Skip to ContentGo to accessibility pageKeyboard shortcuts menu
OpenStax Logo
Intermediate Algebra

3.5 Relations and Functions

Intermediate Algebra3.5 Relations and Functions
Search for key terms or text.

Learning Objectives

By the end of this section, you will be able to:
  • Find the domain and range of a relation
  • Determine if a relation is a function
  • Find the value of a function

Be Prepared 3.5

Before you get started, take this readiness quiz.

  1. Evaluate 3x53x5 when x=−2x=−2.
    If you missed this problem, review Example 1.6.
  2. Evaluate 2x2x32x2x3 when x=a.x=a.
    If you missed this problem, review Example 1.6.
  3. Simplify: 7x14x+5.7x14x+5.
    If you missed this problem, review Example 1.7.

Find the Domain and Range of a Relation

As we go about our daily lives, we have many data items or quantities that are paired to our names. Our social security number, student ID number, email address, phone number and our birthday are matched to our name. There is a relationship between our name and each of those items.

When your professor gets her class roster, the names of all the students in the class are listed in one column and then the student ID number is likely to be in the next column. If we think of the correspondence as a set of ordered pairs, where the first element is a student name and the second element is that student’s ID number, we call this a relation.

(Student name, Student ID #)(Student name, Student ID #)

The set of all the names of the students in the class is called the domain of the relation and the set of all student ID numbers paired with these students is the range of the relation.

There are many similar situations where one variable is paired or matched with another. The set of ordered pairs that records this matching is a relation.

Relation

A relation is any set of ordered pairs,(x,y).(x,y). All the x-values in the ordered pairs together make up the domain. All the y-values in the ordered pairs together make up the range.

Example 3.42

For the relation {(1,1),(2,4),(3,9),(4,16),(5,25)}:{(1,1),(2,4),(3,9),(4,16),(5,25)}:

Find the domain of the relation.

Find the range of the relation.

Try It 3.83

For the relation {(1,1),(2,8),(3,27),(4,64),(5,125)}:{(1,1),(2,8),(3,27),(4,64),(5,125)}:

Find the domain of the relation.

Find the range of the relation.

Try It 3.84

For the relation {(1,3),(2,6),(3,9),(4,12),(5,15)}:{(1,3),(2,6),(3,9),(4,12),(5,15)}:

Find the domain of the relation.

Find the range of the relation.

Mapping

A mapping is sometimes used to show a relation. The arrows show the pairing of the elements of the domain with the elements of the range.

Example 3.43

Use the mapping of the relation shown to list the ordered pairs of the relation, find the domain of the relation, and find the range of the relation.

Try It 3.85

Use the mapping of the relation shown to list the ordered pairs of the relation find the domain of the relation find the range of the relation.

Try It 3.86

Use the mapping of the relation shown to list the ordered pairs of the relation find the domain of the relation find the range of the relation.

A graph is yet another way that a relation can be represented. The set of ordered pairs of all the points plotted is the relation. The set of all x-coordinates is the domain of the relation and the set of all y-coordinates is the range. Generally we write the numbers in ascending order for both the domain and range.

Example 3.44

Use the graph of the relation to list the ordered pairs of the relation find the domain of the relation find the range of the relation.

Try It 3.87

Use the graph of the relation to list the ordered pairs of the relation find the domain of the relation find the range of the relation.

Try It 3.88

Use the graph of the relation to list the ordered pairs of the relation find the domain of the relation find the range of the relation.

Determine if a Relation is a Function

A special type of relation, called a function, occurs extensively in mathematics. A function is a relation that assigns to each element in its domain exactly one element in the range. For each ordered pair in the relation, each x-value is matched with only one y-value.

Function

A function is a relation that assigns to each element in its domain exactly one element in the range.

The birthday example from Example 3.43 helps us understand this definition. Every person has a birthday but no one has two birthdays. It is okay for two people to share a birthday. It is okay that Danny and Stephen share July 24th as their birthday and that June and Liz share August 2nd. Since each person has exactly one birthday, the relation in Example 3.43 is a function.

The relation shown by the graph in Example 3.44 includes the ordered pairs (−3,−1)(−3,−1) and (−3,4).(−3,4). Is that okay in a function? No, as this is like one person having two different birthdays.

Example 3.45

Use the set of ordered pairs to (i) determine whether the relation is a function (ii) find the domain of the relation (iii) find the range of the relation.

{(−3,27),(−2,8),(−1,1),(0,0),(1,1),(2,8),(3,27)}{(−3,27),(−2,8),(−1,1),(0,0),(1,1),(2,8),(3,27)}

{(9,−3),(4,−2),(1,−1),(0,0),(1,1),(4,2),(9,3)}{(9,−3),(4,−2),(1,−1),(0,0),(1,1),(4,2),(9,3)}

Try It 3.89

Use the set of ordered pairs to (i) determine whether the relation is a function (ii) find the domain of the relation (iii) find the range of the function.

{(−3,−6),(−2,−4),(−1,−2),(0,0),(1,2),(2,4),(3,6)}{(−3,−6),(−2,−4),(−1,−2),(0,0),(1,2),(2,4),(3,6)}

{(8,−4),(4,−2),(2,−1),(0,0),(2,1),(4,2),(8,4)}{(8,−4),(4,−2),(2,−1),(0,0),(2,1),(4,2),(8,4)}

Try It 3.90

Use the set of ordered pairs to (i) determine whether the relation is a function (ii) find the domain of the relation (iii) find the range of the relation.

{(27,−3),(8,−2),(1,−1),(0,0),(1,1),(8,2),(27,3)}{(27,−3),(8,−2),(1,−1),(0,0),(1,1),(8,2),(27,3)}

{(7,−3),(−5,−4),(8,−0),(0,0),(−6,4),(−2,2),(−1,3)}{(7,−3),(−5,−4),(8,−0),(0,0),(−6,4),(−2,2),(−1,3)}

Example 3.46

Use the mapping to determine whether the relation is a function find the domain of the relation find the range of the relation.

Try It 3.91

Use the mapping to determine whether the relation is a function find the domain of the relation find the range of the relation.

Try It 3.92

Use the mapping to determine whether the relation is a function find the domain of the relation find the range of the relation.

In algebra, more often than not, functions will be represented by an equation. It is easiest to see if the equation is a function when it is solved for y. If each value of x results in only one value of y, then the equation defines a function.

Example 3.47

Determine whether each equation is a function.

2x+y=72x+y=7 y=x2+1y=x2+1 x+y2=3x+y2=3

Try It 3.93

Determine whether each equation is a function.

4x+y=−34x+y=−3 x+y2=1x+y2=1 yx2=2yx2=2

Try It 3.94

Determine whether each equation is a function.

x+y2=4x+y2=4 y=x27y=x27 y=5x4y=5x4

Find the Value of a Function

It is very convenient to name a function and most often we name it f, g, h, F, G, or H. In any function, for each x-value from the domain we get a corresponding y-value in the range. For the function f, we write this range value y as f(x).f(x). This is called function notation and is read f of x or the value of f at x. In this case the parentheses does not indicate multiplication.

Function Notation

For the function y=f(x)y=f(x)

fis the name of the function xis the domain value f(x)is the range valueycorresponding to the valuexfis the name of the function xis the domain value f(x)is the range valueycorresponding to the valuex

We read f(x)f(x) as f of x or the value of f at x.

We call x the independent variable as it can be any value in the domain. We call y the dependent variable as its value depends on x.

Independent and Dependent Variables

For the function y=f(x),y=f(x),

xis the independent variable as it can be any value in the domain ythe dependent variable as its value depends onxxis the independent variable as it can be any value in the domain ythe dependent variable as its value depends onx

Much as when you first encountered the variable x, function notation may be rather unsettling. It seems strange because it is new. You will feel more comfortable with the notation as you use it.

Let’s look at the equation y=4x5.y=4x5. To find the value of y when x=2,x=2, we know to substitute x=2x=2 into the equation and then simplify.

Let x=2.x=2.

The value of the function at x=2x=2 is 3.

We do the same thing using function notation, the equation y=4x5y=4x5 can be written as f(x)=4x5.f(x)=4x5. To find the value when x=2,x=2, we write:

Let x=2.x=2.

The value of the function at x=2x=2 is 3.

This process of finding the value of f(x)f(x) for a given value of x is called evaluating the function.

Example 3.48

For the function f(x)=2x2+3x1,f(x)=2x2+3x1, evaluate the function.

f(3)f(3) f(−2)f(−2) f(a)f(a)

Try It 3.95

For the function f(x)=3x22x+1,f(x)=3x22x+1, evaluate the function.

f(3)f(3) f(−1)f(−1) f(t)f(t)

Try It 3.96

For the function f(x)=2x2+4x3,f(x)=2x2+4x3, evaluate the function.

f(2)f(2) f(−3)f(−3) f(h)f(h)

In the last example, we found f(x)f(x) for a constant value of x. In the next example, we are asked to find g(x)g(x) with values of x that are variables. We still follow the same procedure and substitute the variables in for the x.

Example 3.49

For the function g(x)=3x5,g(x)=3x5, evaluate the function.

g(h2)g(h2) g(x+2)g(x+2) g(x)+g(2)g(x)+g(2)

Try It 3.97

For the function g(x)=4x7,g(x)=4x7, evaluate the function.

g(m2)g(m2) g(x3)g(x3) g(x)g(3)g(x)g(3)

Try It 3.98

For the function h(x)=2x+1,h(x)=2x+1, evaluate the function.

h(k2)h(k2) h(x+1)h(x+1) h(x)+h(1)h(x)+h(1)

Many everyday situations can be modeled using functions.

Example 3.50

The number of unread emails in Sylvia’s account is 75. This number grows by 10 unread emails a day. The function N(t)=75+10tN(t)=75+10t represents the relation between the number of emails, N, and the time, t, measured in days.

Determine the independent and dependent variable.

Find N(5).N(5). Explain what this result means.

Try It 3.99

The number of unread emails in Bryan’s account is 100. This number grows by 15 unread emails a day. The function N(t)=100+15tN(t)=100+15t represents the relation between the number of emails, N, and the time, t, measured in days.

Determine the independent and dependent variable.

Find N(7).N(7). Explain what this result means.

Try It 3.100

The number of unread emails in Anthony’s account is 110. This number grows by 25 unread emails a day. The function N(t)=110+25tN(t)=110+25t represents the relation between the number of emails, N, and the time, t, measured in days.

Determine the independent and dependent variable.

Find N(14).N(14). Explain what this result means.

Media

Access this online resource for additional instruction and practice with relations and functions.

Section 3.5 Exercises

Practice Makes Perfect

Find the Domain and Range of a Relation

In the following exercises, for each relation find the domain of the relation find the range of the relation.

283.

{ ( 1 , 4 ) , ( 2 , 8 ) , ( 3 , 12 ) , ( 4 , 16 ) , ( 5 , 20 ) } { ( 1 , 4 ) , ( 2 , 8 ) , ( 3 , 12 ) , ( 4 , 16 ) , ( 5 , 20 ) }

284.

{ ( 1 , −2 ) , ( 2 , −4 ) , ( 3 , −6 ) , ( 4 , −8 ) , ( 5 , −10 ) } { ( 1 , −2 ) , ( 2 , −4 ) , ( 3 , −6 ) , ( 4 , −8 ) , ( 5 , −10 ) }

285.

{ ( 1 , 7 ) , ( 5 , 3 ) , ( 7 , 9 ) , ( −2 , −3 ) , ( −2 , 8 ) } { ( 1 , 7 ) , ( 5 , 3 ) , ( 7 , 9 ) , ( −2 , −3 ) , ( −2 , 8 ) }

286.

{ ( 11 , 3 ) , ( −2 , −7 ) , ( 4 , −8 ) , ( 4 , 17 ) , ( −6 , 9 ) } { ( 11 , 3 ) , ( −2 , −7 ) , ( 4 , −8 ) , ( 4 , 17 ) , ( −6 , 9 ) }

In the following exercises, use the mapping of the relation to list the ordered pairs of the relation, find the domain of the relation, and find the range of the relation.

287.


288.


289.

For a woman of height 5454 the mapping below shows the corresponding Body Mass Index (BMI). The body mass index is a measurement of body fat based on height and weight. A BMI of 18.524.918.524.9 is considered healthy.

290.

For a man of height 511511 the mapping below shows the corresponding Body Mass Index (BMI). The body mass index is a measurement of body fat based on height and weight. A BMI of 18.524.918.524.9 is considered healthy.

In the following exercises, use the graph of the relation to list the ordered pairs of the relation find the domain of the relation find the range of the relation.

291.


292.


293.


294.


Determine if a Relation is a Function

In the following exercises, use the set of ordered pairs to determine whether the relation is a function, find the domain of the relation, and find the range of the relation.

295.

{ ( −3 , 9 ) , ( −2 , 4 ) , ( −1 , 1 ) , { ( −3 , 9 ) , ( −2 , 4 ) , ( −1 , 1 ) ,
( 0 , 0 ) , ( 1 , 1 ) , ( 2 , 4 ) , ( 3 , 9 ) } ( 0 , 0 ) , ( 1 , 1 ) , ( 2 , 4 ) , ( 3 , 9 ) }

296.

{ ( 9 , −3 ) , ( 4 , −2 ) , ( 1 , −1 ) , { ( 9 , −3 ) , ( 4 , −2 ) , ( 1 , −1 ) ,
( 0 , 0 ) , ( 1 , 1 ) , ( 4 , 2 ) , ( 9 , 3 ) } ( 0 , 0 ) , ( 1 , 1 ) , ( 4 , 2 ) , ( 9 , 3 ) }

297.

{ ( −3 , 27 ) , ( −2 , 8 ) , ( −1 , 1 ) , { ( −3 , 27 ) , ( −2 , 8 ) , ( −1 , 1 ) ,
( 0 , 0 ) , ( 1 , 1 ) , ( 2 , 8 ) , ( 3 , 27 ) } ( 0 , 0 ) , ( 1 , 1 ) , ( 2 , 8 ) , ( 3 , 27 ) }

298.

{ ( −3 , −27 ) , ( −2 , −8 ) , ( −1 , −1 ) , { ( −3 , −27 ) , ( −2 , −8 ) , ( −1 , −1 ) ,
( 0 , 0 ) , ( 1 , 1 ) , ( 2 , 8 ) , ( 3 , 27 ) } ( 0 , 0 ) , ( 1 , 1 ) , ( 2 , 8 ) , ( 3 , 27 ) }

In the following exercises, use the mapping to determine whether the relation is a function, find the domain of the function, and find the range of the function.

299.


300.


301.


302.


In the following exercises, determine whether each equation is a function.

303.


2x+y=−32x+y=−3
y=x2y=x2
x+y2=−5x+y2=−5

304.


y=3x5y=3x5
y=x3y=x3
2x+y2=42x+y2=4

305.


y3x3=2y3x3=2
x+y2=3x+y2=3
3x2y=63x2y=6

306.


2x4y=82x4y=8
−4=x2y−4=x2y
y2=x+5y2=x+5

Find the Value of a Function

In the following exercises, evaluate the function: f(2)f(2) f(−1)f(−1) f(a).f(a).

307.

f ( x ) = 5 x 3 f ( x ) = 5 x 3

308.

f ( x ) = 3 x + 4 f ( x ) = 3 x + 4

309.

f ( x ) = −4 x + 2 f ( x ) = −4 x + 2

310.

f ( x ) = −6 x 3 f ( x ) = −6 x 3

311.

f ( x ) = x 2 x + 3 f ( x ) = x 2 x + 3

312.

f ( x ) = x 2 + x 2 f ( x ) = x 2 + x 2

313.

f ( x ) = 2 x 2 x + 3 f ( x ) = 2 x 2 x + 3

314.

f ( x ) = 3 x 2 + x 2 f ( x ) = 3 x 2 + x 2

In the following exercises, evaluate the function: g(h2)g(h2) g(x+2)g(x+2) g(x)+g(2).g(x)+g(2).

315.

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

316.

g ( x ) = 5 x 8 g ( x ) = 5 x 8

317.

g ( x ) = −3 x 2 g ( x ) = −3 x 2

318.

g ( x ) = −8 x + 2 g ( x ) = −8 x + 2

319.

g ( x ) = 3 x g ( x ) = 3 x

320.

g ( x ) = 7 5 x g ( x ) = 7 5 x

In the following exercises, evaluate the function.

321.

f(x)=3x25x;f(x)=3x25x; f(2)f(2)

322.

g(x)=4x23x;g(x)=4x23x; g(3)g(3)

323.

F ( x ) = 2 x 2 3 x + 1 ; F ( x ) = 2 x 2 3 x + 1 ;
F ( −1 ) F ( −1 )

324.

G ( x ) = 3 x 2 5 x + 2 ; G ( x ) = 3 x 2 5 x + 2 ;
G ( −2 ) G ( −2 )

325.

h(t)=2|t5|+4;h(t)=2|t5|+4; f(−4)f(−4)

326.

h(y)=3|y1|3;h(y)=3|y1|3; h(−4)h(−4)

327.

f(x)=x+2x1;f(x)=x+2x1; f(2)f(2)

328.

g(x)=x2x+2;g(x)=x2x+2; g(4)g(4)

In the following exercises, solve.

329.

The number of unwatched shows in Sylvia’s DVR is 85. This number grows by 20 unwatched shows per week. The function N(t)=85+20tN(t)=85+20t represents the relation between the number of unwatched shows, N, and the time, t, measured in weeks.

Determine the independent and dependent variable.

Find N(4).N(4). Explain what this result means

330.

Every day a new puzzle is downloaded into Ken’s account. Right now he has 43 puzzles in his account. The function N(t)=43+tN(t)=43+t represents the relation between the number of puzzles, N, and the time, t, measured in days.

Determine the independent and dependent variable.

Find N(30).N(30). Explain what this result means.

331.

The daily cost to the printing company to print a book is modeled by the function C(x)=3.25x+1500C(x)=3.25x+1500 where C is the total daily cost and x is the number of books printed.

Determine the independent and dependent variable.

Find N(0).N(0). Explain what this result means.

Find N(1000).N(1000). Explain what this result means.

332.

The daily cost to the manufacturing company is modeled by the function C(x)=7.25x+2500C(x)=7.25x+2500 where C(x)C(x) is the total daily cost and x is the number of items manufactured.

Determine the independent and dependent variable.

Find C(0).C(0). Explain what this result means.

Find C(1000).C(1000). Explain what this result means.

Writing Exercises

333.

In your own words, explain the difference between a relation and a function.

334.

In your own words, explain what is meant by domain and range.

335.

Is every relation a function? Is every function a relation?

336.

How do you find the value of a function?

Self Check

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Citation/Attribution

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:

    Access for free at https://openstax.org/books/intermediate-algebra/pages/1-introduction

  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:

    Access for free at https://openstax.org/books/intermediate-algebra/pages/1-introduction

Citation information

© Feb 9, 2022 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.