Calculus Homework

i

Precalculus

senior contributing author Jay abramson, arizona state University

i i

Reviewers Nina Alketa, Cecil College Kiran Bhutani, Catholic University of America Brandie Biddy, Cecil College Lisa Blank, Lyme Central School Bryan Blount, Kentucky Wesleyan College Jessica Bolz, The Bryn Mawr School Sheri Boyd, Rollins College Sarah Brewer, Alabama School of Math and Science Charles Buckley, St. Gregory's University Michael Cohen, Hofstra University Kenneth Crane, Texarkana College Rachel Cywinski, Alamo Colleges Nathan Czuba Srabasti Dutta, Ashford University Kristy Erickson, Cecil College Nicole Fernandez, Georgetown University / Kent State University David French, Tidewater Community College Douglas Furman, SUNY Ulster Lance Hemlow, Raritan Valley Community College Erinn Izzo, Nicaragua Christian Academy John Jaffe Jerry Jared, Blue Ridge School Stan Kopec, Mount Wachusett Community College Kathy Kovacs Cynthia Landrigan, Erie Community College Sara Lenhart, Christopher Newport University Wendy Lightheart, Lane Community College Joanne Manville, Bunker Hill Community College Karla McCavit, Albion College Cynthia McGinnis, Northwest Florida State College Lana Neal, University of Texas at Austin Rhonda Porter, Albany State University Steven Purtee, Valencia College William Radulovich, Florida State College Jacksonville Alice Ramos, Bethel College Nick Reynolds, Montgomery Community College Amanda Ross, A. A. Ross Consulting and Research, LLC Erica Rutter, Arizona State University Sutandra Sarkar, Georgia State University Willy Schild, Wentworth Institute of Technology Todd Stephen, Cleveland State University Scott Sykes, University of West Georgia Linda Tansil, Southeast Missouri State University John Thomas, College of Lake County Diane Valade, Piedmont Virginia Community College

Allen Wolmer, Atlanta Jewish Academy

About Our Team Senior Contributing Author

Jay Abramson has been teaching Precalculus for 33 years, the last 14 at Arizona State University, where he is a principal lecturer in the School of Mathematics and Statistics. His accomplishments at ASU include co-developing the university’s first hybrid and online math courses as well as an extensive library of video lectures and tutorials. In addition, he has served as a contributing author for two of Pearson Education’s math programs, NovaNet Precalculus and Trigonometry. Prior to coming to ASU, Jay taught at Texas State Technical College and Amarillo College. He received Teacher of the Year awards at both institutions.

Contributing Authors Valeree Falduto, Palm Beach State College Rachael Gross, Towson University David Lippman, Pierce College Melonie Rasmussen, Pierce College Rick Norwood, East Tennessee State University Nicholas Belloit, Florida State College Jacksonville Jean-Marie Magnier, Springfield Technical Community College Harold Whipple Christina Fernandez

OpenStax Rice University 6100 Main Street MS-375 Houston, Texas 77005

To learn more about OpenStax, visit https://openstax.org. Individual print copies and bulk orders can be purchased through our website.

© 2017 Rice University. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution 4.0 International License. Under this license, any user of this textbook or the textbook contents herein must provide proper attribution as follows:

- If you redistribute this textbook in a digital format (including but not limited to EPUB, PDF, and HTML), then you must retain on every page the following attribution: “Download for free at https://openstax.org/details/books/precal- culus.”

- If you redistribute this textbook in a print format, then you must include on every physical page the following attribution: “Download for free at https:// openstax.org/details/books/precalculus.”

- If you redistribute part of this textbook, then you must retain in every digital format page view (including but not limited to EPUB, PDF, and HTML) and on every physical printed page the following attribution: “Download for free at https://openstax.org/details/books/precalculus.”

- If you use this textbook as a bibliographic reference, please include https:// openstax.org/details/books/precalculus.

For questions regarding this licensing, please contact [email protected].

The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, OpenStax CNX logo, OpenStax Tutor name, Openstax Tutor logo, Con- nexions name, Connexions logo, Rice University name, and Rice University logo are not subject to the license and may not be reproduced without the prior and express written consent of Rice University.

PRINT ISBN-10 1-938168-34-8 PRINT ISBN-13 978-1-938168-34-5 PDF ISBN-10 1-947172-06-9 PDF ISBN-13 978-1-947172-06-7 Revision PR-2015-002(03/17)-BW

Original Publication Year: 2014

i i i

OpenStax OpenStax provides free, peer-reviewed, openly licensed textbooks for introductory college and Advanced Placement® courses and low-cost, personalized courseware that helps students learn. A nonprofit ed tech initiative based at Rice University, we’re committed to helping students access the tools they need to complete their courses and meet their educational goals.

Rice University OpenStax and OpenStax CNX are initiatives of Rice University. As a leading research university with a distinctive commitment to undergraduate education, Rice University aspires to path-breaking research, unsurpassed teaching, and contributions to the betterment of our world. It seeks to fulfill this mission by cultivating a diverse community of learning and discovery that produces leaders across the spectrum of human endeavor.

Foundation Support OpenStax is grateful for the tremendous support of our sponsors. Without their strong engagement, the goal of free access to high-quality textbooks would remain just a dream.

Laura and John Arnold Foundation (LJAF) actively seeks opportunities to invest in organizations and thought leaders that have a sincere interest in implementing fundamental changes that not only yield immediate gains, but also repair broken systems for future generations. LJAF currently focuses its strategic investments on education, criminal justice, research integrity, and public accountability.

The William and Flora Hewlett Foundation has been making grants since 1967 to help solve social and environmental problems at home and around the world. The Foundation concentrates its resources on activities in education, the environment, global development and population, performing arts, and philanthropy, and makes grants to support disadvantaged communities in the San Francisco Bay Area.

Guided by the belief that every life has equal value, the Bill & Melinda Gates Foundation works to help all people lead healthy, productive lives. In developing countries, it focuses on improving people’s health with vaccines and other life-saving tools and giving them the chance to lift themselves out of hunger and extreme poverty. In the United States, it seeks to significantly improve education so that all young people have the opportunity to reach their full potential. Based in Seattle, Washington, the foundation is led by CEO Jeff Raikes and Co-chair William H. Gates Sr., under the direction of Bill and Melinda Gates and Warren Buffett.

The Maxfield Foundation supports projects with potential for high impact in science, education, sustainability, and other areas of social importance.

Our mission at The Michelson 20MM Foundation is to grow access and success by eliminating unnecessary hurdles to affordabilitay. We support the creation, sharing, and proliferation of more effective, more affordable educational content by leveraging disruptive technologies, open educational resources, and new models for collaboration between for-profit, nonprofit, and public entities.

Calvin K. Kazanjian was the founder and president of Peter Paul (Almond Joy), Inc. He firmly believed that the more people understood about basic economics the happier and more prosperous they would be. Accordingly, he established the Calvin K. Kazanjian Economics Foundation Inc, in 1949 as a philanthropic, nonpolitical educational organization to support efforts that enhanced economic understanding.

The Bill and Stephanie Sick Fund supports innovative projects in the areas of Education, Art, Science and Engineering.

THE MICHELSON 20MM FOUNDATION

Access. The future of education.

OpenStax.org

I like free textbooks and I cannot lie.

Give $5 or more to OpenStax and we’ll send you a sticker! OpenStax is a nonprofit initiative, which means that every dollar you give helps us maintain and grow our library of free textbooks.

If you have a few dollars to spare, visit OpenStax.org/give to donate. We’ll send you an OpenStax sticker to thank you for your support!

v

Brief ContentsBrief Contents

1 Functions 1 2 Linear Functions 125 3 Polynomial and Rational Functions 197 4 Exponential and Logarithmic Functions 327 5 Trigonometric Functions 439 6 Periodic Functions 505 7 Trigonometric Identities and Equation 559 8 Further Applications of Trigonometry 643 9 Systems of Equations and Inequalities 757 10 Analytic Geometry 863 11 Sequences, Probability and Counting Theory 937 12 Introduction to Calculus 1017

v i i

Contents Preface xi

1 Functions 1 1.1 Functions and Function Notation 2

1.2 Domain and Range 22

1.3 Rates of Change and Behavior of Graphs 38

1.4 Composition of Functions 51

1.5 Transformation of Functions 64

1.6 Absolute Value Functions 89

1.7 Inverse Functions 100

Chapter 1 Review 113

Chapter 1 Review Exercises 118

Chapter 1 Practice Test 123

2 Linear Functions 125 2.1 Linear Functions 126

2.2 Graphs of Linear Functions 143

2.3 Modeling with Linear Functions 162

2.4 Fitting Linear Models to Data 175

Chapter 2 Review 187

Chapter 2 Review Exercises 190

Chapter 2 Practice Test 194

3 Polynomial and Rational Functions 197 3.1 Complex Numbers 198

3.2 Quadratic Functions 208

3.3 Power Functions and Polynomial Functions 224

3.4 Graphs of Polynomial Functions 239

3.5 Dividing Polynomials 257

3.6 Zeros of Polynomial Functions 266

3.7 Rational Functions 278

3.8 Inverses and Radical Functions 299

3.9 Modeling Using Variation 310

Chapter 3 Review 317

Chapter 3 Review Exercises 322

Chapter 3 Practice Test 325

v i i i

4 Exponential and Logarithmic Functions 327 4.1 Exponential Functions 328

4.2 Graphs of Exponential Functions 343

4.3 Logarithmic Functions 355

4.4 Graphs of Logarithmic Functions 363

4.5 Logarithmic Properties 380

4.6 Exponential and Logarithmic Equations 390

4.7 Exponential and Logarithmic Models 401

4.8 Fitting Exponential Models to Data 416

Chapter 4 Review 429

Chapter 4 Review Exercises 434

Chapter 4 Practice Test 437

5 Trigonometric Functions 439 5.1 Angles 440

5.2 Unit Circle: Sine and Cosine Functions 457

5.3 The Other Trigonometric Functions 473

5.4 Right Triangle Trigonometry 486

Chapter 5 Review 498

Chapter 5 Review Exercises 502

Chapter 5 Practice Test 504

6 Periodic Functions 505 6.1 Graphs of the Sine and Cosine Functions 506

6.2 Graphs of the Other Trigonometric Functions 523

6.3 Inverse Trigonometric Functions 541

Chapter 6 Review 552

Chapter 6 Review Exercises 554

Chapter 6 Practice Test 556

7 Trigonometric Identities and Equations 559 7.1 Solving Trigonometric Equations with Identities 560

7.2 Sum and Difference Identities 570

7.3 Double-Angle, Half-Angle, and Reduction Formulas 584

7.4 Sum-to-Product and Product-to-Sum Formulas 596

7.5 Solving Trigonometric Equations 603

7.6 Modeling with Trigonometric Equations 617

Chapter 7 Review 634

Chapter 7 Review Exercises 639

Chapter 7 Practice Test 642

i x

8 Further Applications of Trigonometry 643 8.1 Non-right Triangles: Law of Sines 644

8.2 Non-right Triangles: Law of Cosines 658

8.3 Polar Coordinates 670

8.4 Polar Coordinates: Graphs 681

8.5 Polar Form of Complex Numbers 697

8.6 Parametric Equations 708

8.7 Parametric Equations: Graphs 719

8.8 Vectors 729

Chapter 8 Review 747

Chapter 8 Review Exercises 752

Chapter 8 Practice Test 755

9 Systems of Equations and Inequalities 757 9.1 Systems of Linear Equations: Two Variables 758

9.2 Systems of Linear Equations: Three Variables 774

9.3 Systems of Nonlinear Equations and Inequalities: Two Variables 785

9.4 Partial Fractions 795

9.5 Matrices and Matrix Operations 805

9.6 Solving Systems with Gaussian Elimination 816

9.7 Solving Systems with Inverses 829

9.8 Solving Systems with Cramer's Rule 843

Chapter 9 Review 854

Chapter 9 Review Exercises 858

Chapter 9 Practice Test 861

10 Analytic Geometry 863 10.1 The Ellipse 864

10.2 The Hyperbola 879

10.3 The Parabola 896

10.4 Rotation of Axis 909

10.5 Conic Sections in Polar Coordinates 922

Chapter 10 Review 931

Chapter 10 Review Exercises 934

Chapter 10 Practice Test 936

x

11 Sequences, Probability and Counting Theory 937 11.1 Sequences and Their Notations 938

11.2 Arithmetic Sequences 951

11.3 Geometric Sequences 961

11.4 Series and Their Notations 969

11.5 Counting Principles 982

11.6 Binomial Theorem 992

11.7 Probability 999

Chapter 11 Review 1008

Chapter 11 Review Exercises 1012

Chapter 11 Practice Test 1015

12 Introduction to Calculus 1017 12.1 Finding Limits: Numerical and Graphical Approaches 1018

12.2 Finding Limits: Properties of Limits 1028

12.3 Continuity 1037

12.4 Derivatives 1051

Chapter 12 Review 1070

Chapter 12 Review Exercises 1073

Chapter 12 Practice Test 1075

Appendix A-1

Try It Answer Section B-1

Odd Answer Section C-1

Index D-1

x i

Preface Welcome to Precalculus, an OpenStax resource. This textbook was w ritten to i ncrease student access to high-quality learning materials, maintaining highest standards of academic rigor at little to no cost.

About OpenStax OpenStax is a nonprofit based at Rice University, and it’s our mission to improve student access to education. Our first openly licensed college textbook was published in 2012, and our library has since scaled to over 25 books for college and AP® courses used by hundreds of thousands of students. Our adaptive learning technology, designed to improve learning outcomes through personalized educational paths, is being piloted in college courses throughout the country. Through our partnerships with philanthropic foundations and our alliance with other educational resource organizations, OpenStax is breaking down the most common barriers to learning and empowering students and instructors to succeed.

About OpenStax’s Resources Customization OpenStax is licensed under a Creative Commons Attribution 4.0 International (CC BY) license, which means that you can distribute, remix, and build upon the content, as long as you provide attribution to OpenStax and its content contributors.

Because our books are openly licensed, you are free to use the entire book or pick and choose the sections that are most relevant to the needs of your course. Feel free to remix the content by assigning your students certain chapters and sections in your syllabus, in the order that you prefer. You can even provide a direct link in your syllabus to the sections in the web view of your book.

Faculty also have the option of creating a customized version of their OpenStax book through the aerSelect platform. The custom version can be made available to students in low-cost print or digital form through their campus bookstore. Visit your book page on openstax.org for a link to your book on aerSelect.

Errata

All OpenStax textbooks undergo a rigorous review process. However, like any professional-grade textbook, errors sometimes occur. Since our books are web based, we can make updates periodically when deemed pedagogically necessary. If you have a correction to suggest, submit it through the link on your book page on openstax.org. Subject matter experts review all errata suggestions. OpenStax is committed to remaining transparent about all updates, so you will also find a list of past errata changes on your book page on openstax.org.

Format You can access this textbook for free in web view or PDF through openstax.org, and for a low cost in print.

x i i

About Precalculus Precalculus is adaptable and designed to fit the needs of a variety of precalculus courses. It is a comprehensive text that covers more ground than a typical one- or two-semester college-level precalculus course. The content is organized by clearly-defined learning objectives, and includes worked examples that demonstrate problem-solving approaches in an accessible way. Coverage and Scope

Precalculus contains twelve chapters, roughly divided into three groups.

Chapters 1-4 discuss various types of functions, providing a foundation for the remainder of the course. Chapter 1: Functions Chapter 2: Linear Functions Chapter 3: Polynomial and Rational Functions Chapter 4: Exponential and Logarithmic Functions

Chapters 5-8 focus on Trigonometry. In Precalculus, we approach trigonometry by first introducing angles and the unit circle, as opposed to the right triangle approach more commonly used in College Algebra and Trigonometry courses.

Chapter 5: Trigonometric Functions Chapter 6: Periodic Functions Chapter 7: Trigonometric Identities and Equations Chapter 8: Further Applications of Trigonometry

Chapters 9-12 present some advanced Precalculus topics that build on topics introduced in chapters 1-8. Most Precalculus syllabi include some of the topics in these chapters, but few include all. Instructors can select material as needed from this group of chapters, since they are not cumulative.

Chapter 9: Systems of Equations and Inequalities Chapter 10: Analytic Geometry Chapter 11: Sequences, Probability and Counting Theory Chapter 12: Introduction to Calculus

All chapters are broken down into multiple sections, the titles of which can be viewed in the Table of Contents.

Development Overview Precalculus is the product of a collaborative effort by a group of dedicated authors, editors, and instructors whose collective passion for this project has resulted in a text that is remarkably unified in purpose and voice. Special thanks is due to our Lead Author, Jay Abramson of Arizona State University, who provided the overall vision for the book and oversaw the development of each and every chapter, drawing up the initial blueprint, reading numerous drafts, and assimilating field reviews into actionable revision plans for our authors and editors.

The first eight chapters are built on the foundation of Precalculus: An Investigation of Functions by David Lippman and Melonie Rasmussen. Chapters 9-12 were written and developed from by our expert and highly experienced author team. All twelve chapters follow a new and innovative instructional design, and great care has been taken to maintain a consistent voice from cover to cover. New features have been introduced to flesh out the instruction, all of the graphics have been redone in a more contemporary style, and much of the content has been revised, replaced, or supplemented to bring the text more in line with mainstream approaches to teaching precalculus.

Accuracy of the Content We have taken great pains to ensure the validity and accuracy of this text. Each chapter’s manuscript underwent at least two rounds of review and revision by a panel of active precalculus instructors. Then, prior to publication, a separate team of experts checked all text, examples, and graphics for mathematical accuracy; multiple reviewers were assigned to each chapter to minimize the chances of any error escaping notice. A third team of experts was responsible for the accuracy of the Answer Key, dutifully reworking every solution to eradicate any lingering errors. Finally, the editorial team conducted a multi-round post-production review to ensure the integrity of the content in its final form was written and developed after the Student Edition, has also been rigorously checked for accuracy following a process similar to that described above. Incidentally, the act of writing out solutions step-by-step served as yet another round of validation for the Answer Key in the back of the Student Edition.

x i i i

Pedagogical Foundations and Features Learning Objectives Each chapter is divided into multiple sections (or modules), each of which is organized around a set of learning objectives. The learning objectives are listed explicitly at the beginning of each section and are the focal point of every instructional element.

Narrative Text Narrative text is used to introduce key concepts, terms, and definitions, to provide real-world context, and to provide transitions between topics and examples. Throughout this book, we rely on a few basic conventions to highlight the most important ideas:

• Key terms are boldfaced, typically when first introduced and/or when formally defined. • Key concepts and definitions are called out in a blue box for easy reference.

Example Each learning objective is supported by one or more worked examples that demonstrate the problem-solving approaches that students must master. Typically, we include multiple Examples for each learning objective in order to model different approaches to the same type of problem, or to introduce similar problems of increasing complexity. All told, there are more than 650 Examples, or an average of about 55 per chapter.

All Examples follow a simple two- or three-part format. First, we pose a problem or question. Next, we demonstrate the Solution, spelling out the steps along the way. Finally (for select Examples), we conclude with an Analysis reflecting on the broader implications of the Solution just shown.

Figures Precalculus contains more than 2000 figures and illustrations, the vast majority of which are graphs and diagrams. Art throughout the text adheres to a clear, understated style, drawing the eye to the most important information in each figure while minimizing visual distractions. Color contrast is employed with discretion to distinguish between the different functions or features of a graph.

y

x

(h, k − a)

(h, k − c)

(h, k) (h + b, k)

(h, k + a)

(h, k + c)

(h − b, k)

Major Axis

Minor Axis

(b)

–6 –4 2 4 6

–4

–2

2

4

6

8

x

y y = x2 + 4x + 3

Vertex Axis of symmetry

x-intercepts –6 –4 –2 2 4 6

–6

–4

–2

2

4

6

x

y

(1, 6)

(0, 1) (6, 1)

(1, 0)

f (x) = 5x3 + 1

y = x

Supporting Features Four small but important features, each marked by a distinctive icon, serve to support Examples.

A “How To” is a list of steps necessary to solve a certain type of problem. A How To typically precedes an Example that proceeds to demonstrate the steps in action.

A “Try It” exercise immediately follows an Example or a set of related Examples, providing the student with an immediate opportunity to solve a similar problem. In the Online version of the text, students can click an Answer link directly below the question to check their understanding. In other versions, answers to the Try-It exercises are located in the Answer Key.

A “Q & A...” may appear at any point in the narrative, but most often follows an Example. This feature pre-empts misconceptions by posing a commonly asked yes/no question, followed by a detailed answer and explanation.

The “Media” links appear at the conclusion of each section, just prior to the Section Exercises. These are a list of links to online video tutorials that reinforce the concepts and skills introduced in the section

While we have selected tutorials that closely align to our learning objectives, we did not produce these tutorials, nor were they specifically produced or tailored to accompany Precalculus.

x i v

Section Exercises Each section of every chapter concludes with a well-rounded set of exercises that can be assigned as homework or used selectively for guided practice. With over 5,900 exercises across the 12 chapters, instructors should have plenty to choose from[i].

Section Exercises are organized by question type, and generally appear in the following order: Verbal questions assess conceptual understanding of key terms and concepts. Algebraic problems require students to apply algebraic manipulations demonstrated in the section. Graphical problems assess students’ ability to interpret or produce a graph. Numeric problems require the student perform calculations or computations. Technology problems encourage exploration through use of a graphing utility, either to visualize or verify algebraic results or to solve problems via an alternative to the methods demonstrated in the section. Extensions pose problems more challenging than the Examples demonstrated in the section. They require students to synthesize multiple learning objectives or apply critical thinking to solve complex problems. Real-World Applications present realistic problem scenarios from fields such as physics, geology, biology, finance, and the social sciences.

Chapter Review Features Each chapter concludes with a review of the most important takeaways, as well as additional practice problems that students can use to prepare for exams.x

Key Terms provides a formal definition for each bold-faced term in the chapter. Key Equations presents a compilation of formulas, theorems, and standard-form equations. Key Concepts summarizes the most important ideas introduced in each section, linking back to the relevant Example(s) in case students need to review. Chapter Review Exercises include 40-80 practice problems that recall the most important concepts from each section. Practice Test includes 25-50 problems assessing the most important learning objectives from the chapter. Note that the practice test is not organized by section, and may be more heavily weighted toward cumulative objectives as opposed to the foundational objectives covered in the opening sections.

Additional Resources

Student and Instructor Resources We’ve compiled additional resources for both students and instructors, including Getting Started Guides, instructor solution manual, and PowerPoint slides. Instructor resources require a verified instructor account, which can be requested on your openstax.org log-in. Take advantage of these resources to supplement your OpenStax book.

Partner Resources OpenStax Partners are our allies in the mission to make high-quality learning materials affordable and accessible to students and instructors everywhere. Their tools integrate seamlessly with our OpenStax titles at a low cost. To access the partner resources for your text, visit your book page on openstax.org.

Online Homework XYZ Homework is built using the fastest-growing mathematics cloud platform. XYZ Homework gives instructors access to the Precalculus aligned problems, organized in the Precalculus Course Template. Instructors have access to thousands of additional algorithmically-generated questions for unparalleled course customization. For one low annual price, students can take multiple classes through XYZ Homework. Learn more at www.xyzhomework.com/openstax.

WebAssign is an independent online homework and assessment solution first launched at North Carolina State University in 1997. Today, WebAssign is an employee-owned benefit corporation and participates in the education of over a million students each year. WebAssign empowers faculty to deliver fully customizable assignments and high quality content to their students in an interactive online environment. WebAssign supports Precalculus with hundreds of problems covering every concept in the course, each containing algorithmically-generated values and links directly to the eBook providing a completely integrated online learning experience.

i. 5,924 total exercises. Includes Chapter Reviews and Practice Tests.

1

1

Functions

1,500 P

y

1,000

500

1970 1975 1980 1985 1990 1995 2000 2005 2010 0

Figure 1 Standard and Poor’s Index with dividends reinvested (credit "bull": modification of work by Prayitno Hadinata; credit "graph": modification of work by MeasuringWorth)

Introduction Toward the end of the twentieth century, the values of stocks of internet and technology companies rose dramatically. As a result, the Standard and Poor’s stock market average rose as well. Figure 1 tracks the value of that initial investment of just under $100 over the 40 years. It shows that an investment that was worth less than $500 until about 1995 skyrocketed up to about $1,100 by the beginning of 2000. That five-year period became known as the “dot-com bubble” because so many internet startups were formed. As bubbles tend to do, though, the dot-com bubble eventually burst. Many companies grew too fast and then suddenly went out of business. The result caused the sharp decline represented on the graph beginning at the end of 2000.

Notice, as we consider this example, that there is a definite relationship between the year and stock market average. For any year we choose, we can determine the corresponding value of the stock market average. In this chapter, we will explore these kinds of relationships and their properties.

CHAPTeR OUTlIne

1.1 Functions and Function notation 1.2 Domain and Range 1.3 Rates of Change and Behavior of Graphs 1.4 Composition of Functions 1.5 Transformation of Functions 1.6 Absolute Value Functions 1.7 Inverse Functions

CHAPTER 1 Functions2

1.1 FUnCTIOnS AnD FUnCTIOn nOTATIOn

A jetliner changes altitude as its distance from the starting point of a flight increases. The weight of a growing child increases with time. In each case, one quantity depends on another. There is a relationship between the two quantities that we can describe, analyze, and use to make predictions. In this section, we will analyze such relationships.

Determining Whether a Relation Represents a Function A relation is a set of ordered pairs. The set of the first components of each ordered pair is called the domain and the set of the second components of each ordered pair is called the range. Consider the following set of ordered pairs. The first numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first.

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

The domain is {1, 2, 3, 4, 5}. The range is {2, 4, 6, 8, 10}.

Note that each value in the domain is also known as an input value, or independent variable, and is often labeled with the lowercase letter x. Each value in the range is also known as an output value, or dependent variable, and is often labeled lowercase letter y.

A function f is a relation that assigns a single value in the range to each value in the domain. In other words, no x-values are repeated. For our example that relates the first five natural numbers to numbers double their values, this relation is a function because each element in the domain, {1, 2, 3, 4, 5}, is paired with exactly one element in the range, {2, 4, 6, 8, 10}.

Now let’s consider the set of ordered pairs that relates the terms “even” and “odd” to the first five natural numbers. It would appear as

{(odd, 1), (even, 2), (odd, 3), (even, 4), (odd, 5)}

Notice that each element in the domain, {even, odd} is not paired with exactly one element in the range, {1, 2, 3, 4, 5}. For example, the term “odd” corresponds to three values from the domain, {1, 3, 5} and the term “even” corresponds to two values from the range, {2, 4}. This violates the definition of a function, so this relation is not a function. Figure 1 compares relations that are functions and not functions.

Relation is a Function Inputs Outputs Inputs Outputs Inputs Outputs

Relation is a Function Relation is NOT a Function

(a) (b) (c)

p

q

r

m

n

p

q

r

x

y

z

p

q

x

y

z

Figure 1 ( a ) This relationship is a function because each input is associated with a single output. note that input q and r both give output n. ( b ) This relationship is also a function. In this case, each input is associated with a single output.

( c ) This relationship is not a function because input q is associated with two different outputs.

function A function is a relation in which each possible input value leads to exactly one output value. We say “the output is a function of the input.” The input values make up the domain, and the output values make up the range.

leARnInG OBjeCTIVeS

In this section, you will: • Determine whether a relation represents a function. • Find the value of a function. • Determine whether a function is one-to-one. • Use the vertical line test to identify functions. • Graph the functions listed in the library of functions.

SECTION 1.1 Functions and Function notation 3

How To… Given a relationship between two quantities, determine whether the relationship is a function.

1. Identify the input values. 2. Identify the output values. 3. If each input value leads to only one output value, classify the relationship as a function. If any input value leads to

two or more outputs, do not classify the relationship as a function.

Example 1 Determining If Menu Price Lists Are Functions

The coffee shop menu, shown in Figure 2 consists of items and their prices.

a. Is price a function of the item? b. Is the item a function of the price?

Menu

Item Price Plain Donut . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.49 Jelly Donut . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.99 Chocolate Donut . . . . . . . . . . . . . . . . . . . . . . . . . . 1.99

Figure 2

Solution a. Let’s begin by considering the input as the items on the menu. The output values are then the prices. See Figure 3.

Menu

Item Price Plain Donut . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.49 Jelly Donut . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.99 Chocolate Donut . . . . . . . . . . . . . . . . . . . . . . . . . . 1.99

Figure 3

Each item on the menu has only one price, so the price is a function of the item.

b. Two items on the menu have the same price. If we consider the prices to be the input values and the items to be the output, then the same input value could have more than one output associated with it. See Figure 4.

Menu

Item Price Plain Donut . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.49 Jelly Donut . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.99 Chocolate Donut . . . . . . . . . . . . . . . . . . . . . . . . . .

Figure 4

Therefore, the item is a not a function of price.

Example 2 Determining If Class Grade Rules Are Functions

In a particular math class, the overall percent grade corresponds to a grade point average. Is grade point average a function of the percent grade? Is the percent grade a function of the grade point average? Table 1 shows a possible rule for assigning grade points.

CHAPTER 1 Functions4

Percent grade 0-56 57-61 62-66 67-71 72-77 78-86 87-91 92-100

Grade point average 0.0 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Table 1

Solution For any percent grade earned, there is an associated grade point average, so the grade point average is a function of the percent grade. In other words, if we input the percent grade, the output is a specific grade point average.

In the grading system given, there is a range of percent grades that correspond to the same grade point average. For example, students who receive a grade point average of 3.0 could have a variety of percent grades ranging from 78 all the way to 86. Thus, percent grade is not a function of grade point average

Try It #1

Table 2[1] lists the five greatest baseball players of all time in order of rank.

Player Rank

Babe Ruth 1

Willie Mays 2

Ty Cobb 3

Walter Johnson 4

Hank Aaron 5

Table 2

a. Is the rank a function of the player name? b. Is the player name a function of the rank?

Using Function Notation

Once we determine that a relationship is a function, we need to display and define the functional relationships so that we can understand and use them, and sometimes also so that we can program them into computers. There are various ways of representing functions. A standard function notation is one representation that facilitates working with functions.

To represent “height is a function of age,” we start by identifying the descriptive variables h for height and a for age. The letters f, g, and h are often used to represent functions just as we use x, y, and z to represent numbers and A, B, and C to represent sets.

h is f of a We name the function f ; height is a function of age. h = f (a) We use parentheses to indicate the function input. f (a) We name the function f ; the expression is read as “f of a.”

Remember, we can use any letter to name the function; the notation h(a) shows us that h depends on a. The value a must be put into the function h to get a result. The parentheses indicate that age is input into the function; they do not indicate multiplication.

We can also give an algebraic expression as the input to a function. For example f (a + b) means “first add a and b, and the result is the input for the function f. ” The operations must be performed in this order to obtain the correct result.

function notation The notation y = f (x) defines a function named f. This is read as “y is a function of x.” The letter x represents the input value, or independent variable. The letter y, or f (x), represents the output value, or dependent variable.

1 http://www.baseball-almanac.com/legendary/lisn100.shtml. Accessed 3/24/2014.

SECTION 1.1 Functions and Function notation 5

Example 3 Using Function Notation for Days in a Month

Use function notation to represent a function whose input is the name of a month and output is the number of days in that month.

Solution The number of days in a month is a function of the name of the month, so if we name the function f, we write days = f (month) or d = f (m). The name of the month is the input to a “rule” that associates a specific number (the output) with each input.

31�f (January)

output input rule

Figure 5

For example, f (March) = 31, because March has 31 days. The notation d = f (m) reminds us that the number of days, d (the output), is dependent on the name of the month, m(the input).

Analysi s Note that the inputs to a function do not have to be numbers; function inputs can be names of people, labels of geometric objects, or any other element that determines some kind of output. However, most of the functions we will work with in this book will have numbers as inputs and outputs.

Example 4 Interpreting Function Notation

A function N = f (y) gives the number of police officers, N, in a town in year y. What does f (2005) = 300 represent?

Solution When we read f (2005) = 300, we see that the input year is 2005. The value for the output, the number of police officers (N), is 300. Remember N = f (y). The statement f (2005) = 300 tells us that in the year 2005 there were 300 police officers in the town.

Try It #2

Use function notation to express the weight of a pig in pounds as a function of its age in days d.

Q & A… Instead of a notation such as y = f (x), could we use the same symbol for the output as for the function, such as y = y (x), meaning “y is a function of x?”

Yes, this is often done, especially in applied subjects that use higher math, such as physics and engineering. However, in exploring math itself we like to maintain a distinction between a function such as f, which is a rule or procedure, and the output y we get by applying f to a particular input x. This is why we usually use notation such as y = f (x), P = W(d), and so on.

Representing Functions Using Tables

A common method of representing functions is in the form of a table. The table rows or columns display the corresponding input and output values. In some cases, these values represent all we know about the relationship; other times, the table provides a few select examples from a more complete relationship.

Table 3 lists the input number of each month (January = 1, February = 2, and so on) and the output value of the number of days in that month. This information represents all we know about the months and days for a given year (that is not a leap year). Note that, in this table, we define a days-in-a-month function f where D = f (m) identifies months by an integer rather than by name.

Month number, m (input) 1 2 3 4 5 6 7 8 9 10 11 12

Days in month, D (output) 31 28 31 30 31 30 31 31 30 31 30 31 Table 3

CHAPTER 1 Functions6

Table 4 defines a function Q = g (n). Remember, this notation tells us that g is the name of the function that takes the input n and gives the output Q.

n 1 2 3 4 5

Q 8 6 7 6 8 Table 4

Table 5 below displays the age of children in years and their corresponding heights. This table displays just some of the data available for the heights and ages of children. We can see right away that this table does not represent a function because the same input value, 5 years, has two different output values, 40 in. and 42 in.

Age in years, a (input) 5 5 6 7 8 9 10

Height in inches, h (output) 40 42 44 47 50 52 54

Table 5

How To… Given a table of input and output values, determine whether the table represents a function.

1. Identify the input and output values. 2. Check to see if each input value is paired with only one output value. If so, the table represents a function.

Example 5 Identifying Tables that Represent Functions

Which table, Table 6, Table 7, or Table 8, represents a function (if any)?

Input Output

2 1

5 3

8 6

Input Output

−3 5

0 1

4 5

Input Output

1 0

5 2

5 4

Solution Table 6 and Table 7 define functions. In both, each input value corresponds to exactly one output value. Table 8 does not define a function because the input value of 5 corresponds to two different output values.

When a table represents a function, corresponding input and output values can also be specified using function notation.

The function represented by Table 6 can be represented by writing

f (2) = 1, f (5) = 3, and f (8) = 6

Similarly, the statements

g (−3) = 5, g (0) = 1, and g (4) = 5

represent the function in table Table 7.

Table 8 cannot be expressed in a similar way because it does not represent a function.

Try It #3

Does Table 9 represent a function? Input Output

1 10

2 100

3 1000

Table 9

Table 6 Table 7 Table 8

SECTION 1.1 Functions and Function notation 7

Finding Input and Output Values of a Function When we know an input value and want to determine the corresponding output value for a function, we evaluate the function. Evaluating will always produce one result because each input value of a function corresponds to exactly one output value.

When we know an output value and want to determine the input values that would produce that output value, we set the output equal to the function’s formula and solve for the input. Solving can produce more than one solution because different input values can produce the same output value.

Evaluation of Functions in Algebraic Forms

When we have a function in formula form, it is usually a simple matter to evaluate the function. For example, the function f (x) = 5 − 3x 2 can be evaluated by squaring the input value, multiplying by 3, and then subtracting the product from 5.

How To… Given the formula for a function, evaluate.

1. Replace the input variable in the formula with the value provided. 2. Calculate the result.

Example 6 Evaluating Functions at Specific Values

Evaluate f(x) = x 2 + 3x − 4 at:

a. 2 b. a c. a + h d. f (a + h) − f (a)

__ h

Solution Replace the x in the function with each specified value. a. Because the input value is a number, 2, we can use simple algebra to simplify.

f (2) = 22 + 3(2) − 4 = 4 + 6 − 4 = 6

b. In this case, the input value is a letter so we cannot simplify the answer any further.

f (a) = a2 + 3a − 4 c. With an input value of a + h, we must use the distributive property.

f (a + h) = (a + h)2 + 3(a + h) − 4 = a2 + 2ah + h2 + 3a + 3h −4

d. In this case, we apply the input values to the function more than once, and then perform algebraic operations on the result. We already found that

f (a + h) = a2 + 2ah + h2 + 3a + 3h − 4 and we know that f(a) = a2 + 3a − 4

Now we combine the results and simplify.

f (a + h) − f(a)

__ h

= (a2 + 2ah + h2 + 3a + 3h − 4) − (a2 + 3a − 4)

____ h

= 2ah + h 2 + 3h ___________

h

= h(2a + h + 3) ___________ h

Factor out h.

= 2a + h + 3 Simplify.

CHAPTER 1 Functions8

Example 7 Evaluating Functions

Given the function h(p) = p2 + 2p, evaluate h(4).

Solution To evaluate h(4), we substitute the value 4 for the input variable p in the given function.

h(p) = p2 + 2p

h(4) = (4)2 +2 (4)

= 16 + 8

= 24

Therefore, for an input of 4, we have an output of 24.

Try It #4

Given the function g(m) = √ —

m − 4 . Evaluate g(5).

Example 8 Solving Functions

Given the function h(p) = p2 + 2p, solve for h(p) = 3.

Solution h(p) = 3

p2 + 2p = 3 Substitute the original function h(p) = p2 + 2p.

p2 + 2p − 3 = 0 Subtract 3 from each side.

(p + 3)(p − 1) = 0 Factor.

If (p + 3)(p − 1) = 0, either (p + 3) = 0 or (p − 1) = 0 (or both of them equal 0). We will set each factor equal to 0 and solve for p in each case. (p + 3) = 0, p = −3

(p − 1) = 0, p = 1

This gives us two solutions. The output h(p) = 3 when the input is either p = 1 or p = −3. We can also verify by graphing as in Figure 6. The graph verifies that h(1) = h(−3) = 3 and h(4) = 24.

35 30 25 20 15 10

5

1 2 3 4 5 p

h(p)

p –3 –2 0 1 4 h(p) 3 0 0 3 24

Figure 6

Try It #5

Given the function g(m) = √ —

m − 4 , solve g(m) = 2.

Evaluating Functions Expressed in Formulas

Some functions are defined by mathematical rules or procedures expressed in equation form. If it is possible to express the function output with a formula involving the input quantity, then we can define a function in algebraic form. For example, the equation 2n + 6p = 12 expresses a functional relationship between n and p. We can rewrite it to decide if p is a function of n.

SECTION 1.1 Functions and Function notation 9

How To… Given a function in equation form, write its algebraic formula.

1. Solve the equation to isolate the output variable on one side of the equal sign, with the other side as an expression that involves only the input variable.

2. Use all the usual algebraic methods for solving equations, such as adding or subtracting the same quantity to or from both sides, or multiplying or dividing both sides of the equation by the same quantity.

Example 9 Finding an Equation of a Function

Express the relationship 2n + 6p = 12 as a function p = f (n), if possible.

Solution To express the relationship in this form, we need to be able to write the relationship where p is a function of n, which means writing it as p = [expression involving n].

2n + 6p = 12

6p = 12 − 2n Subtract 2n from both sides.

p = 12 − 2n _______ 6

Divide both sides by 6 and simplify.

p = 12 __ 6

− 2n ___ 6

p = 2 − 1 __ 3

n

Therefore, p as a function of n is written as

p = f (n) = 2 − 1 __ 3

n

Analysi s It is important to note that not every relationship expressed by an equation can also be expressed as a function with a formula.

Example 10 Expressing the Equation of a Circle as a Function

Does the equation x2 + y2 = 1 represent a function with x as input and y as output? If so, express the relationship as a function y = f (x).

Solution First we subtract x2 from both sides.

y2 = 1 − x2 We now try to solve for y in this equation.

y = ± √ —

1 − x2

= + √ —

1 − x2 and − √ —

1 − x2

We get two outputs corresponding to the same input, so this relationship cannot be represented as a single function y = f (x).

Try It #6

If x − 8y 3 = 0, express y as a function of x.

Q & A… Are there relationships expressed by an equation that do represent a function but which still cannot be represented by an algebraic formula?

Yes, this can happen. For example, given the equation x = y + 2y, if we want to express y as a function of x, there is no simple algebraic formula involving only x that equals y. However, each x does determine a unique value for y, and there are mathematical procedures by which y can be found to any desired accuracy. In this case, we say that the equation gives an implicit (implied) rule for y as a function of x, even though the formula cannot be written explicitly.

CHAPTER 1 Functions1 0

Evaluating a Function Given in Tabular Form

As we saw above, we can represent functions in tables. Conversely, we can use information in tables to write functions, and we can evaluate functions using the tables. For example, how well do our pets recall the fond memories we share with them? There is an urban legend that a goldfish has a memory of 3 seconds, but this is just a myth. Goldfish can remember up to 3 months, while the beta fish has a memory of up to 5 months. And while a puppy’s memory span is no longer than 30 seconds, the adult dog can remember for 5 minutes. This is meager compared to a cat, whose memory span lasts for 16 hours.

The function that relates the type of pet to the duration of its memory span is more easily visualized with the use of a table. See Table 10.[2]

Pet Memory span in hours Puppy 0.008

Adult dog 0.083 Cat 16

Goldfish 2160 Beta fish 3600

Table 10

At times, evaluating a function in table form may be more useful than using equations. Here let us call the function P. The domain of the function is the type of pet and the range is a real number representing the number of hours the pet’s memory span lasts. We can evaluate the function P at the input value of “goldfish.” We would write P(goldfish) = 2160. Notice that, to evaluate the function in table form, we identify the input value and the corresponding output value from the pertinent row of the table. The tabular form for function P seems ideally suited to this function, more so than writing it in paragraph or function form.

How To… Given a function represented by a table, identify specific output and input values.

1. Find the given input in the row (or column) of input values. 2. Identify the corresponding output value paired with that input value. 3. Find the given output values in the row (or column) of output values, noting every time that output value appears. 4. Identify the input value(s) corresponding to the given output value.

Example 11 Evaluating and Solving a Tabular Function

Using Table 11,

a. Evaluate g(3) b. Solve g(n) = 6.

n 1 2 3 4 5 g (n) 8 6 7 6 8

Table 11

Solution

a. Evaluating g (3) means determining the output value of the function g for the input value of n = 3. The table output value corresponding to n = 3 is 7, so g (3) = 7.

b. Solving g (n) = 6 means identifying the input values, n, that produce an output value of 6. Table 11 shows two solutions: 2 and 4. When we input 2 into the function g, our output is 6. When we input 4 into the function g, our output is also 6.

2 http://www.kgbanswers.com/how-long-is-a-dogs-memory-span/4221590. Accessed 3/24/2014.

SECTION 1.1 Functions and Function notation 1 1

Try It #7

Using Table 11, evaluate g(1).

Finding Function Values from a Graph

Evaluating a function using a graph also requires finding the corresponding output value for a given input value, only in this case, we find the output value by looking at the graph. Solving a function equation using a graph requires finding all instances of the given output value on the graph and observing the corresponding input value( s ).

Example 12 Reading Function Values from a Graph

Given the graph in Figure 7,

a. Evaluate f (2). b. Solve f (x) = 4.

4

2

–1 –1–2

2

–3

– –

3

–4–5

1

3

3

21 4

5 6 7

5

f (x)

Figure 7

Solution

a. To evaluate f (2), locate the point on the curve where x = 2, then read the y-coordinate of that point. The point has coordinates (2, 1), so f (2) = 1. See Figure 8.

(2, 1) f (2) = 1

f (x)

4

2

–1 –1–2

2

–3

– –

3

–4–5

1

3

3

21 4

5 6 7

5

Figure 8

b. To solve f (x) = 4, we find the output value 4 on the vertical axis. Moving horizontally along the line y = 4, we locate two points of the curve with output value 4: (−1, 4) and (3, 4). These points represent the two solutions to f (x) = 4: −1 or 3. This means f (−1) = 4 and f (3) = 4, or when the input is −1 or 3, the output is 4. See Figure 9.

4

2

–1 –1–2

2

–3

– –

3

–4–5

1

3

3

21 4

5 6 7

5

(3, 4)(−1, 4)

f (x)

Figure 9

CHAPTER 1 Functions1 2

Try It #8

Using Figure 7, solve f (x) = 1.

Determining Whether a Function is One-to-One Some functions have a given output value that corresponds to two or more input values. For example, in the stock chart shown in Figure 1 at the beginning of this chapter, the stock price was $1,000 on five different dates, meaning that there were five different input values that all resulted in the same output value of $1,000.

However, some functions have only one input value for each output value, as well as having only one output for each input. We call these functions one-to-one functions. As an example, consider a school that uses only letter grades and decimal equivalents, as listed in Table 12.

Letter grade Grade point average A 4.0 B 3.0 C 2.0 D 1.0

Table 12

This grading system represents a one-to-one function, because each letter input yields one particular grade point average output and each grade point average corresponds to one input letter.

To visualize this concept, let’s look again at the two simple functions sketched in Figure 1(a) and Figure 1(b). The function in part ( a) shows a relationship that is not a one-to-one function because inputs q and r both give output n. The function in part (b ) shows a relationship that is a one-to-one function because each input is associated with a single output.

one-to-one function A one-to-one function is a function in which each output value corresponds to exactly one input value.

Example 13 Determining Whether a Relationship Is a One-to-One Function

Is the area of a circle a function of its radius? If yes, is the function one-to-one?

Solution A circle of radius r has a unique area measure given by A = πr 2, so for any input, r, there is only one output, A. The area is a function of radius r.

If the function is one-to-one, the output value, the area, must correspond to a unique input value, the radius. Any area measure A is given by the formula A = πr2. Because areas and radii are positive numbers, there is exactly one solution:

r = √ ___

A __ π

So the area of a circle is a one-to-one function of the circle’s radius.

Try It #9

a. Is a balance a function of the bank account number? b. Is a bank account number a function of the balance? c. Is a balance a one-to-one function of the bank account number?

Try It #10

a. If each percent grade earned in a course translates to one letter grade, is the letter grade a function of the percent grade?

b. If so, is the function one-to-one?

SECTION 1.1 Functions and Function notation 1 3

Using the Vertical line Test As we have seen in some examples above, we can represent a function using a graph. Graphs display a great many input-output pairs in a small space. The visual information they provide often makes relationships easier to understand. By convention, graphs are typically constructed with the input values along the horizontal axis and the output values along the vertical axis.

The most common graphs name the input value x and the output value y, and we say y is a function of x, or y = f (x) when the function is named f. The graph of the function is the set of all points (x, y) in the plane that satisfies the equation y = f (x). If the function is defined for only a few input values, then the graph of the function is only a few points, where the x-coordinate of each point is an input value and the y-coordinate of each point is the corresponding output value. For example, the black dots on the graph in Figure 10 tell us that f (0) = 2 and f (6) = 1. However, the set of all points (x, y) satisfying y = f (x) is a curve. The curve shown includes (0, 2) and (6, 1) because the curve passes through those points.

y

x

Figure 10

The vertical line test can be used to determine whether a graph represents a function. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because a function has only one output value for each input value. See Figure 11.

Function Not a Function Not a Function

Figure 11

How To… Given a graph, use the vertical line test to determine if the graph represents a function.

1. Inspect the graph to see if any vertical line drawn would intersect the curve more than once. 2. If there is any such line, determine that the graph does not represent a function.

CHAPTER 1 Functions1 4

Example 14 Applying the Vertical Line Test

Which of the graphs in Figure 12 represent( s ) a function y = f (x)?

x

f (x) y

(a) (b) (c)

5 4 3 2 1

4 5 6 7 8 9 10 11 12 x x

f (x)

Figure 12

Solution If any vertical line intersects a graph more than once, the relation represented by the graph is not a function. Notice that any vertical line would pass through only one point of the two graphs shown in parts ( a) and (b) of Figure 12. From this we can conclude that these two graphs represent functions. The third graph does not represent a function because, at most x-values, a vertical line would intersect the graph at more than one point, as shown in Figure 13.

x

y

Figure 13

Try It #11

Does the graph in Figure 14 represent a function?

x

y

Figure 14

SECTION 1.1 Functions and Function notation 1 5

Using the Horizontal line Test Once we have determined that a graph defines a function, an easy way to determine if it is a one-to-one function is to use the horizontal line test. Draw horizontal lines through the graph. If any horizontal line intersects the graph more than once, then the graph does not represent a one-to-one function.

How To… Given a graph of a function, use the horizontal line test to determine if the graph represents a one-to-one function.

1. Inspect the graph to see if any horizontal line drawn would intersect the curve more than once. 2. If there is any such line, determine that the function is not one-to-one.

Example 15 Horizontal Line Test

Consider the functions shown in Figure 12(a) and Figure 12(b). Are either of the functions one-to-one?

Solution The function in Figure 12(a) is not one-to-one. The horizontal line shown in Figure 15 intersects the graph of the function at two points (and we can even find horizontal lines that intersect it at three points.)

x

f (x)

Figure 15

The function in Figure 12(b) is one-to-one. Any horizontal line will intersect a diagonal line at most once.

Try It #12

Is the graph shown here one-to-one?

x

y

Identifying Basic Toolkit Functions In this text, we will be exploring functions—the shapes of their graphs, their unique characteristics, their algebraic formulas, and how to solve problems with them. When learning to read, we start with the alphabet. When learning to do arithmetic, we start with numbers. When working with functions, it is similarly helpful to have a base set of building-block elements. We call these our “toolkit functions,” which form a set of basic named functions for which

CHAPTER 1 Functions1 6

we know the graph, formula, and special properties. Some of these functions are programmed to individual buttons on many calculators. For these definitions we will use x as the input variable and y = f (x) as the output variable.

We will see these toolkit functions, combinations of toolkit functions, their graphs, and their transformations frequently throughout this book. It will be very helpful if we can recognize these toolkit functions and their features quickly by name, formula, graph, and basic table properties. The graphs and sample table values are included with each function shown in Table 13.

Toolkit Functions Name Function Graph

Constant f (x) = c, where c is a constant x –2 2 0 2

2 2

x f(x)

f (x)

Identity f (x) = x x

x f(x) –2 –2 0 0

2 2

f (x)

Absolute value f (x) = ∣ x ∣ x

f (x)

–2 2 0 0

2 2

x f(x)

Quadratic f (x) = x2 x

–2 4 –1 1

0 0 1 1 2 4

x f(x)

f (x)

Cubic f (x) = x3 x

–1 –1 –0.5 –0.125

0 0 0.5 0.125 1 1

x f(x)

f (x)

SECTION 1.1 Functions and Function notation 1 7

Reciprocal f (x) = 1 __ x x

–2 –0.5 –1 –1

–0.5 –2 0.5 2 1 1 2 0.5

x f(x)f (x)

Reciprocal squared f (x) = 1 _ x 2

x

–2 0.25 –1 1

–0.5 4 0.5 4 1 1 2 0.25

x f(x)f (x)

Square root f (x) = √ — x x

0 0 1 1

4 2

x f(x)

f (x)

Cube root f (x) = 3 √ — x x

–1 –1 –0.125 –0.5

0 0 0.125 0.5 1 1

x f(x)

f (x)

Table 13

Access the following online resources for additional instruction and practice with functions.

• Determine if a Relation is a Function (http://openstaxcollege.org/l/relationfunction)

• Vertical line Test (http://openstaxcollege.org/l/vertlinetest)

• Introduction to Functions (http://openstaxcollege.org/l/introtofunction)

• Vertical line Test of Graph (http://openstaxcollege.org/l/vertlinegraph)

• One-to-one Functions (http://openstaxcollege.org/l/onetoone)

• Graphs as One-to-one Functions (http://openstaxcollege.org/l/graphonetoone)

1 8 CHAPTER 1 Functions

1.1 SeCTIOn exeRCISeS

VeRBAl

1. What is the difference between a relation and a function?

2. What is the difference between the input and the output of a function?

3. Why does the vertical line test tell us whether the graph of a relation represents a function?

4. How can you determine if a relation is a one-to-one function?

5. Why does the horizontal line test tell us whether the graph of a function is one-to-one?

AlGeBRAIC

For the following exercises, determine whether the relation represents a function.

6. {(a, b), (c, d), (a, c)} 7. {(a, b),(b, c),(c, c)}

For the following exercises, determine whether the relation represents y as a function of x. 8. 5x + 2y = 10 9. y = x 2 10. x = y 2

11. 3x 2 + y = 14 12. 2x + y 2 = 6 13. y = −2x 2 + 40x

14. y = 1 __ x

15. x = 3y + 5 ______ 7y − 1

16. x = √ —

1 − y 2

17. y = 3x + 5 ______ 7x − 1

18. x 2 + y 2 = 9 19. 2xy = 1

20. x = y 3 21. y = x 3 22. y = √ —

1 − x 2

23. x = ± √ —

1 − y 24. y = ± √ —

1 − x 25. y 2 = x 2

26. y 3 = x 2

For the following exercises, evaluate the function f at the indicated values f (−3), f (2), f (−a), −f (a), f (a + h).

27. f (x) = 2x − 5 28. f (x) = −5x 2 + 2x − 1 29. f (x) = √ —

2 − x + 5

30. f (x) = 6x − 1 ______ 5x + 2

31. f (x) = ∣ x − 1 ∣ − ∣ x + 1 ∣

32. Given the function g(x) = 5 − x 2, evaluate g(x + h) − g(x)

__ h

, h ≠ 0

33. Given the function g(x) = x 2 + 2x, evaluate g(x) − g(a)

_ x − a , x ≠ a

34. Given the function k(t) = 2t − 1: a. Evaluate k(2). b. Solve k(t) = 7.

35. Given the function f (x) = 8 − 3x: a. Evaluate f (−2). b. Solve f (x) = −1.

36. Given the function p(c) = c 2 + c: a. Evaluate p(−3). b. Solve p(c) = 2.

37. Given the function f (x) = x2 − 3x a. Evaluate f (5). b. Solve f (x) = 4

38. Given the function f (x) =  √ —

x + 2 : a. Evaluate f (7). b. Solve f (x) = 4

39. Consider the relationship 3r + 2t = 18. a. Write the relationship as a function r = f (t). b. Evaluate f (−3). c. Solve f (t) = 2.

SECTION 1.1 section exercises 1 9

GRAPHICAl

For the following exercises, use the vertical line test to determine which graphs show relations that are functions.

40.

x

y 41.

x

y 42.

x

y

43.

x

y 44. 45.

x

y

46.

x

y 47.

x

y 48.

x

y

49.

x

y 50.

x

y 51.

x

y

x

y

2 0 CHAPTER 1 Functions

52. Given the following graph a. Evaluate f (−1). b. Solve for f (x) = 3.

53. Given the following graph a. Evaluate f (0). b. Solve for f (x) = −3.

54. Given the following graph a. Evaluate f (4). b. Solve for f (x) = 1.

For the following exercises, determine if the given graph is a one-to-one function. 55.

2

–1 –1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

x

y 56.

2

–1 –1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

x

y 57.

2

–1 –1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

x

y

58.

2

–1 –1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

x

y 59.

x

y

π�π

5 4 3 2 1

�1 �2 �3 �4 �5

nUMeRIC For the following exercises, determine whether the relation represents a function.

60. {(−1, −1),(−2, −2),(−3, −3)} 61. {(3, 4),(4, 5),(5, 6)} 62. {(2, 5),(7, 11),(15, 8),(7, 9)}

For the following exercises, determine if the relation represented in table form represents y as a function of x.

63. x 5 10 15 y 3 8 14

64. x 5 10 15 y 3 8 8

65. x 5 10 10 y 3 8 14

For the following exercises, use the function f represented in Table 14 below.

x 0 1 2 3 4 5 6 7 8 9 f (x) 74 28 1 53 56 3 36 45 14 47

Table 14

66. Evaluate f (3). 67. Solve f (x) = 1

2

–1 –1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

x

y

2

–1 –1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

x

y

2

–1 –1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

x

y

SECTION 1.1 section exercises 2 1

For the following exercises, evaluate the function f at the values f (−2), f (−1), f (0), f (1), and f (2).

68. f (x) = 4 − 2x 69. f (x) = 8 − 3x 70. f (x) = 8x2 − 7x + 3

71. f (x) = 3 + √ —

x + 3 72. f (x) = x − 2 _____ x + 3

73. f (x) = 3 x

For the following exercises, evaluate the expressions, given functions f, g, and h:

f (x) = 3x − 2 g(x) = 5 − x2 h(x) = −2x2 + 3x − 1

74. 3f (1) − 4g(−2) 75. f   7 __ 3  − h(−2)

TeCHnOlOGY

For the following exercises, graph y = x2 on the given viewing window. Determine the corresponding range for each viewing window. Show each graph.

76. [−0.1, 0.1] 77. [−10, 10] 78. [−100, 100]

For the following exercises, graph y = x3 on the given viewing window. Determine the corresponding range for each viewing window. Show each graph.

79. [−0.1, 0.1] 80. [−10, 10] 81. [−100, 100]

For the following exercises, graph y = √ — x on the given viewing window. Determine the corresponding range for each

viewing window. Show each graph. 82. [0, 0.01] 83. [0, 100] 84. [0, 10,000]

For the following exercises, graph y = 3 √ — x on the given viewing window. Determine the corresponding range for each

viewing window. Show each graph. 85. [−0.001, 0.001] 86. [−1,000, 1,000] 87. [−1,000,000, 1,000,000]

ReAl-WORlD APPlICATIOnS 88. The amount of garbage, G, produced by a city with

population p is given by G = f (p). G is measured in tons per week, and p is measured in thousands of people. a. The town of Tola has a population of 40,000 and

produces 13 tons of garbage each week. Express this information in terms of the function f.

b. Explain the meaning of the statement f (5) = 2.

89. The number of cubic yards of dirt, D, needed to cover a garden with area a square feet is given by D = g(a). a. A garden with area 5,000 ft2 requires 50 yd3 of

dirt. Express this information in terms of the function g.

b. Explain the meaning of the statement g(100) = 1.

90. Let f (t) be the number of ducks in a lake t years after 1990. Explain the meaning of each statement: a. f (5) = 30 b. f (10) = 40

91. Let h(t) be the height above ground, in feet, of a rocket t seconds after launching. Explain the meaning of each statement: a. h(1) = 200 b. h(2) = 350

92. Show that the function f (x) = 3(x − 5)2 + 7 is not one-to-one.

CHAPTER 1 Functions2 2

1. 2 DOMAIn AnD RAnGe

If you’re in the mood for a scary movie, you may want to check out one of the five most popular horror movies of all time—I am Legend, Hannibal, The Ring, The Grudge, and The Conjuring. Figure 1 shows the amount, in dollars, each of those movies grossed when they were released as well as the ticket sales for horror movies in general by year. Notice that we can use the data to create a function of the amount each movie earned or the total ticket sales for all horror movies by year. In creating various functions using the data, we can identify different independent and dependent variables, and we can analyze the data and the functions to determine the domain and range. In this section, we will investigate methods for determining the domain and range of functions such as these.

350

300

In �a

ti on

-a dj

us te

d G

ro ss

, in

M ill

io ns

o f D

ol la

rs

250

200

150

100

I am Legend (2007)

Hannibal (2001)

Top-Five Grossing Horror Movies for years 2000–2013

e Ring

(2002)

e Grudge (2004)

e Conjuring

(2013)

50 0

8% Market Share of Horror Moveis, by Year

7% 6% 5% 4% 3% 2% 1%

20 00

20 01

20 02

20 03

20 04

20 05

20 06

20 07

20 08

20 09

20 10

20 11

20 12

20 13

0%

Figure 1 Based on data compiled by www.the-numbers.com.[3]

Finding the Domain of a Function Defined by an equation In Functions and Function Notation, we were introduced to the concepts of domain and range. In this section, we will practice determining domains and ranges for specific functions. Keep in mind that, in determining domains and ranges, we need to consider what is physically possible or meaningful in real-world examples, such as tickets sales and year in the horror movie example above. We also need to consider what is mathematically permitted. For example, we cannot include any input value that leads us to take an even root of a negative number if the domain and range consist of real numbers. Or in a function expressed as a formula, we cannot include any input value in the domain that would lead us to divide by 0.

We can visualize the domain as a “holding area” that contains “raw materials” for a “function machine” and the range as another “holding area” for the machine’s products. See Figure 2.

Domain Function machine

a

b c

x y z

Range

Figure 2

We can write the domain and range in interval notation, which uses values within brackets to describe a set of numbers. In interval notation, we use a square bracket [when the set includes the endpoint and a parenthesis (to indicate that the endpoint is either not included or the interval is unbounded. For example, if a person has $100 to spend, he or she would need to express the interval that is more than 0 and less than or equal to 100 and write (0, 100]. We will discuss interval notation in greater detail later. 3 The Numbers: Where Data and the Movie Business Meet. “Box Office History for Horror Movies.” http://www.the-numbers.com/market/genre/Horror. Accessed 3/24/2014

leARnInG OBjeCTIVeS

In this section, you will: • Find the domain of a function defined by an equation. • Graph piecewise-defined functions.

SECTION 1.2 domain and range 2 3

Let’s turn our attention to finding the domain of a function whose equation is provided. Oftentimes, finding the domain of such functions involves remembering three different forms. First, if the function has no denominator or an even root, consider whether the domain could be all real numbers. Second, if there is a denominator in the function’s equation, exclude values in the domain that force the denominator to be zero. Third, if there is an even root, consider excluding values that would make the radicand negative.

Before we begin, let us review the conventions of interval notation:

• The smallest term from the interval is written first.

• The largest term in the interval is written second, following a comma.

• Parentheses, (or), are used to signify that an endpoint is not included, called exclusive.

• Brackets, [or], are used to indicate that an endpoint is included, called inclusive.

See Figure 3 for a summary of interval notation.

Inequality Interval Notation Graph on Number Line Description

x > a (a, ∞) a ( x is greater than a

x < a (−∞, a) a

( x is less than a

x ≥ a [a, ∞) a [ x is greater than or equal to a

x ≤ a (−∞, a] a ] x is less than or equal to a

a < x < b (a, b) a b ( ( x is strictly between a and b

a ≤ x < b [a, b) a b

([ x is between a and b, to include a

a < x ≤ b (a, b] a b

]( x is between a and b, to include b

a ≤ x ≤ b [a, b] a b [ ] x is between a and b, to include a and b

Figure 3

Example 1 Finding the Domain of a Function as a Set of Ordered Pairs

Find the domain of the following function: {(2, 10), (3, 10), (4, 20), (5, 30), (6, 40)}.

Solution First identify the input values. The input value is the first coordinate in an ordered pair. There are no restrictions, as the ordered pairs are simply listed. The domain is the set of the first coordinates of the ordered pairs.

{2, 3, 4, 5, 6}

Try It #1

Find the domain of the function: {(−5, 4), (0, 0), (5, −4), (10, −8), (15, −12)}

How To… Given a function written in equation form, find the domain.

1. Identify the input values. 2. Identify any restrictions on the input and exclude those values from the domain. 3. Write the domain in interval form, if possible.

CHAPTER 1 Functions2 4

Example 2 Finding the Domain of a Function

Find the domain of the function f (x) = x2 − 1.

Solution The input value, shown by the variable x in the equation, is squared and then the result is lowered by one. Any real number may be squared and then be lowered by one, so there are no restrictions on the domain of this function. The domain is the set of real numbers.

In interval form, the domain of f is (−∞, ∞).

Try It #2

Find the domain of the function: f (x) = 5 − x + x 3.

How To… Given a function written in an equation form that includes a fraction, find the domain.

1. Identify the input values. 2. Identify any restrictions on the input. If there is a denominator in the function’s formula, set the denominator equal

to zero and solve for x. If the function’s formula contains an even root, set the radicand greater than or equal to 0, and then solve.

3. Write the domain in interval form, making sure to exclude any restricted values from the domain.

Example 3 Finding the Domain of a Function Involving a Denominator

Find the domain of the function f (x) = x + 1 _____ 2 − x

.

Solution When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve for x.

2 − x = 0 −x = −2 x = 2

Now, we will exclude 2 from the domain. The answers are all real numbers where x < 2 or x > 2. We can use a symbol known as the union, ∪, to combine the two sets. In interval notation, we write the solution: (−∞, 2) ∪ (2, ∞).

–3 –2 –1 0 1 2 3 x < 2 or x > 2

↓ ↓ (−∞, 2) ∪ (2, ∞)

Figure 4

In interval form, the domain of f is (−∞, 2) ∪ (2, ∞).

Try It #3

Find the domain of the function: f (x) = 1 + 4x ______ 2x − 1

.

How To… Given a function written in equation form including an even root, find the domain.

1. Identify the input values. 2. Since there is an even root, exclude any real numbers that result in a negative number in the radicand. Set the

radicand greater than or equal to zero and solve for x. 3. The solution(s) are the domain of the function. If possible, write the answer in interval form.

SECTION 1.2 domain and range 2 5

Example 4 Finding the Domain of a Function with an Even Root

Find the domain of the function f (x) = √ —

7 − x .

Solution When there is an even root in the formula, we exclude any real numbers that result in a negative number in the radicand.

Set the radicand greater than or equal to zero and solve for x. 7 − x ≥ 0 −x ≥ −7 x ≤ 7 Now, we will exclude any number greater than 7 from the domain. The answers are all real numbers less than or equal to 7, or (−∞, 7].

Try It #4

Find the domain of the function f (x) = √ —

5 + 2x .

Q & A… Can there be functions in which the domain and range do not intersect at all?

Yes. For example, the function f (x) = −   1 _ √

— x has the set of all positive real numbers as its domain but the set of all

negative real numbers as its range. As a more extreme example, a function’s inputs and outputs can be completely different categories (for example, names of weekdays as inputs and numbers as outputs, as on an attendance chart), in such cases the domain and range have no elements in common.

Using notations to Specify Domain and Range In the previous examples, we used inequalities and lists to describe the domain of functions. We can also use inequalities, or other statements that might define sets of values or data, to describe the behavior of the variable in set-builder notation. For example, {x | 10 ≤ x < 30} describes the behavior of x in set-builder notation. The braces { } are read as “the set of,” and the vertical bar | is read as “such that,” so we would read {x | 10 ≤ x < 30} as “the set of x-values such that 10 is less than or equal to x, and x is less than 30.”

Figure 5 compares inequality notation, set-builder notation, and interval notation.

Inequality Notation

Set-builder Notation

Interval Notation

5 10 5 < h ≤ 10 {h | 5 < h ≤ 10} (5, 10]

5 10 5 ≤ h < 10 {h | 5 ≤ h < 10} [5, 10)

5 10 5 < h < 10 {h | 5 < h < 10} (5, 10)

5 10 h < 10 {h | h < 10} (−∞, 10)

10 h ≥ 10 {h | h ≥ 10} [10, ∞)

5 10 All real numbers 핉 (−∞, ∞)

Figure 5

CHAPTER 1 Functions2 6

To combine two intervals using inequality notation or set-builder notation, we use the word “or.” As we saw in earlier examples, we use the union symbol, ∪, to combine two unconnected intervals. For example, the union of the sets {2, 3, 5} and {4, 6} is the set {2, 3, 4, 5, 6}. It is the set of all elements that belong to one or the other (or both) of the original two sets. For sets with a finite number of elements like these, the elements do not have to be listed in ascending order of numerical value. If the original two sets have some elements in common, those elements should be listed only once in the union set. For sets of real numbers on intervals, another example of a union is

{x | | x | ≥ 3} = (−∞, −3] ∪ [3, ∞)

set-builder notation and interval notation

Set-builder notation is a method of specifying a set of elements that satisfy a certain condition. It takes the form {x | statement about x} which is read as, “the set of all x such that the statement about x is true.” For example,

{x | 4 < x ≤ 12}

Interval notation is a way of describing sets that include all real numbers between a lower limit that may or may not be included and an upper limit that may or may not be included. The endpoint values are listed between brackets or parentheses. A square bracket indicates inclusion in the set, and a parenthesis indicates exclusion from the set. For example,

(4, 12]

How To… Given a line graph, describe the set of values using interval notation.

1. Identify the intervals to be included in the set by determining where the heavy line overlays the real line. 2. At the left end of each interval, use [with each end value to be included in the set (solid dot) or (for each excluded

end value (open dot). 3. At the right end of each interval, use] with each end value to be included in the set (filled dot) or) for each excluded

end value (open dot). 4. Use the union symbol ∪ to combine all intervals into one set.

Example 5 Describing Sets on the Real-Number Line

Describe the intervals of values shown in Figure 6 using inequality notation, set-builder notation, and interval notation.

–2 –1 0 1 2 3 4 5 6 7

Figure 6

Solution To describe the values, x, included in the intervals shown, we would say, “x is a real number greater than or equal to 1 and less than or equal to 3, or a real number greater than 5.”

Inequality 1 ≤ x ≤ 3 or x > 5

Set-builder notation { x |1 ≤ x ≤ 3 or x > 5 }

Interval notation [1, 3] ∪ (5, ∞)

Remember that, when writing or reading interval notation, using a square bracket means the boundary is included in the set. Using a parenthesis means the boundary is not included in the set.

SECTION 1.2 domain and range 2 7

Try It #5

Given this figure, specify the graphed set in –5 –4 –3 –2 –1 0 1 2 3 4 5

Figure 7 a. words

b. set-builder notation

c. interval notation

Finding Domain and Range from Graphs Another way to identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis. Keep in mind that if the graph continues beyond the portion of the graph we can see, the domain and range may be greater than the visible values. See Figure 8.

2

x

y

–1 –1–2

–2

–3

–3

–4

–4 –5 –6 –7 – –

8 9

–5–6

1

3

3

21 4

4 5 6 7

5 6

Domain

Range

Figure 8

We can observe that the graph extends horizontally from −5 to the right without bound, so the domain is [−5, ∞). The vertical extent of the graph is all range values 5 and below, so the range is (−∞, 5]. Note that the domain and range are always written from smaller to larger values, or from left to right for domain, and from the bottom of the graph to the top of the graph for range.

Example 6 Finding Domain and Range from a Graph

Find the domain and range of the function f whose graph is shown in Figure 9.

f

x

y

–1 –1

–2

–2

–3

3

–4

–

–

4

–5

–5

321

1

4 5

Figure 9

CHAPTER 1 Functions2 8

Solution We can observe that the horizontal extent of the graph is −3 to 1, so the domain of f is (−3, 1].

The vertical extent of the graph is 0 to −4, so the range is [−4, 0]. See Figure 10.

f Range

Domain

x

y

–1 –1

–2

–2

–3

3

–4

–

–

4

–5

–5

321

1

4 5

Figure 10

Example 7 Finding Domain and Range from a Graph of Oil Production

Find the domain and range of the function f whose graph is shown in Figure 11.

� ou

sa nd

b ar

re ls

p er

d ay

1975 1980 1985 1990 1995 2000 2005

Alaska Crude Oil Production

0 200 400 600 800

1000 1200 1400 1600 1800 2000 2200

Figure 11 (credit: modification of work by the U.S. energy Information Administration) [4]

Solution The input quantity along the horizontal axis is “years,” which we represent with the variable t for time. The output quantity is “thousands of barrels of oil per day,” which we represent with the variable b for barrels. The graph may continue to the left and right beyond what is viewed, but based on the portion of the graph that is visible, we can determine the domain as 1973 ≤ t ≤ 2008 and the range as approximately 180 ≤ b ≤ 2010.

In interval notation, the domain is [1973, 2008], and the range is about [180, 2010]. For the domain and the range, we approximate the smallest and largest values since they do not fall exactly on the grid lines.

Try It #6

Given Figure 12, identify the domain and range using interval notation.

M ill

io ns

o f p

eo pl

e

1950 1960 1970 1980 1990 2000

World Population Increase

Year

0 10 20 30 40 50 60 70 80 90

100

Figure 12

Q & A… Can a function’s domain and range be the same?

Yes. For example, the domain and range of the cube root function are both the set of all real numbers.

4 http://www.eia.gov/dnav/pet/hist/Leaf Handler.ashx?n=PET&s=MCRFPAK2&f=A.

SECTION 1.2 domain and range 2 9

Finding Domains and Ranges of the Toolkit Functions

We will now return to our set of toolkit functions to determine the domain and range of each.

x

f (x) = c

f (x) Domain: (−∞, ∞) Range: [c, c]

Figure 13

For the constant function f(x) = c, the domain consists of all real numbers; there are no restrictions on the input. The only output value is the constant c, so the range is the set {c} that contains this single element. In interval notation, this is written as [c, c], the interval that both begins and ends with c.

x

f (x)

Domain: (−∞, ∞) Range: (−∞, ∞)

Figure 14

For the identity function f(x) = x, there is no restriction on x. Both the domain and range are the set of all real numbers.

x

f (x) Domain: (−∞, ∞)

Range: [0, ∞)

Figure 15

For the absolute value function f(x) = ∣ x ∣ , there is no restriction on x. However, because absolute value is defined as a distance from 0, the output can only be greater than or equal to 0.

Domain: (−∞, ∞) Range: [0, ∞)

x

f (x)

Figure 16

For the quadratic function f(x) = x 2, the domain is all real numbers since the horizontal extent of the graph is the whole real number line. Because the graph does not include any negative values for the range, the range is only nonnegative real numbers.

Domain: (−∞, ∞) Range: (−∞, ∞)

x

f (x)

Figure 17

For the cubic function f(x) = x 3, the domain is all real numbers because the horizontal extent of the graph is the whole real number line. The same applies to the vertical extent of the graph, so the domain and range include all real numbers.

CHAPTER 1 Functions3 0

x

f (x)

Domain: (−∞, 0) ∪ (0, ∞) Range: (−∞, 0) ∪ (0, ∞)

Figure 18

For the reciprocal function f(x) = 1 _ x , we cannot divide by 0, so we must exclude 0 from the domain. Further, 1 divided by any value can never be 0, so the range also will not include 0. In set-builder notation, we could also write {x | x ≠ 0}, the set of all real numbers that are not zero.

Domain: (−∞, 0) ∪ (0, ∞) Range: (0, ∞)

x

f (x)

Figure 19

For the reciprocal squared function f(x) = 1 __ x 2 , we cannot divide by 0, so we must exclude 0 from the domain. There is also no x that can give an output of 0, so 0 is excluded from the range as well. Note that the output of this function is always positive due to the square in the denominator, so the range includes only positive numbers.

x

f (x) Domain: [0, ∞) Range: [0, ∞)

Figure 20

For the square root function f(x) = √ — x , we cannot take the

square root of a negative real number, so the domain must be 0 or greater. The range also excludes negative numbers because the square root of a positive number x is defined to be positive, even though the square of the negative number

− √ — x also gives us x.

x

f (x) Domain: (−∞, ∞) Range: (−∞, ∞)

Figure 21

For the cube root function f(x) = 3 √ — x the domain and range

include all real numbers. Note that there is no problem taking a cube root, or any odd-integer root, of a negative number, and the resulting output is negative (it is an odd function).

How To… Given the formula for a function, determine the domain and range.

1. Exclude from the domain any input values that result in division by zero. 2. Exclude from the domain any input values that have nonreal (or undefined) number outputs. 3. Use the valid input values to determine the range of the output values. 4. Look at the function graph and table values to confirm the actual function behavior.

Example 8 Finding the Domain and Range Using Toolkit Functions

Find the domain and range of f (x) = 2x3 − x.

Solution There are no restrictions on the domain, as any real number may be cubed and then subtracted from the result. The domain is (−∞, ∞) and the range is also (−∞, ∞).

SECTION 1.2 domain and range 3 1

Example 9 Finding the Domain and Range

Find the domain and range of f (x) = 2 ____ x + 1

.

Solution We cannot evaluate the function at −1 because division by zero is undefined. The domain is (−∞, −1) ∪ (−1, ∞). Because the function is never zero, we exclude 0 from the range. The range is (−∞, 0) ∪ (0, ∞).

Example 10 Finding the Domain and Range

Find the domain and range of f (x) = 2 √ —

x + 4 .

Solution We cannot take the square root of a negative number, so the value inside the radical must be nonnegative.

x + 4 ≥ 0 when x ≥ −4

The domain of f (x) is [−4, ∞).

We then find the range. We know that f (−4) = 0, and the function value increases as x increases without any upper limit. We conclude that the range of f is [0, ∞).

Analysi s Figure 22 represents the function f .

2

x –1

–1–2–3–4–5

1

3

3

21 4

4

5

5

f

f (x)

Figure 22

Try It #7

Find the domain and range of f (x) = − √ —

2 − x .

Graphing Piecewise-Defined Functions Sometimes, we come across a function that requires more than one formula in order to obtain the given output. For example, in the toolkit functions, we introduced the absolute value function f (x) = |x|. With a domain of all real numbers and a range of values greater than or equal to 0, absolute value can be defined as the magnitude, or modulus, of a real number value regardless of sign. It is the distance from 0 on the number line. All of these definitions require the output to be greater than or equal to 0.

If we input 0, or a positive value, the output is the same as the input.

f (x) = x if x ≥ 0

If we input a negative value, the output is the opposite of the input.

f (x) = −x if x < 0

Because this requires two different processes or pieces, the absolute value function is an example of a piecewise function. A piecewise function is a function in which more than one formula is used to define the output over different pieces of the domain.

We use piecewise functions to describe situations in which a rule or relationship changes as the input value crosses certain “boundaries.” For example, we often encounter situations in business for which the cost per piece of a certain item is discounted once the number ordered exceeds a certain value. Tax brackets are another real-world example of piecewise functions. For example, consider a simple tax system in which incomes up to $10,000 are taxed at 10%, and any additional income is taxed at 20%. The tax on a total income S would be 0.1S if S ≤ $10,000 and $1000 + 0.2(S − $10,000) if S > $10,000.

CHAPTER 1 Functions3 2

piecewise function A piecewise function is a function in which more than one formula is used to define the output. Each formula has its own domain, and the domain of the function is the union of all these smaller domains. We notate this idea like this:

f (x) = formula 1 if x is in domain 1 formula 2 if x is in domain 2 formula 3 if x is in domain 3{

In piecewise notation, the absolute value function is

|x| = x if x ≥ 0

−x if x < 0{

How To… Given a piecewise function, write the formula and identify the domain for each interval.

1. Identify the intervals for which different rules apply. 2. Determine formulas that describe how to calculate an output from an input in each interval. 3. Use braces and if-statements to write the function.

Example 11 Writing a Piecewise Function

A museum charges $5 per person for a guided tour with a group of 1 to 9 people or a fixed $50 fee for a group of 10 or more people. Write a function relating the number of people, n, to the cost, C.

Solution Two different formulas will be needed. For n-values under 10, C = 5n. For values of n that are 10 or greater, C = 50.

C(n) = 5n if 0 < n < 10 50 if n ≥ 10 {

Analysi s The function is represented in Figure 23. The graph is a diagonal line from n = 0 to n = 10 and a constant after that. In this example, the two formulas agree at the meeting point where n = 10, but not all piecewise functions have this property.

C (n)

C (n)

n –

–10 55

10

–

20

–

30

–

40

– 1515 1010 2020 2525

50

Figure 23

Example 12 Working with a Piecewise Function

A cell phone company uses the function below to determine the cost, C, in dollars for g gigabytes of data transfer.

C(g) = 25 if 0 < g < 2 25 + 10(g − 2) if g ≥ 2 {

Find the cost of using 1.5 gigabytes of data and the cost of using 4 gigabytes of data.

Solution To find the cost of using 1.5 gigabytes of data, C(1.5), we first look to see which part of the domain our input falls in. Because 1.5 is less than 2, we use the first formula.

C(1.5) = $25

To find the cost of using 4 gigabytes of data, C(4), we see that our input of 4 is greater than 2, so we use the second formula.

C(4) = 25 + 10(4 − 2) = $45

SECTION 1.2 domain and range 3 3

Analysi s The function is represented in Figure 24. We can see where the function changes from a constant to a shifted and stretched identity at g = 2. We plot the graphs for the different formulas on a common set of axes, making sure each formula is applied on its proper domain.

C (g)

C (g)

g –

–10 1

10

–

20

–

30

–

40

– 3 245

50

53 421

Figure 24

How To… Given a piecewise function, sketch a graph.

1. Indicate on the x-axis the boundaries defined by the intervals on each piece of the domain. 2. For each piece of the domain, graph on that interval using the corresponding equation pertaining to that piece. Do

not graph two functions over one interval because it would violate the criteria of a function.

Example 13 Graphing a Piecewise Function

Sketch a graph of the function.

f (x) = x2 if x ≤ 1 3 if 1 < x ≤ 2 x if x > 2 {

Solution Each of the component functions is from our library of toolkit functions, so we know their shapes. We can imagine graphing each function and then limiting the graph to the indicated domain. At the endpoints of the domain, we draw open circles to indicate where the endpoint is not included because of a less-than or greater-than inequality; we draw a closed circle where the endpoint is included because of a less-than-or-equal-to or greater-than-or-equal-to inequality.

Figure 25 shows the three components of the piecewise function graphed on separate coordinate systems.

(a)

x

f (x)

– –1

1

1

–

2

–

3

–

4

3 24

5

3 421

(b)

x

f (x)

– –1

1

1

–

2

–

3

–

4

3 24

5

3 421

(c)

x

f (x)

– –1

1

1

–

2

–

3

–

4

3 24

5

3 421

Figure 25 ( a) f (x) = x2 if x ≤ 1; ( b) f (x) = 3 if 1 < x ≤ 2; ( c) f (x) = x if x > 2

CHAPTER 1 Functions3 4

Now that we have sketched each piece individually, we combine them in the same coordinate plane. See Figure 26.

x

f (x)

– –1

1

1

–

2

–

3

–

4

3 24

5

3 421

Figure 26

Analysi s Note that the graph does pass the vertical line test even at x = 1 and x = 2 because the points (1, 3) and (2, 2) are not part of the graph of the function, though (1, 1) and (2, 3) are.

Try It #8

Graph the following piecewise function.

f (x) = √

— x

x3 if x < −1 −2 if −1 < x < 4 if x > 4 {

Q & A… Can more than one formula from a piecewise function be applied to a value in the domain?

No. Each value corresponds to one equation in a piecewise formula.

Access these online resources for additional instruction and practice with domain and range.

• Domain and Range of Square Root Functions (http://openstaxcollege.org/l/domainsqroot)

• Determining Domain and Range (http://openstaxcollege.org/l/determinedomain)

• Find Domain and Range Given the Graph (http://openstaxcollege.org/l/drgraph)

• Find Domain and Range Given a Table (http://openstaxcollege.org/l/drtable)

• Find Domain and Range Given Points on a Coordinate Plane (http://openstaxcollege.org/l/drcoordinate)

SECTION 1.2 section exercises 3 5

1.2 SeCTIOn exeRCISeS

VeRBAl

1. Why does the domain differ for different functions? 2. How do we determine the domain of a function defined by an equation?

3. Explain why the domain of f (x) = 3 √ — x is different

from the domain of f (x) = √ — x .

4. When describing sets of numbers using interval notation, when do you use a parenthesis and when do you use a bracket?

5. How do you graph a piecewise function?

AlGeBRAIC

For the following exercises, find the domain of each function using interval notation.

6. f (x) = −2x(x − 1)(x − 2) 7. f (x) = 5 − 2x2 8. f (x) = 3 √ —

x − 2

9. f (x) = 3 − √ —

6 − 2x 10. f (x) = √ —

4 − 3x 11. f (x) = √ —

x 2 + 4

12. f (x) =   3 √ —

1 − 2x 13. f (x) =   3 √ —

x − 1 14. f (x) =   9 _____ x − 6

15. f (x) =  3x + 1 ______ 4x + 2

16. f (x) =   √ —

x + 4 _______ x − 4

17. f (x) =   x − 3 __________

x2 + 9x − 22

18. f (x) =   1 ________ x2 − x − 6

19. f (x) =   2x 3 − 250 __________

x2 − 2x − 15 20. f (x) =  

5 _ √ —

x − 3

21. f (x) = 2x + 1

_ √ —

5 − x 22. f (x) = √

— x − 4 _

√ —

x − 6 23. f (x) = √

— x − 6 _

√ —

x − 4

24. f (x) =  x __ x

25. f (x) = x

2 − 9x _ x2 − 81

26. Find the domain of the function f (x) = √ —

2x 3 − 50x by: a. using algebra. b. graphing the function in the radicand and determining intervals on the x-axis for which the radicand is

nonnegative.

GRAPHICAl

For the following exercises, write the domain and range of each function using interval notation.

27.

4

x

y

–2 –2–4

–4

–6

–6

–8

–8

–10

–10

2

6

6

42 8

8

10

10

28.

4

x

y

–2 –2–4

–4

–6

–6

–8

–8

–10

–10

2

6

6

42 8

8

10

10

29.

2

x

y

–1 –1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

3 6 CHAPTER 1 Functions

30.

2

x

y

–1 –1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

31.

2

x

y

–1 –1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

32.

2

x

y

–1 –1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

33.

2

x

y

–1 –1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

34.

2

x

y

–1 –1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

35.

1 6

–6, – )(

1 6

6, )(

1 6

, )( –6–

1 6

, )( 6

2

x

y

–1 –1–2

–2

–3

–3

–4

–4

–5–6

–5 –6

1

3

3

21 4

4

5 6

5 6

36.

4

x

y

–2 –2–4

–4

–6

–6

–8

–8

–10

–10

2

6

6

42 8

8

10

10

37.

4

x

y

–2 –2–4

–4

–6

–6

–8

–8

–10

–10

2

6

6

42 8

8

10

10

For the following exercises, sketch a graph of the piecewise function. Write the domain in interval notation.

38. f (x) =  x + 1 if x < −2

−2x − 3 if x ≥ −2 { 39. f (x) =  2x − 1 if x < 1 1 + x if x ≥ 1 { 40. f (x) =  x + 1 if x < 0 x − 1 if x > 0 {

41. f (x) =  3 if x < 0 √

— x if x ≥ 0 { 42. f (x) =  x

2 if x < 0 1 − x if x > 0 { 43. f (x) =  x

2 if x < 0 x + 2 if x ≥ 0 {

44. f (x) =  x + 1 if x < 1 x3 if x ≥ 1 { 45. f (x) =  | x | if x < 2 1 if x ≥ 2 {

SECTION 1.2 section exercises 3 7

nUMeRIC For the following exercises, given each function f, evaluate f (−3), f (−2), f (−1), and f (0).

46. f (x) =  x + 1 if x < −2

−2x − 3 if x ≥ −2 { 47. f (x) =  1 if x ≤ −3 0 if x > −3 { 48. f (x) =  −2x 2 + 3 if x ≤ −1

5x − 7 if x > −1 {

For the following exercises, given each function f, evaluate f (−1), f (0), f (2), and f (4).

49. f (x) =  7x + 3 if x < 0 7x + 6 if x ≥ 0{ 50. f (x) =  x

2 − 2 if x < 2 4 + | x − 5 | if x ≥ 2{ 51. f (x) = 

5x if x < 0 3 if 0 ≤ x ≤ 3 x 2 if x > 3{

For the following exercises, write the domain for the piecewise function in interval notation.

52. f (x) =  x + 1 if x < −2

−2x − 3 if x ≥ −2 { 53. f (x) =  x 2 − 2 if x < 1

−x2 + 2 if x > 1 { 54. f (x) =  2x − 3 if x < 0 −3x2 if x ≥ 2 {

TeCHnOlOGY

55. Graph y = 1 __ x 2

on the viewing window [−0.5, −0.1] and [0.1, 0.5]. Determine the corresponding range for the viewing window. Show the graphs.

56. Graph y = 1 _ x on the viewing window [−0.5, −0.1] and [0.1, 0.5]. Determine the corresponding range for the viewing window. Show the graphs.

exTenSIOn 57. Suppose the range of a function f is [−5, 8]. What is the range of | f (x) |?

58. Create a function in which the range is all nonnegative real numbers.

59. Create a function in which the domain is x > 2.

ReAl-WORlD APPlICATIOnS

60. The height h of a projectile is a function of the time t it is in the air. The height in feet for t seconds is given by the function h(t) = −16t 2 + 96t. What is the domain of the function? What does the domain mean in the context of the problem?

61. The cost in dollars of making x items is given by the function C(x) = 10x + 500. a. The fixed cost is determined when zero items are

produced. Find the fixed cost for this item. b. What is the cost of making 25 items? c. Suppose the maximum cost allowed is $1500.

What are the domain and range of the cost function, C(x)?

CHAPTER 1 Functions3 8

1. 3 RATeS OF CHAnGe AnD BeHAVIOR OF GRAPHS

Gasoline costs have experienced some wild fluctuations over the last several decades. Table 1[5] lists the average cost, in dollars, of a gallon of gasoline for the years 2005–2012. The cost of gasoline can be considered as a function of year.

y 2005 2006 2007 2008 2009 2010 2011 2012 C(y) 2.31 2.62 2.84 3.30 2.41 2.84 3.58 3.68

Table 1

If we were interested only in how the gasoline prices changed between 2005 and 2012, we could compute that the cost per gallon had increased from $2.31 to $3.68, an increase of $1.37. While this is interesting, it might be more useful to look at how much the price changed per year. In this section, we will investigate changes such as these.

Finding the Average Rate of Change of a Function The price change per year is a rate of change because it describes how an output quantity changes relative to the change in the input quantity. We can see that the price of gasoline in Table 1 did not change by the same amount each year, so the rate of change was not constant. If we use only the beginning and ending data, we would be finding the average rate of change over the specified period of time. To find the average rate of change, we divide the change in the output value by the change in the input value.

Average rate of change = Change in output

__ Change in input

= ∆y

_ ∆x

= y2 − y1 _ x2 − x1

=   f (x2) − f (x1) __ x2− x1

The Greek letter ∆ (delta) signifies the change in a quantity; we read the ratio as “delta-y over delta-x” or “the change in y divided by the change in x.” Occasionally we write ∆ f instead of ∆y, which still represents the change in the function’s output value resulting from a change to its input value. It does not mean we are changing the function into some other function.

In our example, the gasoline price increased by $1.37 from 2005 to 2012. Over 7 years, the average rate of change was

∆y

_ ∆x

= $1.37 _ 7 years

≈ 0.196 dollars per year

On average, the price of gas increased by about 19.6 each year.

Other examples of rates of change include: • A population of rats increasing by 40 rats per week • A car traveling 68 miles per hour (distance traveled changes by 68 miles each hour as time passes) • A car driving 27 miles per gallon (distance traveled changes by 27 miles for each gallon) • The current through an electrical circuit increasing by 0.125 amperes for every volt of increased voltage • The amount of money in a college account decreasing by $4,000 per quarter

5 http://www.eia.gov/totalenergy/data/annual/showtext.cfm?t=ptb0524. Accessed 3/5/2014.

leARnInG OBjeCTIVeS

In this section, you will: • Find the average rate of change of a function. • Use a graph to determine where a function is increasing, decreasing, or constant. • Use a graph to locate local maxima and local minima. • Use a graph to locate the absolute maximum and absolute minimum.

SECTION 1.3 rates oF change and Behavior oF graphs 3 9

rate of change A rate of change describes how an output quantity changes relative to the change in the input quantity. The units on a rate of change are “output units per input units.”

The average rate of change between two input values is the total change of the function values (output values) divided by the change in the input values.

∆y

_ ∆x

= f (x2) − f (x1) __ x2 − x1

How To… Given the value of a function at different points, calculate the average rate of change of a function for the interval between two values x1 and x2.

1. Calculate the difference y2 − y1 = ∆y.

2. Calculate the difference x2 − x1 = ∆x.

3. Find the ratio ∆y

_ ∆x

.

Example 1 Computing an Average Rate of Change

Using the data in Table 1, find the average rate of change of the price of gasoline between 2007 and 2009.

Solution In 2007, the price of gasoline was $2.84. In 2009, the cost was $2.41. The average rate of change is

∆y ___ ∆x

= y2 − y1 _ x2 − x1

= $2.41 − $2.84 __ 2009 − 2007

= −$0.43 ______ 2 years

= −$0.22 per year

Analysi s Note that a decrease is expressed by a negative change or “negative increase.” A rate of change is negative when the output decreases as the input increases or when the output increases as the input decreases.

Try It #1

Using the data in Table 1 at the beginning of this section, find the average rate of change between 2005 and 2010.

Example 2 Computing Average Rate of Change from a Graph

Given the function g (t) shown in Figure 1, find the average rate of change on the interval [−1, 2].

2

t –1

–1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

g (t)

Figure 1

CHAPTER 1 Functions4 0

Solution At t = −1, Figure 2 shows g (−1) = 4. At t = 2, the graph shows g (2) = 1.

2

t –1

–1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

g (t)

∆g (t) = –3

∆t = 3

Figure 2

The horizontal change ∆ t = 3 is shown by the red arrow, and the vertical change ∆ g(t) = −3 is shown by the turquoise arrow. The output changes by –3 while the input changes by 3, giving an average rate of change of

1 − 4 _______ 2 − (−1)

= −3 ___ 3

= −1

Analysi s Note that the order we choose is very important. If, for example, we use y2 − y1 _ x1 − x2

, we will not get the correct

answer. Decide which point will be 1 and which point will be 2, and keep the coordinates fixed as ( x1 , y1 ) and ( x2 , y2 ).

Example 3 Computing Average Rate of Change from a Table

After picking up a friend who lives 10 miles away, Anna records her distance from home over time. The values are shown in Table 2. Find her average speed over the first 6 hours.

t (hours) 0 1 2 3 4 5 6 7 D(t) (miles) 10 55 90 153 214 240 292 300

Table 2

Solution Here, the average speed is the average rate of change. She traveled 282 miles in 6 hours, for an average speed of

292 − 10 _______ 6 − 0

= 282 ___ 6

= 47

The average speed is 47 miles per hour.

Analysi s Because the speed is not constant, the average speed depends on the interval chosen. For the interval [2, 3], the average speed is 63 miles per hour.

Example 4 Computing Average Rate of Change for a Function Expressed as a Formula

Compute the average rate of change of f (x) = x 2 −  1 __ x

on the interval [2, 4].

Solution We can start by computing the function values at each endpoint of the interval.

f (2) = 22 − 1 __ 2

f(4) = 42 − 1 __ 4

= 4 − 1 __ 2

= 16 − 1 __ 4

= 7 __ 2

= 63 __ 4

Now we compute the average rate of change.

SECTION 1.3 rates oF change and Behavior oF graphs 4 1

Average rate of change = f (4) − f (2)

_ 4 − 2

= 63 __ 4

− 7 __ 2

_

4 − 2

= 49 __ 4

_

2

= 49 __ 8

Try It #2 Find the average rate of change of f (x) = x − 2 √

— x on the interval [1, 9].

Example 5 Finding the Average Rate of Change of a Force

The electrostatic force F, measured in newtons, between two charged particles can be related to the distance between the particles d, in centimeters, by the formula F(d) = 2 _

d 2 . Find the average rate of change of force if the distance

between the particles is increased from 2 cm to 6 cm.

Solution We are computing the average rate of change of F (d) = 2 __ d 2

on the interval [2, 6].

Average rate of change = F(6) − F(2)

__ 6 − 2

= 2 _ 62

− 2 _ 22

_

6 − 2 Simplify.

= 2 __ 36

−   2 __ 4

_

4

= −  16 __

36 _

4 Combine numerator terms.

= −  1 __ 9

Simplify.

The average rate of change is −  1 __ 9

newton per centimeter.

Example 6 Finding an Average Rate of Change as an Expression

Find the average rate of change of g(t) = t2 + 3t + 1 on the interval [0, a]. The answer will be an expression involving a.

Solution We use the average rate of change formula.

Average rate of change = g(a) − g(0)

_ a − 0

Evaluate.

= (a2 + 3a + 1) − (02 + 3(0) + 1)

___ a − 0

Simplify.

= a 2 + 3a + 1 − 1 _____________

a Simplify and factor.

= a(a + 3) _______ a

Divide by the common factor a.

= a + 3

This result tells us the average rate of change in terms of a between t = 0 and any other point t = a. For example, on the interval [0, 5], the average rate of change would be 5 + 3 = 8.

CHAPTER 1 Functions4 2

Try It #3 Find the average rate of change of f (x) = x2 + 2x − 8 on the interval [5, a].

Using a Graph to Determine Where a Function is Increasing, Decreasing, or Constant As part of exploring how functions change, we can identify intervals over which the function is changing in specific ways. We say that a function is increasing on an interval if the function values increase as the input values increase within that interval. Similarly, a function is decreasing on an interval if the function values decrease as the input values increase over that interval. The average rate of change of an increasing function is positive, and the average rate of change of a decreasing function is negative. Figure 3 shows examples of increasing and decreasing intervals on a function.

8

x –1

–4 –2

–8

–3

–12

–4

–16

–5

–20

4

3

12

21 4

16

5

20

f (x)

f (b) > f (a) where b > a

f (b) < f (a) where b > a

f (b) > f (a) where b > a

Increasing

Increasing Decreasing

Figure 3 The function f(x) = x3 − 12x is increasing on (−∞, −2) ∪ (2, ∞) and is decreasing on (−2, 2).

While some functions are increasing (or decreasing) over their entire domain, many others are not. A value of the input where a function changes from increasing to decreasing (as we go from left to right, that is, as the input variable increases) is called a local maximum. If a function has more than one, we say it has local maxima. Similarly, a value of the input where a function changes from decreasing to increasing as the input variable increases is called a local minimum. The plural form is “local minima.” Together, local maxima and minima are called local extrema, or local extreme values, of the function. (The singular form is “extremum.”) Often, the term local is replaced by the term relative. In this text, we will use the term local.

Clearly, a function is neither increasing nor decreasing on an interval where it is constant. A function is also neither increasing nor decreasing at extrema. Note that we have to speak of local extrema, because any given local extremum as defined here is not necessarily the highest maximum or lowest minimum in the function’s entire domain.

For the function whose graph is shown in Figure 4, the local maximum is 16, and it occurs at x = −2. The local minimum is −16 and it occurs at x = 2.

8

x –1

–4 –2

–8

–3

–12

–4

–16

–5

–20

4

3

12

21 4

16

5

20

Local maximum = 16 occurs at x = –2

(–2, 16)

(2, –16) Local minimum = –16 occurs at x = 2

f (x)

Decreasing

Increasing

Increasing

Figure 4

SECTION 1.3 rates oF change and Behavior oF graphs 4 3

To locate the local maxima and minima from a graph, we need to observe the graph to determine where the graph attains its highest and lowest points, respectively, within an open interval. Like the summit of a roller coaster, the graph of a function is higher at a local maximum than at nearby points on both sides. The graph will also be lower at a local minimum than at neighboring points. Figure 5 illustrates these ideas for a local maximum.

f(x)

f(b)

x a b

Local maximum

Increasing function

Decreasing function

c Figure 5 Definition of a local maximum

These observations lead us to a formal definition of local extrema.

local minima and local maxima A function f is an increasing function on an open interval if f (b) > f (a) for every two input values a and b in the interval where b > a.

A function f is a decreasing function on an open interval if f (b) < f (a) for every two input values a and b in the interval where b > a.

A function f has a local maximum at a point b in an open interval (a, c) if f (b) ≥ f (x) for every point x (x does not equal b) in the interval. f has a local minimum at a point b in the interval (a, c) if f (b) ≤ f (x) for every point x (x does not equal both) in the interval.

Example 7 Finding Increasing and Decreasing Intervals on a Graph

Given the function p(t) in Figure 6, identify the intervals on which the function appears to be increasing.

t

2

–1 –1

–2

1

3

3

21 4

4

5 6

5

p

Figure 6

Solution We see that the function is not constant on any interval. The function is increasing where it slants upward as we move to the right and decreasing where it slants downward as we move to the right. The function appears to be increasing from t = 1 to t = 3 and from t = 4 on.

In interval notation, we would say the function appears to be increasing on the interval (1, 3) and the interval (4, ∞).

CHAPTER 1 Functions4 4

Analysi s Notice in this example that we used open intervals (intervals that do not include the endpoints), because the function is neither increasing nor decreasing at t = 1, t = 3, and t = 4. These points are the local extrema (two minima and a maximum).

Example 8 Finding Local Extrema from a Graph

Graph the function f (x) = 2 __ x

+ x __ 3

. Then use the graph to estimate the local extrema of the function and to determine the intervals on which the function is increasing.

Solution Using technology, we find that the graph of the function looks like that in Figure 7. It appears there is a low point, or local minimum, between x = 2 and x = 3, and a mirror-image high point, or local maximum, somewhere between x = −3 and x = −2.

2

x –1

–1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

f (x)

Figure 7

Analysi s Most graphing calculators and graphing utilities can estimate the location of maxima and minima. Figure 8 provides screen images from two different technologies, showing the estimate for the local maximum and minimum.

x Maximum X = –2.449491 Y = –1.632993

y

(b)(a)

2.4494898, 1.6329932

0

2

–2

2 4

4

6

6

Figure 8

Based on these estimates, the function is increasing on the interval (−∞, −2.449) and (2.449, ∞). Notice that, while we expect the extrema to be symmetric, the two different technologies agree only up to four decimals due to the differing approximation algorithms used by each. (The exact location of the extrema is at ± √

— 6 , but determining this requires

calculus.)

Try It #4

Graph the function f (x) = x3 − 6x2 − 15x + 20 to estimate the local extrema of the function. Use these to determine the intervals on which the function is increasing and decreasing.

SECTION 1.3 rates oF change and Behavior oF graphs 4 5

Example 9 Finding Local Maxima and Minima from a Graph

For the function f whose graph is shown in Figure 9, find all local maxima and minima.

4

x –1

–2–2

–4

–3

–6

–4

–8

–5

–10

2

3

6

21 4

8

5

10

y

f

Figure 9

Solution Observe the graph of f. The graph attains a local maximum at x = 1 because it is the highest point in an open interval around x = 1. The local maximum is the y-coordinate at x = 1, which is 2.

The graph attains a local minimum at x = −1 because it is the lowest point in an open interval around x = −1.

The local minimum is the y-coordinate at x = −1, which is −2.

Analyzing the Toolkit Functions for Increasing or Decreasing Intervals

We will now return to our toolkit functions and discuss their graphical behavior in Figure 10, Figure 11, and Figure 12.

Function Increasing/Decreasing Example

Constant Function

f(x) = c Neither increasing

nor decreasing

y

x

Identity Function

f(x) = x Increasing

y

x

Quadratic Function

f(x) = x2

Increasing on (0, ∞)

Decreasing on (−∞, 0)

Minimum at x = 0

y

x

Figure 10

CHAPTER 1 Functions4 6

Function Increasing/Decreasing Example

Cubic Function

f(x) = x3 Increasing

y

x

Reciprocal

f(x) = 1 __ x Decreasing

(−∞, 0) ∪ (0, ∞)

y

x

Reciprocal Squared

f(x) = 1 _ x2

Increasing on (−∞, 0) Decreasing on (0, ∞)

y

x

Figure 11

Function Increasing/Decreasing Example

Cube Root

f(x) = 3 √ — x

Increasing

y

x

Square Root

f(x) = √ — x

Increasing on (0, ∞)

y

x

Absolute Value

f(x) = ∣ x ∣ Increasing on (0, ∞)

Decreasing on (−∞, 0)

y

x

Figure 12

SECTION 1.3 rates oF change and Behavior oF graphs 4 7

Use A Graph to locate the Absolute Maximum and Absolute Minimum There is a difference between locating the highest and lowest points on a graph in a region around an open interval (locally) and locating the highest and lowest points on the graph for the entire domain. The y-coordinates (output) at the highest and lowest points are called the absolute maximum and absolute minimum, respectively.

To locate absolute maxima and minima from a graph, we need to observe the graph to determine where the graph attains it highest and lowest points on the domain of the function. See Figure 13.

2

x f

–1 –1

–2

–2

–3

–3

–4

–4

–5

–5

1

Absolute minimum is f (0) = −2

Absolute maximum is f (2) = 2

3

3

21 4

4

5

5

y

Figure 13

Not every function has an absolute maximum or minimum value. The toolkit function f (x) = x3 is one such function.

absolute maxima and minima The absolute maximum of f at x = c is f (c) where f (c) ≥ f (x) for all x in the domain of f.

The absolute minimum of f at x = d is f (d) where f (d) ≤ f (x) for all x in the domain of f.

Example 10 Finding Absolute Maxima and Minima from a Graph

For the function f shown in Figure 14, find all absolute maxima and minima.

8

x –1

–4–2

–8

–3

–12

–4

–16

–5

–20

4

3

12

21 4

16

5

20

y

f

Figure 14

Solution Observe the graph of f. The graph attains an absolute maximum in two locations, x = −2 and x = 2, because at these locations, the graph attains its highest point on the domain of the function. The absolute maximum is the y-coordinate at x = −2 and x = 2, which is 16.

The graph attains an absolute minimum at x = 3, because it is the lowest point on the domain of the function’s graph. The absolute minimum is the y-coordinate at x = 3, which is −10.

Access this online resource for additional instruction and practice with rates of change.

• Average Rate of Change (http://openstaxcollege.org/l/aroc)

CHAPTER 1 Functions4 8

1.3 SeCTIOn exeRCISeS

VeRBAl

1. Can the average rate of change of a function be constant?

2. If a function f is increasing on (a, b) and decreasing on (b, c), then what can be said about the local extremum of f on (a, c)?

3. How are the absolute maximum and minimum similar to and different from the local extrema?

4. How does the graph of the absolute value function compare to the graph of the quadratic function, y = x 2, in terms of increasing and decreasing intervals?

AlGeBRAIC For the following exercises, find the average rate of change of each function on the interval specified for real numbers b or h.

5. f (x) = 4x 2 − 7 on [1, b] 6. g (x) = 2x 2 − 9 on [4, b]

7. p(x) = 3x + 4 on [2, 2 + h] 8. k(x) = 4x − 2 on [3, 3 + h]

9. f (x) = 2x 2 + 1 on [x, x + h] 10. g(x) = 3x 2 − 2 on [x, x + h]

11. a(t) = 1 ____ t + 4

on [9, 9 + h] 12. b(x) = 1 _____ x + 3

on [1, 1 + h]

13. j(x) = 3x 3 on [1, 1 + h] 14. r(t) = 4t 3 on [2, 2 + h]

15. f (x + h) − f(x)

__ h

given f(x) = 2x2 − 3x on [x, x + h]

GRAPHICAl

For the following exercises, consider the graph of f shown in Figure 15.

16. Estimate the average rate of change from x = 1 to x = 4.

17. Estimate the average rate of change from x = 2 to x = 5.

For the following exercises, use the graph of each function to estimate the intervals on which the function is increasing or decreasing.

18.

2

x

y

–1 –1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5 19.

0

2

x

y

–1 –1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

20.

2

x

y

–1 –1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

Figure 15

x

y

1

5

6

7

3

2

– 7 8321 654

4

SECTION 1.3 section exercises 4 9

21.

2

x

y

–1 –1–2

–2

–3

–3

–4

–4

–5–6

–5 –6

1

3

3

21 4

4

5 6

5 6

For the following exercises, consider the graph shown in Figure 16.

22. Estimate the intervals where the function is increasing or decreasing.

23. Estimate the point(s) at which the graph of f has a local maximum or a local minimum.

40

x

y

–1 –20

–2

–40

–3

–60

–4

–80

–5

–100

20

3

60

21 4

80

5

100

Figure 16

For the following exercises, consider the graph in Figure 17.

24. If the complete graph of the function is shown, estimate the intervals where the function is increasing or decreasing.

25. If the complete graph of the function is shown, estimate the absolute maximum and absolute minimum.

100

x

y

–2 –50

–4

–100

–6

–150

–8

–200

–10

–250

50

6

150

42 8

200

10

250

Figure 17

nUMeRIC

26. Table 3 gives the annual sales (in millions of dollars) of a product from 1998 to 2006. What was the average rate of change of annual sales (a) between 2001 and 2002, and (b) between 2001 and 2004?

Year 1998 1999 2000 2001 2002 2003 2004 2005 2006 Sales (millions of dollars) 201 219 233 243 249 251 249 243 233

Table 3

27. Table 4 gives the population of a town (in thousands) from 2000 to 2008. What was the average rate of change of population (a) between 2002 and 2004, and (b) between 2002 and 2006?

Year 2000 2001 2002 2003 2004 2005 2006 2007 2008 Population (thousands) 87 84 83 80 77 76 78 81 85

Table 4

5 0 CHAPTER 1 Functions

For the following exercises, find the average rate of change of each function on the interval specified. 28. f (x) = x 2 on [1, 5] 29. h(x) = 5 − 2x 2 on [−2, 4]

30. q(x) = x3 on [−4, 2] 31. g (x) = 3x 3 − 1 on [−3, 3]

32. y = 1 _ x on [1, 3] 33. p(t) = (t 2 − 4)(t + 1) ____________

t 2 + 3 on [−3, 1]

34. k (t) = 6t2 + 4 __ t3

on [−1, 3]

TeCHnOlOGY For the following exercises, use a graphing utility to estimate the local extrema of each function and to estimate the intervals on which the function is increasing and decreasing.

35. f(x) = x 4 − 4x 3 + 5 36. h(x) = x 5 + 5x 4 + 10x 3 + 10x 2 − 1

37. g(t) = t √ —

t + 3 38. k(t) = 3 t 2 _ 3 − t

39. m(x) = x 4 + 2x 3 − 12x 2 − 10x + 4 40. n(x) = x 4 − 8x 3 + 18x 2 − 6x + 2

exTenSIOn 41. The graph of the function f is shown in Figure 18.

Maximum X = 1.3333324 Y = 5.1851852

Figure 18

Based on the calculator screen shot, the point (1.333, 5.185) is which of the following?

a. a relative (local) maximum of the function b. the vertex of the function c. the absolute maximum of the function d. a zero of the function

42. Let f (x) = 1 __ x

. Find a number c such that the average rate of change of the function f on the interval (1, c) is −  1 __

4

43. Let f(x) =  1 __ x

. Find the number b such that the average rate of change of f on the interval (2, b) is − 1 __

10 .

ReAl-WORlD APPlICATIOnS 44. At the start of a trip, the odometer on a car read

21,395. At the end of the trip, 13.5 hours later, the odometer read 22,125. Assume the scale on the odometer is in miles. What is the average speed the car traveled during this trip?

45. A driver of a car stopped at a gas station to fill up his gas tank. He looked at his watch, and the time read exactly 3:40 p.m. At this time, he started pumping gas into the tank. At exactly 3:44, the tank was full and he noticed that he had pumped 10.7 gallons. What is the average rate of flow of the gasoline into the gas tank?

46. Near the surface of the moon, the distance that an object falls is a function of time. It is given by d(t) = 2.6667t2, where t is in seconds and d(t) is in feet. If an object is dropped from a certain height, find the average velocity of the object from t = 1 to t = 2.

47. The graph in Figure 19 illustrates the decay of a radioactive substance over t days.

0

8

t

A

6

10

10

5 15 Time (days)A

m ou

nt (m

ill ig

ra m

s)

14 12

20

16

Figure 19

Use the graph to estimate the average decay rate from t = 5 to t = 15.

SECTION 1.4 composition oF Functions 5 1

1. 4 COMPOSITIOn OF FUnCTIOnS

Suppose we want to calculate how much it costs to heat a house on a particular day of the year. The cost to heat a house will depend on the average daily temperature, and in turn, the average daily temperature depends on the particular day of the year. Notice how we have just defined two relationships: The cost depends on the temperature, and the temperature depends on the day.

Using descriptive variables, we can notate these two functions. The function C(T) gives the cost C of heating a house for a given average daily temperature in T degrees Celsius. The function T(d) gives the average daily temperature on day d of the year. For any given day, Cost = C(T(d)) means that the cost depends on the temperature, which in turns depends on the day of the year. Thus, we can evaluate the cost function at the temperature T(d). For example, we could evaluate T(5) to determine the average daily temperature on the 5th day of the year. Then, we could evaluate the cost function at that temperature. We would write C(T(5)).

Cost for the temperature

Temperature on day 5

C(T(5))

By combining these two relationships into one function, we have performed function composition, which is the focus of this section.

Combining Functions Using Algebraic Operations Function composition is only one way to combine existing functions. Another way is to carry out the usual algebraic operations on functions, such as addition, subtraction, multiplication and division. We do this by performing the operations with the function outputs, defining the result as the output of our new function.

Suppose we need to add two columns of numbers that represent a husband and wife’s separate annual incomes over a period of years, with the result being their total household income. We want to do this for every year, adding only that year’s incomes and then collecting all the data in a new column. If w(y) is the wife’s income and h(y) is the husband’s income in year y, and we want T to represent the total income, then we can define a new function.

T(y) = h(y) + w(y)

If this holds true for every year, then we can focus on the relation between the functions without reference to a year and write

T = h + w

Just as for this sum of two functions, we can define difference, product, and ratio functions for any pair of functions that have the same kinds of inputs (not necessarily numbers) and also the same kinds of outputs (which do have to be numbers so that the usual operations of algebra can apply to them, and which also must have the same units or no units when we add and subtract). In this way, we can think of adding, subtracting, multiplying, and dividing functions.

leARnInG OBjeCTIVeS

In this section, you will: • Combine functions using algebraic operations. • Create a new function by composition of functions. • Evaluate composite functions. • Find the domain of a composite function. • Decompose a composite function into its component functions.

CHAPTER 1 Functions5 2

For two functions f (x) and g(x) with real number outputs, we define new functions f + g, f − g, fg, and f _ g by the

relations (f + g)(x) = f (x) + g(x)

(f − g)(x) = f (x) − g(x)

(fg)(x) = f (x)g(x)

  f _ g  (x) = f (x)

_ g(x)

Example 1 Performing Algebraic Operations on Functions

Find and simplify the functions (g − f )(x) and   g _

f  (x), given f (x) = x − 1 and g(x) = x 2 − 1. Are they the same

function?

Solution Begin by writing the general form, and then substitute the given functions.

(g − f )(x) = g(x) − f (x)

(g − f )(x) = x 2 − 1 − (x − 1)

(g − f )(x) = x 2 − x

(g − f )(x) = x(x − 1)

  g _

f  (x) =

g(x) _

f (x)

  g _

f  (x) = x

2 − 1 _ x − 1

where x ≠ 1

  g _

f  (x) = (x + 1)(x − 1) ____________ x − 1 where x ≠ 1

  g _

f  (x) = x + 1 where x ≠ 1

No, the functions are not the same.

Note: For   g _

f  (x), the condition x ≠ 1 is necessary because when x = 1, the denominator is equal to 0, which makes

the function undefined.

Try It #1

Find and simplify the functions ( fg)(x) and ( f − g)(x).

f (x) = x − 1 and g(x) = x 2 − 1 Are they the same function?

Create a Function by Composition of Functions Performing algebraic operations on functions combines them into a new function, but we can also create functions by composing functions. When we wanted to compute a heating cost from a day of the year, we created a new function that takes a day as input and yields a cost as output. The process of combining functions so that the output of one function becomes the input of another is known as a composition of functions. The resulting function is known as a composite function. We represent this combination by the following notation:

( f ∘ g)(x) = f (g(x))

SECTION 1.4 composition oF Functions 5 3

We read the left-hand side as “ f composed with g at x,” and the right-hand side as “ f of g of x.” The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol ∘ is called the composition operator. We use this operator mainly when we wish to emphasize the relationship between the functions themselves without referring to any particular input value. Composition is a binary operation that takes two functions and forms a new function, much as addition or multiplication takes two numbers and gives a new number. However, it is important not to confuse function composition with multiplication because, as we learned above, in most cases f (g(x)) ≠ f (x)g(x).

It is also important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside. In the equation above, the function g takes the input x first and yields an output g(x). Then the function f takes g(x) as an input and yields an output f (g(x)).

x is the input of g

g(x), the output of g is the input of f

(f ° g)(x) = f ( g(x))

In general, f ∘ g and g ∘ f are different functions. In other words, in many cases f ( g(x)) ≠ g( f (x)) for all x. We will also see that sometimes two functions can be composed only in one specific order.

For example, if f (x) = x2 and g(x) = x + 2, then

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

= (x + 2)2

= x2 + 4x + 4 but g( f (x)) = g(x2)

= x2 + 2

These expressions are not equal for all values of x, so the two functions are not equal. It is irrelevant that the expressions happen to be equal for the single input value x = −  1 _

2 .

Note that the range of the inside function (the first function to be evaluated) needs to be within the domain of the outside function. Less formally, the composition has to make sense in terms of inputs and outputs.

composition of functions When the output of one function is used as the input of another, we call the entire operation a composition of functions. For any input x and functions f and g, this action defines a composite function, which we write as f ∘ g such that

( f ∘ g)(x) = f (g(x))

The domain of the composite function f ∘ g is all x such that x is in the domain of g and g(x) is in the domain of f. It is important to realize that the product of functions fg is not the same as the function composition f (g(x)), because, in general, f (x)g(x) ≠ f (g(x)).

Example 2 Determining whether Composition of Functions is Commutative

Using the functions provided, find f (g(x)) and g( f (x)). Determine whether the composition of the functions is commutative.

f (x) = 2x + 1 g(x) = 3 − x

CHAPTER 1 Functions5 4

Solution Let’s begin by substituting g(x) into f (x).

f (g(x)) = 2(3 − x) + 1 = 6 − 2x + 1 = 7 − 2x

Now we can substitute f (x) into g(x). g( f (x)) = 3 − (2x + 1) = 3 − 2x − 1 = − 2x + 2 We find that g( f (x)) ≠ f (g(x)), so the operation of function composition is not commutative.

Example 3 Interpreting Composite Functions

The function c(s) gives the number of calories burned completing s sit-ups, and s(t) gives the number of sit-ups a person can complete in t minutes. Interpret c(s(3)).

Solution The inside expression in the composition is s(3). Because the input to the s-function is time, t = 3 represents 3 minutes, and s(3) is the number of sit-ups completed in 3 minutes.

Using s(3) as the input to the function c(s) gives us the number of calories burned during the number of sit-ups that can be completed in 3 minutes, or simply the number of calories burned in 3 minutes (by doing sit-ups).

Example 4 Investigating the Order of Function Composition

Suppose f (x) gives miles that can be driven in x hours and g(y) gives the gallons of gas used in driving y miles. Which of these expressions is meaningful: f (g(y)) or g( f (x))?

Solution The function y = f (x) is a function whose output is the number of miles driven corresponding to the number of hours driven.

number of miles = f (number of hours)

The function g(y) is a function whose output is the number of gallons used corresponding to the number of miles driven. This means:

number of gallons = g (number of miles)

The expression g(y) takes miles as the input and a number of gallons as the output. The function f (x) requires a number of hours as the input. Trying to input a number of gallons does not make sense. The expression f (g(y)) is meaningless.

The expression f (x) takes hours as input and a number of miles driven as the output. The function g(y) requires a number of miles as the input. Using f (x) (miles driven) as an input value for g(y), where gallons of gas depends on miles driven, does make sense. The expression g( f (x)) makes sense, and will yield the number of gallons of gas used, g, driving a certain number of miles, f (x), in x hours.

Q & A… Are there any situations where f (g(y)) and g( f (x)) would both be meaningful or useful expressions ?

Yes. For many pure mathematical functions, both compositions make sense, even though they usually produce different new functions. In real-world problems, functions whose inputs and outputs have the same units also may give compositions that are meaningful in either order.

Try It #2 The gravitational force on a planet a distance r from the sun is given by the function G(r). The acceleration of a planet subjected to any force F is given by the function a(F). Form a meaningful composition of these two functions, and explain what it means.

SECTION 1.4 composition oF Functions 5 5

evaluating Composite Functions Once we compose a new function from two existing functions, we need to be able to evaluate it for any input in its domain. We will do this with specific numerical inputs for functions expressed as tables, graphs, and formulas and with variables as inputs to functions expressed as formulas. In each case, we evaluate the inner function using the starting input and then use the inner function’s output as the input for the outer function.

Evaluating Composite Functions Using Tables

When working with functions given as tables, we read input and output values from the table entries and always work from the inside to the outside. We evaluate the inside function first and then use the output of the inside function as the input to the outside function.

Example 5 Using a Table to Evaluate a Composite Function

Using Table 1, evaluate f (g(3)) and g( f (3)).

x f (x) g(x) 1 6 3 2 8 5 3 3 2 4 1 7

Table 1

Solution To evaluate f (g(3)), we start from the inside with the input value 3. We then evaluate the inside expression g(3) using the table that defines the function g: g(3) = 2. We can then use that result as the input to the function f, so g(3) is replaced by 2 and we get f (2). Then, using the table that defines the function f, we find that f (2) = 8.

g(3) = 2

f (g(3)) = f (2) = 8

To evaluate g( f (3)), we first evaluate the inside expression f (3) using the first table: f (3) = 3. Then, using the table for g, we can evaluate

g( f (3)) = g(3) = 2

Table 2 shows the composite functions f ∘ g and g ∘ f as tables.

x g(x) f (g(x)) f (x) g( f (x)) 3 2 8 3 2

Table 2

Try It #3

Using Table 1, evaluate f (g(1)) and g(f (4)).

Evaluating Composite Functions Using Graphs

When we are given individual functions as graphs, the procedure for evaluating composite functions is similar to the process we use for evaluating tables. We read the input and output values, but this time, from the x- and y-axes of the graphs.

CHAPTER 1 Functions5 6

How To… Given a composite function and graphs of its individual functions, evaluate it using the information provided by the graphs.

1. Locate the given input to the inner function on the x-axis of its graph. 2. Read off the output of the inner function from the y-axis of its graph. 3. Locate the inner function output on the x-axis of the graph of the outer function. 4. Read the output of the outer function from the y-axis of its graph. This is the output of the composite function.

Example 6 Using a Graph to Evaluate a Composite Function

Using Figure 1, evaluate f (g(1)).

x

g(x)

(a)

321 4 5 6 7

2

–1 –2 –3 –4 –5

1

3 4 5 6 7

x

f(x)

(b)

321 4 5 6 7

2

–1 –2 –3 –4 –5

1

3 4 5 6 7

Figure 1

Solution To evaluate f (g(1)), we start with the inside evaluation. See Figure 2.

x

g(x)

321 4 5 6 7

2

–1 –2 –3 –4 –5

1

3 4 5 6 7

x

f (x)

321 4 5 6 7

2

–1 –2 –3 –4 –5

1

3 4 5 6 7

g(1) = 3 f (3) = 6

(1, 3)

(3, 6)

Figure 2

We evaluate g(1) using the graph of g(x), finding the input of 1 on the x-axis and finding the output value of the graph at that input. Here, g(1) = 3. We use this value as the input to the function f.

f (g(1)) = f (3)

We can then evaluate the composite function by looking to the graph of f (x), finding the input of 3 on the x-axis and reading the output value of the graph at this input. Here, f (3) = 6, so f ( g(1)) = 6.

SECTION 1.4 composition oF Functions 5 7

Analysis Figure 3 shows how we can mark the graphs with arrows to trace the path from the input value to the output value.

4

x

g(x)

–2 –2–4

–4

–6

–6

–8

–8

–10

–10

2

6

6

42 8

8

10

10

4

x

f (x)

–2 –2–4

–4

–6

–6

–8

–8

–10

–10

2

6

6

42 8

8

10

10

Figure 3

Try It #4

Using Figure 1, evaluate g(f (2)).

Evaluating Composite Functions Using Formulas

When evaluating a composite function where we have either created or been given formulas, the rule of working from the inside out remains the same. The input value to the outer function will be the output of the inner function, which may be a numerical value, a variable name, or a more complicated expression.

While we can compose the functions for each individual input value, it is sometimes helpful to find a single formula that will calculate the result of a composition f ( g(x)). To do this, we will extend our idea of function evaluation. Recall that, when we evaluate a function like f (t) = t2 − t, we substitute the value inside the parentheses into the formula wherever we see the input variable.

How To… Given a formula for a composite function, evaluate the function.

1. Evaluate the inside function using the input value or variable provided. 2. Use the resulting output as the input to the outside function.

Example 7 Evaluating a Composition of Functions Expressed as Formulas with a Numerical Input

Given f (t) = t2 − t and h(x) = 3x + 2, evaluate f (h(1)).

Solution Because the inside expression is h(1), we start by evaluating h(x) at 1.

h(1) = 3(1) + 2

h(1) = 5

Then f (h(1)) = f (5), so we evaluate f (t) at an input of 5.

f (h(1)) = f (5)

f (h(1)) = 52 − 5

f (h(1)) = 20

Analysi s It makes no difference what the input variables t and x were called in this problem because we evaluated for specific numerical values.

CHAPTER 1 Functions5 8

Try It #5

Given f (t) = t 2 − t and h(x) = 3x + 2, evaluate

a. h(f (2)) b. h(f (−2))

Finding the Domain of a Composite Function As we discussed previously, the domain of a composite function such as f ∘ g is dependent on the domain of g and the domain of f. It is important to know when we can apply a composite function and when we cannot, that is, to know the domain of a function such as f ∘ g. Let us assume we know the domains of the functions f and g separately. If we write the composite function for an input x as f (g(x)), we can see right away that x must be a member of the domain of g in order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However, we also see that g(x) must be a member of the domain of f, otherwise the second function evaluation in f (g(x)) cannot be completed, and the expression is still undefined. Thus the domain of f ∘ g consists of only those inputs in the domain of g that produce outputs from g belonging to the domain of f. Note that the domain of f composed with g is the set of all x such that x is in the domain of g and g(x) is in the domain of f.

domain of a composite function The domain of a composite function f (g(x)) is the set of those inputs x in the domain of g for which g(x) is in the domain of f.

How To… Given a function composition f (g(x)), determine its domain.

1. Find the domain of g. 2. Find the domain of f. 3. Find those inputs x in the domain of g for which g(x) is in the domain of f. That is, exclude those inputs x from the

domain of g for which g(x) is not in the domain of f. The resulting set is the domain of f ∘ g.

Example 8 Finding the Domain of a Composite Function

Find the domain of

( f ∘ g)(x) where f (x) = 5 ____

x − 1 and g(x) = 4 _____

3x − 2

Solution The domain of g(x) consists of all real numbers except x = 2 __ 3

, since that input value would cause us to divide by 0. Likewise, the domain of f consists of all real numbers except 1. So we need to exclude from the domain of g(x) that value of x for which g(x) = 1.

4 _____ 3x − 2

= 1

4 = 3x − 2 6 = 3x x = 2

So the domain of f ∘ g is the set of all real numbers except 2 __ 3

and 2. This means that

x ≠ 2 __ 3

or x ≠ 2

We can write this in interval notation as

 − ∞, 2 __ 3  ∪   2 __ 3

, 2  ∪ (2, ∞)

SECTION 1.4 composition oF Functions 5 9

Finding the Domain of a Composite Function Involving Radicals Find the domain of ( f ∘ g)(x) where f (x) = √

— x + 2 and g(x) = √

— 3 − x

Solution Because we cannot take the square root of a negative number, the domain of g is (−∞, 3]. Now we check the domain of the composite function

( f ∘ g)(x) = √ ___________

√ —

3 − x +2

For ( f ∘ g)(x) = √ ___________

√ —

3 − x +2 , √ —

3 − x +2 ≥ 0, since the radicand of a square root must be positive. Since the square roots are positive, √

— 3 − x ≥ 0, 3 − x ≥ 0, which gives a domain of (−∞, 3].

Analysi s This example shows that knowledge of the range of functions (specifically the inner function) can also be helpful in finding the domain of a composite function. It also shows that the domain of f ∘ g can contain values that are not in the domain of f, though they must be in the domain of g.

Try It #6

Find the domain of (f ∘ g)(x) where f (x) = 1 ____

x − 2 and g(x) = √

— x + 4

Decomposing a Composite Function into its Component Functions In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions. There may be more than one way to decompose a composite function, so we may choose the decomposition that appears to be most expedient.

Example 9 Decomposing a Function

Write f (x) = √ —

5 − x 2 as the composition of two functions.

Solution We are looking for two functions, g and h, so f (x) = g(h(x)). To do this, we look for a function inside a function in the formula for f (x). As one possibility, we might notice that the expression 5 − x2 is the inside of the square root. We could then decompose the function as

h(x) = 5 − x2 and g(x) = √ — x

We can check our answer by recomposing the functions.

g(h(x)) = g(5 − x2) = √ —

5 − x2

Try It #7

Write f (x) = 4 __ 3 − √

— 4 + x2

as the composition of two functions.

Access these online resources for additional instruction and practice with composite functions.

• Composite Functions (http://openstaxcollege.org/l/compfunction)

• Composite Function notation Application (http://openstaxcollege.org/l/compfuncnot)

• Composite Functions Using Graphs (http://openstaxcollege.org/l/compfuncgraph)

• Decompose Functions (http://openstaxcollege.org/l/decompfunction)

• Composite Function Values (http://openstaxcollege.org/l/compfuncvalue)

6 0 CHAPTER 1 Functions

1.4 SeCTIOn exeRCISeS

VeRBAl

1. How does one find the domain of the quotient of

two functions, f _ g ?

2. What is the composition of two functions, f ∘ g ?

3. If the order is reversed when composing two functions, can the result ever be the same as the answer in the original order of the composition ? If yes, give an example. If no, explain why not.

4. How do you find the domain for the composition of two functions, f ∘ g ?

AlGeBRAIC

5. Given f (x) = x 2 + 2x and g(x) = 6 − x 2, find f + g,

f − g, fg, and f _ g . Determine the domain for each

function in interval notation.

6. Given f (x) = −3x2 + x and g(x) = 5, find f + g, f − g,

fg, and f _ g . Determine the domain for each function

in interval notation.

7. Given f (x) = 2x 2 + 4x and g(x) = 1 _ 2x

, find f + g,

f − g, fg, and f _ g . Determine the domain for each

function in interval notation.

8. Given f (x) = 1 _ x − 4

and g(x) = 1 _ 6 − x

, find

f + g, f − g, fg, and f _ g . Determine the domain for each

function in interval notation.

9. Given f (x) = 3x 2 and g(x) = √ —

x − 5 , find f + g,

f − g, fg, and f _ g . Determine the domain for each

function in interval notation.

10. Given f (x) = √ — x and g(x) = |x − 3|, find

g _

f .

Determine the domain for each function in interval notation.

11. Given f (x) = 2x 2 + 1 and g(x) = 3x − 5, find the following: a. f ( g(2)) b. f ( g(x)) c. g( f (x)) d. ( g ∘ g)(x) e. ( f ∘ f )(−2)

For the following exercises, use each pair of functions to find f (g(x)) and g( f (x)). Simplify your answers.

12. f (x) = x 2 + 1, g(x) = √ —

x + 2 13. f (x) = √ — x + 2, g(x) = x 2 + 3

14. f (x) = |x|, g(x) = 5x + 1 15. f (x) = 3 √— x , g(x) = x + 1 _

x3

16. f (x) = 1 _ x − 6

, g(x) = 7 _ x + 6 17. f (x) = 1 ___

x−4 , g(x) =  2 _ x + 4

For the following exercises, use each set of functions to find f (g(h(x))). Simplify your answers.

18. f (x) = x 4 + 6, g(x) = x − 6, and h(x) = √ — x 19. f (x) = x

2 + 1, g(x) = 1 _ x , and h(x) = x + 3

20. Given f (x) = 1 _ x , and g(x) = x − 3, find the following: a. ( f ∘ g)(x) b. the domain of ( f ∘ g)(x) in interval notation c. ( g ∘ f )(x) d. the domain of ( g ∘ f )(x)

e.  f _ g  x

21. Given f (x) = √ —

2 − 4x and g(x) = −  3 _ x , find the following: a. ( g ∘ f )(x) b. the domain of ( g ∘ f )(x) in interval notation

SECTION 1.4 section exercises 6 1

22. Given the functions f (x) = 1 − x _ x and g(x) = 1 _

1 + x2 ,

find the following:

a. ( g ∘ f )(x) b. ( g ∘ f )(2)

23. Given functions p(x) =   1 _ √

— x and m(x) = x 2 − 4,

state the domain of each of the following functions using interval notation:

a. p(x)

_ m(x)

b. p(m(x)) c. m(p(x))

24. Given functions q(x) = 1 _ √

— x and h(x) = x 2 − 9, state

the domain of each of the following functions using interval notation.

a. q(x)

_ h(x)

b. q(h(x)) c. h(q(x))

25. For f (x) = 1 _ x and g(x) = √ —

x − 1 , write the domain

of ( f ∘ g)(x) in interval notation.

For the following exercises, find functions f (x) and g(x) so the given function can be expressed as h(x) = f (g(x)).

26. h(x) = (x + 2)2 27. h(x) = (x − 5)3 28. h(x) = 3 _ x − 5

29. h(x) = 4 _______ (x + 2)2

30. h(x) = 4 + 3 √ — x 31. h(x) =

3 √

_______

1 ______ 2x − 3

32. h(x) = 1 _

(3x 2 − 4)−3 33. h(x) =

4 √

_______

3x − 2 ______ x + 5

34. h(x) =  8 + x 3 _ 8 − x 3  4

35. h(x) = √ —

2x + 6 36. h(x) = (5x − 1) 3

37. h(x) = 3 √ —

x − 1

38. h(x) = |x 2 + 7| 39. h(x) = 1 _ (x − 2)3

40. h(x) =  1 _ 2x − 3  2

41. h(x) = √ _______

2x − 1 _ 3x + 4

GRAPHICAl

For the following exercises, use the graphs of f, shown in Figure 4, and g, shown in Figure 5, to evaluate the expressions.

f(x)

x –1

1

1–

5

6

3

2

– 7321 654

4 f

Figure 4

f(x)

x –1

1

1–

5

6

3

2

– 7321 654

4 g

Figure 5

42. f ( g(3)) 43. f ( g(1)) 44. g( f (1)) 45. g( f (0))

46. f ( f (5)) 47. f ( f (4)) 48. g( g(2)) 49. g( g(0))

6 2 CHAPTER 1 Functions

For the following exercises, use graphs of f (x), shown in Figure 6, g(x), shown in Figure 7, and h(x), shown in Figure 8, to evaluate the expressions.

f (x)

x

f (x)

– –1

1

1

–

2

–

3

–

4

3 24

5

3 421

Figure 6

g(x)

x

f (x)

– –1

1

1

–

2

–

3

–

4

3 24

5

3 421

Figure 7

h(x)

x

f (x)

– –1

1

1

–

2

–

3

–

4

3 24

5

3 421

Figure 8

50. g( f (1)) 51. g( f (2)) 52. f ( g(4)) 53. f ( g(1))

54. f (h(2)) 55. h( f (2)) 56. f ( g(h(4))) 57. f ( g( f (−2)))

nUMeRIC

For the following exercises, use the function values for f and g shown in Table 3 to evaluate each expression.

x 0 1 2 3 4 5 6 7 8 9 f (x) 7 6 5 8 4 0 2 1 9 3 g(x) 9 5 6 2 1 8 7 3 4 0

Table 3

58. f ( g(8)) 59. f ( g(5)) 60. g( f (5)) 61. g( f (3)) 62. f ( f (4)) 63. f ( f (1)) 64. g( g(2)) 65. g( g(6))

For the following exercises, use the function values for f and g shown in Table 4 to evaluate the expressions.

x −3 −2 −1 0 1 2 3 f (x) 11 9 7 5 3 1 −1 g(x) −8 −3 0 1 0 −3 −8

Table 4

66. ( f ∘ g)(1) 67. ( f ∘ g)(2) 68. ( g ∘ f )(2) 69. ( g ∘ f )(3) 70. ( g ∘ g )(1) 71. ( f ∘ f )(3)

For the following exercises, use each pair of functions to find f (g(0)) and g( f (0)).

72. f (x) = 4x + 8, g(x) = 7 − x2 73. f (x) = 5x + 7, g(x) = 4 − 2x2

74. f (x) = √ —

x + 4 , g(x) = 12 − x3 75. f (x) = 1 _ x + 2

, g(x) = 4x + 3

For the following exercises, use the functions f (x) = 2x2 + 1 and g(x) = 3x + 5 to evaluate or find the composite function as indicated.

76. f ( g(2)) 77. f ( g(x)) 78. g( f ( − 3)) 79. ( g ∘ g )(x)

SECTION 1.4 section exercises 6 3

exTenSIOnS

For the following exercises, use f (x) = x3 + 1 and g(x) = 3 √ —

x − 1 .

80. Find ( f ∘ g)(x) and ( g ∘ f )(x). Compare the two answers. 81. Find ( f ∘ g)(2) and ( g ∘ f )(2).

82. What is the domain of ( g ∘ f )(x) ? 83. What is the domain of ( f ∘ g)(x) ?

84. Let f (x) = 1 __ x

.

a. Find ( f ∘ f )(x). b. Is ( f ∘ f )(x) for any function f the same result as the answer to part (a) for any function ? Explain.

For the following exercises, let F (x) = (x + 1)5, f (x) = x5, and g(x) = x + 1.

85. True or False: ( g ∘ f )(x) = F (x). 86. True or False: ( f ∘ g )(x) = F (x).

For the following exercises, find the composition when f (x) = x2 + 2 for all x ≥ 0 and g(x) = √ —

x − 2 .

87. ( f ∘ g)(6) ; ( g ∘ f )(6) 88. ( g ∘ f )(a) ; ( f ∘ g )(a) 89. ( f ∘ g )(11) ; (g ∘ f )(11)

ReAl-WORlD APPlICATIOnS 90. The function D(p) gives the number of items

that will be demanded when the price is p. The production cost C(x) is the cost of producing x items. To determine the cost of production when the price is $6, you would do which of the following ? a. Evaluate D(C(6)). b. Evaluate C(D(6)). c. Solve D(C(x)) = 6. d. Solve C(D(p)) = 6.

91. The function A(d) gives the pain level on a scale of 0 to 10 experienced by a patient with d milligrams of a pain- reducing drug in her system. The milligrams of the drug in the patient’s system after t minutes is modeled by m(t). Which of the following would you do in order to determine when the patient will be at a pain level of 4 ? a. Evaluate A(m(4)). b. Evaluate m(A(4)). c. Solve A(m(t)) = 4. d. Solve m(A(d)) = 4.

92. A store offers customers a 30 % discount on the price x of selected items. Then, the store takes off an additional 15 % at the cash register. Write a price function P(x) that computes the final price of the item in terms of the original price x. (Hint: Use function composition to find your answer.)

93. A rain drop hitting a lake makes a circular ripple. If the radius, in inches, grows as a function of time in minutes according to r(t) = 25

√ —

t + 2 , find the area of the ripple as a function of time. Find the area of the ripple at t = 2.

94. A forest fire leaves behind an area of grass burned in an expanding circular pattern. If the radius of the circle of burning grass is increasing with time according to the formula r(t) = 2t + 1, express the area burned as a function of time, t (minutes).

95. Use the function you found in the previous exercise to find the total area burned after 5 minutes.

96. The radius r, in inches, of a spherical balloon is related to the volume, V, by r(V) =

3 √

___

   3V ___ 4π

. Air is pumped into the balloon, so the volume after t seconds is given by V(t) = 10 + 20t. a. Find the composite function r(V(t)). b. Find the exact time when the radius reaches

10 inches.

97. The number of bacteria in a refrigerated food product is given by

N(T) = 23T 2 − 56T + 1, 3 < T < 33,

where T is the temperature of the food. When the food is removed from the refrigerator, the temperature is given by T(t) = 5t + 1.5, where t is the time in hours. a. Find the composite function N(T(t)). b. Find the time (round to two decimal places)

when the bacteria count reaches 6,752.

CHAPTER 1 Functions6 4

1. 5 TRAnSFORMATIOn OF FUnCTIOnS

Figure 1 (credit: "Misko"/Flickr)

We all know that a flat mirror enables us to see an accurate image of ourselves and whatever is behind us. When we tilt the mirror, the images we see may shift horizontally or vertically. But what happens when we bend a flexible mirror? Like a carnival funhouse mirror, it presents us with a distorted image of ourselves, stretched or compressed horizontally or vertically. In a similar way, we can distort or transform mathematical functions to better adapt them to describing objects or processes in the real world. In this section, we will take a look at several kinds of transformations.

Graphing Functions Using Vertical and Horizontal Shifts Often when given a problem, we try to model the scenario using mathematics in the form of words, tables, graphs, and equations. One method we can employ is to adapt the basic graphs of the toolkit functions to build new models for a given scenario. There are systematic ways to alter functions to construct appropriate models for the problems we are trying to solve.

Identifying Vertical Shifts

One simple kind of transformation involves shifting the entire graph of a function up, down, right, or left. The simplest shift is a vertical shift, moving the graph up or down, because this transformation involves adding a positive or negative constant to the function. In other words, we add the same constant to the output value of the function regardless of the input. For a function g (x) = f (x) + k, the function f (x) is shifted vertically k units. See Figure 2 for an example.

leARnInG OBjeCTIVeS

In this section, you will: • Graph functions using vertical and horizontal shifts. • Graph functions using reflections about the x-axis and the y-axis. • Determine whether a function is even, odd, or neither from its graph. • Graph functions using compressions and stretches. • Combine transformations.

SECTION 1.5 transFormation oF Functions 6 5

2

x –1–1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

f (x)+1

f (x)

f (x)

Figure 2 Vertical shift by k = 1 of the cube root function f ( x) = 3 √— x .

To help you visualize the concept of a vertical shift, consider that y = f (x). Therefore, f (x) + k is equivalent to y + k. Every unit of y is replaced by y + k, so the y-value increases or decreases depending on the value of k. The result is a shift upward or downward.

vertical shift Given a function f (x), a new function g(x) = f (x) + k, where k is a constant, is a vertical shift of the function f (x). All the output values change by k units. If k is positive, the graph will shift up. If k is negative, the graph will shift down.

Example 1 Adding a Constant to a Function

To regulate temperature in a green building, airflow vents near the roof open and close throughout the day. Figure 3 shows the area of open vents V (in square feet) throughout the day in hours after midnight, t. During the summer, the facilities manager decides to try to better regulate temperature by increasing the amount of open vents by 20 square feet throughout the day and night. Sketch a graph of this new function.

t

V

–4

50

50–

250

300

150

100

– 281284 242016

200

Figure 3

Solution We can sketch a graph of this new function by adding 20 to each of the output values of the original function. This will have the effect of shifting the graph vertically up, as shown in Figure 4.

t

V

–4

50

50–

250

300

150

100

– 281284 242016

200 240

20 Up 20

Figure 4

CHAPTER 1 Functions6 6

Notice that in Figure 4, for each input value, the output value has increased by 20, so if we call the new function S(t), we could write

S(t) = V(t) + 20

This notation tells us that, for any value of t, S(t) can be found by evaluating the function V at the same input and then adding 20 to the result. This defines S as a transformation of the function V, in this case a vertical shift up 20 units. Notice that, with a vertical shift, the input values stay the same and only the output values change. See Table 1.

t 0 8 10 17 19 24 V(t) 0 0 220 220 0 0 S(t) 20 20 240 240 20 20

Table 1

How To… Given a tabular function, create a new row to represent a vertical shift.

1. Identify the output row or column. 2. Determine the magnitude of the shift. 3. Add the shift to the value in each output cell. Add a positive value for up or a negative value for down.

Example 2 Shifting a Tabular Function Vertically

A function f (x) is given in Table 2. Create a table for the function g(x) = f (x) − 3.

x 2 4 6 8 f (x) 1 3 7 11

Table 2

Solution The formula g (x) = f (x) − 3 tells us that we can find the output values of g by subtracting 3 from the output values of f. For example: f (2) = 1 Given

g(x) = f (x) − 3 Given transformation

g(2) = f (2) − 3

= 1 − 3

= −2

Subtracting 3 from each f (x) value, we can complete a table of values for g(x) as shown in Table 3.

x 2 4 6 8 f (x) 1 3 7 11 g(x) −2 0 4 8

Table 3

Analysi s As with the earlier vertical shift, notice the input values stay the same and only the output values change.

Try It #1

The function h(t) = –4.9t2 + 30t gives the height h of a ball (in meters) thrown upward from the ground after t seconds. Suppose the ball was instead thrown from the top of a 10-m building. Relate this new height function b(t) to h(t), and then find a formula for b(t).

SECTION 1.5 transFormation oF Functions 6 7

Identifying Horizontal Shifts

We just saw that the vertical shift is a change to the output, or outside, of the function. We will now look at how changes to input, on the inside of the function, change its graph and meaning. A shift to the input results in a movement of the graph of the function left or right in what is known as a horizontal shift, shown in Figure 5.

2

x –1–1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

f (x+1) f (x)

f (x)

Figure 5 Horizontal shift of the function f (x) = 3 √ — x . note that h = +1 shifts the graph to the left, that is, towards negative values of x.

For example, if f (x) = x2, then g(x) = (x − 2)2 is a new function. Each input is reduced by 2 prior to squaring the function. The result is that the graph is shifted 2 units to the right, because we would need to increase the prior input by 2 units to yield the same output value as given in f.

horizontal shift Given a function f, a new function g(x) = f (x − h), where h is a constant, is a horizontal shift of the function f. If h is positive, the graph will shift right. If h is negative, the graph will shift left.

Example 3 Adding a Constant to an Input

Returning to our building airflow example from Figure 3, suppose that in autumn the facilities manager decides that the original venting plan starts too late, and wants to begin the entire venting program 2 hours earlier. Sketch a graph of the new function.

Solution We can set V(t) to be the original program and F (t) to be the revised program.

V(t) = the original venting plan

F(t) = starting 2 hrs sooner

In the new graph, at each time, the airflow is the same as the original function V was 2 hours later. For example, in the original function V, the airflow starts to change at 8 a.m., whereas for the function F, the airflow starts to change at 6 a.m. The comparable function values are V(8) = F(6). See Figure 6. Notice also that the vents first opened to 220 ft2 at 10 a.m. under the original plan, while under the new plan the vents reach 220 ft2 at 8 a.m., so V(10) = F(8).

In both cases, we see that, because F (t) starts 2 hours sooner, h = −2. That means that the same output values are reached when F (t) = V(t − (−2)) = V(t + 2).

t

V

0–4

50

50–

250

300

150

100

– 281284 242016

200 240

–2 Le� 2

Figure 6

CHAPTER 1 Functions6 8

Analysi s Note that V(t + 2) has the effect of shifting the graph to the left.

Horizontal changes or “ inside changes” affect the domain of a function (the input) instead of the range and often seem counterintuitive. The new function F(t) uses the same outputs as V(t), but matches those outputs to inputs 2 hours earlier than those of V(t). Said another way, we must add 2 hours to the input of V to find the corresponding output for F : F(t) = V(t + 2).

How To… Given a tabular function, create a new row to represent a horizontal shift.

1. Identify the input row or column. 2. Determine the magnitude of the shift. 3. Add the shift to the value in each input cell.

Example 4 Shifting a Tabular Function Horizontally

A function f (x) is given in Table 4. Create a table for the function g (x) = f (x − 3).

x 2 4 6 8 f (x) 1 3 7 11

Table 4

Solution The formula g(x) = f (x − 3) tells us that the output values of g are the same as the output value of f when the input value is 3 less than the original value. For example, we know that f (2) = 1. To get the same output from the function g, we will need an input value that is 3 larger. We input a value that is 3 larger for g (x) because the function takes 3 away before evaluating the function f. g (5) = f (5 − 3) = f (2) = 1

We continue with the other values to create Table 5.

x 5 7 9 11 x − 3 2 4 6 8

f (x − 3) 1 3 7 11 g(x) 1 3 7 11

Table 5

The result is that the function g (x) has been shifted to the right by 3. Notice the output values for g (x) remain the same as the output values for f (x), but the corresponding input values, x, have shifted to the right by 3. Specifically, 2 shifted to 5, 4 shifted to 7, 6 shifted to 9, and 8 shifted to 11.

Analysi s Figure 7 represents both of the functions. We can see the horizontal shift in each point.

4

x

y

–2–2–4

–4

–6

–6

–8

–8

–10–12

–10 –12

2

6

6

42 8

8

10 12

10 12

f (x) g (x) = f(x – 3) Figure 7

SECTION 1.5 transFormation oF Functions 6 9

Example 5 Identifying a Horizontal Shift of a Toolkit Function

Figure 8 represents a transformation of the toolkit function f (x) = x2. Relate this new function g (x) to f (x), and then find a formula for g (x).

6

f (x)

x –1

1

1–

5

3

2

– 7321 654

4

Figure 8

Solution Notice that the graph is identical in shape to the f (x) = x2 function, but the x-values are shifted to the right 2 units. The vertex used to be at (0,0), but now the vertex is at (2,0). The graph is the basic quadratic function shifted 2 units to the right, so g (x) = f (x − 2)

Notice how we must input the value x = 2 to get the output value y = 0; the x-values must be 2 units larger because of the shift to the right by 2 units. We can then use the definition of the f (x) function to write a formula for g (x) by evaluating f (x − 2). f (x) = x2

g (x) = f (x − 2) g (x) = f (x − 2) = (x − 2)2

Ana lysi s To determine whether the shift is +2 or −2, consider a single reference point on the graph. For a quadratic, looking at the vertex point is convenient. In the original function, f (0) = 0. In our shifted function, g (2) = 0. To obtain the output value of 0 from the function f, we need to decide whether a plus or a minus sign will work to satisfy g (2) = f (x − 2) = f (0) = 0. For this to work, we will need to subtract 2 units from our input values.

Example 6 Interpreting Horizontal versus Vertical Shifts

The function G (m) gives the number of gallons of gas required to drive m miles. Interpret G (m) + 10 and G (m + 10).

Solution G (m) + 10 can be interpreted as adding 10 to the output, gallons. This is the gas required to drive m miles, plus another 10 gallons of gas. The graph would indicate a vertical shift.

G (m + 10) can be interpreted as adding 10 to the input, miles. So this is the number of gallons of gas required to drive 10 miles more than m miles. The graph would indicate a horizontal shift.

Try It #2

Given the function f (x) = √ — x , graph the original function f (x) and the transformation g (x) = f (x + 2) on the same axes.

Is this a horizontal or a vertical shift? Which way is the graph shifted and by how many units?

Combining Vertical and Horizontal Shifts

Now that we have two transformations, we can combine them together. Vertical shifts are outside changes that affect the output ( y-) axis values and shift the function up or down. Horizontal shifts are inside changes that affect the input (x-) axis values and shift the function left or right. Combining the two types of shifts will cause the graph of a function to shift up or down and right or left.

CHAPTER 1 Functions7 0

How To… Given a function and both a vertical and a horizontal shift, sketch the graph. 1. Identify the vertical and horizontal shifts from the formula. 2. The vertical shift results from a constant added to the output. Move the graph up for a positive constant and down

for a negative constant. 3. The horizontal shift results from a constant added to the input. Move the graph left for a positive constant and right

for a negative constant. 4. Apply the shifts to the graph in either order.

Example 7 Graphing Combined Vertical and Horizontal Shifts

Given f (x) = ∣ x ∣, sketch a graph of h (x) = f (x + 1) − 3.

Solution The function f is our toolkit absolute value function. We know that this graph has a V shape, with the point at the origin. The graph of h has transformed f in two ways: f (x + 1) is a change on the inside of the function, giving a horizontal shift left by 1, and the subtraction by 3 in f (x + 1) − 3 is a change to the outside of the function, giving a vertical shift down by 3. The transformation of the graph is illustrated in Figure 9.

Let us follow one point of the graph of f (x) = ∣ x ∣.

• The point (0, 0) is transformed first by shifting left 1 unit: (0, 0) → (−1, 0)

• The point (−1, 0) is transformed next by shifting down 3 units: (−1, 0) → (−1, −3)

2

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

y = |x + 1|

y = |x + 1| – 3

y = |x|

Figure 9

Figure 10 shows the graph of h.

2

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

y = |x + 1| – 3

Figure 10

Try It #3

Given f (x) = ∣ x ∣, sketch a graph of h(x) = f (x − 2) + 4.

SECTION 1.5 transFormation oF Functions 7 1

Example 8 Identifying Combined Vertical and Horizontal Shifts

Write a formula for the graph shown in Figure 11, which is a transformation of the toolkit square root function.

h(x)

x –1

1

1–

5

6

3

2

– 7321 654

4

Figure 11

Solution The graph of the toolkit function starts at the origin, so this graph has been shifted 1 to the right and up 2. In function notation, we could write that as

h(x) = f (x − 1) + 2

Using the formula for the square root function, we can write

h(x) = √ —

x − 1 + 2

Analysi s Note that this transformation has changed the domain and range of the function. This new graph has domain [1, ∞) and range [2, ∞).

Try It #4

Write a formula for a transformation of the toolkit reciprocal function f (x) = 1 __ x

that shifts the function’s graph one unit to the right and one unit up.

Graphing Functions Using Reflections about the Axes Another transformation that can be applied to a function is a reflection over the x- or y-axis. A vertical reflection reflects a graph vertically across the x-axis, while a horizontal reflection reflects a graph horizontally across the y-axis. The reflections are shown in Figure 12.

Horizontal re�ection

Original function

Vertical re�ection

x

y

f ( x )

f (–x )

–f ( x )

Figure 12 Vertical and horizontal reflections of a function.

Notice that the vertical reflection produces a new graph that is a mirror image of the base or original graph about the x-axis. The horizontal reflection produces a new graph that is a mirror image of the base or original graph about the y-axis.

CHAPTER 1 Functions7 2

reflections

Given a function f (x), a new function g(x) = −f (x) is a vertical reflection of the function f (x), sometimes called a reflection about (or over, or through) the x-axis.

Given a function f (x), a new function g(x) = f (−x) is a horizontal reflection of the function f (x), sometimes called a reflection about the y-axis.

How To… Given a function, reflect the graph both vertically and horizontally.

1. Multiply all outputs by –1 for a vertical reflection. The new graph is a reflection of the original graph about the x-axis. 2. Multiply all inputs by –1 for a horizontal reflection. The new graph is a reflection of the original graph about the

y-axis.

Example 9 Reflecting a Graph Horizontally and Vertically

Reflect the graph of s(t) = √ — t a. vertically and b. horizontally.

Solution

a. Reflecting the graph vertically means that each output value will be reflected over the horizontal t-axis as shown in Figure 13.

2

t

s(t)

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

2

t

v(t)

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

Figure 13 Vertical reflection of the square root function

Because each output value is the opposite of the original output value, we can write

V(t) = −s(t) or V(t) = − √ — t

Notice that this is an outside change, or vertical shift, that affects the output s(t) values, so the negative sign belongs outside of the function.

b. Reflecting horizontally means that each input value will be reflected over the vertical axis as shown in Figure 14.

2

t

s(t)

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

2

t

H(t)

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

Figure 14 Horizontal reflection of the square root function

SECTION 1.5 transFormation oF Functions 7 3

Because each input value is the opposite of the original input value, we can write

H(t) = s(−t) or H(t) = √ —

−t

Notice that this is an inside change or horizontal change that affects the input values, so the negative sign is on the inside of the function.

Note that these transformations can affect the domain and range of the functions. While the original square root function has domain [0, ∞) and range [0, ∞), the vertical reflection gives the V(t) function the range (−∞, 0] and the horizontal reflection gives the H(t) function the domain (−∞, 0].

Try It #5

Reflect the graph of f (x) = |x − 1| a. vertically and b. horizontally.

Example 10 Reflecting a Tabular Function Horizontally and Vertically

A function f (x) is given as Table 6. Create a table for the functions below.

a. g(x) = −f (x) b. h(x) = f (−x)

x 2 4 6 8 f (x) 1 3 7 11

Table 6

Solution

a. For g(x), the negative sign outside the function indicates a vertical reflection, so the x-values stay the same and each output value will be the opposite of the original output value. See Table 7.

x 2 4 6 8 g (x) –1 –3 –7 –11

Table 7

b. For h(x), the negative sign inside the function indicates a horizontal reflection, so each input value will be the opposite of the original input value and the h(x) values stay the same as the f (x) values. See Table 8.

x –2 –4 –6 –8

h(x) 1 3 7 11

Table 8

Try It #6

A function f (x) is given as Table 9. Create a table for the functions below.

x −2 0 2 4

f(x) 5 10 15 20 Table 9

a. g(x) = −f (x)

b. h(x) = f (−x)

CHAPTER 1 Functions7 4

Example 11 Applying a Learning Model Equation

A common model for learning has an equation similar to k(t) = −2−t + 1, where k is the percentage of mastery that can be achieved after t practice sessions. This is a transformation of the function f (t) = 2t shown in Figure 15. Sketch a graph of k(t).

2

t

f (t)

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

Figure 15

Solution This equation combines three transformations into one equation.

• A horizontal reflection: f (−t) = 2−t

• A vertical reflection: −f (−t) = −2−t

• A vertical shift: −f (−t) + 1 = −2−t + 1

We can sketch a graph by applying these transformations one at a time to the original function. Let us follow two points through each of the three transformations. We will choose the points (0, 1) and (1, 2).

1. First, we apply a horizontal reflection: (0, 1) (−1, 2).

2. Then, we apply a vertical reflection: (0, −1) (1, −2).

3. Finally, we apply a vertical shift: (0, 0) (1, 1).

This means that the original points, (0,1) and (1,2) become (0,0) and (1,1) after we apply the transformations.

In Figure 16, the first graph results from a horizontal reflection. The second results from a vertical reflection. The third results from a vertical shift up 1 unit.

2

t

f (t)

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

2

(a) (b) (c)

1 4

4

5

5

2

t

f (t)

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

2

t

k(t)

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

Figure 16

Analysi s As a model for learning, this function would be limited to a domain of t ≥ 0, with corresponding range [0, 1).

Try It #7

Given the toolkit function f (x) = x 2, graph g(x) = −f (x) and h(x) = f (−x). Take note of any surprising behavior for these functions.

SECTION 1.5 transFormation oF Functions 7 5

Determining even and Odd Functions Some functions exhibit symmetry so that reflections result in the original graph. For example, horizontally reflecting the toolkit functions f (x) = x2 or f (x) = x will result in the original graph. We say that these types of graphs are symmetric about the y-axis. Functions whose graphs are symmetric about the y-axis are called even functions.

If the graphs of f (x) = x3 or f (x) = 1 _ x were reflected over both axes, the result would be the original graph, as shown in Figure 17.

2

x

y y y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

2

(a) (b) (c)

1 4

4

5

5

2

x –1–2–3–4–5 –1

–2 –3 –4 –5

1

3

3

21 4

4

5

5

2

x –1–2–3–4–5 –1

–2 –3 –4 –5

1

3

3

21 4

4

5

5

f (x) = x3 f (−x) −f (−x)

2

x

y y y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

2

(a) (b) (c)

1 4

4

5

5

2

x –1–2–3–4–5 –1

–2 –3 –4 –5

1

3

3

21 4

4

5

5

2

x –1–2–3–4–5 –1

–2 –3 –4 –5

1

3

3

21 4

4

5

5

f (x) = x3 f (−x) −f (−x)

Figure 17 (a) The cubic toolkit function (b) Horizontal reflection of the cubic toolkit function (c) Horizontal and vertical reflections reproduce the original cubic function.

We say that these graphs are symmetric about the origin. A function with a graph that is symmetric about the origin is called an odd function.

Note: A function can be neither even nor odd if it does not exhibit either symmetry. For example, f (x) = 2x is neither even nor odd. Also, the only function that is both even and odd is the constant function f (x) = 0.

even and odd functions A function is called an even function if for every input x: f (x) = f (−x)

The graph of an even function is symmetric about the y-axis. A function is called an odd function if for every input x: f (x) = −f (−x) The graph of an odd function is symmetric about the origin.

How To… Given the formula for a function, determine if the function is even, odd, or neither.

1. Determine whether the function satisfies f (x) = f (−x). If it does, it is even. 2. Determine whether the function satisfies f (x) = −f (−x). If it does, it is odd. 3. If the function does not satisfy either rule, it is neither even nor odd.

Example 12 Determining whether a Function Is Even, Odd, or Neither

Is the function f (x) = x 3 + 2x even, odd, or neither?

Solution Without looking at a graph, we can determine whether the function is even or odd by finding formulas for the reflections and determining if they return us to the original function. Let’s begin with the rule for even functions.

f (−x) = (−x)3 + 2(−x) = −x 3 − 2x This does not return us to the original function, so this function is not even. We can now test the rule for odd functions.

−f (−x) = −(−x 3 − 2x) = x 3 + 2x Because −f (−x) = f (x), this is an odd function.

CHAPTER 1 Functions7 6

Analysi s Consider the graph of f in Figure 18. Notice that the graph is symmetric about the origin. For every point (x, y) on the graph, the corresponding point (−x, −y) is also on the graph. For example, (1, 3) is on the graph of f, and the corresponding point (−1, −3) is also on the graph.

–1

2

x –1–2–3–4–5

–2 –3 –4 –5

1

3

3

21 4

4

5

5

(1, 3)

(−1, −3)

f (x) f

Figure 18

Try It #8

Is the function f (s) = s 4 + 3s 2 + 7 even, odd, or neither?

Graphing Functions Using Stretches and Compressions Adding a constant to the inputs or outputs of a function changed the position of a graph with respect to the axes, but it did not affect the shape of a graph. We now explore the effects of multiplying the inputs or outputs by some quantity.

We can transform the inside (input values) of a function or we can transform the outside (output values) of a function. Each change has a specific effect that can be seen graphically.

Vertical Stretches and Compressions

When we multiply a function by a positive constant, we get a function whose graph is stretched or compressed vertically in relation to the graph of the original function. If the constant is greater than 1, we get a vertical stretch; if the constant is between 0 and 1, we get a vertical compression. Figure 19 shows a function multiplied by constant factors 2 and 0.5 and the resulting vertical stretch and compression.

Vertical stretch

Vertical compression

x

y

2f(x) f(x) 0.5f(x)

Figure 19 Vertical stretch and compression

vertical stretches and compressions Given a function f (x), a new function g(x) = a f (x), where a is a constant, is a vertical stretch or vertical compression of the function f (x).

• If a > 1, then the graph will be stretched.

• If 0 < a < 1, then the graph will be compressed.

• If a < 0, then there will be combination of a vertical stretch or compression with a vertical reflection.

SECTION 1.5 transFormation oF Functions 7 7

How To… Given a function, graph its vertical stretch.

1. Identify the value of a.

2. Multiply all range values by a.

3. If a > 1, the graph is stretched by a factor of a. If 0 < a < 1, the graph is compressed by a factor of a. If a < 0, the graph is either stretched or compressed and also reflected about the x-axis.

Example 13 Graphing a Vertical Stretch

A function P(t) models the population of fruit flies. The graph is shown in Figure 20.

t

P(t)

–1

1

1–

5

6

3

2

– 7321 654

4

Figure 20

A scientist is comparing this population to another population, Q, whose growth follows the same pattern, but is twice as large. Sketch a graph of this population.

Solution Because the population is always twice as large, the new population’s output values are always twice the original function’s output values. Graphically, this is shown in Figure 21.

If we choose four reference points, (0, 1), (3, 3), (6, 2) and (7, 0) we will multiply all of the outputs by 2.

The following shows where the new points for the new graph will be located.

(0, 1) → (0, 2)

(3, 3) → (3, 6)

(6, 2) → (6, 4)

(7, 0) → (7, 0) t

Q(t)

–1

1

1–

5

6

3

2

– 7321 654

4

Figure 21

Symbolically, the relationship is written as

Q(t) = 2P(t)

This means that for any input t, the value of the function Q is twice the value of the function P. Notice that the effect on the graph is a vertical stretching of the graph, where every point doubles its distance from the horizontal axis. The input values, t, stay the same while the output values are twice as large as before.

CHAPTER 1 Functions7 8

How To… Given a tabular function and assuming that the transformation is a vertical stretch or compression, create a table for a vertical compression.

1. Determine the value of a. 2. Multiply all of the output values by a.

Example 14 Finding a Vertical Compression of a Tabular Function

A function f is given as Table 10. Create a table for the function g(x) =  1 __ 2

f (x).

x 2 4 6 8 f (x) 1 3 7 11

Table 10

Solution The formula g(x) =   1 __ 2

f (x) tells us that the output values of g are half of the output values of f with the same inputs. For example, we know that f (4) = 3. Then

g(4) =   1 __ 2

f (4) =   1 __ 2

(3) =   3 __ 2

We do the same for the other values to produce Table 11.

x 2 4 6 8

g(x) 1 __ 2 3 __ 2

7 __ 2 11 __ 2

Table 11

Analysi s The result is that the function g(x) has been compressed vertically by 1 __ 2

. Each output value is divided in half, so the graph is half the original height.

Try It #9

A function f is given as Table 12. Create a table for the function g(x) =   3 __ 4

f (x).

x 2 4 6 8 f(x) 12 16 20 0

Table 12

Example 15 Recognizing a Vertical Stretch

The graph in Figure 22 is a transformation of the toolkit function f (x) = x3. Relate this new function g(x) to f (x), and then find a formula for g(x).

2

x

g(x)

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

Figure 22

SECTION 1.5 transFormation oF Functions 7 9

Solution When trying to determine a vertical stretch or shift, it is helpful to look for a point on the graph that is relatively clear. In this graph, it appears that g(2) = 2. With the basic cubic function at the same input, f (2) = 23 = 8. Based on that, it appears that the outputs of g are 1 __

4 the outputs of the function f because g(2) =   1 __

4 f (2). From this we

can fairly safely conclude that g(x) =   1 __ 4

f (x).

We can write a formula for g by using the definition of the function f.

g(x) =   1 __ 4

f (x) =   1 __ 4

x3

Try It #10

Write the formula for the function that we get when we stretch the identity toolkit function by a factor of 3, and then shift it down by 2 units.

Horizontal Stretches and Compressions

Now we consider changes to the inside of a function. When we multiply a function’s input by a positive constant, we get a function whose graph is stretched or compressed horizontally in relation to the graph of the original function. If the constant is between 0 and 1, we get a horizontal stretch; if the constant is greater than 1, we get a horizontal compression of the function.

x

y

Horizontal stretch

Horizontal compression

–1–2–3–4–5 –1

1

2

3

4

5

6

7

8

9

10

321 4 5

y = (2x)2

y = (0.5x)2

y = x2

Figure 23

Given a function y = f (x), the form y = f (bx) results in a horizontal stretch or compression. Consider the function y = x2. Observe Figure 23. The graph of y = (0.5x)2 is a horizontal stretch of the graph of the function y = x 2 by a factor of 2. The graph of y = (2x)2 is a horizontal compression of the graph of the function y = x 2 by a factor of 2.

horizontal stretches and compressions Given a function f (x), a new function g(x) = f (bx), where b is a constant, is a horizontal stretch or horizontal compression of the function f (x).

• If b > 1, then the graph will be compressed by 1 __ b

.

• If 0 < b < 1, then the graph will be stretched by 1 __ b

.

• If b < 0, then there will be combination of a horizontal stretch or compression with a horizontal reflection.

How To… Given a description of a function, sketch a horizontal compression or stretch.

1. Write a formula to represent the function. 2. Set g(x) = f (bx) where b > 1 for a compression or 0 < b < 1 for a stretch.

CHAPTER 1 Functions8 0

Example 16 Graphing a Horizontal Compression

Suppose a scientist is comparing a population of fruit flies to a population that progresses through its lifespan twice as fast as the original population. In other words, this new population, R, will progress in 1 hour the same amount as the original population does in 2 hours, and in 2 hours, it will progress as much as the original population does in 4 hours. Sketch a graph of this population.

Solution Symbolically, we could write

R(1) = P(2),

R(2) = P(4), and in general,

R(t) = P(2t).

See Figure 24 for a graphical comparison of the original population and the compressed population.

Original population, P(t)

(a)

Transformed population, R(t)

(b)

f (x)

x –1

1

1–

5

6

3

2

– 7321 654

4

f (x)

x –1

1

1–

5

6

3

2

– 7321 654

4

Figure 24 (a) Original population graph (b) Compressed population graph

Analysi s Note that the effect on the graph is a horizontal compression where all input values are half of their original distance from the vertical axis.

Example 17 Finding a Horizontal Stretch for a Tabular Function

A function f (x) is given as Table 13. Create a table for the function g(x) = f   1 __ 2 x  . x 2 4 6 8

f (x) 1 3 7 11

Table 13

Solution The formula g(x) = f   1 __ 2 x  tells us that the output values for g are the same as the output values for the function f at an input half the size. Notice that we do not have enough information to determine g(2) because g(2) = f   1 __ 2 ⋅ 2  = f (1), and we do not have a value for f (1) in our table. Our input values to g will need to be twice as large to get inputs for f that we can evaluate. For example, we can determine g(4).

g(4) = f   1 __ 2 ⋅ 4  = f (2) = 1

We do the same for the other values to produce Table 14.

x 4 8 12 16 g(x) 1 3 7 11

Table 14

SECTION 1.5 transFormation oF Functions 8 1

Figure 25 shows the graphs of both of these sets of points.

4

x

y

–4–8–12–16–20 –2 –4 –6 –8

–10 –12

2

12

6

84 16

(a)

8

20

10 12

4

x

y

–4–8–12–16–20 –2 –4 –6 –8

–10 –12

2

12

6

84 16

(b)

8

20

10 12

Figure 25

Analysi s Because each input value has been doubled, the result is that the function g(x) has been stretched horizontally by a factor of 2.

Example 18 Recognizing a Horizontal Compression on a Graph

Relate the function g(x) to f (x) in Figure 26.

fg

f (x)

x –1

1

1–

5

6

3

2

– 7321 654

4

Figure 26

Solution The graph of g(x) looks like the graph of f (x) horizontally compressed. Because f (x) ends at (6, 4) and g(x) ends at (2, 4), we can see that the x-values have been compressed by 1 __

3 , because 6   1 __ 3  = 2. We might also notice that g(2) = f (6)

and g(1) = f (3). Either way, we can describe this relationship as g(x) = f (3x). This is a horizontal compression by 1 __ 3

.

Analysi s Notice that the coefficient needed for a horizontal stretch or compression is the reciprocal of the stretch or

compression. So to stretch the graph horizontally by a scale factor of 4, we need a coefficient of 1 __ 4

in our function: f   1 __ 4 x  . This means that the input values must be four times larger to produce the same result, requiring the input to be larger, causing the horizontal stretching.

Try It #11

Write a formula for the toolkit square root function horizontally stretched by a factor of 3.

CHAPTER 1 Functions8 2

Performing a Sequence of Transformations When combining transformations, it is very important to consider the order of the transformations. For example, vertically shifting by 3 and then vertically stretching by 2 does not create the same graph as vertically stretching by 2 and then vertically shifting by 3, because when we shift first, both the original function and the shift get stretched, while only the original function gets stretched when we stretch first.

When we see an expression such as 2f (x) + 3, which transformation should we start with? The answer here follows nicely from the order of operations. Given the output value of f (x), we first multiply by 2, causing the vertical stretch, and then add 3, causing the vertical shift. In other words, multiplication before addition.

Horizontal transformations are a little trickier to think about. When we write g(x) = f (2x + 3), for example, we have to think about how the inputs to the function g relate to the inputs to the function f . Suppose we know f (7) = 12. What input to g would produce that output? In other words, what value of x will allow g(x) = f (2x + 3) = 12? We would need 2x + 3 = 7. To solve for x, we would first subtract 3, resulting in a horizontal shift, and then divide by 2, causing a horizontal compression.

This format ends up being very difficult to work with, because it is usually much easier to horizontally stretch a graph before shifting. We can work around this by factoring inside the function.

f (bx + p) = f  b  x + p

_ b

  Let’s work through an example.

f (x) = (2x + 4)2 We can factor out a 2.

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

Now we can more clearly observe a horizontal shift to the left 2 units and a horizontal compression. Factoring in this way allows us to horizontally stretch first and then shift horizontally.

combining transformations When combining vertical transformations written in the form a f (x) + k, first vertically stretch by a and then vertically shift by k.

When combining horizontal transformations written in the form f (bx − h), first horizontally shift by h and then horizontally stretch by 1 __

b .

When combining horizontal transformations written in the form f (b(x − h)), first horizontally stretch by 1 _ b

and then horizontally shift by h.

Horizontal and vertical transformations are independent. It does not matter whether horizontal or vertical transformations are performed first.

Example 19 Finding a Triple Transformation of a Tabular Function

Given Table 15 for the function f (x), create a table of values for the function g(x) = 2f (3x) + 1.

x 6 12 18 24 f (x) 10 14 15 17

Table 15

Solution There are three steps to this transformation, and we will work from the inside out. Starting with the horizontal transformations, f (3x) is a horizontal compression by 1 __

3 , which means we multiply each x-value by 1 __

3 . See Table 16.

x 2 4 6 8 f (3x) 10 14 15 17

Table 16

SECTION 1.5 transFormation oF Functions 8 3

Looking now to the vertical transformations, we start with the vertical stretch, which will multiply the output values by 2. We apply this to the previous transformation. See Table 17.

x 2 4 6 8 2f (3x) 20 28 30 34

Table 17

Finally, we can apply the vertical shift, which will add 1 to all the output values. See Table 18.

x 2 4 6 8 g(x) = 2f (3x) + 1 21 29 31 35

Table 18

Example 20 Finding a Triple Transformation of a Graph

Use the graph of f (x) in Figure 27 to sketch a graph of k(x) = f   1 __ 2 x + 1  − 3.

2

x

f (x)

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

Figure 27

Solution To simplify, let’s start by factoring out the inside of the function.

f   1 __ 2 x + 1  − 3 = f   1 __ 2

(x + 2)  − 3 By factoring the inside, we can first horizontally stretch by 2, as indicated by the 1 __

2 on the inside of the function.

Remember that twice the size of 0 is still 0, so the point (0, 2) remains at (0, 2) while the point (2, 0) will stretch to (4, 0). See Figure 28.

2

x

f (x)

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

Figure 28

CHAPTER 1 Functions8 4

Next, we horizontally shift left by 2 units, as indicated by x + 2. See Figure 29.

2

x

f (x)

–1–2–3–4–5–6 –1 –2 –3 –4 –5

1

3

3

21 4

4

5 6

5

Figure 29

Last, we vertically shift down by 3 to complete our sketch, as indicated by the −3 on the outside of the function. See Figure 30.

2

x

f (x)

–1–2–3–4–5–6 –1 –2 –3 –4 –5

1

3

3

21 4

4

5 6

5

Figure 30

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

• Function Transformations (http://openstaxcollege.org/l/functrans)

SECTION 1.5 section exercises 8 5

1.5 SeCTIOn exeRCISeS

VeRBAl

1. When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal shift from a vertical shift?

2. When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal stretch from a vertical stretch?

3. When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal compression from a vertical compression?

4. When examining the formula of a function that is the result of multiple transformations, how can you tell a reflection with respect to the x-axis from a reflection with respect to the y-axis?

5. How can you determine whether a function is odd or even from the formula of the function?

AlGeBRAIC

6. Write a formula for the function obtained when the graph of f (x) = √

— x is shifted up 1 unit and to the

left 2 units.

7. Write a formula for the function obtained when the graph of f (x) = |x| is shifted down 3 units and to the right 1 unit.

8. Write a formula for the function obtained when the graph of f (x) =   1 _ x is shifted down 4 units and to the right 3 units.

9. Write a formula for the function obtained when the graph of f (x) = 1 _

x2 is shifted up 2 units and to the

left 4 units.

For the following exercises, describe how the graph of the function is a transformation of the graph of the original function f.

10. y = f (x − 49) 11. y = f (x + 43) 12. y = f (x + 3)

13. y = f (x − 4) 14. y = f (x) + 5 15. y = f (x) + 8

16. y = f (x) − 2 17. y = f (x) − 7 18. y = f (x − 2) + 3

19. y = f (x + 4) − 1

For the following exercises, determine the interval(s) on which the function is increasing and decreasing.

20. f (x) = 4(x + 1)2 − 5 21. g(x) = 5(x + 3)2 − 2 22. a(x) = √ —

−x + 4

23. k(x) = −3 √ — x − 1

f2

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

Figure 31

GRAPHICAl For the following exercises, use the graph of f (x) = 2x shown in Figure 31 to sketch a graph of each transformation of f (x).

24. g(x) = 2x + 1 25. h(x) = 2x − 3

26. w(x) = 2x − 1

For the following exercises, sketch a graph of the function as a transformation of the graph of one of the toolkit functions.

27. f (t) = (t + 1)2 − 3 28. h(x) = |x − 1| + 4

29. k(x) = (x − 2)3 − 1 30. m(t) = 3 + √ —

 t + 2

8 6 CHAPTER 1 Functions

nUMeRIC

31. Tabular representations for the functions f, g, and h are given below. Write g(x) and h(x) as transformations of f (x).

x –2 –1 0 1 2 f (x) –2 –1 –3 1 2

x –1 0 1 2 3 g(x) –2 –1 –3 1 2

x –2 –1 0 1 2 h(x) –1 0 –2 2 3

32. Tabular representations for the functions f, g, and h are given below. Write g(x) and h(x) as transformations of f (x).

x –2 –1 0 1 2 f (x) –1 –3 4 2 1

x –3 –2 –1 0 1 g(x) –1 –3 4 2 1

x –2 –1 0 1 2 h(x) –2 –4 3 1 0

For the following exercises, write an equation for each graphed function by using transformations of the graphs of one of the toolkit functions.

33.

f 2

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

34.

f

2

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

35.

f2

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

36.

4

x

f

y

–1–2–3–4–5 –2 –4 –6 –8

–10

2

3

6

21 4

8

5

10

37.

4

x

f

y

–1–2–3–4–5 –2 –4 –6 –8

–10

2

3

6

21 4

8

5

10

38.

2

x

f

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

SECTION 1.5 section exercises 8 7

39.

2

x

f

y

–1–2–3–4–5–6 –1 –2 –3 –4 –5 –6

1

3

3

21 4

4

5 6

5 6

40.

2

x f

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

For the following exercises, use the graphs of transformations of the square root function to find a formula for each of the functions.

41.

2

x

f

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

42.

2

x

f

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

For the following exercises, use the graphs of the transformed toolkit functions to write a formula for each of the resulting functions.

43.

2

x f

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

44.

2

x

f

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

45.

2

x

f

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

46.

2

x

f

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

8 8 CHAPTER 1 Functions

For the following exercises, determine whether the function is odd, even, or neither. 47. f (x) = 3x 4 48. g(x) = √

— x 49. h(x) = 1 __

x + 3x

50. f (x) = (x − 2)2 51. g(x) = 2x4 52. h(x) = 2x − x3

For the following exercises, describe how the graph of each function is a transformation of the graph of the original function f.

53. g(x) = −f (x) 54. g(x) = f (−x) 55. g(x) = 4f (x) 56. g(x) = 6f (x)

57. g(x) = f (5x) 58. g(x) = f (2x) 59. g(x) = f   1 __ 3 x  60. g(x) = f   1 __ 5

x  61. g(x) = 3f (−x) 62. g(x) = −f (3x)

For the following exercises, write a formula for the function g that results when the graph of a given toolkit function is transformed as described.

63. The graph of f (x) = ∣ x ∣ is reflected over the y-axis and horizontally compressed by a factor of 1 __

4 .

64. The graph of f (x) = √ — x is reflected over the x-axis

and horizontally stretched by a factor of 2.

65. The graph of f (x) = 1 __ x2

is vertically compressed by a factor of 1 __

3 , then shifted to the left 2 units and down

3 units.

66. The graph of f (x) = 1 __ x

is vertically stretched by a factor of 8, then shifted to the right 4 units and up 2 units.

67. The graph of f (x) = x2 is vertically compressed by a factor of 1 __

2 , then shifted to the right 5 units and up

1 unit.

68. The graph of f (x) = x2 is horizontally stretched by a factor of 3, then shifted to the left 4 units and down 3 units.

For the following exercises, describe how the formula is a transformation of a toolkit function. Then sketch a graph of the transformation.

69. g(x) = 4(x + 1)2 − 5 70. g(x) = 5(x + 3)2 − 2 71. h(x) = −2 ∣ x − 4 ∣ + 3

72. k(x) = −3 √ — x − 1 73. m(x) =   1 __

2 x3 74. n(x) =  1 __

3 |x − 2|

75. p(x) =      1 __ 3 x  3

− 3 76. q(x) =    1 __ 4 x  3

+ 1 77. a(x) = √ —

−x + 4

For the following exercises, use the graph in Figure 32 to sketch the given transformations.

4

x –2

–2–4

–4

–6

–6

–8

–8

–10

–10

2

6

6

42 8

8

10

10

y

f

Figure 32

78. g(x) = f (x) − 2 79. g(x) = −f (x) 80. g(x) = f (x + 1) 81. g(x) = f (x − 2)

SECTION 1.6 aBsolute value Functions 8 9

1. 6 ABSOlUTe VAlUe FUnCTIOnS

Figure 1 Distances in deep space can be measured in all directions. As such, it is useful to consider distance in terms of absolute values. (credit: “s58y”/Flickr)

Until the 1920s, the so-called spiral nebulae were believed to be clouds of dust and gas in our own galaxy, some tens of thousands of light years away. Then, astronomer Edwin Hubble proved that these objects are galaxies in their own right, at distances of millions of light years. Today, astronomers can detect galaxies that are billions of light years away. Distances in the universe can be measured in all directions. As such, it is useful to consider distance as an absolute value function. In this section, we will investigate absolute value functions.

Understanding Absolute Value Recall that in its basic form f (x) = | x |, the absolute value function, is one of our toolkit functions. The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign.

absolute value function The absolute value function can be defined as a piecewise function

f (x) = ∣ x ∣  = x if x ≥ 0 −x if x < 0{

Example 1 Determine a Number within a Prescribed Distance

Describe all values x within or including a distance of 4 from the number 5.

Solution We want the distance between x and 5 to be less than or equal to 4. We can draw a number line, such as the one in Figure 2, to represent the condition to be satisfied.

5

44

Figure 2

The distance from x to 5 can be represented using the absolute value as ∣ x − 5 ∣ . We want the values of x that satisfy the condition ∣ x − 5 ∣ ≤ 4.

Analysi s Note that −4 ≤ x − 5 x − 5 ≤ 4 1 ≤ x x ≤ 9 So ∣ x − 5 ∣ ≤ 4 is equivalent to 1 ≤ x ≤ 9. However, mathematicians generally prefer absolute value notation.

leARnInG OBjeCTIVeS

In this section you will: • Graph an absolute value function. • Solve an absolute value equation. • Solve an absolute value inequality.

CHAPTER 1 Functions9 0

Try It #1 Describe all values x within a distance of 3 from the number 2.

Example 2 Resistance of a Resistor

Electrical parts, such as resistors and capacitors, come with specified values of their operating parameters: resistance, capacitance, etc. However, due to imprecision in manufacturing, the actual values of these parameters vary somewhat from piece to piece, even when they are supposed to be the same. The best that manufacturers can do is to try to guarantee that the variations will stay within a specified range, often ±1%, ±5%, or ±10%.

Suppose we have a resistor rated at 680 ohms, ±5%. Use the absolute value function to express the range of possible values of the actual resistance.

Solution 5% of 680 ohms is 34 ohms. The absolute value of the difference between the actual and nominal resistance should not exceed the stated variability, so, with the resistance R in ohms,

| R − 680 | ≤ 34

Try It #2 Students who score within 20 points of 80 will pass a test. Write this as a distance from 80 using absolute value notation.

Graphing an Absolute Value Function The most significant feature of the absolute value graph is the corner point at which the graph changes direction. This point is shown at the origin in Figure 3.

2

x

y

–1–1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

y = |x|

5

5

Figure 3

Figure 4 shows the graph of y = 2| x − 3| + 4. The graph of y = | x| has been shifted right 3 units, vertically stretched by a factor of 2, and shifted up 4 units. This means that the corner point is located at (3, 4) for this transformed function.

1 4

6

x

y Vertical stretch

y = 2|x − 3|

Right 3 y = |x − 3|

y = |x|

Up 4 y = 2|x − 3| + 4

–1–1

3

–2

2

–3

1

–4–5–6

4 5

3

7

2

8 9

5 6

10

Figure 4

SECTION 1.6 aBsolute value Functions 9 1

Example 3 Writing an Equation for an Absolute Value Function

Write an equation for the function graphed in Figure 5.

2

x

y

–1–1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

Figure 5

Solution The basic absolute value function changes direction at the origin, so this graph has been shifted to the right 3 units and down 2 units from the basic toolkit function. See Figure 6.

2

x

y

–1 –1

–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

(3, –2)

Figure 6

We also notice that the graph appears vertically stretched, because the width of the final graph on a horizontal line is not equal to 2 times the vertical distance from the corner to this line, as it would be for an unstretched absolute value function. Instead, the width is equal to 1 times the vertical distance as shown in Figure 7.

2

x

y

Ratio 2/1

Ratio 1/1

–1–1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

(3, −2)

Figure 7

From this information we can write the equation

f (x) = 2| x − 3 | − 2, treating the stretch as a vertical stretch, or

f (x) = | 2(x − 3) | − 2, treating the stretch as a horizontal compression.

Analysi s Note that these equations are algebraically equivalent—the stretch for an absolute value function can be written interchangeably as a vertical or horizontal stretch or compression.

CHAPTER 1 Functions9 2

Q & A… If we couldn’t observe the stretch of the function from the graphs, could we algebraically determine it?

Yes. If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor by putting in a known pair of values for x and f (x).

f (x) = a| x − 3 | − 2 Now substituting in the point (1, 2) 2 = a| 1 − 3 | − 2 4 = 2a a = 2

Try It #3

Write the equation for the absolute value function that is horizontally shifted left 2 units, is vertically flipped, and verti- cally shifted up 3 units.

Q & A… Do the graphs of absolute value functions always intersect the vertical axis? The horizontal axis? Yes, they always intersect the vertical axis. The graph of an absolute value function will intersect the vertical axis when the input is zero.

No, they do not always intersect the horizontal axis. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. It is possible for the absolute value function to intersect the horizontal axis at zero, one, or two points (see Figure 8).

2

x

y

–1 –1

–2

–2

–3

–3

–4

–4

–5

1

3

3

21

(a) (b) (c)

4

4

5

5

2

x

y

–1–1–2

–2

–3

–3

–4

–4

–5

1

3

3

21 4

4

5

5

2

x

y

–1–1–2

–2

–3

–3

–4

–4

–5

1

3

3

21 4

4

5

5

Figure 8 (a) The absolute value function does not intersect the horizontal axis. (b) The absolute value function intersects the horizontal axis at one point. (c) The absolute value function intersects the horizontal axis at two points.

Solving an Absolute Value equation Now that we can graph an absolute value function, we will learn how to solve an absolute value equation. To solve an equation such as 8 = | 2x − 6| , we notice that the absolute value will be equal to 8 if the quantity inside the absolute value is 8 or −8. This leads to two different equations we can solve independently.

2x − 6 = 8 or 2x − 6 = −8 2x = 14 2x = −2 x = 7 x = −1 Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point. An absolute value equation is an equation in which the unknown variable appears in absolute value bars. For example,

| x | = 4, | 2x − 1 | = 3 | 5x + 2 | − 4 = 9

SECTION 1.6 aBsolute value Functions 9 3

solutions to absolute value equations For real numbers A and B, an equation of the form | A | = B, with B ≥ 0, will have solutions when A = B or A = −B. If B < 0, the equation | A | = B has no solution.

How To… Given the formula for an absolute value function, find the horizontal intercepts of its graph.

1. Isolate the absolute value term. 2. Use | A | = B to write A = B or −A = B, assuming B > 0. 3. Solve for x.

Example 4 Finding the Zeros of an Absolute Value Function

For the function f (x) = | 4x + 1 | − 7 , find the values of x such that f (x) = 0.

Solution 0 = | 4x + 1 | − 7 Substitute 0 for f (x). 7 = | 4x + 1 | Isolate the absolute value on one side of the equation. 7 = 4x + 1 or −7 = 4x + 1 Break into two separate equations and solve. 6 = 4x −8 = 4x

x = 6 __ 4

= 1.5 x = −8 ___ 4

= −2

The function outputs 0 when x = 1.5 or x = −2. See Figure 9.

4

x

y

f

–0.5 –2

–1

–4

–1.5

–6

–2

–8

–2.5

–10

2

1.5

6

10.5 2

8

2.5

10

(1.5, 0)

x where f(x) = 0

x where f (x) = 0

(−2, 0)

Figure 9

Try It #4

For the function f (x) = | 2x − 1 | − 3, find the values of x such that f (x) = 0.

Q & A… Should we always expect two answers when solving ∣ A ∣ = B?

No. We may find one, two, or even no answers. For example, there is no solution to 2 + | 3x − 5 | = 1.

How To… Given an absolute value equation, solve it.

1. Isolate the absolute value term. 2. Use | A | = B to write A = B or A = −B. 3. Solve for x.

CHAPTER 1 Functions9 4

Example 5 Solving an Absolute Value Equation

Solve 1 = 4| x − 2| + 2.

Solution Isolating the absolute value on one side of the equation gives the following.

1 = 4| x − 2 | + 2

−1 = 4| x − 2 |

−  1 _  4

= | x − 2 | 

The absolute value always returns a positive value, so it is impossible for the absolute value to equal a negative value. At this point, we notice that this equation has no solutions.

Q & A… In Example 5, if f(x) = 1 and g(x) = 4∣ x − 2 ∣ + 2 were graphed on the same set of axes, would the graphs intersect?

No. The graphs of f and g would not intersect, as shown in Figure 10. This confirms, graphically, that the equation 1 = 4| x − 2 | + 2 has no solution.

4

x

y

–1–2–2

–4

–3

–6

–4

–8

–5

–10

2

3

6

21 4

8

f (x) = 1

g(x) = 4|x − 2| + 2

5

10

Figure 10

Try It #5

Find where the graph of the function f (x) = −| x + 2 | + 3 intersects the horizontal and vertical axes.

Solving an Absolute Value Inequality Absolute value equations may not always involve equalities. Instead, we may need to solve an equation within a range of values. We would use an absolute value inequality to solve such an equation. An absolute value inequality is an equation of the form

| A | < B, | A | ≤ B, | A | ≥ B, or | A | ≥ B,

where an expression A (and possibly but not usually B) depends on a variable x. Solving the inequality means finding the set of all x that satisfy the inequality. Usually this set will be an interval or the union of two intervals.

There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two functions. The advantage of the algebraic approach is it yields solutions that may be difficult to read from the graph. For example, we know that all numbers within 200 units of 0 may be expressed as

| x | < 200 or −200 < x < 200 Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of $600. We can solve algebraically for the set of values x such that the distance between x and 600 is less than 200. We represent the distance between x and 600 as | x − 600 |.

| x − 600 | < 200 or −200 < x − 600 < 200 −200 + 600 < x − 600 + 600 < 200 + 600

400 < x < 800

SECTION 1.6 aBsolute value Functions 9 5

This means our returns would be between $400 and $800. Sometimes an absolute value inequality problem will be presented to us in terms of a shifted and/or stretched or compressed absolute value function, where we must determine for which values of the input the function’s output will be negative or positive.

How To… Given an absolute value inequality of the form | x − A | ≤ B for real numbers a and b where b is positive, solve the absolute value inequality algebraically.

1. Find boundary points by solving | x − A | = B. 2. Test intervals created by the boundary points to determine where | x − A | ≤ B. 3. Write the interval or union of intervals satisfying the inequality in interval, inequality, or set-builder notation.

Example 6 Solving an Absolute Value Inequality

Solve | x − 5| < 4.

Solution With both approaches, we will need to know first where the corresponding equality is true. In this case we first will find where | x − 5| = 4. We do this because the absolute value is a function with no breaks, so the only way the function values can switch from being less than 4 to being greater than 4 is by passing through where the values equal 4. Solve | x − 5| = 4.

x − 5 = 4 or x − 5 = −4 x = 9 x = 1 After determining that the absolute value is equal to 4 at x = 1 and x = 9, we know the graph can change only from being less than 4 to greater than 4 at these values. This divides the number line up into three intervals:

x < 1, 1 < x < 9, and x > 9.

To determine when the function is less than 4, we could choose a value in each interval and see if the output is less than or greater than 4, as shown in Table 1.

Interval test x f (x) < 4 or > 4?

x < 1 0 | 0 − 5| = 5 Greater than

1 < x < 9 6 | 6 − 5| = 1 Less than

x > 9 11 | 11 − 5| = 6 Greater than

Table 1

Because 1 ≤ x ≤ 9 is the only interval in which the output at the test value is less than 4, we can conclude that the solution to | x − 5| ≤ 4 is 1 ≤ x ≤ 9, or [1, 9].

To use a graph, we can sketch the function f (x) = | x − 5| . To help us see where the outputs are 4, the line g(x) = 4 could also be sketched as in Figure 11.

g

f

�e graph of f is below or on the graph of g

f (x)

x –1

1

1–

5

6

3

2

– 7 8 9 10321 654

4

Figure 11 Graph to find the points satisfying an absolute value inequality.

CHAPTER 1 Functions9 6

We can see the following:

• The output values of the absolute value are equal to 4 at x = 1 and x = 9.

• The graph of f is below the graph of g on 1 < x < 9. This means the output values of f (x) are less than the output values of g(x).

• The absolute value is less than or equal to 4 between these two points, when 1 ≤ x ≤ 9. In interval notation, this would be the interval [1, 9].

Analysi s For absolute value inequalities,

| x − A | < C, | x − A | > C,

−C < x − A < C, x − A < −C or x − A > C.

The < or > symbol may be replaced by ≤ or ≥.

So, for this example, we could use this alternative approach.

| x − 5| ≤ 4

−4 ≤ x − 5 ≤ 4 Rewrite by removing the absolute value bars.

−4 + 5 ≤ x − 5 + 5 ≤ 4 + 5 Isolate the x.

1 ≤ x ≤ 9

Try It #6

Solve | x + 2| ≤ 6.

How To… Given an absolute value function, solve for the set of inputs where the output is positive (or negative).

1. Set the function equal to zero, and solve for the boundary points of the solution set. 2. Use test points or a graph to determine where the function’s output is positive or negative.

Example 7 Using a Graphical Approach to Solve Absolute Value Inequalities

Given the function f (x) = − 1 __ 2

| 4x − 5 | + 3, determine the x-values for which the function values are negative.

Solution We are trying to determine where f (x) < 0, which is when −  1 __ 2

| 4x − 5| + 3 < 0. We begin by isolating the absolute value.

−  1 __ 2

| 4x − 5| < −3 Multiply both sides by −2, and reverse the inequality. | 4x − 5| > 6

Next we solve for the equality | 4x − 5| = 6.

4x − 5 = 6 or 4x − 5 = −6

4x − 5 = 6 4x = −1

x = 11 __ 4

x = − 1 __ 4

Now, we can examine the graph of f to observe where the output is negative. We will observe where the branches are below the x-axis. Notice that it is not even important exactly what the graph looks like, as long as we know that

it crosses the horizontal axis at x = − 1 __ 4

and x = 11 __ 4

and that the graph has been reflected vertically. See Figure 12.

SECTION 1.6 aBsolute value Functions 9 7

2

x

y

–1–1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4 5

5

Below x-axis

x = −0.25 x = −2.75

Below x-axis

4

Figure 12

We observe that the graph of the function is below the x-axis left of x = − 1 __ 4

and right of x = 11 __ 4

This means the

function values are negative to the left of the first horizontal intercept at x = −  1 __ 4

, and negative to the right of the

second intercept at x = 11 __ 4

.This gives us the solution to the inequality.

x < − 1 __ 4

or x > 11 __ 4

In interval notation, this would be (−∞, −0.25) ∪ (2.75, ∞).

Try It #7

Solve −2| k − 4| ≤ −6.

Access these online resources for additional instruction and practice with absolute value.

• Graphing Absolute Value Functions (http://openstaxcollege.org/l/graphabsvalue)

• Graphing Absolute Value Functions 2 (http://openstaxcollege.org/l/graphabsvalue2)

• equations of Absolute Value Function (http://openstaxcollege.org/l/findeqabsval)

• equations of Absolute Value Function 2 (http://openstaxcollege.org/l/findeqabsval2)

• Solving Absolute Value equations (http://openstaxcollege.org/l/solveabsvalueeq)

9 8 CHAPTER 1 Functions

1.6 SeCTIOn exeRCISeS

VeRBAl

1. How do you solve an absolute value equation? 2. How can you tell whether an absolute value function has two x-intercepts without graphing the function?

3. When solving an absolute value function, the isolated absolute value term is equal to a negative number. What does that tell you about the graph of the absolute value function?

4. How can you use the graph of an absolute value function to determine the x-values for which the function values are negative?

5. How do you solve an absolute value inequality algebraically?

AlGeBRAIC

6. Describe all numbers x that are at a distance of 4 from the number 8. Express this using absolute value notation.

7. Describe all numbers x that are at a distance of 1 __ 2

from the number −4. Express this using absolute value notation.

8. Describe the situation in which the distance that point x is from 10 is at least 15 units. Express this using absolute value notation.

9. Find all function values f (x) such that the distance from f (x) to the value 8 is less than 0.03 units. Express this using absolute value notation.

For the following exercises, solve the equations below and express the answer using set notation.

10. | x + 3 | = 9 11. | 6 − x | = 5 12. | 5x − 2 | = 11

13. | 4x − 2 | = 11 14. 2| 4 − x | = 7 15. 3| 5 − x | = 5

16. 3| x + 1 | − 4 = 5 17. 5| x − 4 | − 7 = 2 18. 0 = −| x − 3 | + 2

19. 2| x − 3 | + 1 = 2 20. | 3x − 2 | = 7 21. | 3x − 2 | = −7

22.    1 __ 2 x − 5    = 11 23.     1 __ 3

x + 5  = 14 24. −    1 __ 3 x + 5   + 14 = 0

For the following exercises, find the x- and y-intercepts of the graphs of each function.

25. f (x) = 2| x +1 | − 10 26. f (x) = 4| x − 3 | + 4

27. f (x) = −3| x − 2 | − 1 28. f (x) = −2| x +1 | + 6

For the following exercises, solve each inequality and write the solution in interval notation.

29. | x − 2 | > 10 30. 2| v − 7 | − 4 ≥ 42 31. | 3x − 4 | ≤ 8

32. | x − 4 | ≥ 8 33. | 3x − 5 | ≥ 13 34. | 3x − 5 | ≥ −13

35.    3 __ 4 x − 5   ≥ 7 36.     3 __ 4

x − 5    + 1 ≤ 16

SECTION 1.6 section exercises 9 9

GRAPHICAl

For the following exercises, graph the absolute value function. Plot at least five points by hand for each graph.

37. y = | x − 1 | 38. y = | x + 1 |  39. y = | x | + 1

For the following exercises, graph the given functions by hand.

40. y = | x | − 2 41. y = −| x |  42. y = −| x | − 2

43. y = −| x − 3 | − 2 44. f (x) = −| x − 1 | − 2 45. f (x) = −| x + 3 | + 4

46. f (x) = 2| x + 3 | + 1 47. f (x) = 3| x − 2 | + 3 48. f (x) = | 2x − 4 | − 3

49. f (x) = | 3x + 9 | + 2 50. f (x) = −| x − 1 | − 3 51. f (x) = −| x + 4 | −3

52. f (x) = 1 __ 2

| x + 4 | − 3

TeCHnOlOGY

53. Use a graphing utility to graph f (x) = 10| x − 2| on the viewing window [0, 4]. Identify the corresponding range. Show the graph.

54. Use a graphing utility to graph f (x) = −100| x| + 100 on the viewing window [−5, 5]. Identify the corresponding range. Show the graph.

For the following exercises, graph each function using a graphing utility. Specify the viewing window.

55. f (x) = −0.1| 0.1(0.2 − x)| + 0.3 56. f (x) = 4 × 109 ∣ x − (5 × 109) ∣  + 2 × 109

exTenSIOnS

For the following exercises, solve the inequality.

57. | −2x − 2 __ 3

(x + 1) | + 3 > −1 58. If possible, find all values of a such that there are no x-intercepts for f (x) = 2| x + 1| + a.

59. If possible, find all values of a such that there are no y-intercepts for f (x) = 2| x + 1| + a.

ReAl-WORlD APPlICATIOnS

60. Cities A and B are on the same east-west line. Assume that city A is located at the origin. If the distance from city A to city B is at least 100 miles and x represents the distance from city B to city A, express this using absolute value notation.

61. The true proportion p of people who give a favorable rating to Congress is 8% with a margin of error of 1.5%. Describe this statement using an absolute value equation.

62. Students who score within 18 points of the number 82 will pass a particular test. Write this statement using absolute value notation and use the variable x for the score.

63. A machinist must produce a bearing that is within 0.01 inches of the correct diameter of 5.0 inches. Using x as the diameter of the bearing, write this statement using absolute value notation.

64. The tolerance for a ball bearing is 0.01. If the true diameter of the bearing is to be 2.0 inches and the measured value of the diameter is x inches, express the tolerance using absolute value notation.

CHAPTER 1 Functions1 0 0

1.7 InVeRSe FUnCTIOnS

A reversible heat pump is a climate-control system that is an air conditioner and a heater in a single device. Operated in one direction, it pumps heat out of a house to provide cooling. Operating in reverse, it pumps heat into the building from the outside, even in cool weather, to provide heating. As a heater, a heat pump is several times more efficient than conventional electrical resistance heating.

If some physical machines can run in two directions, we might ask whether some of the function “machines” we have been studying can also run backwards. Figure 1 provides a visual representation of this question. In this section, we will consider the reverse nature of functions.

x

x

f ?

y

y

Figure 1 Can a function “machine” operate in reverse?

Verifying That Two Functions Are Inverse Functions Suppose a fashion designer traveling to Milan for a fashion show wants to know what the temperature will be. He is not familiar with the Celsius scale. To get an idea of how temperature measurements are related, he asks his assistant, Betty, to convert 75 degrees Fahrenheit to degrees Celsius. She finds the formula

C = 5 __ 9

(F − 32)

and substitutes 75 for F to calculate

5 __ 9

(75 − 32) ≈ 24°C.

Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the week’s weather forecast from Figure 2 for Milan, and asks her to convert all of the temperatures to degrees Fahrenheit.

26°C | 19°C 29°C | 19°C 30°C | 20°C 26°C | 18°C

Mon Tue Web Thu

Figure 2

leARnInG OBjeCTIVeS

In this section, you will: • Verify inverse functions. • Determine the domain and range of an inverse function, and restrict the domain of a function to make it one-to-one. • Find or evaluate the inverse of a function. • Use the graph of a one-to-one function to graph its inverse function on the same axes.

SECTION 1.7 inverse Functions 1 0 1

At first, Betty considers using the formula she has already found to complete the conversions. After all, she knows her algebra, and can easily solve the equation for F after substituting a value for C. For example, to convert 26 degrees Celsius, she could write

26 =   5 __ 9

(F − 32)

26 ⋅ 9 __ 5

= F − 32

F = 26 ⋅ 9 __ 5

+ 32 ≈ 79

After considering this option for a moment, however, she realizes that solving the equation for each of the temperatures will be awfully tedious. She realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one that takes the Celsius temperature and outputs the Fahrenheit temperature.

The formula for which Betty is searching corresponds to the idea of an inverse function, which is a function for which the input of the original function becomes the output of the inverse function and the output of the original function becomes the input of the inverse function.

Given a function f (x), we represent its inverse as f −1(x), read as “ f inverse of x.” The raised −1 is part of the notation. It is not an exponent; it does not imply a power of −1. In other words, f −1(x) does not mean 1 ___

f (x) because 1 ___

f (x) is the

reciprocal of f and not the inverse.

The “exponent-like” notation comes from an analogy between function composition and multiplication: just as a−1 a = 1 (1 is the identity element for multiplication) for any nonzero number a, so f −1 ∘ f equals the identity function, that is,

(f −1 ∘ f )(x) = f −1( f (x)) = f −1 (y) = x

This holds for all x in the domain of f. Informally, this means that inverse functions “undo” each other. However, just as zero does not have a reciprocal, some functions do not have inverses.

Given a function f (x), we can verify whether some other function g (x) is the inverse of f (x) by checking whether either g ( f (x)) = x or f (g (x)) = x is true. We can test whichever equation is more convenient to work with because they are logically equivalent (that is, if one is true, then so is the other.)

For example, y = 4x and y =   1 __ 4

x are inverse functions.

(f −1 ∘ f )(x) = f −1 (4x) =   1 __ 4

(4x) = x and

(f ∘ f −1)(x) = f   1 __ 4 x  = 4   1 __ 4

x  = x

A few coordinate pairs from the graph of the function y = 4x are (−2, −8), (0, 0), and (2, 8). A few coordinate pairs from the graph of the function y = 1 __

4 x are (−8, −2), (0, 0), and (8, 2). If we interchange the input and output of each

coordinate pair of a function, the interchanged coordinate pairs would appear on the graph of the inverse function.

inverse function For any one-to-one function f (x) = y, a function f −1 (x) is an inverse function of f if f −1(y) = x. This can also be written as f −1( f (x)) = x for all x in the domain of f . It also follows that f ( f −1(x)) = x for all x in the domain of f −1 if f −1 is the inverse of f .

The notation f −1 is read “ f inverse.” Like any other function, we can use any variable name as the input for f −1, so we will often write f −1(x), which we read as “ f inverse of x.” Keep in mind that

f −1(x) ≠ 1 ___ f (x)

and not all functions have inverses.

CHAPTER 1 Functions1 0 2

Example 1 Identifying an Inverse Function for a Given Input-Output Pair

If for a particular one-to-one function f (2) = 4 and f (5) = 12, what are the corresponding input and output values for the inverse function?

Solution The inverse function reverses the input and output quantities, so if

f (2) = 4, then f −1 (4) = 2;

f (5) = 12, then f −1 (12) = 5.

Alternatively, if we want to name the inverse function g, then g (4) = 2 and g (12) = 5.

Analysi s Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed. See Table 1.

(x, f (x)) (x, g (x)) (2, 4) (4, 2) (5, 12) (12, 5)

Table 1

Try It #1

Given that h−1(6) = 2, what are the corresponding input and output values of the original function h?

How To… Given two functions f (x) and g (x), test whether the functions are inverses of each other.

1. Determine whether f (g(x)) = x or g(f (x)) = x. 2. If both statements are true, then g = f −1 and f = g −1. If either statement is false, then both are false, and g ≠ f −1

and f ≠ g−1.

Example 2 Testing Inverse Relationships Algebraically

If f (x) = 1 ____ x + 2

and g (x) =   1 __ x

− 2, is g = f −1?

Solution g ( f (x)) = 1 _   1 _ x + 2 

− 2

= x + 2 − 2

= x so

g = f −1 and f = g −1

This is enough to answer yes to the question, but we can also verify the other formula.

f (g (x)) = 1 _ 1 __ x

− 2 + 2

= 1 _ 1 __ x

= x

Analysi s Notice the inverse operations are in reverse order of the operations from the original function.

Try It #2 If f (x) = x3 − 4 and g (x) =  

3 √ —

x − 4 , is g = f −1?

SECTION 1.7 inverse Functions 1 0 3

Example 3 Determining Inverse Relationships for Power Functions

If f (x) = x3 (the cube function) and g (x) =   1 __ 3

x, is g = f −1?

Solution f (g (x)) = x 3 __

27 ≠ x

No, the functions are not inverses.

Analysi s The correct inverse to the cube is, of course, the cube root 3 √ — x = x1/3 that is, the one-third is an exponent,

not a multiplier.

Try It #3

If f (x) = (x − 1)3 and g (x) = 3 √ —

x + 1, is g = f −1?

Finding Domain and Range of Inverse Functions The outputs of the function f are the inputs to f −1, so the range of f is also the domain of f −1. Likewise, because the inputs to f are the outputs of f −1, the domain of f is the range of f −1. We can visualize the situation as in Figure 3.

Domain of f Range of f

ba

Range of f –1 Domain of f –1 f –1(x)

f (x)

Figure 3 Domain and range of a function and its inverse

When a function has no inverse function, it is possible to create a new function where that new function on a limited domain does have an inverse function. For example, the inverse of f (x) = √

— x is f −1(x) = x2, because a square “undoes”

a square root; but the square is only the inverse of the square root on the domain [0, ∞), since that is the range of f (x) = √

— x .

We can look at this problem from the other side, starting with the square (toolkit quadratic) function f (x) = x2. If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and –3. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the “inverse” is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function.

In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function f (x) = x2 with its domain limited to [0, ∞), which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).

If f (x) = (x − 1)2 on [1, ∞), then the inverse function is f −1(x) = √ — x + 1.

• The domain of f = range of f −1 = [1, ∞). • The domain of f −1 = range of f = [0, ∞).

Q & A… Is it possible for a function to have more than one inverse?

No. If two supposedly different functions, say, g and h, both meet the definition of being inverses of another function f, then you can prove that g = h. We have just seen that some functions only have inverses if we restrict the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading to different inverses. However, on any one domain, the original function still has only one unique inverse.

CHAPTER 1 Functions1 0 4

domain and range of inverse functions The range of a function f (x) is the domain of the inverse function f −1(x). The domain of f (x) is the range of f −1(x).

How To… Given a function, find the domain and range of its inverse.

1. If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse.

2. If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function.

Example 4 Finding the Inverses of Toolkit Functions

Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are reviewed in Table 2. We restrict the domain in such a fashion that the function assumes all y-values exactly once.

Constant Identity Quadratic Cubic Reciprocal

f (x) = c f (x) = x f (x) = x 2 f (x) = x 3 f (x) = 1 _ x

Reciprocal squared Cube root Square root Absolute value

f (x) = 1 _ x2

f (x) = 3 √ — x f (x) = √

— x f (x) = ∣x∣

Table 2

Solution The constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-to-one, so the constant function has no meaningful inverse.

The absolute value function can be restricted to the domain [0, ∞), where it is equal to the identity function.

The reciprocal-squared function can be restricted to the domain (0, ∞).

Analysi s We can see that these functions (if unrestricted) are not one-to-one by looking at their graphs, shown in Figure 4. They both would fail the horizontal line test. However, if a function is restricted to a certain domain so that it passes the horizontal line test, then in that restricted domain, it can have an inverse.

2

x

f (x)

–1–1–2

–2

–3

–3

–4

–4

–5

1

3

3

21 4

4

5

(a) (b)

5

x

f (x)

–1–1–2

–2

–3

–3

–4

–4

–5

1

321 4

4

5

5

3 2

Figure 4 (a ) Absolute value (b ) Reciprocal squared

Try It #4

The domain of function f is (1, ∞) and the range of function f is (−∞, −2). Find the domain and range of the inverse function.

SECTION 1.7 inverse Functions 1 0 5

Finding and evaluating Inverse Functions Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases.

Inverting Tabular Functions

Suppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and range.

Each row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function.

Example 5 Interpreting the Inverse of a Tabular Function

A function f (t) is given in Table 3, showing distance in miles that a car has traveled in t minutes. Find and interpret f −1(70).

t (minutes) 30 50 70 90

f (t) (miles) 20 40 60 70

Table 3

Solution The inverse function takes an output of f and returns an input for f. So in the expression f −1(70), 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function f, 90 minutes, so f −1(70) = 90. The interpretation of this is that, to drive 70 miles, it took 90 minutes.

Alternatively, recall that the definition of the inverse was that if f (a) = b, then f −1(b) = a. By this definition, if we are given f −1(70) = a, then we are looking for a value a so that f (a) = 70. In this case, we are looking for a t so that f (t) = 70, which is when t = 90.

Try It #5

Using Table 4, find and interpret a. f (60), and b. f −1(60).

t (minutes) 30 50 60 70 90

f (t) (miles) 20 40 50 60 70

Table 4

Evaluating the Inverse of a Function, Given a Graph of the Original Function

We saw in Functions and Function Notation that the domain of a function can be read by observing the horizontal extent of its graph. We find the domain of the inverse function by observing the vertical extent of the graph of the original function, because this corresponds to the horizontal extent of the inverse function. Similarly, we find the range of the inverse function by observing the horizontal extent of the graph of the original function, as this is the vertical extent of the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function’s graph.

How To… Given the graph of a function, evaluate its inverse at specific points.

1. Find the desired input on the y-axis of the given graph. 2. Read the inverse function’s output from the x-axis of the given graph.

CHAPTER 1 Functions1 0 6

Example 6 Evaluating a Function and Its Inverse from a Graph at Specific Points

A function g(x) is given in Figure 5. Find g(3) and g−1(3).

g(x)

x –1

1

1–

5

6

3

2

– 7321 654

4

Figure 5

Solution To evaluate g(3), we find 3 on the x-axis and find the corresponding output value on the y-axis. The point (3, 1) tells us that g(3) = 1.

To evaluate g −1(3), recall that by definition g −1(3) means the value of x for which g(x) = 3. By looking for the output value 3 on the vertical axis, we find the point (5, 3) on the graph, which means g(5) = 3, so by definition, g −1(3) = 5. See Figure 6.

g(x)

x –1

1

1–

5

6

3

2

– 7321 654

4

(3, 1)

(5, 3)

Figure 6

Try It #6

Using the graph in Figure 6, a. find g−1(1), and b. estimate g−1(4).

Finding Inverses of Functions Represented by Formulas

Sometimes we will need to know an inverse function for all elements of its domain, not just a few. If the original function is given as a formula— for example, y as a function of x— we can often find the inverse function by solving to obtain x as a function of y.

How To… Given a function represented by a formula, find the inverse.

1. Make sure f is a one-to-one function. 2. Solve for x. 3. Interchange x and y.

Example 7 Inverting the Fahrenheit-to-Celsius Function

Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature.

C = 5 __ 9

(F − 32)

SECTION 1.7 inverse Functions 1 0 7

Solution C = 5 __ 9

(F − 32)

C ⋅ 9 __ 5

= F − 32

F = 9 __ 5

C + 32

By solving in general, we have uncovered the inverse function. If

C = h(F) = 5 __ 9

(F − 32),

then

F = h−1(C) = 9 __ 5

C + 32.

In this case, we introduced a function h to represent the conversion because the input and output variables are descriptive, and writing C−1 could get confusing.

Try It #7

Solve for x in terms of y given y = 1 __ 3

(x − 5)

Example 8 Solving to Find an Inverse Function

Find the inverse of the function f (x) = 2 ____ x − 3

+ 4.

Solution y = 2 ____ x − 3

+ 4 Set up an equation.

y − 4 = 2 _____ x − 3

Subtract 4 from both sides.

x − 3 = 2 ____ y − 4

Multiply both sides by x − 3 and divide by y − 4.

x = 2 ____ y − 4

+ 3 Add 3 to both sides.

So f −1 (y) = 2 _____ y − 4

+ 3 or f −1 (x) = 2 ____ x − 4

+ 3.

Analysi s The domain and range of f exclude the values 3 and 4, respectively. f and f −1 are equal at two points but are not the same function, as we can see by creating Table 5.

x 1 2 5 f −1(y) f (x) 3 2 5 y

Table 5

Example 9 Solving to Find an Inverse with Radicals

Find the inverse of the function f (x) = 2 + √ —

x − 4 .

Solution y = 2 + √ —

x − 4

(y − 2)2 = x − 4

x = (y − 2)2 + 4

So f −1 (x) = (x − 2)2 + 4.

The domain of f is [4, ∞). Notice that the range of f is [2, ∞), so this means that the domain of the inverse function f −1 is also [2, ∞).

Analysi s The formula we found for f −1(x) looks like it would be valid for all real x. However, f −1 itself must have an inverse (namely, f) so we have to restrict the domain of f −1 to [2, ∞) in order to make f −1 a one-to-one function. This domain of f −1 is exactly the range of f.

CHAPTER 1 Functions1 0 8

Try It #8

What is the inverse of the function f (x) = 2 − √ — x ? State the domains of both the function and the inverse function.

Finding Inverse Functions and Their Graphs Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function f (x) = x2 restricted to the domain [0, ∞), on which this function is one-to-one, and graph it as in Figure 7.

2

x

f (x)

–1–1–2

–2

–3–4–5

1

3

3

21 4

4

5

5

Figure 7 Quadratic function with domain restricted to [0, ∞).

Restricting the domain to [0, ∞) makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain.

We already know that the inverse of the toolkit quadratic function is the square root function, that is, f −1(x) = √ — x .

What happens if we graph both f and f −1 on the same set of axes, using the x-axis for the input to both f and f −1 ?

We notice a distinct relationship: The graph of f −1(x) is the graph of f (x) reflected about the diagonal line y = x, which we will call the identity line, shown in Figure 8.

2

x

y

–1–1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5 f (x) y = x

f –1(x)

Figure 8 Square and square-root functions on the non-negative domain

This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes.

SECTION 1.7 inverse Functions 1 0 9

Example 10 Finding the Inverse of a Function Using Reflection about the Identity Line

Given the graph of f (x) in Figure 9, sketch a graph of f −1(x).

2

x

y

–1–1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

Figure 9

Solution This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of (0, ∞) and range of (−∞, ∞), so the inverse will have a domain of (−∞, ∞) and range of (0, ∞).

If we reflect this graph over the line y = x, the point (1, 0) reflects to (0, 1) and the point (4, 2) reflects to (2, 4). Sketching the inverse on the same axes as the original graph gives Figure 10.

2

x

y

–1–1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

f (x)

y = xf –1(x)

Figure 10 The function and its inverse, showing reflection about the identity line

Try It #9

Draw graphs of the functions f and f −1 from Example 8.

Q & A… Is there any function that is equal to its own inverse?

Yes. If f = f −1, then f ( f (x)) = x, and we can think of several functions that have this property. The identity function does, and so does the reciprocal function, because

1 _ 1 _ x

= x

Any function f (x) = c − x, where c is a constant, is also equal to its own inverse.

Access these online resources for additional instruction and practice with inverse functions. • Inverse Functions (http://openstaxcollege.org/l/inversefunction)

• Inverse Function Values Using Graph (http://openstaxcollege.org/l/inversfuncgraph)

• Restricting the Domain and Finding the Inverse (http://openstaxcollege.org/l/restrictdomain)

1 1 0 CHAPTER 1 Functions

1.7 SeCTIOn exeRCISeS

VeRBAl

1. Describe why the horizontal line test is an effective way to determine whether a function is one-to-one?

2. Why do we restrict the domain of the function f (x) = x2 to find the function’s inverse?

3. Can a function be its own inverse? Explain. 4. Are one-to-one functions either always increasing or always decreasing? Why or why not?

5. How do you find the inverse of a function algebraically?

AlGeBRAIC

6. Show that the function f (x) = a − x is its own inverse for all real numbers a.

For the following exercises, find f −1(x) for each function.

7. f (x) = x + 3 8. f (x) = x + 5 9. f (x) = 2 − x

10. f (x) = 3 − x 11. f (x) = x _____ x + 2

12. f (x) = 2x + 3 ______ 5x + 4

For the following exercises, find a domain on which each function f is one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of f restricted to that domain.

13. f (x) = (x + 7)2 14. f (x) = (x − 6)2 15. f (x) = x2 − 5

16. Given f (x) = x __ 2 + x

and g(x) = 2x _____ 1 − x

:

a. Find f (g(x)) and g (f (x)). b. What does the answer tell us about the relationship between f (x) and g(x)?

For the following exercises, use function composition to verify that f (x) and g(x) are inverse functions.

17. f (x) = 3 √ —

x − 1 and g (x) = x3 + 1 18. f (x) = −3x + 5 and g (x) = x − 5 _____ −3

GRAPHICAl

For the following exercises, use a graphing utility to determine whether each function is one-to-one.

19. f (x) = √ — x 20. f (x) =

3 √ —

3x + 1 21. f (x) = −5x + 1 22. f (x) = x3 − 27

For the following exercises, determine whether the graph represents a one-to-one function.

23.

10

x

y

f

–5 –5–10

–10

–15

–15

–20

–20

–25

–25

5

15

15

105 20

20

25

25

24.

4

x

y

f –2–4–6–8–10 –2

–4 –6 –8

–10

2

6

6

42 8

8

10

10

SECTION 1.7 section exercises 1 1 1

For the following exercises, use the graph of f shown in Figure 11.

2

x

y

f

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

Figure 11

25. Find f (0).

26. Solve f (x) = 0.

27. Find f −1(0).

28. Solve f −1(x) = 0.

For the following exercises, use the graph of the one-to-one function shown in Figure 12.

4

x

y

f

–2–4–6–8–10 –2 –4 –6 –8

–10

2

6

6

42 8

8

10

10

Figure 12

29. Sketch the graph of f −1.

30. Find f (6) and f −1(2).

31. If the complete graph of f is shown, find the domain of f.

32. If the complete graph of f is shown, find the range of f.

nUMeRIC

For the following exercises, evaluate or solve, assuming that the function f is one-to-one.

33. If f (6) = 7, find f −1(7). 34. If f (3) = 2, find f −1(2).

35. If f −1(−4) = −8, find f (−8). 36. If f −1(−2) = −1, find f (−1).

For the following exercises, use the values listed in Table 6 to evaluate or solve.

x 0 1 2 3 4 5 6 7 8 9

f (x) 8 0 7 4 2 6 5 3 9 1 Table 6

37. Find f (1). 38. Solve f (x) = 3. 39. Find f −1(0). 40. Solve f −1(x) = 7.

41. Use the tabular representation of f in Table 7 to create a table for f −1(x).

x 3 6 9 13 14

f (x) 1 4 7 12 16 Table 7

1 1 2 CHAPTER 1 Functions

TeCHnOlOGY

For the following exercises, find the inverse function. Then, graph the function and its inverse.

42. f (x) = 3 _____ x − 2

43. f (x) = x 3 − 1

44. Find the inverse function of f (x) = 1 _ x − 1

. Use a graphing utility to find its domain and range. Write the domain and range in interval notation.

ReAl-WORlD APPlICATIOnS

45. To convert from x degrees Celsius to y degrees Fahrenheit, we use the formula f (x) = 9 __

5 x + 32.

Find the inverse function, if it exists, and explain its meaning.

46. The circumference C of a circle is a function of its radius given by C(r) = 2πr. Express the radius of a circle as a function of its circumference. Call this function r(C). Find r(36π) and interpret its meaning.

47. A car travels at a constant speed of 50 miles per hour. The distance the car travels in miles is a function of time, t, in hours given by d(t) = 50t. Find the inverse function by expressing the time of travel in terms of the distance traveled. Call this function t(d). Find t(180) and interpret its meaning.

CHAPTER 1 review 1 1 3

CHAPTeR 1 ReVIeW

Key Terms absolute maximum the greatest value of a function over an interval

absolute minimum the lowest value of a function over an interval

absolute value equation an equation of the form ∣A∣ = B, with B ≥ 0; it will have solutions when A = B or A = −B

absolute value inequality a relationship in the form ∣A∣ < B, ∣A∣ ≤ B, ∣A∣ > B, or ∣A∣ ≥ B

average rate of change the difference in the output values of a function found for two values of the input divided by the difference between the inputs

composite function the new function formed by function composition, when the output of one function is used as the input of another

decreasing function a function is decreasing in some open interval if f (b) < f (a) for any two input values a and b in the given interval where b > a

dependent variable an output variable

domain the set of all possible input values for a relation

even function a function whose graph is unchanged by horizontal reflection, f (x) = f (−x), and is symmetric about the y-axis

function a relation in which each input value yields a unique output value

horizontal compression a transformation that compresses a function’s graph horizontally, by multiplying the input by a constant b > 1

horizontal line test a method of testing whether a function is one-to-one by determining whether any horizontal line intersects the graph more than once

horizontal reflection a transformation that reflects a function’s graph across the y-axis by multiplying the input by −1

horizontal shift a transformation that shifts a function’s graph left or right by adding a positive or negative constant to the input

horizontal stretch a transformation that stretches a function’s graph horizontally by multiplying the input by a constant 0 < b < 1

increasing function a function is increasing in some open interval if f (b) > f (a) for any two input values a and b in the given interval where b > a

independent variable an input variable

input each object or value in a domain that relates to another object or value by a relationship known as a function

interval notation a method of describing a set that includes all numbers between a lower limit and an upper limit; the lower and upper values are listed between brackets or parentheses, a square bracket indicating inclusion in the set, and a parenthesis indicating exclusion

inverse function for any one-to-one function f (x), the inverse is a function f −1(x) such that f −1(f (x)) = x for all x in the domain of f; this also implies that f ( f −1(x)) = x for all x in the domain of f −1

local extrema collectively, all of a function’s local maxima and minima

local maximum a value of the input where a function changes from increasing to decreasing as the input value increases.

local minimum a value of the input where a function changes from decreasing to increasing as the input value increases.

odd function a function whose graph is unchanged by combined horizontal and vertical reflection, f (x) = − f (−x), and is symmetric about the origin

one-to-one function a function for which each value of the output is associated with a unique input value

output each object or value in the range that is produced when an input value is entered into a function

piecewise function a function in which more than one formula is used to define the output

range the set of output values that result from the input values in a relation

1 1 4 CHAPTER 1 Functions

rate of change the change of an output quantity relative to the change of the input quantity

relation a set of ordered pairs

set-builder notation a method of describing a set by a rule that all of its members obey; it takes the form {x ∣ statement about x}

vertical compression a function transformation that compresses the function’s graph vertically by multiplying the output by a constant 0 < a < 1

vertical line test a method of testing whether a graph represents a function by determining whether a vertical line intersects the graph no more than once

vertical reflection a transformation that reflects a function’s graph across the x-axis by multiplying the output by −1

vertical shift a transformation that shifts a function’s graph up or down by adding a positive or negative constant to the output

vertical stretch a transformation that stretches a function’s graph vertically by multiplying the output by a constant a > 1

Key equations

Constant function f (x) = c, where c is a constant

Identity function f (x) = x

Absolute value function f (x) = ∣x∣

Quadratic function f (x) = x2

Cubic function f (x) = x3

Reciprocal function f (x) = 1 _ x

Reciprocal squared function f (x) = 1 _ x2

Square root function f (x) = √ — x

Cube root function f (x) = 3 √ — x

Average rate of change ∆y

_ ∆x

= f (x2) − f (x1) _ x2 − x1

Composite function ( f ∘ g)(x) = f (g(x))

Vertical shift g(x) = f (x) + k (up for k > 0)

Horizontal shift g(x) = f (x − h) (right for h > 0)

Vertical reflection g(x) = −f (x)

Horizontal reflection g(x) = f (−x)

Vertical stretch g(x) = af (x) (a > 0)

Vertical compression g(x) = af (x) (0 < a < 1)

Horizontal stretch g(x) = f (bx) (0 < b < 1)

Horizontal compression g(x) = f (bx) (b > 1)

CHAPTER 1 review 1 1 5

Key Concepts

1.1 Functions and Function Notation • A relation is a set of ordered pairs. A function is a specific type of relation in which each domain value, or input,

leads to exactly one range value, or output. See Example 1 and Example 2.

• Function notation is a shorthand method for relating the input to the output in the form y = f (x). See Example 3 and Example 4.

• In tabular form, a function can be represented by rows or columns that relate to input and output values. See Example 5.

• To evaluate a function, we determine an output value for a corresponding input value. Algebraic forms of a function can be evaluated by replacing the input variable with a given value. See Example 6 and Example 7.

• To solve for a specific function value, we determine the input values that yield the specific output value. See Example 8.

• An algebraic form of a function can be written from an equation. See Example 9 and Example 10.

• Input and output values of a function can be identified from a table. See Example 11.

• Relating input values to output values on a graph is another way to evaluate a function. See Example 12.

• A function is one-to-one if each output value corresponds to only one input value. See Example 13.

• A graph represents a function if any vertical line drawn on the graph intersects the graph at no more than one point. See Example 14.

• The graph of a one-to-one function passes the horizontal line test. See Example 15.

1.2 Domain and Range • The domain of a function includes all real input values that would not cause us to attempt an undefined

mathematical operation, such as dividing by zero or taking the square root of a negative number.

• The domain of a function can be determined by listing the input values of a set of ordered pairs. See Example 1.

• The domain of a function can also be determined by identifying the input values of a function written as an equation. See Example 2, Example 3, and Example 4.

• Interval values represented on a number line can be described using inequality notation, set-builder notation, and interval notation. See Example 5.

• For many functions, the domain and range can be determined from a graph. See Example 6 and Example 7.

• An understanding of toolkit functions can be used to find the domain and range of related functions. See Example 8, Example 9, and Example 10.

• A piecewise function is described by more than one formula. See Example 11 and Example 12.

• A piecewise function can be graphed using each algebraic formula on its assigned subdomain. See Example 13.

1.3 Rates of Change and Behavior of Graphs • A rate of change relates a change in an output quantity to a change in an input quantity. The average rate of change is

determined using only the beginning and ending data. See Example 1.

• Identifying points that mark the interval on a graph can be used to find the average rate of change. See Example 2.

• Comparing pairs of input and output values in a table can also be used to find the average rate of change. See Example 3.

• An average rate of change can also be computed by determining the function values at the endpoints of an interval described by a formula. See Example 4 and Example 5.

• The average rate of change can sometimes be determined as an expression. See Example 6.

• A function is increasing where its rate of change is positive and decreasing where its rate of change is negative. See Example 7.

• A local maximum is where a function changes from increasing to decreasing and has an output value larger (more positive or less negative) than output values at neighboring input values.

1 1 6 CHAPTER 1 Functions

• A local minimum is where the function changes from decreasing to increasing (as the input increases) and has an output value smaller (more negative or less positive) than output values at neighboring input values.

• Minima and maxima are also called extrema.

• We can find local extrema from a graph. See Example 8 and Example 9.

• The highest and lowest points on a graph indicate the maxima and minima. See Example 10.

1.4 Composition of Functions • We can perform algebraic operations on functions. See Example 1.

• When functions are combined, the output of the first (inner) function becomes the input of the second (outer) function.

• The function produced by combining two functions is a composite function. See Example 2 and Example 3.

• The order of function composition must be considered when interpreting the meaning of composite functions. See Example 4.

• A composite function can be evaluated by evaluating the inner function using the given input value and then evaluating the outer function taking as its input the output of the inner function.

• A composite function can be evaluated from a table. See Example 5.

• A composite function can be evaluated from a graph. See Example 6.

• A composite function can be evaluated from a formula. See Example 7.

• The domain of a composite function consists of those inputs in the domain of the inner function that correspond to outputs of the inner function that are in the domain of the outer function. See Example 8 and Example 9.

• Just as functions can be combined to form a composite function, composite functions can be decomposed into simpler functions.

• Functions can often be decomposed in more than one way. See Example 10.

1.5 Transformation of Functions • A function can be shifted vertically by adding a constant to the output. See Example 1 and Example 2.

• A function can be shifted horizontally by adding a constant to the input. See Example 3, Example 4, and Example 5.

• Relating the shift to the context of a problem makes it possible to compare and interpret vertical and horizontal shifts. See Example 6.

• Vertical and horizontal shifts are often combined. See Example 7 and Example 8.

• A vertical reflection reflects a graph about the x-axis. A graph can be reflected vertically by multiplying the output by –1.

• A horizontal reflection reflects a graph about the y-axis. A graph can be reflected horizontally by multiplying the input by –1.

• A graph can be reflected both vertically and horizontally. The order in which the reflections are applied does not affect the final graph. See Example 9.

• A function presented in tabular form can also be reflected by multiplying the values in the input and output rows or columns accordingly. See Example 10.

• A function presented as an equation can be reflected by applying transformations one at a time. See Example 11.

• Even functions are symmetric about the y-axis, whereas odd functions are symmetric about the origin.

• Even functions satisfy the condition f (x) = f (−x).

• Odd functions satisfy the condition f (x) = −f (−x).

• A function can be odd, even, or neither. See Example 12.

• A function can be compressed or stretched vertically by multiplying the output by a constant. See Example 13, Example 14, and Example 15.

• A function can be compressed or stretched horizontally by multiplying the input by a constant. See Example 16, Example 17, and Example 18.

CHAPTER 1 review 1 1 7

• The order in which different transformations are applied does affect the final function. Both vertical and horizontal transformations must be applied in the order given. However, a vertical transformation may be combined with a horizontal transformation in any order. See Example 19 and Example 20.

1.6 Absolute Value Functions • The absolute value function is commonly used to measure distances between points. See Example 1.

• Applied problems, such as ranges of possible values, can also be solved using the absolute value function. See Example 2.

• The graph of the absolute value function resembles a letter V. It has a corner point at which the graph changes direction. See Example 3.

• In an absolute value equation, an unknown variable is the input of an absolute value function.

• If the absolute value of an expression is set equal to a positive number, expect two solutions for the unknown variable. See Example 4.

• An absolute value equation may have one solution, two solutions, or no solutions. See Example 5.

• An absolute value inequality is similar to an absolute value equation but takes the form ∣A∣ < B, ∣A∣ ≤ B, ∣A∣ > B, or ∣A∣ ≥ B. It can be solved by determining the boundaries of the solution set and then testing which segments are in the set. See Example 6.

• Absolute value inequalities can also be solved graphically. See Example 7.

1.7 Inverse Functions • If g(x) is the inverse of f (x), then g(f (x)) = f (g(x)) = x. See Example 1, Example 2, and Example 3.

• Each of the toolkit functions has an inverse. See Example 4.

• For a function to have an inverse, it must be one-to-one (pass the horizontal line test).

• A function that is not one-to-one over its entire domain may be one-to-one on part of its domain.

• For a tabular function, exchange the input and output rows to obtain the inverse. See Example 5.

• The inverse of a function can be determined at specific points on its graph. See Example 6.

• To find the inverse of a formula, solve the equation y = f (x) for x as a function of y. Then exchange the labels x and y. See Example 7, Example 8, and Example 9.

• The graph of an inverse function is the reflection of the graph of the original function across the line y = x. See Example 10.

1 1 8 CHAPTER 1 Functions

CHAPTeR 1 ReVIeW exeRCISeS

FUnCTIOnS AnD FUnCTIOn nOTATIOn

For the following exercises, determine whether the relation is a function.

1. {(a, b), (c, d), (e, d)}

2. {(5, 2), (6, 1), (6, 2), (4, 8)}

3. y 2 + 4 = x, for x the independent variable and y the dependent variable

4. Is the graph in Figure 1 a function?

f 10

y

–5–10–15–20–25 –5 –10 –15 –20 –25

5

15

15

05 20

20

25

Figure 1

For the following exercises, evaluate the function at the indicated values: f (−3); f (2); f (−a); −f (a); f (a + h).

5. f (x) = −2x 2 + 3x 6. f ( x) = 2∣3 x − 1∣

For the following exercises, determine whether the functions are one-to-one.

7. f (x) = −3x + 5 8. f (x) = ∣x − 3∣

For the following exercises, use the vertical line test to determine if the relation whose graph is provided is a function.

9.

2

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

10.

2

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

11.

2

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

For the following exercises, graph the functions.

12. f (x) = ∣x + 1∣ 13. f (x) = x 2 − 2

25

1 x

CHAPTER 1 review 1 1 9

For the following exercises, use Figure 2 to approximate the values.

2

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

Figure 2

14. f (2)

15. f (−2)

16. If f (x) = −2, then solve for x.

17. If f (x) = 1, then solve for x.

For the following exercises, use the function h(t) = −16t 2 + 80t to find the values.

18. h(2) − h(1)

__ 2 − 1

19. h(a) − h(1)

__ a − 1

DOMAIn AnD RAnGe

For the following exercises, find the domain of each function, expressing answers using interval notation.

20. f (x) = 2 _ 3x + 2

21. f (x) = x − 3 ___________ x 2 − 4x − 12

22. f (x) =   √ —

x − 6 _ √ —

x − 4

23. Graph this piecewise function: f (x) = { x + 1 x < −2 −2x − 3 x ≥ −2

RATeS OF CHAnGe AnD BeHAVIOR OF GRAPHS

For the following exercises, find the average rate of change of the functions from x = 1 to x = 2.

24. f (x) = 4x − 3 25. f (x) = 10x 2 + x 26. f (x) = −  2 _ x 2

For the following exercises, use the graphs to determine the intervals on which the functions are increasing, decreasing, or constant.

27.

4

x

y

–1–2–3–4–5 –2 –4 –6 –8

–10

2

3

6

21 4

8

5

10

28.

2

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

29.

2

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

30. Find the local minimum of the function graphed in Exercise 27.

31. Find the local extrema for the function graphed in Exercise 28.

1 2 0 CHAPTER 1 Functions

32. For the graph in Figure 3, the domain of the function is [−3, 3]. The range is [−10, 10]. Find the absolute minimum of the function on this interval.

4

x

y

–1–2–3–4–5 –2 –4 –6 –8

–10

2

3

6

21 4

8

5

10

Figure 3

33. Find the absolute maximum of the function graphed in Figure 3.

COMPOSITIOn OF FUnCTIOnS

For the following exercises, find ( f ∘ g)(x) and (g ∘ f )(x) for each pair of functions.

34. f (x) = 4 − x, g(x) = −4x 35. f (x) = 3x + 2, g(x) = 5 − 6x 36. f (x) = x 2 + 2x, g(x) = 5x + 1

37. f (x) = √ —

x + 2 , g(x) = 1 _ x 38. f (x) = x + 3 _____

2 , g(x) = √

— 1 − x

For the following exercises, find ( f ∘ g) and the domain for ( f ∘ g)(x) for each pair of functions.

39. f (x) = x + 1 _ x + 4

, g(x) = 1 _ x 40. f (x) = 1 _

x + 3 , g(x) = 1 _

x − 9 41. f (x) = 1 _ x , g(x) = √

— x

42. f (x) = 1 _ x2 − 1

, g(x) = √ —

x + 1

For the following exercises, express each function H as a composition of two functions f and g where H(x) = ( f ∘ g)(x).

43. H(x) = √ _______

2x − 1 ______ 3x + 4

44. H(x) = 1 _ (3x 2 − 4)−3

TRAnSFORMATIOn OF FUnCTIOnS

For the following exercises, sketch a graph of the given function.

45. f (x) = (x − 3)2 46. f (x) = (x + 4)3 47. f (x) = √ — x + 5

48. f (x) = −x 3 49. f (x) = 3 √ —

−x 50. f (x) = 5 √ —

−x − 4

51. f (x) = 4[∣x − 2∣ − 6] 52. f (x) = −(x + 2)2 − 1

For the following exercises, sketch the graph of the function g if the graph of the function f is shown in Figure 4.

2

x

y

–1–2–3–4–5 –1 –2

1

3

3

21 4

4

5

5

Figure 4

53. g(x) = f (x − 1)

54. g(x) = 3f (x)

CHAPTER 1 review 1 2 1

For the following exercises, write the equation for the standard function represented by each of the graphs below.

55.

2

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

56.

2

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

For the following exercises, determine whether each function below is even, odd, or neither.

57. f (x) = 3x 4 58. g(x) = √ — x 59. h(x) = 1 _ x + 3x

For the following exercises, analyze the graph and determine whether the graphed function is even, odd, or neither.

60.

10

x

y

–5–10–15–20–25 –5 –10 –15 –20 –25

5

15

15

105 20

20

25

25

61.

10

x

y

–2–4–6–8–10 –5 –10 –15 –20 –25

5

6

15

42 8

20

10

25

62.

4

x

y

–2–4–6–8–10 –2 –4 –6 –8

–10

2

6

6

42 8

8

10

10

ABSOlUTe VAlUe FUnCTIOnS

For the following exercises, write an equation for the transformation of f (x) = | x |.

63.

2

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

64.

2

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

65.

2

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

For the following exercises, graph the absolute value function.

66. f (x) = | x − 5 | 67. f (x) = −| x − 3 | 68. f (x) = | 2x − 4 |

1 2 2 CHAPTER 1 Functions

For the following exercises, solve the absolute value equation.

69. | x + 4 | = 18 70.    1 __ 3 x + 5   =     3 __ 4

x − 2 

For the following exercises, solve the inequality and express the solution using interval notation.

71. | 3x − 2 | < 7 72.   1 __ 3 x − 2   ≤ 7

InVeRSe FUnCTIOnS

For the following exercises, find f −1(x) for each function.

73. f (x) = 9 + 10x 74. f (x) = x _ x + 2

For the following exercise, find a domain on which the function f is one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of f restricted to that domain.

75. f (x) = x 2 + 1

76. Given f (x) = x 3 − 5 and g(x) = 3 √ —

x + 5 : a. Find f (g(x)) and g(f (x)). b. What does the answer tell us about the relationship between f (x) and g(x)?

For the following exercises, use a graphing utility to determine whether each function is one-to-one.

77. f (x) = 1 _ x 78. f (x) = −3x 2 + x 79. If f (5) = 2, find f −1(2).

80. If f (1) = 4, find f −1(4).

1 2 3CHAPTER 1 practice test

CHAPTeR 1 PRACTICe TeST

For the following exercises, determine whether each of the following relations is a function.

1. y = 2x + 8 2. {(2, 1), (3, 2), (−1, 1), (0, −2)}

For the following exercises, evaluate the function f (x) = −3x 2 + 2x at the given input.

3. f (−2) 4. f (a)

5. Show that the function f (x) = −2(x − 1)2 + 3 is not one-to-one.

6. Write the domain of the function f (x) = √ —

3 − x in interval notation.

7. Given f (x) = 2x 2 − 5x, find f (a + 1) − f (1). 8. Graph the function f (x) = { x + 1 if −2 < x < 3−x if x ≥ 3

9. Find the average rate of change of the function f (x) = 3 − 2x 2 + x by finding f (b) − f (a)

_ b − a

.

For the following exercises, use the functions f (x) = 3 − 2x 2 + x and g(x) = √ — x to find the composite functions.

10. (g ∘ f )(x) 11. (g ∘ f )(1)

12. Express H(x) = 3 √ —

5x 2 − 3x as a composition of two functions, f and g, where (f ∘ g)(x) = H(x).

For the following exercises, graph the functions by translating, stretching, and/or compressing a toolkit function.

13. f (x) = √ —

x + 6 − 1 14. f (x) = 1 _ x + 2

− 1

For the following exercises, determine whether the functions are even, odd, or neither.

15. f (x) = − 5 _ x2

+ 9x 6 16. f (x) = − 5 _ x 3

+ 9x 5

17. f (x) = 1 _ x 18. Graph the absolute value function

f (x) = −2| x − 1 | + 3.

19. Solve | 2x − 3 | = 17. 20. Solve −   1 _ 3 x − 3  ≥ 17. Express the solution in interval notation.

For the following exercises, find the inverse of the function.

21. f (x) = 3x − 5 22. f (x) = 4 _ x + 7

1 2 4 CHAPTER 1 Functions

For the following exercises, use the graph of g shown in Figure 1.

2

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

Figure 1

23. On what intervals is the function increasing?

24. On what intervals is the function decreasing?

25. Approximate the local minimum of the function. Express the answer as an ordered pair.

26. Approximate the local maximum of the function. Express the answer as an ordered pair.

For the following exercises, use the graph of the piecewise function shown in Figure 2.

2

x

y

f

–1–2–3–4–5 –1 –2 –3 –4 –5

1

3

3

21 4

4

5

5

Figure 2

27. Find f (2).

28. Find f (−2).

29. Write an equation for the piecewise function.

For the following exercises, use the values listed in Table 1.

x 0 1 2 3 4 5 6 7 8 F (x) 1 3 5 7 9 11 13 15 17

Table 1

30. Find F (6). 31. Solve the equation F (x) = 5.

32. Is the graph increasing or decreasing on its domain? 33. Is the function represented by the graph one-to-one?

34. Find F −1(15). 35. Given f (x) = −2x + 11, find f −1(x).

2

1 2 5

Linear Functions

Figure 1 A bamboo forest in China (credit: “JFXie”/Flickr)

Introduction Imagine placing a plant in the ground one day and finding that it has doubled its height just a few days later. Although it may seem incredible, this can happen with certain types of bamboo species. These members of the grass family are the fastest-growing plants in the world. One species of bamboo has been observed to grow nearly 1.5 inches every hour.[6] In a twenty-four hour period, this bamboo plant grows about 36 inches, or an incredible 3 feet! A constant rate of change, such as the growth cycle of this bamboo plant, is a linear function.

Recall from Functions and Function Notation that a function is a relation that assigns to every element in the domain exactly one element in the range. Linear functions are a specific type of function that can be used to model many real-world applications, such as plant growth over time. In this chapter, we will explore linear functions, their graphs, and how to relate them to data.

6 http://www.guinnessworldrecords.com/records-3000/fastest-growing-plant/

CHAPTeR OUTlIne

2.1 linear Functions 2.2 Graphs of linear Functions 2.3 Modeling with linear Functions 2.4 Fitting linear Models to Data

CHAPTER 2 linear Functions1 2 6

2.1 lIneAR FUnCTIOnS

Figure 1 Shanghai Maglev Train (credit: “kanegen”/Flickr)

Just as with the growth of a bamboo plant, there are many situations that involve constant change over time. Consider, for example, the first commercial maglev train in the world, the Shanghai MagLev Train (Figure 1). It carries passengers comfortably for a 30-kilometer trip from the airport to the subway station in only eight minutes.[7]

Suppose a maglev train were to travel a long distance, and that the train maintains a constant speed of 83 meters per second for a period of time once it is 250 meters from the station. How can we analyze the train’s distance from the station as a function of time? In this section, we will investigate a kind of function that is useful for this purpose, and use it to investigate real-world situations such as the train’s distance from the station at a given point in time.

Representing linear Functions The function describing the train’s motion is a linear function, which is defined as a function with a constant rate of change, that is, a polynomial of degree 1. There are several ways to represent a linear function, including word form, function notation, tabular form, and graphical form. We will describe the train’s motion as a function using each method.

Representing a Linear Function in Word Form

Let’s begin by describing the linear function in words. For the train problem we just considered, the following word sentence may be used to describe the function relationship.

• The train’s distance from the station is a function of the time during which the train moves at a constant speed plus its original distance from the station when it began moving at constant speed.

The speed is the rate of change. Recall that a rate of change is a measure of how quickly the dependent variable changes with respect to the independent variable. The rate of change for this example is constant, which means that it is the same for each input value. As the time (input) increases by 1 second, the corresponding distance (output) increases by 83 meters. The train began moving at this constant speed at a distance of 250 meters from the station.

7 http://www.chinahighlights.com/shanghai/transportation/maglev-train.htm

leARnInG OBjeCTIVeS

In this section, you will:

• Represent a linear function.

• Determine whether a linear function is increasing, decreasing, or constant.

• Calculate and interpret slope.

• Write the point-slope form of an equation.

• Write and interpret a linear function.

SECTION 2.1 linear Functions 1 2 7

Representing a Linear Function in Function Notation

Another approach to representing linear functions is by using function notation. One example of function notation is an equation written in the form known as the slope-intercept form of a line, where x is the input value, m is the rate of change, and b is the initial value of the dependent variable.

Equation form y = mx + b

Equation notation f (x) = mx + b

In the example of the train, we might use the notation D(t) in which the total distance D is a function of the time t. The rate, m, is 83 meters per second. The initial value of the dependent variable b is the original distance from the station, 250 meters. We can write a generalized equation to represent the motion of the train.

D(t) = 83t + 250

Representing a Linear Function in Tabular Form

A third method of representing a linear function is through the use of a table. The relationship between the distance from the station and the time is represented in Figure 2. From the table, we can see that the distance changes by 83 meters for every 1 second increase in time.

0 250

1 333

2 416

3 499

t D(t)

1 second 1 second 1 second

83 meters 83 meters 83 meters

Figure 2 Tabular representation of the function D showing selected input and output values

Q & A… Can the input in the previous example be any real number?

No. The input represents time, so while nonnegative rational and irrational numbers are possible, negative real numbers are not possible for this example. The input consists of non-negative real numbers.

Representing a Linear Function in Graphical Form

Another way to represent linear functions is visually, using a graph. We can use the function relationship from above, D(t) = 83t + 250, to draw a graph, represented in Figure 3. Notice the graph is a line. When we plot a linear function, the graph is always a line.

The rate of change, which is constant, determines the slant, or slope of the line. The point at which the input value is zero is the vertical intercept, or y-intercept, of the line. We can see from the graph in Figure 3 that the y-intercept in the train example we just saw is (0, 250) and represents the distance of the train from the station when it began moving at a constant speed.

100 200 300 400 500

D is

ta nc

e (m

)

Time (s)

0 321 4 5

Figure 3 The graph of D(t) = 83t + 250. Graphs of linear functions are lines because the rate of change is constant.

Notice that the graph of the train example is restricted, but this is not always the case. Consider the graph of the line f (x) = 2x + 1. Ask yourself what numbers can be input to the function, that is, what is the domain of the function? The domain is comprised of all real numbers because any number may be doubled, and then have one added to the product.

CHAPTER 2 linear Functions1 2 8

linear function A linear function is a function whose graph is a line. Linear functions can be written in the slope-intercept form of a line

f (x) = mx + b

where b is the initial or starting value of the function (when input, x = 0), and m is the constant rate of change, or slope of the function. The y-intercept is at (0, b).

Example 1 Using a Linear Function to Find the Pressure on a Diver

The pressure, P, in pounds per square inch (PSI) on the diver in Figure 4 depends upon her depth below the water surface, d, in feet. This relationship may be modeled by the equation, P(d) = 0.434d + 14.696. Restate this function in words.

Figure 4 (credit: Ilse Reijs and jan-noud Hutten)

Solution To restate the function in words, we need to describe each part of the equation. The pressure as a function of depth equals four hundred thirty-four thousandths times depth plus fourteen and six hundred ninety-six thousandths.

Analysi s The initial value, 14.696, is the pressure in PSI on the diver at a depth of 0 feet, which is the surface of the water. The rate of change, or slope, is 0.434 PSI per foot. This tells us that the pressure on the diver increases 0.434 PSI for each foot her depth increases.

Determining Whether a linear Function Is Increasing, Decreasing, or Constant The linear functions we used in the two previous examples increased over time, but not every linear function does. A linear function may be increasing, decreasing, or constant. For an increasing function, as with the train example, the output values increase as the input values increase. The graph of an increasing function has a positive slope. A line with a positive slope slants upward from left to right as in Figure 5(a). For a decreasing function, the slope is negative. The output values decrease as the input values increase. A line with a negative slope slants downward from left to right as in Figure 5(b). If the function is constant, the output values are the same for all input values so the slope is zero. A line with a slope of zero is horizontal as in Figure 5(c).

Increasing function Decreasing function f (x)

f

f

f

f (x)f (x)

x x x

Constant function

(c)(b)(a) Figure 5

SECTION 2.1 linear Functions 1 2 9

increasing and decreasing functions The slope determines if the function is an increasing linear function, a decreasing linear function, or a constant function. • f (x) = mx + b is an increasing function if m > 0.

• f (x) = mx + b is an decreasing function if m < 0.

• f (x) = mx + b is a constant function if m = 0.

Example 2 Deciding Whether a Function Is Increasing, Decreasing, or Constant

Some recent studies suggest that a teenager sends an average of 60 texts per day.[8] For each of the following scenarios, find the linear function that describes the relationship between the input value and the output value. Then, determine whether the graph of the function is increasing, decreasing, or constant.

a. The total number of texts a teen sends is considered a function of time in days. The input is the number of days, and output is the total number of texts sent. b. A teen has a limit of 500 texts per month in his or her data plan.

The input is the number of days, and output is the total number of texts remaining for the month.

c. A teen has an unlimited number of texts in his or her data plan for a cost of $50 per month. The input is the number of days, and output is the total cost of texting each month.

Solution Analyze each function.

a. The function can be represented as f (x) = 60x where x is the number of days. The slope, 60, is positive so the function is increasing. This makes sense because the total number of texts increases with each day.

b. The function can be represented as f (x) = 500 − 60x where x is the number of days. In this case, the slope is negative so the function is decreasing. This makes sense because the number of texts remaining decreases each day and this function represents the number of texts remaining in the data plan after x days.

c. The cost function can be represented as f (x) = 50 because the number of days does not affect the total cost. The slope is 0 so the function is constant.

Calculating and Interpreting Slope In the examples we have seen so far, we have had the slope provided for us. However, we often need to calculate the slope given input and output values. Given two values for the input, x1 and x2, and two corresponding values for the output, y1 and y2 —which can be represented by a set of points, (x1, y1) and (x2, y2) —we can calculate the slope m, as follows

m = change in output (rise)

__ change in input (run)

= Δy

_ Δx

= y2 − y1 _ x2 − x1

where ∆y is the vertical displacement and ∆x is the horizontal displacement. Note in function notation two corresponding values for the output y1 and y2 for the function f, y1 = f (x1) and y2 = f (x2), so we could equivalently write

m = f (x2) − f (x1) _ x2 − x1

Figure 6 indicates how the slope of the line between the points, (x1, y1) and (x2, y2), is calculated. Recall that the slope measures steepness. The greater the absolute value of the slope, the steeper the line is.

8 http://www.cbsnews.com/8301-501465_162-57400228-501465/teens-are-sending-60-texts-a-day-study-says/

CHAPTER 2 linear Functions1 3 0

x

y

–1

5 6 7 8

1 2 3 4

9 10

(x1 , y1)

(x2 , y2) x2 – x1 y2 – y1

0 321 4

m = = ∆y ∆x

y2 – y1 x2 – x1

Figure 6 The slope of a function is calculated by the change in y divided by the change in x. It does not matter which coordinate is used as the (x

2 , y

2 ) and which is the (x

1 , y

1 ), as long as each calculation is started with the elements from the same coordinate pair.

Q & A… Are the units for slope always

units for the output __

units for the input ?

Yes. Think of the units as the change of output value for each unit of change in input value. An example of slope could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for the input.

calculate slope The slope, or rate of change, of a function m can be calculated according to the following:

m = change in output (rise)

__ change in input (run)

= Δy

_ Δx

= y2 − y1 _ x2 − x1

where x1 and x2 are input values, y1 and y2 are output values.

How To… Given two points from a linear function, calculate and interpret the slope.

1. Determine the units for output and input values. 2. Calculate the change of output values and change of input values. 3. Interpret the slope as the change in output values per unit of the input value.

Example 3 Finding the Slope of a Linear Function

If f (x) is a linear function, and (3, −2) and (8, 1) are points on the line, find the slope. Is this function increasing or decreasing?

Solution The coordinate pairs are (3, −2) and (8, 1). To find the rate of change, we divide the change in output by the change in input.

m = change in output

__ change in input

= 1 − (−2) ________ 8 − 3

= 3 __ 5

We could also write the slope as m = 0.6. The function is increasing because m > 0.

Analysi s As noted earlier, the order in which we write the points does not matter when we compute the slope of the line as long as the first output value, or y-coordinate, used corresponds with the first input value, or x-coordinate, used.

Try It #1

If f (x) is a linear function, and (2, 3) and (0, 4) are points on the line, find the slope. Is this function increasing or decreasing?

SECTION 2.1 linear Functions 1 3 1

Example 4 Finding the Population Change from a Linear Function

The population of a city increased from 23,400 to 27,800 between 2008 and 2012. Find the change of population per year if we assume the change was constant from 2008 to 2012.

Solution The rate of change relates the change in population to the change in time. The population increased by 27,800 − 23,400 = 4,400 people over the four-year time interval. To find the rate of change, divide the change in the number of people by the number of years.

4,400 people

__ 4 years

= 1,100 people

__ year

So the population increased by 1,100 people per year.

Analysi s Because we are told that the population increased, we would expect the slope to be positive. This positive slope we calculated is therefore reasonable.

Try It #2

The population of a small town increased from 1,442 to 1,868 between 2009 and 2012. Find the change of population per year if we assume the change was constant from 2009 to 2012.

Writing the Point-Slope Form of a linear equation Up until now, we have been using the slope-intercept form of a linear equation to describe linear functions. Here, we will learn another way to write a linear function, the point-slope form.

y − y1 = m(x − x1)

The point-slope form is derived from the slope formula.

m = y − y1 _ x − x1

Assuming x ≠ x1

m(x − x1) = y − y1 _ x − x1

(x − x1) Multiply both sides by (x − x1).

m(x − x1) = y − y1 Simplify.

y − y1 = m(x − x1) Rearrange.

Keep in mind that the slope-intercept form and the point-slope form can be used to describe the same function. We can move from one form to another using basic algebra. For example, suppose we are given an equation in point-slope form, y − 4 = − 1 __

2 (x − 6).

We can convert it to the slope-intercept form as shown.

y − 4 = − 1 __ 2

(x − 6)

y − 4 = − 1 __ 2

x + 3 Distribute the − 1 __ 2

.

y = − 1 __ 2

x + 7 Add 4 to each side.

Therefore, the same line can be described in slope-intercept form as y = − 1 __ 2

x + 7.

point-slope form of a linear equation The point-slope form of a linear equation takes the form

y − y1 = m(x − x1)

where m is the slope, x1 and y1 are the x- and y-coordinates of a specific point through which the line passes.

CHAPTER 2 linear Functions1 3 2

Writing the Equation of a Line Using a Point and the Slope

The point-slope form is particularly useful if we know one point and the slope of a line. Suppose, for example, we are told hat a line has a slope of 2 and passes through the point (4, 1). We know that m = 2 and that x1 = 4 and y1 = 1. We can substitute these values into the general point-slope equation. y − y1 = m(x − x1) y − 1 = 2(x − 4)

If we wanted to then rewrite the equation in slope-intercept form, we apply algebraic techniques. y − 1 = 2(x − 4) y − 1 = 2x − 8 Distribute the 2. y = 2x − 7 Add 1 to each side.

Both equations, y − 1 = 2(x − 4) and y = 2x − 7, describe the same line. See Figure 7.

x

y

–2–4–6–8–10 –2 –4 –6 –8

–10

2 4 6 8

10

642 8 10

Figure 7

Example 5 Writing Linear Equations Using a Point and the Slope

Write the point-slope form of an equation of a line with a slope of 3 that passes through the point (6, −1). Then rewrite it in the slope-intercept form.

Solution Let’s figure out what we know from the given information. The slope is 3, so m = 3. We also know one point, so we know x1 = 6 and y1 = −1. Now we can substitute these values into the general point-slope equation.

y − y1 = m(x − x1)

y − (−1) = 3(x − 6) Substitute known values.

y + 1 = 3(x − 6) Distribute −1 to find point-slope form.

Then we use algebra to find the slope-intercept form.

y + 1 = 3(x − 6)

y + 1 = 3x − 18 Distribute 3.

y = 3x − 19 Simplify to slope-intercept form.

Try It #3

Write the point-slope form of an equation of a line with a slope of −2 that passes through the point (−2, 2). Then rewrite it in the slope-intercept form.

Writing the Equation of a Line Using Two Points

The point-slope form of an equation is also useful if we know any two points through which a line passes. Suppose, for example, we know that a line passes through the points (0, 1) and (3, 2). We can use the coordinates of the two points to find the slope.

SECTION 2.1 linear Functions 1 3 3

m = y2 − y1 _ x2 − x1

= 2 − 1 _ 3 − 0

= 1 __ 3

Now we can use the slope we found and the coordinates of one of the points to find the equation for the line. Let use (0, 1) for our point. y − y1 = m(x − x1)

y − 1 = 1 __ 3

(x − 0)

As before, we can use algebra to rewrite the equation in the slope-intercept form.

y − 1 = 1 __ 3

(x − 0)

y − 1 = 1 __ 3

x Distribute the 1 __ 3

.

y = 1 __ 3

x + 1 Add 1 to each side.

Both equations describe the line shown in Figure 8.

2

x

y

–1–2–3–4–5–6 –1 –2 –3 –4 –5 –6

1

3

3

21 4

4

5 6

5 6

Figure 8

Example 6 Writing Linear Equations Using Two Points

Write the point-slope form of an equation of a line that passes through the points (5, 1) and (8, 7). Then rewrite it in the slope-intercept form.

Solution Let’s begin by finding the slope.

m = y2 − y1 _ x2 − x1

= 7 − 1 _ 8 − 5

= 6 __ 3

= 2

So m = 2. Next, we substitute the slope and the coordinates for one of the points into the general point-slope equation. We can choose either point, but we will use (5, 1).

y − y1 = m(x − x1)

y − 1 = 2(x − 5)

The point-slope equation of the line is y2 − 1 = 2(x2 − 5). To rewrite the equation in slope-intercept form, we use algebra. y − 1 = 2(x − 5)

y − 1 = 2x − 10

y = 2x − 9

The slope-intercept equation of the line is y = 2x − 9.

CHAPTER 2 linear Functions1 3 4

Try It #4

Write the point-slope form of an equation of a line that passes through the points (−1, 3) and (0, 0). Then rewrite it in the slope-intercept form.

Writing and Interpreting an equation for a linear Function Now that we have written equations for linear functions in both the slope-intercept form and the point-slope form, we can choose which method to use based on the information we are given. That information may be provided in the form of a graph, a point and a slope, two points, and so on. Look at the graph of the function f in Figure 9.

x

f y

–2–4–6–8–10 –2 –4

2 4 6 8

10

6

(4, 4) (0, 7)

42 8 10

Figure 9

We are not given the slope of the line, but we can choose any two points on the line to find the slope. Let’s choose (0, 7) and (4, 4). We can use these points to calculate the slope.

m = y2 − y1 _ x2 − x1

= 4 − 7 _ 4 − 0

= −  3 __ 4

Now we can substitute the slope and the coordinates of one of the points into the point-slope form.

y − y1 = m(x − x1)

y − 4 = −  3 __ 4

(x − 4)

If we want to rewrite the equation in the slope-intercept form, we would find

y − 4 = −  3 __ 4

(x − 4)

y − 4 = −  3 __ 4

x + 3

y = −  3 __ 4

x + 7

If we wanted to find the slope-intercept form without first writing the point-slope form, we could have recognized that the line crosses the y-axis when the output value is 7. Therefore, b = 7. We now have the initial value b and the slope m so we can substitute m and b into the slope-intercept form of a line.

f (x) = mx + b ↑ ↑ −  3 __

4 7

f (x) = −  3 __ 4

x + 7

So the function is f (x) = −  3 __ 4

x + 7, and the linear equation would be y = −  3 __ 4

x + 7.

SECTION 2.1 linear Functions 1 3 5

How To… Given the graph of a linear function, write an equation to represent the function.

1. Identify two points on the line. 2. Use the two points to calculate the slope. 3. Determine where the line crosses the y-axis to identify the y-intercept by visual inspection. 4. Substitute the slope and y-intercept into the slope-intercept form of a line equation.

Example 7 Writing an Equation for a Linear Function

Write an equation for a linear function given a graph of f shown in Figure 10.

x

f

y

–2–4–6–8–10 –2 –4 –6 –8

–10

2 4 6 8

10

642 8 10

Figure 10

Solution Identify two points on the line, such as (0, 2) and (−2, −4). Use the points to calculate the slope.

m = y2 − y1 _ x2 − x1

= −4 − 2 _ −2 − 0

= −6 _ −2

= 3 Substitute the slope and the coordinates of one of the points into the point-slope form.

y − y1 = m(x − x1)

y − (−4) = 3(x − (−2))

y + 4 = 3(x + 2)

We can use algebra to rewrite the equation in the slope-intercept form.

y + 4 = 3(x + 2)

y + 4 = 3x + 6

y = 3x + 2

Analysi s This makes sense because we can see from Figure 11 that the line crosses the y-axis at the point (0, 2), which is the y-intercept, so b = 2.

x

y

–2–4–6–8–10 –2 –4 –6 –8

–10

2 4 6 8

10

642 8 10

(0, 2)

(−2, −4)

Figure 11

CHAPTER 2 linear Functions1 3 6

Example 8 Writing an Equation for a Linear Cost Function

Suppose Ben starts a company in which he incurs a fixed cost of $1,250 per month for the overhead, which includes his office rent. His production costs are $37.50 per item. Write a linear function C where C(x) is the cost for x items produced in a given month.

Solution The fixed cost is present every month, $1,250. The costs that can vary include the cost to produce each item, which is $37.50 for Ben. The variable cost, called the marginal cost, is represented by 37.5. The cost Ben incurs is the sum of these two costs, represented by C(x) = 1250 + 37.5x.

Analysi s If Ben produces 100 items in a month, his monthly cost is represented by

C(100) = 1250 + 37.5(100) = 5000 So his monthly cost would be $5,000.

Example 9 Writing an Equation for a Linear Function Given Two Points

If f is a linear function, with f (3) = −2 , and f (8) = 1 , find an equation for the function in slope-intercept form.

Solution We can write the given points using coordinates.

f (3) = −2 → (3, −2)

f (8) = 1 → (8, 1)

We can then use the points to calculate the slope.

m = y2 − y1 _ x2 − x1

= 1 − (−2) ________ 8 − 3

=  3 __ 5

Substitute the slope and the coordinates of one of the points into the point-slope form.

y − y1 = m(x − x1)

y − (−2) =   3 __ 5

(x − 3)

We can use algebra to rewrite the equation in the slope-intercept form.

y + 2 =   3 __ 5

(x − 3)

y + 2 =   3 __ 5

x −   9 __ 5

y =  3 __ 5

x − 19 __ 5

Try It #5

If f (x) is a linear function, with f (2) = −11, and f (4) = −25, find an equation for the function in slope-intercept form.

Modeling Real-World Problems with linear Functions In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know, or can figure out, the initial value and the rate of change of a linear function, we can solve many different kinds of real-world problems.

SECTION 2.1 linear Functions 1 3 7

How To… Given a linear function f and the initial value and rate of change, evaluate f (c).

1. Determine the initial value and the rate of change (slope). 2. Substitute the values into f (x) = mx + b. 3. Evaluate the function at x = c.

Example 10 Using a Linear Function to Determine the Number of Songs in a Music Collection

Marcus currently has 200 songs in his music collection. Every month, he adds 15 new songs. Write a formula for the number of songs, N, in his collection as a function of time, t, the number of months. How many songs will he own in a year?

Solution The initial value for this function is 200 because he currently owns 200 songs, so N(0) = 200, which means that b = 200.

The number of songs increases by 15 songs per month, so the rate of change is 15 songs per month. Therefore we know that m = 15. We can substitute the initial value and the rate of change into the slope-intercept form of a line.

f (x) = mx + b ↑ ↑ 15 200

N(t) = 15t + 200 Figure 12

We can write the formula N(t) = 15t + 200.

With this formula, we can then predict how many songs Marcus will have in 1 year (12 months). In other words, we can evaluate the function at t = 12. N(12) = 15(12) + 200

= 180 + 200

= 380

Marcus will have 380 songs in 12 months.

Analysi s Notice that N is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well.

Example 11 Using a Linear Function to Calculate Salary Plus Commission

Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya’s weekly income, I, depends on the number of new policies, n, he sells during the week. Last week he sold 3 new policies, and earned $760 for the week. The week before, he sold 5 new policies and earned $920. Find an equation for I(n), and interpret the meaning of the components of the equation.

Solution The given information gives us two input-output pairs: (3,760) and (5,920). We start by finding the rate of change.

m = 920 − 760 ________ 5 − 3

= $160 _______ 2 policies

= $80 per policy

Keeping track of units can help us interpret this quantity. Income increased by $160 when the number of policies increased by 2, so the rate of change is $80 per policy. Therefore, Ilya earns a commission of $80 for each policy sold during the week.

CHAPTER 2 linear Functions1 3 8

We can then solve for the initial value. I(n) = 80n + b

760 = 80(3) + b when n = 3, I(3) = 760

760 − 80(3) = b

520 = b

The value of b is the starting value for the function and represents Ilya’s income when n = 0, or when no new policies are sold. We can interpret this as Ilya’s base salary for the week, which does not depend upon the number of policies sold.

We can now write the final equation. I(n) = 80n + 520

Our final interpretation is that Ilya’s base salary is $520 per week and he earns an additional $80 commission for each policy sold.

Example 12 Using Tabular Form to Write an Equation for a Linear Function

Table 1 relates the number of rats in a population to time, in weeks. Use the table to write a linear equation.

w, number of weeks 0 2 4 6

P(w), number of rats 1,000 1,080 1,160 1,240 Table 1

Solution We can see from the table that the initial value for the number of rats is 1000, so b = 1000.

Rather than solving for m, we can tell from looking at the table that the population increases by 80 for every 2 weeks that pass. This means that the rate of change is 80 rats per 2 weeks, which can be simplified to 40 rats per week.

P(w) = 40w + 1000

If we did not notice the rate of change from the table we could still solve for the slope using any two points from the table. For example, using (2, 1080) and (6, 1240)

m = 1240 − 1080 __________ 6 − 2

= 160 ___ 4

= 40

Q & A… Is the initial value always provided in a table of values like Table 1?

No. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of 0, then the initial value would be the corresponding output. If the initial value is not provided because there is no value of input on the table equal to 0, find the slope, substitute one coordinate pair and the slope into f (x) = mx + b, and solve for b.

Try It #6

A new plant food was introduced to a young tree to test its effect on the height of the tree. Table 2 shows the height of the tree, in feet, x months since the measurements began. Write a linear function, H(x), where x is the number of months since the start of the experiment.

x 0 2 4 8 12

H(x) 12.5 13.5 14.5 16.5 18.5

Table 2

Access this online resource for additional instruction and practice with linear functions. • linear Functions (http://openstaxcollege.org/l/linearfunctions)

SECTION 2.1 section exercises 1 3 9

2.1 SeCTIOn exeRCISeS

VeRBAl

1. Terry is skiing down a steep hill. Terry’s elevation, E(t), in feet after t seconds is given by E(t) = 3000 − 70t. Write a complete sentence describing Terry’s starting elevation and how it is changing over time.

2. Maria is climbing a mountain. Maria’s elevation, E(t), in feet after t minutes is given by E(t) = 1200 + 40t. Write a complete sentence describing Maria’s starting elevation and how it is changing over time.

3. Jessica is walking home from a friend’s house. After 2 minutes she is 1.4 miles from home. Twelve minutes after leaving, she is 0.9 miles from home. What is her rate in miles per hour?

4. Sonya is currently 10 miles from home and is walking farther away at 2 miles per hour. Write an equation for her distance from home t hours from now.

5. A boat is 100 miles away from the marina, sailing directly toward it at 10 miles per hour. Write an equation for the distance of the boat from the marina after t hours.

6. Timmy goes to the fair with $40. Each ride costs $2. How much money will he have left after riding n rides?

AlGeBRAIC

For the following exercises, determine whether the equation of the curve can be written as a linear function.

7. y =   1 __ 4

x + 6 8. y = 3x − 5 9. y = 3x 2 − 2

10. 3x + 5y = 15 11. 3x 2 + 5y = 15 12. 3x + 5y 2 = 15

13. −2x 2 + 3y2 = 6 14. −  x − 3 ______ 5

= 2y

For the following exercises, determine whether each function is increasing or decreasing.

15. f (x) = 4x + 3 16. g(x) = 5x + 6 17. a(x) = 5 − 2x

18. b(x) = 8 − 3x 19. h(x) = −2x + 4 20. k(x) = −4x + 1

21. j(x) =   1 __ 2

x − 3 22. p(x) =   1 __ 4

x − 5 23. n(x) = −  1 __ 3

x − 2

24. m(x) = −  3 __ 8

x + 3

For the following exercises, find the slope of the line that passes through the two given points.

25. (2, 4) and (4, 10) 26. (1, 5) and (4, 11) 27. (−1, 4) and (5, 2)

28. (8, −2) and (4, 6) 29. (6, 11) and (−4, 3)

1 4 0 CHAPTER 2 linear Functions

For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible.

30. f (−5) = −4, and f (5) = 2 31. f (−1) = 4 and f (5) = 1

32. (2, 4) and (4, 10) 33. Passes through (1, 5) and (4, 11)

34. Passes through (−1, 4) and (5, 2) 35. Passes through (−2, 8) and (4, 6)

36. x-intercept at (−2, 0) and y-intercept at (0, −3) 37. x-intercept at (−5, 0) and y-intercept at (0, 4)

GRAPHICAl

For the following exercises, find the slope of the lines graphed.

38.

x

y

–1–2–3–4–5–6 –1 –2 –3 –4 –5

1 2 3 4 5

0 321 4 5 6

39.

x

y

–1–2–3–4–5–6 –1 –2 –3 –4 –5 –6

1 2 3 4 5 6

0 321 4 5 6

40.

x

y

–1–2–3–4–5–6 –1 –2 –3 –4 –5 –6

1 2 3 4 5 6

0 321 4 5 6

For the following exercises, write an equation for the lines graphed.

41.

x

y

–1–2–3–4–5–6 –1 –2 –3 –4 –5 –6

1 2 3 4 5 6

0 321 4 5 6

42.

x

y

–1–2–3–4–5–6 –1 –2 –3 –4 –5 –6

1 2 3 4 5 6

0 321 4 5 6

43.

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5 –6

1 2 3 4

6 5

0 321 4 5

44.

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5 –6

1 2 3 4

6 5

0 321 4 5

45.

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1 2 3 4 5

0 321 4 5 6–6

46.

x

y

–1–2–3–4–5 –1 –2 –3 –4 –5

1 2 3 4 5

0 321 4 5 6–6

SECTION 2.1 section exercises 1 4 1

nUMeRIC

For the following exercises, which of the tables could represent a linear function? For each that could be linear, find a linear equation that models the data.

47. x 0 5 10 15 g(x) 5 −10 −25 −40

48. x 0 5 10 15 h(x) 5 30 105 230

49. x 0 5 10 15 f (x) −5 20 45 70

50. x 5 10 20 25 k(x) 13 28 58 73

51. x 0 2 4 6 g(x) 6 −19 −44 −69

52. x 2 4 6 8 f (x) 13 23 43 53

53. x 2 4 6 8 f (x) −4 16 36 56

54. x 0 2 6 8 k(x) 6 31 106 231

TeCHnOlOGY

55. If f is a linear function, f (0.1) = 11.5, and f (0.4) = −5.9, find an equation for the function.

56. Graph the function f on a domain of [−10, 10] : f (x) = 0.02x − 0.01. Enter the function in a graphing utility. For the viewing window, set the minimum value of x to be −10 and the maximum value of x to be 10.

57. Graph the function f on a domain of [−10, 10] : f (x) = 2,500x + 4,000

58. Table 3 shows the input, w, and output, k, for a linear function k. a. Fill in the missing values of the table. b. Write the linear function k, round to 3 decimal places.

w −10 5.5 67.5 b

k 30 −26 a −44

Table 3

59. Table 4 shows the input, p, and output, q, for a linear function q. a. Fill in the missing values of the table. b. Write the linear function k.

p 0.5 0.8 12 b

q 400 700 a 1,000,000

Table 4

60. Graph the linear function f on a domain of [−10, 10] for the function whose slope is 1 __ 8

and y-intercept is 31 __ 16

. Label the points for the input values of −10 and 10.

61. Graph the linear function f on a domain of [−0.1, 0.1] for the function whose slope is 75 and y-intercept is −22.5. Label the points for the input values of −0.1 and 0.1.

62. Graph the linear function f where f (x) = ax + b on the same set of axes on a domain of [−4, 4] for the following values of a and b. a. a = 2; b = 3 b. a = 2; b = 4 c. a = 2; b = −4 d. a = 2; b = −5

exTenSIOnS

63. Find the value of x if a linear function goes through the following points and has the following slope: (x, 2), (−4, 6), m = 3

64. Find the value of y if a linear function goes through the following points and has the following slope: (10, y), (25, 100), m = −5

65. Find the equation of the line that passes through the following points: (a, b) and (a, b + 1)

66. Find the equation of the line that passes through the following points: (2a, b) and (a, b + 1)

67. Find the equation of the line that passes through the following points: (a, 0) and (c, d)

1 4 2 CHAPTER 2 linear Functions

ReAl-WORlD APPlICATIOnS 68. At noon, a barista notices that she has $20 in her

tip jar. If she makes an average of $0.50 from each customer, how much will she have in her tip jar if she serves n more customers during her shift?

69. A gym membership with two personal training sessions costs $125, while gym membership with five personal training sessions costs $260. What is cost per session?

70. A clothing business finds there is a linear relationship between the number of shirts, n, it can sell and the price, p, it can charge per shirt. In particular, historical data shows that 1,000 shirts can be sold at a price of $30, while 3,000 shirts can be sold at a price of $22. Find a linear equation in the form p(n) = mn + b that gives the price p they can charge for n shirts.

71. A phone company charges for service according to the formula: C(n) = 24 + 0.1n, where n is the number of minutes talked, and C(n) is the monthly charge, in dollars. Find and interpret the rate of change and initial value.

72. A farmer finds there is a linear relationship between the number of bean stalks, n, she plants and the yield, y, each plant produces. When she plants 30 stalks, each plant yields 30 oz of beans. When she plants 34 stalks, each plant produces 28 oz of beans. Find a linear relationship in the form y = mn + b that gives the yield when n stalks are planted.

73. A city’s population in the year 1960 was 287,500. In 1989 the population was 275,900. Compute the rate of growth of the population and make a statement about the population rate of change in people per year.

74. A town’s population has been growing linearly. In 2003, the population was 45,000, and the population has been growing by 1,700 people each year. Write an equation, P(t), for the population t years after 2003.

75. Suppose that average annual income (in dollars) for the years 1990 through 1999 is given by the linear function: I(x) = 1,054x + 23,286, where x is the number of years after 1990. Which of the following interprets the slope in the context of the problem? a. As of 1990, average annual income was $23,286. b. In the ten-year period from 1990–1999, average

annual income increased by a total of $1,054. c. Each year in the decade of the 1990s, average

annual income increased by $1,054. d. Average annual income rose to a level of $23,286

by the end of 1999.

76. When temperature is 0 degrees Celsius, the Fahrenheit temperature is 32. When the Celsius temperature is 100, the corresponding Fahrenheit temperature is 212. Express the Fahrenheit temperature as a linear function of C, the Celsius temperature, F (C). a. Find the rate of change of Fahrenheit temperature

for each unit change temperature of Celsius. b. Find and interpret F (28). c. Find and interpret F (−40).

SECTION 2.2 graphs oF linear Functions 1 4 3

2.2 GRAPHS OF lIneAR FUnCTIOnS

Two competing telephone companies offer different payment plans. The two plans charge the same rate per long distance minute, but charge a different monthly flat fee. A consumer wants to determine whether the two plans will ever cost the same amount for a given number of long distance minutes used. The total cost of each payment plan can be represented by a linear function. To solve the problem, we will need to compare the functions. In this section, we will consider methods of comparing functions using graphs.

Graphing linear Functions In Linear Functions, we saw that the graph of a linear function is a straight line. We were also able to see the points of the function as well as the initial value from a graph. By graphing two functions, then, we can more easily compare their characteristics.

There are three basic methods of graphing linear functions. The first is by plotting points and then drawing a line through the points. The second is by using the y-intercept and slope. And the third is by using transformations of the identity function f (x) = x.

Graphing a Function by Plotting Points

To find points of a function, we can choose input values, evaluate the function at these input values, and calculate output values. The input values and corresponding output values form coordinate pairs. We then plot the coordinate pairs on a grid. In general, we should evaluate the function at a minimum of two inputs in order to find at least two points on the graph. For example, given the function, f (x) = 2x, we might use the input values 1 and 2. Evaluating the function for an input value of 1 yields an output value of 2, which is represented by the point (1, 2). Evaluating the function for an input value of 2 yields an output value of 4, which is represented by the point (2, 4). Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error.

How To… Given a linear function, graph by plotting points.

1. Choose a minimum of two input values. 2. Evaluate the function at each input value. 3. Use the resulting output values to identify coordinate pairs. 4. Plot the coordinate pairs on a grid. 5. Draw a line through the points.

Example 1 Graphing by Plotting Points

Graph f (x) = −  2 __ 3

x + 5 by plotting points.

Solution Begin by choosing input values. This function includes a fraction with a denominator of 3, so let’s choose multiples of 3 as input values. We will choose 0, 3, and 6.

leARnInG OBjeCTIVeS

In this section, you will:

• Graph linear functions.

• Write the equation for a linear function from the graph of a line.

• Given the equations of two lines, determine whether their graphs are parallel or perpendicular.

• Write the equation of a line parallel or perpendicular to a given line.

• Solve a system of linear equations.

CHAPTER 2 linear Functions1 4 4

Evaluate the function at each input value, and use the output value to identify coordinate pairs.

x = 0 f (0) = −  2 __ 3

(0) + 5 = 5 (0, 5)

x = 3 f (3) = −  2 __ 3

(3) + 5 = 3 (3, 3)

x = 6 f (6) = −  2 __ 3

(6) + 5 = 1 (6, 1)

Plot the coordinate pairs and draw a line through the points. Figure 1 represents the graph of the function f (x) = − 2 __

3 x + 5.

f

(6, 1)

(3, 3)

(0, 5)

f(x)

x –1

1

5

6

3

2

– 7321 654

4

Figure 1 The graph of the linear function f (x) = −  2 __ 3 x + 5.

Analysis The graph of the function is a line as expected for a linear function. In addition, the graph has a downward slant, which indicates a negative slope. This is also expected from the negative constant rate of change in the equation for the function.

Try It #1

Graph f (x) = −  3 __ 4

x + 6 by plotting points.

Graphing a Function Using y-intercept and Slope

Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. The first characteristic is its y-intercept, which is the point at which the input value is zero. To find the y-intercept, we can set x = 0 in the equation.

The other characteristic of the linear function is its slope m, which is a measure of its steepness. Recall that the slope is the rate of change of the function. The slope of a function is equal to the ratio of the change in outputs to the change in inputs. Another way to think about the slope is by dividing the vertical difference, or rise, by the horizontal difference, or run. We encountered both the y-intercept and the slope in Linear Functions.

Let’s consider the following function. f (x) =   1 __

2 x + 1

The slope is 1 __ 2

. Because the slope is positive, we know the graph will slant upward from left to right. The y-intercept is the point on the graph when x = 0. The graph crosses the y-axis at (0, 1). Now we know the slope and the y-intercept. We can begin graphing by plotting the point (0, 1). We know that the slope is rise over run, m = rise ___

run .

From our example, we have m =   1 __ 2

, which means that the rise is 1 and the run is 2. So starting from our y-intercept (0, 1), we can rise 1 and then run 2, or run 2 and then rise 1. We repeat until we have a few points, and then we draw a line through the points as shown in Figure 2.

f

(0, 1)

y-intercept

↑Run = 2 ←Rise = 1

y

x –1–2

5

3

2

7321 654

4

Figure 2

SECTION 2.2 graphs oF linear Functions 1 4 5

graphical interpretation of a linear function In the equation f (x) = mx + b

• b is the y-intercept of the graph and indicates the point (0, b) at which the graph crosses the y-axis.

• m is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope:

m = change in output (rise)

__ change in input (run)

= Δy

_ Δx

= y2 − y1 _ x2 − x1

Q & A… Do all linear functions have y-intercepts?

Yes. All linear functions cross the y-axis and therefore have y-intercepts. (Note: A vertical line parallel to the y-axis does not have a y-intercept, but it is not a function.)

How To… Given the equation for a linear function, graph the function using the y-intercept and slope.

1. Evaluate the function at an input value of zero to find the y-intercept. 2. Identify the slope as the rate of change of the input value. 3. Plot the point represented by the y-intercept.

4. Use rise ___ run

to determine at least two more points on the line.

5. Sketch the line that passes through the points.

Example 2 Graphing by Using the y-intercept and Slope

Graph f (x) = −  2 __ 3

x + 5 using the y-intercept and slope.

Solution Evaluate the function at x = 0 to find the y-intercept. The output value when x = 0 is 5, so the graph will cross the y-axis at (0, 5).

According to the equation for the function, the slope of the line is −  2 __ 3

. This tells us that for each vertical decrease in the “rise” of −2 units, the “run” increases by 3 units in the horizontal direction. We can now graph the function by first plotting the y-intercept on the graph in Figure 3. From the initial value (0, 5) we move down 2 units and to the right 3 units. We can extend the line to the left and right by repeating, and then draw a line through the points.

f(x)

f

x –1

1

5

6

3

2

– 7321 654

4

Figure 3

Analysi s The graph slants downward from left to right, which means it has a negative slope as expected.

Try It #2

Find a point on the graph we drew in Example 2 that has a negative x-value.

CHAPTER 2 linear Functions1 4 6

Graphing a Function Using Transformations

Another option for graphing is to use transformations of the identity function f (x) = x. A function may be transformed by a shift up, down, left, or right. A function may also be transformed using a reflection, stretch, or compression.

Vertical Stretch or Compression

In the equation f (x) = mx, the m is acting as the vertical stretch or compression of the identity function. When m is negative, there is also a vertical reflection of the graph. Notice in Figure 4 that multiplying the equation of f (x) = x by m stretches the graph of f by a factor of m units if m > 1 and compresses the graph of f by a factor of m units if 0 < m < 1. This means the larger the absolute value of m, the steeper the slope.

–1–2

–2 –1

1

2

x

y

–3

–3

–4

–4

–5 –6

–5–6

1

3

3

2 4

4

5 6

5 6

Figure 4 Vertical stretches and compressions and reflections on the function f ( x ) = x.

Vertical Shift

In f (x) = mx + b, the b acts as the vertical shift, moving the graph up and down without affecting the slope of the line. Notice in Figure 5 that adding a value of b to the equation of f (x) = x shifts the graph of f a total of b units up if b is positive and ∣b∣ units down if b is negative.

4

x

y

–2–4–6–8–10 –2 –4 –6 –8

–10

2

6

6

42 8

8

10

10

f(x) = x + 4 f(x) = x + 2

f(x) = x – 2 f(x) = x – 4

f(x) = x

Figure 5 This graph illustrates vertical shifts of the function f ( x ) = x.

Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of linear functions. Although this may not be the easiest way to graph this type of function, it is still important to practice each method.

How To… Given the equation of a linear function, use transformations to graph the linear function in the form f (x) = mx + b.

1. Graph f (x) = x. 2. Vertically stretch or compress the graph by a factor m. 3. Shift the graph up or down b units.

SECTION 2.2 graphs oF linear Functions 1 4 7

Example 3 Graphing by Using Transformations

Graph f (x) =  1 __ 2

x − 3 using transformations.

Solution The equation for the function shows that m =   1 __ 2

so the identity function is vertically compressed by 1 __ 2

. The equation for the function also shows that b = −3 so the identity function is vertically shifted down 3 units. First, graph the identity function, and show the vertical compression as in Figure 6.

Then show the vertical shift as in Figure 7.

2

x

y = x y

–1–2–3–4–5–6–7 –1 –2 –3 –4

–5

3

3

21 4

4

5 6 7

y = x12

Figure 6 The function, y = x 1__

compressed by a factor of 2 .

y = x − 312 2

x –1–2–3–4–5–6–7 –1

–2 –3 –4

1

3

3

21 4

4

5 6 7

5 y = x12

Figure 7 The function y =   __1 2

x, shifted down 3 units.

Try It #3 Graph f (x) = 4 + 2x, using t ransformations. Q & A… In Example 3, could we have sketched the graph by reversing the order of the transformations? No. The order of the transformations follows the order of operations. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. This is why we performed the compression first. For example, following the order: Let the input be 2.

f (2) =   1 __ 2

(2) − 3

= 1 − 3 = −2

Writing the equation for a Function from the Graph of a line Recall that in Linear Functions, we wrote the equation for a linear function from a graph. Now we can extend what we know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at Figure 8. We can see right away that the graph crosses the y-axis at the point (0, 4) so this is the y-intercept.

4

x

fy

–2–4–6–8–10 –2 –4 –6 –8

–10

2

6

6

42 8

8

10

10

Figure 8

Then we can calculate the slope by finding the rise and run. We can choose any two points, but let’s look at the point (−2, 0). To get from this point to the y-intercept, we must move up 4 units (rise) and to the right 2 units (run). So the slope must be

m = rise ___ run

=   4 __ 2

= 2

Substituting the slope and y-intercept into the slope-intercept form of a line gives

y = 2x + 4

1

y

CHAPTER 2 linear Functions1 4 8

How To… Given a graph of linear function, find the equation to describe the function.

1. Identify the y-intercept of an equation. 2. Choose two points to determine the slope. 3. Substitute the y-intercept and slope into the slope-intercept form of a line.

Example 4 Matching Linear Functions to Their Graphs

Match each equation of the linear functions with one of the lines in Figure 9.

a. f (x) = 2x + 3 b. g(x) = 2x − 3 c. h(x) = −2x + 3 d. j(x) =   1 __ 2

x + 3

I II III

IV

2

x

y

–1–2–3–4–5–6–7 –1 –2 –3 –4 –5

1

3

3

21 4

4

5 6 7

5

Figure 9

Solution Analyze the information for each function.

a. This function has a slope of 2 and a y-intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. We can use two points to find the slope, or we can compare it with the other functions listed. Function g has the same slope, but a different y-intercept. Lines I and III have the same slant because they have the same slope. Line III does not pass through (0, 3) so f must be represented by line I.

b. This function also has a slope of 2, but a y-intercept of −3. It must pass through the point (0, −3) and slant upward from left to right. It must be represented by line III.

c. This function has a slope of −2 and a y-intercept of 3. This is the only function listed with a negative slope, so it must be represented by line IV because it slants downward from left to right.

d. This function has a slope of 1 __ 2

and a y-intercept of 3. It must pass through the point (0, 3) and slant upward

from left to right. Lines I and II pass through (0, 3), but the slope of j is less than the slope of f so the line for j must be flatter. This function is represented by Line II.

Now we can re-label the lines as in Figure 10.

h(x) = −2x + 3f (x) = 2x + 3

g(x) = 2x − 3

j(x) = x + 312 2

x

y

–1–2–3–4–5–6–7 –1 –2 –3 –4 –5

1

3

3

21 4

4

5 6 7

5

SECTION 2.2 graphs oF linear Functions 1 4 9

Figure 10

Finding the x-intercept of a Line So far, we have been finding the y-intercepts of a function: the point at which the graph of the function crosses the y-axis. A function may also have an x-intercept, which is the x-coordinate of the point where the graph of the function crosses the x-axis. In other words, it is the input value when the output value is zero.

To find the x-intercept, set a function f (x) equal to zero and solve for the value of x. For example, consider the function shown.

f (x) = 3x − 6

Set the function equal to 0 and solve for x. 0 = 3x − 6 6 = 3x 2 = x x = 2

The graph of the function crosses the x-axis at the point (2, 0).

Q & A… Do all linear functions have x-intercepts?

No. However, linear functions of the form y = c, where c is a nonzero real number are the only examples of linear functions with no x-intercept. For example, y = 5 is a horizontal line 5 units above the x-axis. This function has no x-intercepts, as shown in Figure 11.

2

x

y

–1–1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

y = 5

5

5

Figure 11

x-intercept The x-intercept of the function is value of x when f (x) = 0. It can be solved by the equation 0 = mx + b.

Example 5 Finding an x-intercept

Find the x-intercept of f (x) =   1 __ 2

x − 3.

Solution Set the function equal to zero to solve for x.

0 =   1 __ 2

x − 3

3 =   1 __ 2

x

6 = x

x = 6

CHAPTER 2 linear Functions1 5 0

The graph crosses the x-axis at the point (6, 0).

Analysi s A graph of the function is shown in Figure 12. We can see that the x-intercept is (6, 0) as we expected.

4

x

y

–2–4–6–8–10 –2 –4 –6 –8

–10

2

642 8 10

Figure 12 The graph of the linear function f (x) = 1 __ 2 x − 3.

Try It #4

Find the x-intercept of f (x) =  1 __ 4

x − 4.

Describing Horizontal and Vertical Lines

There are two special cases of lines on a graph—horizontal and vertical lines. A horizontal line indicates a constant output, or y-value. In Figure 13, we see that the output has a value of 2 for every input value. The change in outputs between any two points, therefore, is 0. In the slope formula, the numerator is 0, so the slope is 0. If we use m = 0 in the equation f (x) = mx + b, the equation simplifies to f (x) = b. In other words, the value of the function is a constant. This graph represents the function f (x) = 2.

x −4 −2 0 2 4 y 2 2 2 2 2

2

x

y

–1–2–3–4–5

1

3

3

21 4

4 f

5

5

Figure 13 A horizontal line representing the function f (x) = 2.

A vertical line indicates a constant input, or x-value. We can see that the input value for every point on the line is 2, but the output value varies. Because this input value is mapped to more than one output value, a vertical line does not represent a function. Notice that between any two points, the change in the input values is zero. In the slope formula, the denominator will be zero, so the slope of a vertical line is undefined.

m = change of output

__ change of input

← Non-zero real number ← 0

Notice that a vertical line, such as the one in Figure 14, has an x-intercept, but no y-intercept unless it’s the line x = 0. This graph represents the line x = 2.

2

x

y

–1–1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

x 2 2 2 2 2 y −4 −2 0 2 4

SECTION 2.2 graphs oF linear Functions 1 5 1

Figure 14 The vertical line, x = 2, which does not represent a function.

horizontal and vertical lines Lines can be horizontal or vertical.

A horizontal line is a line defined by an equation in the form f (x) = b.

A vertical line is a line defined by an equation in the form x = a.

Example 6 Writing the Equation of a Horizontal Line

Write the equation of the line graphed in Figure 15.

x

f

y

–2–4–6–8–10 –2 –4 –6 –8

–10

642 8 10

Figure 15

Solution For any x-value, the y-value is −4, so the equation is y = −4.

Example 7 Writing the Equation of a Vertical Line

Write the equation of the line graphed in Figure 16.

4

x

f

y

–2–4–6–8–10 –2 –4 –6 –8

–10

2

6

6

42 8

8

10

10

Figure 16

Solution The constant x-value is 7, so the equation is x = 7.

Determining Whether lines are Parallel or Perpendicular The two lines in Figure 17 are parallel lines: they will never intersect. Notice that they have exactly the same steepness, which means their slopes are identical. The only difference between the two lines is the y-intercept. If we shifted one line vertically toward the y-intercept of the other, they would become the same line.

CHAPTER 2 linear Functions1 5 2

2

x

y

–1–2–3–4–5–6 –1 –2 –3

1

3

3

21 4

4

6 5

7 8

5 6

Figure 17 Parallel lines

We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the y-intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel.

f (x) = −2x + 6 } f (x) = 3x + 2 } f (x) = −2x − 4 parallel f (x) = 2x + 2 not parallel Unlike parallel lines, perpendicular lines do intersect. Their intersection forms a right, or 90-degree, angle. The two lines in Figure 18 are perpendicular.

2

x

y

–1–2–3–4–5–6 –1 –2 –3 –4 –5 –6 –7 –8

1

3

3

21 4

4

5 6

Figure 18 Perpendicular lines

Perpendicular lines do not have the same slope. The slopes of perpendicular lines are different from one another in a specific way. The slope of one line is the negative reciprocal of the slope of the other line. The product of a number and its reciprocal is 1. So, if m1 and m2 are negative reciprocals of one another, they can be multiplied together to yield −1.

m1m2 = −1

To find the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is 1 __ 8

, and the reciprocal of 1 __ 8

is 8. To find the negative reciprocal, first find the reciprocal and then change the sign.

As with parallel lines, we can determine whether two lines are perpendicular by comparing their slopes, assuming that the lines are neither horizontal nor perpendicular. The slope of each line below is the negative reciprocal of the other so the lines are perpendicular.

f (x) =   1 __ 4

x + 2 negative reciprocal of 1 __ 4

is −4

f (x) = −4x + 3 negative reciprocal of −4 is 1 __ 4

The product of the slopes is −1. −4  1 __ 4  = −1

parallel and perpendicular lines Two lines are parallel lines if they do not intersect. The slopes of the lines are the same.

f (x) = m1x + b1 and g(x) = m2x + b2 are parallel if m1 = m2.

SECTION 2.2 graphs oF linear Functions 1 5 3

If and only if b1 = b2 and m1 = m2, we say the lines coincide. Coincident lines are the same line.

Two lines are perpendicular lines if they intersect at right angles.

f (x) = m1x + b1 and g(x) = m2x + b2 are perpendicular if m1m2 = −1, and so m2 = −   1 _ m1

Example 8 Identifying Parallel and Perpendicular Lines

Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines. f (x) = 2x + 3 h(x) = −2x + 2

g(x) =   1 __ 2

x − 4 j(x) = 2x − 6

Solution Parallel lines have the same slope. Because the functions f (x) = 2x + 3 and j(x) = 2x − 6 each have a slope of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because −2 and 1 __

2 are negative

reciprocals, the equations, g(x) =   1 __ 2

x − 4 and h(x) = −2x + 2 represent perpendicular lines.

Analysi s A graph of the lines is shown in Figure 19.

4

x

y

–2–4–6–8–10 –2 –4 –6 –8

–10

2

6

6

42 8

8

10

10

h(x) = −2x + 2 f (x) = 2x + 3

j(x) = 2x − 6

g(x) = x − 412

Figure 19

The graph shows that the lines f (x) = 2x + 3 and j(x) = 2x − 6 are parallel, and the lines g(x) =   1 __ 2

x − 4 and h(x) = −2x + 2 are perpendicular.

Writing the equation of a line Parallel or Perpendicular to a Given line If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line.

Writing Equations of Parallel Lines

Suppose for example, we are given the following equation.

f (x) = 3x + 1

We know that the slope of the line formed by the function is 3. We also know that the y-intercept is (0, 1). Any other line with a slope of 3 will be parallel to f (x). So the lines formed by all of the following functions will be parallel to f (x).

g(x) = 3x + 6

h(x) = 3x + 1

p(x) = 3x +   2 __ 3

Suppose then we want to write the equation of a line that is parallel to f and passes through the point (1, 7). We already know that the slope is 3. We just need to determine which value for b will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form.

CHAPTER 2 linear Functions1 5 4

y − y1 = m(x − x1) y − 7 = 3(x − 1) y − 7 = 3x − 3 y = 3x + 4 So g(x) = 3x + 4 is parallel to f (x) = 3x + 1 and passes through the point (1, 7).

How To… Given the equation of a function and a point through which its graph passes, write the equation of a line parallel to the given line that passes through the given point.

1. Find the slope of the function. 2. Substitute the given values into either the general point-slope equation or the slope-intercept equation for a line. 3. Simplify.

Example 9 Finding a Line Parallel to a Given Line

Find a line parallel to the graph of f (x) = 3x + 6 that passes through the point (3, 0).

Solution The slope of the given line is 3. If we choose the slope-intercept form, we can substitute m = 3, x = 3, and f (x) = 0 into the slope-intercept form to find the y-intercept.

g(x) = 3x + b

0 = 3(3) + b

b = −9

The line parallel to f (x) that passes through (3, 0) is g(x) = 3x − 9.

Analysi s We can confirm that the two lines are parallel by graphing them. Figure 20 shows that the two lines will never intersect.

2

x

y

–1–2–3–4–5–6 –1 –2 –3 –4 –5

1

3

3

2

Up 3

Up 3

Right 1

Right 1

1 4

4

5 6

5

y = 3x + 6

y = 3x − 9

Figure 20

Writing Equations of Perpendicular Lines

We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following function:

f (x) = 2x + 4

The slope of the line is 2, and its negative reciprocal is −  1 __ 2

. Any function with a slope of −  1 __ 2

will be perpendicular to f (x). So the lines formed by all of the following functions will be perpendicular to f (x).

g(x) = −  1 __ 2

x + 4 h(x) = −  1 __ 2

x + 2 p(x) = −  1 __ 2

x −  1 __ 2

As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to f (x) and passes through the point (4, 0). We already know that the slope is −  1 __

2 . Now we can use the point to find the y-intercept by substituting

the given values into the slope-intercept form of a line and solving for b.

SECTION 2.2 graphs oF linear Functions 1 5 5

g (x) = mx + b

0 = −  1 __ 2

(4) + b

0 = −2 + b

2 = b

b = 2 The equation for the function with a slope of −  1 __

2 and a y-intercept of 2 is

g(x) = −  1 __ 2

x + 2.

So g(x) = −  1 __ 2

x + 2 is perpendicular to f (x) = 2x + 4 and passes through the point (4, 0). Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.

Q & A… A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not −1. Doesn’t this fact contradict the definition of perpendicular lines? No. For two perpendicular linear functions, the product of their slopes is −1. However, a vertical line is not a function so the definition is not contradicted.

How To… Given the equation of a function and a point through which its graph passes, write the equation of a line perpendicular to the given line.

1. Find the slope of the function. 2. Determine the negative reciprocal of the slope. 3. Substitute the new slope and the values for x and y from the coordinate pair provided into g(x) = mx + b. 4. Solve for b. 5. Write the equation for the line.

Example 10 Finding the Equation of a Perpendicular Line

Find the equation of a line perpendicular to f (x) = 3x + 3 that passes through the point (3, 0).

Solution The original line has slope m = 3, so the slope of the perpendicular line will be its negative reciprocal, or −  1 __

3 . Using this slope and the given point, we can find the equation for the line.

g(x) = −  1 __ 3

x + b

0 = −  1 __ 3

(3) + b

1 = b b = 1

The line perpendicular to f (x) that passes through (3, 0) is g(x) = −  1 __ 3

x + 1.

Analysi s A graph of the two lines is shown in Figure 21.

Figure 21

–1–2–3–4–5–6 –1 –2 –3

2

x

y

1

3

3

21 4

4

5 6

5

7 6

8 9

f (x) = 3x + 6

g(x) = − x + 113

CHAPTER 2 linear Functions1 5 6

Try It #5

Given the function h(x) = 2x − 4, write an equation for the line passing through (0, 0) that is a. parallel to h(x) b. perpendicular to h(x)

How To… Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point.

1. Determine the slope of the line passing through the points. 2. Find the negative reciprocal of the slope. 3. Use the slope-intercept form or point-slope form to write the equation by substituting the known values. 4. Simplify.

Example 11 Finding the Equation of a Line Perpendicular to a Given Line Passing through a Point

A line passes through the points (−2, 6) and (4, 5). Find the equation of a perpendicular line that passes through the point (4, 5).

Solution From the two points of the given line, we can calculate the slope of that line.

m1 = 5 − 6 _______

4 − (−2)

= −1 ___ 6

= −  1 __ 6

Find the negative reciprocal of the slope.

m2 = −1 _ −  1 __

6

= −1  −  6 __ 1  = 6

We can then solve for the y-intercept of the line passing through the point (4, 5).

g (x) = 6x + b

5 = 6(4) + b

5 = 24 + b

−19 = b

b = −19

The equation for the line that is perpendicular to the line passing through the two given points and also passes through point (4, 5) is y = 6x − 19

Try It #6

A line passes through the points, (−2, −15) and (2,−3). Find the equation of a perpendicular line that passes through the point, (6, 4).

SECTION 2.2 graphs oF linear Functions 1 5 7

Solving a System of linear equations Using a Graph A system of linear equations includes two or more linear equations. The graphs of two lines will intersect at a single point if they are not parallel. Two parallel lines can also intersect if they are coincident, which means they are the same line and they intersect at every point. For two lines that are not parallel, the single point of intersection will satisfy both equations and therefore represent the solution to the system.

To find this point when the equations are given as functions, we can solve for an input value so that f (x) = g(x). In other words, we can set the formulas for the lines equal to one another, and solve for the input that satisfies the equation.

Example 12 Finding a Point of Intersection Algebraically

Find the point of intersection of the lines h(t) = 3t − 4 and j(t) = 5 − t.

Solution Set h(t) = j(t). 3t − 4 = 5 − t

4t = 9

t =   9 __ 4

This tells us the lines intersect when the input is 9 __ 4

.

We can then find the output value of the intersection point by evaluating either function at this input.

j  9 __ 4  = 5 −   9 __ 4

= 11 __ 4

These lines intersect at the point  9 __ 4 , 11 __ 4

 .

Analysi s Looking at Figure 22, this result seems reasonable.

j(t)

h(t)

9 4

11 4

, 2

x

y

–1–1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

Figure 22

Q & A… If we were asked to find the point of intersection of two distinct parallel lines, should something in the solution process alert us to the fact that there are no solutions?

Yes. After setting the two equations equal to one another, the result would be the contradiction “0 = non-zero real number.”

CHAPTER 2 linear Functions1 5 8

Try It #7

Look at the graph in Figure 22 and identify the following for the function j(t): a. y-intercept b. x-intercept(s) c. slope d. Is j(t) parallel or perpendicular to h(t) (or neither)? e. Is j(t) an increasing or decreasing function (or neither)? f. Write a transformation description for j(t) from the identity toolkit function f (x) = x.

Example 13 Finding a Break-Even Point

A company sells sports helmets. The company incurs a one-time fixed cost for $250,000. Each helmet costs $120 to produce, and sells for $140.

a. Find the cost function, C, to produce x helmets, in dollars. b. Find the revenue function, R, from the sales of x helmets, in dollars. c. Find the break-even point, the point of intersection of the two graphs C and R.

Solution

a. The cost function in the sum of the fixed cost, $250,000, and the variable cost, $120 per helmet.

C(x) = 120x + 250,000 b. The revenue function is the total revenue from the sale of x helmets, R(x) = 140x. c. The break-even point is the point of intersection of the graph of the cost and revenue functions. To find the

x-coordinate of the coordinate pair of the point of intersection, set the two equations equal, and solve for x.

C(x) = R(x) 250,000 + 120x = 140x 250,000 = 20x 12,500 = x x = 12,500

To find y, evaluate either the revenue or the cost function at 12,500. R(x) = 140(12,500) = $1,750,000

The break-even point is (12,500, 1,750,000).

Analysi s This means if the company sells 12,500 helmets, they break even; both the sales and cost incurred equaled 1.75 million dollars. See Figure 23.

y

x0 0

1,000,000

20,000

2,000,000 Cost < Sales

Company loses money

C R

Sales > Cost Company makes a pro�t

(12,500, 1,750,000)

Figure 23

Access these online resources for additional instruction and practice with graphs of linear functions.

• Finding Input of Function from the Output and Graph (http://openstaxcollege.org/l/findinginput)

• Graphing Functions Using Tables (http://openstaxcollege.org/l/graphwithtable)

SECTION 2.2 section exercises 1 5 9

2.2 SeCTIOn exeRCISeS

VeRBAl

1. If the graphs of two linear functions are parallel, describe the relationship between the slopes and the y-intercepts.

2. If the graphs of two linear functions are perpendicular, describe the relationship between the slopes and the y-intercepts.

3. If a horizontal line has the equation f (x) = a and a vertical line has the equation x = a, what is the point of intersection? Explain why what you found is the point of intersection.

4. Explain how to find a line parallel to a linear function that passes through a given point.

5. Explain how to find a line perpendicular to a linear function that passes through a given point.

AlGeBRAIC

For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither parallel nor perpendicular:

6. 4x − 7y = 10 7x + 4y = 1

7. 3y + x = 12 −y = 8x + 1

8. 3y + 4x = 12 −6y = 8x + 1

9. 6x − 9y = 10

3x + 2y = 1

10. y =   2 __ 3

x + 1

3x + 2y = 1

11. y =   3 __ 4

x + 1

−3x + 4y = 1

For the following exercises, find the x- and y-intercepts of each equation

12. f (x) = −x + 2 13. g(x) = 2x + 4 14. h(x) = 3x − 5

15. k(x) = −5x + 1 16. −2x + 5y = 20 17. 7x + 2y = 56

For the following exercises, use the descriptions of each pair of lines given below to find the slopes of Line 1 and Line 2. Is each pair of lines parallel, perpendicular, or neither?

18. Line 1: Passes through (0, 6) and (3, −24) Line 2: Passes through (−1, 19) and (8, −71)

19. Line 1: Passes through (−8, −55) and (10, 89) Line 2: Passes through (9, −44) and (4, −14)

20. Line 1: Passes through (2, 3) and (4, −1) Line 2: Passes through (6, 3) and (8, 5)

21. Line 1: Passes through (1, 7) and (5, 5) Line 2: Passes through (−1, −3) and (1, 1)

22. Line 1: Passes through (0, 5) and (3, 3) Line 2: Passes through (1, −5) and (3, −2)

23. Line 1: Passes through (2, 5) and (5, −1) Line 2: Passes through (−3, 7) and (3, −5)

24. Write an equation for a line parallel to f (x) = −5x − 3 and passing through the point (2, −12).

25. Write an equation for a line parallel to g(x) = 3x − 1 and passing through the point (4, 9).

26. Write an equation for a line perpendicular to h(t) = −2t + 4 and passing through the point (−4, −1).

27. Write an equation for a line perpendicular to p(t) = 3t + 4 and passing through the point (3, 1).

1 6 0 CHAPTER 2 linear Functions

28. Find the point at which the line f (x) = −2x − 1 intersects the line g(x) = −x.

29. Find the point at which the line f (x) = 2x + 5 intersects the line g(x) = −3x − 5.

30. Use algebra to find the point at which the line

f (x) = −  4 __ 5

x + 274 ___ 25

intersects h(x) =  9 __ 4

x + 73 __ 10

.

31. Use algebra to find the point at which the line

f (x) =  7 __ 4

x + 457 ___ 60

intersects g(x) =   4 __ 3

x + 31 __ 5

.

GRAPHICAl

For the following exercises, match the given linear equation with its graph in Figure 24.

2

–1–1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5 A B

C

D E

F Figure 24

32. f (x) = −x − 1

33. f (x) = −2x − 1

34. f (x) = −  1 __ 2

x − 1

35. f (x) = 2

36. f (x) = 2 + x

37. f (x) = 3x + 2

For the following exercises, sketch a line with the given features.

38. An x-intercept of (−4, 0) and y-intercept of (0, −2) 39. An x-intercept of (−2, 0) and y-intercept of (0, 4)

40. A y-intercept of (0, 7) and slope −  3 __ 2

41. A y-intercept of (0, 3) and slope 2 __ 5

42. Passing through the points (−6, −2) and (6, −6) 43. Passing through the points (−3, −4) and (3, 0)

For the following exercises, sketch the graph of each equation.

44. f (x) = −2x − 1 45. g(x) = −3x + 2 46. h(x) =   1 __ 3

x + 2

47. k(x) =   2 __ 3

x − 3 48. f (t) = 3 + 2t 49. p(t) = −2 + 3t

50. x = 3 51. x = −2 52. r(x) = 4

53. q(x) = 3 54. 4x = −9y + 36 55. x __ 3

− y _ 4

= 1

56. 3x − 5y = 15 57. 3x = 15 58. 3y = 12

59. If g(x) is the transformation of f (x) = x after a vertical compression by 3 __

4 , a shift right by 2, and

a shift down by 4 a. Write an equation for g(x). b. What is the slope of this line? c. Find the y-intercept of this line.

60. If g(x) is the transformation of f (x) = x after a vertical compression by 1 __

3 , a shift left by 1, and a

shift up by 3 a. Write an equation for g(x). b. What is the slope of this line? c. Find the y-intercept of this line.

SECTION 2.2 section exercises 1 6 1

For the following exercises, write the equation of the line shown in the graph.

61.

2

x

y

–1–1–2

–2

–3

–3

–4

–4

–5–6

–5 –6

1

3

3

21 4

4

5 6

5 6

62.

2

x

y

–1–1–2

–2

–3

–3

–4

–4

–5–6

–5 –6

1

3

3

21 4

4

5 6

5 6

63.

2

x

y

–1–1–2

–2

–3

–3

–4

–4

–5–6

–5 –6

1

3

3

21 4

4

5 6

5 6

64.

2

x

y

–1–1–2

–2

–3

–3

–4

–4

–5

–5

1

3

3

21 4

4

5

5

For the following exercises, find the point of intersection of each pair of lines if it exists. If it does not exist, indicate that there is no point of intersection.

65. y =   3 __ 4

x + 1

−3x + 4y = 12

66. 2x − 3y = 12

5y + x = 30

67. 2x = y − 3

y + 4x = 15

68. x − 2y + 2 = 3 x − y = 3

69. 5x + 3y = − 65 x − y = − 5

exTenSIOnS 70. Find the equation of the line parallel to the line

g(x) = −0.01x + 2.01 through the point (1, 2). 71. Find the equation of the line perpendicular to the

line g(x) = −0.01x + 2.01 through the point (1, 2).

For the following exercises, use the functions f (x) = −0.1x + 200 and g(x) = 20x + 0.1. 72. Find the point of intersection of the lines f and g. 73. Where is f (x) greater than g(x)? Where is g(x)

greater than f (x)?

ReAl-WORlD APPlICATIOnS 74. A car rental company offers two plans for renting a

car. Plan A: $30 per day and $0.18 per mile Plan B: $50 per day with free unlimited mileage How many miles would you need to drive for plan B to save you money?

75. A cell phone company offers two plans for minutes. Plan A: $20 per month and $1 for every one

hundred texts. Plan B: $50 per month with free unlimited texts. How many texts would you need to send per month for plan B to save you money?

76. A cell phone company offers two plans for minutes. Plan A: $15 per month and $2 for every 300 texts. Plan B: $25 per month and $0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?

CHAPTER 2 linear Functions1 6 2

2.3 MODelInG WITH lIneAR FUnCTIOnS

Figure 1 (credit: eeK Photography/Flickr)

Emily is a college student who plans to spend a summer in Seattle. She has saved $3,500 for her trip and anticipates spending $400 each week on rent, food, and activities. How can we write a linear model to represent her situation? What would be the x-intercept, and what can she learn from it? To answer these and related questions, we can create a model using a linear function. Models such as this one can be extremely useful for analyzing relationships and making predictions based on those relationships. In this section, we will explore examples of linear function models.

Identifying Steps to Model and Solve Problems When modeling scenarios with linear functions and solving problems involving quantities with a constant rate of change, we typically follow the same problem strategies that we would use for any type of function. Let’s briefly review them:

1. Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system.

2. Carefully read the problem to identify important information. Look for information that provides values for the variables or values for parts of the functional model, such as slope and initial value.

3. Carefully read the problem to determine what we are trying to find, identify, solve, or interpret.

4. Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table, or even finding a formula for the function being used to model the problem.

5. When needed, write a formula for the function.

6. Solve or evaluate the function using the formula.

7. Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically.

8. Clearly convey your result using appropriate units, and answer in full sentences when necessary.

leARnInG OBjeCTIVeS

In this section, you will:

• Identify steps for modeling and solving.

• Build linear models from verbal descriptions.

• Build systems of linear models.

SECTION 2.3 modeling with linear Functions 1 6 3

Building linear Models Now let’s take a look at the student in Seattle. In her situation, there are two changing quantities: time and money. The amount of money she has remaining while on vacation depends on how long she stays. We can use this information to define our variables, including units.

• Output: M, money remaining, in dollars • Input: t, time, in weeks

So, the amount of money remaining depends on the number of weeks: M(t)

We can also identify the initial value and the rate of change.

• Initial Value: She saved $3,500, so $3,500 is the initial value for M. • Rate of Change: She anticipates spending $400 each week, so −$400 per week is the rate of change, or slope.

Notice that the unit of dollars per week matches the unit of our output variable divided by our input variable. Also, because the slope is negative, the linear function is decreasing. This should make sense because she is spending money each week.

The rate of change is constant, so we can start with the linear model M(t) = mt + b. Then we can substitute the intercept and slope provided.

M(t) = mt + b ↑ ↑ −400 3500

M(t) = − 400t + 3500

To find the x-intercept, we set the output to zero, and solve for the input.

0 = −400t + 3500

t = 3500 ____ 400

= 8.75

The x-intercept is 8.75 weeks. Because this represents the input value when the output will be zero, we could say that Emily will have no money left after 8.75 weeks.

When modeling any real-life scenario with functions, there is typically a limited domain over which that model will be valid—almost no trend continues indefinitely. Here the domain refers to the number of weeks. In this case, it doesn’t make sense to talk about input values less than zero. A negative input value could refer to a number of weeks before she saved $3,500, but the scenario discussed poses the question once she saved $3,500 because this is when her trip and subsequent spending starts. It is also likely that this model is not valid after the x-intercept, unless Emily will use a credit card and goes into debt. The domain represents the set of input values, so the reasonable domain for this function is 0 ≤ t ≤ 8.75.

In the above example, we were given a written description of the situation. We followed the steps of modeling a problem to analyze the information. However, the information provided may not always be the same. Sometimes we might be provided with an intercept. Other times we might be provided with an output value. We must be careful to analyze the information we are given, and use it appropriately to build a linear model.

Using a Given Intercept to Build a Model

Some real-world problems provide the y-intercept, which is the constant or initial value. Once the y-intercept is known, the x-intercept can be calculated. Suppose, for example, that Hannah plans to pay off a no-interest loan from her parents. Her loan balance is $1,000. She plans to pay $250 per month until her balance is $0. The y-intercept is the initial amount of her debt, or $1,000. The rate of change, or slope, is −$250 per month. We can then use the slope- intercept form and the given information to develop a linear model.

f (x) = mx + b

= −250x + 1000

CHAPTER 2 linear Functions1 6 4

Now we can set the function equal to 0, and solve for x to find the x-intercept.

0 = −250x + 1000

1000 = 250x

4 = x

x = 4

The x-intercept is the number of months it takes her to reach a balance of $0. The x-intercept is 4 months, so it will take Hannah four months to pay off her loan.

Using a Given Input and Output to Build a Model

Many real-world applications are not as direct as the ones we just considered. Instead they require us to identify some aspect of a linear function. We might sometimes instead be asked to evaluate the linear model at a given input or set the equation of the linear model equal to a specified output.

How To… Given a word problem that includes two pairs of input and output values, use the linear function to solve a problem.

1. Identify the input and output values. 2. Convert the data to two coordinate pairs. 3. Find the slope. 4. Write the linear model. 5. Use the model to make a prediction by evaluating the function at a given x-value. 6. Use the model to identify an x-value that results in a given y-value. 7. Answer the question posed.

Example 1 Using a Linear Model to Investigate a Town’s Population

A town’s population has been growing linearly. In 2004 the population was 6,200. By 2009 the population had grown to 8,100. Assume this trend continues.

a. Predict the population in 2013.

b. Identify the year in which the population will reach 15,000.

Solution The two changing quantities are the population size and time. While we could use the actual year value as the input quantity, doing so tends to lead to very cumbersome equations because the y-intercept would correspond to the year 0, more than 2,000 years ago!

To make computation a little nicer, we will define our input as the number of years since 2004:

• Input: t, years since 2004

• Output: P(t), the town’s population

To predict the population in 2013 (t = 9), we would first need an equation for the population. Likewise, to find when the population would reach 15,000, we would need to solve for the input that would provide an output of 15,000. To write an equation, we need the initial value and the rate of change, or slope.

To determine the rate of change, we will use the change in output per change in input.

m = change in output

__ change in input

The problem gives us two input-output pairs. Converting them to match our defined variables, the year 2004 would correspond to t = 0, giving the point (0, 6200). Notice that through our clever choice of variable definition, we have

“given” ourselves the y-intercept of the function. The year 2009 would correspond to t = 5, giving the point (5, 8100).

The two coordinate pairs are (0, 6200) and (5, 8100). Recall that we encountered examples in which we were provided two points earlier in the chapter. We can use these values to calculate the slope.

SECTION 2.3 modeling with linear Functions 1 6 5

m = 8100 − 6200 __________ 5 − 0

= 1900 ____ 5

= 380 people per year

We already know the y-intercept of the line, so we can immediately write the equation:

P(t) = 380t + 6200

To predict the population in 2013, we evaluate our function at t = 9.

P(9) = 380(9) + 6,200 = 9,620

If the trend continues, our model predicts a population of 9,620 in 2013.

To find when the population will reach 15,000, we can set P(t) = 15000 and solve for t. 15000 = 380t + 6200 8800 = 380t t ≈ 23.158

Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere around the year 2027.

Try It #1

A company sells doughnuts. They incur a fixed cost of $25,000 for rent, insurance, and other expenses. It costs $0.25 to produce each doughnut.

a. Write a linear model to repre sent the cost C of the company as a function of x, the number of doughnuts produced. b. Find and interpret the y-intercept.

Try It #2

A city’s population has been growing linearly. In 2008, the population was 28,200. By 2012, the population was 36,800. Assume this trend continues.

a. Predict the population in 2014.

b. Identify the year in which the population will reach 54,000.

Using a Diagram to Model a Problem

It is useful for many real-world applications to draw a picture to gain a sense of how the variables representing the input and output may be used to answer a question. To draw the picture, first consider what the problem is asking for. Then, determine the input and the output. The diagram should relate the variables. Often, geometrical shapes or figures are drawn. Distances are often traced out. If a right triangle is sketched, the Pythagorean Theorem relates the sides. If a rectangle is sketched, labeling width and height is helpful.

Example 2 Using a Diagram to Model Distance Walked

Anna and Emanuel start at the same intersection. Anna walks east at 4 miles per hour while Emanuel walks south at 3 miles per hour. They are communicating with a two-way radio that has a range of 2 miles. How long after they start walking will they fall out of radio contact?

Solution In essence, we can partially answer this question by saying they will fall out of radio contact when they are 2 miles apart, which leads us to ask a new question:

“How long will it take them to be 2 miles apart?”

CHAPTER 2 linear Functions1 6 6

In this problem, our changing quantities are time and position, but ultimately we need to know how long will it take for them to be 2 miles apart. We can see that time will be our input variable, so we’ll define our input and output variables.

• Input: t, time in hours.

• Output: A(t), distance in miles, and E(t), distance in miles

Because it is not obvious how to define our output variable, we’ll start by drawing a picture such as Figure 2.

Anna walking east, 4 miles/hour

Distance between them

Emanuel walking south, 3 miles/hour

Figure 2

Initial Value: They both start at the same intersection so when t = 0, the distance traveled by each person should also be 0. Thus the initial value for each is 0.

Rate of Change: Anna is walking 4 miles per hour and Emanuel is walking 3 miles per hour, which are both rates of change. The slope for A is 4 and the slope for E is 3.

Using those values, we can write formulas for the distance each person has walked.

A(t) = 4t

E(t) = 3t

For this problem, the distances from the starting point are important. To notate these, we can define a coordinate system, identifying the “starting point” at the intersection where they both started. Then we can use the variable, A, which we introduced above, to represent Anna’s position, and define it to be a measurement from the starting point in the eastward direction. Likewise, can use the variable, E, to represent Emanuel’s position, measured from the starting point in the southward direction. Note that in defining the coordinate system, we specified both the starting point of the measurement and the direction of measure.

We can then define a third variable, D, to be the measurement of the distance between Anna and Emanuel.

Showing the variables on the diagram is often helpful, as we can see from Figure 3.

Recall that we need to know how long it takes for D, the distance between them, to equal 2 miles. Notice that for any given input t, the outputs A(t), E(t), and D(t) represent distances.

A

D

E

Figure 3

Figure 3 shows us that we can use the Pythagorean Theorem because we have drawn a right angle.

Using the Pythagorean Theorem, we get:

SECTION 2.3 modeling with linear Functions 1 6 7

D(t)2 = A(t)2 + E(t)2

= (4t)2 + (3t)2

= 16t2 + 9t2

= 25t2

D(t) = ± √ —

25t2 Solve for D(t) using the square root. = ±5 | t | In this scenario we are considering only positive values of t, so our distance D(t) will always be positive. We can simplify this answer to D(t) = 5t. This means that the distance between Anna and Emanuel is also a linear function. Because D is a linear function, we can now answer the question of when the distance between them will reach 2 miles. We will set the output D(t) = 2 and solve for t. D(t) = 2

5t = 2

t =   2 __ 5

= 0.4

They will fall out of radio contact in 0.4 hours, or 24 minutes.

Q & A… Should I draw diagrams when given information based on a geometric shape?

Yes. Sketch the figure and label the quantities and unknowns on the sketch.

Example 3 Using a Diagram to Model Distance Between Cities

There is a straight road leading from the town of Westborough to Agritown 30 miles east and 10 miles north. Partway down this road, it junctions with a second road, perpendicular to the first, leading to the town of Eastborough. If the town of Eastborough is located 20 miles directly east of the town of Westborough, how far is the road junction from Westborough?

Solution It might help here to draw a picture of the situation. See Figure 4. It would then be helpful to introduce a coordinate system. While we could place the origin anywhere, placing it at Westborough seems convenient. This puts Agritown at coordinates (30, 10), and Eastborough at (20, 0).

Agritown

EastboroughWestborough 20 miles (0, 0)

(30, 10)

(20, 0)

Figure 4

Using this point along with the origin, we can find the slope of the line from Westborough to Agritown:

m = 10 − 0 ______ 30 − 0

=   1 __ 3

The equation of the road from Westborough to Agritown would be

W(x) =   1 __ 3

x

From this, we can determine the perpendicular road to Eastborough will have slope m = −3. Because the town of Eastborough is at the point (20, 0), we can find the equation:

E(x) = −3x + b 0 = −3(20) + b Substitute in (20, 0). b = 60 E(x) = −3x + 60

CHAPTER 2 linear Functions1 6 8

We can now find the coordinates of the junction of the roads by finding the intersection of these lines. Setting them equal,

1 __ 3

x = −3x + 60

10 __ 3

x = 60

10x = 180

x = 18 Substitute this back into W(x).

y = W(18)

=  1 __ 3

(18)

= 6

The roads intersect at the point (18, 6). Using the distance formula, we can now find the distance from Westborough to the junction. distance =  √

— (x2 − x1)

2 + (y2 − y1) 2

= √ ——

(18 − 0)2 + (6 − 0)2

≈ 18.974 miles

Analysi s One nice use of linear models is to take advantage of the fact that the graphs of these functions are lines. This means real-world applications discussing maps need linear functions to model the distances between reference points.

Try It #3

There is a straight road leading from the town of Timpson to Ashburn 60 miles east and 12 miles north. Partway down the road, it junctions with a second road, perpendicular to the first, leading to the town of Garrison. If the town of Garrison is located 22 miles directly east of the town of Timpson, how far is the road junction from Timpson?

Building Systems of linear Models Real-world situations including two or more linear functions may be modeled with a system of linear equations. Remember, when solving a system of linear equations, we are looking for points the two lines have in common. Typically, there are three types of answers possible, as shown in Figure 5.

y y y

x g f

x gf

x

Exactly one solution

In�nitely many solutions

No solutions

(a) (b) (c)

Figure 5

How To… Given a situation that represents a system of linear equations, write the system of equations and identify the solution.

1. Identify the input and output of each linear model. 2. Identify the slope and y-intercept of each linear model. 3. Find the solution by setting the two linear functions equal to one another and solving for x, or find the point of

intersection on a graph.

SECTION 2.3 modeling with linear Functions 1 6 9

Example 4 Building a System of Linear Models to Choose a Truck Rental Company

Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mile[9]. When will Keep on Trucking, Inc. be the better choice for Jamal?

Solution The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions.

Input d, distance driven in miles

Outputs K(d): cost, in dollars, for renting from Keep on Trucking M(d) cost, in dollars, for renting from Move It Your Way

Initial Value Up-front fee: K(0) = 20 and M(0) = 16

Rate of Change K(d) = $0.59/mile and P(d) = $0.63/mile

Table 1

A linear function is of the form f (x) = mx + b. Using the rates of change and initial charges, we can write the equations

K(d) = 0.59d + 20

M(d) = 0.63d + 16

Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when K(d) < M(d). The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the K(d) function is smaller.

These graphs are sketched in Figure 6, with K(d) in red.

d

$

0 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160

10 20 30 40 50 60 70 80 90

100 110 120

(100, 80) K(d) = 0.59d + 20

M(d) = 0.63d + 16

Figure 6

To find the intersection, we set the equations equal and solve:

K(d) = M(d)

0.59d + 20 = 0.63d + 16

4 = 0.04d

100 = d

d = 100

This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that K(d) is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is d > 100.

Access this online resources for additional instruction and practice with linear function models. • Interpreting a linear Function (http://openstaxcollege.org/l/interpretlinear)

9 Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/

1 7 0 CHAPTER 2 linear Functions

2.3 SeCTIOn exeRCISeS

VeRBAl

1. Explain how to find the input variable in a word problem that uses a linear function.

2. Explain how to find the output variable in a word problem that uses a linear function.

3. Explain how to interpret the initial value in a word problem that uses a linear function.

4. Explain how to determine the slope in a word problem that uses a linear function.

AlGeBRAIC

5. Find the area of a parallelogram bounded by the y-axis, the line x = 3, the line f (x) = 1 + 2x, and the line parallel to f (x) passing through (2, 7).

6. Find the area of a triangle bounded by the x-axis, the line f (x) = 12 – 1 __

3 x, and the line perpendicular

to f (x) that passes through the origin.

7. Find the area of a triangle bounded by the y-axis, the line f (x) = 9 – 6 __

7 x, and the line perpendicular to

f (x) that passes through the origin.

8. Find the area of a parallelogram bounded by the x-axis, the line g (x) = 2, the line f (x) = 3x, and the line parallel to f (x) passing through (6, 1).

For the following exercises, consider this scenario: A town’s population has been decreasing at a constant rate. In 2010 the population was 5,900. By 2012 the population had dropped 4,700. Assume this trend continues.

9. Predict the population in 2016. 10. Identify the year in which the population will reach 0.

For the following exercises, consider this scenario: A town’s population has been increased at a constant rate. In 2010 the population was 46,020. By 2012 the population had increased to 52,070. Assume this trend continues.

11. Predict the population in 2016. 12. Identify the year in which the population will reach 75,000.

For the following exercises, consider this scenario: A town has an initial population of 75,000. It grows at a constant rate of 2,500 per year for 5 years.

13. Find the linear function that models the town’s population P as a function of the year, t, where t is the number of years since the model began.

14. Find a reasonable domain and range for the function P.

15. If the function P is graphed, find and interpret the x-and y-intercepts.

16. If the function P is graphed, find and interpret the slope of the function.

17. When will the output reached 100,000? 18. What is the output in the year 12 years from the onset of the model?

SECTION 2.3 section exercises 1 7 1

For the following exercises, consider this scenario: The weight of a newborn is 7.5 pounds. The baby gained one-half pound a month for its first year.

19. Find the linear function that models the baby’s weight, W, as a function of the age of the baby, in months, t.

20. Find a reasonable domain and range for the function W.

21. If the function W is graphed, find and interpret the x- and y-intercepts.

22. If the function W is graphed, find and interpret the slope of the function.

23. When did the baby weight 10.4 pounds? 24. What is the output when the input is 6.2? Interpret your answer.

For the following exercises, consider this scenario: The number of people afflicted with the common cold in the winter months steadily decreased by 205 each year from 2005 until 2010. In 2005, 12,025 people were afflicted.

25. Find the linear function that models the number of people inflicted with the common cold, C, as a function of the year, t.

26. Find a reasonable domain and range for the function C.

27. If the function C is graphed, find and interpret the x-and y-intercepts.

28. If the function C is graphed, find and interpret the slope of the function.

29. When will the output reach 0? 30. In what year will the number of people be 9,700?

GRAPHICAl

For the following exercises, use the graph in Figure 7, which shows the profit, y, in thousands of dollars, of a company in a given year, t, where t represents the number of years since 1980.

t

y

0 0

130

10 15 20 25

135 140 145 150

155

Figure 7

31. Find the linear function y, where y depends on t, the number of years since 1980.

32. Find and interpret the y-intercept.

33. Find and interpret the x-intercept.

34. Find and interpret the slope.

1 7 2 CHAPTER 2 linear Functions

For the following exercises, use the graph in Figure 8, which shows the profit, y, in thousands of dollars, of a company in a given year, t, where t represents the number of years since 1980.

t

y

0 0

100

10 15 20 25

200

300

400

500

50

150

250

350

450

Figure 8

35. Find the linear function y, where y depends on t, the number of years since 1980.

36. Find and interpret the y-intercept.

37. Find and interpret the x-intercept.

38. Find and interpret the slope.

nUMeRIC

For the following exercises, use the median home values in Mississippi and Hawaii (adjusted for inflation) shown in Table 2. Assume that the house values are changing linearly.

Year Mississippi Hawaii 1950 $25,200 $74,400 2000 $71,400 $272,700

Table 2

39. In which state have home values increased at a higher rate?

40. If these trends were to continue, what would be the median home value in Mississippi in 2010?

41. If we assume the linear trend existed before 1950 and continues after 2000, the two states’ median house values will be (or were) equal in what year? (The answer might be absurd.)

For the following exercises, use the median home values in Indiana and Alabama (adjusted for inflation) shown in Table 3. Assume that the house values are changing linearly.

Year Indiana Alabama 1950 $37,700 $27,100 2000 $94,300 $85,100

Table 3

42. In which state have home values increased at a higher rate?

43. If these trends were to continue, what would be the median home value in Indiana in 2010?

44. If we assume the linear trend existed before 1950 and continues after 2000, the two states’ median house values will be (or were) equal in what year? (The answer might be absurd.)

SECTION 2.3 section exercises 1 7 3

ReAl-WORlD APPlICATIOnS

45. In 2004, a school population was 1,001. By 2008 the population had grown to 1,697. Assume the population is changing linearly. a. How much did the population grow between the

year 2004 and 2008? b. How long did it take the population to grow from

1,001 students to 1,697 students? c. What is the average population growth per year? d. What was the population in the year 2000? e. Find an equation for the population, P, of the

school t years after 2000. f. Using your equation, predict the population of

the school in 2011.

46. In 2003, a town’s population was 1,431. By 2007 the population had grown to 2,134. Assume the population is changing linearly. a. How much did the population grow between the

year 2003 and 2007? b. How long did it take the population to grow from

1,431 people to 2,134 people? c. What is the average population growth per year? d. What was the population in the year 2000? e. Find an equation for the population, P of the

town t years after 2000. f. Using your equation, predict the population of

the town in 2014.

47. A phone company has a monthly cellular plan where a customer pays a flat monthly fee and then a certain amount of money per minute used on the phone. If a customer uses 410 minutes, the monthly cost will be $71.50. If the customer uses 720 minutes, the monthly cost will be $118. a. Find a linear equation for the monthly cost of

the cell plan as a function of x, the number of monthly minutes used.

b. Interpret the slope and y-intercept of the equation.

c. Use your equation to find the total monthly cost if 687 minutes are used.

48. A phone company has a monthly cellular data plan where a customer pays a flat monthly fee of $10 and then a certain amount of money per megabyte (MB) of data used on the phone. If a customer uses 20 MB, the monthly cost will be $11.20. If the customer uses 130 MB, the monthly cost will be $17.80. a. Find a linear equation for the monthly cost of the

data plan as a function of x , the number of MB used.

b. Interpret the slope and y-intercept of the equation.

c. Use your equation to find the total monthly cost if 250 MB are used.

49. In 1991, the moose population in a park was measured to be 4,360. By 1999, the population was measured again to be 5,880. Assume the population continues to change linearly. a. Find a formula for the moose population,

P since 1990. b. What does your model predict the moose

population to be in 2003?

50. In 2003, the owl population in a park was measured to be 340. By 2007, the population was measured again to be 285. The population changes linearly. Let the input be years since 1990. a. Find a formula for the owl population, P. Let the

input be years since 2003. b. What does your model predict the owl

population to be in 2012?

51. The Federal Helium Reserve held about 16 billion cubic feet of helium in 2010 and is being depleted by about 2.1 billion cubic feet each year. a. Give a linear equation for the remaining federal

helium reserves, R, in terms of t, the number of years since 2010.

b. In 2015, what will the helium reserves be? c. If the rate of depletion doesn’t change, in what

year will the Federal Helium Reserve be depleted?

52. Suppose the world’s oil reserves in 2014 are 1,820 billion barrels. If, on average, the total reserves are decreasing by 25 billion barrels of oil each year: a. Give a linear equation for the remaining oil

reserves, R, in terms of t, the number of years since now.

b. Seven years from now, what will the oil reserves be?

c. If the rate at which the reserves are decreasing is constant, when will the world’s oil reserves be depleted?

1 7 4 CHAPTER 2 linear Functions

53. You are choosing between two different prepaid cell phone plans. The first plan charges a rate of 26 cents per minute. The second plan charges a monthly fee of $19.95 plus 11 cents per minute. How many minutes would you have to use in a month in order for the second plan to be preferable?

54. You are choosing between two different window washing companies. The first charges $5 per window. The second charges a base fee of $40 plus $3 per window. How many windows would you need to have for the second company to be preferable?

55. When hired at a new job selling jewelry, you are given two pay options: • Option A: Base salary of $17,000 a year with a

commission of 12% of your sales • Option B: Base salary of $20,000 a year with a

commission of 5% of your sales How much jewelry would you need to sell for option A to produce a larger income?

56. When hired at a new job selling electronics, you are given two pay options: • Option A: Base salary of $14,000 a year with a

commission of 10% of your sales • Option B: Base salary of $19,000 a year with a

commission of 4% of your sales How much electronics would you need to sell for option A to produce a larger income?

57. When hired at a new job selling electronics, you are given two pay options: • Option A: Base salary of $20,000 a year with a

commission of 12% of your sales • Option B: Base salary of $26,000 a year with a

commission of 3% of your sales How much electronics would you need to sell for option A to produce a larger income?

58. When hired at a new job selling electronics, you are given two pay options: • Option A: Base salary of $10,000 a year with a

commission of 9% of your sales • Option B: Base salary of $20,000 a year with a

commission of 4% of your sales How much electronics would you need to sell for option A to produce a larger income?

SECTION 2.4 Fitting linear models to data 1 7 5

2.4 FITTInG lIneAR MODelS TO DATA

A professor is attempting to identify trends among final exam scores. His class has a mixture of students, so he wonders if there is any relationship between age and final exam scores. One way for him to analyze the scores is by creating a diagram that relates the age of each student to the exam score received. In this section, we will examine one such diagram known as a scatter plot.

Drawing and Interpreting Scatter Plots A scatter plot is a graph of plotted points that may show a relationship between two sets of data. If the relationship is from a linear model, or a model that is nearly linear, the professor can draw conclusions using his knowledge of linear functions. Figure 1 shows a sample scatter plot.

80 90

100

70 60 50 40 30 20 10

Fi na

l E xa

m S

co re

Final Exam Score vs. Age

Age 100 20 30 40 50

0

Figure 1 A scatter plot of age and final exam score variables.

Notice this scatter plot does not indicate a linear relationship. The points do not appear to follow a trend. In other words, there does not appear to be a relationship between the age of the student and the score on the final exam.

Example 1 Using a Scatter Plot to Investigate Cricket Chirps

Table 1 shows the number of cricket chirps in 15 seconds, for several different air temperatures, in degrees Fahrenheit[10]. Plot this data, and determine whether the data appears to be linearly related.

Chirps 44 35 20.4 33 31 35 18.5 37 26 Temperature 80.5 70.5 57 66 68 72 52 73.5 53

Table 1

Solution Plotting this data, as depicted in Figure 2 suggests that there may be a trend. We can see from the trend in the data that the number of chirps increases as the temperature increases. The trend appears to be roughly linear, though certainly not perfectly so.

10 Selected data from http://classic.globe.gov/fsl/scientistsblog/2007/10/. Retrieved Aug 3, 2010.

leARnInG OBjeCTIVeS

In this section, you will:

• Draw and interpret scatter plots.

• Find the line of best fit.

• Distinguish between linear and nonlinear relations.

• Use a linear model to make predictions.