Functions & Relations Intro

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CXC MATHEMATICSTutorial

Functions & Relations

Author: John Spencer MBA (Dist), M. Sc, B. Sc.

Former Senior Lecturer and Head of Section- University of Technology - Jamaica

Former Lecturer John Donaldson Technical Institute - Trinidad

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First printed October 2009
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Functions & Relations

Consider two sets of information where the elements in

one set can be linked or has a connection to the

elements in the other set of information.

Mathematically we can say that a relation exist betweenthe two sets.

Some examples are :

Menu items and prices. Names and ages of students in a class

Books and their location in a library

Names and phone numbers in a directory

Input and output elements of functions

Each element is the input set called the Domain, is

paired or mapped to one or more elements in the other

set called the Range. This pairing or mapping is done

based on a specific rule.

The domain elementx is sometimes said to be mapped

unto the range element.

x 5x1

This notation is called a mapping notation and is used

to describe the specific rule used to map x unto 5x+1.

We can also define x as an input variable and y as the

output so that:

y = 5x + 1

For the above equation, the relation between x and y for

values of x between -2 and +2 is shown in the table

below.

Each value ofxresults in an output y based on a

specific rule ofy = 5x+1.

The last row of the table, shows the (x, y) pairs. Note

that the pairs are ordered, which is to say thatxis

specified as the first element in the pair, andy as the

last element. By ordering the pairs, we can easily show

the relation on a graph.



A relation can be defined as

set of ordered pairs where each pair

has a specific rule of assignment.

x -2 -1 0 1 2

5x -10 -5 0 5 10

+1 1 1 1 1 1

y -9 -4 1 6 11

( x, y ) ( -2, -9 ) ( -1, -4 ) ( 0, 1 ) ( 1, 6 ) ( 2, 11 )


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Types of relations:

One to One Relation

If each value of x maps to only one value of y then we

have a one to one relation.

Many to one Relation

Here we see a case where more than one

input value x, maps to the same output y.

Example:x= 1.41 and x= - 1.41 both map to the

same output y = 2.

This is an example of a many to one relation.

One to Many Relation

Here we see a case where one input value x, maps to

more than one out values of y.

Example: x = 4, maps to two possible outputs:

y = 2, or y = - 2.

This is an example of a one to many relation.

Many to Many Relation

Here we have the case where x is paired with one or

more element in y, and also y is paired with one or

more element in x giving rise to a many to manyrelation.



y = x2

1

2

3

4

5

11

22

33

34

45

y = x + 1


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Functions

A function is a special type of relation.

We define a function as a relation where each element

in the input set maps to one and only one element in

the output set.

From the three relations discussed earlier, we can seethat the one to one relation and the many to onerelation are functions. However, in the one to manyrelation, an element in the input set can be mapped to

more than one output element, so the one to many

relation is not a function.

An example of a function is the exchange rate between theCanadian dollar and the US dollar

A simple test for a function is called a vertical line test.If a vertical line cuts a graph at more that one point,then it is Not a function.

Not a function

Working with Functions

We can visualize the functionfrelating two variablesx

andy as a block diagram as shown below.

Herexrepresents the domain elements also called the

independent variable or the input to the function

block.

Similarlyy represents the range elements, also called

the dependent variable or in this case, the output.

So: y = f(x)

where: f(x) is pronounced: fofx

n.bf(0) means that you substitute x with 0 in f(x)

Example: 5.1

Given f(x) = x2 - 4x +5,

What is the value of:

a)f(0)b)f(- 1)

c)f(3/2)

d) (3x)

e)f(1 x)

f)f(a)

Solution:

a) f(0) = (0)2 4(0) + 5 = 5

b) f(-1) = (-1)2 4(-1) + 5 = 10

c) f(3/2) = (3/2)2 4(3/2) + 5 = 5/4

d) f(3x) = (3x)2 4(3x) + 5 = 9x2 12x + 5

e) f(1- x) = (1- x)2 4(1- x) + 5

=(1- x)(1-x) 4(1-x) + 5

=1 x x + x2 - 4 +4x + 5

= x2 + 2x+2

f) f(a) = a2 4a + 5

Exercise: 5.2

Iff(x) = 2x2 + 4x 9 :

find:

a)f(2)

b)f(- 2)

c)f(1/2)

d)f(5x)

e)f(4 -x)



fx y = f(x)


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Composite Functions

If the input to a functionf, is another functiong, then

the output combination of the two functions is called a

composite function.

This can be visualized using the block diagram below.

Note that the input to the functionf isg(x), so the

output will bef(g(x)). So by putting g(x) into f(x),

the resulting composite function isy = f (g(x)) which

can be written as fg(x).

This is pronounced f of g of x.

*******************************************************

Example: 5.2

Given f(x) = 2x + 1 andg(x) =x2 - 1

what is the value of:

a) f g(x)

b) gf(x)c) fg(0)

d) fg(-1)

e) gf(-1)

Solution:

a) f(x) = 2x+1 and g(x) = x2 1

so fg(x) put g into f

= 2(x2 1) + 1

= 2x2 2 + 1

i.e. f g(x) = 2x2 1

*******************************************************

b) now gf(x) put f(x) into g

= (2x + 1)2 - 1

= (2x + 1) (2x + 1) 1

= 4x2 + 2x +2x+ 1 1

=4x2 + 4x

so: gf(x) = 4(x 2 + x)

*******************************************************

now f g(x) = 2x2 1 ( obtained from (a) above)

(c) fg(0) = 2(0)2 1

= - 1

(d) fg(-1) =2(-1)2 1

= 2(1) 1 = 1

e) gf(x) =4(x2 + x) .. obtained from (b)

so gf(-1) = 4((-1)2 + (-1))

= 4(1 + (-1) )

= 0

************************************************

Exercise: 5.3

Iff(x) = 2x - 1,

g(x) = 3x + 2, and h(x) = 5x,

what is the value of:

a)fg(x)

b)fh(x)

c)gh(x)

d) gfh(x)

e)ghf(x)



gx g(x) y = f(g(x))f


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Inverse of a Function

The inverse of a function is another function that

reverses the operation of the original function.

Mathematically we say that a function is the inverse of

another function if:

f1

fx=x

Wheref-1(x) is defined as the inverse off(x)

*******************************************************

The steps to find f-1(x) are:

1. Introduce a new variable y = f(x)

2. Interchange x and y

3. Transpose the equation to make y the subject

4. The new result for y is now the inversefunctionf-1(x)

********************************************************

Example 5.23: Algebraic manipulation

Given fx=3x1 ; find f1x

Solution:

step 1 define: y = f(x)

y = 3x 1

step 2 Interchange x and y x = 3y-1

step 3 Make y the subject

y=x13

step 4 replace y withf-1(x)

f-1(x)=x13

*********************************************************

Example 5.24:

Given fx=5x

23 ; findf-1(x)

Solution:

let y=5x

2 3

then x=5y

23 interchange x and y

make y the subject:

x3=5y

2

y=2 x3

5

replace y withf-1(x)

f -1(x) =2 x3

5

******************************************************

Exercise: 5.7Find the inverse of the following functions:

i) fx=x1

ii) fx=x11

3

iii) fx=x22x hint: first, complete the square




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Inverse of a composite functionExample 5.3

Given fx=x1x2

; and gx =x2

Find:

1. f1

gx2. fg1 x

3. f1

g1 x

4. What is the value of x for which f(x) is undefined

Solution:

f1

gx put gx into f 1x

First, we need to find: f1x

let y=f x y=x1x2

Interchange x and y x=y1y2

Make y the subject:

x y2= y1

2xxy=y1

2x+1=yxy

2x+1=y (1x)

y=2x+11x

replace y with f1(x)

gives: f1(x) =

2x+11x

now we must put g(x) into f 1

(x)

f1g(x) =2x

2+11x2

******************************************************

2. To find: ( fg)1 (x)

This requires two steps:

step1: put g into f to get fg(x)step2: find the inverse of fg(x)

Step1:

fg(x) = f(x)=x

21x2+2

.. ... g into f

Step2: Find the inverse:

let: y=

x21

x2+2

x=y

21y2+2

Interchange x and y :

x y22=y21 Make y the subject:

2x1=y21x

y2=

2x11x

y=2x11x

replace y with fg1 x

fg1 x = 2x11x

3.

f1

g1x inverse ofginto the inverse off

inverse of g = g1 x = x

inverse of f = f1x = 2x1

1x

so : f1

g1x =

2 x11x

4. f(x) is undefined for values of x which makes the

denominator = 0.

if x = - 2 , then f(x) is undefined

****************************************

Exercise 5.8 Given:

f:x 3x+1 ; g:x 2x and h : x x1

Find

i) gf1x ; fgh1 x

ii) show that gf1x =f 1g1 x

iii) gf11 ; fgh1 17




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Using the inverse function to solve equations.

Example 5.5

1. Solve the equation below using the inverse function:

5x

2

x4 =9

2. What is the value of x for which the function = 0

3. What is the value of x for which the function isundefined

Note: if fx=9 ; then x = f1 9

Proof:

Given:f(x) = 9

then: f1f x=f1 . 9 ..multiply both sides by f-1 (x)

but: f1

fx=x

x = f1 9

This means that the solution x, is found substituting the value 9 inthe inverse of the function.

Solution:

let fx=5x2x4

Two steps are needed to solve:Step1 find the inverse of the function

Step2 substitute the value 9

Step1. (find the inverse algebraic method)

Let y=5x2x4

Interchange x and y x=5y2y4

Make y the subject:

xy4x=5y2

4x2=y 5x

y=4x25x

f1x=

4x25x

Step2:

substitute 9 into the inverse function

4 9259

= 384

= 9.5

2. For the function to be zero the numerator mustbe zero.

5x2=0 x=2 /5

3. For the function to be undefined, thedenominator must be zero.

x4=0 or when x= 4

Exercise 5.9

given h x= 7x1x3

What is value of x for which the function:

1) is equal to zero

2) is undefined

Use the inverse function method to solve the equation

7x1x3

=8

**********************************************

Answers

Exercise: 5.2

a =7, b = - 9, c = - 6.5d = 50x2 + 20x 9e = 2x2 - 20x + 39

Activity: 5.3

a = 6x+3, b=10x-1e= 15x+2, d = 30x - 1, e = 30x-13

Activity: 5.7

i) 1x2 ii) x31 iii) 1x1

Activity: 5.8

gf1

x=x26 ; fgh

1

x=x76

gf1 1 =1 /2 fgh1 17=5 /3

Activity: 5.9

h1x=

3x17x

h18 =25 ;

h x=0 if x = 1/7; h x is undefined if x = -3



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Functions & Relations Intro