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Pre-calculus IMTH-119ContentsEssex County CollegeDivision of MathematicsLMS ProjectClass Notes1 Review 62 Course Web Page 73 This Guide 84 Syllabus 95 Introduction 156 mth.119.AB.04 166.1 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166.1.1 Properties of Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . 166.1.2 Examples from MTH100 . . . . . . . . . . . . . . . . . . . . . . . . . 166.1.3 Absolute Value Inequalities . . . . . . . . . . . . . . . . . . . . . . . 176.1.4 Absolute Value Inequality Examples . . . . . . . . . . . . . . . . . . 176.1.5 Polynomial Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . 186.1.6 Polynomial Inequality Examples . . . . . . . . . . . . . . . . . . . . . 186.1.7 Rational Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . 196.1.8 Rational Inequality Examples . . . . . . . . . . . . . . . . . . . . . . 206.2 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 mth.119.01.02 287.1 Introductory Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287.1.1 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297.3 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358 mth.119.01.03 368.1 Functions Continued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368.1.1 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368.1.2 Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388.3 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 469 mth.119.01.04 479.1 Shifting, Reflecting, and Stretching Functions . . . . . . . . . . . . . . . . . 479.1.1 Rigid Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . 479.1.2 Non-Rigid Transformations . . . . . . . . . . . . . . . . . . . . . . . 479.2 Families . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 519.4 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56ron.bannon@mathography.orgL A TEX 2ε

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes10 mth.119.01.05 5710.1 Combinations and Compostions of Functions . . . . . . . . . . . . . . . . . . 5710.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5710.3 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6011 mth.119.01.06 6111.1 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6111.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6411.3 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6512 mth.119.02.01 6612.1 Polynomial Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6612.1.1 Graphing Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6612.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6712.3 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6813 mth.119.02.02 6913.1 Higher Degree Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6913.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6913.3 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7114 mth.119.02.03–mth.119.02.05 7214.1 Overview of Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7214.1.1 Introduction to Complex Numbers . . . . . . . . . . . . . . . . . . . 7214.1.2 Operations on Complex Numbers . . . . . . . . . . . . . . . . . . . . 7214.1.3 Fundamental Theorem of Algebra . . . . . . . . . . . . . . . . . . . . 7314.1.4 Rational Root Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 7414.1.5 Not All Roots are Rational . . . . . . . . . . . . . . . . . . . . . . . . 7414.1.6 Not All Roots are Real . . . . . . . . . . . . . . . . . . . . . . . . . . 7514.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7614.3 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8115 mth.119.02.06 8215.1 Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8215.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8515.3 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8616 mth.119.02.07 8716.1 Rationals Continued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8716.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9116.3 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92Page 2 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes17 mth.119.03.01 9317.1 Exponential Functions and Their Graphs . . . . . . . . . . . . . . . . . . . . 9317.1.1 Exponentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9317.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9417.3 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9918 mth.119.03.02 10018.1 Logarithmic Functions and Their Graphs . . . . . . . . . . . . . . . . . . . . 10018.1.1 Inverses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10018.1.2 Special Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10018.1.3 Graphing Logarithmic Functions . . . . . . . . . . . . . . . . . . . . 10118.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10218.3 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10819 mth.119.03.03 10919.1 Properties of Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10919.1.1 Basic Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10919.1.2 Changing Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10919.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11019.3 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11420 mth.119.03.04 11520.1 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11520.1.1 Exact and Approximate . . . . . . . . . . . . . . . . . . . . . . . . . 11520.1.2 Solving Exponential Equations . . . . . . . . . . . . . . . . . . . . . 11520.1.3 Solving Logarithmic Equations . . . . . . . . . . . . . . . . . . . . . 11520.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11620.3 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12221 mth.119.03.05 12321.1 Exponential and Logarithmic Models . . . . . . . . . . . . . . . . . . . . . . 12321.1.1 Exponential Decay Model . . . . . . . . . . . . . . . . . . . . . . . . 12321.1.2 Exponential Growth Model . . . . . . . . . . . . . . . . . . . . . . . 12321.1.3 Natural Logarithmic Model . . . . . . . . . . . . . . . . . . . . . . . 12421.1.4 Common Logarithmic Model . . . . . . . . . . . . . . . . . . . . . . . 12421.1.5 Logistic Growth Model . . . . . . . . . . . . . . . . . . . . . . . . . . 12521.1.6 Gaussian Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12521.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12521.3 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12722 mth.119.07.04 12822.0.1 Linear Systems Reviewed . . . . . . . . . . . . . . . . . . . . . . . . . 12822.0.2 Introduction to Non-linear Systems . . . . . . . . . . . . . . . . . . . 12822.0.3 Introduction to Linear Algebra: Linear 2 × 2 System . . . . . . . . . 129Page 3 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes22.1 Linear 3 × 3 System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13122.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13222.3 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13623 mth.119.07.05 13723.1 Introduction to Matrix Mathematics . . . . . . . . . . . . . . . . . . . . . . 13723.1.1 Something Old . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13723.1.2 Something New . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13723.1.3 Equal Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14023.1.4 Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14023.1.5 Simple Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14123.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14123.3 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14424 mth.119.07.06 14524.1 Inverse of a Square Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14524.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14524.3 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14825 mth.119.07.07 14925.1 Determinants of a Square Matrix . . . . . . . . . . . . . . . . . . . . . . . . 14925.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14925.3 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15026 mth.119.07.07 15126.1 Application of Matrices and Determinants . . . . . . . . . . . . . . . . . . . 15126.1.1 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15126.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15126.3 Assignment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15227 Final Review 153Page 4 of 169

Pre-calculus IMTH-1191 ReviewEssex County CollegeDivision of MathematicsLMS ProjectClass NotesIt is absolutely imperative that you have a fairly good grasp of the mathematics covered inprior courses including Algebra I (MTH 100). You may visithttp://mth100.mathography.org/to see what materials are covered in MTH-100. Although I may spend some time in classdiscussing material covered in prior mathematics courses, you can not expect that this prerequisitematerials will be covered sufficiently well enough to remediate inadequate preparation.Being prepared is the first step!Page 6 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes2 Course Web PagePlease visit http://mth119.mathography.org/ to learn more about how this course will berun. On this website you will be able to view the course outline, section specific syllabi, thisguide, 1 and information about accessing online homework. Visit this page often!1 This is what I’d like my notes to look like If I were a student in this class. You should make someattempt to write your own notes, or at least annotate mine. Please try to follow what is being done in class,and do not do the homework until you understand what was done in class.Page 7 of 169

Pre-calculus IMTH-1193 This GuideEssex County CollegeDivision of MathematicsLMS ProjectClass NotesThis guide is being provided as a way to structure MTH 119 across sections and semesters.Section headings that follow correspond to actual WebAssign assignments, and the materialin these sections address materials in these assignments. This guide is not a textbook andit should not be used as such—it just provides structure for teachers and students.1. Please come to all classes, these notes are not to be used as an excuse not to attend.2. These notes will not be read as a script, but teachers/students are encouraged to followthe content of these notes. Your teacher may or may not do all the problems that arein the notes. You should make every attempt to follow what your teacher is doing.Page 8 of 169

Pre-calculus IMTH-1194 SyllabusEssex County CollegeDivision of MathematicsLMS ProjectClass NotesLecturer: Ron BannonEssex County CollegeMathematics and Physics DivisionSyllabus for MTH 119 §004 — Pre-calculus ISpring 2013 Class SyllabusOffice: 2210Phone: 973-877-1886Email:bannon@essex.eduCourse Website:http://mth119.mathography.org/Regular Office Hours: Tuesday & Thursday 8:30 a.m.–10:45 a.m.;Appointment Office Hours: Monday & Wednesday 11:30 a.m.–12:50 p.m.;Class Meetings: Monday, Wednesday & Friday 10:00 a.m.–11:20 a.m., 2107;First Class: Monday, January 7, 2013;Last Class: Monday, April 22, 2013.• Course Description: Topics covered include absolute value inequalities and polynomialinequalities; relations and functions; polynomials and rational functions and theirgraphs; logarithmic and exponential functions; determinants and matrices. A graphingcalculator may be required. This course is designed for students who plan to take MTH121. (NOTE: Both MTH 119 & MTH 120 are required prerequisites of MTH 121.)• Required Materials: Textbook Pre-Calculus: Real Mathematics, Real People, 6thedition, by Larson; published by Cengage/Brooks/Cole, 2012 (ISBN:1-133-294-154).Make sure your book comes package with a WebAssign access card. Students shouldalso have access to a graphics calculator, notebook, and pencil or pen.• Course Prerequisites Grade of “C” or better in MTH 100 or placement.• Course Prerequisites & Course Co-requisites None.• General Education GoalsMTH 121 is affirmed in the following General Education Foundation Category: QuantitativeKnowledge and Skills. The corresponding General Education Goal is as follows:Students will use appropriate mathematical and statistical concepts and operations tointerpret data and to solve problems.• Course Goals Upon successful completion of this course, students should be able todo the following:1. demonstrate knowledge of the fundamental concepts and theories from calculus;2. utilize various problem-solving and critical-thinking techniques to set up and solveapplied problems in engineering, sciences, business and technology fields;Page 9 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes3. communicate accurate mathematical terminology and notation in written and/ororal form in order to explain strategies to solve problems as well as to interpretfound solutions; and4. use appropriate technology, such as graphing calculators and computer software,effectively as a tool to solve such problems as those described above.• Measurable Course Performance Objectives (MPOs) Upon successful completionof this course, students should specifically be able to do the following:1. Demonstrate knowledge of the fundamental concepts and theories from pre-calculus:1.1 evaluate and graph polynomial, rational, piecewise, composite, combination,exponential, logarithmic, and multi-variable functions;1.2 solve linear, quadratic, exponential, and logarithmic equations;1.3 identify the key characteristics of a given function and use them to graph thefunction in an appropriate coordinate system;1.4 write the equations of linear functions, logarithmic functions, exponentialfunctions, and polynomial and rational functions based on their properties;1.5 identify and describe function transformations;1.6 calculate difference quotients;1.7 find the zeroes of polynomial and rational functions;1.8 use the concepts of matrices and determinants to solve systems of equations;1.9 find the determinant and the inverse of a matrix; and1.10 find the domain and range of polynomial, rational, radical, exponential, andlogarithmic functions2. Utilize various pre-calculus problem-solving and critical-thinking techniques toset up and solve applied problems in engineering, sciences, and other fields:2.1 solve compound interest problems;2.2 solve current and voltage in a circuit applications using simultaneous equations;2.3 solve optimization problems (in geometry, finance, inventory control, etc.)including those involving functions of several variables; and2.4 solve growth and decay problems (in finance, biology, chemistry, physics, etc.)using exponential functions3. 3. Communicate accurate mathematical terminology and notation in writtenand/or oral form in order to explain strategies to solve problems as well as tointerpret found solutions:3.1 write and explain solutions to application problems including growth anddecay problemsPage 10 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes4. Use graphing calculators 2 effectively as a tool to solve such problems as thosedescribed above:4.1 use the GRAPH feature to display polynomial, piecewise, composite, exponentialand logarithmic functions;4.2 use the TABLE feature to determine account balances for given compoundinterest problem;4.3 use the TABLE feature with an exponential function to approximate thevalue of the e-constant;4.4 use the ZERO feature to find zeroes of polynomial and rational functions;4.5 use the MINIMUM and MAXIMUM features to find relative minimum andmaximum values of polynomial functions;4.6 use the GRAPH feature to observe the domain and range of polynomial,rational, exponential and logarithmic functions; and4.7 use the MATRIX feature to find the inverse and determinant of a matrix• Methods of InstructionInstruction will consist of a combination of lectures, class discussion, individual study,and computer lab work.• Class Requirements All students are required to:1. Read the textbook 3 and do the suggested review problems in a timely manner.2. Be an active participant in all classes.3. Complete all written and/or electronic homework and adhere to assignment deadlines.4. Take exams/quizzes in class and adhere to the exam/quiz schedule. 4• Grading Criteria1. Homework is worth 20%.2. Exams/quizzes (approximately 33 problems total) and projects (approximately 6problems total), equally weighed, is worth 55%.3. Cumulative final exam is worth 25%.Tentative grade breakdown:A if average ≥ 94%;B+ if 88% ≤ average < 94%;2 You may decide to do this on any computational device, and it does not necessarily have to be a TIcalculator. I will discuss your options in class.3 The e-book is available through WebAssign.4 I will give students advanced notice in class when an assessment is scheduled.Page 11 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesB if 82% ≤ average < 88%;C+ if 76% ≤ average < 82%;C if 70% ≤ average < 76%;D if 64% ≤ average < 70%;F if average < 64%;• Academic Integrity Dishonesty disrupts the search for truth that is inherent in thelearning process and so devalues the purpose and the mission of the College. Academicdishonesty includes, but is not limited to, the following:plagiarism — the failure to acknowledge another writers words or ideas or to giveproper credit to sources of information;cheating — knowingly obtaining or giving unauthorized information on any test/examor any other academic assignment;interference — any interruption of the academic process that prevents others fromthe proper engagement in learning or teaching; andfraud — any act or instance of willful deceit or trickery.Violations of academic integrity will be dealt with by imposing appropriate sanctions.Sanctions for acts of academic dishonesty could include the resubmission of an assignment,failure of the test/exam, failure in the course, probation, suspension from theCollege, and even expulsion from the College.• Student Code of Conduct All students are expected to conduct themselves asresponsible and considerate adults who respect the rights of others. Disruptive behaviorwill not be tolerated. All students are also expected to attend and be on time allclass meetings. No cell phones or similar electronic devices are permitted in class.Please refer to the Essex County College student handbook, Lifeline, for more specificinformation about the Colleges Code of Conduct and attendance requirements.• Tentative Class Schedule The course will follow, in order, the following sectionsand suggested problems. 5 Prepared notes are available 6 that will be used in class tohelp guide the student trough the material. We will move quickly, and students will betested on materials assigned. To remain current a student should come to each class.1. Appendix B.4 (on web), Solving Inequalities Algebraically and Graphically:Handout Assignment2. Appendix B.4 (on web), Solving Inequalities Algebraically and Graphically:Handout Assignment3. Appendix B.4 (on web), Solving Inequalities Algebraically and Graphically:Handout Assignment5 The corresponding WebAssign format is chapter.section.exercise from Larson’s textbook.6 http://mth119.mathography.org/Page 12 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes4. Chapter 1, §2, Functions: Do problems: 1, 3, 5, 7, 9, 13, 15, 17, 19, 27, 29,33, 35, 37, 39, 53, 55, 59, 61, 675. Chapter 1, §3, Graphs of Functions: Do problems: 1, 3, 5, 7, 9, 15, 17, 21,23, 27, 37, 39, 43, 49, 59, 61, 75, 776. Chapter 1, §4, Shifting, Reflecting, and Stretching Graphs: Do problems:3, 7, 9, 11, 13, 17 19, , 21, 23, 27, 31, 43, 47, 49, 517. Chapter 1, §5, Combinations of Functions: Do problems: 5, 7, 9, 13, 15,19, 35, 37, 49, 51, 55, 57, 65, 698. Chapter 1, §6, Inverse Functions: Do problems: 1, 3, 5, 7, 9, 11, 17, 19, 25,29, 35, 39, 43, 59, 61, 79, 1139. Chapter 2, §1, Quadratic Functions: Do problems: 1, 3, 9, 11, 13, 17, 21,23, 25, 27, 29, 31, 35, 39, 41, 45, 53, 55, 57, 5910. Chapter 2, §2, Polynomial Functions of Higher Degree: Do problems: 1,3, 5, 7, 9, 15, 17, 19, 23, 27, 29, 31, 37, 41, 49, 51, 53, 69, 71, 75, 9111. Chapter 2, §3, Real Zeros of Polynomial Functions: Do problems: 7, 9,13, 15, 19, 21, 25, 31, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 61, 63, 67, 71, 73,7512. Chapter 2, §4, Complex Numbers: Do problems:3, 5, 11, 15, 21, 29, 31, 39,45, 47, 49, 51, 53, 65, 71, 7513. Chapter 2, §5, The Fundamental Theorem of Algebra: Do problems: 1,3, 5, 7, 13, 15, 17, 23, 29, 33, 37, 39, 41, 51, 53, 55, 61, 6314. Chapter 2, §6, Rational Functions and Asymptotes: Do problems: 7, 9,11, 13, 17, 19, 23, 31, 39, 49, 5115. Chapter 2, §7, Graphs of Rational Functions: Do problems:9, 13, 15, 17,21, 23, 25, 27, 43, 47, 49, 5316. Chapter 3, §1, Exponential Functions and Their Graphs: Do problems:1, 3, 7, 11, 13, 15, 19, 23, 27, 31, 41, 55, 57, 67, 6917. Chapter 3, §2, Logarithmic Functions and Their Graphs: Do problems:1, 3, 5, 7, 17, 25, 33, 35, 47, 49, 51, 53, 55, 73, 91, 9318. Chapter 3, §3, Properties of Logarithms: Do problems: 1, 13, 31, 35, 37,39, 41, 43, 45, 47, 51, 53, 55, 59, 61, 63, 73, 81, 85, 87, 89, 9119. Chapter 3, §4, Solving Exponential and Logarithmic Equations: Doproblems:1, 17, 21, 23, 25, 29, 37, 39, 43, 45, 49, 53, 57, 85, 87, 89, 91, 93, 97,101, 103, 13720. Chapter 3, §5, Exponential and Logarithmic Models: Do problems: 1, 3,5, 28, 29, 33(b), 35, 39, 4021. Chapter 7, §4, Matrices and Systems of Equations: Do problems: 1, 3,5, 7, 9, 11, 13, 15, 17, 19, 21, 41, 43, 47, 49, 55, 59, 65Page 13 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes22. Chapter 7, §5, Operations with Matrices: Do problems: 1, 3, 5, 7, 9, 13,15, 21, 25, 27, 29, 31, 33, 35, 3923. Chapter 7, §6, The Inverse of a Square Matrix: Do problems: 1, 5, 7, 11,13, 17, 19, 21, 29, 43, 45, 49, 6524. Chapter 7, §7, The Determinant of a Square Matrix: Do problems: 1, 3,5, 7, 9, 11, 3325. Chapter 7, §8, Applications of Matrices and Determinants: Do problems:1, 3, 15, 16, 18, 21, 22, 25Page 14 of 169

Pre-calculus IMTH-1195 IntroductionEssex County CollegeDivision of MathematicsLMS ProjectClass NotesYou should make every effort to attend all scheduled classes. The notes that follow are notself-contained and you will need to attend class in order to make sense out of the contentthat follows. The notes just give structure to the classroom lectures, and will keep you ontrack with your assignments. You should note that almost every section that follow will referto an assignment in WebAssign. It is incredibly important that you visithttp://mth119.mathography.organd get started with WebAssign immediately! You should also print out the complete notes(available on the course webpage) or purchase a copy (link may be provided online).If you have any questions or concerns about WebAssign or the notes, you need to contactRon Bannon as soon as possible.Page 15 of 169

Pre-calculus IMTH-1196 mth.119.AB.04Essex County CollegeDivision of MathematicsLMS ProjectClass NotesThis material is not in your physical textbook, the publisher has instead made this contentavailable online. This material will be covered in class, but I do encourage everyone to readthe online content provided by the publisher. Please see your textbook for more informationabout online content.6.1 Inequalities6.1.1 Properties of InequalitiesThis subsection is a review of the material covered in MTH100.Let a, b, c, and d be real numbers.1. Transitive Property.a < b and b < c ⇒ a < c2. Addition of Inequalities.a < b and c < d ⇒ a + c < b + d3. Addition of Constants.a < b ⇒ a + c < b + c4. Multiplying by Constants.(a) If c > 0.(b) If c < 0.a < b ⇒ ac < bca < b ⇒ ac > bc6.1.2 Examples from MTH100Solve for the variable, graph, and express the solution set using interval notation.1. 6x ≤ 19Solution: Work will be done in class.2. 2x + 7 > 3 (1 − 2x)Page 16 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.3. 3 + 2 7 x < 2x − 5Solution: Work will be done in class.4. 0 < 2 − 3 (x + 1) < 20Solution: Work will be done in class.6.1.3 Absolute Value InequalitiesLet x be a variable of an algebraic expression and let a be a real number such that a ≥ 0.1. The solution of |x| < a are the values of x that lie between −a and a.|x| < a if and only if − a < x < a.2. The solution of |x| > a are the values of x that are less than −a or greater than a.|x| > a if and only if x < −a or x > a.6.1.4 Absolute Value Inequality ExamplesSolve for the variable, graph, and express the solution set using interval notation.1. |x| ≤ 7Solution: Work will be done in class.2. |x| > 7Solution: Work will be done in class.3. |x − 10| > 7Solution: Work will be done in class.4. |x − 10| ≤ 7Page 17 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.5. |3 − 2x| ≤ 4Solution: Work will be done in class.6. |3 − 2x| > 4Solution: Work will be done in class.7. 5 |2x − 3| < 20Solution: Work will be done in class.6.1.5 Polynomial InequalitiesTypically we will only be dealing with easily factorable polynomials, and you will need tounderstand that linear factors change signs at their zeros. For example, the factor (x − 3)changes sign at x = 3; that is, to the right of three the factor is always positive, and tothe left of three the factor is always negative. This, in fact, is the central idea of linearfactor analysis. What is key here, is that our inequalities will now be completely factorablepolynomials, and we need to invariably have them compared to zero for this factor analysisto work.Here’s the method for solving polynomial inequalities.• First and foremost you will need to have one side of your inequality set equal to zero.• Factor the polynomial completely.• Find all zeros of the polynomial, and arrange the zeros in increasing order on a numberline. Draw a small circle at each zero directly on the number line.• Find a test value on each of the intervals to test in the factored form of the polynomial.It will either be positive or negative, and you’ll need to indicate this directly on thenumber line using either ± sign.• Shade the region(s) that satisfy the inequality/equality.• Write the final answer in interval notation.6.1.6 Polynomial Inequality Examples1. x 2 − x − 6 ≤ 0Page 18 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.2. 2x 2 + 5x − 12 > 0Solution: Work will be done in class.3. 2x 3 − 3x 2 − 32x + 48 ≥ 0Solution: Work will be done in class.4. (x − 3) 2 ≤ 1Solution: Work will be done in class.5. x 2 − 6x + 9 > 16Solution: Work will be done in class.6. x 4 (x − 2) ≥ 0Solution: Work will be done in class.6.1.7 Rational InequalitiesYou need to have a good understanding of polynomial inequalities before attempting rationalinequalities. Typically we will only be dealing with easily factorable ratios of polynomials,and once again you will need to understand that linear factors change signs at their zeros.Central to solving rational inequalities, is that our inequalities will now be completely factorableratio of polynomials, and we need to invariably have them compared to zero for thisfactor analysis to work. Big idea here is that we will only have one rational expression,completely factored, and compared to zero.Here’s the method for solving rational inequalities.• First and foremost you will need to have one side of your inequality set equal to zero.• If you have more than one rational expression, you will need to combine them into onerational expression.• Factor the polynomials, both numerator and denominator, completely.• Find all zeros of the numerator and denominator, and arrange the zeros in increasingorder on a number line. Draw a small circle at each zero directly on the number line.Page 19 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes• Find a test value on each of the intervals to test in the factored form of the rationalexpression. It will either be positive or negative, and you’ll need to indicate this directlyon the number line using either ± sign.• Shade the region(s) that satisfy the inequality/equality. Be very careful that you don’tdivide by zero.• Write the final answer in interval notation.6.1.8 Rational Inequality Examples1.(x − 1) (x + 2)(x − 2) (x + 1) < 0Solution: Work will be done in class.2.(2x − 1) (3x + 5)(x − 1) (x + 1)≥ 0Solution: Work will be done in class.3.1x ≥ 1Solution: Work will be done in class.4.1x ≤ xSolution: Work will be done in class.5.1x − 2 ≥ 1x + 3Solution: Work will be done in class.6.1x ≤ 1 x 3Solution: Work will be done in class.7.1x 3 ≤ 1 x 2 Page 20 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.6.2 AssignmentDirections: Solve each inequality algebraically and express your solution set two ways: asa graph on a number line 7 and in interval notation. Partial answers are provided, but nowork is being provided as a guide. Yes, you need to be able to do these problems and youwill be tested. You need to get help if you cannot do these problems exactly as was done inclass!1. x (x − 7) > 8Solution: (−∞, −1) ∪ (8, ∞)2. −5 < 3x − 2 < 1Solution: (−1, 1)3. x 2 − 1 ≥ 0Solution: (−∞, −1] ∪ [1, ∞)4. x 2 + 7x ≤ −12Solution: [−4, −3]5. 4 ≤ 2x + 2 ≤ 10Solution: [1, 4]6.3x − 5x + 2 < 2Solution: (−2, 9)7.1x − x ≥ 07 Please go to http://www.wolframalpha.com/ and graph these inequalities. A video is available at:http://screencast.com/t/vrcXRWtt. These URLs are ‘clickable’!Page 21 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: (−∞, −1] ∪ (0, 1]8.12 ≤ x + 13< 3 4Solution: [1/2, 5/4)9.1x ≤ 4Solution: (−∞, 0) ∪ [ 14 , ∞)10. −2 (x + 3) < 8Solution: (−7, ∞)11. −3 < 2x − 14< 0Solution: (−11/2, 1/2)12. |5x − 1| < 9Solution: (−8/5, 2)13. 9 + |5x| < 24Solution: (−3, 3)14.x + 6x + 1 ≥ 2Solution: (−1, 4]15. |6 + 4x| ≥ 10Solution: (−∞, −4] ∪ [1, ∞)16.x + 12x + 2 < 3 Page 22 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: (−∞, −2) ∪ (3, ∞)17. x 2 + 4x + 4 ≥ 9Solution: (−∞, −5] ∪ [1, ∞)18. 3x 2 − 11x − 4 ≤ 0Solution: [ − 1 3 , 4]19. 2x 3 + 5x 2 > 6x + 9 [Hint, x + 1 is a factor.]Solution: (−3, −1) ∪ ( 32 , ∞)20. x 3 < x 2Solution: (−∞, 0) ∪ (0, 1)21. −1 < 3 − x2≤ 1Solution: 1 ≤ x < 522. |10 − x| ≥ 1Solution: (−∞, 9] ∪ [11, ∞)23. 3 |4 − 5x| > 12( ) 8Solution: (−∞, 0) ∪5 , ∞24. |2x + 11| − 3 ≤ 4Solution: [−9, −2]25. −8 ≤ 1 − 3 (x − 2) ≤ 13Page 23 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: [−2, 5]26. 3.6 + 1.5x > 2.3x − 5.2Solution: x < 1127. −5 ≤ 4 − 3x ≤ 2Solution: [2/3, 3]28. −3 < 3 − 2x < 9Solution: (−3, 3)29. x 2 − 4x > 0Solution: (−∞, 0) ∪ (4, ∞)30. x 2 + 8x ≤ 0Solution: [−8, 0]31. x 2 − 9 ≤ 032.Solution: [−3, 3]∣ 3x + 2 5∣ ≥ 433.34.Solution: (−∞, −22/15] ∪ [6/5, /∞)4x − 3∣ 5 ∣ < 1Solution: (−1/2, 2)∣ x 2 − 3 ∣ ∣∣ > 7Page 24 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: (−∞, −8) ∪ (20, ∞)35. (x − 5) (x + 2) > 0Solution: (−∞, −2) ∪ (5, ∞)36. |y − 5| ≤ 3Solution: [2, 8]37. |1 + 3x| + 4 < 538.Solution: (−2/3, 0)x − 3∣ 2 ∣ ≤ 5Solution: [−7, 13]39. |3x| − 8 > 1Solution: (−∞, −3) ∪ (3, ∞)40. 4 − 3 (1 − x) < 3Solution: (−∞, 2/3)41. 2x 2 < 5x + 3Solution: (−1/2, 3)42.x + 3x − 2 < 0Solution: (−3, 2)43.x − 5x − 2 ≥ 0 Page 25 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: (−∞, 2) ∪ [5, ∞)44.x − 5x − 6 ≥ 0Solution: (−∞, 5] ∪ (6, ∞)45.4x + 5x + 2 ≥ 3Solution: (−∞, −2) ∪ [1, ∞)46. x 2 + x > 12Solution: (−∞, −4) ∪ (3, ∞)47.x2 ≤ 1 − x 4Solution: (−∞, 4/3]48. (x − 5) (x + 2) < 0Solution: (−2, 5)49. 1 − 2x ≤ 3Solution: [−1, ∞)50. 3x − 1 ≤ 3 + xSolution: (−∞, 2]51.52.x3 ≥ 2 + x 6Solution: [12, ∞)∣ x 3 + 6 ∣ ∣∣ − 8 > −5Page 26 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: (−∞, −27) ∪ (−9, ∞)53.x + 4x − 2 < 1Solution: (−∞, 2)Page 27 of 169

Pre-calculus IMTH-1197 mth.119.01.02Essex County CollegeDivision of MathematicsLMS ProjectClass Notes7.1 Introductory FunctionsNotationally, we often write y = f (x), which is read as, “y is a function of x.” We areoften asked to evaluate functions for given values, for example, if f (x) = 3x 2 − 2x + 1, thenf (1) = 3 − 2 + 1 = 2, and f (2 + s) = 3 (2 + s) 2 − 2 (2 + s) + 1 = 3s 2 + 10s + 9. We’re just“plugging in” and then simplifying—some students have trouble with this, but it’s really nothard once you try!Definition of a Function: A function f from a set A to a set B is a relation thatassigns to each element x in the set A exactly one element y in the set B. The set A is thedomain of the function f, and the set B contains the range. You should note that the set Bis not necessarily the range, it just contains the range.Characteristics of a function from set A to set B 8• Each element of A must be matched with an element from B.• An element from A cannot be matched with two different elements of B.Not all relationships are functional and you’ll need to evaluate when a relationship isfunctional. We’ll keep it very simple!Here, you may need to be reminded that you should never divide by zero! So if you’reasked to find the domain of functionf (x) = 5x − 3 ,you’re basically being asked what values of x are allowed? I hope you can see that x = 3will result in a division by zero—so the domain of this function is all real numbers exceptthree. This is often written: R, x ≠ 3, but you should also be able to write the domain ininterval notation.Also, even roots, when dealing with real numbers, need to be non-negative. For exampleif f (x) = √ 4 − x 2 , then the domain is: [−2, 2].Don’t get your brain into a bunch, we’ll do this material by example—there’s very littleto remember and much of this was already covered in MTH 100.7.1.1 EquationsIn algebra we usually use equations to represent relationships. For example, you alreadyknow many such relationships. Here’s a partial list.Lines: y = mx + bParabolas: y = Ax 2 + Bx + CCircles: (x − h) 2 + (y − k) 2 = r 28 We’ll discuss using graphs (vertical line test) and diagrams to determine if we have a function or not.Page 28 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesIf you can 9 solve for y in terms of x, you’ve got a function where the elements in x is thedomain and the elements in y is the range. Here we often say that y depends on x or is afunction of x.7.2 Examples1. Let f (x) = 2x − 3, find each of the following.(a) f (1) =Solution: Work will be done in class.(b) f (−2) =Solution: Work will be done in class.(c) f (a) =Solution: Work will be done in class.(d) f (2x) =Solution: Work will be done in class.(e) f (x + 1) =Solution: Work will be done in class.(f) f (x + h) =Solution: Work will be done in class.(g) f (x + h) − f (x) =Solution: Work will be done in class.f (x + h) − f (x)(h) For h ≠ 0, =hSolution: Work will be done in class.2. Is 2x + 3y = 7 a function, that is, is y a function of x?Solution: Work will be done in class.3. Is y 2 − x = 5 a function, that is, is y a function of x?9 This is not entirely true, for exampley (x − 1) = x 2 − 1can easily be solved for y, but y is not a function of x.anywhere near this level of understanding.However, most examples in the book don’t goPage 29 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.4. Is 5y − 2x 3 − 4x 2 + 7x = 9 a function, that is, is y a function of x?Solution: Work will be done in class.5. Is xy − x 2 y = 9 a function, that is, is y a function of x?Solution: Work will be done in class.6. Given that A is the set of people in this classroom, and B ∈ [0, 10] measured in feet. Isthis relationship functional, and if it is, what is its range?Solution: Work will be done in class.7. Given that A = {1, 2, 3}, and B = {0, 1, 2, 3}, which set of ordered pairs represent afunction from A to B? If the relationship is functional, what is its range?(a) {(1, 1) , (3, 2) , (3, 3) , (3, 0)}Solution: Work will be done in class.(b) {(1, 1) , (2, 2) , (3, 3)}Solution: Work will be done in class.(c) {(1, 1) , (0, 1) , (2, 2) , (3, 3)}Solution: Work will be done in class.(d) {(3, 0) , (2, 0) , (1, 3)}Solution: Work will be done in class.8. Determine whether the equation x = y 2 + 49 represents y as a function of x.Solution: Work will be done in class.9. Determine whether the equation x = −y + 8 represents y as a function of x.Solution: Work will be done in class.10. Given the following pictures are relationships between a set A and a set B, which relationshipsrepresent a function. If the relationship is functional, what is its range?(a) A ∈ [−1, 2], B ∈ R.Page 30 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes654321-1 0 1 2Figure 1: A relationship from A to BSolution: Work will be done in class.(b) C ∈ [−2, 4], D ∈ R.321-4 -3 -2 -1 0 1 2 3 4 5-1-2-3-4Figure 2: A relationship from C to DSolution: Work will be done in class.11. Evaluate the function at each specified value of the independent variable and simplify.g (y) = 4 − 3y(a)g (0) =Page 31 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.(b)( 4g =3)Solution: Work will be done in class.(c)g (s + 1) =Solution: Work will be done in class.12. Evaluate the function at each specified value of the independent variable and simplify.f (x) = x 2 − 2x(a)f (−2) =(b)Solution:f (1.1) =(c)Solution: Work will be done in class.f (x + 2) =Solution: Work will be done in class.13. Evaluate the function (this is a piecewise defined function) at each specified value of theindependent variable and simplify.{ 4x + 5, x ≤ 0f (x) =4x 2 + 4, x > 0Page 32 of 169

Pre-calculus IMTH-119(a)Essex County CollegeDivision of MathematicsLMS ProjectClass Notesf (−2) =Solution: Work will be done in class.(b)f (0) =Solution: Work will be done in class.(c)f (2) =Solution: Work will be done in class.14. Find all values of x such that f (x) = 0.f (x) = 8x − 99Solution: Work will be done in class.15. Find the domain of the function.s (y) =7yy + 9Solution: Work will be done in class.16. Find the domain of the function.g (x) = 4√ x 2 + 7xPage 33 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.17. Find the domain of the function.f (x) = x + 2 √x2 − 1Solution: Work will be done in class.18. Find the domain of the function.f (x) =√x2 − 1x + 2Solution: Work will be done in class.19. Consider the function below.f (x) = 3x − 2Find the difference quotient below (where h ≠ 0) and simplify your answer.f (x + h) − f (x)hSolution: Work will be done in class.20. Consider the function below.f (x) = x 2 − 2x + 1Find the difference quotient below (where h ≠ 0) and simplify your answer.f (x + h) − f (x)hPage 34 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.7.3 AssignmentYou should read 1.2 and do the WebAssign assignment 01.02.Page 35 of 169

Pre-calculus IMTH-1198 mth.119.01.03Essex County CollegeDivision of MathematicsLMS ProjectClass Notes8.1 Functions Continued . . .8.1.1 Terminology• Increasing Function: A function f is increasing on an interval if, for any x 1 and x 2in the interval, x 1 < x 2 implies that f (x 1 ) < f (x 2 ).• Decreasing Function: A function f is decresing on an interval if, for any x 1 and x 2in the interval, x 1 < x 2 implies that f (x 1 ) > f (x 2 ).• Constant Function: A function f is constant on an interval if, for any x 1 and x 2 inthe interval, that f (x 1 ) = f (x 2 ).• Relative Minimum of a Function: A function value f (a) is called a relative minimumof f if there exists an interval (x 1 , x 2 ) such thatf (a) ≤ f (x) , ∀ x ∈ (x 1 , x 2 ) .• Relative Maximum of a Function: A function value f (a) is called a relativemaximum of f if there exists an interval (x 1 , x 2 ) such thatf (a) ≥ f (x) , ∀ x ∈ (x 1 , x 2 ) .• A Function with Even Symmetry: A function f (x) is said to have even symmetryif for each x in f’s domain, f (x) = f (−x).• A Function with Odd Symmetry: A function f (x) is said to have odd symmetryif for each x in f’s domain, f (x) = −f (−x).• Special Points: A function f (x) has a y-intercept if 0 is in its domain, and the y-intercept is (0, f (0)). A function f (x) has an x-intercept if 0 is in its range, and thex-intercept is (a, f (a) = 0). An x-intercept is often referred to as a root or a zero ofthe function.8.1.2 GraphingMake sure you have access to a graphic calculator. You may need to read your calculator’smanual especially as it relates to identifying relative extrema on a graph. It’s importantthat you can graph simple functions and identify intervals and points as they relate to theterminology above. For example, supposed you are asked to graphf (x) = x 3 − 3x 2 − x + 3and discuss this graph as it relates to the terminology above. Using your own calculator youshould be able to get a graph similar to mine. 10 And, also using your calculator you should10 Yes, I am using a calculator to graph.Page 36 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes321-3 -2 -1 0 1 2 3 4 5 6-1-2-3Figure 3: f (x) = x 3 − 3x 2 − x + 3be able to identify—with a good deal of accurarcy—the relative minimum and maximum onthis graph. Here’s what I’m getting.• Increasing: On the interval (−∞, −0.155) ∪ (2.15, ∞)• Decreasing: On the interval (−0.155, 2.15)• Constant: NA• Relative Minimum of a Function: The point is (2.15, −3.08)• Relative Maximum of a Function: The point is (−0.155, 3.08)• A Function with Even Symmetry: Not even.• A Function with Odd Symmetry: Not odd.• Special Points: The y-intercept is (0, 3);. The x-intercepts are (−1, 0), (1, 0), and(3, 0).Again, make sure you can use your calculator. Each calculator model varies, but I amalmost 100% positive that all graphic calculators available nowadays will be able to do theseproblems. And, in the future you should be able to graph many functions without the aid ofa calculator. Graphing takes time and having a good graphic calculator will certainly speedup the learning process. Let’s get to work!Page 37 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes8.2 Examples1. I want to point out that I’ve indicated two points on this function, one point is (−2, 104)and the other point is (8, −396). These points are given because they are difficult todetermine by inspection. However, if the points are obvious, for example (0, 52), it willnot be given.-10 0 10 20-400(a) Domain in interval notation?Figure 4: f (x) = x 3 − 9x 2 − 48x + 52Solution: Work will be done in class.(b) Range in interval notation?Solution: Work will be done in class.(c) The interval where f is increasing?Solution: Work will be done in class.(d) The interval where f is decreasing?Solution: Work will be done in class.(e) The y-intercept.Solution: Work will be done in class.(f) How many x intercepts are there? Can you find their exact values or approximatethem?Solution: Work will be done in class.(g) Symmetry?Solution: Work will be done in class.(h) Relative maximum?Page 38 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.(i) Relative minimum?Solution: Work will be done in class.2. You really don’t need a calculator for this one.963-3 0 3 6 9(a) Domain in interval notation?Figure 5: f (x) = x 2 − 6x + 9Solution: Work will be done in class.(b) Range in interval notation?Solution: Work will be done in class.(c) The interval where f is increasing?Solution: Work will be done in class.(d) The interval where f is decreasing?Solution: Work will be done in class.(e) The y-intercept.Solution: Work will be done in class.(f) How many x intercepts are there? Can you find their exact values or approximatethem?Solution: Work will be done in class.(g) Symmetry?Page 39 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.(h) Relative maximum?Solution: Work will be done in class.(i) Relative minimum?Solution: Work will be done in class.3. Two points are indicated here, and they occur at x = ±√52 .54321-2 -1 0 1 2-1-2-3(a) Domain in interval notation?Figure 6: f (x) = x 4 − 5x 2 + 4Solution: Work will be done in class.(b) Range in interval notation?Solution: Work will be done in class.(c) The interval where f is increasing?Solution: Work will be done in class.(d) The interval where f is decreasing?Solution: Work will be done in class.(e) The y-intercept.Solution: Work will be done in class.(f) How many x intercepts are there? Can you find their exact values or approximatethem?Page 40 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.(g) Symmetry?Solution: Work will be done in class.(h) Relative maximum?Solution: Work will be done in class.(i) Relative minimum?Solution: Work will be done in class.4. This is a piecewise defined function.963-3 -2 -1 0 1 2 3-3-6-9Figure 7: f (x)f (x) ={ x 2 + 1 x < 1−x 2 + −1 x ≥ 1(a) Domain in interval notation?Solution: Work will be done in class.(b) Range in interval notation?Solution: Work will be done in class.(c) The interval where f is increasing?Solution: Work will be done in class.(d) The interval where f is decreasing?Solution: Work will be done in class.(e) The y-intercept.Page 41 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.(f) How many x intercepts are there? Can you find their exact values or approximatethem?Solution: Work will be done in class.(g) Symmetry?Solution: Work will be done in class.(h) Relative maximum?Solution: Work will be done in class.(i) Relative minimum?Solution: Work will be done in class.5. This graph may look deceptive if you don’t yet understand domain.15010050-12 -9 -6 -3 0 3 6 9 12-50-100-150Figure 8: f (x) = x √ x 2 − 9(a) Domain in interval notation?Solution: Work will be done in class.(b) Range in interval notation?Solution: Work will be done in class.(c) The interval where f is increasing?Solution: Work will be done in class.(d) The interval where f is decreasing?Solution: Work will be done in class.Page 42 of 169

Pre-calculus IMTH-119(e) The y-intercept.Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.(f) How many x intercepts are there? Can you find their exact values or approximatethem?Solution: Work will be done in class.(g) Symmetry?Solution: Work will be done in class.(h) Relative maximum?Solution: Work will be done in class.(i) Relative minimum?Solution: Work will be done in class.6. Try looking at the function and graph and see if you can make sense out of this.321-3 -2 -1 0 1 2 3 4-1-2-3Figure 9: f (x) = x √ 4 − x(a) Domain in interval notation?Solution: Work will be done in class.(b) Range in interval notation?Solution: Work will be done in class.(c) The interval where f is increasing?Solution: Work will be done in class.(d) The interval where f is decreasing?Page 43 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.(e) The y-intercept.Solution: Work will be done in class.(f) How many x intercepts are there? Can you find their exact values or approximatethem?Solution: Work will be done in class.(g) Symmetry?Solution: Work will be done in class.(h) Relative maximum?Solution: Work will be done in class.(i) Relative minimum?Solution: Work will be done in class.7. Again, try looking at the function and graph and see if you can make sense out of this.1-2 -1 0 1 2-1Figure 10: f (x) = −x |x + 1|(a) Domain in interval notation?Solution: Work will be done in class.(b) Range in interval notation?Solution: Work will be done in class.(c) The interval where f is increasing?Solution: Work will be done in class.Page 44 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes(d) The interval where f is decreasing?Solution: Work will be done in class.(e) The y-intercept.Solution: Work will be done in class.(f) How many x intercepts are there? Can you find their exact values or approximatethem?Solution: Work will be done in class.(g) Symmetry?Solution: Work will be done in class.(h) Relative maximum?Solution: Work will be done in class.(i) Relative minimum?Solution: Work will be done in class.8. Again, try looking at the function and graph and see if you can make sense out of this.{1 − (x − 1)2x ≤ 2f (x) = √ x − 2 x > 2321-1 0 1 2 3 4 5 6-1-2-3Figure 11: f (x)(a) Domain in interval notation?Solution: Work will be done in class.(b) Range in interval notation?Page 45 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.(c) The interval where f is increasing?Solution: Work will be done in class.(d) The interval where f is decreasing?Solution: Work will be done in class.(e) The y-intercept.Solution: Work will be done in class.(f) How many x intercepts are there? Can you find their exact values or approximatethem?Solution: Work will be done in class.(g) Symmetry?Solution: Work will be done in class.(h) Relative maximum?Solution: Work will be done in class.(i) Relative minimum?Solution: Work will be done in class.8.3 AssignmentYou should read 1.3 and do the WebAssign assignment 01.03.Page 46 of 169

Pre-calculus IMTH-1199 mth.119.01.04Essex County CollegeDivision of MathematicsLMS ProjectClass Notes9.1 Shifting, Reflecting, and Stretching Functions9.1.1 Rigid TransformationsVertical and Horizontal Shifts: Let c > 0 and f (x) be a function, and h (x) be a functionrelated to f by a simple vertical or horizontal shift.Vertical shift c units upward: h (x) is simply f (x) shifted c units upward.h (x) = f (x) + cVertical shift c units downward: h (x) is simply f (x) shifted c units downward.h (x) = f (x) − cHorizontal shift c units right: h (x) is simply f (x) shifted c units rightward.h (x) = f (x − c)Horizontal shift c units left: h (x) is simply f (x) shifted c units leftward.h (x) = f (x + c)Reflections in the Coordinate Axes: Reflections in the coordinate axes of the graph off (x), are as follows:Reflections in the x-axis: h (x) is simply f (x) rotated about the x-axis.h (x) = −f (x)Reflections in the y-axis: h (x) is simply f (x) rotated about the y-axis.h (x) = f (−x)The rigid transformation should be someone reasonable, but I still think that we need todevelop a technique that does not depend on memorizing the above. More on that later!9.1.2 Non-Rigid TransformationsWhen we’re dealing with a constant multiple of f or x we’ll get a stretching or compressingbehavior—both vertical and horizontal. These are perhaps more difficult to understand,and things can get pretty messy when we start combine rigid and non-rigid transformations.I will essentially avoid memorizing results and will instead use a simple point translationmethod. However, it is nonetheless important that you are able to see what’s happening,even if it takes graphing. Let’s take a simple example to see what a non rigid transformationlooks like. The parent is f (x), and we will see what happens when we do the following:Page 47 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes1. 2 · f (x). Give it some thought and I think you’ll see that we’ll get a vertical stretchby a factor of 2.2. 0.5 · f (x). Give it some thought and I think you’ll see that we’ll get a vertical compressionby a factor of 2.3. f (2 · x). Give it some thought and I think you’ll see that we’ll get a horizontal compressionby a factor of 2.4. f (0.5 · x). Give it some thought and I think you’ll see that we’ll get a horizontalexpansion by a factor of 2.Here’s the parent graph.1-1 0 1 2 3-1Figure 12: f (x)And here are f’s children. Please make sure you make note of the scale in each case andpay attention to how the characteristic points are changing as we multiply f or x.321-5 -4 -3 -2 -1 0 1 2 3 4 5-1-2-3Figure 13: 2 · f (x)Page 48 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes1-2 -1 0 1 2-1Figure 14: 0.5 · f (x)1-2 -1 0 1 2-1Figure 15: f (2x)1-2 -1 0 1 2-1Figure 16: f (0.5x)Page 49 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes9.2 FamiliesTypically you will be given a function that you either know well, for example:1. A simple line: f (x) = x;1-2 -1 0 1 2-1Figure 17: f (x) = x2. A simple parabola: f (x) = x 2 ;321-2 -1 0 1 2Figure 18: f (x) = x 23. A simple square root: f (x) = √ x.3210 1 2 3 4Figure 19: f (x) = √ xCertainly there are other simple curves to know (often called parents) and you’ll startrecognizing them (often called families) as you do more-and-more problems—not unlikelife itself. However, you will often be given graphs that you will not recognize, but thesePage 50 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notesgraphs too will have characteristic points. If you are given a graph you should try todetermine several characteristic points on f that will be easy to shift, and once shiftedyou should be able to connect-the-dots in such a way that the graphs have similarshapes.72645648403224168-4 -3 -2 -1 0 1 2 3-8Figure 20: f (x)During class we’ll identify four key points on this graph that should be clear to all.Then, doing the following examples we will develop a reasonable process to translatethese characteristic points. It should become easy!9.3 ExamplesAgain, you’ll either be given a function that you know well or a graph with easily identifiablecharacteristic points. For simple problems, I want to suggest that the rules (see above) areeasy to do, but more complicated examples will invariable confuse the best students. So,I want to strongly suggest that you try to reason through the translation of points. Don’tpanic! Please try to follow what we’re doing in class.1. Given the following graph. Graph each of the following.72645648403224168-4 -3 -2 -1 0 1 2 3-8Figure 21: f (x)(a) h (x) = f (x − 2) + 1Solution: Work will be done in class.(b) h (x) = f (x − 1) − 2Page 51 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.(c) h (x) = f (x + 3) − 2Solution: Work will be done in class.(d) h (x) = f (x + 5) + 2Solution: Work will be done in class.Page 52 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes2. Given the following graph. Graph each of the following.1-1 0 1 2 3-1Figure 22: f (x)(a) h (x) = −f (x)Solution: Work will be done in class.(b) h (x) = f (−x)Solution: Work will be done in class.(c) h (x) = −f (x) + 3Solution: Work will be done in class.(d) h (x) = f (−x) − 2Solution: Work will be done in class.Page 53 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes3. Given the following graph. Answer each of the following.321-5 -4 -3 -2 -1 0 1 2 3 4-1-2-3-4-5Figure 23: f (x)(a) What are the characteristic points on f (x)?Solution: Work will be done in class.(b) How are the characteristic points on on f (x) connected?Solution: Work will be done in class.(c) Graph h (x) = −2f (0.5x − 1) + 2Solution: Work will be done in class.0Figure 24: h (x) = −2f (0.5x − 1) + 2Page 54 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes(d) Graph h (x) = 3f (3 − x) + 1Solution: Work will be done in class.0Figure 25: h (x) = 3f (3 − x) + 1Page 55 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes9.4 AssignmentYou should read 1.4 and do the WebAssign assignment 01.04.Page 56 of 169

Pre-calculus IMTH-11910 mth.119.01.05Essex County CollegeDivision of MathematicsLMS ProjectClass Notes10.1 Combinations and Compostions of FunctionsSums, Differences, Products, and Quotients of Functions: Let f and g be twofunctions with overlapping domains. Then, for all x common to both domains, the sum,difference, product, and quotient of f and g are defined as follows.Sum: The domain is the intersection of the domains f and g.(f + g) (x) = f (x) + g (x)Difference: The domain is the intersection of the domains f and g.(f − g) (x) = f (x) − g (x)Product: The domain is the intersection of the domains f and g.(f · g) (x) = f (x) · g (x)Quotient: The domain is the intersection of the domains f and g, but we must exclude thex’s in g’s domain where g (x) = 0.( fg)(x) = f (x)g (x) , g (x) ≠ 0The composition of functions is a bit more difficult, and here you’re going to have to workover some pretty vexing concepts. Don’t despair, you’ll get it if you try! The Composition:of the function f with the function g is(f ◦ g) (x) = f (g (x))The domain of f ◦ g is the set of all x in the domain of g such that g (x), i.e. the range ofg, is in the domain of f. That sounds scary, but we’ll see in the examples to follow that itjust requires little patience.10.2 Examples1. Given f (x) = 2x − 3 and g (x) = 3x 2 + 2x − 5, find each of the following.(a) The domain of f (x).Solution: Work will be done in class.(b) The domain of g (x).Page 57 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.(c) f (x) + g (x) and its domain.Solution: Work will be done in class.(d) f (x) − g (x) and its domain.Solution: Work will be done in class.(e) f (x) · g (x) and its domain.Solution: Work will be done in class.(f) f (x) /g (x) and its domain.Solution: Work will be done in class.2. Given f (x) = √ 4 − x 2 and g (x) = √ 1 − x, find each of the following.(a) The domain of f (x).Solution: Work will be done in class.(b) The domain of g (x).Solution: Work will be done in class.(c) f (x) + g (x) and its domain.Solution: Work will be done in class.(d) f (x) − g (x) and its domain.Solution: Work will be done in class.(e) f (x) · g (x) and its domain.Solution: Work will be done in class.(f) f (x) /g (x) and its domain.Solution: Work will be done in class.3. Given f (x) = x 2 − 1 and g (x) = √ 1 − x 2 , find each of the following.(a) The domain of f (x).Solution: Work will be done in class.(b) The domain of g (x).Solution: Work will be done in class.(c) The range of f (x).Page 58 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.(d) The range of g (x).Solution: Work will be done in class.(e) (f ◦ g) (x) and its domain.Solution: Work will be done in class.(f) (g ◦ f) (x) and its domain.Solution: Work will be done in class.4. Given f (x) = 3 and g (x) = x + 1, find each of the following. A graph of f is givenx 2 − 1(following page) to help determine f’s range.654321-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10-1-2-3-4-5-6Figure 26: A partial graph of f.(a) The domain of f (x).Solution: Work will be done in class.(b) The domain of g (x).Solution: Work will be done in class.(c) The range of f (x).Solution: Work will be done in class.(d) The range of g (x).Solution: Work will be done in class.(e) (f ◦ g) (x) and its domain.Solution: Work will be done in class.Page 59 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes(f) (g ◦ f) (x) and its domain.Solution: Work will be done in class.10.3 AssignmentYou should read 1.5 and do the WebAssign assignment 01.05.Page 60 of 169

Pre-calculus IMTH-11911 mth.119.01.06Essex County CollegeDivision of MathematicsLMS ProjectClass Notes11.1 Inverse FunctionsDefinition: A function f is called one-to-one if it never takes on the same value twice; thatisf (x 1 ) ≠ f (x 2 )whenever x 1 ≠ x 2 . We can often visually verified that a function is one-to-one by using thehorizontal line test, which states, “A function is one-to-one if an only if no horizontal lineintersects its graph more than once. For example, the following function is not one-to-one.321-2 -1 0 1 2However, this function is one-to-one.Figure 27: f (x) = x 2√ x 3 + 5321-2 -1 0 1 2Figure 28: f (x) = √ x 3 + 1Some however may be mislead by the horizontal line test, because it is hard to detectfrom the visual alone that f (x) = √ x 3 + 1 is one-to-one. It should be clear though thatf (x) = x 2√ x 3 + 5 is not one-to-one from the graph alone. To show that f (x) = √ x 3 + 1 isone-to-one we proceed as follows. Since for all x 1 , x 2 ≥ −1, and x 1 ≠ x 2 , we have x 3 1 ≠ x 3 2,which further implies that x 3 1 + 1 ≠ x 3 2 + 1, and finally that √ x 3 1 + 1 ≠ √ x 3 2 + 1.Definition: Let f be a one-to-one function with domain A and range B. Then its inversefunction f −1 has a domain B and range A and is defined byf −1 (y) = x ⇔ f (x) = y.Page 61 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesTo find the inverse of a one-to-one function, follow these steps.1. In the equation for f (x), replace f (x) by y.2. Interchange the roles of x and y, and solve for y.3. Replace y by f −1 (x) in the new equation.4. Although not necessary, you may want to verify thatf ( f −1 (x) ) = f −1 (f (x)) = x.Example: Let’s start with a simple example. Given a one-to-one function, f (x) = 2x−1,find its domain, range, and inverse.Work: First off we know f is one-to-one with domain x ∈ R and range f (x) ∈ R, so itsinverse, denoted f −1 (x), has a domain x ∈ R and range f −1 (x) ∈ R. Using simple algebra(method above) to find f −1 (x).f (x) = 2x − 1y = 2x − 1 write y = f (x)x = 2y − 1 interchange x and yx + 1= y solve for y2x + 1= f −1 (x)24321-5 -4 -3 -2 -1 0 1 2 3 4 5-1-2-3-4Figure 29: f (x), and f −1 (x) in blue.You should observe the symmetry along the line y = x.Example: Here’s a more difficult example, however it is still algebraically manageable.Using f (x) = √ x 3 + 1, let’s find its inverse.Work: First off we know f is one-to-one with domain x ≥ −1 and range f (x) ≥ 0, so itsinverse, denoted f −1 (x), has a domain x ≥ 0 and range f −1 (x) ≥ −1. Using simple algebrato find f −1 (x).Page 62 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notesf (x) = √ x 3 + 1y = √ x 3 + 1 write y = f (x)x = √ y 3 + 1 interchange x and yx 2 = y 3 + 1 solve for yx 2 − 1 = y 33√x2 − 1 = y3√x2 − 1 = f −1 (x) write y = f −1 (x)54321-4 -3 -2 -1 0 1 2 3 4 5-1-2Figure 30: f (x), and f −1 (x) in blue.You should observe the symmetry along the line y = x.Example: An even more difficult example, given,f (x) = 1 − √ x1 + √ x .find f −1 and its domain and range. A graph is provided as a visual aid.Page 63 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes321-1 0 1 2 3Figure 31: f (x), and f −1 (x) in blue.Work:f (x) = 1 − √ x1 + √ xy = 1 − √ x1 + √ xx = 1 − √ y1 + √ yx + x √ y = 1 − √ y√ y + x√ y = 1 − x√ y (1 + x) = 1 − x√ 1 − xy =1 + x( 1 − xy =1 + xWe finally have,( ) 2 1 − xf −1 (x) = ,1 + x) 2and its domain is (−1 1] and its range is [0, ∞).11.2 Examples1. Consider the following functions.f (x) = 1 − x 2 , x ≥ 0g (x) = √ 1 − x, x ≤ 1Page 64 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesShow that f and g are inverse functions algebraically and graphically.Solution: Work will be done in class.2. Determine algebraically whether the function is one-to-one. Verify you answer graphically.f (x) = √ x − 4Solution: Work will be done in class.3. Find the inverse function of f algebraically.f (x) = √ 16 − x 2 , x ∈ [−4, 0]Use a graphing utility to graph both f and f −1 in the same viewing window. Describethe relationship between the graphs.Solution: Work will be done in class.4. Find the inverse function of f algebraically.f (x) =1√ x − 1Use a graphing utility to graph both f and f −1 in the same viewing window. Describethe relationship between the graphs.Solution: Work will be done in class.5. Use the functions f (x) = x/8 − 4 and g (x) = x 3 to find the indicated value.(g −1 ◦ f −1) (−4)Solution: Work will be done in class.11.3 AssignmentYou should read 1.6 and do the WebAssign assignment 01.06.Page 65 of 169

Pre-calculus IMTH-11912 mth.119.02.01Essex County CollegeDivision of MathematicsLMS ProjectClass Notes12.1 Polynomial FunctionsThis section covers the basics of polynomial graphs. The graph of some polynomial functionsis already familiar, these polynomials include lines and parabolas. However, higher degreepolynomials are more difficult to graph and will require more time and effort.A polynomial function of degree n ∈ Z + has the formf (x) = a n x n + a n−1 x n−1 + · · · + a 2 x 2 + a 1 x + a 0 ,where a n ≠ 0.12.1.1 Graphing Parabolas21-6 -5 -4 -3 -2 -1 0 1 2 3-1-2-3-4Figure 32: Partial graph of f (x) = y = x 2 + 2x − 3, with important features indicated.You should be able to graph a simple case of the general form of a parabola,y = Ax 2 + Bx + C.I strongly suggest you start by creating a table with simple points, at least six. I do expectthat you are capable to finding the following key features though:• x-intercepts by setting y = 0. They’re not always there, but in general they’re worthfinding. In our example above we would do the following.y = x 2 + 2x − 30 = x 2 + 2x − 30 = (x + 3) (x − 1)So the x-intercepts are: (−3, 0) and (1, 0).Page 66 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes• y-intercept by setting x = 0. That’s just too easy! The y-intercept in the exampleabove is: (0, −3).• The vertex, which is the point (at least in our examples) that will either be the highest(maximum) or lowest (minimum) point on our graph. This point is dead-center of thex values of the x-intercepts. If you use the quadratic formula and take the averageyou’ll get a nice formula for the vertex.(− B (2A , f − B ))2ASo in our example above we have (−1, −4) as the vertex.• The axis-of-symmetry which is the dashed-line in our graph above. It should be notedthat this line is a folding line of symmetry. The equation of this line is x = −1.The examples that follow are really not difficult and are representative of what you’ll see inthe homework and on exams.12.2 Examples1. Find the coordinates of the vertex and the equation of the axis of symmetry for theparabola given by the equation. Then graph the equation.y = x 2 − 4x − 1Solution: Work will be done in class.2. Find the coordinates of the vertex and the equation of the axis of symmetry for theparabola given by the equation. Then graph the equation. 11y = − (x + 1) 2 − 5Solution: Work will be done in class.3. Find two quadratic functions, one that opens upward and one that opens downward,whose graphs have the given x-intercepts.(−2, 0) , (3, 0)11 This form is referred to as standard f (x) = A (Bx + C) 2 + D and can easily be seen to relate to theparent f (x) = x 2 . Yes, we can use what we learned in prior sections, particularly transformation/translation.Page 67 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.4. Find the x and y-intercepts of the graph of the parabola given by the equation.y = x 2 − 5x + 6Solution: Work will be done in class.5. Find the values of b (there’s two) such that the function has the given maximum valueof 52. Graph both cases to verify that the maximum does indeed occur where we expectit to.f (x) = −x 2 + bx − 12Solution: Work will be done in class.6. Find the zeros (x coordinate of the x-intercepts) of f.f (x) = 12 + 7x + x 2Solution: Work will be done in class.7. Find the two positive real numbers whose sum is 86 and whose product is a maximum.Solution: Work will be done in class.12.3 AssignmentYou should read 2.1 and do the WebAssign assignment 02.01.Page 68 of 169

Pre-calculus IMTH-11913 mth.119.02.02Essex County CollegeDivision of MathematicsLMS ProjectClass Notes13.1 Higher Degree Polynomialsin this sections we will use simple factoring (what you did in MTH 100) to find roots ofhigher degree polynomials. By example I hope to intuitively show that a polynomial ofdegree n has at most n real roots, and at most n − 1 relative extrema. We’ll also talk aboutrepeated 12 roots, and how the leading coefficient (the constant multiple of the highest degreeterm) affects the graph. Of particular importance will be how the even and odd degreepolynomials differ. Again, simple factoring and reasoning skills will prove useful.13.2 Examples1. Find the zeros algebraically and use your calculator to graph the function.f (x) = x 5 + 2x 4 − 3x 3Solution: Work will be done in class.2. Find the zeros algebraically and use your calculator to graph the function.f (x) = −5x 5 + 20x 3 − 20xSolution: Work will be done in class.3. Find the zeros algebraically and use your calculator to graph the function.f (x) = 48 − 16x − 3x 2 + x 3Solution: Work will be done in class.4. Find the zeros algebraically and use your calculator to graph the function.f (x) = x 4 − 5x 2 + 412 Referred to as multiplicity.Page 69 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.5. Find the zeros algebraically and use your calculator to graph the function.f (x) = x 5 − 5x 3 + 4xSolution: Work will be done in class.6. Find the zeros algebraically and use your calculator to graph the function.f (x) = x 4 − 4x 2 + 1Solution: Work will be done in class.7. Use your calculator to find all roots and relative extrema off (x) = 4x 5 − 3x 2 − x + 1.Solution: Work will be done in class.8. Find a polynomial of degree 4 that has roots 0, 1, 2.Solution: Work will be done in class.9. Find a polynomial of any degree that has roots 7, 5, and 1 ± √ 3.Solution: Work will be done in class.10. Find a polynomial of degree 4 with leading coefficient −2 that has the given zeros withmultiplicities: zero is 1 with multiplicity two; zero is 7, and zero is −1.Solution: Work will be done in class.11. Intermediate Value Theorem: Karl Weierstrass was a German mathematician whoproved the Intermediate Value Theorem (IVT), which was not an easy task. Some havesaid that he proved the obvious, but it nonetheless had to be proved before it couldbecome a theorem. In MTH-119 we will use this theorem to show the existence of roots.Here’s what the theorem states:Page 70 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSuppose f is continuous on the closed interval [a, b] and W is any numberbetween f (a) and f (b), where f (a) ≠ f (b). Then, there is a number c ∈(a, b) for which f (c) = W .Now consider the following function.f (x) = x 4 − 13x 3 + 17x 2 − 3Use the Intermediate Value Theorem and a graphing utility to find graphically anyintervals of length 1 in which the polynomial function is guaranteed to have at least onezero. Can you find an interval that contains at most two roots? Three roots? Fourroots?Solution: Work will be done in class.13.3 AssignmentYou should read 2.2 and do the WebAssign assignment 02.02.Page 71 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes14 mth.119.02.03–mth.119.02.0514.1 Overview of Roots14.1.1 Introduction to Complex NumbersFact is, you’ve been told in MTH-100 that not all numbers are real and that some areimaginary. You may have further been told that they were invented and don’t really exists.That’s hogwash—they not only exists, but were a necessary consequence of solving a certainclass of problems—that is, they were discovered and not invented. I won’t dwell on this, andwe’ll stick to some very basic concepts.The unit imaginary is denoted by i, wherei = √ −1.A complex number is written as a linear combination of a real and imaginary part, and isusually denoted by the letter z, wherez = a + bi.Both a and b are real numbers, and i is just the imaginary unit. You should note thatif i = √ −1 then i 2 = −1. That’s about it! However, since i is really a root you need tofollow the same conventions that were used in radical simplifications. And you should alwayssimplify your radicals first, that is, if you’re asked to multiple √ −8 by the √ −25, you’d dothis:√ √ √−25 · −8 = 5i · 2i 2= 10i 2√ 2= −10 √ 214.1.2 Operations on Complex NumbersAddition/Subtraction: just as you did with any algebraic expression. For example ifyou’re asked to add 3 + 2i to 7 − 3i you’ll get 10 − i. And if you’re asked to subtract2 − 7i from 5 − 3i you’ll get 3 + 4i. Really simple!Multiplication: again, just as you did with any algebraic expression. However, if you geti 2 in your multiplications you need to rewrite this as −1. Here goes:(2 − 3i) (3 + 2i) = 6 + 4i − 9i − 6i 2= 6 − 5i + 6= 12 − 5iPage 72 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesDivision: just as you did with radical divisions! For example:13 + 2i = 13 + 2i · 3 − 2i3 − 2i= 3 − 2i9 − 4i 2= 3 − 2i13= 3 13 − 2 13 iI won’t bore you with a bunch of examples, and I think it is best to review this on yourown. Yes, you’ll have a simple homework set to complete and you will need to get help ifyou do not understand this material.14.1.3 Fundamental Theorem of AlgebraEvery polynomial equation having complex coefficients and degree at least one has at leastone complex root. This theorem was first proven by Gauss. It is equivalent to the statementthat a polynomial P (z) of degree n has n values z i (some of them possibly degenerate) forwhich P (z i ) = 0. Such values are called polynomial roots. An example of a polynomial witha single root of multiplicity greater than one is z 2 − 2z + 1, which has as a root z = 1 ofmultiplicity 2.The essential crux os this theorem is that a polynomial of degree n has n-roots. However,it does not say how to get them. We’ll use all kinds of tricks to find the roots. Sometimesthey’ll be simple rational numbers, and other times they will be irrational, but they canalways be written as complex a + bi. One trick 13 you’ll learn is that complex roots occur inconjugate pairs if the polynomials have real coefficients. For example if 6 − 3i is a root of apolynomial, then so is 6 + 3i. This is useful. For example suppose we are asked to factorx 4 − 2x 3 − 3x 2 + 10x − 10,and you’re given that x = 1 − i is a root. Here’s what I’d do . . .1. If x = 1−i is a root then so is x = 1+i, this certainly means that (x − 1 + i) (x − 1 + i)is a factor of x 4 − 2x 3 − 3x 2 + 10x − 10.2. Expand and simplify (x − 1 + i) (x − 1 + i) and you’ll get x 2 − 2x + 2.3. Divide x 4 − 2x 3 − 3x 2 + 10x − 10 by x 2 − 2x + 2 and you’ll get x 2 − 5.4. Factor x 2 − 5 and you’ll get ( x − √ 5 ) ( x + √ 5 ) .5. Finally the factorization of x 4 − 2x 3 − 3x 2 + 10x − 10 is((x − 1 + i) (x − 1 + i) x − √ ) (5 x + √ )5 .13 In mathematics, the complex conjugate root theorem states that if P is a polynomial in one variablewith real coefficients, and a + bi is a root of P with a and b real numbers, then its complex conjugate a − biis also a root of P .Page 73 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes14.1.4 Rational Root TheoremIt will be necessary to find key points on a polynomial graph, including the y-intercept, whichcan easily be found by setting x = 0. On the other hand the x-intercept(s) are typicallymore difficult to find, but essentially involve setting f (x) = 0 and then factoring. Factoringpolynomials in general is very difficult, if not impossible, however the examples I will givewill move forward smoothly. And once again, you should also be aware that x-intercepts areoften referred to as roots or zeros of the polynomial.The approach I will take in factoring polynomials will typically involve the Rational RootTheorem, which states, that if a polynomialf (x) = a n x n + a n−1 x n−1 + · · · + a 2 x 2 + a 1 x + a 0has integer coefficients, every rational zero of f has the form± factors of a 0factors of a n.This is identical to the method that was used to factor quadratics and was often referred toas trial and error. Let’s proceed with an example.1. Given f (x) = 2x 3 + 3x 2 − 8x + 3, find:(a) candidate rational roots;Solution: Work will be done in class.(b) test these roots to see which ones are roots;Solution: Work will be done in class.(c) factor f (x) completely.Solution: Work will be done in class.14.1.5 Not All Roots are RationalSometimes we will be presented with a polynomial function that has no rational roots. Ifthe roots are real, but not rational, we say they are irrational. Irrational numbers are simplyreal numbers that cannot be written as a ratio of integers. For example:√2,√11,3 √ 7, . . .are just a few examples. In fact there’s an infinite number of irrational numbers. Again,let’s proceed with an example.1. Given f (x) = 10x 3 − 15x 2 − 16x + 12, find:(a) candidate rational roots;Page 74 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Work will be done in class.(b) test these roots to see which ones are roots;Solution: Work will be done in class.(c) partially factor f (x);Solution: Work will be done in class.(d) factor the remaining quadratic factor using the quadratic formula;Solution: Work will be done in class.(e) factor f.Solution: Work will be done in class.Rational roots are the simplest to deal with and you should look for them first! Typicallythe irrational roots are found later.14.1.6 Not All Roots are RealSometimes we will be presented with a polynomial function that has non-real roots. Herewe say their roots are non-real, or complex. From MTH-100 you should recall the followingtwo facts:i = √ −1 and i 2 = −1.Here’s an example that will hopefully remind you of what you did in MTH-100.1. Given f (x) = x 2 − 2x + 10, find:(a) candidate rational roots;Solution: Work will be done in class.(b) test these roots to see which ones are roots;Solution: Work will be done in class.(c) use the quadratic formula to find the roots of f;Solution: Work will be done in class.(d) factor f.Solution: Work will be done in class.Page 75 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes14.2 Examples1. Perform the operation and write the result in standard form.4 + ii− 829 − iSolution: This answer will be discussed in class.4 + ii− 829 − i= −8 − 5i2. Solve the quadratic equation.x 2 + 2x + 9 = 0Solution: This answer will be discussed in class.x = −1 ± 2i √ 23. Simplify the complex number 9i 2 − 2i 3 and write it in standard form.Solution: This answer will be discussed in class.−9 + 2i4. Simplify the complex number 3i 60 − 2i 150 − 5i 82 + 6i 109 − 2i 61 and write it in standardform.Solution: This answer will be discussed in class.10 + 4i5. Three of the zeros of a fourth-degree polynomial function f are 7, 6, and 2 − 7i. Whatis the other zero of f?Page 76 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: This answer will be discussed in class.2 + 7i6. Find all zeros of f (z) = 625z 4 − 81.Solution: This answer will be discussed in class.± 3 5 , ±3 5 i7. Find and expand a third degree polynomial function f (x) with real coefficients that hasroots x = 2 and x = −3i.Solution: This answer will be discussed in class.(x − 2) (x + 3i) (x − 3i) = x 3 − 2x 2 + 9x − 188. Verify that x = 8i is a root of f (x) = x 3 + x 2 + 64x + 64 and then factor this functioncompletely.Solution: This answer will be discussed in class.f (8i) = 0f (x) = (x − 8i) (x + 8i) (x + 1)9. Find and expand a fourth degree polynomial function f (x) with real coefficients thathas roots x = ± √ 2 and x = 1 ± i.Solution: This answer will be discussed in class.(x − √ ) (2 x + √ )2 (x − 1 − i) (x − 1 + i) = x 4 − 2x 3 + 4x − 4Page 77 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes10. Givenf (x) = 6x 5 − 17x 4 + 28x 3 − 8x 2 − 24x,and its graph. Use this information to answer the following questions.10-1 0 1 2-10Figure 33: Partial graph of f (x) = 6x 5 − 17x 4 + 28x 3 − 8x 2 − 24x.(a) What is the y-intercept?Solution: This work will be discussed in class.Just set x = 0 and you’ll get (0, 0).(b) Using the Rational Root Theorem, what are the candidate rational roots?Solution: This work will be discussed in class.First you will need to factor out the GCF.f (x) = x ( 6x 4 − 17x 3 + 28x 2 − 8x − 24 )Then you’ll need to write the ratios out.1, 2, 3, 4, 6, 8, 12, 24±1, 2, 3, 6Then you’ll need to figure out the set.{±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24, ± 1 }2 , ±3 2 , ±1 3 , ±2 3 , ±4 3 , ±8 3 , ±1 6Page 78 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes(c) Using the graph, and your answers from (b), what are reasonable candidates forrational roots. Test them to see if they actually are.Solution: This work will be discussed in class.You should be able to roughly determine the values of the two real roots by lookingat the graph, and 3/2 and −2/3 are good candidates if they’re rational. Testingthem shows that all are.f (3/2) = f (−2/3) = 0(d) Using (c), factor f (x) completely.Solution: This work will be discussed in class.Now we know that 2x − 3, and 3x + 2 are factors of f. Just use long division tofind the missing quadratic factor. We will certainly discuss the factorization of(x 2 − 2x + 4) into ( x − 1 + i √ 3 ) ( x − 1 − i √ 3 ) —it’ really pretty simple!f (x) = x ( 6x 4 − 17x 3 + 28x 2 − 8x − 24 )= x (2x − 3) (3x + 2) ( x 2 − 2x + 4 )(= x (2x − 3) (3x + 2) x − 1 + i √ ) (3 x − 1 − i √ )3(e) List all roots, real and imaginary.Solution: This work will be discussed in class.Again, as in the above factorization you’ll need to use the quadratic formula tosolve x 2 − 2x + 4 = 0.{0, −2/3, 3/2, 1 − i √ 3, 1 + i √ }3Page 79 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes11. Givenf (x) = 36 + 18x − 28x 2 − 13x 3 + 5x 4 + 2x 5 ,and its graph. Use this information to answer the following questions.5045403530252015105-3 -2 -1 0 1 2 3-5Figure 34: Partial graph of f (x) = 36 + 18x − 28x 2 − 13x 3 + 5x 4 + 2x 5 .(a) What is the y-intercept?Solution: This work will be discussed in class.Just set x = 0 and you’ll get (0, 36).(b) Using the Rational Root Theorem, what are the candidate rational roots?Solution: This work will be discussed in class.Writing the ratios out.1, 2, 3, 4, 6, 9, 18, 36±1, 2Then you’ll need to figure out the set.{±1, ±2, ±3, ±4, ±6, ±9, ±18, ±36, ± 1 }2 , ±3 2 , ±9 2(c) Using the graph, and your answers from (b), what are reasonable candidates forrational roots. Test them to see if they actually are.Solution: This work will be discussed in class.You should be able to roughly determine the values of the candidate rational rootsby looking at the graph, and −3, 2, −3/2 and 3/2 are good candidates if they’rerational. Testing them shows that only three of these are.f (−3) = f (2) = f (−3/2) = 0Page 80 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes(d) Using (c), factor f (x) completely.Solution: This work will be discussed in class.Now we know that x + 3, 2x + 3, x − 2 are factors of f. Just use long division tofind the missing quadratic factor.f (x) = 2x 5 + 5x 4 − 13x 3 − 28x 2 + 18x + 36= (x + 3) (2x + 3) (x − 2) ( x 2 − 2 )(= (x + 3) (2x + 3) (x − 2) x + √ ) (2 x − √ )2(e) List all roots, real and imaginary.Solution: This work will be discussed in class.{2, −3, −3/2, √ 2, − √ }2Here’s a close-up of the region between −2 and −1. You should notice the two roots,what are they? 14 You should also look back at the original graph.-2 -1Figure 35: Partial graph of f (x) = 36 + 18x − 28x 2 − 13x 3 + 5x 4 + 2x 5 .14.3 AssignmentYou should read 2.3–2.5 and do the WebAssign assignment 02.03–05. Yes, it is three sections,and some of it should look a lot like the material that was covered in MTH-100. However,please feel free to ask questions if you have any concerns about the material presented hereor in your textbook.14 One is − √ 2 and the other is −3/2. They’re very close together and actually appear as one in the originalgraph. Don’t be deceived by appearances.Page 81 of 169

Pre-calculus IMTH-11915 mth.119.02.06Essex County CollegeDivision of MathematicsLMS ProjectClass Notes15.1 Rational FunctionsHere we will be dealing with a ratio of polynomials and their graphs. We need to be especiallycareful that we do not allow divisions by zero—but this is generally near where we want to be,mainly because we love danger and there’s nothing in mathematics that’s more dangerousthan dividing by zero. We’re also going to see behavior in these graphs that appears linear,that is, somewhere along these curves it will look nearly, if not exactly, like a line. The ideathat a curve may come arbitrarily close to a line without actually becoming the same iscalled asymptotic behavior. The details will be explained by examples, and I do encourageeveryone to use their graphic calculators to help aid their desire to understand the differencebetween something that looks linear and something that is linear. Perceptibly there maybe no difference between them, but conceptually there is. That kind of differentiation takesconsiderable time to discern. Let’s start with an example . . .Givenf (x) = 2x2 + 7x − 4x 2 + x − 2and its partial graph.=(2x − 1) (x + 4)(x − 1) (x + 2) ,15105-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6-5-10-15Figure 36: Partial graph of f (x) with asymptotes indicated in red.Use this information to answer the following questions.1. What is the y-intercept?Page 82 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: This answer will be discussed in class.Just set x = 0 and you’ll get (0, 2).2. What are the x-intercepts?Solution: This answer will be discussed in class.Just set f (x) = y = 0 and you’ll get (−4, 0) and (1/2, 0).3. What is the equation of the horizontal asymptote?Solution: This answer will be discussed in class.By inspection, we get y = 2.4. What are the equations of the two vertical asymptotes?Solution: This answer will be discussed in class.By inspection, we get x = −2 and x = 1.5. What is the domain of f (x)?Solution: This answer will be discussed in class.By inspection, we get all real numbers except −2 and 1.6. What is the range of f (x)?Solution: This answer will be discussed in class.By inspection, we get all real numbers, this is often denoted R.Answering questions about rational functions is often aided by having a good graph.However, you should not rely solely on graphs and you should try to develop a skill set thatdoes not heavily rely on using a graphic calculator.Now try this one. 15f (x) = y = x2 − 3x − 42x 2 + 4x=(x + 1) (x − 4)2x (x + 2)1. What is/are the y-intercept(s)?15 No graph is provided and now is a good time to see if you can graph this on your graphic calculator.Page 83 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: This answer will be discussed in class.Just set x = 0 and you’ll see the dreaded division by zero, so there is none.2. What is/are the x-intercept(s)?Solution: This answer will be discussed in class.Just set y = 0 and you’ll get (−1, 0) and (4, 0).3. What is the domain of f (x)?Solution: This answer will be discussed in class.All real numbers except 0 and −2.4. What is/are the equation(s) of the horizontal asymptote(s)?Solution: This answer will be discussed in class.We will need to look at two limits to determine this.andx 2 − 3x − 4limx→∞ 2x 2 + 4x= lim 1 − 3/x − 4/x 2x→∞ 2 + 4/xx 2 − 3x − 4limx→−∞ 2x 2 + 4x = lim 1 − 3/x − 4/x 2x→−∞ 2 + 4/xSo the horizontal asymptote is y = 1/2.= 1/2,= 1/2.5. What is/are the equation(s) of the vertical asymptote(s)?Solution: This answer will be discussed in class.You’ll need to look for divisions by zero, and then look at limits to see what happens.We will discuss this in class.x 2 − 3x − 4limx→0 − 2x 2 + 4xx 2 − 3x − 4limx→0 + 2x 2 + 4xx 2 − 3x − 4limx→−2 − 2x 2 + 4xx 2 − 3x − 4limx→−2 + 2x 2 + 4x= ∞= −∞= ∞= −∞Page 84 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSo the vertical asymptotes occur at x = 0 and x = −2.6. Graph a precalculus sketch of f (x)?Solution: This answer will be discussed in class.Simple sign analysis along with all the information above produces the followinggraph. Again we will discuss this in class. It’s really not difficult and you’ll haveplenty of practice, just do the homework.15105-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6-5-10-15Figure 37: Partial graph of f (x) with asymptotes indicated in red.7. What is the range of f (x)?Solution: This answer will be discussed in class.You’ll need the graph, and then it should be clear that the range is R.15.2 Examples1. Givenf (x) =2(x − 1) 3 ,find all asymptotes, key points, and graph.Page 85 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: This will be done in class.2. Givenf (x) = x2 − 9xx 2 − 9 ,find all asymptotes, key points, and graph.Solution: This will be done in class.3. Givenf (x) = x2 + 2x + 13x 2 − x − 4 ,find all asymptotes, key points, and graph.Solution: This will be done in class.4. Givenf (x) = x3 − 8x 2 + 7 ,find all asymptotes, key points, and graph.Solution: This will be done in class.5. Find the real zeros of the rational function.f (x) = 1 + 13x 2 + 1Solution: This will be done in class.15.3 AssignmentYou should read 2.6 and do the WebAssign assignment 02.06.Page 86 of 169

Pre-calculus IMTH-11916 mth.119.02.07Essex County CollegeDivision of MathematicsLMS ProjectClass Notes16.1 Rationals Continued . . .Here we’ll be introducing a third type of linear asymptote called a slant or oblique asymptote.They occur when the degree of the numerator is one more than the degree of the denominator.Let’s start with a visual example.Givenf (x) = x2 − 4x − 5x − 3and its graph.=(x + 1) (x − 5)x − 3= x − 1 − 8x − 3 ,105-15 -10 -5 0 5 10 15-5-10Figure 38: Partial graph of f (x) with asymptotes indicated in red.Use this information to answer the following questions.1. What is the y-intercept?Solution: This answer will be discussed in class.Just set x = 0 and you’ll get (0, 5/3).2. What are the x-intercepts?Page 87 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: This answer will be discussed in class.Just set y = 0 and you’ll get (−1, 0) and (5, 0).3. What is the equation of the vertical asymptote?Solution: This answer will be discussed in class.By inspection, we get x = 3.4. What is the equation of the slant (oblique) asymptote?Solution: This answer will be discussed in class.By inspection, we get y = x − 1.5. What is the domain of f (x)?Solution: This answer will be discussed in class.By inspection, we get R, x ≠ 3.6. What is the range of f (x)?Solution: This answer will be discussed in class.By inspection, we get R.Again, you should be able to do problems without a graph, but learning how requirespractice and that may involve using a graphic calculator to help out.Page 88 of 169

Pre-calculus IMTH-119Now try this one.f (x) = y = x2 + 2xx − 1Essex County CollegeDivision of MathematicsLMS ProjectClass Notes1. First rewrite the function as it was done in the first example.Solution: This answer will be discussed in class.You will need to do both a long division and a factorization.f (x) = y = x2 + 2xx − 1=x (x + 2)x − 1= x + 3 + 3x − 12. What is the y-intercept?Solution: This answer will be discussed in class.Just set x = 0 and you’ll get (0, 0).3. What are the x-intercepts?Solution: This answer will be discussed in class.Just set y = 0 and you’ll get (−2, 0) and (0, 0).4. What is the equation of the vertical asymptote?Solution: This answer will be discussed in class.You will need to take a look a limits. Limits are not easy to understand, and we willdiscuss several approaches in class.x 2 + 2xlimx→1 − x − 1x 2 + 2xlimx→1 + x − 1= −∞= ∞So the vertical asymptote is x = 1.5. What is the equation of the slant (oblique) asymptote?Page 89 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: This answer will be discussed in class.When x → ±∞ the function is asymptotic with y = x + 3.6. Use the information above to make a sketch of f (x).105-15 -10 -5 0 5 10 15-5-10Figure 39: Partial graph of f (x).Solution: This answer will be discussed in class.Your graph should roughly look like mine!105-15 -10 -5 0 5 10 15-5-10Figure 40: Partial graph of f (x).7. What is the domain of f (x)?Page 90 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: This answer will be discussed in class.R, x ≠ 1.8. What is the range of f (x)?Solution: This answer will be discussed in class.By inspection, we clearly see that it is not all real numbers, however we cannotdetermine the range with the math we know now. 1616.2 Examples1. Use long division to find the slant asymptote.f (x) = x3 − 2x 2 + 3x − 5x 2 + x − 1Solution: This answer will be discussed in class.y = x − 32. Graphy = x2 − 9x + 3 ,and explain what happens at x = −3Solution: I will do this is class.3. Graphf (x) = x2 − 2x − 1 − 2,and indicate what you have learned so far about the graphs of rational functions.Page 91 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: I will do this is class.4. Graphf (x) = 3x4 − 5x + 3x 4 + 2and indicate what you have learned so far about the graphs of rational functions.Solution: I will do this is class.5. Using a graphic calculator, graphy =x√x2 + 1 .You should try to make sense out of this graph by looking at key points such as x andy-intercepts, sign-analysis, limits, and domain. From this graph try to determine therange.Solution: I will do this is class.16.3 AssignmentYou should read 2.7 and do the WebAssign assignment 02.07.Page 92 of 169

Pre-calculus IMTH-11917 mth.119.03.01Essex County CollegeDivision of MathematicsLMS ProjectClass Notes17.1 Exponential Functions and Their Graphs17.1.1 ExponentialsThe function f (x) = a x , where x ∈ R, a > 0 and a ≠ 1, is called the exponential function,with base a. Since a is a positive real number we have a variety of possible graphs. Hereare some examples. You should be able to graph a variety of exponential functions, and I987654321-2 -1 0 1 2 3Figure 41: y = 2 x , y = 5 x , y = 20 xcan only suggest that creating a table of values can often lead to a pattern that is easy topredict. Be sure to select easy values. Here’s an example to try.f (x) = −2 1−xBe sure to create a table of values first, then plot the points and try to connect-the-dots.21-2 -1 0 1 2-1-2-3-4-5-6-7Figure 42: f (x) = −2 1−xAgain, a > 0 and you’ll see a variety of numbers used. Of special note is the use of theletter e which represents Euler’s number. Everyone should be able to locate this symbol onPage 93 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notesyour calculator. It is often referred to as a natural base and we’ll see this quite often inmathematics.As we do more graphs you’ll see that some increase as x increases, and some decrease as xincreases. We’ll do some application problems too, typically the growth of money (compoundand continuous interest) and you’ll see some problems related to decay (radioactive) in yourtextbook. What’s important to remember is that these problems are exponential.17.2 Examples1. What type of transformation of the graph of f (x) = 5 x is the graph of f (x + 1)?Solution: We’ll do this is class.The graph of f (x + 1) is a horizontal shift one unit to the left of f (x) = 5 x .2. Graph the exponential function by hand. Identify any asymptotes and intercepts anddetermine whether the graph of the function is increasing or decreasing.( x 2f (x) =3)Solution: We’ll do this is class.This is a decreasing function. The horizontal asymptote is y = 0. The y-intercept is 1.321-4 -3 -2 -1 0 1 2 3 4 5-1( x 2Figure 43: f (x) =3)3. Graph the exponential function by hand. Identify any asymptotes and intercepts anddetermine whether the graph of the function is increasing or decreasing.f (x) = 3 1−x2 Page 94 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll do this is class.Your graph should look like mine.321-3 -2 -1 0 1 2 3Figure 44: f (x) = 3 1−x24. Graph the exponential function by hand. Identify any asymptotes and intercepts anddetermine whether the graph of the function is increasing or decreasing.( x−1 3f (x) =2)Solution: We’ll do this is class.Your graph should look like mine.4321-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4-1Figure 45: f (x) = ( )3 x−125. Graph the exponential function by hand. Identify any asymptotes and intercepts anddetermine whether the graph of the function is increasing or decreasing.f (x) = 2 x − 3Page 95 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll do this is class.Your graph should look like mine.21-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5-1-2-3-46. The formula A = Pinterest?Figure 46: f (x) = 2 x − 3(1 + r n) ntgives the balance A of an account earning what type ofSolution: We’ll do this is class.(The formula A = P 1 +n) r ntgives the balance A of an account earning interestcompounded discretely where n indicates the number of times per year that the interestis computed. When interest is computed daily, use n = 365.7. The formula A = P e rt gives the balance A of an account earning what type of interest?Solution: We’ll do this is class.The formula A = P e rt gives the balance A of an account earning interest compoundedcontinuously.8. Use a calculator to evaluate the function at the indicated value of x. Round your resultto four decimal places.f (x) = 2.3 x , x = 2 3Solution: We’ll do this is class.( 2f = 2.33)2/3 ≈ 1.7424Page 96 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes9. Show that the value of f (x) approaches the value of g (x) as x increases without boundgraphically and numerically.(f (x) = 1 + 3 ) x, g (x) = e 3xSolution: We’ll do this is class.You need to graph both f and g on the same coordinate axis and verify as x increasesthe two graphs appear to merge. You also need to create a table of values for both fand g for large values of x and verify the difference between f and g tend to zero as xincreases.20151050 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150-5Figure 47: f (x) and g (x)10. Complete the balance A for $1,100 invested at rate r = 3% for t = 41 years andcompounded n = 1, 2, 12, 365 times per year.(A = P 1 + r ) ntnSolution: We’ll do this is class.(A = 1100 1 + 0.03 ) 1·41≈ $3695.891(A = 1100 1 + 0.03 ) 2·41≈ $3729.142(A = 1100 1 + 0.03 ) 12·41≈ $3757.5812(A = 1100 1 + 0.03 ) 365·41≈ $3763.1‘6365Page 97 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes11. Use a graphing utility to graph the function and find any asymptotes numerically bycreating a table of values for the function.f (x) =31 + e −0.5/xSolution: We’ll do this is class.Your graph should look like mine.321-2 -1 0 1 2Figure 48: f (x) =31 + e −0.5/x12. You build an annuity by investing P dollars every month at interest rate r, compoundedmonthly. Find the amount A accrued after n months using the formula[ (1 + r/12) n ]− 1A = Pr/12where r is in decimal form. P = $119, r = 0.09, n = 52 months.Solution: We’ll do this is class.7,534.0213. Choose the correct symbol (< or>) between the pair of numbers.8 3/4 ?( 34) 8Solution: We’ll do this is class.( 38 3/4 >4) 8Page 98 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes17.3 AssignmentYou should read 3.1 and do the WebAssign assignment 03.01.Page 99 of 169

Pre-calculus IMTH-11918 mth.119.03.02Essex County CollegeDivision of MathematicsLMS ProjectClass Notes18.1 Logarithmic Functions and Their Graphs18.1.1 InversesHere’s the graph y = 2 x and its inverse. You may want to review inverses at this point. The54321-6 -5 -4 -3 -2 -1 0 1 2 3 4 5-1-2-3Figure 49: y = 2 x and x = 2 yfunction, x = 2 y , is the inverse if y = 2 x . However, since f (x) = 2 x , and we know it has aninverse, we need to know what f −1 (x) is. This inverse function is written asf −1 (x) = log 2 x.So we have this basic relationship to consider.y = log a x ⇒ a y = xAs before, a > 0 and a ≠ 1, and y ∈ R, so x > 0.18.1.2 Special BasesLogarithms, base 10, are common, and were once used extensively to simplify computation.In fact the discovery of logarithms was a substantial event in speeding up computation,similar to the invention of the digital computer. In fact, once people learned to use logarithmsthey were able to rapidly compute, but now we use computers to rapidly compute. However,logarithms are still used, just not for computation. The other special base is Euler’s number,represented by the letter e. Logarithms with this base are called natural. Certainly e is astrange enough number and can be written as a limit.e = lim(1 + 1 ) x≈ 2.718281828x→∞ xPage 100 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes210 5 10 15 20Figure 50: y = ( 1 + 1 x) xand y = eHere’s the graph.In the textbook we are using, and on your calculators, the common logarithms arewritten aslog x,and the natural are written asln x.This, I believe, is due to the French (Mercator, although German, may have coined the termin France), although European logarithms originated in Switzerland and Scotland.18.1.3 Graphing Logarithmic FunctionsAs you may recall, we graphed both y = 2 x and its inverse on the same axis.And we now know (or should be familiar with) that x = 2 y can be rewritten as y = log 2 x.You should also recall that when dealing with logarithmic functions, a basic relationshipshould always be consider. That basic relationship is:y = log a x ⇔ a y = xAs before, a > 0 and a ≠ 1, and y ∈ R, so x > 0.Okay, let’s graph a simple logarithmic function.y = f (x) = log 3 (x − 2) + 1First, I strongly suggest that you rewrite the logarithmic function as an exponential, or atleast until you become more experienced with logarithmic functions. This typically bestPage 101 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes54321-6 -5 -4 -3 -2 -1 0 1 2 3 4 5-1-2-3Figure 51: Partial graph of y = 2 x and x = 2 y .done by solving for x in simple steps, as follows.y = log 3 (x − 2) + 1y − 1 = log 3 (x − 2)3 y−1 = x − 23 y−1 + 2 = xNow, pick simple values for y and evaluate to find x. For example, if y = 1 we have:3 1−1 + 2 = 3 0 + 2 = 1 + 2 = 3 = x.Continue to pick easy points and you’ll eventually see a pattern. And the pattern shouldlook like this graph.210 1 2 3 4-1Figure 52: Partial graph of y = log 3 (x − 2) + 1.18.2 Examples1. What exponential equation is equivalent to the logarithmic equation log a b = c?Page 102 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss in class.a c = b2. Use the definition of logarithmic function to evaluate log 2 8.Solution: We’ll discuss this in class.log 2 8 = 33. Use the definition of logarithmic function to evaluate log 10 100.Solution: We’ll discuss this in class.log 10 100 = 24. Use the definition of logarithmic function to evaluate log 9 3.Solution: We’ll discuss this in class.log 9 3 = 1 25. Use the definition of logarithmic function to evaluate log 6 1.Solution: We’ll discuss this in class.log 6 1 = 0Page 103 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes6. Use the definition of logarithmic function to evaluate the function at the indicated valueof x without using a calculator.f (x) = log 36 x, x = 1 6Solution: We’ll discuss this in class.( 1f6)= log 3616 = −1 27. Graphy = f (x) = 1 − 1 2 log 2 (1 − x) ,being sure to indicate key features including asymptote(s).Solution: This will be discussed in class.Your graph should look like mine.321-3 -2 -1 0 1Figure 53: Partial graph of y = 1 − 1 2 log 2 (1 − x).8. Evaluate each of the following.(a) ln ePage 104 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss this is class.ln e = 1(b) ln e 2Solution: We’ll discuss this is class.ln e 2 = 2(c) ln e 3Solution: We’ll discuss this is class.ln e 3 = 3(d) ln e nSolution: We’ll discuss this is class.ln e n = n(e) log 10Solution: We’ll discuss this is class.log 10 = 1(f) log 10 2Solution: We’ll discuss this is class.log 10 2 = 2(g) log 10 3 Page 105 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss this is class.log 10 3 = 3(h) log 10 nSolution: We’ll discuss this is class.log 10 n = n(i) e ln 2Solution: We’ll discuss this is class.e ln 2 = 29. The modelt = 16.625 lnxx − 750 , x > 750approximates the length of a home mortgage of $150, 000 at 6% in terms of the monthlypayment. In the model, t is the length of the mortgage in years and x is the monthlypayment in dollars.(a) Use the model to approximate the lengths of a $150, 000 mortgage at 6% when themonthly payment is $897.72 and when the monthly payment is $1, 659.21. (Roundyour answers to the nearest whole number.)Solution: We’ll discuss this in class.897.72t = 16.625 ln897.72 − 750 = 301659.21t = 16.625 ln1659.21 − 750 = 10(b) Approximate the total amounts paid over the term of the mortgage with a monthlypayment of $897.72 and with a monthly payment of $1,659.21. (Round your answersto nearest dollar.)Solution: We’ll discuss this in class.Page 106 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesT 30 = 30 · 12 · 897.72 = 323179T 10 = 10 · 12 · 1659.21 = 199105(c) What amount of the total is interest costs for each payment? (Round your answersto nearest dollar.)Solution: We’ll discuss this in class.I 30 = 323179 − 150000 = 173179I 10 = 199105 − 150000 = 4910510. Find the domain, x-intercept, and vertical asymptote of the logarithmic function andsketch its graph.f (x) = ln (x + 3)Solution: We’ll discuss in class.Domain is x > −3; x-intercept is −2; vertical asymptote is x = −3. Your graph shouldlook like mine.321-5 -4 -3 -2 -1 0 1 2 3 4 5-1-2-3Figure 54: Partial graph of f (x) = ln (x + 3).11. Use a graphing utility to find the domain, range , x and y-intercept, any asymptotes ofthe logarithmic function and sketch its graph.y = 12 ln (x2 + 1)x − 2Page 107 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss in class.Domain is all reals x ≠ 2; x and y-intercept are both 0; vertical asymptote is x = 2.Your graph should look like mine.252015105-35 -30 -25 -20 -15 -10 -5 0 5 10 15 20 25 30 35 40-5-10-15-20-25Figure 55: Partial graph of y = 12 ln (x2 + 1).x − 218.3 AssignmentYou should read 3.2 and do the WebAssign assignment 03.02.Page 108 of 169

Pre-calculus IMTH-11919 mth.119.03.03Essex County CollegeDivision of MathematicsLMS ProjectClass Notes19.1 Properties of Logarithms19.1.1 Basic RulesProduct Rule: For any positive numbers M and N, and any logarithmic base a,log a (M · N) = log a M + log a N.Proof: Letlog a M = x and log a N = y.Now rewrite these as exponentials.M = a x and N = a y .We also know thatM · N = a x · a y = a x+yConverting this back to logarithmM · N = a x+y ⇒ log a (M · N) = x + ySo we have, after looking back,log a (M · N) = log a M + log a N.Q.E.D.This product rule can easily be extended to include powers. The Power Rule states,that for any positive numbers M, and any logarithmic base a, and any real number p,log a M p = p · log a .Now using the power and product rules, we can include subtraction. The SubtractionRule states, that for any positive numbers M and N, and any logarithmic base a,log aMN = log a M − log a N.19.1.2 Changing BasesYou will have to evaluate logs for a variety of bases, and on occasion you will need to useyour calculator to carry out the computation. However your calculator most likely only hasPage 109 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notesnatural (base e) and common logs (base 10), so what should you do if you’re given anotherbase? Here goes . . .log a x = yx = a yln x = ln a yln x = y ln ay = ln xln aThis leads us to a simple change of base formula.log a x = ln xln a = log xlog a19.2 Examples1. Evaluate the logarithm using the change-of-base formula. Round your result to threedecimal places.log 11 4Solution: We’ll discuss this in class.log 11 4 = ln 4ln 11 ≈ 0.5782. Evaluate the logarithm using the change-of-base formula. Round your result to threedecimal places.log 637Solution: We’ll discuss this in class.log 637=ln 3/7ln 6 ≈ −0.473 Page 110 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes3. Rewrite the expression in terms of ln 4 and ln 5. Verify the equivalence using yourcalculator.ln 1600Solution: We’ll discuss this in class.ln 1600 = ln ( 5 2 4 3) = 2 ln 5 + 3 ln 44. Use the change-of-base formula and a graphing utility to graph the function.f (x) = log 4 (x − 1)Solution: We’ll discuss this in class.f (x) = log 4 (x − 1) =ln (x − 1)ln 4=log (x − 1)log 4Your graph should look like mine.21-1 0 1 2 3 4 5 6 7 8-1-2-3Figure 56: Partial graph of f (x) = log 4 (x − 1).5. Use the properties of logarithms to rewrite and simplify the logarithmic expression.log 9 27Page 111 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss this in class.log 9 27 = 3 log 9 3 = 3 log 9√9 =326. Expand.log 7x 2 y 3z 4Solution: We’ll discuss this in class.log 7x 2 y 3z 4= 2 log 7 x + 3 log 7 y − 4 log 7 z7. Expand.√log 113x 2 z 3y 5Solution: We’ll discuss this in class.√log 113x 2 z 3y 5= 1 2 [2 log 11 x + 3 log 11 z − 5 log 11 y] = log 11 x + 3 2 log 11 z − 5 2 log 11 y8. Condense.5 log x − 3 log y + 1 2 log z Page 112 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss this in class.5 log x − 3 log y + 1 2 log z = log x5√ zy 39. Condense.12 log 2 x + 4 log 2 y − 3 log 2 zSolution: We’ll discuss this in class.12 log y 4√ x2 x + 4 log 2 y − 3 log 2 z = log 2z 310. Consider the following.y 1 = 2 [ ln 5 + ln ( x 2 + 1 )] ,y 2 = ln[25 ( x 2 + 1 ) ] 2(a) Use a graphing utility to graph the two equations in the same viewing window.Solution: We’ll discuss this in class.Graphs are identical.2322212019181716151413121110987654321-20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19-1-2Figure 57: Partial graph of y 1 and y 2 .(b) Are these two graphs identical? Explain in English.Page 113 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss this in class.The statements are equivalent. The process of simplification has no restrictions.This is an identity for all real numbers.11. Consider the following.y 1 = 4 (ln 2 + ln x) , y 2 = ln 16x 4(a) Use a graphing utility to graph the two equations in the same viewing window.Solution: We’ll discuss this in class.These graphs are not identical. The curve is y 1 is only to the right of the y-axisand y 2 lies on top of this, as well as the reflection about the y axis.2322212019181716151413121110987654321-20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19-1-2Figure 58: Partial graph of y 1 and y 2 .(b) Are these two graphs identical? Explain in English.Solution: We’ll discuss this in class.The domain for y 1 is x > 0 and the domain for y 2 is all real numbers x ≠ 0. Theequivalence only works for x > 0 and that’s why the graphs only agree there.19.3 AssignmentYou should read 3.3 and do the WebAssign assignment 03.03.Page 114 of 169

Pre-calculus IMTH-11920 mth.119.03.04Essex County CollegeDivision of MathematicsLMS ProjectClass Notes20.1 Equations20.1.1 Exact and ApproximateThere’s a big difference between exact and approximate answers. For example the exactsolution of the equationisx 2 = 2x = − √ 2 or x = √ 2.However, if I were to ask for an approximate solution to this equation I would first have totell you how precise I want the approximate to be. For example, if I want the answer to thenearest tenth, then the solution would be written as:x ≈ −1.4 or x ≈ 1.4but if I instead ask for six decimal places, then the solution would be written as:x ≈ −1.414214 or x ≈ 1.414214You should pay careful attention to this matter.20.1.2 Solving Exponential EquationsFor any a > 0, and a ≠ 1,a x = a y ⇔ x = y.This should be somewhat obvious. However if a ≠ b will will need to use logs. For anya, b > 0, and a, b ≠ 1,a x = b ⇔ x = ln bln a .This for most students is not obvious and I will discuss this in detail in class.20.1.3 Solving Logarithmic EquationsFor any a > 0, and a ≠ 1, (recall that x and y must be positive)log a x = log a y ⇔ x = y.This should be somewhat obvious. Furthermore, you should not forget how to rewrite alogarithmic equation as an exponential.Page 115 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesAs before, you should be especially mindful about the difference between exact andapproximate solutions. And please don’t forget to check your solutions—some may be extraneous.For example, here’s a problem from MTH-100 that has an extraneous solution.√x + 2 = x square both sidesx + 2 = x 2 solve for zero0 = x 2 − x − 2 factor0 = (x + 1) (x − 2)The last line above has the solution x = −1 or x = 2; however, if you check these solutionsin the original problem you should quickly note that x = −1 does not work! Hence we saythat x = −1 is an extraneous solution. Finally the solution to the equation√x + 2 = xis x = 2 only.20.2 Examples1. Solve for x.2 x = 16Solution: We’ll discuss this is class.x = 42. Solve for x.log 5 (2x − 1) = log 5 (3x − 4)Solution: We’ll discuss this is class.x = 3Page 116 of 169

Pre-calculus IMTH-1193. Solve for x.4 3x−5 = 16Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss this is class.x = 7 24. Solve for x.ln x = −2Solution: We’ll discuss this is class.x = e −25. Solve for x.ln 2x = 3Solution: We’ll discuss this is class.x = e326. Solve for x.3 + 2 log 10 x = 5Page 117 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss this is class.x = 107. Solve for x.3 x2 +4x = 1 27Solution: We’ll discuss this is class.x = −1, −38. Solve for x.ln (x + 1) 2 = 0Solution: We’ll discuss this is class.x = 0, −29. Solve for x.log 4 x − log 4 (x − 1) = 1 2Solution: We’ll discuss this is class.x = 2Page 118 of 169

Pre-calculus IMTH-11910. Solve for x.27 = 3 5x 9 x2Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss this is class.x = 1 2 , −311. Solve for x.log 3 x + log 3 (x − 8) = 2Solution: We’ll discuss this is class.x = 9Note that x = −1 does not work; it is extraneous.12. Solve for x. Both exact and an approximate (4 decimal places).2 x = 5Solution: We’ll discuss this is class.x = ln 5ln 2 = log 5log 2 ≈ 2.321913. Solve for x.ln (x + 1) + ln (x − 2) = ln xPage 119 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss this is class.x = 1 + √ 3Note that x = 1 − √ 3 does not work; it is extraneous.14. Use a graphing utility to graph f (x) = 8 log 4 x and g (x) = 3 in the same viewingwindow. Approximate the point of intersection of the graphs of f and g. Then solve theequation of f (x) = g (x) algebraically. (Round your answer to three decimal places.)Solution: We’ll discuss this is class.x = 4 3/8 ≈ 1.682Your graph should look like mine.Figure 59: Partial graph of f and g with point of intersection indicated..15. Solve for x. Both exact and and approximate (4 decimal places).e 2x − 5e x + 6 = 0Page 120 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution:x = ln 2, ln 3 ≈ 0/6931, 1.0986We’ll discuss this is class.16. Solve for x. Both exact and and approximate (4 decimal places).525= 3251 + e−x Solution: We’ll discuss this is class.x = ln 138 ≈ 0.485517. Solve for x.−x 2 e −x + 3xe −x = 0Solution: We’ll discuss this is class.x = 0, 318. This one’s a bit tricky—try it though. Graphing may help!5 x − 5 −x5 x + 5 −x = 2 3Page 121 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss this is class.x = 1 219. This one’s a bit tricky too—try it though. Both exact and an approximate (4 decimalplaces).e x − 6e −x = 1Solution:x = ln 3 ≈ 1.0986We’ll discuss this is class.20. Find the time t required for a $1050 investment to double at interest rate r = 5%,compounded continuously. Round your results to two decimal places.Solution: We’ll discuss this is class.2 · 1050 = 1050 · e 0.05t , t = ln 20.05 ≈ 13.8620.3 AssignmentYou should read 3.4 and do the WebAssign assignment 03.04.Page 122 of 169

Pre-calculus IMTH-11921 mth.119.03.05Essex County CollegeDivision of MathematicsLMS ProjectClass Notes21.1 Exponential and Logarithmic ModelsThere’s no need to memorize the different models. Most questions in this section are justtesting to see if you can apply what you’ve learned so far. As you study mathematics furtheryou will see that exponentials and logarithmic models abound.21.1.1 Exponential Decay Modely = ae −bx , b > 0Figure 60: General shape.21.1.2 Exponential Growth Modely = ae bx , b > 0Figure 61: General shape.Page 123 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes21.1.3 Natural Logarithmic Modely = a + b ln xFigure 62: General shape.21.1.4 Common Logarithmic Modely = a + b log 10 xFigure 63: General shape.Page 124 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes21.1.5 Logistic Growth Modely =a1 + be −rx Figure 64: General shape.21.1.6 Gaussian Modely = ae −(x−b)2 /cFigure 65: General shape.21.2 Examples1. Find the exponential growth model y = ae bx , b > 0 that fits the points: ( 0, 1 3)and (5, 7).(Round your value for b to four decimal places.)Page 125 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss this in class.a = 1 ln 21, b =3 5≈ 0.60892. The populations P (in thousands) of a certain city from 2000 through 2008 can bemodeled byP = 1115.4e kt ,where t is the year, with t = 0 corresponding to 2000. In 2002, the population wasabout 1.3 million. Find k (five decimal places) and then use this modell to predict thepopulation in 2015.Solution: We’ll discuss this in class.k = ln√13001115.4 ≈ 0.07658, P (15) = 1115.4e0.07658·15 ≈ 3.5 million3. Carbon-14 dating assumes that the carbon dioxide on Earth today has the same radioactivecontent as it did centuries ago. If this is true, the amount of Carbon-14 absorbed bya tree that grew several centuries ago should be the same as the amount of Carbon-14absorbed by a tree growing today. A piece of ancient charcoal contains only 20% as muchradioactive carbon as a piece of modern charcoal. How long ago was the tree burned tomake the ancient charcoal given that the half-life of Carbon-14 is 5700 years? (Roundyour answer to the nearest whole number.)Solution: We’ll discuss this in class.It’s an exponential decay model withy = ae −bx , b > 01 = 2e −5700bb = ln 25700 ≈ 0.000122When the tree died it stops taking in Carbon-14, so what it has starts to decay andwe’re told it only has 20%.0.2 = e −0.000122tt = 13235Page 126 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes4. The IQ scores for adults roughly follow the normal distributiony = 0.0266e −(x−101)2 /450 , 70 ≤ x ≤ 115,where x is the IQ score. Use a graphing utility to graph the function, and determine thevalue of x where the graph is maximal—this is the average IQ.Solution: We’ll discuss this in class.Average IQ is 101.Your graph should look like mine.50 60 70 80 90 100 110 120 130 140 150Figure 66: Partial graph of IQ model.21.3 AssignmentYou should read 3.5 and do the WebAssign assignment 03.05.Page 127 of 169

654321-4 -3 -2 -1 0 1 2 3 4 5-1-254321-3 -2 -1 0 1 2 3-1-2-3-4-5Pre-calculus IMTH-11922 mth.119.07.04Essex County CollegeDivision of MathematicsLMS ProjectClass Notes22.0.1 Linear Systems ReviewedAs you may recall from MTH-100, you solved linear systems with two variables—usually xand y, but other variables may have been used—of the following form.{ 2x + 3y = 55x − 2y = −16The methods that you used were: elimination, substitution, and graphical. We will reviewthese three methods in class, and you should be very familiar with these three methods.Here’s a graph of this linear system to start us off.Figure 67: Linear system, 2x + 3y = 5 and 5x − 2y = −16.22.0.2 Introduction to Non-linear SystemsThe method of elimination works best for linear systems. However, graphical along withsubstitution generally works best for non-linear systems. Here’s an example, where we’reasked to solve a non-linear system for x and y.{ x + y = 0x 3 − 5x = yAgain, here’s a graph of this non-linear system. Unlike linear system, having a visual isvery important. Here, we at least know that there are three points of intersection.Figure 68: Linear system, x + y = 0 and x 3 − 5x = y.Page 128 of 169

21-1 0 1Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesYou should be able to visually determine the points of intersection and then verify them.We’ll also do the algebra. Certainly we will do this work in class!Here’s another one.{ 2 − 2x2= y2x 4 − 4x 2 = y − 2Again, here’s a graph of this non-linear system. Unlike linear system, having a visual isvery important. Here, we at least know that there are three points of intersection.Figure 69: Linear system, 2 − 2x 2 = y and 2x 4 − 4x 2 = y − 2.You should be able to visually determine the points of intersection and then verify them.We’ll also do the algebra. Certainly we will do this work in class!22.0.3 Introduction to Linear Algebra: Linear 2 × 2 SystemHere we will have two linear equations, with two unknowns. Graphically we know that thelines may intersect once, always intersect (same line), or possibly never intersect (differentparallel lines). Here’s some typical terms that apply to linear systems:Consistent: At least one solution.Inconsistent: No solution.Dependent: Infinite number of solutions.Independent: Not dependent.The basics of Gaussian elimination was extensively covered in MTH-100, and the methodhere is no different. It essentially relies on the following.1. Interchange any two equations.2. Multiply both sides of one equation by a non-zero constant.3. Add a nonzero multiple of one equation to another equation.Page 129 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesAlthough the method I’ll present today is no different, the essential structure will need tobe adapted to allow for extending the system to more variables and equations. I assure youhowever, that the method is not different from what you learned in MTH-100. For example,you might be asked to solve.{ 2x − 3y = 7x + 4y = −2But now to speed the process we will write this system using matrix notation, as[ 2 −3] 71 4 −2This matrix is called an augmented matrix. It’s dimension is 2 × 3, that is we reportthe dimension of the augmented matrix to be the number of rows × the number ofcolumns. For example, the coefficient matrix of this example is 2 × 2 (this is referred to asa square matrix 17 )and looks like[ 2 −31 4].The constant matrix of this example is a single column and looks like[ ]7,−2and its dimension is 2 × 1.You should clearly make the connection between augmented matrix and the originalsystem of equations. It is also important to note that these two forms are identical ininformation, as long as we know the first column corresponds to x; the second columncorresponds to y; the vertical bar corresponds to an equal sign; and finally the last columncorresponds to the constants. Now, instead of operating on equations, we’re operating onrows. The allowable operations (mention above) are now called row-equivalent operations,and that’s all we’re going to use to rewrite our systems into more readable formats.1. Interchange any two rows.2. Multiply each entry of a row by a non-zero constant.3. Add a nonzero multiple of one row to another row.Let’s proceed using these operations. (This will be done in class.)[ ] [ ] [ ]2 −3 7 2 −3 7 1 0 2∼∼1 4 −2 0 1 −1 0 1 −1The last matrix is no different than that of the others, but it is very readable. The lastmatrix simple states that x = 2 and y = −1.A matrix that has undergone this process of Gaussian elimination is said to be inreduced row echelon form. Such a matrix has the following characteristics:17 Whenever the number of rows equal the number of columns.Page 130 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes1. All zero rows are at the bottom of the matrix.2. The leading entry of each nonzero row after the first occurs to the right of the leadingentry of the previous row.3. The leading entry in any nonzero row is 1.4. All entries in the column above and below a leading 1 are zero.Another common definition is row echelon form and that only requires zeros below theleading ones, while the above definition also requires them above the leading ones. By theway, your calculators are quite capable of getting a matrix into either reduced row echelonform (rref) or row echelon form (ref). 1822.1 Linear 3 × 3 SystemNow that we’ve just done a 2×2 system, you should be ready to follow the method forward. 19Be aware that you will have solve this problem—and others to follow—by using an augmentedmatrix and elementary row operations. Here’s a 3 × 3 system to solve. Again we’ll do thisin class, and I will slowly show the details.⎧⎨⎩2x + 3y − z = −73x − 3y + z = 122x + 4y + z = −3Solution: Augmented matrix form of the system:⎡2 3 −1⎤−7⎣ 3 −3 1 12 ⎦2 4 1 −3Elementary row operations, in order given:R 1 + R 2 → R 2R 1 + R 3 → R 3Produces:⎡2 3 −1⎤−7⎡⎣ 3 −3 1 12 ⎦ ∼ ⎣2 4 1 −32 3 −1 −75 0 0 54 7 0 −1018 I strongly suggest that you look at your calculator’s manual to figure this out.19 Don’t worry, I’ll be reasonable.⎤⎦Page 131 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesThe second row gives x = 1; using x = 1 in row three, gives y = −2; finally, using x = 1 andy = −2 in row one, gives z = 3. Now that we know the solution we can actually go one stepfurther and write the matrix in a more readable format.⎡⎣2 3 −1 −73 −3 1 122 4 1 −3⎤⎡⎦ ∼ ⎣2 3 −1 −75 0 0 54 7 0 −10⎤⎡⎦ ∼ ⎣1 0 0 10 1 0 −20 0 1 3This final form is called reduced row echelon form, and your calculator can actually doit. However, you may be required to do this by hand, and you must show work to get fullcredit. At first it looks impossible, but with practice you will be able to fly through theseproblems.⎤⎦22.2 Examples1. Solve for x and y. Use both a graph and algebra.y = x − 20 = x 2 + y 2 − 4x + 6y + 4Solution: We’ll discuss this in class.You’ve got a circle and a line that intersect at: (2, 0) and (−1, −3). Your graph shouldlook like mine.1-4 -3 -2 -1 0 1 2 3 4 5 6 7 8-1-2-3-4-5-6Figure 70: Partial graph of y = x − 2 and 0 = x 2 + y 2 − 4x + 6y + 4.2. Solve for x and y. Use both a graph and algebra.xy − 2 = 03x − 2y = −4Page 132 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss this in class.You’ve got a line and a rational that intersect at: (2/3, 3) and (−2, −1). Your graphshould look like mine.4321-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6-1-2-3Figure 71: Partial graph of xy − 2 = 0 and 3x − 2y = −4.3. Solve for x and y. Use both a graph and algebra.y = 2x − 1y = √ x + 1Solution: We’ll discuss this in class.You’ve got a line and a square root function that intersect at: (5/4, 3/2). Your graphshould look like mine.21-2 -1 0 1 2 3 4-1Figure 72: Partial graph of y = 2x − 1 and y = √ x + 1.4. Identify the elementary row operation performed to obtain the new row-equivalentmatrix.[]]3 −1 −4−4 7 9∼[ 3 −1 −48 3 −7Page 133 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss this in class.Add 4 times Row 1 to Row 2.5. Use the matrix capabilities of a graphing utility to write the matrix in reduced rowechelon form.⎡⎤1 3 2⎣ 20 60 36 ⎦2 6 10Solution: We’ll discuss this in class.⎡⎣1 3 00 0 10 0 0⎤⎦6. Write the system of linear equations represented by the augmented matrix. Then useback-substitution to solve. (Use the variables x, y, and z, if applicable.)⎡⎣1 2 −2 30 1 1 10 0 1 3⎤⎦Solution: We’ll discuss this in class.x + 2y − 2z = 3y + z = 1z = 3Using back-substitution, z = 3, y = −2, and x = −13.7. Solve the following system using a augmented matrix and row-equivalent operations.Your final augmented matrix should be in reduced row echelon form.⎧⎨⎩x + 2y − 3z = −82x − y + 2z = 53x − y − 4z = −7Page 134 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss this in class.x = 0, y = −1, and z = 2.⎡1 0 0⎤0⎣ 0 1 0 −1 ⎦0 0 1 28. Solve the following system using a augmented matrix and row-equivalent operations.Your final augmented matrix should be in reduced row echelon form.⎧⎨⎩2x − y + 4z = 11x − 2y − 10z = −173x + 4z = 11Solution: We’ll discuss this in class.x = 1, y = −1, and z = 2.⎡1 0 0⎤1⎣ 0 1 0 −1 ⎦0 0 1 29. Use matrices to solve the system of equations, if possible. Use Gauss-Jordan elimination.(If there is no solution, enter No solution. If the system is dependent, set z = a andsolve for x and y in terms of a.)⎧⎨⎩4x − 3y + 7z = 38x − 9y + 15z = −212x + 3z = 15Solution: We’ll discuss this in class.15 − 3ax =2y = a + 273z = aPage 135 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes10. Use a system of equations to find the quadratic function f (x) = Ax 2 + Bx + C thatsatisfies the equations. Solve the system using matrices.f (−2) = −15, f (1) = −6, f (2) = −23Solution: We’ll discuss this in class.f (x) = −5x 2 − 2x + 111. A corporation borrowed $1.49 million to expand its line of shoes. Some of the money wasborrowed at 3%, some at 4%, and some at 6%. Use a system of equations to determinehow much was borrowed at each rate given that the annual interest was $72, 200 and theamount borrowed at 6% was four times the amount borrowed at 3%. Solve the systemusing matrices.Solution: We’ll discuss this in class.$180, 000 at 3%; $590, 000 at 4%; $720, 000 at 6%22.3 AssignmentYou should read 7.4 and do the WebAssign assignment 07.04.Page 136 of 169

Pre-calculus IMTH-11923 mth.119.07.05Essex County CollegeDivision of MathematicsLMS ProjectClass Notes23.1 Introduction to Matrix Mathematics23.1.1 Something Old . . .Suppose you’re asked to solve this system{ 2x − 3y = 7x + 4y = −2,but not as we did in MTH-100. Instead used matrix notation, along with elementary rowoperations to rewrite the augmented matrix into reduced row echelon form.[ ] [ ] [ ]2 −3 7 2 −3 7 1 0 2∼∼1 4 −2 0 1 −1 0 1 −1Again, the last matrix is no different than the others, but it is very readable. The last matrixsimple states that x = 2 and y = −1. This final matrix is the reduced row echelon form.23.1.2 Something New . . .As before, we are now going to rewrite our original system in terms of matrices alone, butwe’re actually going to use three separate matrices. The three matrices are as follows.Coefficient Matrix Here we are going to just list the coefficients of the variables.[ ] 2 −3A =1 4Variable Matrix Here we are going to just list the variables.[ ] xX =yConstant Matrix Here we are going to just list the constants.[ ]7B =−2Our equation is now of the form:A · X = B,or, if we use the matrices we have:[ ] [ ] [ ]2 −3 x 7· = .1 4 y −2Page 137 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesIn class I will show that this matrix multiplication 20 is equivalent to the original problem.Our matrix multiplications will be no different than this, and we will do other examples.What’s most interesting about this form is that we can now easily solve for X, by doing thefollowing.A · X = BX = A −1 · BThis new matrix A −1 is called the inverse of A. Computing this inverse is not difficult, butit can become quite involved for larger matrices. For now, I will tell you what the inverse is.[ ] −1 []2 −3A −1 4/11 3/11==1 4 −1/11 2/11Now try this multiplication (we’ll do it in class).[ ] [ ] [ ]2 −3 x 7· =1 4 y −2[ ] [ ] −1 [ ]x 2 −3 7=·y 1 4 −2[ ] [] [ ]x 4/11 3/11 7=·y −1/11 2/11 −2[ ] [ ]x 2=y −1Now we will verify thatA · X = B.That is, verify[ 2 −31 4] [·2−1] [=7−2]We’ll certainly discuss the basics of matrix multiplication in class. You’ll need to be able tomultiply matrices by hand and when necessary, a calculator.Now let’s take an even bigger example.⎧⎨⎩2x + 3y − z = −73x − 3y + z = 122x + 4y + z = −3Here goes . . .⎡2 3 −1⎤−7⎡⎣ 3 −3 1 12 ⎦ ∼ ⎣2 4 1 −32 3 −1 −75 0 0 54 7 0 −10⎤⎡⎦ ∼ ⎣1 0 0 10 1 0 −20 0 1 320 We’ll be doing this by example and I also suggest that your learn how to do this on your calculators.⎤⎦Page 138 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesAgain x = 1; y = −2; and z = 3.Now however, I want to rewrite it using matrix multiplication. As before, we are nowgoing to rewrite our original system in terms of matrices alone, but we’re actually going touse three separate matrices. The three matrices are as follows.Coefficient Matrix Here we are going to just list the coefficients of the variables.⎡2 3⎤−1A = ⎣ 3 −3 1 ⎦2 4 1Variable Matrix Here we are going to just list the variables.⎡ ⎤xX = ⎣ y ⎦zConstant Matrix Here we are going to just list the constants.⎡ ⎤−7B = ⎣ 12 ⎦−3Our equation is now of the form:A · X = B,or, if we use the matrices we have:⎡⎤ ⎡ ⎤ ⎡2 3 −1 x −7⎣ 3 −3 1 ⎦ · ⎣ y ⎦ = ⎣ 122 4 1 z −3⎤⎦ .In class I will show that this matrix multiplication is equivalent to the original problem.Our matrix multiplications will be no different than this, and we will do other examples.What’s most interesting about this form is that we can now easily solve for X, by doing thefollowing.A · X = BX = A −1 · BThis new matrix A −1 is called the inverse of A. Computing this inverse is not difficult, butit can become quite involved for larger matrices. For now, I will tell you what the inverse is.⎡A −1 = ⎣2 3 −13 −3 12 4 1⎤⎦−1⎡= ⎣1/5 1/5 01/35 −4/35 1/7−18/35 2/35 3/7⎤⎦Page 139 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesNow try this multiplication (we’ll do it in class).⎡⎤ ⎡ ⎤ ⎡ ⎤2 3 −1 x −7⎣ 3 −3 1 ⎦ · ⎣ y ⎦ = ⎣ 12 ⎦2 4 1 z −3⎡⎣xyz⎤⎦ =⎡⎣2 3 −13 −3 12 4 1Yes, I know this looks difficult, but you can do this.⎡ ⎤x⎡2 3⎤−1−1⎡ ⎤−7⎣ y ⎦ = ⎣ 3 −3 1 ⎦ · ⎣ 12 ⎦z 2 4 1 −3⎡⎣⎡⎣xyzxyz⎤⎦ =⎤⎡⎣⎡⎦ = · ⎣1/5 1/5 01/35 −4/35 1/7−18/35 2/35 3/71−23And once again you’ll need to verify thatA · X = B.⎤⎦23.1.3 Equal Matrices⎤⎡⎦ · ⎣Two matrices are equal if they are of the same dimension and corresponding entries areequal. For example⎡⎤ ⎡⎤2 3 −1 2 3 −1⎣ 3 −3 1 ⎦ = ⎣ 3 −3 1 ⎦ ,2 4 1 2 4 1are equal. If you’re given⎡⎤2x 3 −1⎡⎣ 3 y − 2 1 ⎦ = ⎣2 4 −6⎤⎦−712−3x − 1 3 −13 3 − y 12 4 x − 2y−1⎤⎦⎡· ⎣−712−3then x = −1 and y = 5/2 is a necessary condition for equality.23.1.4 ArithmeticYou’ll be asked to add matrices, and to multiply matrices by scalars 21 and matrices 22 . We’llkeep these operations simple. No need to memorize properties and you should see matrixmultiplication as a natural extension of a system of equations.21 In linear algebra, real numbers are called scalar. They’re used to scale matrices through multiplication.22 A rectangular array of numbers, symbols, or expressions, arranged in rows and columns.⎤⎦ ,⎤⎦Page 140 of 169

Pre-calculus IMTH-11923.1.5 Simple PropertiesEssex County CollegeDivision of MathematicsLMS ProjectClass NotesGiven that A, B, and C are matrices, and c and d are scalars. We have the following:1. Associative Property of Scalar Multiplication(cd) A = c (dA)2. Commutative Property of Matrix AdditionA + B = B + A3. Scalar Identity1A = A4. Distributive Propertyc (A + B) = cA + cB5. Associative Property of Matrix AdditionA + (B + C) = (A + B) + C6. Left Distributive Property of Matrix MultiplicationA (B + C) = AB + AC7. Associative Property of Scalar Multiplicationc (AB) = (cA) B = A (cB)8. Associative Property of Matrix MultiplicationA (BC) = (AB) C9. Right Distributive Property of Matrix Multiplication23.2 Examples(A + B) C = AC + BC1. In general, matrix multiplication is not commutative. Given⎡⎤ ⎡⎤1 −2 30 1 −2A = ⎣ −4 3 2 ⎦ , B = ⎣ −3 4 3 ⎦ ,1 2 −32 −1 0verifyA · B ≠ B · A.Page 141 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss this in class.2. What is the dimension of AB given A is a 3 × 2 matrix and B is a 2 × 5 matrix?Solution: We’ll discuss this in class.3 × 53. Find x and y.⎡x + 2 −4⎤−3⎡⎣ −2 2y 2x ⎦ = ⎣−8 −1 y + 22x + 6 −4 −3−2 8 −8−8 −1 6⎤⎦Solution: We’ll discuss this in class.x = −4, y = 44. Find the product.⎡[ ] 3 1 0· ⎣4 −2 50 1 2−1 −5 −10 1 −1⎤⎦Solution:[ 3 1 04 −2 5⎡]· ⎣0 1 2−1 −5 −10 1 −1⎤⎦ =[ −1 −2 52 19 5]5. Find, if possible, A + B, A − B, 3A, and 3A − 2B. Use the matrix capabilities of agraphing utility to verify your results.[ ] [ ]−1 2 −32 3 −5A =, B =5 0 40 1 5Page 142 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss this in class.A + B =A − B =3A =3A − 2B =[ ] 1 5 −85 1 9[ ]−3 −1 25 −1 −1[ ]−3 6 −915 0 12[ ]−7 0 115 −2 26. Solve using matrix multiplication and inverses. 23⎧⎨ 2x + 3y + z = −13x + 3y + z = 1⎩2x + 4y + z = −2Solution: We’ll discuss this in class.x = 2, y = −1, z = −27. Solve using matrix multiplication and inverses. 24⎧⎨ x + y − z = 0x − y + z = 2⎩x + y + z = 6Solution: We’ll discuss this in class.x = 1, y = 2, z = 323 Inverse will be given in class.24 Inverse will be given in class.Page 143 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes8. Solve using matrix multiplication and inverses. 25⎧⎨ 5x + 4y − 3z = 82x − y = 2⎩x + 4y + 7z = −6Solution: We’ll discuss this in class.x = 1, y = 0, z = −123.3 AssignmentYou should read 7.5 and do the WebAssign assignment 07.05.25 Inverse will be given in class.Page 144 of 169

Pre-calculus IMTH-11924 mth.119.07.06Essex County CollegeDivision of MathematicsLMS ProjectClass Notes24.1 Inverse of a Square MatrixIf there exists an n × n matrix A −1 such that AA −1 = I n = A −1 A, then A −1 is called theinverse or inverse matrix of A. I n is an n×n matrix whose entries are: a i=j = 1 (diagonal)and a i≠j = 0, and is called the identity matrix. I n has the property that I n A n×n = A n×n I nIf a matrix A has an inverse, it is called invertible or nonsingular; if it does not have aninverse, it is called singular. Not all n × n matrices are invertible.In this section we will cover how to verify if two n × n matrices are inverses of oneanother. And we will also develop a simple technique to find inverses of nonsingularmatrices. However, in general, finding inverses can be quite painful and I strongly suggestyou learn how to find inverses using a calculator.24.2 Examples1. Given⎡A = ⎣−2 3 −54 −1 1−7 6 0verify AI 3 = I 3 A = A.⎤⎦ ,Solution: We’ll discuss this in class.2. Show that B is the inverse of A. To show that B is the inverse of A, we need to showthat AB = I = BA.[ ] [ ]3 −52 5A =, B =−1 21 3Solution: We’ll discuss this in class.AB = BA =[ 1 00 1]3. Show that B is the inverse of A. To show that B is the inverse of A, we need to showthat AB = I = BA.⎡⎤⎡⎤1 0 −1A = ⎣ −1 2 0 ⎦ , B = 1 0 −5 2⎣ 0 1 1 ⎦71 5 0−7 −5 2Page 145 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss this in class.⎡AB = BA = ⎣1 0 00 1 00 0 1⎤⎦4. Find the inverse, if possible.[ ] 1 34 13Solution: This method along with using your calculator will be throughly discussedin class. I know it’s difficult, and I will try to go easy on you with regards to findinginverses on exams. Nonetheless, a calculator certainly makes this process much easier.[ 1 3 1 04 13 0 1So the inverse is[ 13 −3−4 1].]∼[ 1 0 13 −30 1 −4 1]5. Find the inverse, if possible.⎡1 1⎤1⎣ 3 5 4 ⎦3 6 5Solution: This method along with using your calculator will be throughly discussedin class. I know it’s difficult, and I will try to go easy on you with regards to findinginverses on exams. Nonetheless, a calculator certainly makes this process much easier.⎡⎣1 1 1 1 0 03 5 4 0 1 03 6 5 0 0 1⎤⎡⎦ ∼ ⎣1 0 0 1 1 −10 1 0 −3 2 −10 0 1 3 −3 2⎤⎦Page 146 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSo the inverse is⎡1 1⎤−1⎣ −3 2 −1 ⎦ .3 −3 26. The formula for finding the inverse of a 2 × 2 matrix[ ] a bA =c disA −1 =[1ad − bcd−c−ba],provided ad − bc ≠ 0. Use this formula to find the inverse of[ ]7 6A =.−5 −12Solution: We’ll discuss this in class.[A −1 =2/9 1/9−5/54 −7/54].7. If possible, find the value of the constant k such that B = A −1 .[ ] [ ]−4 6−5/38 3/19A = , B =3 5k 2/19Solution: We’ll discuss this in class.k = 3 38Page 147 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes8. Solve using matrix multiplication and inverses.⎧⎨ 4x − 2y + 3z = 62x + 2y + 5z = 6⎩8x − 5y − 2z = 52Solution: We’ll discuss this in class.x = 10, y = 8, z = −69. Consider a company that specializes in potting soil. Each bag of potting soil for seedlingsrequires 2 units of sand, 1 unit of loam, and 1 unit of peat moss. Each bag of pottingsoil for general potting requires 1 unit of sand, 2 units of loam, and 1 unit of peat moss.Each bag of potting soil for hardwood plants requires 2 units of sand, 2 units of loam,and 2 units of peat moss. Find the numbers of bags of the three types of potting soilthat the company can produce with the given amounts of raw materials: 750 units ofsand; 850 units of loam; 600 units of peat moss.Solution: We’ll discuss this in class.150 bags of seedling soil; 250 bags of general soil; and 100 bags for hardwood plants.24.3 AssignmentYou should read 7.6 and do the WebAssign assignment 07.06.Page 148 of 169

Pre-calculus IMTH-11925 mth.119.07.07Essex County CollegeDivision of MathematicsLMS ProjectClass Notes25.1 Determinants of a Square MatrixYou may recall that the formula for finding the inverse of a 2 × 2 matrix[ ] a bA =c disA −1 =[1ad − bcd−c−ba],provided ad − bc ≠ 0.Here the matrix A is invertible if and only if ad − bc ≠ 0. We called this number thedeterminant of A. It certainly would be nice to have a similar result for bigger squarematrices. This determinant has many uses, but for now we’re just interested in using thedeterminant to determine if a square matrix is invertible or not. You may also recall thata square matrix that is not invertible is called singular (degenerate). A square matrix issingular if and only if its determinant is 0. 26 First let us use the following notation for thedeterminant.[ ]d −bdet=d −b−c a ∣ −c a ∣ = ad − bcTo find the determinant of a bigger square matrix you will have to resort to variety oftechniques, many are outlined in your textbook, but I generally just want you to learn howto do so on your calculators.25.2 Examples1. Show for a 2 × 2 that det (AB) = det (A) det (B).Solution: We’ll discuss this in class.([ ] [ ])a1 adet2 b1 b 2a 3 a 4 b 3 b 4det (AB) = det (A) det (B)= det([ ]) ([ ])a1 a 2b1 bdet2a 3 a 4 b 3 b 426 A very good intuitive reason is that det (AB) = det (A) det (B). If AB = I then det (A) det (B) = 1not 0 so neither det (A) nor det (B) can be 0.Page 149 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes2. Solve for x([detx 4−1 x])= 20Solution: We’ll discuss this in class.x = ±43. Find the determinate of⎡1 1⎤−1⎣ −3 2 −1 ⎦ .3 −3 2Solution: We’ll discuss this in class.125.3 AssignmentYou should read 7.7 and do the WebAssign assignment 07.07.Page 150 of 169

Pre-calculus IMTH-11926 mth.119.07.07Essex County CollegeDivision of MathematicsLMS ProjectClass Notes26.1 Application of Matrices and Determinants26.1.1 AreaGiven the vertices of a triangle in the coordinate plane you can find the area using the threevertices entered into a 3 × 3 matrix (see below) and its corresponding determinant. Thistechnique also works for for a parallelogram because the parallelogram is composed to twocongruent triangles.For the there points (x a , y a ), (x b , y b ), (x c , y c ), you can just put the x and y values inas the first two rows of the matrix, then fill the third row with ones:⎡⎣x a x b x cy a y b y c1 1 1⎤⎦ .The absolute value of the determinant is the area of the parallelogram, and 1/2 that valueis the area of the triangle.26.2 Examples1. Use a determinant to find the area of the figure with the given vertices.(−2, 3) , (2, 3) , (2, −6)Solution: We’ll discuss this in class.A = ± 1 2∣−2 2 23 3 −61 1 1∣ = ±1 [−36] = ±182The area is 18 square units. This one is also easy to do using simple geometry.2. Find y such that the triangle has an area of 13 square units.(−4, 2) , (−3, 5) , (−2, y)Page 151 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: We’ll discuss this in class.A = ± 1 2∣−4 −3 −22 5 y1 1 1∣ = ±1 [y − 11] = 132This results in two equations: y − 11 = 26 and y − 11 = −26.y = −15, 373. Use a determinant to find the area of the figure with the given vertices.(−7, −1) , (0, 3) , (4, 8) , (−3, 4)Solution: We’ll discuss this in class.−7 0 4A = ±−1 3 8= ± [19] = ±19∣ 1 1 1 ∣The area is 19 square units.26.3 AssignmentYou should read 7.8 and do the WebAssign assignment 07.08.Page 152 of 169

Pre-calculus IMTH-11927 Final ReviewEssex County CollegeDivision of MathematicsLMS ProjectClass NotesAt Essex County College you should be prepared to show all work clearly and in order,ending your work by boxing the answer. Furthermore, justify your answers algebraicallywhenever possible. These questions are for review and will not be covered in class.1. Find the domain off (x) = x2 − 49√x2 + 9 − 5 .Solution: x 2 + 9 is always positive so we don’t have to worry about the square root.However, to find the domain we need to solve √ x 2 + 9 − 5 = 0.√x2 + 9 − 5 = 0√x2 + 9 = 5x 2 + 9 = 25x 2 = 16x = ±4So, the domain is R, x ≠ ±4 .2. Solve the inequality.|14 − x| − 3 < 17Solution:|14 − x| − 3 < 17|14 − x| < 20We know that for a > 0, that |x| < a is equivalent to −a < x < a, so|14 − x| < 20−20 < 14 − x < 20−34 < −x < 634 > x > −6−6 < x < 34Here the proper interval is: (−6, 34) .Page 153 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes3. Solve by using an augmented matrix and elementary row operations.⎧⎨ 2x + 3y − z = −73x − 3y + z = 12⎩2x + 4y + z = −3Solution: Augmented matrix form of the system:⎡2 3 −1⎤−7⎣ 3 −3 1 12 ⎦2 4 1 −3Elementary row operations, in order given:R 1 + R 2 → R 2R 1 + R 3 → R 3Produces:⎡2 3 −1⎤−7⎡⎣ 3 −3 1 12 ⎦ ∼ ⎣2 4 1 −32 3 −1 −75 0 0 54 7 0 −10The second row gives x = 1 ; using x = 1 in row three, gives y = −2 ; finally, usingx = 1 and y = −2 in row one, gives z = 3 .⎤⎦4. Use long division to find the quotient and remainder when x 4 − 4x 2 + 2x + 5 is dividedby x − 2.Solution:A remainder of 9 and a quotient of x 3 + 2x 2 + 2 .5. Is x = −1 a root of the polynomial function f (x) = 2x 3 − 5x 2 − 4x + 3?Solution: Yes , because f (−1) = 0.6. Factor 27 f (x) = 2x 3 − 5x 2 − 4x + 3.27 Previous problem may be helpful.Page 154 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Using the fact that x = −1 is a root, we can divide f (x) by x + 1, getting2x 3 − 5x 2 − 4x + 3x + 1= 2x 2 − 7x + 3 = (2x − 1) (x − 3) ,so the complete factorization of f (x) is(x + 1) (2x − 1) (x − 3) .7. Find the area of the triangle with vertices (−1, 0), (2, 1) and (1, −2).Solution:A = ± 1 2∣−1 0 12 1 11 −2 1∣ = ±1 [−1 (3) − 0 (1) + 1 (−5)] = ±42The area is 4 square units.8. Solve each inequality, and use interval notation to express the solution.(a) 1 x ≥ 1 x 3Solution:1x ≥ 1 x 31x − 1 ≥ 0x 3x 2 − 1x 3 ≥ 0(x − 1) (x + 1)x 3 ≥ 0Using a number line, you’ll get [−1, 0) ∪ [1, ∞) .(b) (1 − x) 2 ≤ 10 − 2xSolution:Page 155 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes(1 − x) 2 ≤ 10 − 2x1 − 2x + x 2 ≤ 10 − 2xx 2 − 9 ≤ 0(x − 3) (x + 3) ≤ 0Using a number line, you’ll get [−3, 3] .9. Solve for x.(a) log 5 x = 2Solution:log 5 x = 2 ⇔ x = 5 2 ⇒ x = 25(b) log x 64 = 3Solution:log x 64 = 3 ⇔ x 3 = 64 ⇒ x = 4(c) log 319 = xSolution:log 319 = x ⇔ 3x = 1 9⇒x = −210. Factorf (x) = ( x 2 + x + 1 ) ( 6x 2 + 5x − 6 )into linear factors.Solution: You’ll need to use the quadratic formula on x 2 + x + 1 = 0, which gives:x = −1 ± √ −32= −1 ± √ 3i,2so x 2 + x + 1 factors into two linear factors, as follows(x 2 + x + 1 = x − −1 + √ ) (3ix − −1 − √ )3i22= 1 (2x + 1 − √ ) (3i 2x + 1 + √ )3i .4Page 156 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesYou can also use the quadratic formula to factor 6x 2 + 5x − 6, but I think it is easierto use intelligent trial and error, as follows6x 2 + 5x − 6 = (2x + 3) (3x − 2) .Finally, as requested, the complete factorization isf (x) = 1 (2x4 (2x + 3) (3x − 2) + 1 − √ ) (3i 2x + 1 + √ )3i11. A total of $17,500 is invested at an annual rate of 9.3%, compounded weekly. Find thebalance after 35 years.Solution:17500 ·(1 + 0.093 ) (52·35)≈ 452276.4652So, the balance is $452, 276.46 .12. Solve the inequality.3 − 2xx − 1 + 2 ≥ 0Solution:3 − 2xx − 1 + 2 ≥ 03 − 2x + 2 (x − 1)x − 1≥ 03 − 2x + 2x − 2≥ 0x − 11x − 1 ≥ 0Using simple sign analysis, the proper interval is: (1, ∞) .13. Solve for x.log 3 (2x + 1) + log 3 (2x − 1) = 1Page 157 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution:log 3 (2x + 1) + log 3 (2x − 1) = 1log 3 (2x + 1) (2x − 1) = 1log 3(4x 2 − 1 ) = 14x 2 − 1 = 34x 2 − 4 = 0x 2 − 1 = 0x = ±1The solution 28 is x = 1 and you should check this. I will also give full credit forx = ±1.14. Find the product.⎡[ ] 3 1 0· ⎣4 −2 50 1 2−1 −5 −10 1 −1⎤⎦Solution:[ 3 1 04 −2 5⎡]· ⎣0 1 2−1 −5 −10 1 −1⎤⎦ =[ −1 −2 52 19 5]15. Givenf (x) = 2x − 12 − 3x ,find and simplify the difference quotientf (x + h) − f (x), h ≠ 0.hPage 158 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Do I need to say it? Yes, the following is true if and only if h ≠ 0.f (x + h) − f (x)h==2x + 2h − 12 − 3x − 3h + 2x − 12 − 3xh(2x + 2h − 1) (2 − 3x) − (2x − 1) (2 − 3x − 3h)h (2 − 3x − 3h) (2 − 3x)= 4x + 4h − 2 − 6x2 − 6xh + 3x − 4x + 2 + 6x 2 − 3x + 6xh − 3hh (2 − 3x − 3h) (2 − 3x)h=h (2 − 3x − 3h) (2 − 3x)=1(2 − 3x − 3h) (2 − 3x)16. Find the inverse, if possible.⎡1 1⎤1⎣ 3 5 4 ⎦3 6 5Solution:⎡1 1 1 1 0⎤0⎡⎣ 3 5 4 0 1 0 ⎦ ∼ ⎣3 6 5 0 0 1So the inverse is⎡1 1⎤−1⎣ −3 2 −1 ⎦ .3 −3 21 0 0 1 1 −10 1 0 −3 2 −10 0 1 3 −3 2⎤⎦17. Solve the system, any method is fine.⎧⎨ x + y − z = 4−3x + 2y − z = −6⎩3x − 3y + 2z = 5Page 159 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: You may have noticed that you found the inverse of the coefficient matrixin the prior problem. So I’ll solve using the inverse.⎡⎣xyz⎤⎡⎣1 1 −1−3 2 −13 −3 2⎡⎦ = ⎣1 1 13 5 43 6 5⎤⎤⎡⎦ · ⎣⎡⎦ · ⎣xyz4−65⎤⎦ =⎤⎦ =⎡⎣⎡⎣4−65321⎤⎦⎤⎦I get x = 3 , y = 2 and z = 1 which can be easily checked.18. Identify the elementary row operation performed on the original matrix (left) to obtainedthe new row-equivalent matrix (right).[ ] [ ]−2 5 1 13 0 −39∼3 −1 −8 3 −1 −8Solution: By inspection:5R 2 + R 1 → R 1 .19. Find the domain off (x) = x + 2 √2x2 − 18 .Solution: Finding the domain requires solving 2x 2 − 18 > 0 for x.2x 2 − 18 > 0x 2 − 9 > 0(x − 3) (x + 3) > 0Simple sign analysis gives (−∞, −3) ∪ (3, ∞) .20. Givenf (x) = (x + 3) 2 , x ≥ −3,find f −1 (x). Graphing may be helpful, but is not required.Page 160 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: The domain of f (x) is [−3, ∞) and the range is [0, ∞), so the domain off −1 (x) is [0, ∞) and the range is [−3, ∞).f (x) = (x + 3) 2y = (x + 3) 2x = (y + 3) 2± √ x = y + 3± √ x − 3 = yHowever, since y ≥ −3 we havef −1 (x) = √ x − 3, x ≥ 0 .21. Givenf (x) = √ x + 6 and g (x) = x 2 − 5,answer each of the following questions.(a) Find (f ◦ g) (x)Solution:(f ◦ g) (x) = f (g (x)) = f ( x 2 − 5 ) = √ x 2 − 5 + 6 = √ x 2 + 1(b) The domain of (f ◦ g) (x)Solution: R .(c) Find (g ◦ f) (x)Solution:( √x ) ( √x ) 2(g ◦ f) (x) = g (f (x)) = g + 6 = + 6 − 5 = x + 6 − 5 = x + 1(d) The domain of (g ◦ f) (x)Solution: x ≥ −6 .22. Givenf (x) = 2x 4 + 7x 3 − 4x 2 − 27x − 18, and f (2) = f (−3) = 0answer the following questions.Page 161 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes(a) Use the given roots, and long division, to completely factor f (x).Solution: Since we are given two roots, we know two factors of f (x) are (x − 2)and (x + 3). Dividing f (x) by the product of these two factors gives2x 2 + 5x + 3 = (2x + 3) (x + 1) .So the complete factorization of f (x) is(2x + 3) (x + 1) (x − 2) (x + 3)(b) Graph f (x) using the roots, y-intercept and sign-analyses.Solution: Your graph does not need to have such precise detail as mine, but itshould still reflect the key pre-calculus analysis.10-4 -3 -2 -1 0 1 2 3 4-10-20-30-40Figure 73: Graph of f (x) = 2x 4 + 7x 3 − 4x 2 − 27x − 18.23. Givenf (x) = x3 + 2x 2x 2 + 1 ,answer the following questions.(a) x-intercept(s) in point form.Solution: Set f (x) = 0 and solve for x.0 = x3 + 2x 2x 2 + 10 = x2 (x + 2)x 2 + 1Clearly, (−2, 0) and (0, 0) .Page 162 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes(b) y-intercept in point form.Solution: Set x = 0 and evaluate. Clearly, (0, 0) .(c) All linear asymptotes in equation form.Solution: Since the degree of the numerator is one more than the denominator,we have the possibility of getting a slant asymptote. The long division givesf (x) = x3 + 2x 2x 2 + 1 = x + 2 − x + 2x 2 + 1 ,so the slant asymptote is y = x + 2 .(d) Graph f (x) using the information above and sign-analyses.Solution: Your graph does not need to have such precise detail as mine, but itshould still reflect the key pre-calculus analysis.54321-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9-1-2-3-4-5Figure 74: Partial graph of f (x) = x3 + 2x 2x 2 + 1 = x + 2 − x + 2x 2 + 1 .24. Givenf (x) = x 2 − x + 1,find and simplify the difference quotientf (x + h) − f (x), h ≠ 0.hPage 163 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: Do I need to say it? Yes, the following is true if and only if h ≠ 0.f (x + h) − f (x)h= (x + h)2 − (x + h) + 1 − (x 2 − x + 1)h= x2 + 2xh + h 2 − x − h + 1 − x 2 + x − 1h= 2xh + h2 − hh= 2x + h − 125. Solve the inequality.|14 − 3x| − 5 > −2Solution:|14 − 3x| − 5 > −2|14 − 3x| > 3We know that |x| > a is equivalent to x > a or x < −a, so|14 − 3x| > 3 ⇒ 14 − 3x > 3 or 14 − 3x < −3x < 11/3 or 17/3 < xHere the proper intervals are: (−∞, 11/3) ∪ (17/3, ∞) .26. Write the system of linear equations represented by the augmented matrix. Then useback-substitution to find the solution. (Use the variables x, y, and z.)⎡1 −1 2⎤4⎣ 0 1 −1 2 ⎦0 0 1 −2Solution:⎧⎨ x − y + 2z = 4⎩y − z = 2z = −2Page 164 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesThe last line gives z = −2 , then using the value for z in line two, we get y = 0 , finallyusing these two values in line one we get x = 8 .27. Condense the expression to the logarithm of a single quantity.1 [ (2 ln (x + 3) + ln x − ln x 2 − 1 )]3Solution:1 [ (2 ln (x + 3) + ln x − ln x 2 − 1 )] = 1 [ln (x + 3) 2 + ln x − ln ( x 2 − 1 )]33 [= 1 3ln]x (x + 3)2x 2 − 1= ln 3 √x (x + 3) 2x 2 − 128. Given the following functional graph. Answer the following questions.87654321-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10-1-2-3-4-5-6-7-8Figure 75: Graph of f (x).(a) Is f (x) invertible? Explain your answer.Solution:No . It fails the horizontal line test.(b) Graph2 · f(1 − x )− 12Page 165 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: The points translated, in order are{(10, 3) , (6, 5) , (2, −3) , (−6, −5)} .87654321-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10-1-2-3-4-5-6-7Figure 76: Graph of 2 · f-8(1 − x )− 1.229. Evaluate the expression, if possible.⎡ ⎤−1 6 [ ]⎣ −4 5 ⎦ 2 3·0 90 3Solution:⎡ ⎤−1 6⎣ −4 5 ⎦ ·0 3[ 2 30 9]=⎡⎣−2 51−8 330 27⎤⎦30. Sketch the graph of the function.f (x) = 3 x+2Solution: Your graph should look similar to mine.Page 166 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass Notes12.5107.552.5-15 -12.5 -10 -7.5 -5 -2.5 0 2.5-2.5Figure 77: Graph of f (x).31. Solve for x.( ) x 2= 813 16Solution:( ) x 23= 81( 23) x=16( 4 32)( 23) x=( 23) −4So, x = −4 .32. Solve the inequality.x + 6x + 1 − 2 < 0 Page 167 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution:x + 6x + 1 − 2 < 0x + 6 2 (x + 1)−x + 1 x + 1< 04 − xx + 1 < 0Using simple sign analysis, the proper intervals are: (−∞, −1) ∪ (4, ∞) .33. Solve algebraically (exact answer) and then approximate to three decimal places.−2 + 2 ln 3x = 17Solution:−2 + 2 ln 3x = 172 ln 3x = 19ln 3x = 1923x = e 19/2x = e19/23≈ 4, 453.24234. Givenf (x) = x2 − 3x + 22x 2 + 5x + 3answer the following questions:(a) x-intercepts in point form.Solution: (1, 0) ; (2, 0)(b) y-intercept in point form.Solution: (0, 2/3)=(x − 1) (x − 2)(2x + 3) (x + 1) ,(c) Equation of the horizontal asymptote.Page 168 of 169

Pre-calculus IMTH-119Essex County CollegeDivision of MathematicsLMS ProjectClass NotesSolution: y = 1/2(d) Equation of the vertical asymptotes.Solution: x = −1; x = −3/2(e) Graph f (x) using the above information and sign analysis.10-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10-10-20-30-40-50-60-70Figure 78: Graph of f (x).Page 169 of 169