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CXC MATHEMATICS Tutorial 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 © All rights reserved. No part of this document must be reproduced stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior written permission of the author. CXCDirect Institute Email: [email protected] Website: www.cxcDirect.org Math club : http://cxclink.50.forumer.com/ Telephone: 876 469-2775, 876 462-6139 First printed October 2009

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last element. By ordering the pairs, we can easily show the relation on a graph.

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 between the two sets.

A relation can be defined as set of ordered pairs where each pair has a specific rule of assignment.

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 element x is sometimes said to be mapped unto the range element. x  5x 1

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. x 5x +1

-2 -10 1

-1 -5 1

0 0 1

1 5 1

2 10 1

y

-9

-4

1

6

11

( x, y )

( -2, -9 )

( -1, -4 )

( 0, 1 )

( 1, 6 )

( 2, 11 )

Each value of x results in an output y based on a specific rule of y = 5x+1. The last row of the table, shows the (x, y) pairs. Note that the pairs are ordered, which is to say that x is specified as the first element in the pair, and y as the © cxcDirect Institute - 876 469-2775 Email: [email protected] website: www.cxcdirect.org

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Types of relations:

One to Many Relation

One to One Relation

y=x+1

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

y = x2

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Many to Many Relation

1

11

2

22

3

33

4

34

5

45

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 many relation.

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n.b

Functions

f (0) means that you substitute x with 0 in f(x)

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 see that the one to one relation and the many to one relation are functions. However, in the one to many relation, 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.

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)

An example of a function is the exchange rate between the Canadian 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.

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 Not a function

=(1- x)(1-x) – 4(1-x) + 5

Working with Functions

=1 –x –x + x2 - 4 +4x + 5 = x2 + 2x+2

We can visualize the function f relating two variables x and y as a block diagram as shown below. x

f

y = f(x)

Here x represents the domain elements also called the independent variable or the input to the function block. Similarly y represents the range elements, also called the dependent variable or in this case, the output. So:

y = f(x)

where: f(x) is pronounced: “ f of x”

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f) f(a) = a2 – 4a + 5

Exercise: 5.2 2 If f (x) = 2x + 4x – 9 : find: a) f (2) b) f (- 2) c) f (1/2) d) f (5x) e) f (4 - x)

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Composite Functions

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

If the input to a function f, is another function g, then the output combination of the two functions is called a composite function.

so fg(x)

… put g into f

⇒

= 2(x2 – 1) + 1 = 2x2 – 2 + 1 i.e.

f g(x) = 2x2 – 1

This can be visualized using the block diagram below. *******************************************************

x

g

g(x)

f

y = f( g(x) )

b)

now gf(x)

⇒

… put f(x) into g

= (2x + 1)2 - 1

Note that the input to the function f is g(x), so the output will be f(g(x)). So by putting g(x) into f(x), the resulting composite function is y = f (g(x)) which can be written as fg(x). This is pronounced “f of g of x”.

= (2x + 1) (2x + 1) – 1 = 4x2 + 2x +2x+ 1 – 1 =4x2 + 4x so:

gf(x)

= 4(x2 + x)

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

now f g(x) *******************************************************

(c)

fg(0)

a) f g(x) b) gf(x) c) fg(0) d) fg(-1) e) gf(-1)

( obtained from (a) above)

2

= 2(0) – 1 =-1

Example: 5.2 Given f (x) = 2x + 1 and g(x) = x2 - 1 what is the value of:

= 2x2 – 1

(d)

fg(-1)

=2(-1)2 – 1 = 2(1) – 1 = 1

e)

gf(x)

=4(x2 + x) ..

so

gf(-1)

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

obtained from (b)

= 4(1 + (-1) )

=0 ************************************************

Exercise: 5.3 If f (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) © cxcDirect Institute - 876 469-2775 Email: [email protected] website: www.cxcdirect.org

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Inverse of a Function

Example 5.24:

The inverse of a function is another function that reverses the operation of the original function.

Given

f  x=

5x − 3 ; find f -1(x) 2

Solution:

Mathematically we say that a function is the inverse of another function if: f

−1

f  x= x

let

y=

5x −3 2

then

x=

5y −3 2

interchange x and y

Where f -1(x) is defined as the inverse of f(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 inverse function f -1(x)

make y the subject:

f

−1

 x

2 x 3 5

replace y with f -1(x)

f

Example 5.23: Algebraic manipulation f  x= 3x− 1 ; find

5y 2

y=

⇒

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

Given

x 3=

⇒

-1

2 x 3 5

(x) =

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

Solution: step 1

⇒ step 2

⇒ step 3

⇒

step 4

⇒

define:

y = f(x) y = 3x – 1

Interchange x and y x = 3y-1 Make y the subject y=

x1 3

Exercise: 5.7 Find the inverse of the following functions: i)

f  x=  x−1

ii)

f  x= x 1 3

iii)

1

2 f  x= x − 2x hint: first, complete the square

replace y with f -1(x)

f -1(x)=

x1 3

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

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Step1:

Inverse of a composite function Example 5.3 f  x=

Given

x−1 x 2

; and

g  x= x

2

−1

f

2.

 fg   x

3.

f

y=

x −1 x2 + 2

⇒

x=

y −1 y2 + 2

⇒

2 2 x  y  2= y −1 Make y the subject:

let:

g  x −1

−1

2

−1

g  x

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

g  x

put

⇒

First, we need to find:

f

−1

g  x into f

−1

x− 1 x 2

Interchange x and y ⇒

x=

y− 1 y 2

⇒

x  y 2= y−1

⇒

2x  xy= y −1 2x +1 = y− xy

⇒

2x +1= y (1− x)

replace y with

f

−1

f

2x 1 1 −x

⇒

y=



2

2x 1 1− x

−1

replace y with

 fg   x



−1  fg   x =

⇒

f

⇒

y=

y=

2x 1 1− x

3.

Make y the subject:

⇒

⇒

 x y=

gives:

2

x

⇒

y= f  x

Interchange x and y :

2x 1= y 1− x

⇒

Solution:

let

.. ... g into f

Step2: Find the inverse:

1.

−1

x −1 x 2 +2

2

Find:

f

2

f ( x)=

fg ( x) =

2x +1 1− x

−1

−1

−1

g  x

inverse of g into the inverse of f

⇒

inverse of g =

−1 g  x =

inverse of f =

f

so :

f

−1

−1

x 2x 1 1− x

 x =

2  x1 1−  x

−1 g  x =

( x) f ( x)

4.

2x+ 1 1− x

( x) =

is undefined for values of x which makes the

denominator = 0. ⇒

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

**************************************** now we must put

g ( x) into f

−1

( x)

2

⇒

f

−1

g ( x) =

2x + 1 1− x 2

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

2. To find:

−1

( fg ) ( x)

This requires two steps: step1: put g into f to get step2: find the inverse of

fg ( x) fg ( x)

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Exercise 5.8 Given: g : x  2x and

f : x → 3x +1 ;

h : x  x 1

Find i)  gf −1  x 

;

ii) show that

 gf   x = f

iii)

−1

 fgh  x −1

−1  gf  −1 ;

−1

g−1  x

−1

 fgh 17

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Using the inverse function to solve equations.

Step2: substitute 9 into the inverse function

Example 5.5

4  92 5 −9

⇒

= 38 = −9.5 −4

1. Solve the equation below using the inverse function: 5x − 2 =9 x 4

2.

2. What is the value of x for which the function = 0 3. What is the value of x for which the function is undefined

For the function to be zero the numerator must be zero. 5x – 2 =0 x=2/5

⇒ ⇒

3.

f  x= 9

Note: if

;

then

x= f

−1

For the function to be undefined, the denominator must be zero. x 4 =0 or when

⇒

9

x=− 4

Proof:

Exercise 5.9

Given: f(x) = 9 then:

f

−1

but:

f

−1

⇒

f  x= f f  x= x

x= f

−1

−1

. 9

-1

..multiply both sides by f (x)

Solution:

Step1. (find the inverse – algebraic method)

Interchange x and y

⇒

x=

5y − 2 y4

Make y the subject: ⇒

xy  4x= 5y− 2

⇒

4x 2 = y 5 − x

⇒

f

−1

7x −1 =8 x 3

Exercise: 5.2 a =7, b = - 9, c = - 6.5 d = 50x2 + 20x – 9 e = 2x2 - 20x + 39 Activity: 5.3 a = 6x+3, b=10x-1 e= 15x+2, d = 30x - 1, e = 30x-13

5x− 2 x 4

⇒

Use the inverse function method to solve the equation

Answers

Step1 find the inverse of the function Step2 substitute the value 9

y=

7x −1 x 3

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

5x −2 f  x= x 4

Two steps are needed to solve:

Let

h  x=

What is value of x for which the function: 1) is equal to zero 2) is undefined

9

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

let

given

y=

4x 2 5−x

 x=

4x  2 5− x

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Activity: 5.7

i)

1 x

2

ii)

3 x − 1 iii)

1  x1

Activity: 5.8

 gf −1  x= −1

x −2 6

 gf  −1 =−1 /2

;

−1

 fgh  x=

x−7 6

−1

 fgh 17= 5/ 3

Activity: 5.9

3x 1 7− x −1 h 8 =−25 ; h  x=0 if x = 1/7; h  x −1

h  x=

is undefined if x = -3

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