Calculus Better Explained 2018 PDF

MMTH011/MAH101M: MATHS MADE EASY EXAM PRACTICE BOOK PREPARED BY: MR E. CHAUKE It is illegal to photocopy any pages from this book without the written permission of the copyright holder.  Author: Emmanuel Chauke...

MMTH011/MAH101M: MATHS MADE EASY EXAM PRACTICE BOOK PREPARED BY: MR E. CHAUKE It is illegal to photocopy any pages from this book without the written permission of the copyright holder.  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS TABLE OF CONTENTS: Compiled by Mr. E. Chauke: OVERVIEW:…………….…………………………………………...……… 1  Algebra Refresher………………………………………………………………………………………………..…….. 5  About the Author………………………………………………………………………………………….………….... 8  Acknowledgements…………………………………………………………………………………….…………...... 9  Purpose of the book…………………………………….…………………………………………….……...………. 10  Massage to the students……………………………………………………………….……………………………. 10 CHAPTER 1: ………………………...………………..………..…..….. 11 ABSOLUTE VALUES/MODULUS. CHAPTER 2: ……………………………………………………………… 17 THE RELTIONSHIP BETWEEN RADIANS AND DEGREES. CHAPTER 3: ……………………………………………………………… 21 SOLVING LOGARITHMIC AND TRIGONOMETRIC EQUATIONS. CHAPTER 4: …………………….……………………………………….. 34 FUNCTIONS. CHAPTER 5: ………………………………………………………….….. 53 THE LIMIT OF A FUNCTIONS. CHAPTER 6: ……………………………………….………………….…. 94 DERIVATIVES OF ORDINARY FUNCTIONS. 2|Page  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS CHAPTER 7: …………………………………………………...………. 120 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS. CHAPTER 8: ……………………………………………………….…… 128 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS. CHAPTER 9: ……………………………………………………………. 140 IMPLICIT AND HIGHER ORDER DIFFERENTIATION. CHAPTER 10: …………………………………………………………. 148 DERIVATIVES OF THE INVERSE TRIGONOMETRIC FUNCTIONS. CHAPTER 11…………………………………………………………… 153 L’HOPITAL’S RULE. CHAPTER 12: …………………………………………………………. 159 ROLLE’S & MEAN THEOREM. CHAPTER 13: ……………………………………………………….… 168 APPLICATION OF CALCULUS. CHAPTER 14: ………………………………………………………… 178 INTEGRATION. CHAPTER 15: ………………………………………………………… 244 THE AREAS BETWEEN THE CURVES. 3|Page  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS CHAPTER 16: …..………………………………………….…………. 249 THE ARC LENGTH. SUMMERY AND THE FORMULAS: ….………………………….. 256  FORMULAS.  PROPERTIES OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS.  TRIG-INVERSE FUNCTIONS.  DIFFEENTIATION TOOL-BOX.  TOOL-BOX FOR INTEGRATION. CHECK YOURSELF QUESTIONS: ………………………………. 263 TESTBANKS & EXAMS PAPAERS: ………………………….… 281 MEMORANDUMS: …………………………………………… 306-324 4|Page  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Trigonometry Tool-Box Trigonometric ratios in right-angled triangle: 𝑦 𝑟 1. sin 𝑥 = 4. cosec 𝑥 = 𝑦 𝑟 𝑟 𝑦 𝑥 𝑟 2. cos 𝑥 = 𝑟 5. sec 𝑥 = 𝑥 𝜃 𝑦 𝑥 3. tan 𝑥 = 𝑥 6. cot 𝑥 = 𝑦 𝑥 Measured in Radians Using special angles to get exact values: 𝝅 𝝅 𝜽 Radians 𝐬𝐢𝐧 𝜽 𝐜𝐨𝐬 𝜽 𝐭𝐚𝐧 𝜽 𝟔 𝟒 0° 0 0 1 0 𝟐 √𝟐 30° 𝜋⁄6 1⁄2 √3⁄2 1⁄√3 √𝟑 𝟏 45° 𝜋⁄4 1⁄√2 1⁄√2 1 𝝅 𝝅 60° 𝜋⁄3 √3⁄2 1⁄2 √3 𝟑 𝟒 90° 𝜋⁄2 1 0 − 𝟏 𝟏 Pythagorean Identities: Double Angle Identities: 1. sin2 𝜃 + cos2 𝜃 = 1 1. sin 2𝜃 = 2 sin 𝜃 cos 𝜃 2. 1 + tan2 𝜃 = sec 2 𝜃 3. 1 + cot 2 𝜃 = cosec 2 𝜃 2. cos 2𝜃 = cos2 𝜃 − sin2 𝜃 = 1 − 2sin2 𝜃 = 2cos2 𝜃 − 1 2 tan 𝜃 3. tan 2𝜃 = 1−tan2 𝜃 Compound Angle Identities: Half Angle Identities: 1−cos 2𝜃 𝜃 1−cos 𝜃 1. sin(𝜃 + 𝛽) = sin 𝜃 cos 𝛽 + cos 𝜃 sin 𝛽 1. sin2 𝜃 = 4. sin 2 = √ 2 2 2. sin(𝜃 − 𝛽) = sin 𝜃 cos 𝛽 − cos 𝜃 sin 𝛽 3. cos(𝜃 + 𝛽) = cos 𝜃 cos 𝛽 − sin 𝜃 sin 𝛽 1+cos 2𝜃 𝜃 1+cos 𝜃 4. cos(𝜃 − 𝛽) = cos 𝜃 cos 𝛽 + sin 𝜃 sin 𝛽 2. cos 2 𝜃 = 5. cos 2 = √ 2 2 tan 𝜃+tan 𝛽 5. tan(𝜃 + 𝛽) = 1−tan 𝜃 tan 𝛽 1−cos 2𝜃 𝜃 sin 𝜃 1−cos 𝜃 tan 𝜃−tan 𝛽 6. tan(𝜃 − 𝛽) = 1+tan 𝜃 tan 𝛽 3. tan2 𝜃 = 1+cos 2𝜃 6. tan 2 = 1+cos 𝜃 = sin 𝜃 5|Page  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Fundamental Identities / Negative Angles: Co-functions Identities: 𝜋 1. sin(−𝜃) = − sin 𝜃 1. sin ( 2 − 𝜃) = cos 𝜃 2. cos(−𝜃) = cos 𝜃 𝜋 2. sin ( 2 + 𝜃) = cos 𝜃 3. tan(−𝜃) = − tan 𝜃 𝜋 4. cosec(−𝜃) = − cosec 𝜃 3. cos (2 − 𝜃) = sin 𝜃 5. sec(−𝜃) = sec 𝜃 𝜋 4. cos (2 + 𝜃) = − sin 𝜃 6. cot(−𝜃) = − cot 𝜃 𝜋 5. tan ( 2 − 𝜃) = cot 𝜃 𝜋 6. cot (2 − 𝜃) = tan 𝜃 𝜋 7. sec ( 2 − 𝜃) = cosec 𝜃 𝜋 8. cosec ( 2 − 𝜃) = sec 𝜃 Tangents and Cotangents Formulas: Reciprocal Identities: sin 𝜃 1 1. tan 𝜃 = cos 𝜃 1. cosec 𝜃 = sin 𝜃 cos 𝜃 1 2. cot 𝜃 = 2. sec 𝜃 = cos 𝜃 sin 𝜃 1 3. cot 𝜃 = tan 𝜃 Sum to product formulas: Product to sum formulas: 𝜃+𝛽 𝜃−𝛽 1 1. sin 𝜃 + sin 𝛽 = 2 sin ( ) cos ( ) 1. sin 𝜃 cos 𝛽 = 2 [ sin(𝜃 + 𝛽) + sin(𝜃 − 𝛽) ] 2 2 𝜃+𝛽 𝜃−𝛽 1 2. sin 𝜃 − sin 𝛽 = 2 cos ( ) sin ( ) 2. cos 𝜃 sin 𝛽 = 2 [ sin(𝜃 + 𝛽) − sin(𝜃 − 𝛽) ] 2 2 𝜃+𝛽 𝜃−𝛽 1 3. cos 𝜃 + cos 𝛽 = 2 cos ( ) cos ( ) 3. cos 𝜃 cos 𝛽 = 2 [ cos(𝜃 + 𝛽) + cos(𝜃 − 𝛽) ] 2 2 1 4. cos 𝜃 − cos 𝛽 = −2 sin ( 𝜃+𝛽 𝜃−𝛽 ) sin ( ) 4. sin 𝜃 sin 𝛽 = 2 [ cos(𝜃 − 𝛽) − cos(𝜃 + 𝛽) ] 2 2 6|Page  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Periodic Formulas: Trigonometric Quadrants: If 𝒏 is an integer. 1. sin(𝜃 + 2𝜋𝑛) = sin 𝜃 S A 2. cos(𝜃 + 2𝜋𝑛) = cos 𝜃 3. tan(𝜃 + 2𝜋𝑛) = tan 𝜃 4. cosec(𝜃 + 2𝜋𝑛) = cosec 𝜃 T C 5. sec(𝜃 + 2𝜋𝑛) = sec 𝜃 6. cot(𝜃 + 2𝜋𝑛) = cot 𝜃 Quadrant ll: Quadrant lll: 1. sin(𝜋 − 𝜃) = + sin 𝜃 1. sin(𝜋 + 𝜃) = − sin 𝜃 2. cos(𝜋 − 𝜃) = − cos 𝜃 2. cos(𝜋 + 𝜃) = − cos 𝜃 3. tan(𝜋 − 𝜃) = − tan 𝜃 3. tan(𝜋 + 𝜃) = + tan 𝜃 Quadrant lV: Negative Angles: 1. sin(2𝜋 − 𝜃) = − sin 𝜃 1. sin(−𝜋 − 𝜃) = + sin 𝜃 2. cos(2𝜋 − 𝜃) = + cos 𝜃 2. sin(−𝜋 + 𝜃) = − sin 𝜃 3. tan(2𝜋 − 𝜃) = − tan 𝜃 3. cos(−𝜋 − 𝜃) = − cos 𝜃 4. cos(−𝜋 + 𝜃) = − cos 𝜃 5. tan(−𝜋 − 𝜃) = − tan 𝜃 6. tan(−𝜋 + 𝜃) = + tan 𝜃 Area rule: Sine Rule: 1 𝑎 𝑏 𝑐 1. ∆𝐴𝐵𝐶 = 2 𝑎𝑏 sin 𝐶̂ 1. = sin 𝐵̂ = sin 𝐶̂ sin 𝐴̂ 1 sin 𝐴̂ sin 𝐵̂ sin 𝐶̂ 2. ∆𝐴𝐵𝐶 = 2 𝑏𝑐 sin 𝐴̂ 2. = = 𝑎 𝑏 𝑐 1 3. ∆𝐴𝐵𝐶 = 2 𝑎𝑐 sin 𝐵̂ 7|Page  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS ABOUT THE AUTHOR: CHAUKE EMMANUEL EMMANUEL CHAUKE is a young Post Graduate in BSc (Mathematical Science) who is passionate about education, more especially in Maths, Statistics and Applied Maths. He has been giving extra lessons in both Calculus and Algebra for the past 3 years. He is a very informative person who helps students who love math and logic, but struggle to have real world, concise explanations of these subjects. As an extra lesson Tutor, Emmanuel has established the critical areas where students struggle and has written the Comprehensive guide for Introductory Algebra with these areas in mind. The explanation are very clear to comprehend and it makes one to love the subjects at hand. The books are written in a clear, simple, visual and logical manner. The colour coding facilitates explanations, definition, formulas, and recaps of previous work, hints and ideas. They are easy to read, easy to understand and easy to apply what has been learnt. They work in conjunction with all other Calculus Books. Emmanuel’s objective is for the Maths Handbook and Study Guides to demystify Maths and help students to reach their potential in this challenging module/subject. The subtitles of the books are ‘Maths made Easy’ and this is what he aims to do. Emmanuel ensures that his work is up to date at all times and that it is suitable for any Calculus Courses and National Curriculum students. 8|Page  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS ACKNOWLEDGEMENTS: Words have no power to express and convery my heart full thanks to the Almighty God, the one whom nothing before is impossible for giving me strength in fulfilling this work. It gives me great pleasure to express my deep sense of gratitude and respect to my beloved Mentor, Mr. Chauke T.E for infusing confidence and a sense of excitement and inspiring me in my work during the course of study, through his constant encouragement and guidance. My sincere and heartful thanks to him for his valuable suggestions. It is with great pride and pleasure that I submit this dissertation work as his student.The preparation of this book has involved much time spent reading the reasoned (but sometimes contradictory) advice from a large number of astute reviewers. I greatly appreciate the time they spent to understand my motivation for the approach taken. I have learned something from each of them.  I would like to extend my deep gratitude’s to my past and present lecturers for making me the better person that I am today. I highly appreciate the knowledge they have imparted in me, Namely: Prof Gopal Raja , Dr K. Adem , Dr PWN. Chin , Mrs. D Vijayasenan and Mr. J.L Thabane. My respectable gratitude to my past grade 11 and 12 teachers, Mrs. Mthamayendza and Mr. Shirinda for making Mathematics come alive in their outstanding Maths classes. I am thankful to my Principal Mr. Khosa for his kind of support and encouragement during course of study.my sincere and heart ful thanks to him for his valuable suggestions. I express my thanks to all my teachers who have taught me since my childhood: My high school teachers, my undergraduate teachers, my graduate teachers and post graduate teachers. Lastly but not least I would like to extend my deep gratitude to my very own blood ( Wombmate ) Chauke Welma, for her Love, affection and blessing and moral support without which this work would remain Unfinished. I WOULD LIKE TO DEDICATE THIS BOOK TO ALL MY PAST AND PRESENT STUDENTS, WISHING THEM THE BEST OF LUCK IN MATHEMATICS AND IN LIFE. 9|Page  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS PURPOSE OF THE BOOK: CALCULUS: Written to improve algebra and problem solving skills of students taking a Calculus course, every chapter in this companion is keyed to a calculus topic, providing conceptual background and specific algebra techniques needed to understand and solve calculus problems related to that topic. It is designed for calculus courses that integrate the review of precalculus concepts or individual use. ALGEBRA: This comprehensive book, designed to supplement the calculus course, provides an introduction to and review of the basic ideas of Linear algebra. TO THE STUDENTS: Reading a calculus textbook is different from reading a newspaper or novel, or even a physics book. Don’t be discouraged if you have to read a passage more than once in order to understand it. You should have a pencil and paper and calculator at hand to sketch a diagram or make a calculations. Some students start by trying their homework problems and read the text only if they get stuck on an exercise. I suggest that a far better plan is to read and understand a section of the text before attempting the exercises. In particular, you should look at the definition to see the exact meanings of the terms. And before you read the example, I suggest that you cover up the solution and try solving the problem yourself. You’ll get a lot more from looking at the solution if you do so. The solutions to the exercises and the previous tests and exams are provided at the back of the book. Some exercises ask for a verbal expression or interpretation or description. In such cases there is no single correct way of expressing the answer, so don’t worry that you haven’t found the definitive answer. In addition, there are often several different forms in which to express a numerical or algebraic answer, so if your answer differs from mine, don’t immediately assume you’re wrong. From example, if the answer given in the back of the book is √2 − 1 and you obtain 1⁄(1 + √2) , then you’re right and rationalizing the denominator will show that the answers are equivalent. I recommend that you keep this book for reference purpose after you finish the course. Because you will likely forget some of the specific details of calculus, the book will serve as a useful reminder when you need to use calculus in subsequent courses. And, because this book contains more materials than can be covered in any one course, it can also serve as a valuable resource for working in scientist or engineer. Calculus is an exciting subject, justly considered to be one of the greatest achievements of the human intellect. I hope you will discover that it is not only useful but also intrinsically beautiful. 10 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS CHAPTER ONE: ABSOLUTE VALUES / MODULUS: MMTH011 / MAH101M DIFFERENTIAL AND INTEGRAL CALCULUS: 11 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter one: Absolute Values / Modulus. Definition 1.1: Absolute Value: is the distance between the number and the origin. Or Absolute Value: describes the distance of a number on the number line from the origin without considering which direction from zero the number lies. About the Absolute Value:  The absolute value of a number is never negative.  The absolute value makes a negative number positive.  We denote the absolute value by, | | which means modulus. Definition 1.2. The Absolute value or Magnitude of a real number 𝑎 is denoted by |𝑎| and is defined by |𝑎| = { 𝑎 if 𝑎 ≥ 0 −𝑎 if 𝑎 < 0 Inequalities and Absolute Properties of an Absolute Values/Modulus. Values/Modulus.  |𝑥| = 0, if and only if 𝑥 = 0.  |𝑥| = 𝑎 means 𝑥 = 𝑎 𝑜𝑟 𝑥 = −𝑎  |𝑥| ≥ 0  |𝑥| < 𝑎 means −𝑎 < 𝑥 < 𝑎  |−𝑥| = |𝑥|, A number and its negative have the same  |𝑥| > 𝑎 means 𝑥 < −𝑎 𝑜𝑟 𝑥 > 𝑎 absolute value  |𝑥| = |𝑥| means (𝑥)2 = (𝑥)2  |𝑥|2 = 𝑥 2  |𝑥| > |𝑥| means (𝑥)2 > (𝑥)2  √𝑥 2 = |𝑥| , For any real number 𝑎  |𝑥| < |𝑥| means (𝑥)2 < (𝑥)2  |𝑥𝑦| = |𝑥||𝑦| , The absolute value of a product is the product of the absolute values  |𝑥 ± 𝑦|2 = (𝑥 ± 𝑦)2 𝑥 |𝑥|  |𝑦| = |𝑦| , The absolute value of a ratio is the ratio of the absolute values.  |𝑥 − 𝑦| = |𝑦 − 𝑥|  |𝑥 + 𝑦| ≤ |𝑥| + |𝑦| , (Triangle Inequality) 12 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter one: Absolute Values / Modulus. Worked Examples: Example 1: Evaluate the following Modulus: 1. |7| 2. |−11| 3. |8 − 12| 4. |2(−5) + 1| Solution: 1. |7| = 7 2. |−11| = 11 , The absolute value makes a negative number positive. 3. |8 − 12| = |−4| =4 4. |2(−5) + 1| = |−10 + 1| = |−9| =9 NB: Note that the effect of taking the absolute value of a number is to strip away the minus sign if the number is negative and to leave the number unchanged if it is nonnegative. 13 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter one: Absolute Values / Modulus. Theorem: Triangle Inequality. Prove the Triangle Inequality: |𝑥 + 𝑦| ≤ |𝑥| + |𝑦| Proof: 𝐿𝐻𝑆 = |𝑥 + 𝑦| 2 = (𝑥 + 𝑦)2 = 𝑥 2 + 2𝑥𝑦 + 𝑦 2 = |𝑥|2 + 2𝑥𝑦 + |𝑦|2 ≤ |𝑥|2 + 2|𝑥||𝑦| + |𝑦|2 2 ≤ ||𝑥| + |𝑦|| ∴ |𝑥 + 𝑦| ≤ |𝑥| + |𝑦| SOLVING MODULUS INEQUALITIES: Worked Examples: Example 2: Solve for 𝑥 in the following modulus inequality: 1. |4𝑥 − 3| ≥ 5 2. |3𝑥 + 2| < 4 3. |3𝑥 + 9| = |2𝑥 + 1| 4. |2𝑥 − 3| ≥ |𝑥 + 3| 5. 𝑥 + 6 > |3𝑥 + 2| 6. 4 − 2|2𝑥 + 1| < 5 14 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter one: Absolute Values / Modulus. Solution: 1. |4𝑥 − 3| ≥ 5 Using the properties of inequalities and absolute value, we obtain the following results, ∴ 4𝑥 − 3 ≤ −5 𝑜𝑟 4𝑥 − 3 ≥ 5 ∴ 4𝑥 ≤ −2 𝑜𝑟 4𝑥 ≥ 8 Simplify the inequalities. 1 Thus, 𝑥 ≤ − 2 𝑜𝑟 𝑥 ≥ 2 Inequality Notation. 2. |3𝑥 + 2| < 4 Using the same procedure above, we have the following, ∴ −4 ≤ 3𝑥 + 2 < 4 ∴ −6 < 3𝑥 < 2 Adding −2 to both sides of the equations. 2 Thus, −2 < 𝑥 < 3 Inequality Notation. 3. |3𝑥 + 9| = |2𝑥 + 1| This is slightly different from the above problems, but it’s very easy if one can recall the properties very well, And hence we going to square both sides to remove the modulus sign. ∴ (3𝑥 + 9)2 = (2𝑥 + 1)2 ∴ 9𝑥 2 + 54𝑥 + 81 = 4𝑥 2 + 4𝑥 + 1 Removing the brackets in both sides. 2 ∴ 5𝑥 + 50𝑥 + 80 = 0 Rearranging the equation. ∴ 𝑥 2 + 10𝑥 + 16 = 0 Divide both sides by 5. ∴ (𝑥 + 2)(𝑥 + 8) = 0 Factorize the quadratic equation. Thus, 𝑥 = −2 𝑜𝑟 𝑥 = −8 4. |2𝑥 − 3| ≥ |𝑥 + 3| Remember modulus signs are removed by squaring both sides of the equation, and thus we have the following results, ∴ (2𝑥 − 3)2 ≥ (𝑥 + 3)2 ∴ 4𝑥 2 − 12𝑥 + 9 ≥ 𝑥 2 + 6𝑥 + 9 Removing the brackets in both sides of the equations. ∴ 3𝑥 2 − 18𝑥 ≥ 0 Rearranging the equation. ∴ 3𝑥(𝑥 − 6) ≥ 0 Taking out the common factor of the equation. ∴ 𝐶𝑉: 𝑥 = 0 𝑜𝑟 𝑥 = 6 Critical Values. Thus, 𝑥 ≤ 0 𝑜𝑟 𝑥 ≥ 6 Inequality Notation. 15 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter one: Absolute Values / Modulus. 5. 𝑥 + 6 > |3𝑥 + 2| The best way to solve this kind of a problem is to square both sides of the equations to deal with the modulus sign, and thus obtain the following results, ∴ (𝑥 + 6)2 > (3𝑥 + 2)2 ∴ −8𝑥 2 > −32 Removing the brackets in both sides of the equation. ∴ −8𝑥 2 + 32 > 0 Rearranging the equation. 2 ∴ 𝑥 −4 < 0 Divide both sides by −8 , taking into an account the minus sign. ∴ (𝑥 + 2)(𝑥 − 2) < 0 This is a different of two squares factorization. ∴ 𝐶𝑉: 𝑥 = ±2 Critical Values. Thus, −2 < 𝑥 < 2 Inequality Notation. 6. 4 − 2|2𝑥 + 1| < 5 This is tricky to most of the most students, but let’s now think of Algebra manipulation, ∴ −2|2𝑥 + 1| < 1 Transposing 4 to the right side of the equation. 1 ∴ |2𝑥 + 1| > − 2 Divide both sides by −2, taking into an account the minus sign. 1 1 ∴ 2𝑥 + 1 < − (− 2) 𝑜𝑟 2𝑥 + 1 > − 2 Applying Modulus properties introduced earlier. 1 3 ∴ 2𝑥 < − 𝑜𝑟 2𝑥 > − Adding −1 to both sides of the equation. 2 2 1 3 Thus, 𝑥 < −4 𝑜𝑟 𝑥 > −4 Inequality Notation. 16 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS CHAPTER TWO: THE RELATIONSHIP BETWEEN RADIANS AND DEGREES: MMTH011 / MAH101M DIFFERENTIAL AND INTEGRAL CALCULUS: 17 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Two: The relationship between radians and degrees. THE RELATIONSHIP BETWEEN RADIANS AND DEGREES: At high school we usually lean to measure an angle in degrees. However, there are other ways of measuring an angle. One that we are going to have a look at here is measuring angles in units called Radians. In many scientific and engineering calculators radians are used in preference to degrees. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. Conversions between Degrees and Radians:  𝜋 𝑟𝑎𝑑 = 180° 𝑎𝑛𝑑 2𝜋 𝑟𝑎𝑑 = 360° 𝜋 1. To convert degrees to radians, multiply degrees by 180° 180° 2. To convert radians to degrees, multiply radians by 𝜋 𝜋 180° 3. 1° = 𝑎𝑛𝑑 1 𝑟𝑎𝑑 = 180° 𝜋 1° 4. 1′ = 𝑜𝑛𝑒 𝑚𝑖𝑛𝑢𝑡𝑒 = 60 1° 5. 1′′ = 𝑜𝑛𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 = 3600 18 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Two: The relationship between radians and degrees. Worked Examples: Example 3: Change the following angle sizes to Radians: a) 270° b) 120° c) 540° d) 315° e) 135° Solution: Change the following angle sizes to Radians: a) 270° 𝜋 = 270° × 180° 3𝜋 = 2 b) 120° 𝜋 = 120° × 180° 2𝜋 = 3 c) 540° 𝜋 = 540° × 180° = 3𝜋 d) 315° 𝜋 = 315° × 180° 7𝜋 = 4 e) 135° 𝜋 = 135° × 180° 3𝜋 = 4 19 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Two: The relationship between radians and degrees. Worked Examples: Example 4: Change the following angle sizes to Degrees: a) − 𝜋 ⁄2 b) 9𝜋⁄2 c) 3𝜋 d) 4𝜋⁄5 e) − 5𝜋⁄6 Solution: 𝜋 a) − 2 𝜋 180° =− × 2 𝜋 = −90° 9𝜋 b) 2 9𝜋 180° = 2 × 𝜋 = 810° c) 3𝜋 180° = 3𝜋 × 𝜋 = 540° 4𝜋 d) 5 4𝜋 180° = × 5 𝜋 = 144° 5𝜋 e) − 6 5𝜋 180° =− × 6 𝜋 = −150° 20 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS CHAPTER THREE: SOLVING LOGARITHMIC AND TRIGNOMETRIC EQUATIONS: MMTH011 / MAH101M DIFFERENTIAL AND INTEGRAL CALCULUS: 21 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Three: Solving Logarithmic Equations. Properties of the Logarithmic and Exponential functions: 1. LOGARITHMIC LAWS: 2. PROPERTIES OF LOGARITHMIC WITH A BASE "𝒂" Definition: 𝑦 = log 𝑎 𝑥 if and only if 𝑥 = 𝑎 𝑦 where 𝑎 > 0. In other words, Logarithmic are exponents. a) log 𝑎 𝑎 = 1 for all 𝑎 > 0 and b) log 10 = 1 Remarks: c) log 𝑎 1 = 0 for all 𝑎 > 0  log 𝑥 always refers to log base 10, d) log 𝑎 𝑥𝑦 = log 𝑎 𝑥 + log 𝑎 𝑦 𝑥 i.e. log 𝑥 = log10 𝑥 e) log 𝑎 𝑦 = log 𝑎 𝑥 − log 𝑎 𝑦  ln 𝑥 Is called the natural logarithmic and is used to f) log 𝑎 𝑥 𝑦 = 𝑦 log 𝑎 𝑥 represent log 𝑒 𝑥 , where the irrational number g) log 𝑎 𝑎 𝑦 = 𝑦 log 𝑎 𝑎 = 𝑦(1) = 𝑦 𝑒 ≈ 2.71828128. Therefore, ln 𝑥 = 𝑦 if and only if h) 𝑎log𝑎 𝑥 = 𝑥 𝑥 = 𝑒𝑦. i) log 𝑒 𝑥 = ln 𝑥  Change of base formulas: ln 𝑎 log 𝑎 log 𝑏 𝑎 = ln 𝑏 = log 𝑏 3. PROPERTIES OF NATURAL LOGARITHMIC: 4. PROPERTIES OF EXPONENTS: If 𝑚 and 𝑛 are integers then the following a) ln 𝑒 = 1 holds: b) ln 1 = 0 c) ln 𝑥 = 𝑦 ⟺ 𝑒 𝑦 = 𝑥 a) 𝑎𝑚 𝑎𝑛 = 𝑎𝑚+𝑛 d) ln 𝑥𝑦 = ln 𝑥 + ln 𝑦 b) (𝑎𝑚 )𝑛 = 𝑎𝑚𝑛 𝑥 e) ln 𝑦 = ln 𝑥 − ln 𝑦 c) (𝑎𝑏)𝑚 = 𝑎𝑚 𝑏 𝑚 𝑎𝑚 f) ln 𝑥 𝑦 = 𝑦 ln 𝑥 d) 𝑎𝑛 = 𝑎𝑚−𝑛 g) ln 𝑒 𝑥 = 𝑥 ln 𝑒 = 𝑥(1) = 𝑥 , 𝑥∈ℝ 𝑎 𝑚 𝑎𝑚 e) (𝑏 ) = 𝑏𝑚 , where 𝑏 ≠ 0 h) 𝑒 ln 𝑥 = 𝑥 , 𝑥 > 0 1 −𝑚 f) 𝑎 = 𝑎𝑚 , where 𝑎 ≠ 0 1 𝑛 g) 𝑎𝑛 = √𝑎 , where 𝑎 ≠ 0 𝑚 𝑚 𝑛 𝑛 h) 𝑎 = √𝑎𝑚 = ( √𝑎) 𝑛 i) 𝑎0 = 1 , where 𝑎 ≠ 0. 22 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Three: Solving Logarithmic Equations. Worked Examples: Example 5: Solve for 𝑥 in the following examples: a) 3𝑥+4 = 32𝑥−1 b) 3𝑥+4 = 5𝑥−6 c) 3𝑥 = 25 d) 𝑥 2 2𝑥 − 2𝑥 = 0 e) 𝑥 2 𝑒 𝑥 − 5𝑥𝑒 𝑥 − 6𝑒 𝑥 = 0 Solution: a) 3𝑥+4 = 32𝑥−1 Copy the original equation. ∴ 𝑥 + 4 = 2𝑥 − 1 Equate the powers of the exponents ∴𝑥=5 b) 3𝑥+4 = 5𝑥−6 Copy the original equation. ∴ (𝑥 + 4) log 3 = (𝑥 − 6) log 5 Logarithmic property ∴ 𝑥 log 3 + 4 log 3 = 𝑥 log 5 − 6 log 5 Remove the brackets ∴ 𝑥(log 3 − log 5) = −6 log 5 − 4 log 3 6 log 5+4 log 3 −(6 log 5+4 log 3) Thus, 𝑥 = − [ log 3−log 5 ] 𝑜𝑟 𝑥= log 3−log 5 c) 3𝑥 = 25 Copy the original equation. ∴ 𝑥 ln 3 = ln 25 Notice that we could have used the Log property if we wanted to. ln 25 ∴𝑥= ln 3 ≈ 2.930 23 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Three: Solving Logarithmic Equations. d) 𝑥 2 2𝑥 − 2𝑥 = 0 Copy the original equation. ∴ 2𝑥 (𝑥 2 − 1) = 0 Take out the common factor of the equation. ∴ 2𝑥 = 0 𝑜𝑟 𝑥 2 − 1 = 0 ∴ (𝑥 − 1)(𝑥 + 1) = 0 𝑏𝑢𝑡 2𝑥 ≠ 0 Factorize the quadratic equation. Thus, 𝑥 = ±1 e) 𝑥 2 𝑒 𝑥 − 5𝑥𝑒 𝑥 − 6𝑒 𝑥 = 0 Copy the original equation. ∴ 𝑥 2 − 5𝑥 − 6 = 0 Divide throughout by 𝑒 𝑥 ∴ (𝑥 + 1)(𝑥 − 6) = 0 Factorize the quadratic equation. ∴ 𝑥 = −1 𝑜𝑟 𝑥 = 6 Worked Examples: Example 6: Solve for 𝑥 in the following examples: a) 𝑒 2𝑥+3 − 7 = 0 b) ln(5 − 2𝑥) = −3 10 c) 1+𝑒 −𝑥 =2 d) 4 + 3𝑥+1 = 8 e) 𝑒 2𝑥 − 3𝑒 𝑥 + 2 = 0 24 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Three: Solving Logarithmic Equations. Solution: a) 𝑒 2𝑥+3 − 7 = 0 Copy the original equation. ∴ 𝑒 2𝑥+3 = 7 Transpose 7 to the right side of the equation. ∴ ln(𝑒 2𝑥+3 ) = ln 7 Take the ln in both sides of the equation. ∴ (2𝑥 + 3) ln 𝑒 = ln 7 Applying the Natural Log property. ∴ 2𝑥 + 3 = ln 7 Note: ln 𝑒 = 1 1 ∴ 𝑥 = (ln 7 − 3) 2 b) ln(5 − 2𝑥) = −3 Copy the original equation. ∴ 𝑒 ln(5−2𝑥) = 𝑒 −3 Exponentiate both sides of the equation. ∴ (5 − 2𝑥)𝑒 ln = 𝑒 −3 From the properties. ∴ 5 − 2𝑥 = 𝑒 −3 Note: 𝑒 ln = 1 1 Thus, 𝑥 = − (𝑒 −3 − 5) ≈ 2.475 2 10 c) 1+𝑒 −𝑥 =2 Copy the original equation. ∴ 2 + 2𝑒 −𝑥 = 10 Cross Multiplication. ∴ 𝑒 −𝑥 = 4 Simplify the equation. ∴ −𝑥 = ln 4 Take the ln in both sides of the equation. ∴ 𝑥 = − ln 4 ≈ −1.386 d) 4 + 3𝑥+1 = 8 Copy the original equation. ∴ 3𝑥+1 = 4 Transpose 7 to the right side of the equation. ∴ (𝑥 + 1) ln 3 = ln 4 Take the ln in both sides of the equation. ln 4 ∴ 𝑥 = (ln 3) − 1 Remove the brackets and solve the equation. Thus, 𝑥 = 0.262 e) 𝑒 2𝑥 − 3𝑒 𝑥 + 2 = 0 Copy the original equation. Let 𝑘 = 𝑒 𝑥 and 𝑘 2 = 𝑒 2𝑥 = 𝑒 𝑥 𝑒 𝑥 ∴ 𝑘 2 − 3𝑘 + 2 = 0 Replace, 𝑘 = 𝑒 𝑥 and 𝑘 2 = 𝑒 2𝑥 ∴ (𝑘 − 1)(𝑘 − 2) = 0 Factorize the equation. ∴ 𝑘 = 1 𝑜𝑟 𝑘 = 2 Hence, since 𝑘 = 𝑒 𝑥 = 1 Or 𝑘 = 𝑒 𝑥 = 2 ∴ 𝑒 𝑥 = 1 Or 𝑒 𝑥 = 2 ∴ 𝑥 = ln 1 = 0 𝑜𝑟 𝑥 = ln 2 ≈ 0.693 25 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Three: Solving Logarithmic Equations. Worked Examples: Example 6: Solve for 𝑥 in the following examples: a) log 3 (7𝑥 + 3) = log 3(5𝑥 + 9) b) log 2 (5𝑥 + 7) = 5 c) log 4 𝑥 + log 4 (𝑥 − 12) = 3 d) log(𝑥 − 3) + log(𝑥) = log 18 e) log 4 (2𝑥 + 1) = log 4(𝑥 + 2) − log 4 3 Solution: a) log 3 (7𝑥 + 3) = log 3(5𝑥 + 9) Copy the original equation. ∴ 7𝑥 + 3 = 5𝑥 + 9 Drop the Logs since the base are the same. ∴𝑥=3 b) log 2 (5𝑥 + 7) = 5 Copy the original equation. ∴ 5𝑥 + 7 = 25 By the Log definition. ∴ 5𝑥 + 7 = 32 ∴𝑥=5 c) log 4 𝑥 + log 4 (𝑥 − 12) = 3 Copy the original equation. ∴ log 4 [ 𝑥(𝑥 − 12)] = 3 By the Log property. ∴ 𝑥(𝑥 − 12) = 43 ∴ 𝑥 2 − 12𝑥 = 64 ∴ 𝑥 2 − 12𝑥 − 64 = 0 ∴ (𝑥 − 16)(𝑥 + 4) = 0 Factorize the quadratic equation. ∴ 𝑥 = 16 𝑜𝑟 𝑥 = −4 Thus, 𝑥 = 16 𝑏𝑢𝑡 𝑥 ≠ −4 d) log(𝑥 − 3) + log(𝑥) = log 18 Copy the original equation. ∴ log(𝑥 − 3)(𝑥) = log 18 By the Log property. ∴ 𝑥 2 − 3𝑥 = 18 Drop the Logarithmic. ∴ (𝑥 − 6)(𝑥 + 3) = 0 Factorize the quadratic equation. ∴ 𝑥 = 6 𝑜𝑟 𝑥 = −3 Thus, 𝑥 = 6 𝑏𝑢𝑡 𝑥 ≠ −3 26 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Three: Solving Logarithmic Equations. e) log 4 (2𝑥 + 1) = log 4(𝑥 + 2) − log 4 3 Copy the original equation. 𝑥+2 ∴ log 4 (2𝑥 + 1) = log 4 (3 ) By the Log property. 𝑥+2 ∴ 2𝑥 + 1 = 3 Drop the Logs since the base are the same. ∴ 6𝑥 + 3 = 𝑥 + 2 Multiply both sides by 3 1 ∴𝑥=− 5 Thus, there is no solution for this problem. Since we can’t take the logarithm of a negative number. 27 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Solving equations involving Trigonometry: Recap: When solving trig equations you have to find a reference angle (𝑅𝐴) and remember the sin 𝜃 and cos 𝜃 functions have a period of 2𝜋 while tan 𝜃 has a period of 𝜋. 1. If sin 𝜃 = 𝑚 And −1 ≤ 𝑚 ≤ 1 then: 𝜃 = 𝑅𝐴 + 2𝜋𝑘 Or 𝜃 = (𝜋 − 𝑅𝐴) + 2𝜋𝑘 , 𝑘 ∈ ℤ 2. If cos 𝜃 = 𝑚 And −1 ≤ 𝑚 ≤ 1 then: 𝜃 = 𝑅𝐴 + 2𝜋𝑘 Or 𝜃 = (2𝜋 − 𝑅𝐴) + 2𝜋𝑘 , 𝑘 ∈ ℤ Alternative method: If cos 𝜃 = 𝑚 And −1 ≤ 𝑚 ≤ 1 then: 𝜃 = ±𝑅𝐴 + 2𝜋𝑘 , 𝑘 ∈ ℤ 3. If tan 𝜃 = 𝑚 And 𝑚 ∈ ℝ then: 𝜃 = 𝑅𝐴 + 𝜋𝑘 , 𝑘 ∈ ℤ Or 𝜃 = (𝜋 − 𝑅𝐴) + 2𝜋𝑘 , 𝑘 ∈ ℤ NOTE: The general solution must be used when no interval is given. The general solution is used to find the values of the angles that falls within the given interval [𝑎, 𝑏]. 4. When sin 𝜃 = 0 , write 𝜃 = 0 + 𝜋𝑘 (𝑁𝑂𝑇 2𝜋). 𝜋 5. When cos 𝜃 = 0 , write 𝜃 = + 𝜋𝑘 (𝑁𝑂𝑇 2𝜋). 2 6. For all other values of sin 𝜃 and cos 𝜃 add 2𝜋𝑘 to the solution. 7. For all the solution of tan 𝜃 including tan 𝜃 = 0 add 𝜋𝑘. 8. For sin 𝜃 and cos 𝜃 , any value greater than 1 or less than −1 has NO solution. Solving negative angles: 9. If sin 𝜃 = −𝑚 , then 𝜃 = sin−1 (𝑚) + 2𝜋𝑘 = 𝑅𝐴 + 2𝜋𝑘 , 𝑘 ∈ ℤ Or 𝜃 = (𝜋 + 𝑅𝐴) + 2𝜋𝑘 , 𝑘 ∈ ℤ Or 𝜃 = (2𝜋 − 𝑅𝐴) + 2𝜋𝑘 , 𝑘 ∈ ℤ 10. If cos 𝜃 = −𝑚 , then 𝜃 = cos −1 (−𝑚) + 2𝜋𝑘 = 𝑅𝐴 + 2𝜋𝑘 , 𝑘 ∈ ℤ Or 𝜃 = (𝜋 − 𝑅𝐴) + 2𝜋𝑘 , 𝑘 ∈ ℤ Or 𝜃 = (𝜋 + 𝑅𝐴) + 2𝜋𝑘 , 𝑘 ∈ ℤ Alternative method: If cos 𝜃 = −𝑚 , then 𝜃 = cos−1(−𝑚) + 2𝜋𝑘 = ±𝑅𝐴 + 2𝜋𝑘 11. If tan 𝜃 = −𝑚 , then 𝜃 = tan−1 (𝑚) + 𝜋𝑘 = 𝑅𝐴 + 𝜋𝑘 , 𝑘 ∈ ℤ Or 𝜃 = (𝜋 − 𝑅𝐴) + 𝜋𝑘 , 𝑘 ∈ ℤ Or 𝜃 = (2𝜋 − 𝑅𝐴) + 𝜋𝑘 , 𝑘 ∈ ℤ 28 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Three: Solving Trigonometric Equations. Worked Examples: Example 6: Solve for 𝑥 in the following examples: a) √3 tan 𝑥 = 1 𝜋 b) √2 sin (𝑥 − ) − 1 = 0 2 𝜋 𝜋 √3 c) sin 2𝑥 cos 3 − sin 3 cos 2𝑥 = 2 d) 2sin2 𝑥 + 3 sin 𝑥 − 2 = 0 e) cos 2𝑥 + 3 cos 𝑥 − 1 = 0 Solution: a) √3 tan 𝑥 = 1 Copy the original equation. 1 ∴ tan 𝑥 = Divide throughout by √3. √3 1 ∴𝑥= tan−1 ( 3 ) Take the arctangent in both sides. √ 𝜋 ∴𝑥= 6 𝜋 b) √2 sin (𝑥 − ) − 1 = 0 Copy the original equation. 2 𝜋 ∴ √2 sin (𝑥 − 2 ) = 1 Transpose 1 to the right side of the equation 𝜋1 ∴ sin (𝑥 − ) = Divide throughout by √2. 2√2 𝜋 1 ∴ 𝑥 − = sin−1 ( ) Take the arsine in both sides. 2 √2 𝜋 𝜋 ∴ 𝑥−2=4 Then solve the equation. 3𝜋 ∴𝑥= 4 29 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Three: Solving Trigonometric Equations. 𝜋 𝜋 √3 c) sin 2𝑥 cos 3 − sin 3 cos 2𝑥 = 2 Copy the original equation. 𝜋 √3 ∴ sin (2𝑥 − ) = Recall: sin(𝜃 − 𝛽) = sin 𝜃 cos 𝛽 − cos 𝜃 sin 𝛽 3 2 𝜋 √3 ∴ 2𝑥 − 3 = sin−1 ( 2 ) Take the arsine in both sides. 𝜋 𝜋 ∴ 2𝑥 − 3 = 3 Then solve the equation. 2𝜋 ∴ 2𝑥 = 3 𝜋 ∴ 𝑥= 3 d) 2sin2 𝑥 + 3 sin 𝑥 − 2 = 0 Copy the original equation. ∴ (2 sin 𝑥 − 1)(sin 𝑥 + 2) = 0 Factorize the quadratic equation. ∴ 2 sin 𝑥 − 1 = 0 𝑜𝑟 sin 𝑥 + 2 = 0 1 ∴ sin 𝑥 = 2 𝑜𝑟 sin 𝑥 = −2 (No solution) 𝜋 ∴𝑥= 6 e) cos 2𝑥 + 3 cos 𝑥 − 1 = 0 Copy the original equation. ∴ (2cos2 𝑥 − 1) − 3 cos 𝑥 − 1 = 0 Double angle identity: cos 2𝑥 = 2cos 2 𝑥 − 1 2 ∴ 2cos 𝑥 − 3 cos 𝑥 − 2 = 0 ∴ (2 cos 𝑥 + 1)(cos 𝑥 − 2) = 0 Factorize the quadratic equation. ∴ 2 cos 𝑥 + 1 = 0 𝑜𝑟 cos 𝑥 − 2 = 0 1 ∴ cos 𝑥 = − 2 𝑜𝑟 cos 𝑥 = 2 (No solution) 2𝜋 ∴𝑥=± 3 30 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Three: Solving Trigonometric Equations. Worked Examples: Example 6: Solve for 𝑥 in the following examples: a) cos 2𝑥 − 3 sin 𝑥 = −1 −2𝜋 ≤ 𝑥 ≤ 2𝜋 b) sin 2𝑥 + sin 𝑥 = 6 cos 𝑥 + 3 −𝜋 ≤ 𝑥 ≤ 𝜋 5𝜋 𝜋 c) sin (3𝑥 + 18 ) + cos (2𝑥 − 18) = 0 −𝜋 ≤ 𝑥 ≤ 𝜋 d) 4cos2 𝑥 + sin 2𝑥 − 1 = 0 0 ≤ 𝑥 ≤ 2𝜋 e) sin 2𝑥 + 2 sin 𝑥 + cos 2 𝑥 + cos 𝑥 = 0 0 ≤ 𝑥 ≤ 2𝜋 Solution: a) cos 2𝑥 − 3 sin 𝑥 = −1 −2𝜋 ≤ 𝑥 ≤ 2𝜋 ∴ (1 − 2sin2 𝑥) − 3 sin 𝑥 + 1 = 0 Double angle identity: cos 2𝑥 = 1 − 2sin2 𝑥 ∴ 2sin2 𝑥 + 3 sin 𝑥 − 2 = 0 Rearrange the equation in standard form. ∴ (2 sin 𝑥 − 1)(sin 𝑥 + 2) = 0 Factorize the quadratic equation. ∴ 2 sin 𝑥 − 1 = 0 𝑜𝑟 sin 𝑥 + 2 = 0 1 ∴ sin 𝑥 = 2 𝑜𝑟 sin 𝑥 = −2 (No solution) 𝜋 𝜋 ∴ 𝑥 = 6 + 2𝜋𝑘 , 𝑘 ∈ ℤ OR 𝑥 = (𝜋 − 6 ) + 2𝜋𝑘 , 𝑘 ∈ ℤ 5𝜋 = 6 + 2𝜋𝑘 , 𝑘 ∈ ℤ 5𝜋 7𝜋 𝜋 5𝜋 Thus, 𝑥 = − ;− ; ; 3 6 6 6 31 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Three: Solving Trigonometric Equations. b) sin 2𝑥 + sin 𝑥 = 6 cos 𝑥 + 3 −𝜋 ≤ 𝑥 ≤ 𝜋 ∴ 2 sin 𝑥 cos 𝑥 + sin 𝑥 = 6 cos 𝑥 + 3 Change: sin 2𝑥 = 2 sin 𝑥 cos 𝑥 ∴ 2 sin 𝑥 cos 𝑥 + sin 𝑥 − 6 cos 𝑥 − 3 = 0 Subtract 6 cos 𝑥 and 3 in both sides. ∴ sin 𝑥 (2 cos 𝑥 + 1) − 3(2 cos 𝑥 + 1) = 0 Take out the common factor. ∴ (2 cos 𝑥 + 1)(sin 𝑥 − 3) = 0 Factorize the quadratic equation. ∴ 2 cos 𝑥 + 1 = 0 𝑜𝑟 cos 𝑥 − 3 = 0 1 ∴ cos 𝑥 = − 2 𝑜𝑟 cos 𝑥 = 3 (No solution) 𝜋 𝜋 ∴ 𝑥 = 3 + 2𝜋𝑘 , 𝑘 ∈ ℤ Or 𝑥 = (𝜋 − 3 ) + 2𝜋𝑘 , 𝑘 ∈ ℤ 𝜋 2𝜋 Or 𝑥 = (𝜋 + 3 ) + 2𝜋𝑘 , 𝑘 ∈ ℤ = 3 + 2𝜋𝑘 , 𝑘 ∈ ℤ 4𝜋 = + 2𝜋𝑘 , 𝑘 ∈ ℤ 3 𝜋 2𝜋 Thus, 𝑥 = , ± 3 3 5𝜋 𝜋 c) sin (3𝑥 + 18 ) + cos (2𝑥 − 18) = 0 −𝜋 ≤ 𝑥 ≤ 𝜋 5𝜋 𝜋 𝜋 ∴ sin (3𝑥 + 18 ) = − cos (2𝑥 − 18) Transpose cos (2𝑥 − 18) to the right side 5𝜋 𝜋 𝜋 ∴ sin (3𝑥 + ) = − sin [ − (2𝑥 − )] Co-functions identities. 18 2 18 5𝜋 5𝜋 ∴ sin (3𝑥 + 18 ) = − sin ( 9 − 2𝑥) 5𝜋 5𝜋 ∴ sin (3𝑥 + 18 ) = sin (2𝑥 − 9 ) 5𝜋 5𝜋 5𝜋 5𝜋 ∴ 3𝑥 + 18 = 2𝑥 − 9 + 2𝜋𝑘 OR 3𝑥 + 18 = 𝜋 − (2𝑥 − 9 )+ 2𝜋𝑘 5𝜋 23𝜋 2𝜋 ∴ 𝑥 = − 6 + 2𝜋𝑘 , 𝑘 ∈ ℤ 𝑥= 90 + 5 𝑘 , 𝑘∈ℤ 23𝜋 59𝜋 13𝜋 49𝜋 17𝜋 5𝜋 Thus, 𝒙 = 90 ; 90 ;− 90 ;− 90 ;− 18 ; − 6 32 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Three: Solving Trigonometric Equations. d) 4cos2 𝑥 + sin 2𝑥 − 1 = 0 0 ≤ 𝑥 ≤ 2𝜋 ∴ 4cos2 𝑥 + 2 sin 𝑥 cos 𝑥 − (cos 2 𝑥 + sin2 𝑥) = 0 [cos2 𝑥 + sin2 𝑥 = 1]. ∴ 3cos2 𝑥 + 2 sin 𝑥 cos 𝑥 − sin2 𝑥 = 0 Simplify the equation. ∴ (3 cos 𝑥 − sin 𝑥)(sin 𝑥 + cos 𝑥) = 0 Factorize the quadratic equation. ∴ 3 cos 𝑥 − sin 𝑥 = 0 𝑜𝑟 sin 𝑥 + cos 𝑥 = 0 ∴ 3 cos 𝑥 = sin 𝑥 𝑜𝑟 sin 𝑥 = − cos 𝑥 ∴ tan 𝑥 = 3 𝑜𝑟 tan 𝑥 = −1 Divide by sin 𝑥 in both sides of the equation 𝜋 ∴ 𝑥 = 1.23 𝑟𝑎𝑑 + 𝜋𝑘 , 𝑘 ∈ ℤ OR 𝑥 = − + 𝜋𝑘 , 𝑘 ∈ ℤ 4 3𝜋 7𝜋 ∴𝑥= + 𝜋𝑘 , 𝑘 ∈ ℤ OR 𝑥 = + 𝜋𝑘 , 𝑘 ∈ ℤ 4 4 3𝜋 7𝜋 Thus, 𝑥 = 1.23 𝑟𝑎𝑑 , , 4 4 e) sin 2𝑥 + 2 sin 𝑥 + cos2 𝑥 + cos 𝑥 = 0 0 ≤ 𝑥 ≤ 2𝜋 ∴ 2 sin 𝑥 cos 𝑥 + 2 sin 𝑥 + cos2 𝑥 + cos 𝑥 = 0 Change: sin 2𝑥 = 2 sin 𝑥 cos 𝑥 ∴ 2 sin 𝑥 (cos 𝑥 + 1) + cos 𝑥 (cos 𝑥 + 1) = 0 Take out the common factor. ∴ (cos 𝑥 + 1)(2 sin 𝑥 + cos 𝑥) = 0 Factorize the quadratic equation. ∴ cos 𝑥 + 1 = 0 𝑜𝑟 2 sin 𝑥 + cos 𝑥 = 0 1 ∴ cos 𝑥 = −1 𝑜𝑟 tan 𝑥 = − 2 ∴ 𝑥 = 0 + 2𝜋𝑘 , 𝑘 ∈ ℤ Or 𝑥 = 0.464 𝑟𝑎𝑑 + 𝜋𝑘 , 𝑘 ∈ ℤ Or 𝑥 = 𝜋 + 2𝜋𝑘 , 𝑘 ∈ ℤ Or 𝑥 = 2.678 𝑟𝑎𝑑 + 𝜋𝑘 , 𝑘 ∈ ℤ Or 𝑥 = 5.819 + 𝜋𝑘 , 𝑘 ∈ ℤ Thus, 𝑥 = 0 ; 0.464 ; 2.678 ; 5.819 ; 𝜋 33 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS CHAPTER FOUR: FUNCTIONS: MMTH011 / MAH101M DIFFERENTIAL AND INTEGRAL CALCULUS: 34 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Four: Functions. Definition: A function 𝑓 is a rule that assigns to each element 𝑥 in a set 𝐷 exactly one element, called 𝑓(𝑥) in a set 𝐸. The properties of a function:  Domain  Range  Asymptotes  Period  Amplitudes Types of functions and its Format: Types of Format to which it takes: functions: 1. Linear function 𝑓(𝑥) = 𝑚𝑥 + 𝑐 2. Quadratic Function 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 3. Cubic function 𝑓(𝑥) = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 4. Exponential function 𝑓(𝑥) = 𝑏 𝑥 5. Power function 𝑓(𝑥) = 𝑥 𝑛 6. Hyperbolic function 𝑎 𝑓(𝑥) = +𝑞 𝑥±𝑝 7. Rational function 𝑃(𝑥) 𝑓(𝑥) = 𝑄(𝑥) 8. Radical function 𝑓(𝑥) = √𝑔(𝑥) 9. Trigonometric function 𝑓(𝑥) = 𝑎 sin 𝑏𝑥 , 𝑎 cos 𝑏𝑥 and tan 𝑏𝑥 10. Composite function. ℎ(𝑥) = 𝑓 ∘ 𝑔(𝑥) = 𝑓(𝑔(𝑥)) 11. Piecewise 𝑥 𝑖𝑓 𝑥 ≥ 0 𝑓(𝑥) = { −𝑥 𝑖𝑓 𝑥 < 0 12. Inverse function 𝑦 = 𝑓 −1 (𝑥) 13. Absolute function 𝑓(𝑥) = |𝑥| 35 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Four: Functions. Graphs: Define the following terms:  Domain: is the set of all inputs over which the function has defined outputs.  Range: is the set of all values that the function takes when 𝑥 takes values in the domain. Worked Examples: Example 7: Sketch and find the domain and the range of the following functions: 1. 𝑓(𝑥) = 2𝑥 − 1 ; −1 ≤ 𝑥 ≤ 1 2. 𝑓(𝑥) = 𝑥 2 ; −1 ≤ 𝑥 ≤ 3 3. 𝑓(𝑥) = √𝑥 + 4 1 4. 𝑓(𝑥) = 2𝑥 + 2𝑥 ; 1≤𝑥≤4 Solution: 1. 𝑓(𝑥) = 2𝑥 − 1 ; −1 ≤ 𝑥 ≤ 1 𝑦 Note: This is a linear function, where our domain is the 𝑥 restricted domains and the range, we just substituted the values of the domain into the original function. i) 𝐷𝑓 = [−1,1] ii) 𝑅𝑓 = [−3,1] 36 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Four: Functions. 2. 𝑓(𝑥) = 𝑥 2 ; −1 ≤ 𝑥 ≤ 3 𝑦 𝑥 i) 𝐷𝑓 = [−1,3] ii) 𝑅𝑓 = [1,9] 3. 𝑓(𝑥) = √𝑥 + 4 ∴𝑥+4≥0 Thus, 𝑥 ≥ −4 , this will give us a clear picture of how our grah will look like. i) 𝐷𝑓 = [−4, ∞) ii) 𝑅𝑓 = [0, ∞) 37 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Four: Functions. 1 4. 𝑓(𝑥) = 2𝑥 + ; 1≤𝑥≤4 2𝑥 i) 𝐷𝑓 = [1,4] ii) 𝑅𝑓 = [2.5 , 8.125] Worked Examples: Example 8: Sketch and find the domain and the asymptotes of the following functions: 𝑥 1. 𝑓(𝑥) = 𝑥−2 𝑥+1 2. 𝑓(𝑥) = 𝑥−2 𝑥+1 3. 𝑓(𝑥) = 𝑥−1 𝑥 2 +1 4. 𝑓(𝑥) = 𝑥 2 −1 𝑥 2 +𝑥−6 5. 𝑓(𝑥) = 𝑥−3 38 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Four: Functions. Solution: 𝑥 1. 𝑓(𝑥) = 𝑥−2 (𝑥−2)+2 2 ∴ 𝑓(𝑥) = 𝑥−2 = 𝑥−2 + 1 𝑦=1 𝑥=2 i) 𝐷𝑓 = ℝ − {2} ii) 𝐷𝑓 = ℝ − {1} iii) Vertical asymptotes: ∴𝑥=2 iv) Horizontal asymptotes: ∴𝑦=1 v) Oblique asymptotes: No Oblique asymptotes. There are no oblique/slant asymptotes because the degree of the numerator and the denominator are the same, note that if we have the horizontal asymptotes we won have the oblique asymptotes. 39 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Four: Functions. 𝑥+1 2. 𝑓(𝑥) = 𝑥−2 (𝑥−2)+3 3 ∴ 𝑓(𝑥) = = +1 𝑥−2 𝑥−2 𝑦=1 𝑥=2 i) 𝐷𝑓 = ℝ − {2} ii) 𝐷𝑓 = ℝ − {1} iii) Vertical asymptotes: ∴𝑥=2 iv) Horizontal asymptotes: ∴𝑦=1 v) Oblique asymptotes: No Oblique asymptotes. 𝑥+1 3. 𝑓(𝑥) = 𝑥−1 (𝑥−1)+2 2 ∴ 𝑓(𝑥) = 𝑥−1 = 𝑥−1 + 1 𝑦=1 𝑥=1 i) 𝐷𝑓 = ℝ − {1} ii) 𝐷𝑓 = ℝ − {1} iii) Vertical asymptotes: 𝑥 = 2 iv) Horizontal asymptotes: 𝑦 = 1 v) Oblique asymptotes: No Oblique asymptotes. 40 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Four: Functions. 𝑥 2 +1 4. 𝑓(𝑥) = 𝑥 2 −1 (𝑥 2 −1)+2 2 ∴ 𝑓(𝑥) = 𝑥 2 −1 = 𝑥 2 −1 + 1 Vertical asymptotes: Oblique asymptotes: No Oblique asymptotes. ∴ 𝑥 2 − 1 = 0 ⟹ 𝑥 = ±1 Horizontal asymptotes: ∴𝑦=1 𝑦=1 𝑥 = −1 𝑥=1 41 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Four: Functions. 𝑥 2 +𝑥−6 5. 𝑓(𝑥) = 𝑥−3 This is slightly different from the previous example because here, the degree of the numerator is higher than the degree of the denominator by 1. Which tells us that we are going to have the Oblique asymptotes, of which we going to use long division method to find the Oblique asymptotes, and if we have the oblique then we won’t have the horizontal asymptotes. Vertical asymptotes: ∴𝑥=3 Oblique asymptotes: 𝑥+4 𝑥−3 𝑥2 + 𝑥 − 6 −(𝑥 2 − 3𝑥) 4𝑥 − 6 −(4𝑥 − 12) 6 Thus, the Oblique asymptotes by, 𝑦 = 𝑥 + 4 𝑦 = 𝑥+4 𝑥=3 42 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Four: Functions. Piecewise Function: Definition: Piecewise function: is a function that is defined on a sequence of intervals. Or Piecewise function: Is a pieces of different functions all on one graph. Worked Examples: Example 9: Find the domain and the range of the following functions: 𝑥+3 ; 𝑥≤0 1. 𝑓(𝑥) = { 3 ; 02 2𝑥 + 1 ; 𝑥 ≤ −1 2. 𝑓(𝑥) = { 𝑥2 − 2 ; 𝑥 > −1 1−𝑥 ; 𝑥≤1 3. 𝑓(𝑥) = { 𝑥2 ; 𝑥>1 43 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Four: Functions. Solution: 𝑥+3 ; 𝑥≤0 1. 𝑓(𝑥) = { 3 ; 02 Note: Label each function as follows 𝑦 = 𝑥 + 3 , 𝑦=3 And 𝑦 = 2𝑥 + 1 , the domains are given corresponding to each function. To get the range of 𝑓, substitute the values of the restricted domains independtely. i) Domain: 𝐷𝑓 = (−∞, 0] ∪ (0,2] ∪ (2, ∞) ii) Range: 𝑅𝑓 = (−∞, 3] ∪ {3} ∪ (5, ∞) 44 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Four: Functions. 2𝑥 + 1 ; 𝑥 ≤ −1 2. 𝑓(𝑥) = { 𝑥2 − 2 ; 𝑥 > −1 Here we notice that we only have two pieces of functions drawn in one function, which is called Piecewise function. So to draw the function 𝑓, we let 𝑦 = 2𝑥 + 1 which is a linear function with the gradient of 2. And 𝑦 = 𝑥 2 − 2 is a parabolic function with the turning point 𝑦 = −2, but the graph is greater than −1, which is our domain for the parabola function. i) Domain: 𝐷𝑓 = (−∞, −1] ∪ (−1, ∞) ii) Range: 𝑅𝑓 = (−∞, −1] ∪ (−1, ∞) 45 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Four: Functions. 1−𝑥 ; 𝑥≤1 3. 𝑓(𝑥) = { 𝑥2 ; 𝑥>1 Let 𝑦 = 1 − 𝑥 with 𝑥 ≤ 1 , and 𝑦 = 𝑥 2 where 𝑥 > 1 , to get the range of the original function 𝑓, Substitute the restricted domains in each function respectively and write them as an interval notation. But the sketched graph will also give you the domains and the range. i) Domain: 𝐷𝑓 = (−∞, 1] ∪ (1, ∞) ii) Range: 𝑅𝑓 = (−∞, 0] ∪ (1, ∞) 46 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Four: Functions. EVALUATION OF A PIECEWISE FUNCTIONS: Worked Examples: Example 10: Consider the given piecewise functions and answer the below. 𝑥−2 ; 𝑥≥2 1. Let 𝑓(𝑥) = { 𝑥2 ; 𝑥 0 , there exist a 𝛿 > 0 such that if 0 < |𝑥 − 1| < 𝛿 , then |2𝑥 + 1 − 3| < 𝜀. ∴ Compare |𝑥 − 1| and |2𝑥 + 1 − 3|. Simply: |2𝑥 + 1 − 3| = |2𝑥 − 2| = |2(𝑥 − 1)| = |2||𝑥 − 1| ∴ 2|𝑥 − 1| < 𝜀 𝜀 ∴ |𝑥 − 1| < 2 𝜀 Thus, 𝛿 = 2 2. lim (4 − 3𝑥) = −2 𝑥→2 For every > 0 , there exist a 𝛿 > 0 such that if 0 < |𝑥 − 2| < 𝛿 , then |4 − 3𝑥 + 2| < 𝜀. ∴ Compare |𝑥 − 2| and |4 − 3𝑥 + 2|. Simply: |4 − 3𝑥 + 2| = |6 − 3𝑥| = |−3(𝑥 − 2)| = |−3||𝑥 − 2| ∴ 3|𝑥 − 2| < 𝜀 𝜀 ∴ |𝑥 − 2| < 3 𝜀 Thus, 𝛿 = 3 3. lim (5 − 2𝑥) = 1 𝑥→2 For every > 0 , there exist a 𝛿 > 0 such that if 0 < |𝑥 − 2| < 𝛿 , then |5 − 2𝑥 − 1| < 𝜀. ∴ Compare |𝑥 − 2| and |5 − 2𝑥 − 1|. Simply: |5 − 2𝑥 − 1| = |4 − 2𝑥| = |−2(𝑥 − 2)| = |−2||𝑥 − 2| ∴ 2|𝑥 − 2| < 𝜀 𝜀 ∴ |𝑥 − 2| < 2 𝜀 Thus, 𝛿 = 2 68 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Five: The Limit of a functions. 4. lim (3𝑥 − 1) = 2 𝑥→1 For every > 0 , there exist a 𝛿 > 0 such that if 0 < |𝑥 − 1| < 𝛿 , then |3𝑥 − 1 − 2| < 𝜀. ∴ Compare |𝑥 − 1| and |3𝑥 − 1 − 2|. Simply: |3𝑥 − 1 − 2| = |3𝑥 − 3| = |3(𝑥 − 1)| = |3||𝑥 − 1| ∴ 3|𝑥 − 1| < 𝜀 𝜀 ∴ |𝑥 − 1| < 3 𝜀 Thus, 𝛿 = 3 5. lim (1 − 3𝑥) = −5 𝑥→2 For every > 0 , there exist a 𝛿 > 0 such that if 0 < |𝑥 − 2| < 𝛿 , then |1 − 3𝑥 + 5| < 𝜀. ∴ Compare |𝑥 − 2| and |1 − 3𝑥 + 5|. Simply: |1 − 3𝑥 + 5| = |6 − 3𝑥| = |−3(𝑥 − 2)| = |−3||𝑥 − 2| ∴ 3|𝑥 − 2| < 𝜀 𝜀 ∴ |𝑥 − 2| < 3 𝜀 Thus, 𝛿 = 3 6. lim |𝑥 + 4| = 1 𝑥→−3 For every > 0 , there exist a 𝛿 > 0 such that if 0 < |𝑥 + 3| < 𝛿 , then ||𝑥 + 4| − 1| < 𝜀. ∴ Compare |𝑥 + 3| and ||𝑥 + 4| − 1|. Simply: ||𝑥 + 4| − 1| < 𝜀. ⟹ |𝑥 + 4| − |1| < ||𝑥 + 4| − 1| < 𝜀. ⟹ |𝑥 + 4| − |1| < 𝜀. ⟹ |𝑥 + 4| − 1 < 𝜀. ⟹ |𝑥 + 4| < 𝜀 + 1 ⟹ −𝜀 − 1 < 𝑥 + 4 < 𝜀 + 1 From the modulus inequality property. ⟹ −𝜀 − 2 < 𝑥 + 3 < 𝜀 < 𝜀 + 2 Subtract 1 in both sides of the equations. ⟹ −𝜀 − 2 < 𝑥 + 3 < 𝜀 + 2 ⟹ |𝑥 + 3| < 𝜀 + 2 Thus, 𝛿 = 𝜀 + 2 69 | P a g e  Author: Emmanuel Chauke DIFFERENTIAL AND INTEGRAL CALCULUS Chapter Five: The Limit of a functions. ONE SIDED LIMIT: THEOREM: lim 𝑓(𝑥) = 𝐿 if and only if lim 𝑓(𝑥) = lim 𝑓(𝑥) = 𝐿 𝑥→𝑎 𝑥→𝑎 − 𝑥→𝑎 + In other words, the limit Exist if and only if the limit from the right is equal to the limit from the left. Note:  When computing one-sided limits, we use the fact that the Limit Laws also holds for one-sided limits.  If the limit computed from the right and the left are not equal, it means that the limit doesn’t exist. Worked Examples:  Example 16: 1. Determine whether the lim |𝑥| exists or not. 𝑥→0 |𝑥| 2. Show that, lim does not exist 𝑥→0 𝑥 3. Consider the piecewise function and answer the questions below. −1 − 𝑥 ; 𝑥 ≤ −7 𝑓(𝑥) = {3𝑥 2 + 1 ; 𝑥