2x = = and we have 2x y= = . If then2y > 4 2x y=

section 3c inverse functions

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