Intermediate Algebra (11th edition)

(d)

By definition, yis a function of xif every value of xleads to exactly one value

of y. Here, a particular value of x, say, 1, corresponds to many values of y. The or-

dered pairs

and so on

all satisfy the inequality. Thus, this relation does not define a function. Any number

can be used for x, so the domain is the set of all real numbers,

(e)

Given any value of xin the domain, we find yby subtracting 1 and then dividing

the result into 5. This process produces exactly one value of yfor each value in the

domain, so the given equation defines a function.

The domain includes all real numbers except those which make the denominator 0.

We find these numbers by setting the denominator equal to 0 and solving for x.

Add 1.

The domain includes all real numbers except1, written 1 - q, 1 2 ́ 1 1, q 2.

x= 1

x- 1 = 0

y=

5

x- 1

1 - q, q 2.

1 1, 0 2 , 1 1, - 12 , 1 1, - 22 , 1 1, - 32 ,

y...x- 1

SECTION 3.5 Introduction to Relations and Functions 187

Concept Check Express each relation using a different form. There is more than one cor-

rect way to do this. See Objective 2.

5. 51 0, 2 2 , 1 2, 4 2 , 1 4, 6 26 6. 7.

In summary, we give three variations of the definition of a function.

Variations of the Definition of a Function

1. A functionis a relation in which, for each value of the first component of the

ordered pairs, there is exactly one value of the second component.

2. A functionis a set of distinct ordered pairs in which no first component is

repeated.

3. A functionis a correspondence or rule that assigns exactly one range value

to each domain value.

Complete solution available
on the Video Resources on DVD

3.5 EXERCISES

1.In your own words, define a function and give an example.

2.In your own words, define the domain of a function and give an example.

3.Concept Check In an ordered pair of a relation, is the first element the independent or

the dependent variable?

4.Concept Check Give an example of a relation that is not a function and that has domain

5 - 3, 2, 6 6 and range 5 4, 6 6. (There are many possible correct answers.)

NOW TRY
EXERCISE 5
Decide whether each relation
defines yas a function of x,
and give the domain.

(a)

(b)

(c)

(d)y 63 x+ 1

y=

1

x- 2

y= 22 x- 4

y= 4 x- 3

NOW TRY ANSWERS

5. (a)yes;
   (b)yes;
   (c)yes;
   (d)no; 1 - q, q 2

1 - q, 2 2 ́ 1 2, q 2

3 2, q 2

1 - q, q 2

–3

2

–4

1

0

xy

13

01

11

33

• 
  


• 
  
  
  
  • 
    
  
  
  
  

NOW TRY

8.Concept Check Does the relation given in Exercise 7define a function? Why or why not?