§ 2.1

21
§ 2.1 Introduction to Functions and sets from 1.1

Upload: alexis

Post on 19-Jan-2016

35 views

Category:

Documents


0 download

DESCRIPTION

§ 2.1. Introduction to Functions. Relation p 96. Definition of a Relation A relation is any set of ordered pairs. The set of all first components of the ordered pairs is called the domain of the relation - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: § 2.1

§ 2.1

Introduction to Functions and sets from 1.1

Page 2: § 2.1

Blitzer, Intermediate Algebra, 5e – Slide #2 Section 1.1

Sets p 6

The braces, { }, indicate that we are representing a set. This form of representation, called the roster method, uses commas to separate the elements of the set. The ellipsis indicates that there is no final element and that

the listing goes on forever.

The objects in a set are called the elements The objects in a set are called the elements of the set. Such as:of the set. Such as:

A set is a collection of objects whose contents A set is a collection of objects whose contents can be clearly determined.can be clearly determined.

,5,4,3,2,1

Page 3: § 2.1

Blitzer, Intermediate Algebra, 5e – Slide #3 Section 1.1

Set-Builder Notation p 6

{x | x is a real number and greater than 10}

Express x > 10 using set-builder notation

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

The set of all x Such that X is a real number greater than 10

Page 4: § 2.1

Blitzer, Intermediate Algebra, 5e – Slide #4 Section 1.1

Symbols p 7 and

The symbol is used to indicate that a number or object is ina particular set. Here is an example:

7 {1,2,5,7,9}

The symbol is used to indicate that a number or object is not in a particular set. For example:

3 {4,6}

Page 5: § 2.1

Blitzer, Intermediate Algebra, 5e – Slide #5 Section 2.1

Relation p 96

Definition of a Relation

A relation is any set of ordered pairs.

The set of all first components of the ordered pairs is called the domain of the relation

and the set of all second components is called the range of the relation.

Page 6: § 2.1

Blitzer, Intermediate Algebra, 5e – Slide #6 Section 2.1

Relation p 96

Check Point 1Find the domain and the range of the relation: {(0, 9.1), (10, 6.7), (20, 10.7), (30, 13.2), (34, 15.5)}

Domain: {0, 10, 20, 30, 34}

Range: {9.1, 6.7, 10.7, 13.2, 15.5}

What is the rule that assigned the “inputs” in the domain to the “outputs” in the range? For example, for the ordered pair (30, 13.2), how does the data in Figure 2.1(a), which shows the percentage of first-year US college students claiming no religious affiliation.

010

3020

6.710.7

9.1

13.2

Domain Range34 15.5

Page 7: § 2.1

Blitzer, Intermediate Algebra, 5e – Slide #7 Section 2.1

Functions p 98

Definition of a Function

A function is a correspondence from a first set, called the domain, to a second set, called the range, such that each element in the domain corresponds to exactly one element

in the range.

(Each x corresponds to exactly one y in a function. Values of x have to be “faithful” – so to speak – that is, each x goes to exactly one y. However, a y value

may be the image of more than one x – so y’s are not required to be “faithful”)

For example, the following relation is a function:

{(1,2), (2,3), (3,5), (4,3)} Note that the y value 3 is the image of two x values.

Page 8: § 2.1

Blitzer, Intermediate Algebra, 5e – Slide #8 Section 2.1

Basic Functions 98

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Determine whether the following is a function.

12

43

guitarviolin

drums

flute

Yes, because none of the members of the domain correspond to more than one member of the range.

Domain Range

Page 9: § 2.1

Blitzer, Intermediate Algebra, 5e – Slide #9 Section 2.1

Basic Functions 98

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Determine whether the following is a function.

38

95

beetlescrickets

ants

moths

No, because one of the members of the domain, 9, corresponds to more than one member of the range.

Domain Range

Page 10: § 2.1

Blitzer, Intermediate Algebra, 5e – Slide #10 Section 2.1

Functions 99

Example: Determine whether each relation is a function:

(a) {(2,3), (4,5), (6,5), (9,10)}

(a)Yes – a function

(b) {(1,2), (3,3), (6,8), (1,10)}

(b) No – not a function since 1 is mapped to 2 and 10

(c) {(1,5), (4,5), (2,5), (3,5)}

(c) Yes – a function

Page 11: § 2.1

Blitzer, Intermediate Algebra, 5e – Slide #11 Section 2.1

Relation p 99

Check Point 2Determine whether each relation is a function:

a. {(1,2), (3, 4), (5, 6), (5, 7)}

b. {(1, 2), (3, 4), (6, 5), (7, 5)}

Not a function

function

Page 12: § 2.1

Blitzer, Intermediate Algebra, 5e – Slide #12 Section 2.1

Functions as Equations 99

Functions are usually given in terms of equations rather than as sets of ordered pairs.

For example, consider the function y = 2x – 3

For each value of x, there is one and only one value of y.

The variable x is called the independent variable and the variable y is called the dependent variable.

Page 13: § 2.1

Blitzer, Intermediate Algebra, 5e – Slide #13 Section 2.1

Function Notation 100

If an equation in x and y gives only one value of y for each value of x, then the variable y is a function of the variable x.

When an equation represents a function, the function is often named by a letter such as f, g, h, F, G, or H.

The output value, the y value, is often denoted by f(x), read

“f of x” or “f at x”

The notation f(x) does not mean “f times x”. The notation describes the value of the function f at x.

Page 14: § 2.1

Blitzer, Intermediate Algebra, 5e – Slide #14 Section 2.1

Basic Functions 100

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find the indicated function value:

724for 3 3 xxxff

732343 3 f

7322743 f

761083 f

1073 f

Replace x with 3

Evaluate the exponent

Multiply

Add and Subtract

Page 15: § 2.1

Blitzer, Intermediate Algebra, 5e – Slide #15 Section 2.1

Basic Functions 101

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find the indicated function value:

52 63for 4 zzzQQ

52 46434 Q

102461634 Q

6144484 Q

61924 Q

Replace z with -4

Evaluate the exponents

Multiply

Add

Page 16: § 2.1

Blitzer, Intermediate Algebra, 5e – Slide #16 Section 2.1

Basic Functions 101

Check Point 3a.Check Point 3a.

SOLUTIONSOLUTION

Find the indicated function value:

54for 6 xxff

5646 f

524

29

Replace x with 6

Multiply

Add and Subtract

Page 17: § 2.1

Blitzer, Intermediate Algebra, 5e – Slide #17 Section 2.1

Basic Functions 100

Check Point 3b.Check Point 3b.

SOLUTIONSOLUTION

Find the indicated function value:

103gfor 5 2 xxg

10535 2 g

10253

1075

65

Replace x with -5

Evaluate the exponent

Multiply

Add and Subtract

Page 18: § 2.1

Blitzer, Intermediate Algebra, 5e – Slide #18 Section 2.1

Basic Functions 100

Check Point 3c.Check Point 3c.

SOLUTIONSOLUTION

Find the indicated function value:

27for 4 2 rrrhh

24744 2 h

2)4(716

22816

46

Replace r with -4

Evaluate the exponent

Multiply

Add and Subtract

Page 19: § 2.1

Blitzer, Intermediate Algebra, 5e – Slide #19 Section 2.1

Basic Functions 101

Check Point 3d.Check Point 3d.

SOLUTIONSOLUTION

Find the indicated function value:

96Ffor xxhaF

96 hahaF

966 ha

Replace x with a+h

Multiply

NOTE: THIS CANNOT BE SIMPLIFIED ANY FURTHER!!!

Page 20: § 2.1

DONE

Page 21: § 2.1

Blitzer, Intermediate Algebra, 5e – Slide #21 Section 2.1

Basic Functions 101

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find the indicated function value:

wwwLyxL 63for 2

yxyxyxL 63 2

yxyxyxyxL 63

yxyxyxyxyxL 63 22

Replace w with x+y

Rewrite exponent

Multiply

Add yxyxyxyxL 623 22

yxyxyxyxL 6663 22 Distribute

NOTE: THIS CANNOT BE SIMPLIFIED ANY FURTHER!!!