Chapter 3 Section 6 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.
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Transcript of Chapter 3 Section 6 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.
Chapter Chapter 33Section Section 66
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Introduction to Functions
11
44
33
22
66
55
3.63.63.63.6Understand the definition of a relation.Understand the definition of a function.Decide whether an equation defines a function.Find domains and ranges.Use function notation. Apply the function concept in an application.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Objective 11
Understand the definition of a relation.
Slide 3.6 - 3
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Understand the definition of a relation.
In an ordered pair (x, y), x and y are called the components of the ordered pair.
Slide 3.6 - 4
Any set of ordered pairs is called a relation.
The set of all first components of the ordered pairs of a relation is the domain of the relation, and the set of all second components of the ordered pairs is the range of the relation.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 1
Solution:
Identifying Domains and Ranges of Relations Defined by Ordered Pairs
Use ordered pairs to define the relation.
4, , 6, , 7, , 3,A B B C
Slide 3.6 - 5
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Objective 22
Understand the definition of a function.
Slide 3.6 - 6
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Understand the definition of a function.A very important type of relation called a function.
A function is a set of ordered pairs in which each first component corresponds to exactly one second component.
Slide 3.6 - 7
By definition, the relation in the following order pairs is not a function, because the same first component, 3, corresponds to more then one second component.
5 63 3 3, , , ,7 3,8If the ordered pairs from this example were interchanged, giving the
relation
the result would be a function. In that case, each domain (first component) corresponds to exactly one range element (second component).
5 6 7, , , ,3 3 3 8,3 ,
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 2
Solution: function
Determining Whether Relations Are Functions
Determine whether each relation is a function.
2,8 , 1,1 , 0,0 , 1,1 , 2,8 ,
Slide 3.6 - 8
5,2 , 5,1 , 5,0
Solution: not a function
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Objective 33
Decide whether an equation defines a function.
Slide 3.6 - 9
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Decide whether an equation defines a function.
Slide 3.6 - 10
Given the graph of an equation, the definition of a function can be used to decide whether or not the graph represents a function. By the definition of a function , each x-value must lead to exactly one y-value.
The way to determine if a graph is a function is the vertical line test. If a vertical line intersects a graph in more than one point, then the graph is not the graph of a function.
Any nonvertical line is the graph of a function. For this reason, any linear equation of the form y = mx + b defines a function. (Recall that a vertical line has an undefined slope.)
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 3Deciding Whether Relations Define Functions
Determine whether each relation is a function.
Slide 3.6 - 11
Solution: functionSolution: not a function
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Objective 44
Find domains and ranges.
Slide 3.6 - 12
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
By the definitions of domain and range given for
relations, the set of all numbers that can be used as
replacements for x in a function is the domain of the
function. The set of all possible values of y is the range
of the function.
Slide 3.6 - 13
Find domains and ranges.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 4
Find the domain and range of the function y = x2 + 4.
Solution:
Domain:
Finding the Domain and Range of Functions
,
Slide 3.6 - 14
Range: 4,
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Objective 55
Use function notation.
Slide 3.6 - 15
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Use function notation.
The letters f, g, and h are commonly used to name functions. For example, the function y = 3x + 5 may be written
where f (x) is read “f of x.” The notation f (x) is another way of writing y in a function. For the function defined by f (x) = 3x + 5, if x = 7, then
Read this result, f (7) = 26, as “f of 7 equals 26.” The notation f (7) means the values of y when x is 7. The statement f (7) = 26 says that the value of y = 26 when x is 7. It also indicates that the point (7,26) lies on the graph of f.
Slide 3.6 - 16
3 5,f x x
7 73 5f
21 5 26
The notation f(x) does not mean f times x; f(x) means the value of x for the function f. It represents the y –value that corresponds to x.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Function NotationIn the notation f(x),
f is the name of the function,
x is the domain value,
and f(x) is the range value y for the domain value x.
Slide 3.6 - 17
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 5
Solution:
Using Function Notation
For the function f (x) = 6x − 2, find f (−1).
6 21 1f
6 21f
1 8f
Slide 3.6 - 18
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Objective 66
Apply the function concept in an application.
Slide 3.6 - 19
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 6Applying the Function Concept to Population
The number of U.S. students ages 3 – 21served by educational programs for students with disabilities for selected years are given in the table.
Slide 3.6 - 20
a) Write a set of ordered pairs that defines a function f for these data.
b) Give the domain and range of f.
c) Find f (1998).
d) In which year did the number of students equal 5.7 million? That is, for what value of x does f (x) = 5.7million?
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 6Applying the Function Concept to Population (cont’d)
Solution:
The number of U.S. students ages 3 – 21served by educational programs for students with disabilities for selected years are given in the table.
1994 1996 1998
2
{ , ,
000
4.9 5.2 5.5
5. 2
, ,
0
, ,
027, }9, ,5.
f
1994 1996
1998
domain : { , ,
, , 2000 2002}
5.5 million
2000Slide 3.6 - 21
4.9 5.range : { , , 2 5.5 5., 7,5.9}
a)
b)
c)
d)