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General Mathematics-Senior Highschool Teaching Guide

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General Mathematics-Senior Highschool Teaching Guide

  1. 1. TEACHING GUIDE FOR SENIOR HIGH SCHOOL General Mathematics CORE SUBJECT This Teaching Guide was collaboratively developed and reviewed by educators from public and private schools, colleges, and universities. We encourage teachers and other education stakeholders to email their feedback, comments, and recommendations to the Commission on Higher Education, K to 12 Transition Program Management Unit - Senior High School Support Team at k12@ched.gov.ph. We value your feedback and recommendations. Commission on Higher Education in collaboration with the Philippine Normal University
  2. 2. This Teaching Guide by the Commission on Higher Education is licensed under a Creative Commons Attribution- NonCommercial-ShareAlike 4.0 International License. This means you are free to: Share — copy and redistribute the material in any medium or format Adapt — remix, transform, and build upon the material. The licensor, CHED, cannot revoke these freedoms as long as you follow the license terms. However, under the following terms: Attribution — You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use. NonCommercial — You may not use the material for commercial purposes. ShareAlike — If you remix, transform, or build upon the material, you must distribute your contributions under the same license as the original. Printed in the Philippines by EC-TEC Commercial, No. 32 St. Louis Compound 7, Baesa, Quezon City, ectec_com@yahoo.com Published by the Commission on Higher Education, 2016
 Chairperson: Patricia B. Licuanan, Ph.D. Commission on Higher Education
 K to 12 Transition Program Management Unit
 Office Address: 4th Floor, Commission on Higher Education, 
 C.P. Garcia Ave., Diliman, Quezon City
 Telefax: (02) 441-1143 / E-mail Address: k12@ched.gov.ph DEVELOPMENT TEAM Team Leader: Debbie Marie B. Verzosa, Ph.D. Writers: Leo Andrei A. Crisologo, Lester C. Hao, 
 Eden Delight P. Miro, Ph.D., Shirlee R. Ocampo, Ph.D., 
 Emellie G. Palomo, Ph.D., Regina M. Tresvalles, Ph.D. Technical Editors: Mark L. Loyola, Ph.D.,
 Christian Chan O. Shio, Ph.D. Copy Reader: Sheena I. Fe Typesetters: Juan Carlo F. Mallari, Regina Paz S. Onglao Illustrator: Ma. Daniella Louise F. Borrero Cover Artists: Paolo Kurtis N. Tan, Renan U. Ortiz CONSULTANTS THIS PROJECT WAS DEVELOPED WITH THE PHILIPPINE NORMAL UNIVERSITY.
 University President: Ester B. Ogena, Ph.D.
 VP for Academics: Ma. Antoinette C. Montealegre, Ph.D.
 VP for University Relations & Advancement: Rosemarievic V. Diaz, Ph.D. Ma. Cynthia Rose B. Bautista, Ph.D., CHED
 Bienvenido F. Nebres, S.J., Ph.D., Ateneo de Manila University
 Carmela C. Oracion, Ph.D., Ateneo de Manila University
 Minella C. Alarcon, Ph.D., CHED
 Gareth Price, Sheffield Hallam University
 Stuart Bevins, Ph.D., Sheffield Hallam University SENIOR HIGH SCHOOL SUPPORT TEAM
 CHED K TO 12 TRANSITION PROGRAM MANAGEMENT UNIT Program Director: Karol Mark R. Yee Lead for Senior High School Support: Gerson M. Abesamis Lead for Policy Advocacy and Communications: Averill M. Pizarro Course Development Officers:
 Danie Son D. Gonzalvo, John Carlo P. Fernando Teacher Training Officers:
 Ma. Theresa C. Carlos, Mylene E. Dones Monitoring and Evaluation Officer: Robert Adrian N. Daulat Administrative Officers: Ma. Leana Paula B. Bato, 
 Kevin Ross D. Nera, Allison A. Danao, Ayhen Loisse B. Dalena
  3. 3. Contents Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . i DepEd General Mathematics Curriculum Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii Chapter 1 Functions Lesson 1: Functions as Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Lesson 2: Evaluating Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Lesson 3: Operations on Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Chapter 2 Rational Functions Lesson 4: Representing Real-Life Situations Using Rational Functions . . . . . . . . . . . . . . 23 Lesson 5: Rational Functions, Equations, and Inequalities . . . . . . . . . . . . . . . . . . . . . . . . 28 Lesson 6: Solving Rational Equations and Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 Lesson 7: Representations of Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 Lesson 8: Graphing Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 Chapter 3 One-to-One and Inverse Functions Lesson 9: One-to-One Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 Lesson 10: Inverse of One-to-One Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 Lesson 11: Graphs of Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 Chapter 4 Exponential Functions Lesson 12: Representing Real-Life Situations Using Exponential Functions . . . . . . . . . . 88 Lesson 13: Exponential Functions, Equations, and Inequalities . . . . . . . . . . . . . . . . . . . . 94 Lesson 14: Solving Exponential Equations and Inequalities . . . . . . . . . . . . . . . . . . . . . . . 96 Lesson 15: Graphing Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 Lesson 16: Graphing Transformations of Exponential Functions . . . . . . . . . . . . . . . . . . . 107 Chapter 5 Logarithmic Functions Lesson 17: Introduction to Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 Lesson 18: Logarithmic Functions, Equations, and Inequalities . . . . . . . . . . . . . . . . . . . . 125
  4. 4. Lesson 19: Basic Properties of Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 Lesson 20: Laws of Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 Lesson 21: Solving Logarithmic Equations and Inequalities . . . . . . . . . . . . . . . . . . . . . . . 136 Lesson 22: The Logarithmic Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 Chapter 6 Simple and Compound Interest Lesson 23: Illustrating Simple and Compound Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 Lesson 24: Simple Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 Lesson 25: Compound Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 Lesson 26: Compounding More than Once a Year . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 Lesson 27: Finding Interest Rate and Time in Compound Interest . . . . . . . . . . . . . . . . . 185 Chapter 7 Annuities Lesson 28: Simple Annuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 Lesson 29: General Annuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 Lesson 30: Deferred Annuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224 Chapter 8 Basic Concepts of Stocks and Bonds Lesson 31: Stocks and Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234 Lesson 32: Market Indices for Stocks and Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 Lesson 33: Theory of Efficient Markets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 Chapter 9 Basic Concepts of Loans Lesson 34: Business Loans and Consumer Loans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 Lesson 35: Solving Problems on Business and Consumer Loans (Amortization and Mortgage) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252 Chapter 10 Logic Lesson 36: Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 Lesson 37: Logical Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268 Lesson 38: Constructing Truth Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280 Lesson 39: Logical Equivalence and Forms of Conditional Propositions . . . . . . . . . . . . . . 285 Lesson 40: Valid Arguments and Fallacies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 Lesson 41: Methods of Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306 Biographical Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319
  5. 5. i Introduction As the Commission supports DepEd’s implementation of Senior High School (SHS), it upholds the vision and mission of the K to 12 program, stated in Section 2 of Republic Act 10533, or the Enhanced Basic Education Act of 2013, that “every graduate of basic education be an empowered individual, through a program rooted on...the competence to engage in work and be productive, the ability to coexist in fruitful harmony with local and global communities, the capability to engage in creative and critical thinking, and the capacity and willingness to transform others and oneself.” To accomplish this, the Commission partnered with the Philippine Normal University (PNU), the National Center for Teacher Education, to develop Teaching Guides for Courses of SHS. Together with PNU, this Teaching Guide was studied and reviewed by education and pedagogy experts, and was enhanced with appropriate methodologies and strategies. Furthermore, the Commission believes that teachers are the most important partners in attaining this goal. Incorporated in this Teaching Guide is a framework that will guide them in creating lessons and assessment tools, support them in facilitating activities and questions, and assist them towards deeper content areas and competencies. Thus, the introduction of the SHS for SHS Framework. The SHS for SHS Framework The SHS for SHS Framework, which stands for “Saysay-Husay-Sarili for Senior High School,” is at the core of this book. The lessons, which combine high-quality content with flexible elements to accommodate diversity of teachers and environments, promote these three fundamental concepts: SAYSAY: MEANING Why is this important? Through this Teaching Guide, teachers will be able to facilitate an understanding of the value of the lessons, for each learner to fully engage in the content on both the cognitive and affective levels. HUSAY: MASTERY How will I deeply understand this? Given that developing mastery goes beyond memorization, teachers should also aim for deep understanding of the subject matter where they lead learners to analyze and synthesize knowledge. SARILI: OWNERSHIP What can I do with this? When teachers empower learners to take ownership of their learning, they develop independence and self- direction, learning about both the subject matter and themselves.
  6. 6. ii The Parts of the Teaching Guide This Teaching Guide is mapped and aligned to the DepEd SHS Curriculum, designed to be highly usable for teachers. It contains classroom activities and pedagogical notes, and integrated with innovative pedagogies. All of these elements are presented in the following parts: 1. INTRODUCTION • Highlight key concepts and identify the essential questions • Show the big picture • Connect and/or review prerequisite knowledge • Clearly communicate learning competencies and objectives • Motivate through applications and connections to real-life 2. MOTIVATION • Give local examples and applications • Engage in a game or movement activity • Provide a hands-on/laboratory activity • Connect to a real-life problem 3. INSTRUCTION/DELIVERY • Give a demonstration/lecture/simulation/hands-on activity • Show step-by-step solutions to sample problems • Give applications of the theory • Connect to a real-life problem if applicable 4. PRACTICE • Provide easy-medium-hard questions • Give time for hands-on unguided classroom work and discovery • Use formative assessment to give feedback 5. ENRICHMENT • Provide additional examples and applications • Introduce extensions or generalisations of concepts • Engage in reflection questions • Encourage analysis through higher order thinking prompts • Allow pair/small group discussions • Summarize and synthesize the learnings 6. EVALUATION • Supply a diverse question bank for written work and exercises • Provide alternative formats for student work: written homework, journal, portfolio, group/ individual projects, student-directed research project
  7. 7. iii On DepEd Functional Skills and CHED’s College Readiness Standards As Higher Education Institutions (HEIs) welcome the graduates of the Senior High School program, it is of paramount importance to align Functional Skills set by DepEd with the College Readiness Standards stated by CHED. The DepEd articulated a set of 21st century skills that should be embedded in the SHS curriculum across various subjects and tracks. These skills are desired outcomes that K to 12 graduates should possess in order to proceed to either higher education, employment, entrepreneurship, or middle-level skills development. On the other hand, the Commission declared the College Readiness Standards that consist of the combination of knowledge, skills, and reflective thinking necessary to participate and succeed - without remediation - in entry-level undergraduate courses in college. The alignment of both standards, shown below, is also presented in this Teaching Guide - prepares Senior High School graduates to the revised college curriculum which will initially be implemented by AY 2018-2019. College Readiness Standards Foundational Skills DepEd Functional Skills Produce all forms of texts (written, oral, visual, digital) based on: 1. Solid grounding on Philippine experience and culture; 2. An understanding of the self, community, and nation; 3. Application of critical and creative thinking and doing processes; 4. Competency in formulating ideas/arguments logically, scientifically, and creatively; and 5. Clear appreciation of one’s responsibility as a citizen of a multicultural Philippines and a diverse world; Visual and information literacies Media literacy Critical thinking and problem solving skills Creativity Initiative and self-direction Systematically apply knowledge, understanding, theory, and skills for the development of the self, local, and global communities using prior learning, inquiry, and experimentation Global awareness Scientific and economic literacy Curiosity Critical thinking and problem solving skills Risk taking Flexibility and adaptability Initiative and self-direction Work comfortably with relevant technologies and develop adaptations and innovations for significant use in local and global communities; Global awareness Media literacy Technological literacy Creativity Flexibility and adaptability Productivity and accountability Communicate with local and global communities with proficiency, orally, in writing, and through new technologies of communication; Global awareness Multicultural literacy Collaboration and interpersonal skills Social and cross-cultural skills Leadership and responsibility Interact meaningfully in a social setting and contribute to the fulfilment of individual and shared goals, respecting the fundamental humanity of all persons and the diversity of groups and communities Media literacy Multicultural literacy Global awareness Collaboration and interpersonal skills Social and cross-cultural skills Leadership and responsibility Ethical, moral, and spiritual values
  8. 8. iv The General Mathematics Teaching Guide Implementing a new curriculum is always subject to a new set of challenges. References are not always available, and training may be too short to cover all the required topics. Under these circumstances, providing teachers with quality resource materials aligned with the curricular competencies may be the best strategy for delivering the expected learning outcomes. Such is the rationale for creating a series of teaching guides for several Grade 11 and 12 subjects. The intention is to provide teachers a complete resource that addresses all expected learning competencies, as stated in the Department of Education’s offi︎cial curriculum guide. This resource is a teaching guide for General Mathematics. The structure is quite unique, re︎flective of the wide scope of General Mathematics: functions, business mathematics, and logic. Each lesson begins with an introductory or motivational activity. The main part of the lesson presents important ideas and provides several solved examples. Explanations to basic properties, the rationale for mathematical procedures, and the derivation of important formulas are also provided. The goal is to enable teachers to move learners away from regurgitating information and towards an authentic understanding of, and appreciation for, the subject matter. The chapters on functions are an extension of the functions learned in Junior High School, where the focus was primarily on linear, quadratic, and polynomial functions. In Grade 11, learners will be exposed to other types of functions such as piecewise, rational, exponential, and logarithmic functions. Related topics such as solving equations and inequalities, as well as identifying the domain, range, intercepts, and asymptotes are also included.
  9. 9. v The chapters on business mathematics in Grade 11 may be learners' ︎first opportunity to be exposed to topics related to ︎financial literacy. Here, they learn about simple and compound interest, annuities, loans, stocks, and bonds. These lessons can hopefully prepare learners to analyze business-related problems and make sound ︎financial decisions. The ︎final chapter on logic exposes learners to symbolic forms of propositions (or statements) and arguments. Through the use of symbolic logic, learners should be able to recognize equivalent propositions, identify fallacies, and judge the validity of arguments. The culminating lesson is an application of the rules of symbolic logic, as learners are taught to write their own justifications to mathematical and real-life statements. This Teaching Guide is intended to be a practical resource for teachers. It includes activities, explanations, and assessment tools. While the beginning teacher may use this Teaching Guide as a “script,” more experienced teachers can use this resource as a starting point for writing their own lesson plans. In any case, it is hoped that this resource, together with the Teaching Guide for other subjects, can support teachers in achieving the vision of the K to 12 Program.
  10. 10. vi Hour 1 Hour 2 Hour 3 Hour 4 Week a Lesson 1 Lesson 1, 2 Lesson 3 Lesson 3 Week b Lesson 4 Lesson 5, 6 Lesson 6 Lesson 7 Week c Lesson 7 Lesson 8 Lesson 8 Review/Exam Week d Lesson 9 Lesson 10 Lesson 10 Lesson 11 Week e Lesson 11 Review/Exam Lesson 12 Lesson 12, 13 Week f Lesson 14 Lesson 14 Lesson 15 Lesson 15 Week g Lesson 16 Lesson 16 Review/Exam Review/Exam Week h Lesson 17 Lesson 17 Lesson 18, 19 Lesson 19, 20 Week i Lesson 20 Lesson 21 Lesson 21 Lesson 21 Week j Lesson 22 Lesson 22 Review/Exam Review/Exam First Quarter Hour 1 Hour 2 Hour 3 Hour 4 Week a Lesson 23 Lesson 24 Lesson 25 Lesson 25, 26 Week b Lesson 26 Lesson 27 Lesson 27 / Review Review/Exam Week c Lesson 28 Lesson 28 Lesson 29 Lesson 29 Week d Lesson 29 Lesson 30 Lesson 30 Review/Exam Week e Lesson 31 Lesson 31 Lesson 32 Lesson 33 Week f Lesson 34 Lesson 35 Lesson 35 Review/Exam Week g Lesson 36 Lesson 36 Lesson 37 Lesson 37 Week h Lesson 38 Lesson 39 Lesson 39 Lesson 39 Week i Lesson 40 Lesson 40 Lesson 40 Lesson 41 Week j Lesson 41 Lesson 41 Review/Exam Review/Exam Second Quarter
  11. 11. K to 12 BASIC EDUCATION CURRICULUM SENIOR HIGH SCHOOL – CORE SUBJECT K to 12 Senior High School Core Curriculum – General Mathematics December 2013 Page 1 of 5 Grade: 11 Semester: First Semester Core Subject Title: General Mathematics No. of Hours/Semester: 80 hours/semester Prerequisite (if needed): Core Subject Description: At the end of the course, the students must know how to solve problems involving rational, exponential and logarithmic functions; to solve business-related problems; and to apply logic to real-life situations. CONTENT CONTENT STANDARDS PERFORMANCE STANDARDS LEARNING COMPETENCIES CODE Functions and Their Graphs The learner demonstrates understanding of... 1. key concepts of functions. The learner is able to... 1. accurately construct mathematical models to represent real-life situations using functions. The learner... 1. represents real-life situations using functions, including piece-wise functions. M11GM-Ia-1 2. evaluates a function. M11GM-Ia-2 3. performs addition, subtraction, multiplication, division, and composition of functions M11GM-Ia-3 4. solves problems involving functions. M11GM-Ia-4 2. key concepts of rational functions. 2. accurately formulate and solve real-life problems involving rational functions. 5. represents real-life situations using rational functions. M11GM-Ib-1 6. distinguishes rational function, rational equation, and rational inequality. M11GM-Ib-2 7. solves rational equations and inequalities. M11GM-Ib-3 8. represents a rational function through its: (a) table of values, (b) graph, and (c) equation. M11GM-Ib-4 9. finds the domain and range of a rational function. M11GM-Ib-5 10. determines the: (a) intercepts (b) zeroes; and (c) asymptotes of rational functions M11GM-Ic-1 11. graphs rational functions. M11GM-Ic-2 12. solves problems involving rational functions, equations, and inequalities. M11GM-Ic-3
  12. 12. K to 12 BASIC EDUCATION CURRICULUM SENIOR HIGH SCHOOL – CORE SUBJECT K to 12 Senior High School Core Curriculum – General Mathematics December 2013 Page 2 of 5 CONTENT CONTENT STANDARDS PERFORMANCE STANDARDS LEARNING COMPETENCIES CODE 3. key concepts of inverse functions, exponential functions, and logarithmic functions. 3. apply the concepts of inverse functions, exponential functions, and logarithmic functions to formulate and solve real-life problems with precision and accuracy. 1. represents real-life situations using one-to one functions. M11GM-Id-1 2. determines the inverse of a one-to-one function. M11GM-Id-2 3. represents an inverse function through its: (a) table of values, and (b) graph. M11GM-Id-3 4. finds the domain and range of an inverse function. M11GM-Id-4 5. graphs inverse functions. M11GM-Ie-1 6. solves problems involving inverse functions. M11GM-Ie-2 7. represents real-life situations using exponential functions. M11GM-Ie-3 8. distinguishes between exponential function, exponential equation, and exponential inequality. M11GM-Ie-4 9. solves exponential equations and inequalities. M11GM-Ie-f-1 10. represents an exponential function through its: (a) table of values, (b) graph, and (c) equation. M11GM-If-2 11. finds the domain and range of an exponential function. M11GM-If-3 12. determines the intercepts, zeroes, and asymptotes of an exponential function. M11GM-If-4 13. graphs exponential functions. M11GM-Ig-1 14. solves problems involving exponential functions, equations, and inequalities. M11GM-Ig-2 15. represents real-life situations using logarithmic functions. M11GM-Ih-1 16. distinguishes logarithmic function, logarithmic equation, and logarithmic inequality. M11GM-Ih-2 17. illustrates the laws of logarithms. M11GM-Ih-3 18. solves logarithmic equations and inequalities. M11GM-Ih-i-1 19. represents a logarithmic function through its: (a) table of values, (b) graph, and (c) equation. M11GM-Ii-2 20. finds the domain and range of a logarithmic function. M11GM-Ii-3 21. determines the intercepts, zeroes, and asymptotes of logarithmic functions. M11GM-Ii-4 22. graphs logarithmic functions. M11GM-Ij-1 23. solves problems involving logarithmic functions, equations, and inequalities. M11GM-Ij-2
  13. 13. K to 12 BASIC EDUCATION CURRICULUM SENIOR HIGH SCHOOL – CORE SUBJECT K to 12 Senior High School Core Curriculum – General Mathematics December 2013 Page 3 of 5 CONTENT CONTENT STANDARDS PERFORMANCE STANDARDS LEARNING COMPETENCIES CODE Basic Business Mathematics The learner demonstrates understanding of... 1. key concepts of simple and compound interests, and simple and general annuities. The learner is able to... 1. investigate, analyze and solve problems involving simple and compound interests and simple and general annuities using appropriate business and financial instruments. 24. illustrates simple and compound interests. M11GM-IIa-1 25. distinguishes between simple and compound interests. M11GM-IIa-2 26. computes interest, maturity value, future value, and present value in simple interest and compound interest environment. M11GM-IIa-b-1 27. solves problems involving simple and compound interests. M11GM-IIb-2 28. illustrates simple and general annuities. M11GM-IIc-1 29. distinguishes between simple and general annuities. M11GM-IIc-2 30. finds the future value and present value of both simple annuities and general annuities. M11GM-IIc-d-1 31. calculates the fair market value of a cash flow stream that includes an annuity. M11GM-IId-2 32. calculates the present value and period of deferral of a deferred annuity. M11GM-IId-3 2. basic concepts of stocks and bonds. 2. use appropriate financial instruments involving stocks and bonds in formulating conclusions and making decisions. 33. illustrate stocks and bonds. M11GM-IIe-1 34. distinguishes between stocks and bonds. M11GM-IIe-2 35. describes the different markets for stocks and bonds. M11GM-IIe-3 36. analyzes the different market indices for stocks and bonds. M11GM-IIe-4 37. interprets the theory of efficient markets. M11GM-IIe-5 3. basic concepts of business and consumer loans. 3. decide wisely on the appropriateness of business or consumer loan and its proper utilization. 38. illustrates business and consumer loans. M11GM-IIf-1 39. distinguishes between business and consumer loans. M11GM-IIf-2 40. solves problems involving business and consumer loans (amortization, mortgage). M11GM-IIf-3 Logic The learner demonstrates understanding of... 1. key concepts of propositional logic; syllogisms and fallacies. The learner is able to... 1. judiciously apply logic in real-life arguments. 41. illustrates a proposition. M11GM-IIg-1 42. symbolizes propositions. M11GM-IIg-2 43. distinguishes between simple and compound propositions. M11GM-IIg-3 44. performs the different types of operations on propositions. M11GM-IIg-4 45. determines the truth values of propositions. M11GM-IIh-1 46. illustrates the different forms of conditional propositions. M11GM-IIh-2 47. illustrates different types of tautologies and fallacies. M11GM-IIi-1
  14. 14. K to 12 BASIC EDUCATION CURRICULUM SENIOR HIGH SCHOOL – CORE SUBJECT K to 12 Senior High School Core Curriculum – General Mathematics December 2013 Page 4 of 5 CONTENT CONTENT STANDARDS PERFORMANCE STANDARDS LEARNING COMPETENCIES CODE 48. determines the validity of categorical syllogisms. M11GM-IIi-2 49.establishes the validity and falsity of real-life arguments using logical propositions, syllogisms, and fallacies. M11GM-IIi-3 2. key methods of proof and disproof. 2. appropriately apply a method of proof and disproof in real-life situations. 50. illustrates the different methods of proof (direct and indirect) and disproof (indirect and by counterexample). M11GM-IIj-1 51. justifies mathematical and real-life statements using the different methods of proof and disproof. M11GM-IIj-2
  15. 15. (x, y) (x, y) x y
  16. 16. f = {(1, 2), (2, 2), (3, 5), (4, 5)} g = {(1, 3), (1, 4), (2, 5), (2, 6), (3, 7)} h = {(1, 3), (2, 6), (3, 9), . . . , (n, 3n), . . .} f h x y g (1, 3) (1, 4) x y f g
  17. 17. h f g x 2 X y 2 Y h X y x = 7 y = 11 13 x = 2 y = 17 19 x = a (a, b) (a, c) y x = a
  18. 18. (a) (b) (c) (d) (e) x y x x y y = 2x + 1 y = x2 2x + 2 x2 + y2 = 1 y = p x + 1 y = 2x + 1 x 1
  19. 19. y = bxc + 1 bxc x y x = 0 y +1 1 x R R [ 1, 1] [ 1, +1) ( 1, 1) [ (1, +1) R y f(x) y x f f(x) = 2x + 1 q(x) = x2 2x + 2 g(x) = p x + 1 r(x) = 2x + 1 x 1 F(x) = bxc + 1 bxc C x 40 40 C(x) = 40x
  20. 20. A x A = xy x x + 2y = 100 y = (100 x)/2 = 50 0.5x A(x) = x(50 0.5x) = 50x 0.5x2 300 1 m t(m) t(m) = ( 300 0 < m  100 300 + m m > 100 8.00 1.50 d
  21. 21. F(d) F(d) = ( 8 0 < d  4 8 + 1.5bdc d > 4 bdc d d b4.1c = b4.9c = 4 • 25 • 0 • 0 • 100 • 100 • 100 T(x) T(x)
  22. 22. 15 1, 000 400 f(x) = 8 < : 1000 0  x  3 1000 + 400dx 3e x > 3 700
  23. 23. f(x) = 700dx 4 e x 2 N 150 130 110 100 f(x) = 8 >>>>>>< >>>>>>: 150x 0  x  20 130x 21  x  50 110x 51  x  100 100x x > 100 x 2 N
  24. 24. x f a a f f(a) x = 1.5 f(x) = 2x + 1 q(x) = x2 2x + 2 g(x) = p x + 1 r(x) = 2x + 1 x 1 F(x) = bxc + 1 bxc 1.5 x f(1.5) = 2(1.5) + 1 = 4 q(1.5) = (1.5)2 2(1.5) + 2 = 2.25 3 + 2 = 1.25 g(1.5) = p 1.5 + 1 = p 2.5 r(1.5) = 2x + 1 x 1 = 2(1.5) + 1 (1.5) 1 = 3 + 1 0.5 = 8 F(1.5) = bxc + 1 = b1.5c + 1 = 1 + 1 = 2 g( 4) r(1) g r 4 g(x) r(x) f q f(3x 1) q(2x + 3)
  25. 25. f(3x 1) x f(x) = 2x + 1 (3x 1) f(3x 1) = 2(3x 1) + 1 = 6x 2 + 1 = 6x 1 q(3x + 3) x q(x) = x2 2x + 2 (2x + 3) q(2x + 3) = (2x + 3)2 2(2x + 3) + 2 = (4x2 + 12x + 9) 4x 6 + 2 = 4x2 + 8x + 5 f(x) = x 2 f(0) 2 f(3) f( 1) f(⇡) ⇡ 2 f(x + 1) x 1 f(3x) 3x 2 f(x) = 4 x f(1) 4 f(2) 2 f( 1) 4 f( p 2) 2 p 2 f(1/x) 4x f(2x) 2/x f(x) = p x 3 f(3) 0 f(4) 1 f(12) 3 f(x 3) p x 6 f ✓ 1 1 x ◆ 3x 2 1 x f(x2 + 4x + 7) p x2 + 4x + 4 |x + 2|
  26. 26. 200 25 C(x) = 25x + 200 x C(x) 2700 3950 t s(t) = 5t2 + 100 g = 10m/s2
  27. 27. 1 3 2 5 1 3 + 2 5 = 5 15 + 6 15 = 5 + 6 15 = 11 15 1 x 3 2 x 5 (x 3)(x 5) (x2 8x + 15) 1 x 3 + 2 x 5 = x 5 x2 8x + 15 + 2(x 3) x2 8x + 15 = x 5 + 2x 6 x2 8x + 15 = 3x 11 x2 8x + 15 10 21 15 8 10 21 · 15 8 = 2 · 5 3 · 7 · 3 · 5 2 · 2 · 2 = 6 2 · 5· 6 3 · 5 6 3 · 7· 6 2 · 2 · 2 = 25 28
  28. 28. x2 4x 5 x2 3x + 2 x2 5x + 6 x2 3x 10 x2 4x 5 x2 3x + 2 · x2 5x + 6 x2 3x 10 = (x + 1)(x 5) (x 2)(x 1) · (x 2)(x 3) (x 5)(x + 2) = (x + 1)⇠⇠⇠⇠(x 5)⇠⇠⇠⇠(x 2)(x 3) ⇠⇠⇠⇠(x 2)(x 1)⇠⇠⇠⇠(x 5)(x + 2) = (x + 1)(x 3) (x 1)(x + 2) = x2 2x 3 x2 + x 2 f g f + g (f + g)(x) = f(x) + g(x) f g (f g)(x) = f(x) g(x) f · g (f · g)(x) = f(x) · g(x) f/g (f/g)(x) = f(x)/g(x) x g(x) = 0
  29. 29. • f(x) = x + 3 • p(x) = 2x 7 • v(x) = x2 + 5x + 4 • g(x) = x2 + 2x 8 • h(x) = x + 7 2 x • t(x) = x 2 x + 3 (v + g)(x) (f · p)(x) (f + h)(x) (p f)(x) (v/g)(x) (v + g)(x) = x2 + 5x + 4 + x2 + 2x 8 = 2x2 + 7x 4 (f · p)(x) = (x + 3) (2x 7) = 2x2 x 21 (f + h)(x) = (x + 3) + x + 7 2 x = (x + 3) · 2 x 2 x + x + 7 2 x = (x + 3)(2 x) + (x + 7) 2 x = = 6 x x2 + x + 7 2 x = 13 x2 2 x = 13 x2 2 x · 1 1 = x2 13 x 2 (p f)(x) = (2x 7) (x + 3) = 2x 7 x 3 = x 10 (v/g)(x) = (x2 + 5x + 4) ÷ (x2 + 2x 8) = x2 + 5x + 4 x2 + 2x 8 • f(x) = 2x + 1 • q(x) = x2 2x + 2 • r(x) = 2x + 1 x 1 f1(x) = x2 + 3 q(x) f(x) x2 + 3 q(x) + f(x) = (x2 2x + 2) + (2x + 1) = x2 + 3 = f1(x) f1(x) = q(x) + f(x) f2(x) = x2 4x+1
  30. 30. q(x) f(x) x2 4x + 1 q(x) f(x) = (x2 2x + 2) (2x + 1) = x2 4x + 1 = f2(x) f2(x) = q(x) f(x) f3(x) = 2x2 + x x 1 2x2 + x x 1 x 1 r(x) = 2x + 1 x 1 f(x) r(x) f(x) + r(x) = 2x + 1 + 2x + 1 x 1 = (2x + 1)(x 1) x 1 + 2x + 1 x 1 = (2x + 1)(x 1) + (2x + 1) x 1 = (2x2 x 1) + (2x + 1) x 1 = 2x2 + x x 1 = f3(x) (f + g)(x) = f(x) + g(x) f1(x) = q(x) + f(x) = (q + f)(x) f2(x) = q(x) f(x) = (q f)(x) f3(x) = f(x) + r(x) = (f + r)(x) g1(x) = 2x3 3x2 + 2x + 2
  31. 31. 2x3 3x2 + 2x + 2 f(x) q(x) f(x) · q(x) = (2x + 1)(x2 2x + 2) = (2x)(x2 2x + 2) + (x2 2x + 2) = (2x3 4x2 + 4x) + (x2 2x + 2) = 2x3 3x2 + 2x + 2 = g1(x) g1(x) = f(x) · q(x) g2(x) = x 1 r(x) = 2x + 1 x 1 x 1 2x + 1 f(x) r(x) f(x) r(x) = (2x + 1) ÷ 2x + 1 x 1 = (2x + 1) · x 1 2x + 1 = 2x + 1 2x + 1 · (x 1) = x 1 = g2(x) g2(x) = f(x) r(x) g3(x) = 1 x 1 g3(x) = 1 x 1 r(x) = 2x + 1 x 1 2x + 1 r(x) f(x) = 2x + 1 r(x) f(x) = 2x + 1 x 1 ÷ (2x + 1) = 2x + 1 x 1 · 1 2x + 1 = 1 x 1 = g3(x) g3(x) = r(x) f(x)
  32. 32. f(x) = 2x + 1 q(x) = x2 2x + 2 (2x + 1)2 2(2x + 1) + 2 f g (f g) (f g)(x) = f(g(x)). • f(x) = 2x + 1 • q(x) = x2 2x + 2 • r(x) = 2x + 1 x 1 • g(x) = p x + 1 • F(x) = bxc + 1 (g f)(x) (g f)(x) = g(f(x)) = p f(x) + 1 = p (2x + 1) + 1 = p 2x + 2 (q f)(x) (f q)(x)
  33. 33. (q f)(x) = q(f(x)) = [f(x)]2 2 [f(x)] + 2 = (2x + 1)2 2(2x + 1) + 2 = (4x2 + 4x + 1) (4x + 2) + 2 = 4x2 + 1 (f q)(x) = f(q(x)) = 2(x2 2x + 2) + 1 = 2x2 4x + 5 (q f)(x) (f q)(x) (f r)(x) (f r)(x) = f(r(x)) = 2r(x) + 1 = 2  2x + 1 x 1 + 1 = 4x + 2 x 1 + 1 = (4x + 2) + (x 1) x 1 = 5x + 1 x 1 (F r)(5) (F r)(5) = F(r(5)) = br(5)c + 1 = 2(5) + 1 5 1 ⌫ + 1 = 11 4 ⌫ + 1 = 2 + 1 = 3
  34. 34. f g f + g f g f · g f/g g/f f(x) = x + 2 g(x) = x2 4 x2 + x 2 x2 + x + 6 x3 + 2x2 4x 8 1 x 2 x 2 f(x) = p x 1 g(x) = x2 + 4 p x 1 + x2 + 4 p x 1 x2 4 p x 1(x2 + 4) p x 1 x2 + 4 x 2 f(x) = x 2 x + 2 g(x) = 1 x x 2 x + 2 + 1 x x 2 x + 2 1 x x 2 x(x + 2) x(x 2) x + 2 x + 2 x(x 2) f(x) = 1 x + 2 g(x) = x 2 x 1 x + 2 + x 2 x 1 x + 2 x 2 x x 2 x(x + 2) x (x + 2)(x 2) (x + 2)(x 2) x f(x) = 1 x2 g(x) = p x 1 x2 + p x 1 x2 p x p x x2 1 x2 p x x2p x f(x) = x2 + 3x g(x) = x 2 f g g f f f g g x2 x 2 x2 + 3x 2 x4 + 6x3 + 12x2 + 9x x 4 (f g) = x (g f) = x f(x) = 3x 2 g(x) = 1 3 (x + 2) f(x) = x 2 x g(x) = 2x x 1 f(x) = (x 1)3 + 2 g(x) = 3 p x 2 + 1
  35. 35. p(x) = anxn + an 1xn 1 + an 2xn 2 + · · · + a1x + a0 a0, a1, . . . , an 2 R an 6= 0 n a0, a1, a2, . . . , an an anxn a0 100, 000 y x y = 100, 000 x x y 750 g(x) g(x) = 100, 000 x + 750 x y
  36. 36. f(x) = p(x) q(x) p(x) q(x) q(x) q(x) 6⌘ 0 f(x) x q(x) 6= 0 v v(t) t v t t v v(t) = 10 t v t c(t) = 5t t2 + 1 t c(t) t = 1, 2, 5, 10 t c(t)
  37. 37. • • • • • a • b • c • x • y
  38. 38. x y y = a x b + c x P(x)
  39. 39. P(x) = 2x2 + 800 x b(t) = 50t t + 1 0  t  20 t b(t) t = 1, 2, 5, 10, 15, 20
  40. 40. x2 + 3x + 2 x + 4 1 3x2 x2 + 4x 3 2 p x + 1 x3 1 1 x + 2 x 2 1 (x + 2)(x 2) f(x) = p(x) q(x) p(x) q(x) q(x) 2 x 3 2x = 1 5 5 x 3  2 x f(x) = x2 + 2x + 3 x + 1 y = x2 + 2x + 3 x + 1 x x y
  41. 41. 15 p x 1 5x4 6x7 + 1 5 x3 x y = 5x3 2x + 1 8 x 8 = x 2x 1 p x 2 = 4 x 1 x + 1 = x3 y = 7x3 4 p x + 1 x2 + 3 6x 5 x + 3 0
  42. 42. x + 1 2x = 10 x + 1 2x  10 x 2 x 3 2x = 1 5 10x 10x
  43. 43. 10x ✓ 2 x ◆ 10x ✓ 3 2x ◆ = 10x ✓ 1 5 ◆ 20 15 = 2x 5 = 2x x = 5 2 x x x + 2 1 x 2 = 8 x2 4 x x + 2 1 x 2 = 8 (x 2)(x + 2) (x 2)(x+2) (x 2)(x + 2) · x x + 2 (x 2)(x + 2) · 1 x 2 = [(x 2)(x + 2)] ✓ 8 (x 2)(x + 2) ◆ (x 2)x (x + 2) = 8 x2 3x 10 = 0 x2 3x 10 = 0 (x + 2)(x 5) = 0 x + 2 = 0 x 5 = 0 x = 2 x = 5 x = 2 x = 5
  44. 44. x x 12 + x 25 + x 12 + x 25 + x = 0.6 25 + x 12 + x 25 + x = 0.6 12 + x = 0.6(25 + x) 12 + x = 0.6(25) + 0.6x x 0.6x = 15 12 0.4x = 3 x = 7.5 x 60% v = d t v = d t t = d v v v + 10 5 v 5 v+10 4 3 5 v + 5 v + 10 = 4 3 3v(v + 10) 5 v + 5 v + 10 = 4 3
  45. 45. 3v(v + 10) · 5 v + 3v(v + 10) · 5 v + 10 = 3v(v + 10) · 4 3 15(v + 10) + 15v = 4v(v + 10) 30v + 150 = 4v2 + 40v 4v2 + 10v 150 = 0 2v2 + 5v 75 = 0 (2v + 15)(v 5) = 0 v = 15 2 v = 5 v (a, b) {x|a < x < b} a b [a, b] {x|a  x  b} a b [a, b) {x|a  x < b} a b (a, b] {x|a < x  b} a b (a, 1) {x|a < x} a [a, 1) {x|a  x} a ( 1, b) {x|x < b} b ( 1, b] {x|x  b} b ( 1, 1) R
  46. 46. x • • 2x x + 1 1 2x x + 1 1 0 2x (x + 1) x + 1 0 x 1 x + 1 0 x = 1 x = 1 x = 1 x = 1 1
  47. 47. 1 1 x 1 x + 1 x < 1 1 < x < 1 x > 1 x = 2 x = 0 x = 2 x 1 + x + 1 + + x 1 x + 1 + + x < 1 x 1 1 {x 2 R|x < 1 x 1} ( 1, 1) [ [1, 1) 3 x 2 < 1 x 3 x 2 1 x < 0 3x (x 2) x(x 2) < 0 2x + 2 x(x 2) < 0 2(x + 1) x(x 2) < 0 x = 1 0 1
  48. 48. x < 1 1 < x < 0 0 < x < 2 x > 2 x = 2 x = 1 2 x = 1 x = 3 2(x + 1) + + + x + + x 2 + 2(x + 1) x(x 2) + + {x 2 R|x < 1 0 < x < 2} x h x h h x 8 = x2 h h x h = 8 x2 h > x 8 x2 > x
  49. 49. 8 x2 > x 8 x2 x > 0 8 x3 x2 > 0 (2 x)(x2 + 2x + 4) x2 > 0 x = 16 x = 28 x = 0 x = 4 4 x < 0 0 < x < 2 x > 2 x = 1 x = 1 x = 3 2 x + + x2 + 2x + 4 + + + x2 + + + (2 x)(x2+2x+4) x2 + + 0 < x < 2 x < 0 x x
  50. 50. 1120 x 1600 x + 4 1600 x + 4 1120 x 10 1600 x + 4 1120 x 10 160 x + 4 112 x 1 160 x + 4 112 x 1 0 160x 112(x + 4) (x2 + 4x) x(x + 4) 0 160x 112x 448 x2 4x x(x + 4) 0 x2 44x + 448 x(x + 4)  0 (x 16)(x 28) x(x + 4)  0 x = 16 x = 28 x = 0 x = 4 x < 4 4 < x < 0 0 < x < 16 16 < x < 28 x > 28 x = 5 x = 1 x = 10 x = 20 x = 30 x 16 + + x 28 + x + + + x + 4 + + + + (x 16)(x 28) x(x + 4) + + +
  51. 51. 4 < x < 0 16 < x < 28 x = 16 x = 28 3 x + 1 = 2 x 3 2x x + 1 + 5 2x = 2 5 x2 10 x 1 = 14 5x x 1 4, 1 x2 4x x 2 = 14 9x x 2 7 (x + 3)(x 2) (x + 2)(x 1) 0 ( 1, 3] [ [2, 1) (x + 4)(x 3) (x 2)(x2 + 2) 0 [ 4, 2) [ [3, 1) x + 1 x + 3  2 ( 1, 5] [ ( 3, 1) x 2 x2 3x 10 < 0 ( 1, 2) [ (2, 5) x 3 + x 6 + x
  52. 52. t
  53. 53. f(x) = p(x) q(x) p(x) q(x) q(x) q(x) 6⌘ 0 f(x) x q(x) 6= 0 s = d t s = 100 t
  54. 54. x s(x) s(x) = 100 x s = d t x x s(x) s(x) = 100 x x s(x) s(x) = 100 x
  55. 55. x
  56. 56. f(x) = x 1 x + 1 x x 10 10 x 10 8 6 4 2 f(x) 1 1 E F
  57. 57. f(x) = x 1 x + 1 E F f(x) = x 1 x + 1 x = 1 E F x = 1 E F x 1
  58. 58. f(x) = x2 3x 10 x x x = 0 f 6  x  10 x 6= 0 x 5 4 3 2 1 f(x) 6 4.5 2.67 12 6 x 3 4 5 6 7 8 9 10 f(x) 3.33 1.5
  59. 59. x = 0 x  1 x 1 x = 2 x = 5 x x
  60. 60. x p p(x) = 12 + x 25 + x p(x) x p(x) 25 + x P(t) = 60(t + 1) t + 6 ⌫ P t b·c
  61. 61. t = 5 P(5) = 60(5 + 1) 5 + 6 ⌫ = b32.726c = 32 P(x) t P(t) t I V R I = V R R I f(x) = x 3 x + 4 6  x  2 x x x 6 5 4 3 2 1 f(x) 6 2.5 1.33 0.75 0.4 0.167 f x = 4 f(x) = x2 + x 6 x2 + x 20 x = 4, 5 6  x  2 x
  62. 62. x 6 5 4 3 2 1 f(x) 0.75 0.22 0.3 0.3 0.22 0 x = 3, 2
  63. 63. x f(x) x x y x = 0 x y x
  64. 64. f(x) = x 2 x + 2 f(x) {x 2 R | x 6= 2} x = 2 x = 2 f(x) x x f(x) y 1 x x x 2 x = 2 x = 2 f(x) x y f(0) f(0) = 2 2 = 1 f(x) x x x = 2 x 2 x < 2 2 x 2 x > 2 2+ x 2 x 3 2.5 2.1 2.01 2.001 2.0001 x 2 f(x) f(x) f(x) ! +1 x ! 2 f(x) x 2 x 2+ x 1 1.5 1.9 1.99 1.999 1.9999 x 2+ f(x) 3 7 39 399 3999 39999 f(x)
  65. 65. f(x) ! 1 x ! 2+ f(x) x 2 x 2 x = 2 x = a f f x a a x = a
  66. 66. x f(x) x x ! +1 f(x) x ! +1 x x ! +1 f(x) f(x) 1 f(x) x x ! 1 f(x) x ! 1 x 5 10 100 1000 10000 x ! 1 f(x) f(x) 1+ x f(x) y = 1 y = b f f(x) b x x ! +1 x ! 1 y = b b
  67. 67. x 2 x ! +1 x ! 1 x x = 2 x = 2 x < 2 2 < x < 2 x > 2 x = 3 x = 0 x = 3 x 2 + x + 2 + + x 2 x + 2 + + x x x x x = 2 y f(x) < 1 x ! +1 f(x) > 1 x ! 1 (2, 0) f(x) x ! 2 f(x) x ! 2+ f(x)
  68. 68. y = x 2 x + 2 y = 1 f(x)
  69. 69. f(x) ( 1, 1) [ (1, +1) f(x) = 4x2 + 4x + 1 x2 + 3x + 2 x x ! 1 x ! +1 x x = 1000 4x2 +4x+1 4, 004, 001 4x2 4, 000, 000 x x2 + 3x + 2 x2 x f(x) 4x2 x2 = 4 f(x) x y = 4 f(x) = 2x2 5 3x2 + x 7 2x2 5 3x2 + x 7 2x2 3x2 = 2 3 x y = 2 3 f(x) = 3x + 4 2x2 + 3x + 1 3x + 4 2x2 + 3x + 1 3x 2x2 = 3 2x x x 3 2x 0 y = 0 f(x) = 4x3 1 3x2 + 2x 5 x 4x3 1 3x2 + 2x 5 4x3 3x2 = 4x 3 x 4x 3 x y
  70. 70. n m • n < m y = 0 • n = m y = a b a b • n > m x y y y x = 0 x x x f(x) = 3x2 8x 3 2x2 + 7x 4 x f(x) ( 1, 4) [ ( 4, 1 2 ) [ (1 2 , +1) f(x) f(x) = 3x2 8x 3 2x2 + 7x 4 = (3x + 1)(x 3) (2x 1)(x + 4) • y f(0) = 0 0 3 0 + 0 4 = 3 4 • x 3x + 1 = 0 ) x = 1 3 x 3 = 0 ) x = 3
  71. 71. • 2x 1 = 0 ) x = 1 2 x + 4 = 0 ) x = 4 • y = 3 2 f(x) x 4, 1 3 , 1 2 3 x < 4 4 < x < 1 3 1 3 < x < 1 2 1 2 < x < 3 x > 3 x = 10 x = 2 x = 0 x = 1 x = 10 3x + 1 + + + x 3 + 2x 1 + + x + 4 + + + + (3x + 1)(x 3) (2x 1)(x + 4) + + + x x x x x
  72. 72. y x
  73. 73. R x y x
  74. 74. f(x) = 2 x + 1 f(x) = 2 x2 + 2x + 1 f(x) = 3x x + 3 f(x) = 2x + 3 4x 7 f(x) = (4x 3)(x 1) (2x + 1)(x + 1) f(x) = (5x 2)(x 2) (3x 4)(x + 2) f(x) = x2 x + 6 x2 6x + 8 f(x) = x2 4x 5 x 4 f(x) = x 1 x3 4x f(x) = x2 9 x2 + 4 N(t) t N(t) = 75t t + 5 t 0. N N(t) t ! 1 t ! 1 N(t) ! 75 c t c(t) = 20t t2 + 2 t 0. c(t) c(t) t ! 1 t ! 1 c(t) ! 0 x = 3 x = 3 y = 1 x 5 y 5 9 f(x) = (x 5)2(x2 + 1) (x + 3)(x 3)(x2 + 5)
  75. 75. p = 5125000V 2 449000V + 19307 125V 2(1000V 43) p V p V V = 0 V = 43/1000 p = 0 p V V = 0.043
  76. 76. T C = 5 9 (T F 32) T F = 9 5 T C + 32 F C
  77. 77. 200 F 93.33 C 120 C 248 F f x1 x2 f f(x1) 6= f(x2) y x 2 • • • •
  78. 78. • d F(d) = ( 8.00 0 < d  4 (8.00 + 1.50 bdc) d > 4 bdc d F(3) = 8 F(3) = F(2) = F(3.5) = 8 F y x
  79. 79. y x 2 y = x2 4
  80. 80. y = 2x 1 x 4 3 2 1 y 9 7 5 3 1 y x x y x 9 7 5 3 1 y 4 3 2 1 x y
  81. 81. x 4 3 2 1 y 1 1 1 1 0 y x y = 1 x = 1, 2, 3, 4 x y x 1 1 1 1 0 y 4 3 2 1 x = 1 y • x y • x y f A B f f 1 B A f 1(y) = x f(x) = y y B x y
  82. 82. y x y = f(x) x y y x f(x) = 3x + 1 y = 3x + 1 x y x = 3y + 1 y x x = 3y + 1 x 1 = 3y x 1 3 = y =) y = x 1 3 f(x) = 3x + 1 f 1(x) = x 1 3 f(f 1(x)) f 1(f(x))
  83. 83. f(x) f 1(x) f 1(x) f(x) f(f 1(x)) = x x f 1 f 1(f(x)) = x x f g(x) = x3 2 y = x3 2 x y x = y3 2 y x x = y3 2 x + 2 = y3 3 p x + 2 = y =) y = 3 p x + 2 g(x) = x3 2 g 1(x) = 3 p x + 2 f(x) = 2x + 1 3x 4 y = 2x + 1 3x 4 x y x = 2y + 1 3y 4 y x x = 2y + 1 3y 4 x(3y 4) = 2y + 1 3xy 4x = 2y + 1 3xy 2y = 4x + 1 y
  84. 84. y y(3x 2) = 4x + 1 y = 4x + 1 3x 2 f(x) f 1(x) = 4x + 1 3x 2 f(x) = x2 + 4x 2 y = x2 + 4x 2 x y x = y2 + 4y 2 y x x = y2 + 4y 2 x + 2 = y2 + 4y x + 2 + 4 = y2 + 4y + 4 x + 6 = (y + 2)2 ± p x + 6 = y + 2 ± p x + 6 2 = y =) y = ± p x + 6 2 y = ± p x + 6 2 x y x = 3 y 1 5 f(x) = x2 + 4x 2 f(x) = |3x| y = |3x|
  85. 85. f 1 f f(1) = f( 1) = 3 x 1 y f y = |4x| x y x = |4y| y x x = |4y| x = p (4y)2 |x| = p x2 x2 = 4y2 x2 4 = y2 ± r x2 4 = y =) y = ± r x2 4 x = 2 y = 1 y = 1 y = ± q x2 4 f(x) = |3x| k(t) = 5 9 (t 32)+273.15 t k = 5 9 (t 32) + 273.15 k t t k k = 5 9 (t 32) + 273.15 k 273.15 = 5 9 (t 32) 9 5 (k 273.15) = t 32 9 5 (k 273.15) + 32 = t =) t = 9 5 (k 273.15) t(k) = 9 5 (k 273.15) k
  86. 86. f(x) = 1 2 x + 4 f 1(x) = 2x 8 f(x) = (x + 3)3 f 1(x) = 3 p x 3 f(x) = 3 x 4 f 1(x) = 4x + 3 x f(x) = x + 3 x 3 f 1(x) = 3x + 3 x 1 f(x) = 2x + 1 4x 1 f 1(x) = x + 1 4x 2 f(x) = |x 1|
  87. 87. y = x y = x y = x y = x
  88. 88. y = x • • y = x x y f f 1 f 1(f(x)) = x f 1(x) y f(x) x x f(x) ! y f 1(y) ! x y = x y = f 1(x) y = f(x) = 2x + 1 {x | 2  x  1.5} f(x)
  89. 89. y = 2x + 1 y = x {y 2 R | 3  y  4} f 1(x) = x 1 2 f 1 (x) = [ 3, 4] f 1 (x) = [ 2, 1.5]
  90. 90. f(x) f 1(x) [ 2, 1.5] [ 3, 4] [ 3, 4] [ 2, 1.5] f(x) = 1 x f(x) = 1 x y = x y = x f(x) f 1(x) = f(x)
  91. 91. f 1(x) = f(x) = 1 x f(x) = 3 p x + 1 f(x) y = x
  92. 92. f(x) = 3 p x + 1 y = x3 1 f 1(x) = x3 1
  93. 93. f(x) = 5x 1 x + 2 f(x) = ( 1, 2) [ (2, 1) f(x) = ( 1, 5) [ ( 5, 1) x = 2 y = 5 y = x
  94. 94. x y y = x x = 5 y = 2 f(x) f 1(x) ( 1, 2) [ (2, 1) ( 1, 5) [ ( 5, 1) ( 1, 5) [ ( 5, 1) ( 1, 2) [ (2, 1) y = x y = x f 1 1 (x) = f(x)
  95. 95. f(x) = (x + 2)2 · 3 ÷ 2 = 3(x + 2)2 2 x 0 x 0 x y x = 3(y + 2)2 2 , y 0 y x
  96. 96. x = 3(y + 2)2 2 2x 3 = (y + 2)2 r 2x 3 = y + 2 y 2 r 2x 3 r 2x 3 2 = y =) y = r 2x 3 2 x = 54 f 1 (54) = r 2(54) 3 2 = r 108 3 2 = p 36 2 = 6 2 = 4 t d t(d) = ✓ 12.5 d ◆3 t = ✓ 12.5 d ◆3 d t d t t = ✓ 12.5 d ◆3 3 p t = 12.5 d d = 12.5 3 p t d(t) = 12.5 3 p t t = 6.5 d(6.5) = 22.5 3 p 6.5 = 12.06 12.06
  97. 97. f(x) = x2 + 1 {0, 0.5, 1, 1.5, 2, 2.5, 3} x 0 0.5 1 1.5 2 2.5 3 f(x) 1 1.25 2 3.25 5 7.25 10 f x 1 1.25 2 3.25 5 7.25 10 f 1(x) 0 0.5 1 1.5 2 2.5 3 A = {( 4, 4), ( 3, 2), ( 2, 1), (0, 1), (1, 3), (2, 5)} A 1 = {(4, 4), (2, 3), (1, 2), ( 1, 0), ( 3, 1), ( 5, 2)} f(x) = 2 p x 2 + 3 [2, 1) [3, 1) [2, 1) f(x) = 3x + 2 x 4 ( 1, 3) [ (3, 1) ( 1, 4) [ (4, 1) x = a f 1(a) = 0 1/2
  98. 98. w l w = (3.24 ⇥ 10 3)l2 l 0 l = p w/(3.4 ⇥ 10 3)
  99. 99. n s n = 2s b f(x) = bx y = bx b > 0 b 6= 1 x = 3, 2, 1, 0, 1, 2 3 y = 1 3 x y = 10x y = (0.8)x
  100. 100. x 3 2 1 y = 1 3 x 1 3 1 9 1 27 y = 10x 1 1000 1 100 1 10 y = (0.8)x f(x) = 3x f(2) f( 2) f 1 2 f(0.4) f(⇡) f(2) = 32 = 9 f( 2) = 3 2 = 1 32 = 1 9 f ✓ 1 2 ◆ = 31/2 = p 3 f(0.4) = 30.4 = 32/5 = 5 p 32 = 5 p 9 ⇡ ⇡ 3.14159 3⇡ f(⇡) = 3⇡ 33.14 33.14159 3⇡ ⇡ 3⇡ b b g(x) = a · bx c + d a c d
  101. 101. t t = 0 t t = 0 t = 100 20(2) t = 200 = 20(2)2 t = 300 = 20(2)3 t = 400 = 20(2)4 y = 20(2)t/100 y T y0 y t y = y0(2)t/T t t = 0 t = 10 t = 20 t = 30
  102. 102. y = 10 1 2 t/10 100, 000 6% t t = 0 = 100, 000 t = 1 = 100, 000(1.06) = 106, 000 t = 2 = 106, 000(1.06) = 112, 360 t = 3 = 112, 360(1.06) ⇡ 119, 101.60 t = 4 = 119, 101.60(1.06) ⇡ 126, 247.70 t = 5 = 26, 247.70(1.06) ⇡ 133, 822.56 y = 100, 000(1.06)t P r t A = P(1 + r)t y = 100000(1.06)t t t t = 8 y = 100, 000(1.06)8 ⇡ 159, 384.81 t = 10 y = 100, 000(1.06)10 ⇡ 179, 084.77 200, 000
  103. 103. e ⇡ 2.71828 e e e f(x) = ex T t T = 170165e0.006t t t T 50, 000 A = 50, 000(1.044)t t t = 18 A = 50, 000(1.044)18 ⇡ 108, 537.29 100, 000 A = 20, 000(1.05)t t t = 10, A = 20, 000(1.05)10 ⇡ 32, 577.89 32, 577.89
  104. 104. y = 100 1 2 x/250 x = 0 x = 500 y = 100 1 2 500/250 = 100 1 2 2 = 25 y = 1, 000(3)x/80 x = 0 x = 100 y = 1, 000(3)100/80 = 3, 948.22 ⇡ 3, 948 10, 000 A = 10, 000(1.02)t t A = 10, 000(1.02)12 = 12, 682.42 12, 682.42
  105. 105. 4x 1 = 16x y = 2x 2x 26 a · bx c + d b > 0 b 6= 1 f(x) = bx b > 0 b 6= 1 72x x2 = 1 343 52x 5x+1  0 f(x) = (1.8)x y = (1.8)x x x y f(x) = 2x3 f(x) = 2x y = ex 22(5x+1) = 500 625 5x+8
  106. 106. a 6= 0 a0 = 1 a n = 1 an r s aras = ar+s ar as = ar s (ar)s = ars (ab)r = arbr ⇣a b ⌘r = ar br
  107. 107. 49 = 7x+1 7 = 2x + 3 3x = 32x 1 5x 1 = 125 8x = x2 9 x2 = 3x3 + 2x 1 2x + 3 > x 1 2x 2 > 8 x1 6= x2 bx1 6= bx2 bx1 = bx2 x1 = x2 4x 1 = 16 4x 1 = 16 4x 1 = 42 x 1 = 2 x = 2 + 1 x = 3 4x 1 = 16 (22 )x 1 = 24 22(x 1) = 24 2(x 1) = 4 2x 2 = 4 2x = 6 x = 3
  108. 108. x = 3 43 1 = 42 = 16 125x 1 = 25x+3 125x 1 = 25x+3 (53 )x 1 = (52 )x+3 53(x 1) = 52(x+3) 3(x 1) = 2(x + 3) 3x 3 = 2x + 6 x = 9 9x2 = 3x+3 (32 )x2 = 3x+3 32x2 = 3x+3 2x2 = x + 3 2x2 x 3 = 0 (2x 3)(x + 1) = 0 2x 3 = 0 x + 1 = 0 x = 3 2 x = 1
  109. 109. b > 1 y = bx x bx < by x < y 0 < b < 1 y = bx x bx > by x < y bm < bn m < n m > n b 3x < 9x 2 3x < (32 )x 2 3x < 32(x 2) 3x < 32x 4 3 > 1 x < 2x 4 4 < 2x x 4 < x (4, +1] x = 5 x = 4 ✓ 1 10 ◆x+5 ✓ 1 100 ◆3x 1 100 = 1 10 2 1 10 ✓ 1 10 ◆x+5 ✓ 1 100 ◆3x ✓ 1 10 ◆x+5 ✓ 1 102 ◆3x ✓ 1 10 ◆x+5 ✓ 1 10 ◆6x
  110. 110. 1 10 < 1 x + 5  6x 5  6x x 5  5x 1  x [1, +1) x = 1 x = 0 1 y0 1 256 t y = y0 1 2 t/2.45 y0 1 2 t/2.45 = 1 256 y0 ✓ 1 2 ◆t/2.45 = 1 256 ✓ 1 2 ◆t/2.45 = ✓ 1 2 ◆8 t 2.45 = 8 t = 19.6 t = 0 (0.6)x 3 > (0.36) x 1
  111. 111. (0.6)x 3 > (0.36) x 1 (0.6)x 3 > (0.62 ) x 1 (0.6)x 3 > (0.6)2( x 1) (0.6)x 3 > (0.6) 2x 2 x 3 > 2x 2 3x > 1 x > 1 3 (0.6)x 3 > (0.36) x 1 (0.6)x 3 > (0.62 ) x 1 (0.6)x 3 > (0.6)2( x 1) (0.6)x 3 > (0.6) 2x 2 x 3 < 2x 2 3x < 1 x < 1 3 (0.6)x 3 > (0.6) 2x 2 x 3 > 2x 2 x > 1 3 x 162x 3 = 4x+2 x = 8 3 ✓ 1 2 ◆2x = 23 x 1 2 2 1 x = 3 42x+7  322x 3  29 6 , +1 ◆ ✓ 2 5 ◆5x 1 25 4 25 4 ✓ 2 5 ◆ 2 ( 1, 1 5 ] x 7x+4 = 492x 1 x = 2 4x+2 = 82x x = 1 ✓ 2 3 ◆5x+2 = ✓ 3 2 ◆2x ✓ 2 7 , +1 ◆
  112. 112. f(x) = bx b > 1 f(x) = bx 0 < b < 1 f(x) = 2x f(x) x 4 3 2 1 f(x) 1 16 1 8 1 4 1 2
  113. 113. f(x) = 2x f(x) = 2x x y x y = 0 g(x) = ✓ 1 2 ◆x g(x) x 3 2 1 f(x) 1 2 1 4 1 8 1 16
  114. 114. g(x) = ✓ 1 2 ◆x g(x) = ✓ 1 2 ◆x x x y = 0 b > 1 0 < b < 1 f(x) = bx b > 1 0 < b < 1
  115. 115. f(x) = bx b > 0 b 6= 1 R (0, +1) y x y = 0 x f(x) = 2x g(x) = 3x x 4 3 2 1 f(x) g(x) x 4 y 1 f(x) g(x) y f(x) = ✓ 1 2 ◆x g(x) = ✓ 1 3 ◆x x 4 3 2 1 f(x) g(x) x 4 y 1 f(x) g(x)
  116. 116. y f(x) = 5 x y 5 x = 1 5x = ✓ 1 5 ◆x
  117. 117. • x 3 • x 4 y 1 • y = 2x y = 2x y = 2x y = 3 · 2x y = 2 5 · 2x y = 2x + 1 y = 2x 1 y = 2x+1 y = 2x 1
  118. 118. x 3 2 1 y = 2x y = 2x y = 2x 0.125 .25 .5 1 2 4 8 y 1 y = 3 · 2x y y = 2 5 · 2x y 2/5 y = 2x + 1 y = 2x 1 0.875 0.75 0.5 y = 2x+1 y = 2x 1 y = 2x y = 2x y = 2 x y x 3 2 1 y = 2x y = 2x 0.125 0.25 0.5 1 2 4 8 y = 2 x y y = 2x y y = 2x y = 2x y = 2x x y = 2 x x y = 2x x y = 2 x y = 2x y
  119. 119. y = f(x) x y = f(x) y = f( x) y y = f(x) y = 2x y = 3(2x) y = 0.4(2x) y x 3 2 1 y = 2x y = 3(2x) y = 0.4(2x) y y = 3(2x) y y = 2x y y = 0.4(2x) y y = 2x
  120. 120. R y y y = 3(2x) y y = 0.4(2x) y = 0 (0, +1) a > 0 y = af(x) y y = f(x) a a > 1 0 < a < 1 y = f(x) y = 2x y = 2x 3 y = 2x + 1 y x 3 2 1 y = 2x y = 2x 3 2.875 2.75 2.5 2 1 y = 2x + 1
  121. 121. R y = 2x + 1 (1, +1) y = 2x 3 ( 3, +1) y y y = 2x y = 2x y = 0 y = 2x + 1 y = 1 y = 2x 3 y = 3 d y = f(x) + d d d > 0 d d < 0 y = f(x) y = 2x y = 2x 2 y = 2x+4 y
  122. 122. x 3 2 1 y = 2x y = 2x 2 y = 2x+4 R (0, +1) y x = 0 y y = 2x+4 24 = 16 y y = 2x 2 22 = 0.25 y = 0 c y = f(x c) c c > 0 c c < 0 y = f(x)
  123. 123. f(x) = a · bx c + d • b b > 1 0 < b < 1 • |a| a x • d d > 0 d d < 0 • c c > 0 c c < 0 y = bx y y = 3x 4 y = ✓ 1 2 ◆x + 2 y = 2x 5 y = (0.8)x+1 y = 2 ✓ 1 3 ◆x y = 0.25(3x) y = 2x 3 + 1 y = ✓ 1 3 ◆x 1 2
  124. 124. x 24 = x 43 = x 5 1 = x 16 1 2 = x 1 5 1 4
  125. 125. 5x = 625 3x = 1 9 7x = 0 10x = 100, 000 x x b logb x b x log3 81 = 4 34 = 81 log2 32 = 5 25 = 32 log5 1 = 0 50 = 1 log6 ✓ 1 6 ◆ = ( 1) 6 1 = 1 6 log2 32 log9 729 log5 5 log1 2 16 log7 1 log5 1 p 5 4 1 p 5 1/2
  126. 126. a b b 6= 1 a b logb a blogb a = a logb a b a log2 32 = 5 25 = 32 log9 729 = 3 93 = 729 log5 5 = 1 51 = 5 log1/2 16 = 4 1 2 4 = 16 log7 1 = 0 70 = 1 log5 1p 5 = 1 2 5 1/2 = 1p 5 logb a = c bc = a • b • c c = logb a • logb a a log2( 8) • logb x log5 1 125 = 3 5 3 = 1 125 log x log10 x e e ln ln x loge x
  127. 127. 53 = 125 7 2 = 1 49 102 = 100 ✓ 2 3 ◆2 = 4 9 (0.1) 4 = 10, 000 40 = 1 7b = 21 e2 = x ( 2)2 = 4 log5 125 = 3 log7 1 49 = 2 log 100 = 2 log2 3 ✓ 4 9 ◆ = 2 log0.1 10, 000 = 4 log4 1 = 0 log7 21 = b b = 3 73 6= 21 b 71.5645 ln x = 2 log m = n log3 81 = 4 logp 5 5 = 2 log3 4 64 27 = 3 log4 2 = 1 2 log10 0.001 = 3 ln 8 = a
  128. 128. 10n = m 34 = 81 ( p 5)2 = 5 ✓ 3 4 ◆ 3 = 64 27 41/2 = 2 10 3 = 0.001 10 3 = 1 1, 000 ea = 8 1031 log 1031 = 31 R = 2 3 log E 104.40 E 104.40 1012
  129. 129. E = 1012 R = 2 3 log 1012 104.40 = 2 3 log 107.6 log107.6 = 7.6 107.6 log 107.6 = 7.6 R = 2 3 (7.6) ⇡ 5.1 1012/104.40 = 107.6 ⇡ 39810717
  130. 130. D = 10 log I 10 12 I 2 10 12 2 106
  131. 131. 60 85 90 100 10 6 m2 D = log 10 6 10 12 = 10 log 106 log 106 106 log 106 = 6 D = 10(6) = 60 10 6 10 12 = 106 = 100, 000
  132. 132. [H+] pH = log[H+ ] pH = log 1 [H+] 10 5 log 10 5 log 10 5 10 5 log 10 5 = 5 log 10 5 = ( 5) = 5
  133. 133. log3 243 5 log6 ✓ 1 216 ◆ 3 log0.25 16 2 49x = 7 log49 7 = x 6 3 = 1 216 log6 216 = 3 102 = 100 log 100 = 2 log11 2 ✓ 4 121 ◆ = 2 ✓ 11 2 ◆ 2 = 4 121 ln 3 = y ey = 3 log 0.001 = 3 10 3 = 0.001
  134. 134. log3(x 2) = 5 ln x 9 y = log1 2 x logx 2 = 4 ln x2 > (ln x)2 g(x) = log3 x x
  135. 135. x y • logx 2 = 4 • log2 x = 4 • log2 4 = x • logx 2 = 4 x • log2 x = 4 4 • log2 4 = x x log 2 x = 4
  136. 136. f(x) = 3x x f(x) 4 1 g(x) = log3 x x x g(x) 1 81 1 3 log2 x = 4 logx 16 = 2 log 1000 = x log2 x = 4 24 = x x = 16 logx 16 = 2 x2 = 16 x2 16 = 0 (x + 4)(x 4) = 0 x = 4, +4 x = 4 log 1000 = x
  137. 137. log 1000 3 = x x = 3 log3 81 34 = 81 log4 1 log3 3 log4 42 log4 3 = 0 40 = 1 log3 3 = 1 31 = 3 log4 42 = 2 42 = 42
  138. 138. b x b > 0 b 6= 1 logb 1 = 0 logb bx = x x > 0 blogb x = x logb 1 b b? = 1 logb bx b bx x logb x b x b x log2 14 23 = 8 24 = 16 23.8074 ⇡ 14.000 2log2 14 = 14 logb 1 = 0 logb bx = x blogb x = x log 10 ln e3 log4 64 log5 ✓ 1 125 ◆ 5log5 2 log 1 log 10 = log10 101 = 1 ln e3 = loge e3 = 3 log4 64 = log4 43 = 3 log5 ✓ 1 125 ◆ = log5 5 3 = 3
  139. 139. 5log5 2=2 log 1 = 0 10 2 m2 D = 10 log ✓ I I0 ◆ D = 10 log ✓ 10 2 10 12 ◆ D = 10 log 1010 D = 10 · 10 D = 100 pH = log[H+] 3.0 = log[H+] 3.0 = log[H+] 10 3.0 = 10log H+ 10 3.0 = [H+] 10 3.0 log7 49 log27 3 1 3 ln e
  140. 140. log7 73 · 78 log7 73 + log7 78 log7 ✓ 49 7 ◆ log7 49 log7 7 log7 75 5 · log7 7 log2 ✓ 24 210 ◆ log2 24 log2 210 6 log3(27 · 81) log3 27 + log3 81 b > 0, b 6= 1 n 2 R u > 0, v > 0 logb(uv) = logb u + logb v logb ⇣u v ⌘ = logb u logb v logb(un) = n logb u
  141. 141. logb(uv) = logb u + logb v log7 73 · 78 = log7 73 + log7 78 logb ⇣u v ⌘ = logb u logb v log7 ✓ 49 7 ◆ = log7 49 log7 7 log2 ✓ 24 210 ◆ = log2 24 log2 210 logb(un) = n logb u log7 75 = 5 · log7 7 log3(27 · 81) = log3 27 + log3 81 r = logb u s = logb v u = br v = bs logb(uv) = logb(br bs ) ) logb(uv) = logb br+s ) logb(uv) = r + s ) logb(uv) = logb u = logb v r = logb u u = br un = brn un = brn ) logb(un ) = logb(brn ) ) logb(un ) = rn ) logb(un ) = n logb u logb un (logb u)n n u n logb u
  142. 142. log2(5 + 2) 6= log2 5 + log2 2 log2(5 + 2) 6= (log2 5)(log2 2) log2(5 2) 6= log2 5 log2 2 log2(5 2) 6= log2 5 log2 2 log2(52 · 2) 6= 2 log2(5 · 2) log(ab2) log3 ✓ 3 x ◆3 ln[x(x 5)] log(ab2) = log a + log b2 = log a + 2 log b log3 ✓ 3 x ◆3 = 3 log3 ✓ 3 x ◆ = 3(log3 3 log3 x) = 3(1 log3 x) = 3 3 log3 x ln[x(x 5)] = ln x + ln(x 5) log 2 + log 3 2 ln x ln y log5(x2) 3 log5 x 2 log 5 log 2 + log 3 = log(2 · 3) = log 6 2 ln x ln y = ln(x2) ln y = ln ✓ x2 y ◆ log5(x2) 3 log5 x = log5(x2) log5(x3) = log5 ✓ x2 x3 ◆ = log5 ✓ 1 x ◆ = log5(x 1) = log5 x 2 log 5 2 = 2(1) = 2(log 10) 1 = logb b = log 102 n logb u = logb un = log 100 2 log 5 = log 100 = log ✓ 100 5 ◆ = log 20
  143. 143. log3 729 6 log9 729 3 log27 729 2 log1/27 729 2 log729 729 1 log81 729 3/2 log81 729 811/4 = 3 36 = 729 (811/4)6 = 729 816/4 = 813/2 = 729 log81 729 = 3 2 log3n 729 = 6 n a b x a 6= 1, b 6= 1 logb x = loga x loga b log3n 729 = 6 n log3n 729 = log3 729 log3 3n = 6 n . log8 32 log243 1 27 log25 1 p 5
  144. 144. log8 32 = log2 32 log2 8 = 5 3 log243 1 27 = log3 1 27 log3 243 = 3 5 log25 1 p 5 = log5 1 p 5 log5 25 = 1/2 2 = 1 4 log6 4 log1 2 2 e log6 4 = log2 4 log2 6 = 2 log2 6 log1 2 2 = ln 2 ln ✓ 1 2 ◆ = ln 2 ln 1 ln 2 = ln 2 0 ln 2 = ln 2 ln 2 = 1 log ✓ x3 2 ◆ 3 log x log 2 ln(2e)2 (2e)2 = 4e3 2 ln 2 + 2 log4(16a) 2 + log4 a log(x + 2) + log(x 2) log(x2 4) 2 log3 5 + 1 log3 75 2 ln ✓ 3 2 ◆ ln 4 ln ✓ 9 16 ◆ (log3 2)(log3 4) = log3 8 (log3 2)(log3 4) = log3 6
  145. 145. log 22 = (log 2)2 log4(x 4) = log4 x log4 4 3 log9 x2 = 6 log9 x 3(log9 x)2 = 6 log9 x log3 2x2 = log3 2 + 2 log3 x log3 2x2 = 2 log3 2x log5 2 ⇡ 0.431 log5 8, log5 ✓ 1 16 ◆ , log5 p 2, log25 2 log25 8 1.2920, 1.7227, 0.2153, 0.2153, 0.6460 log 6 ⇡ 0.778 log 4 ⇡ 0.602 log6 4, log 24, log4 6, log ✓ 2 3 ◆ log ✓ 3 2 ◆ 0.7737, 1.3802, 1.2925, 0.1761, 0.1761 b x logb x
  146. 146. • b • x • logb x • • • f(x) = logb x logb u = logb v u = v ab = 0 a = 0 b = 0 x log4(2x) = log4 10 log3(2x 1) = 2 logx 16 = 2 log2(x + 1) + log2(x 1) = 3 log x2 = 2 (log x)2 + 2 log x 3 = 0 log4(2x) = log4 10 2x = 10 x = 5
  147. 147. log4(2 · 5) = log4(10) log3(2x 1) = 2 2x 1 = 32 2x 1 = 9 2x = 10 x = 5 log3(2 · (5)1) = log3(9) = 2 logx 16 = 2 x2 = 16 x2 16 = 0 (x + 4)(x 4) = 0 a2 b2 = (a + b)(a b) x = 4, 4 log4(16) = 2 4 log 4(16) log2(x + 1) + log2(x 1) = 3 log2[(x + 1)(x 1)] = 3 logb u + logb v = logb(uv) (x + 1)(x 1) = 23 x2 1 = 8 x2 9 = 0 (x + 3)(x 3) = 0 a2 b2 = (a + b)(a b) x = 3, 3 log2(3 + 1) log2(31 3 log2(3 + 1) = log2(2) x = 3 log x2 = 2 log x2 = 2 x2 = 102 x2 = 100
  148. 148. x2 100 = 0 (x + 10)(x 10) = 0 x = 10, 10 log(10)2 = 2 log(10)2 = 2 log x2 = 2 log x2 = log 102 2 = 2(1) = 2(log 10) = log 102 x2 = 102 x2 100 = 0 (x + 10)(x 10) = 0 x = 10, 10 log(10)2 = 2 log(10)2 = 2 logb un = n logb u log x2 = 2 2 log x = 2 log x = 1 x = 10 log x2 = 2 log x x > 0 (log x)2 + 2 log x 3 = 0 log x = A A2 + 2A 3 = 0 (A + 3)(A 1) = 0 A = 3 A = 1 log x = 3 log x = 1 x = 10 3 = 1 1000 x = 10 log ✓ 1 1000 ◆ log 10 x 2x = 3 2x = 3
  149. 149. log 2x = log 3 x log 2 = log 3 logb un = n logb u x = log 3 log 2 ⇡ 1.58496 x log1 2 x 1 8 1 4 1 2 x log2 x 1 8 1 4 1 2 x log1 2 x 1 8 1 4 1 2 1 2 3 x log2 x 1 8 3 1 4 2 1 2 1 1 2 log1 2 x x log1 2 x log2 x x log2 x
  150. 150. logb x 0 < b < 1 x1 < x2 logb x1 > logb x2 b > 1 x1 < x2 logb x1 < logb x2 • < > b • x log3(2x 1) > log3(x + 2) log0.2 > 3 2 < log x < 2 log3(2x 1) > log3(x + 2) 2x 1 > 0 x + 2 > 0 2x 1 > 0 x > 1 2 x + 2 > 0 x > 2 x > 1 2 x > 1 2 x 2 log3(2x 1) > log3(x + 2) 2x 1 > x + 2 x > 3 x ) x > 3 (3, +1) log0.2 > 3 x > 0
  151. 151. 3 1 5 3 = log1 5 ✓ 1 5 ◆ 3 log1 5 x > log1 5 ✓ 1 5 ◆ 3 0.2 = 1 5 log1 5 x > log1 5 ✓ 1 5 ◆ 3 x < ✓ 1 5 ◆ 3 = 125 x 0 < x < 125 (0, 125) 2 < log x < 2 x > 0 2 log 10 2 log 102 log 10 2 < log x < log 102 log 10 2 < log x log x < log 102 10 2 < x x < 102 1 100 < x < 100 ✓ 1 100 , 100 ◆ EB EN EB EN 7.2 = 2 3 log EB 104.4 6.7 = 2 3 log EN 104.4
  152. 152. EB 7.2 ✓ 3 2 ◆ = log EB 104.4 10.8 = log EB 104.4 1010.8 = EB 104.4 EB = 1010.8 · 104.4 = 1015.2 EN 6.7 ✓ 3 2 ◆ = log EN 104.4 10.05 = log EN 104.4 1010.05 = EN 104.4 EB = 1010.05 · 104.4 = 1014.45 EB EN = 1015.2 1014.45 = 100.75 ⇡ 5.62 n n + 1 E1 E2 n n+1 E2 E1 n = 2 3 log E1 104.4 n + 1 = 2 3 log E2 104.4 E1 3 2 n = log E1 104.4 103n/2 = E1 104.4 E1 = 103n/2 · 104.4 = 10 3n 2 +4.4 E2 3 2 (n + 1) = log E2 104.4 103(n+1)/2 = E2 104.4 E2 = 103(n+1)/2 · 104.4 = 10 3(n+1) 2 +4.4 E2 E1 = 10 3(n+1) 2 +4.4 10 3n 2 +4.4 = 10 3 2 ⇡ 31.6
  153. 153. A = P(1 + r)n A P r n A = 2P r = 2.5% = 0.025 A = P(1 + r)n 2P = P(1 + 0.025)n 2 = (1.025)n log 2 = log(1.025)n log 2 = n log(1.025) n = log 2 log 1.025 ⇡ 28.07 log ln ln 2 ln 1.025 P(x) = 20, 000, 000 · e0.0251x x x = 0 P(x) = 200, 000, 000 200, 000, 000 = 20, 000, 000 · e0.0251x 10 = e0.0251x ln 10 = ln e0.0251x ln 10 = 0.0251x(ln e) ln 10 = 0.0251x x = ln 10 0.0251 ⇡ 91
  154. 154. f(t) = Aekt t A k f(0) = 5, 000 f(90) = 12, 000 f(0) = Aek(0) = A = 5, 000 f(90) = 5, 000ek(90) = 12, 000 ) e90k = 12 5 ln e90k = ln 12 5 ) 90k = ln 12 5 ) k ⇡ 0.00973 f(t) = 5, 000 · e0.00973t f(180) = 5, 000 · e0.00973(180) ⇡ 28, 813 y = ex + e x 2 y = 4
  155. 155. 4 = ex + e x 2 8 = ex + e x 8 = ex + e x 8 = ex + 1 ex 8ex = e2x + 1 e2x 8ex + 1 = 0 u = ex u2 = e2x u2 8u + 1 = 0 u = 4 ± p 15 u = ex 4 + p 15 = ex 4 p 15 = ex ln(4 + p 15) = x ln(4 p 15) = x y = 4 x ln(4 + p 15) ln(4 p 15) ⇡ 4.13 log5(x 1) + log5(x + 3) 1 = 0 log3 x + log3(x + 2) = 1 3x+1 = 10 1 log 3 log 3 ⇡ 1.0959 ln x > 1 e ⇡ 2.7183 (e, +1)
  156. 156. log0.5(4x + 1) < log0.5(1 4x) (0, 1/4) log2 [log3(log4 x)] = 0 f(x) = bx f(x) = bx y = f(x) = bx y = bx x = by x y y = logb x f 1(x) = logb x
  157. 157. y = log2 x y = log2 x x 1 16 1 8 1 4 1 2 y 4 3 2 1 y = log2 x y = log2 x x > 0 x x = 0
  158. 158. y = 2x y = log2 x y = x y = log1 2 x y = log1 2 x x 1 16 1 8 1 4 1 2 y 4 3 2 1 1 2 3
  159. 159. y = log1 2 x y = log1 2 x x > 0 x x = 0 y = log2 x y = log1 2 x y = logb x b y = logb x(b > 1) y = logb x(0 < b < 1) y = logb x b > 1 0 < b < 1
  160. 160. {x 2 R|x > 0} x logb x x y x = 0 y y = x y = bx y = logb x(b > 1) y = bx y = logb x(0 < b < 1) y = log2 x y = log1 2 x y = 2 log2 x x y = 2 log2 x y = 2 log2 x y
  161. 161. x 1 16 1 8 1 4 1 2 log2 x 4 3 2 1 y = 2 log2 x 8 6 4 2 {x|x 2 R, x > 0} {y|y 2 R} x = 0 x y = log3 x 1 y = log3 x > 1 y = log3 x (1, 0), (3, 1) (9, 2) (1, 1), (3, 0) (9, 1)
  162. 162. {x|x 2 R, x > 0} {y|y 2 R} x = 0 x x x y = 0 0 = log3 x 1 log3 x = 1 x = 31 = 3 y = log0.25(x + 2) y = log0.25 x 0 < 0.25 < 1 y = log0.25[x ( 2)] 2 y = log0.25 x (1, 0), (4, 1), (0.25, 1) ( 1, 0), (2, 1), ( 1.75, 1)
  163. 163. {x|x 2 R, x > 2} x + 2 log0.25(x + 2) x 2 {y|y 2 R} x = 2 x 1 f(x) = a · logb(x c) + d • b b > 1 0 < b < 1 • a a x • f(x) = a · logbx d d > 0 d d < 0 c c > 0 c c < 0
  164. 164. y = logb x y y = logx(x + 3) y = log1 3 (x 1) y = (log5 x) + 6 y = (log0.1 x) 2 y = log2 5 (x 4) + 2 y = log6(x + 1) + 5 log 2 ⇡ 0.3010 log 3 ⇡ 0.4771 log 5 ⇡ 0.6990 log 7 ⇡ 0.8451 21/3 51/4 21/3 51/4 n = 21/3 51/4 log n = 1 3 log 2 1 4 log 5 log n log n ⇡ 1 3 (0.3010) 1 4 (0.6990) ⇡ 0.0744 0.0744 n n log n ⇡ 0.0744 n ⇡ 10 0.0744
  165. 165. • • • • • t • P • r • I • Is • Ic • F t
  166. 166. r t P t r t t t
  167. 167. Is = Prt Is P r t
  168. 168. P = 1, 000, 000 r = 0.25% = 0.0025 t = 1 Is Is = Prt Is = (1, 000, 000)(0.0025)(1) Is = 2, 500 P = 50, 000 r = 10% = 0.10 t = 9 12 Is M t = M 12 Is = Prt Is = (50, 000)(0.10) ✓ 9 12 ◆ Is = (50, 000)(0.10)(0.75) Is = 3, 750
  169. 169. P r t P = Is rt = 1, 500 (0.025)(4) P = 15, 000 r = Is Pt = 4, 860 (36, 000)(1.5) r = 0.09 = 9% t = Is Pr = 275 (250, 000)(0.005) t = 0.22 Is = Prt = (500, 000)(0.125)(10) Is = 625, 000
  170. 170. r = 7% = 0.07 t = 2 IS = 11, 200 P P = Is rt = 11, 200 (0.07)(2) P = 80, 000 P = 500, 000 Is = 157, 500 t = 3 r r = Is Pt = 157, 500 (500, 000)(3) r = 0.105 = 10.5% P r = 5% = 0.05 Is = 1 2 P = 0.5P t
  171. 171. t = Is Pr = 0.5P (P)(0.05) t = 10 t r F F = P + Is F P Is Is Prt F = P + Prt F = P(1 + rt) F = P(1 + rt) F P r t P = 1, 000, 000, r = 0.25% = 0.0025 F F
  172. 172. Is P F = P + Is. F = P(1 + rt) t = 1 Is = Prt Is = (1, 000, 000)(0.0025)(1) Is = 2, 500 F = P + Is F = 1, 000, 000 + 2, 500 F = 1, 002, 500 F F = P(1 + rt) F = (1, 000, 000)(1 + 0.0025(1)) F = 1, 002, 500 t = 5 Is = Prt Is = (1, 000, 000)(0.0025)(5) Is = 12, 500 F = P + Is F = 1, 000, 000 + 12, 500 F = 1, 012, 500 F = P(1 + rt) F = (1, 000, 000)(1 + 0.0025(5)) F = 1, 012, 500
  173. 173. P r t I P r t I 1 2 I F
  174. 174. I F P F r = 9.5% I = 9, 500 P = 300, 000 t = 5 I = 16, 250 F = 1, 016, 250 1 4
  175. 175. P r P r
  176. 176. P(1 + r) = P(1 + r) 100, 000 · 1.05 = 105, 000 P(1 + r)(1 + r) = P(1 + r)2 105, 000·, 1.05 = 110, 250 P(1 + r)2(1 + r) = P(1 + r)3 110, 250 · 1.05 = 121, 550.63 P(1 + r)3(1 + r) == P(1 + r)4 121, 550.63 · 1.05 = 127, 628.16 (1 + r) 1 + r r F = P(1 + r)t P F r t Ic Ic = F P P = 10, 000 r = 2% = 0.02 t = 5 F Ic F = P(1 + r)t F = (10, 000)(1 + 0.02)5
  177. 177. F = 11, 040.081 Ic = F P Ic = 11, 040.81 10, 000 Ic = 1, 040.81 F P = 50, 000 r = 5% = 0.05 t = 8 F Ic F = P(1 + r)t F = (50, 000)(1 + 0.05)8 F = 73, 872.77 Ic = F P Ic = 73, 872.77 50, 000 Ic = 23, 872.77 F
  178. 178. P = 10, 000 r = 0.5% = 0.005 t = 12 F F F = P(1 + r)t F = (10, 000)(1 + 0.005)12 F = 10, 616.78 F t r F = P(1 + r)t P P(1 + r)t = F P(1 + r)t (1 + r)t = F (1 + r)t P = F (1 + r)t P = F(1 + r) t P P = F (1 + r)t = F(1 + r) t P F r t
  179. 179. F = 50, 000 r = 10% = 0.1 t = 7 P P P = F (1 + r)t P = 50, 000 (1 + 0.1)7 P = 25, 657.91 F = 200, 000 r = 1.1% = 0.011 t = 6 P P P = F (1 + r)t P = 200, 000 (1 + 0.011)6 P = 187, 293.65 P r t Ic
  180. 180. P r t Ic F Fc = 23, 820.32 Ic = 3, 820.32 Fc = 25, 250.94 Ic = 250.94 Fc = 90, 673.22 Ic = 2, 673.22 P = 89, 632.37 t t r Ic t
  181. 181. 1 2 1 2 1 2 1 2 1 2
  182. 182. 1 2 1 2 1 2 1 2 1 2 • • m • i(m) • j j = i(m) m = • n n = tm = ⇥
  183. 183. r i(m) j r, i(m) , j i(m) i(1) = 0.02 0.02 1 = 0.02 = 2% i(2) = 0.02 0.02 2 = 0.01 = 1% i(3) = 0.02 0.02 4 = 0.005 = 0.5% i(12) = 0.02 0.02 12 = 0.0016 = 0.16% i(365) = 0.02 0.02 365 F = P(1 + j)t • j = i(m) m • t mt
  184. 184. F = P 1 + i(m) m !mt F = P = i(m) = m = t = F = P(1 + j)t F = P 1 + i(m) m !mt j i(m) m t mt
  185. 185. P = 10, 000 i(4) = 0.02 t = 5 m = 4 F P j = i(4) m = 0.02 4 = 0.005 n = mt = (4)(5) = 20 . F = P(1 + j)n = (10, 000)(1 + 0.005)20 F = 11, 048.96 Ic = F P = 11, 048.96 10, 000 = 1, 048.96 P = 10, 000 i(12) = 0.02 t = 5 m = 12 F P
  186. 186. j = i(12) m = 0.02 12 = 0.0016 n = mt = (12)(5) = 60 . F = P(1 + j)n = (10, 000)(1 + 0.0016)60 F = 11, 050.79 Ic = F P = 11, 050.79 10, 000 = 1, 050.79 P = 50, 000 i(12) = 0.12 t = 6 m = 12 F
  187. 187. F = P 1 + i(12) m !tm F = (50, 000) ✓ 1 + 0.12 12 ◆(6)(12) F = (50, 000)(1.01)72 F = 102, 354.97 P = F 1 + i(m) m !mt F = P = i(m) = m = t = F = 50, 000 t = 4 i(2) = 0.12 P j = i(2) m = 0.12 2 = 0.06
  188. 188. n = tm = (4)(2) = 8 P = F (1 + j)n P = 50, 000 (1 + 0.06)8 = 50, 000 (1.06)8 = 31, 370.62 F = 25, 000 t = 2 1 2 i(4) = 0.10 P j = i(4) m = 0.10 4 = 0.025 n = tm = (2 1 2 )(4) = 10 P = F (1 + j)n P = 25, 000 (1 + 0.025)10 = 25, 000 (1.025)10 = 19, 529.96
  189. 189. i(m) F F
  190. 190. 3%
  191. 191. i(m) x = 1 j = m i(m) m = x(i(m)) m ! 1 x = m i(m) ! 1 m x = m i(m) F = P 1 + i(m) m !mt F = P ✓ 1 + 1 x ◆xi(m)t F = P ✓ 1 + 1 x ◆x i(m)t x ✓ 1 + 1 x ◆x x ! 1 ✓ 1 + 1 x ◆x e P i(m) F t F = Pei(m)t
  192. 192. P = 20, 000 i(m) = 0.03 t = 6 F F = Pei(m)t F = Pei(m)t = 20, 000e(0.03)(6) = 20, 000e0.18 = 23, 944.35
  193. 193. P = 3, 000 F = 3, 500 i(12) = 0.25% = 0.0025 m = 12 j = i(12) m = 0.0025 12 t F = P(1 + j)n 3, 500 = 3, 000 ✓ 1 + 0.0025 12 ◆n 3, 500 3, 000 = ✓ 1 + 0.0025 12 ◆n
  194. 194. n log ✓ 3, 500 3, 000 ◆ = log ✓ 1 + 0.0025 12 ◆n log(1.166667) = n log ✓ 1 + 0.0025 12 ◆ n = 740.00 t = n m = 740 12 = 61.67 1, 000 300 12% F = 1, 300 m = 2 i(2) = 0.12 j = i(2) 2 = 0.12 2 = 0.06 n t F = P(1 + j)n 1, 300 = 1, 000(1 + 0.06)n 1.3 = (1.06)n log(1.3) = log(1.06)n log(1.3) = n log(1.06) n = log 1.3 log(1.06) = 4.503 300 n = 5 t = n m = 5 2 = 2.5 1, 000 300
  195. 195. n n = 5 n = 4.503 t F = 15, 000 P = 10, 000 t = 10 m = 2 n = mt = (2)(10) = 20 i(2) F = P(1 + j)n 15, 000 = 10, 000(1 + j)20 15, 000 10, 000 = (1 + j)20 1.5 = (1 + j)20 (1.5) 1 20 = 1 + j (1.5) 1 20 1 = j j = 0.0205 j = i(m) m 0.0205 = i(2) 2 i(2) = (0.0205)(2) i(2) = 0.0410 4.10%
  196. 196. F = 2P t = 10 s m = 4 n = mt = (4)(10) = 40 i(4) F = P(1 + j)n 2P = P(1 + j)n 2 = (1 + i)40 (2)1/40 = 1 + j (2)1/40 1 = j j = 0.0175 1.75% j = i(4) m 0.0175 = i(4) 4 i(4) = (0.0175)(4) i(4) = 0.070 7.00% •
  197. 197. • • i(1) i(4) = 0.10 m = 4 i(1) F1 = F2 P(1 + i(1) )t = P 1 + i(4) m !mt P (1 + i(1) )t = 1 + i(4) m !mt 1 t (1 + i(i) ) = ✓ 1 + 0.10 4 ◆4 i(i) = ✓ 1 + 0.10 4 ◆4 1 = 0.103813 10.38%
  198. 198. j j = (1.025)4 1 F1 = F2 t t = 1 P t i(12) = 0.12 m = 12 P t i(1) = m = 1 P t F1 F2 F1 = F2 P 1 + i(1) 1 !(1)t = P 1 + i(12) 12 !12t 1 + i(1) 1 ! = ✓ 1 + 0.12 12 ◆12 i(1) = (1.01)12 1 i(1) = 0.126825% i(2) = 0.08 m = 2 P t i(4) = m = 4 P t
  199. 199. F1 F2 F1 = F2 P 1 + i(4) 4 !(4)t = P 1 + i(2) 2 !(2)t 1 + i(4) 4 !4 = ✓ 1 + 0.08 2 ◆2 1 + i(4) 4 !4 = (1.04)2 1 + i(4) 4 = [(1.04)2 ](1/4) 1 + i(4) 4 = (1.04)1/2 1 + i(4) 4 = 1.019804 i(4) 4 = 1.019804 1 i(4) 4 = 0.019804 i(4) = (0.019804)(4) i(4) = 0.079216% i(12) = 0.12 m = 12 P t i(2) = m = 2 P t F1 F2
  200. 200. F1 = F2 P 1 + i(2) 2 !(2)t = P 1 + i(12) 12 !(12)t 1 + i(2) 2 !2 = ✓ 1 + 0.12 12 ◆12 1 + i(2) 2 !2 = (1.01)12 1 + i(2) 42 = [(1.01)12 ](1/2) 1 + i(2) 2 = (1.01)6 1 + i(2) 2 = 1.061520 i(2) 2 = 1.061520 1 i(2) 2 = 0.061520 i(2) = (0.061520)(2) i(2) = 0.12304 12.304% F = 2, 000, P = 1, 750, m = 2, t = j i(m) j = 0.016831504 1.68%, i(m) = 0.033663008 3.37% F = 100, 000, P = 10, 000, t = i(m) j j = 0.024275 2.43%, i(m) = 0.2913 29.13% F = 30, 000, P = 10, 000, i(m) = 16% j n t j = 0.04, n = 28 , t = 7 F = 18, 000, P = 12, 000, i(m) = 12% j n t j = 0.06, n = 7 , t = 3.48
  201. 201. j j j P
  202. 202. • • • t • R • F • P
  203. 203. • • P F R R R R R · · · R 0 1 2 3 4 5 · · · n P F 3, 000 3, 000 3, 000 3, 000 3, 000 3, 000 0 1 2 3 4 5 6 6 R = 3, 000 t = 6 = 1
  204. 204. P F R R R R R · · · R 0 1 2 3 4 5 · · · n • • R = 3, 000 t = 6 i(12) = 0.09 m = 12 j = 0.09 12 = 0.0075 F 3, 000 3, 000 3, 000 3, 000 3, 000 3, 000 0 1 2 3 4 5 6 t = 6 3, 000 3, 000 3, 000 3, 000 3, 000 3, 000 0 1 2 3 4 5 6 3, 000 3, 000(1 + 0.0075) 3, 000(1 + 0.0075)2 3, 000(1 + 0.0075)3 3, 000(1 + 0.0075)4 3, 000(1 + 0.0075)5
  205. 205. 3, 000 = 3, 000 (3, 000)(1 + 0.0075) = 3, 022.5 (3, 000)(1 + 0.0075)2 = 3, 045.169 (3, 000)(1 + 0.0075)3 = 3, 068.008 (3, 000)(1 + 0.0075)4 = 3, 091.018 (3, 000)(1 + 0.0075)5 = 3, 114.20 F = 18, 340.89 t = 1 t = 2 t = 3 3, 000 ⇥ 6 = 18, 000 t = 1 t = 2 F
  206. 206. F R R · · · R R 0 1 2 · · · n 1 n R R(1 + j) R(1 + j)n 2 R(1 + j)n 1 F = R + R(1 + j) + R(1 + j)2 + · · · + R(1 + j)n 2 + R(1 + j)n 1 1 + j F(1 + j) = R(1 + j) + R(1 + j)2 + R(1 + j)3 + · · · + R(1 + j)n 1 + R(1 + j)n F(1 + j) F = R(1 + j)n R F[(1 + j) 1] = R[(1 + j)n 1] F(j) = R[(1 + j)n 1] F = R (1 + j)n 1 j (1 + j)n 1 j sn s n R R R R R · · · R 0 1 2 3 4 5 · · · n F = Rsn = R (1 + j)n 1 j , R j n
  207. 207. F = R (1 + j)n 1 j = 3, 000 (1 + 0.0075)6 1 0.0075 = 18, 340.89 F R = 200 m = 12 i(12) = 0.250% = 0.0025 j = 0.0025 12 = 0.0002083 t = 6 n = tm = 6(12) = 72 F F = R (1 + j)n 1 j = 200 (1 + 0.0002083)72 1 0.0002083 = 14, 507.02
  208. 208. R = 3, 000 t = 6 i(12) = 0.09 m = 12 j = 0.09 12 = 0.0075 P P = F 1 + i(m) m !mt = 3, 000 1.0075t = 3, 000(1.0075) t . 3, 000 3, 000 3, 000 3, 000 3, 000 3, 000 0 1 2 3 4 5 6 3, 000(1.0075) 1 3, 000(1.0075) 2 3, 000(1.0075) 3 3, 000(1.0075) 4 3, 000(1.0075) 5 3, 000(1.0075) 6 (3, 000)(1.0075) 1 = 2, 977.667 (3, 000)(1.0075) 2 = 2, 955.501 (3, 000)(1.0075) 3 = 2, 933.50 (3, 000)(1.0075) 4 = 2, 911.663 (3, 000)(1.0075) 5 = 2, 889.988 (3, 000)(1.0075) 6 = 2, 868.474 P = 17, 536.79
  209. 209. P = F (1 + j)n = F 1 + i(m) m !tm = 18, 340.89 ✓ 1 + .09 12 ◆6 = 17, 536.79. F = P(1 + j)n P = F (1 + j)n P P = F (1 + j)n = F(1 + j) n . R R · · · R R 0 1 2 · · · n 1 n R(1 + j) 1 R(1 + j) 2 R(1 + j) (n 1) R(1 + j) n P = R(1 + j) 1 + R(1 + j) 2 + · · · + R(1 + j) (n 1) + R(1 + j) n P = R (1 + j)1 + R (1 + j)2 + R (1 + j)3 + · · · + R (1 + j)n 1 + R (1 + j)n
  210. 210. 1 1 + j P 1 + j = R (1 + j)2 + R (1 + j)3 + · · · + R (1 + j)n + R (1 + j)n+1 P P 1 1 + j = R 1 + j R (1 + j)n P ✓ 1 1 1 + j ◆ = R 1 + j ✓ 1 1 (1 + j)n ◆ = P ✓ 1 + j 1 1 + j ◆ = R 1 + j 1 (1 + j) n P ✓ j 1 + j ◆ = R 1 + j 1 (1 + j) n Pj = R 1 (1 + j) n P = R 1 (1 + j) n j 1 (1 + j) n j an a n P P = Ran = R 1 (1 + j) n j . F = R (1 + j)n 1 j . P = F (1 + j)n P = F (1 + j)n = R(1+j)n 1 j (1 + j)n = R (1 + j)n 1 j (1 + j) n = R 1 (1 + j) n j .
  211. 211. R R R R R · · · R 0 1 2 3 4 5 · · · n P = Rn = R 1 (1 + j) n j , R j n P P = R 1 (1 + j) n j = 3, 000 1 (1 + 0.0075) 6 0.0075 = 17, 536.79 = 200, 000 R = 16, 200 i(12) = 0.105 j = 0.105 12 = 0.00875 t = 5 n = mt = 12(5) = 60
  212. 212. P =? 16, 200 16, 200 16, 200 · · · 16, 200 0 1 2 3 · · · 60 P = R 1 (1 + j) n j = 16, 200 1 (1 + 0.00875) 60 0.00875 = 753, 702.20 = + = 200, 000 + 753, 702.20 = 953, 702.20 P = 100, 000 i(1) = 0.08 m = 1 j = 0.08 t = 3 n = mt = 1(3) = 3 R P = 10, 000 R =? R =? R =? 0 1 2 3
  213. 213. P = R 1 (1 + j) n j R = P 1 (1+j) n j = 100, 000 1 (1+0.08) 3 0.08 = 38, 803.35 F = Rsn = R (1 + j)n 1 j P = Ran = R 1 (1 + j) n j P F 0.2 365 30 30⇥365
  214. 214. • 12% = 10% =
  215. 215. • • • R R R R R · · · R 0 1 2 3 4 5 · · · n
  216. 216. j F R R R R R · · · R 0 1 2 3 4 5 · · · n F = R (1 + j)n 1 j , R j n R = 1, 000 n = 12(15) = 180 i(4) = 0.06 m = 4 F F 1, 000 1, 000 1, 000 · · · 1, 000 1, 000 0 1 2 3 · · · 179 180
  217. 217. F1 = F2 P 1 + i(12) 12 !12t = P 1 + i(4) 4 !4t 1 + i(12) 12 !12t = 1 + i(4) 4 !4t 1 + i(12) 12 !12 = (1.015)4 1 + i(12) 12 = [(1.015)4 ]1/12 i(12) 12 = (1.015)1/3 1 i(12) 12 = 0.004975 = j F = R (1 + j)n 1 j = 1, 000 (1 + 0.004975)180 1 0.004975 = 290, 076.28 R = 5, 000 n = 2(10) = 20 i(12) = 0.25% = 0.0025 m = 12 F
  218. 218. 5, 000 5, 000 5, 000 · · · 5, 000 5, 000 0 1 2 3 · · · 19 20 F1 = F2 P 1 + i(2) 2 !2t = P 1 + i(12) 12 !12t 1 + i(2) 2 !2 = 1 + i(12) 12 !12 1 + i(2) 2 !2 = (1.00020833)12 1 + i(2) 2 = [(1.00020833)12 ]1/2 i(2) 2 = (1.00020833)6 1 i(2) 2 = 0.00125063 = j F = R (1 + j)n 1 j = 5, 000 (1 + 0.00125063)20 1 0.00125063 = 101, 197.06 j
  219. 219. P R R R R · · · R 0 1 2 3 4 5 · · · n P = R 1 (1 + j) n j , R j n R = 38, 973.76 i(4) = 0.08 m = 4 n = 3 P P =? R = 38, 973.76 R = 38, 973.76 R = 38, 973.76 0 1 2 3
  220. 220. F1 = F2 P 1 + i(1) 1 !(1)t = P 1 + i(4) 4 !4t 1 + i(1) 1 ! = ✓ 1 + 0.08 4 ◆4 i(1) 1 = (1.02)4 1 i(1) 1 = j = 0.082432 = 8.24% j P = R  1 (1 + j) n j = 38, 973.76  1 (1 + 0.082432) 3 0.082432 = 100, 000.00 R = 3, 000 i(2) = 0.09 m = 2 n = 6 P
  221. 221. F1 = F2 P 1 + i(12) 12 !(12)t = P 1 + i(2) 2 !(2)t 1 + i(12) 12 !12 = ✓ 1 + 0.09 2 ◆2 1 + i(12) 12 !12 = (1 + 0.045)2 1 + i(12) 12 = [(1.045)2 ]1/12 i(12) 12 = (1.045)1/6 1 i(12) 12 = 0.00736312 = j j P = R ✓ 1 (1 + j) n j ◆ = 3, 000 ✓ 1 (1 + 0.00736312) 6 0.00736312 ◆ = 17, 545.08
  222. 222. 50, 000 1, 000, 000 0 1 2 3 · · · 20 50, 000 40, 000 40, 000 40, 000 · · · 40, 000 0 1 2 3 · · · 20 t = 0 P = F(1 + j) n = 1, 000, 000(1 + 0.05) 5 = 783, 526.17 = 50, 000 + 783, 526.17 = 833, 526.17
  223. 223. F1 = F2 P 1 + i(4) 4 !(4)(t) = P 1 + i(1) 1 !(1)(t) 1 + i(4) 4 !4 = ✓ 1 + 0.05 1 ◆1 1 + i(4) 4 = (1.05)1/4 i(4) 4 = (1.05)1/4 1 i(4) = 0.012272 P = R 1 (1 + j) n j = 40, 000 1 (1 + 0.012272) 20 0.012272 = 705, 572.68 = + = 50, 000 + 705, 572.68 = 755, 572.68 833, 526.17 755, 572.68 = 77, 953.49 t = 5
  224. 224. F = P(1 + j)n = 50, 000(1 + 0.05)5 = 63, 814.08 = 63, 814.08 + 1, 000, 000 = 1, 063, 814.08 F = R (1 + j)n 1 j = 40, 000 (1 + 0.012272)20 1 0.012272 = 900, 509.40 = 63, 814.08 + 900, 509.40 = 964, 323.48 1, 063, 814.08 964, 323.48 = 99, 490.60. P = 99, 490.60(1 + 0.05) 5 = 77, 953.49
  225. 225. 0 1 2 3 4 5 · · · 0 1 2 3 · · · 20 P1 = F(1 + j) n = 150, 000(1 + 0.04) 6 = 118, 547.18 P2 = F(1 + j) n = 300, 000(1 + 0.04) 10 = 202, 669.25 = P1 + P2 = 118, 547.18 + 202, 669.25 = 321, 216.43
  226. 226. F1 = F2 P 1 + i(4) 4 !4(5) = P 1 + i(2) 2 !2(5) 1 + i(4) 4 !20 = ✓ 1 + 0.08 2 ◆10 1 + i(4) 4 = (1.04)1/2 i(4) 4 = (1.04)1/2 1 i(4) 4 = 0.019803903 P = R 1 (1 + j) n j = 25, 000 1 (1 + 0.019803903) 20 0.019803903 = 409, 560.4726 P F
  227. 227. • • 15, 000 = 2, 000 1 (1 + j) 8 j j
  228. 228. • • • • • • •
  229. 229. P R R R R R · · · R 0 1 2 3 4 5 · · · n P = Ran = R 1 (1 + j) n j , R j n R = 2, 500 i(12) = 0.09 t = 1 m = 12 P P =? 2, 500 2, 500 · · · R 0 1 2 · · · n j = i(12) m = 0.09 12 = 0.0075 n = mt = 12(1) = 12 P = Ran = 2, 500 1 (1 + 0.0075) 12 0.0075 = 28, 587.28.
  230. 230. n j P =? 2, 500 2, 500 · · · R 0 1 2 3 4 5 · · · P 0 =? 2, 500 2, 500 2, 500 2, 500 2, 500 · · · R 0 1 2 3 4 5 · · · P0 P0 = Ran = R 1 (1 + j) n j = 2, 500 1 (1 + 0.0075) 15 0.0075 = 35, 342.49 P⇤ = Ran = R 1 (1 + j) n j = 2, 500 1 (1 + 0.0075) 3 0.0075 = 7, 388.89 P = P0 P⇤ = 35, 342.49 7, 388.89 = 27, 953.60
  231. 231. k R⇤ R⇤ R⇤ · · · R⇤ R R · · · R 0 1 2 · · · k k + 1 k + 2 · · · k + n P R R · · · R 0 1 2 · · · k k + 1 k + 2 · · · k + n P = Rak+n Rk = R 1 (1 + j) (k+n) j R 1 (1 + j) k j , R i n k
  232. 232. R = 10, 000 i(4) = 0.08 t = 5 m = 4 P k = mt = 4(20) = 80 n = mt = 4(5) = 20 j = i(4) m = 0.08 4 = 0.02 P 10, 000 10, 000 · · · 10, 000 0 1 2 · · · 80 81 82 · · · 100 P = Rak+n Rak = R 1 (1 + j) (k+n) j R 1 (1 + j) k j = 10, 000 1 (1 + 0.02) 100 0.02 10, 000 1 (1 + 0.02) 80 0.02 = 33, 538.38
  233. 233. R = 4, 000 i(12) = 0.10 t = 2 m = 12 P k = 2 n = mt = 12(2) = 24 j = i(12) m = 0.10 12 = 0.00833 k + n = 2 + 24 = 26 P 4, 000 4, 000 · · · 4, 000 0 1 2 3 4 · · · 26 P = R 1 (1 + j) (k+n) j R 1 (1 + j) k j = 4, 000 1 (1 + 0.00833) 26 0.00833 4, 000 1 (1 + 0.00833) 2 0.00833 = 85, 260.53
  234. 234.
  235. 235. • • •
  236. 236. • • •
  237. 237. • • • • = 30, 000, 000 = 700, 000 = = 30, 000, 000 700, 000 = 42.86 = 3% = 500 = 200 500 ⇥ 0.03 = 15 15 ⇥ 200 = 3, 000.
  238. 238. = ⇥ ⇥ = 0.03(500)(200) = 3, 000 = 8 = 52 = = 8 52 = 0.1538 = 15.38% = 12 = 95 = = 12 95 = 0.1263 = 12.63% • • • r • P
  239. 239. • F P = F P < F P > F • • F = 300, 000 r = 10% 300, 000(0.10) = 30, 000 30, 000 1 2 = 15, 000 F = 100, 000 r = 5% = 10 = 2(10) = 20 = 4%
  240. 240. 100, 000 0.05 2 = 2, 500 t = 10 P = F (1 + j)n = 100, 000 (1 + 0.04)10 = 67, 556.42 (1 + 0.04)1 = 1 + i(2) 2 !2 i(2) 2 = 0.019804 P = R 1 (1 + j) n j = 2, 500 1 (1 + 0.019804) 20 0.019804 = 40, 956.01, = 67, 556.42 + 40, 956.01 = 108, 512.43. = ⇥ ⇥
  241. 241. • • • • • • • • •
  242. 242. • 57.29 0.10 = 57.19 • • • •
  243. 243. • • • • •
  244. 244. • • • •
  245. 245. • • • • •
  246. 246. P = 1, 000, 000 j = 0.07 n = 1 F F = P(1 + j)n = 1, 000, 000(1 + 0.07) = 1, 070, 000 P = 1, 200, 000 = 31, 000 = (31, 000)(12 )(5 ) = 1, 860, 000 = 1, 860, 000 1, 200, 000 = 660, 000
  247. 247. = ( ) ⇥ ( ) = 0.20(3, 000, 000) = 600, 000 = ( ) ( ) = 3, 000, 000 600, 000 = 2, 400, 000 (0.80)( 3, 000, 000) = 2, 400, 000. P = 400, 000 i(12) = 0.09 j = i(12) 12 = 0.09 12 = 0.005 n = 36 R R = P  1 (1 + j) n j = 400, 000  1 (1 + 0.0075) 36 0.0075 = 12, 719.89
  248. 248. n k Bk n k k P R R · · · R R R · · · R 0 1 2 · · · k k + 1 k + 2 · · · k + n Bk pOBk R = 11, 122.22 i(12) = 0.12 j = i(12) 12 = 0.12 12 = 0.01 k = 12 n k = 48 Bk = R " 1 (1 + j) (n k) j # = 11, 122.22  1 (1.01) 48 0.01 = 422, 354.73
  249. 249. P = 3, 200, 000 i(12) = 0.12 j = i(12) 12 = 0.12 12 = 0.01 n = mt = (12)(20) = 240 R P = R  1 (1 + j) n j R = P  1 (1 + j) n j = 3, 200, 000  1 (1 + 0.01) 240 0.01 = 35, 234.76 P = 3, 200, 000 R = 35, 234.76 n = 240 240⇥ 35, 234.76 = 8, 456, 342.40. = = 8, 456, 342.40 3, 200, 000 = 5, 256, 342.40
  250. 250. P = 3, 200, 000 i(12) = 0.12 j = i(12) 12 = 0.12 12 = 0.01 n = mt = (12)(20) = 240 R = 35, 234.76 B50 = R  1 (1 + j) 190 j = 35, 234.76  1 (1 + 0.01) 190 0.01 = 2, 991, 477.63 PR51 PR51 = R I50 = 35, 234.76 29, 914.78 = 5, 319.98
  251. 251. F = (1 + j)n = 100, 000, (1 + 0.08)3 = 251, 942.40 B3 = R  1 (1 + j) 2 j = 26, 379.75  1 (1 + 0.10) 2 0.10 = 45, 783.04
  252. 252. B1 = R  1 (1 + j) 5 j = 183, 026.37  1 (1.06) 5 0.06 = 770, 973.65 R = P  1 (1 + j) n j = 30, 000  1 (1 + 0.0075) 12 0.0075 = 2, 623.54 R = P  1 (1 + j) n j = 750, 000  1 (1 + 0.02) 8 0.02 = 102, 382.35 R = P  1 (1 + j) n j = 500, 000  1 (1 + 0.01) 36 0.01 = 16, 607.15 B2 = R  1 (1 + j) 34 j = 16, 607.15  1 (1 + 0.01) 34 0.01 = 476, 669.63 I3 = i(P2) = (0.01)(476, 669.63) = 4, 766.00 PR51 = R I50 = 16, 607.15 4, 766.7 = 11, 840.45
  253. 253. F = P(1 + j)n = 800, 000(1 + 0.08)2 = 933, 120 R = P  1 (1 + j) n j = 700, 000  1 (1 + 0.01) 48 0.01 = 18, 433.68 R = P  1 (1 + j) n j = 300, 000  1 (1 + 0.025) 20 0.025 = 19, 244.14 B6 = R  1 (1 + j) 18 j = 29, 994.20 2 6 6 6 4 1 ✓ 1 + 0.10 12 ◆ 18 0.10 12 3 7 7 7 5 = 499, 428.21 B8 = R  1 (1 + j) 12 j = 32, 073.56  1 (1.025) 12 0.025 = 329, 003.03
  254. 254. P1 = R  1 (1 + j) 3 j = 11, 485.35 2 6 6 6 4 1 ✓ 1 + 0.10 12 ◆ 3 0.10 12 3 7 7 7 5 = 33, 889.65 P4 = R  1 (1 + j) 1 j = 25, 045.65  1 (1 + 0.08) 1 0.08 = 23, 190.42 I5 = j(P4) = (0.08)(23, 190.42) = 1, 855.23 PR5 = R I5 = 25, 045.65 1, 855.23 = 23, 190.42
  255. 255. T F p p p1, p2, ... p q r s t 3 + 2 = 5 u f(x) = p x x + 1 v w p1 p 2
  256. 256. p2 p3 p4 p p q r r s s t 3 + 2 = 5 t 2 2 N 100 2 Z ⇡ /2 Q ⇡ ⇡ 2 R ⇡ p 2 < 2 N ⇢ Z ⇢ Q ⇢ R u f u v w w w
  257. 257. w w p1 p 2 p1 p2 p2 p3 p4 2x = 1 (x + y)2 = x2 + y2 x x y p q p p q p q p q < · >
  258. 258. p r t u p1 p 2 p2 p3 p4 p1 p2 p3 p4 p1 r p 2 p2 f i b p3 h g p4 d l p1 r p2 f i b p3 h g p4 d l l d
  259. 259. 2.5
  260. 260. p p T F p q p q p q 22 = 4 p q p q T T T F F T F F p q r 23 = 8
  261. 261. p q r T T T T T F T F T T F F F T T F T F F F T F F F n 2n ⇠ p ⇠ p : p, ) p ⇠ p T F F T p ⇠ p p ⇠ p n1 p(x) = x 1 x+2 n2 2 n3 n4
  262. 262. ⇠ n1 p(x) = x 1 x+2 p(x) = x 1 x+2 ⇠ n2 2 2 ⇠ n3 ⇠ n4 ⇠ p ⇠ p p ⇠ n2 n2 ⇠ p p ^ p q p ^ q : (p q, ) p q p ^ q T T T T F F F T F F F F p q p^q p q p q
  263. 263. p q ⇡ > 3 p ^ q p ^ (⇠ q) ⇡  3 ⇡ > 3 p ^ q ⇡ > 3 p ^ (⇠ q) ⇡  3 ⇡  3 (⇠ p) ^ (⇠ q) (⇠ p) ^ q q p ^ q p p ^ (⇠ q) (⇠ p) ^ (⇠ q) ⇠ q p ^ q p : q :
  264. 264. _ p q p _ q : (p q, ) p q p _ q T T T T F T F T T F F F p q p _ q p q p _ q p q p q r p q r p _ q q _ (⇠ r) p _ (q _ r)
  265. 265. p _ q q _ (⇠ r) p _ (q _ r) (p _ q) _ r (p ^ q) _ r p _ (q ^ r) (p ^ q) _ (p ^ r) p _ (q _ r) (p _ q) _ r p _ (q _ r) (p _ q) _ r p _ q _ r p ^ (q ^ r) (p ^ q) ^ r p ^ q ^ r p q r (⇠ p) _ (q ^ r) r q q ^ r p (⇠ p) (⇠ p) _ (q ^ r) p q r ⇠ p q ^ r (⇠ p) _ (q ^ r) F F T T F T
  266. 266. p q p ! q : ( p, q, ) p q p ! q T T T T F F F T T F F T p ! q p q p q p1 p2 p3 p1 p1 p2 p2 p3 p3
  267. 267. p ! q p ! q p q p q p ! q p ! q p ! q p ! q p p ! q q 2 > 0 2 > 0
  268. 268. 2 < 0 2 2 > 0 2 < 0 2 > 0 2 < 0 2 > 0 p q p $ q : (p q, ) p q p $ q T T T T F F F T F F F T p q p q
  269. 269. p q p1 p2 p3 p1 p1 p2 p3 p3 p q r p q r s
  270. 270. p ^ (⇠ q) p ^ (⇠ p) ⇠ (q _ r) ((⇠ p) _ q) ^ r (p ^ (⇠ q)) _ (r ^ s) ⇠ (q ! r) (p ^ (⇠ q)) ! (r ^ s) p $ (⇠ p) ((⇠ p) ! q) $ r p q r s T F T T
  271. 271. X X X X X X X X X X X X X X X X X X X X X X X X X X X X
  272. 272. p ⇠ p p q p ^ q p q p _ q p q p ! q p q p $ q p q p q (p ! q) ^ (q ! p) p q p q p q T T T F F T F F
  273. 273. (p ! q) (q ! p) p q p ! q q ! p T T T T T F F T F T T F F F T T (p ! q) ^ (q ! p) p q p ! q q ! p (p ! q) ^ (q ! p) T T T T T T F F T F F T T F F F F T T T (p ! q) ^ (q ! p) p $ q p q (p ! q) ^ (q ! p) p $ q T T T T T F F F F T F F F F T T $ [(p ! r) ^ (q ! r)] ! [(p _ q) ! r] (p ! r) ^ (q ! r) (p _ q) ! r p, q r 23 = 8 p, q r p, q r
  274. 274. (p ! r) ^ (q ! r) p ! r q ! r (p ! r)^(q ! r) p q r p ! r q ! r (p ! r) ^ (q ! r) T T T T T T T T F F F F T F T T T T T F F F T F F T T T T T F T F T F F F F T T T T F F F T T T (p_q) ! r p_q (p _ q) ! r p q r p ! r q ! r (p ! r) ^ (q ! r) p _ q (p _ q) ! r T T T T T T T T T T F F F F T F T F T T T T T T T F F F T F T F F T T T T T T T F T F T F F T F F F T T T T F T F F F T T T F T s : [(p ! r) ^ (q ! r)] ! [(p _ q) ! r] p q r p ! r q ! r (p ! r) ^ (q ! r) p _ q (p _ q) ! r s T T T T T T T T T T T F F F F T F T T F T T T T T T T T F F F T F T F T F T T T T T T T T F T F T F F T F T F F T T T T F T T F F F T T T F T T
  275. 275. T [(p ! r) ^ (q ! r)] ! [(p _ q) ! r] p, q, r ⌧ p q p _ ⌧ p ^ p ! (p _ q) (p ^ (⇠ q)) ^ (p ^ q) ⌧ T p _ ⌧ p p ⌧ p _ ⌧ T T T F T T F p ^ p p ^ p p ^ T F F F F F p ! (p _ q) p q p q p _ q p ! (p _ q) T T T T T F T T F T T T F F F T (p^ ⇠ q) ^ (p ^ q) p q
  276. 276. p q ⇠ q p ^ (⇠ q) p ^ q (p ^ (⇠ q)) ^ (p ^ q) T T F F T F T F T T F F F T F F F F F F T F F F ((p ! q) ^ q) ! p ((p ! q) ^ (⇠ p)) !⇠ q ((p _ q) ^ p) ! (⇠ q) (p ! q) ! (q ! p) (⇠ (p ^ q) ^ (⇠ p)) ! q (p ! q) ! ((⇠ p) ! (⇠ q)) (p ^ q) ! p p ! (p _ q) (p ^ q) ! (p ^ q) ((p ! q) ^ p) ! q ((p ! q) ^ (⇠ q)) !⇠ p ((p ! q) ^ (q ! r)) ! (p ! r) ((p _ q) ^ (⇠ p)) ! q ((⇠ p) ! ) ! p ((p ! r) ^ (q ! r)) ! ((p _ q) ! r)
  277. 277. (p ! q) ^ (q ! p) p $ q p q (p ! q) ^ (q ! p) p $ q T T T T T F F F F T F F F F T T p q p , q p , q p q p $ q (p ! q) , [(⇠ p) _ q]
  278. 278. (p ! q) (⇠ p) _ q p q p ! q ⇠ p (⇠ p) _ q T T T F T T F F F F F T T T T F F T T T (p ! q) (⇠ p) _ q (p ! q) , [(⇠ p) _ q] (p ! q) $ [(⇠ p) _ q] p q p ! q ⇠ p (⇠ p) _ q (p ! q) $ [(⇠ p) _ q] T T T F T T T F F F F T F T T T T T F F T T T T (p ! q) $ [(⇠ p) _ q] p q r (p ^ ⌧) , p (p _ ) , p (p _ ⌧) , ⌧ (p ^ ) , (p _ p) , p (p ^ p) , p (p _ [⇠ p]) , ⌧ (p ^ [⇠ p]) , ⇠ (⇠ p) , p p _ (q _ r) , (p _ q) _ r p ^ (q ^ r) , (p ^ q) ^ r p _ q , q _ p p ^ q , q ^ p p _ (q ^ r) , (p _ q) ^ (p _ r) p ^ (q _ r) , (p ^ q) _ (p ^ r) ⇠ (p _ q) , (⇠ p) ^ (⇠ q) ⇠ (p ^ q) , (⇠ p) _ (⇠ q) p _ (p ^ q) , p p ^ (p _ q) , p
  279. 279. ⇠ (p ! q) , [p ^ (⇠ q)] ⇠ (p ! q) p ^ (⇠ q) ⇠ (p ! q) p ^ (⇠ q) ⇠ (p ! q) , ⇠ ((⇠ p) _ q) , ⇠ (⇠ p) ^ (⇠ q) , p ^ (⇠ q) p q p ! q q ! p ⇠ p !⇠ q ⇠ q !⇠ p p q p ! q q ! p ⇠ p ⇠ q ⇠ p !⇠ q ⇠ q !⇠ p T T T T F F T T T F F T F T T F F T T F T F F T F F T T T T T T (p ! q) ,⇠ q !⇠ p (q ! p) , (⇠ p !⇠ q)
  280. 280. p q p ! q q ! p ⇠ q !⇠ p ⇠ p !⇠ q (p ! q) ,⇠ q !⇠ p ⇠ q !⇠ p , ⇠ (⇠ q)_ ⇠ p , q_ ⇠ p , ⇠ p _ q , p ! q (q ! p) , (⇠ p !⇠ q) ⇠ p !⇠ q , ⇠ (⇠ p) _ (⇠ q)) , p ^ (⇠ q) , (⇠ q) ^ p , q ! p p ! q :
  281. 281. q ! p : ⇠ q !⇠ p : ⇠ p !⇠ q : (⇠ q !⇠ p) p ! q p ! q p q ⇠ p ⇠ q
  282. 282. ⇠ (p _ ((⇠ p) ^ q)) , ((⇠ p) ^ (⇠ q)) ((p ^ q) ! (p _ q)) , ⌧ ⇠ (p ! (⇠ q)) , (p ^ q) ⇠ (p ! (⇠ q)) , ⇠ (⇠ p_ ⇠ q) , ⇠ (⇠ p)^ ⇠ (⇠ q) , p ^ q (p ! (q ^ r)) , ((p ! q) ^ (p ! r)) (p ! (q ^ r)) , ⇠ p _ (q ^ r) , (sin p _ q) ^ (⇠ p _ r) , ((p ! q) ^ (p ! r)) (p _ q) , ((⇠ p) ! q) ⇠ (p ! q) , (p ^ (⇠ q)) ((p ! r) ^ (q ! r)) , ((p _ q) ! r) ((p ! q) _ (p ! r)) , (p ! (q _ r)) ((p ! r) _ (q ! r)) , ((p ^ q) ! r)
  283. 283. (p1 ^ p2 ^ . . . ^ pn) ! q. p1, p2, . . . , pn q p1 p2 pn ) q
  284. 284. p1 : p2 : q : (p1 ^ p2) ! q, p1 p2 ) q A A0 A A0 p ! q p ! q p q ) q ) p A p ! q p q p ! q p p p ! q q
  285. 285. p q p ! q T T T T F F F T T F F T A A0 p ! q q p p ! q q p ! q q p p q p ! q T T T T F F F T T F F T A0 p ! q ! q ) p ) q p1, p2, . . . , pn (p1 ^ p2 ^ . . . ^ pn) ! q
  286. 286. ((p ! q) ^ p) ! q ((p ! q) ^ p) ! q p q p ! q (p ! q) ^ p ((p ! q) ^ p) ! q T T T T T T F F F T F T T F T F F T F T ((p ! q) ^ p) ! q ((p ! q) ^ p) ! q p ! q p ) q
  287. 287. p q r (p ^ q) ! p p ^ q ) p p ! (p _ q) p ) p _ q (p ^ q) ! (p ^ q) p q ) p ^ q ((p ! q) ^ p) ! q p ! q p ) q ((p ! q) ^ (⇠ q)) !⇠ p p ! q ⇠ q )⇠ p ((p ! q) ^ (q ! r)) ! (p ! r) p ! q q ! r ) p ! r ((p _ q) ^ (⇠ p)) ! q p _ q ⇠ p ) q ((⇠ p) ! ) ! p (⇠ p) ! ) p ((p ! r) ^ (q ! r)) ! ((p _ q) ! r) p ! r q ! r ) (p _ q) ! r
  288. 288. p : q : p ! q ⇠ q )⇠ p p : q : p ^ q ) p
  289. 289. p : q : p ) p _ q, p : q : r : p ! q q ! r ) p ! r (p1 ^ p2 ^ . . . ^ pn) ! q, p1, p2, . . . , pn q (p1 ^ p2 ^ . . . ^ pn) ! q
  290. 290. ((p ! q) ^ q) ! p ((p ! q) ^ q) ! p p q p ! q (p ! q) ^ q ((p ! q) ^ q) ! p T T T T T T F F F T F T T T F F F T F T q p ! q p p q p q p q p ! q (p ! q) ^ q ((p ! q) ^ q) ! p F T T T F p ! q q ) p
  291. 291. A0 B0 p q r ((p ! q) ^ q) ! p p ! q q ) p ((p ! q) ^ (⇠ p)) ! (⇠ q) p ! q ⇠ p )⇠ q ((p _ q) ^ p) ! (⇠ q) p _ q p )⇠ q (p ! q) ! (q ! p) p ! q ) q ! p (⇠ (p ^ q) ^ (⇠ p)) ! q ⇠ (p ^ q) ⇠ p ) q (p ! q) ! ((⇠ p) ! (⇠ q)) p ! q ) (⇠ p) ! (⇠ q)
  292. 292. p : q : p _ q p )⇠ q [(p _ q) ^ p] !⇠ q p q p q [(p _ q) ^ p] ⇠ q ⇠ q q p p q ⇠ q p _ q (p _ q) ^ p [(p _ q) ^ p] !⇠ q T T F T T F p _ q ⇠ q ) p (p ^ q) , (q ^ p) q _ p ⇠ q ) p ⇠ (p ^ q) ⇠ p ) q p q p q
  293. 293. p ⇠ p q p ^ q ⇠ (p ^ q) ⇠ (p ^ q)^ ⇠ p [⇠ (p ^ q)^ ⇠ p] ! q F T F F T T F p ! q p ) q p ! q p q
  294. 294. (p ^ q) ! r p ^ q ) r (p ^ q) ! r r ) p ^ q
  295. 295. p !⇠ q q )⇠ p p !⇠ q ⇠ p ) q
  296. 296. (p ^ (⇠ q)) ! r s ! p q ! (⇠ u) u ^ s ) r (((p ^ (⇠ q)) ! r) ^ (s ! p) ^ (q ! (⇠ u)) ^ (u ^ s)) ! r 24 = 16 p ! (r ^ s) ⇠ r )⇠ p ⇠ p
  297. 297. ⇠ r ⇠ r ⇠ p p ! (r^s) ⇠ (r^s) ⇠ r (⇠ r)_(⇠ s) [(⇠ r)_(⇠ s)] ,⇠ (r ^s) ⇠ r (⇠ r) _ (⇠ s) ⇠ (r ^ s) p ! (r ^ s) ⇠ p (p ^ r) ! (⇠ q) (⇠ q) ! r ⇠ r )⇠ (p ^ r) (p ^ r) ! (⇠ q) (⇠ q) ! r (p ^ r) ! r ⇠ r ⇠ (p ^ r) (p ^ r) ! r ⇠ r (p ^ r) ! (⇠ q) (⇠ q) ! r (p ^ r) ! r ⇠ r ⇠ (p ^ r)
  298. 298. ⇠ r ⇠ r (⇠ q) ! r ⇠ (⇠ q) (p ^ r) ! (⇠ q) ⇠ (p ^ r) ⇠ r (⇠ q) ! r ⇠ (⇠ q) (p ^ r) ! (⇠ q) ⇠ (p ^ r) p _ r (⇠ r) _ (⇠ s) s ) p s ⇠ r_ ⇠ s ⇠ s s ⇠ r p _ r p s ⇠ (⇠ s) (⇠ r) _ (⇠ s) ⇠ r p _ r p
  299. 299. p_r (⇠ r)_(⇠ s) ⇠ p ! r r !⇠ s p _ r ⇠ p ! r (⇠ r) _ (⇠ s) r !⇠ s ⇠ p !⇠ s s ⇠ (⇠ s) ⇠ (⇠ p) p b, e, c, d b : e : c : d : b ! e c ! d b _ c ) e _ d
  300. 300. b _ c ⇠ (⇠ b) _ c (⇠ b) ! c c ! d (⇠ b) ! d b ! e (⇠ e) ! (⇠ b) (b ! e) , (⇠ e) ! (⇠ b) (⇠ e) ! d ⇠ (⇠ e) _ d e _ d p ! q r ! s p _ r ) q _ s t : d : s :
  301. 301. t _ d t ! s ) d ! s d ! s d s t t _ d t ! s t d s t d s t _ d t ! s d ! s F T F T T F p _ q ⇠ q ) p p _ q ⇠ q p
  302. 302. p q p _ q ⇠ q ((p _ q)^ ⇠ q) ((p _ q)^ ⇠ q) ! p T T T F F T T F T T T T F T T F F T F F F T F T ((p _ q)^ ⇠ q) ! p p ⇠ q q p q p _ q p _ q ⇠ (p _ q) ^ (p _ q) , m m = 2k k m m = 2k k m m = 2k k x y x + y
  303. 303. x x x = 2k1 k1 y y = 2k2 k2 x + y = 2k1 + 2k2 = 2(k1 + k2) k1 + k2 x + y ABCD mA = 90 mB = 90 mC = 85 ABCD ABCD D = 90 360 mA + mB + mC + mD = 360 . 90 + 90 + 85 + mD = 360 . mD = 95 ABCD A B ABCD C A B D mA + mB + mC + mD = 360 , 90 + 90 + mC + 90 = 360 mC = 90 C 85 ABCD ABCD
  304. 304. (p _ q) ! r q ) r q p _ q r (p ^ q) ! r q ) r p, r q p ! (q ^ r) ) (p ! q) ^ (p ! r) p ! (q ^ r) ⇠ p _ (q ^ r) (⇠ p _ q) ^ (⇠ p _ r) (p ! q) ^ (p ! r) p ! (q _ r) r ! s ) p ! s p, q r, s p ! (r _ t) ⇠ r )⇠ p ⇠ r ⇠ r_ ⇠ t ⇠ (r ^ t) ⇠ p (p _ q) ! r ⇠ r ) ⇠ p
  305. 305. ⇠ r ⇠ (p _ q) ⇠ p^ ⇠ q ⇠ p (p ^ q) ! r ⇠ r ) ⇠ p p q, r m 3 m = 3k k n 6 n = 6k k 3 6 m 3 m = 3k k n 6 n = 6k k 6 3 m = 6k k m = 3 · (2k) 2k m 3 a, b c a2 + b2 = c2 ABC 6 p 2 ABC a2 + 62 = (6 p 2)2 a = 6 ABCD

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