Rancho Cucamonga High School - Ontario, CA Algebra and Trigonometry Mathematics Mathematics Resource Guide Guide Resource 2016−2017 The vision of the United States Academic Decathlon® is to provide students the opportunity to excel academically through team competition. Toll Free: 866-511-USAD (8723) • Direct: 712-326-9589 • Fax: 712-366-3701 • Email: info@usad.org • Website: www.usad.org This material may not be reproduced or transmitted, in whole or in part, by any means, including but not limited to photocopy, print, electronic, or internet display (public or private sites) or downloading, without prior written permission from USAD. Violators may be prosecuted. Copyright © 2016 by United States Academic Decathlon®. All rights reserved. Table of Contents SECTION 1: Basic Properties of Real Numbers..................................4 SECTION 2: Linear and Quadratic Equations..............................................7 2.1 Linear Equations.........................................7 Section 2.1: Exercises................................ 10 2.2 Quadratic Equations.................................. 11 2.2.1 Equations of the Form x2 – p = 0.............. 11 2.2.2 Equations of the Form k(x + r)2 – p = 0, Where k ≠ 0.......... 12 2.2.3 Equations of the Form ax2 + bx + c = 0, Where a ≠ 0......... 12 2.2.4 The Discriminant.......................................... 15 Section 2.2: Exercises................................. 16 SECTION 3: Polynomial Equations..... 17 3.1 When Are Two Polynomials Equal?............ 18 3.2 H ow Do We Add and Subtract Polynomials?................................ 18 3.3 How Do We Multiply Polynomials?............ 19 3.4 How Do We Divide Polynomials?..............20 3.5 H ow Does the Division of Polynomials Help Us with Polynomial Equations?.......... 24 3.6 Proof of the Rational Root Theorem.............29 3.7 Proof of the Factor Theorem.........................30 SEcTionS 3.1–3.7: Exercises......................... 31 3.8 Complex Numbers...................................... 32 2 3.8.1 How Do We Add Complex Numbers?......................................33 3.8.2 How Do We Multiply Complex Numbers?.....................................34 3.8.3 How Do We Divide Complex Numbers?.....................................35 SEcTion 3.8: Exercises.................................38 SECTION 4: Functions........................................... 39 4.1 Preliminaries............................................. 39 4.2 Definition of a Function.............................40 4.3 Many-to-One Functions versus One-to-One Functions..................... 43 4.4 Inverse Functions.......................................44 Section 4: Exercises....................................48 SECTION 5: Graphing............................................. 49 5.1 What Does the Graph of a Linear Function y = ax + b Look Like?........49 5.2 What Does the Graph of a Quadratic Function y = ax2 + bx + c Look Like?........... 51 5.2.1 The Case y = x2. .............................................51 5.2.2 The General Case y = ax2 + bx + c...........53 5.3 What Do the Graphs of Some Polynomials Look Like?.................... 56 5.4 What Does the Graph of the Exponential Function y = ax Look Like?...... 57 5.5 What Does the Graph of the Logarithmic Function y = logax Look Like?.................... 59 Sections 5.1–5.5: Exercises......................... 61 5.6 Transformations of Graphs..........................62 5.6.1 Graphing y = f(x + c) from the Graph of y = f(x)..........................62 USAD Mathematics Resource Guide • 2016–2017 Rancho Cucamonga High School - Ontario, CA part I Algebra...............................4 5.6.2 Graphing y = f(x) + C from the Graph of y = f(x).........................63 8.3 Lines....................................................... 112 5.6.3 Graphing y = f(ax) from the Graph of y = f(x).........................65 8.3.2 Point-Point Form ....................................... 116 8.3.1 Slope Form.................................................... 115 5.6.4 Graphing y = Af(x) from the Graph of y = f(x)..........................67 8.3.3 Slope-Point Form....................................... 118 Section 5.6: Exercises.................................68 8.4 Circles......................................................120 8.5 Solving Geometry Problems Using Coordinate Geometry..................... 121 SEcTion 8: Exercises..................................124 part II Trigonometry. ..126 Non-Polynomial Equations........................................... 69 6.1 Rational Equations.................................... 69 6.1.1 How Do We Solve Rational Equations?......................................70 SECTION 6.1.1 Exercises...........................................75 6.1.2 What Do the Graphs of Some Rational Functions Look Like?..................76 SECTION 6.1.2 Exercises...........................................81 6.2 Exponential Equations.............................. 81 6.2.1 Basic Properties.............................................81 6.2.2 How Do We Solve Exponential Equations?.............................82 Section 6.2: Exercises.................................84 6.3 Logarithmic Equations...............................84 6.3.1 Basic Properties.............................................84 6.3.2 How Do We Solve Logarithmic Equations?.............................85 Section 6.3: Exercises.................................89 6.4 Radical Equations......................................90 6.4.1 Method 1.........................................................90 6.4.2 Method 2.........................................................92 Section 6.4: Exercises.................................92 SECTION 7: Inequalities................................... 94 7.1 Linear Inequalities.....................................94 SECTION 7.1: Exercises...............................100 7.2 Quadratic Inequalities..............................100 7.2.1 Inequalities of the Form ax2 + bx + c > 0 and a > 0.............101 7.2.2 Inequalities of the Form ax2 + bx + c < 0 and a > 0............ 104 Section 7.2: Exercises............................... 106 SECTION 8: Coordinate Geometry... 108 8.1 The Pythagorean Theorem......................... 108 8.2 Points...................................................... 110 SECTION 9: Trigonometric Functions. ......................................126 9.1 The Sine Function for Acute Angles............127 9.2 The Tangent Function for Acute Angles . ...129 9.3 The Cosine and Cotangent Functions for Acute Angles.......................130 9.4 Relations Among Trigonometric Functions........................... 131 9.5 Trigonometric Functions of Special Angles...................................... 133 9.6 Trigonometric Functions of Angles of Any Measure.........................134 9.6.1 Definitions and Properties .....................134 9.6.2 Negative Angles..........................................136 9.7 Trigonometric Identities............................ 137 9.7.1 Sum and Difference Identities................137 9.7.2 Double-Angle Identities...........................140 9.7.3 Half-Angle Identities..................................142 9.7.4 Sum-to-Product Identities.......................143 9.7.5 Product-to-Sum Identities.......................146 SEcTionS 9.1–9.7: Exercises....................... 147 9.8 Graphs of Trigonometric Functions........... 148 9.9 Inverse Trigonometric Functions............... 155 9.10 Trigonometric Equations........................ 163 SEcTionS 9.8–9.10 Exercises..................... 168 9.11 The Law of Sines and Cosines.................. 169 SEcTion 9.11: Exercises............................. 174 9.12 Radians................................................. 174 SEcTion 9.12: Exercises............................. 176 USAD Mathematics Resource Guide Instructor’s Manual • 2016–2017 3 Rancho Cucamonga High School - Ontario, CA SECTION 6: 8.3.4 Mutual Positions of Lines......................... 119 part TI R A P 1 algebra i: Part Algeb rA aLGeBRa Section 1 Basic Properties of Real Numbers SsecTion ecTIon 1: Basic Properties of numbers real numbers 1: Basic Properties of Real We will begin this resource guide with a discussion of the real numbers—the numbers with which you addition, + , and multiplication, ⋅ . When two real numbers are added, the result is a real number. We often express this fact by saying that is closed under addition. Likewise, when two real numbers are multiplied, the result is a real number. Thus, is also closed under multiplication. Let us now turn to the basic properties of addition and multiplication. These properties are very important for understanding the structure of the real numbers. In high school mathematics, however, they are mainly used to simplify computations, especially computations that involve algebraic expressions. The numbers 0 and 1 are special—they have the following properties: 1. For any real number x , x + 0 = x 2. For any real number x , x ⋅ 1 = x No real numbers other than 0 and 1 have these properties. The rest of the properties are as follows: 3. Additive Inverse. For any real number x there is a unique number, denoted by −x , such that x + (−x) = 0 . 4. Multiplicative Inverse. For any real number x different from 0 , there is a unique number, denoted by x −1 , such that x ⋅ x −1 = 1 . 5. Associative Law of Addition. For all real numbers x , y , and z , x + ( y + z ) = ( x + y ) + z 6. Commutative Law of Addition. For all real numbers x , y , x + y = y + x 7. Associative Law of Multiplication. For all real numbers x , y , and z , x ⋅ ( y ⋅ z) = (x ⋅ y) ⋅ z 8. Commutative Law of Multiplication. For all real numbers x , y , x ⋅ y = y ⋅ x Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 4 4 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MATheMATics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA are most familiar. The set of the real numbers is denoted by . This set is equipped with two operations: 9. Distributive Law. For all real numbers x , y , and z , x ⋅ ( y + z ) = x ⋅ y + x ⋅ z 10. Zero Factor Property. For all real numbers x and y , x ⋅ y = 0 if and only if x = 0 or y = 0 or both x and y are equal to 0. Note that the last property consists of two properties. The first property is: If x ⋅ y = 0 , then x = 0 or yy ==00 or bothsecond x and property y are equal second = 0 or y = 0property y =If0x. = 0 and/or y = 0 , then x . y = 0. . The is: to If 0.x The , then x ⋅ is: You might be wondering, what about the operations subtraction and division? These two operations are defined in terms of addition and multiplication, respectively, as follows: definition And if y ≠ 0 , x is the real number x ⋅ y −1 . y , the set of all real numbers, includes the integers: 0,±1,± 2,± 3, . The set of all integers is denoted by . The set of all positive integers is denoted by + , and the set of all negative integers is denoted by − . The real numbers also include the rational numbers. A rational number is a ratio, or division, of two 2 −4 −23 3 integers. For example, , , , ,2.13,− 0.98123 are all rational numbers. Any integer n is rational 3 17 42 −18 n because it can be written as , which is a ratio of two integers. The set of all rational numbers is denoted 1 by . You can see from this description that: a. + and − are subsets of , b. is a subset of , and c. is a subset of . Figure 1–1 illustrates these relations: Figure 1–1 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 ® AcAdemic decAthlon mAthemAtics ResouRce Guide MATheMATics ResouRce Guide Resource USAD Mathematics Guide • 2016–2017 55 Rancho Cucamonga High School - Ontario, CA If x and y are real numbers, their difference x − y is the real number x + ( − y ) . Another important subset of the real numbers is the set of irrational numbers. A real number that is not rational is called irrational. For example, 2 is irrational. Do you know why 2 is irrational? theorem 2 is an irrational number. Proof 1. Assume that 2= 2 is rational. Then by definition n ≠ 0. m , where m and n are integers and n m , then we should be able to reduce this fraction to the lowest term. n m is a reduced fraction. 3. Let m and n be so that n m (Why?) 4. Since 2 = , 2n 2 = m 2 . This implies that m 2 is even, and, hence, m is also even .(why?) n 5. Since m is even, we can write m = 2k for some integer k . 2= ( ) 6. Substituting m = 2k in 2n 2 = m 2 , we get 2n2 = 2k , and so n 2 = 2k 2 . This implies that n 2 is 2 even, and, hence, n is also even. m is not a reduced fraction—a contradiction. This shows that 7. Since m and n are both even, n our assumption that 2 is rational is not true. Hence 2 is irrational. We will conclude this section with another set of properties that concern inequalities between real numbers: 11. For any two real numbers x and y , one and only one of the following is true: x < y , x = y , or xy >< y.x . 12. For any three real numbers x , y , z , if x < y , y < z , then x < z . 13. For any three real numbers x , y , z , if x < y then x + z < y + z . 14. For any three real numbers x , y , z , if x < y and z > 0 , then zx < zy . 15. For any three real numbers x , y , z , if x < y and z < 0 , then zx > zy . SsecTion ecTIon 2: Linear Quadratic equations 2: Linear andand Quadratic equations 2 . Linear Equations 2 .1 1 linear equations The simplest equations are those in the form ax + b = 0 . These equations are called linear equations, 6 6 6 Unauthorized Duplication is Prohibited Outside the Terms of Your• License Agreement USAD Mathematics Resource Guide 2016–2017 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MATheMATics ResouRce Guide –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA 2. If indeed We will conclude this section with another set of properties that concern inequalities between real numbers: 11. For any two real numbers x and y , one and only one of the following is true: x < y , x = y , or y < x . 12. For any three real numbers x , y , z , if x < y , y < z , then x < z . 13. For any three real numbers x , y , z , if x < y then x + z < y + z . 14. For any three real numbers x , y , z , if x < y and z > 0 , then zx < zy . Section 2 15. For any three real numbers x , y , z , if x < y and z < 0 , then zx > zy . Linear and Quadratic SsecTion ecTIon 2: Linear Quadratic equations 2: Linear andand Quadratic equations Equations The simplest equations are those in the form ax + b = 0 . These equations are called linear equations, because, as we will see in Section 8, they are associated with straight lines. The following equations are all linear: Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 6 6 (1) 2 x + 6 = 0 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MATheMATics ResouRce Guide –2014 2013 2010–2011 (2) −3x + 4 = 0 2 (3) − x + 3.1 = 0 5 Equations (1), (2), and (3) have the form ax + b = 0 . In equation (1), a = 2 and b = 6 ; in equation (2), 2 a = −3 and b = 4 ; and in equation (3) a = − and b = 3.1 . Hence, equations (1), (2), and (3) are linear. 5 Now, let’s consider the following equations: (4) 4 x = 17 (5) 2 x − 9 = 34 (6) −6 + 2x = 4x + 9 These equations do not have the form ax + b = 0 , but they are equivalent to equations of this form. They, too, are considered linear equations. When are two equations equivalent? Two equations are equivalent if they have the same solution set, that is to say, if each solution of one equation is also a solution of the second equation, and vice versa. ExamplE 2.1a: The equations 2 x − 9 = 34 and 2 x − 43 = 0 are equivalent. REason: The solution set of each equation consists of the same number: x = 21.5 . USAD Mathematics Resource Guide • 2016–2017 Now, let’s solve a few linear equations. 7 Rancho Cucamonga High School - Ontario, CA 2 . 11 Linear Equations 2 . linear equations 2.1 Linear Equations second equation, equation, and and vice vice versa. versa. second second equation, and vice versa. ExamplE 2 xx − − 99 = = 34 34 and 2 xx − − 43 43 = = 00 are ExamplE 2.1a: 2.1a: The The equations equations 2 and 2 are equivalent. equivalent. ExamplE 2.1a: The equations 2 x − 9 = 34 and 2 x − 43 = 0 are equivalent. 21.5 .. REason: REason: The The solution solution set set of of each each equation equation consists consists of of the the same same number: number: xx == 21.5 REason: The solution set of each equation consists of the same number: x = 21.5 . Now, Now, let’s let’s solve solve aa few few linear linear equations. equations. Now, let’s solve a few linear equations. solution: We We can can add add 34 34 to to both both sides sides of of the the equation equation without without changing changing the the solution solution set set of of solution: solution: We can add 34 to both sides of the equation without changing the solution set of the the equation. equation. And, And, in in doing doing so, so, we we get get aa simpler simpler equation, equation, an an equation equation that that is is easier easier to to solve. solve. Thus: Thus: the equation. And, in doing so, we get a simpler equation, an equation that is easier to solve. Thus: 88 xx − − 34 34 + + 34 34 = = 15 15 + + 34 34 8 x − 34 + 34 = 15 + 34 88 xx + + 00 = = 49 49 The is equivalent to the original equation and is easier to solve. To get x, we simply 8 x +last 0 = equation 49 88 xx = 49 = 49 divide 8 x = 49both sides of the equation by 8. The last equation is equivalent to the original equation and is easier to solve. To get x, we simply 8 x 49 Unauthorized Duplication Prohibited Outside the the Terms of of Your License License Agreement Unauthorized Duplication divide of the equation byisisis8.Prohibited = both sides Unauthorized Duplication Prohibited Outside Outside the Terms Terms of Your Your License Agreement Agreement 8 8 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement ® ® mAthemAtics ResouRce Guide 2013 –2014 AcAdemic decAthlon 2010–2011 MATheMATics MATheMATics ResouRcedecAthlon Guide 2013 –2014 AcAdemic ® mAthemAtics ResouRce Guide 2010–2011 ResouRce Guide 2013 –2014 AcAdemic decAthlon mAthemAtics ResouRce Guide 2010–2011 MATheMATics ResouRce Guide 8 x 49 ® 2013 –2014 MATheMATics AcAdemic mAthemAtics ResouRce Guide 2010–2011 ResouRcedecAthlon Guide x = =6.125 8 8 x = 6.125 The solution set of the given equation 8 x − 34 = 15 consists of one number: x = 6.125 . Let’s check that x = 6.125 is indeed a solution to this equation: The solution set of the given equation 8 x − 34 = 15 consists of one number: x = 6.125 . 8(6.125) − 34 = 49 − 34 = 15 Let’s check that x = 6.125 is indeed a solution to this equation: 8(6.125) − 34 = 49 − 34 = 15 4 1 = 3x − . 7 7 4 1 ExamplE 2.1c: Solve the equation −5 x + = 3x − . 7 7 solution: ExamplE 2.1c: Solve the equation −5 x + 8 4 4 1 4 solution: −5x + − = 3x − − 7 7 7 7 4 4 5 1 4 −5x ++ 0 =−3x =− 3x − − −5x 7 7 7 7 7 5− 5 −5x + 0 = 3x −5x = 3x − 7 7 5 −5x −= 3x 3x = − 3x − 5 − 3x −5x 7 7 5 −5x =− − 3x5= 3x − − 3x −8x 7 7 4 from both sides of the equation. 7 4 Subtract from both sides of the equation. 7 Subtract Subtract 3x from both sides of the equation. USAD Mathematics Resource Guide • 2016–2017 Subtract 3x from both sides of the equation. 7 7 7 7 Rancho Cucamonga High School - Ontario, CA ExamplE 8 xx − − 34 34 = = 15 15 .. ExamplE 2.1b: 2.1b: Solve Solve the the equation equation 8 ExamplE 2.1b: Solve the equation 8 x − 34 = 15 . −5x + 0 = 3x − −5x = 3x − 5 7 5 7 −5x − 3x = 3x − −8x = − 8x = 5 − 3x 7 5 7 5 7 Multiply both sides of the equation by −1 . 5 x 8x 7 = 8 8 Divide both sides of the equation by 8 . 5 56 In the next series of examples, we will present the solution process without a detailed explanation as was given in the previous examples. Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 8 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MATheMATics ResouRce Guide ExamplE 2.1d: Solve the equation 2( x − 4) + 3 = 2 x − 5 . –2014 2013 2010–2011 solution: 2(x − 4) + 3 = 2x − 5 2x − 8 + 3 = 2x − 5 2x − 5 = 2x − 5 Any number x is a solution to the last equation because with any number that we substitute for x , we get a true statement. Hence, our original equation, 2( x − 4) + 3 = 2 x − 5 , has infinitely many solutions. ExamplE 2.1e: Solve the equation x − 7 = 2( x + 3) − x . solution: x + 7 = 2( x + 3) − x x + 7 = 2x + 6 − x x+7= x+6 USAD Mathematics Resource Guide • 2016–2017 9 Rancho Cucamonga High School - Ontario, CA x= Subtract 3x from both sides of the equation. ExamplE 2.1e: Solve the equation x − 7 = 2( x + 3) − x . ExamplE . ExamplE 2.1e: 2.1e: Solve Solve the the equation equation xx −− 77 == 2( 2(xx ++ 3) 3) −− xx . solution: x + 7 = 2( x + 3) − x solution: solution: x + 7 = 2x + 6 − x xx ++ 77 == 2( 2(xx ++ 3) 3) −− xx xx ++ 77 == x2 x+ +6 6 − x x + 7 = 2x + 6 − x xx ++ 77 ==isxxno ++ 66number x that satisfies the last equation because if there were such a number, then There we would get 7 = 6 , which is absurdity. Hence, the original equation x − 7 = 2( x + 3) − x has no There There is is no no number number xx that that satisfies satisfies the the last last equation equation because because ifif there there were were such such aa number, number, then then solution. has no we which is is absurdity. absurdity. Hence, Hence, the the original original equation equation xx −− 77 == 2( we would would get get 77 == 66 ,, which 2(xx ++ 3) 3) −− xx has no solution. solution. Section 2 .2.1 1: section 2 .1: exercises Section Exercises :: k k k k k k k k k k k k k k k k kk k k k k k k k 2 . kk k k kk k kSection k k k k 2 . k 11k Exercises Section 2 . section 1 : exercises section 2 .1: exercises kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk Exercises Exercises kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 1. Solve the following equations: 1 4 1. Solve equations: (x 8) = 1. a. Solve the the+following following equations: 3 57 44 = 11 2 1 a. (x + 8) 3 a. 257 b. − + 8)+=33x + + 4 x − = 9x +17 x(x 57 4 3 3 5 3 2 1 b. 22xx −− 3++ 33xx ++ 2++ 44xx −− 1 == 9x +17 b. 8x − 3 4 5x +2 3 5 9x +17 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement − = 1 c. 4 3 5 2 11 ® 2013 –2014 MATheMATics AcAdemic mAthemAtics ResouRce Guide 2010–2011 ResouRcedecAthlon Guide 7 x +11 x + 3 x −Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement Duplication is Prohibited Outside the Terms of Your License Agreement d. + Unauthorized = ® 2013 –2014 6 AcAdemic 2010–2011 MATheMATics ResouRcedecAthlon Guide 3 2 ® mAthemAtics ResouRce Guide 2013 –2014 MATheMATics AcAdemic mAthemAtics ResouRce Guide 2010–2011 ResouRcedecAthlon Guide e. x + 2x + 3x ++ 33x = 51 2. Sam took a number, multiplied it by 12, added 60 to the result, and got 372. What was Sam’s initial number? 4 2 from it. Then, he divided 3 by the result and got 3. Tom took a number and subtracted 2 15 3 5 3 . What was Tom’s initial number? 13 4. John left town A at the same time that Mary left town B. They walked toward each other and met 40 minutes after their departure. John arrived at town B 32 minutes after he met Mary. How long did it take Mary to arrive at town A? 5. Cities A and B are 70 miles apart. A biker leaves town A at the same time that a bus leaves town B. They travel toward each other and meet 84 minutes after their departure at a point between A 10 and B. The bus arrives at City B, Mathematics stays there forResource 20 minutes, and then heads back to City A. The USAD Guide • 2016–2017 bus meets the biker again 2 hours and 41 minutes after their first meeting. What are the speeds 9 9 9 Rancho Cucamonga High School - Ontario, CA kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 5. Cities A and B are 70 miles apart. A biker leaves town A at the same time that a bus leaves town 5. Cities A and B are 70 miles apart. A biker leaves town A at the same time that a bus leaves town B. They travel toward each other and meet 84 minutes after their departure at a point between A B. They travel toward each other and meet 84 minutes after their departure at a point between A and B. The bus arrives at City B, stays there for 20 minutes, and then heads back to City A. The and B. The bus arrives at City A, B, stays there for 20 minutes, and then heads back to City B. A. The bus meets the biker again 2 hours and 41 minutes after their first meeting. What are the speeds bus meets the biker again 2 hours and 41 minutes after their first meeting. What are the speeds of the bus and the biker? of the bus and the biker? kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk We have learned how to solve linear equations. In this section, we will learn how to solve a new kind of We have learned how to solve linear equations. In this section, we will learn how to solve a new kind of equation, quadratic equations. These are equations of the form ax 2 + bx + c = 0 , where a ≠ 0 . We add equation, quadratic equations. These are equations of the form ax 2 + bx + c = 0 , where a ≠ 0 . We add the condition a ≠ 0 because if a = 0 , then we would just have a linear equation, bx + c = 0 , which we the condition a ≠ 0 because if a = 0 , then we would just have a linear equation, bx + c = 0 , which we have already discussed. When we say “quadratic equation,” we always assume that the coefficient of x 2 have already discussed. When we say “quadratic equation,” we always assume that the coefficient of x 2 is nonzero. We will begin with two special cases of quadratic equations, and then we will deal with the is nonzero. We will begin with two special cases of quadratic equations, and then we will deal with the general case. general case. 2 .2 .1 eQuATIonS The form 2 .2 . eQuations the 2.2.11Equations ofof theof Form x2form –xp22p0 = 0x 22p0 2 .2 .1 of of Thethe formform x p0 2 .2 .1 eQuATIonS eQuations x p0 In this case, a = 1 , b = 0 , c = − p . Equations of this form are easy to solve: In2 this case, a = 1 , b = 0 , c = − p . Equations of this form are easy to solve: x2 − p = 0 x2 − p = 0 x2 = p x =p x = p or x = − pUnauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 10 10 Or, we just write 10 10 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement ® –2014 decAthlon mAthemAtics ResouRce MATheMATics Guide 2013 2010–2011 x = ± AcAdemic p . Note, however, that p must be non-negative. IfGuide is no real solution pResouRce < 0 , there 2013 –2014 AcAdemic decAthlon ® mAthemAtics ResouRce Guide MATheMATics ResouRce Guide 2010–2011 to the given equation. ExamplE 2.2a: Solve the equation x 2 − 16 = 0. solution: x 2 − 16 = 0 x 2 = 16 x = ± 16 x = ±4 2 2 .2 .2eQuATIonS eQuations k0 2 .2 .2 of of The the formform k(xr)k(xr) p0,2p0, WhereWhere k0 This type of equationUSAD can beMathematics reduced to the format we just •discussed. To do this, we need to divide both 11 Resource Guide 2016–2017 p p p sides of the equation by k . Then, we get ( x + r ) 2 = (remember that k ≠ 0 ). If ≥ 0 , then x + r = ± , Rancho Cucamonga High School - Ontario, CA 2 .2 Quadratic equations 2 .2 Quadratic Equations 2.2 Quadratic Equations 2 .2 Quadratic equations 2 .2 Quadratic Equations 2p = 0, 2Where 2 .2 .2 eQuations the k0 2.2.2 Equations ofof theof form k(xform + r)2 – k(xr) k≠ 0 2 .2 .2 eQuATIonS The form k(xr) p0,p0, WhereWhere k0 This type of equation can be reduced to the format we just discussed. To do this, we need to divide both p p p sides of the equation by k . Then, we get ( x + r ) 2 = (remember that k ≠ 0 ). If ≥ 0 , then x + r = ± , k k k p p and so x = − r ± . If < 0 , the equation has no real solution. k k ExamplE 2.2b: Solve the equation 8( x + 3) 2 − 72 = 0 . solution: 8(x + 3)2 = 72 72 72 2 ((xx ++ 1) 3)2 == 88 22 ((xx ++ 1) 3) == 99 −− xx +1 + 3==±±√99 3 xx + + 13== ±±3 −1 ±± 33 xx = = −3 xx = = 02 or −4 or xx==−6 2.2.3 Equations ofof theThe form axform + 2bx +axc2bxc0, = 0, Where aa0 ≠ 0 a0 2 .2 .3 eQuations of the 2 .2 .3 eQuATIonS form ax bxc0, WhereWhere Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2 2013 –2014 2010–2011 AcAdemic MATheMATics ResouRcedecAthlon Guide ® 11 11 mAthemAtics ResouRce Guide To solve this equation, we first must change it into the form we just discussed, k ( x + r ) − p = 0 , which we 2 already know how to solve. Let’s begin with a specific example: ExamplE 2.2c: Solve 2 x 2 − 3x + 1 = 0 . solution: We solve this equation by making it an equivalent equation of the form k ( x + r )2 − p = 0 . When opening the parentheses in the last equation, we get kx 2 + 2krx + kr 2 − p = 0 . Comparing the coefficients of this equation with the coefficients of the given equation (see the figure below), 12 we get: k = 2 2kr = −3 2 kr − p = 1 USAD Mathematics Resource Guide • 2016–2017 2 2 Rancho Cucamonga High School - Ontario, CA 8(x + 3)2 − 72 = 0 When opening the parentheses in the last equation, we get kx 2 + 2krx + kr 2 − p = 0 . Comparing the coefficients of this equation with the coefficients of the given equation (see the figure below), we get: k = 2 2kr = −3 2 kr − p = 1 2 −3 By substituting k = 2 in 2kr = −3 , we get r = . 4 2 −3 1 −3 2 in kr − p = 1 , we get p = 2 ⋅ −1 = . And, by substituting k = 2 and r = 4 8 4 2 −3 1 Thus, our equation can be written as: 2 x + − = 0 . 4 8 2 Now, we have arrived at an equation of the form k(x + r)2 − p = 0 , which we know how to solve. −3 1 2 x + = 4 8 Rancho Cucamonga High School - Ontario, CA 2 −3 1 x + = 4 16 2 x+ −3 1 =± 4 4 x= 3 1 ± 4 4 x = 1 or x = 1 2 Thus, the solution set of our equation consists of two numbers: x = 1 and x = 1 . 2 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 12 12 ExamplE 2.2d: Guide AcAdemic decAthlon ® mAthemAtics ResouRce MATheMATics ResouRce Guide –2014 2013 2010–2011 solution: This process is applicable to the general case ax 2 + bx + c = 0 , where a ≠ 0 . As before, our aim is to bring this equation into the form k(x + r)2 − p = 0 . This form, as has just been shown, is kx 2 + 2krx + kr 2 − p = 0 . k = a Comparing the coefficients of the two equations, we get 2kr = b kr 2 − p = c b Substituting k = a in 2kr = b , we get r = . 2a 2 b b 2 in kr − p = c , we get p = a ⋅ − c . Substituting k = a and r = 2a 2a Thus, our equation can be written as: USAD Mathematics Resource Guide • 2016–2017 2 b b 2 a x + − a − c = 0 . 13 Comparing the coefficients of the two equations, we get ° NU " E ²NU S " F ± E Substituting N " D in NU " E , we get U " . D 2 b E in NU S " F , we get p = a ⋅ − c . Substituting N " D and U " D 2a Thus, our equation can be written as: 2 b b 2 a x + − a − c = 0 . 2a 2a Now we can solve for [ : 2 b b a x + = a − c 2a 2a 2 b a ⋅ − c 2 b 2a x + = a 2a Rancho Cucamonga High School - Ontario, CA 2 b b 2 c x + = − 2a 2a a 2 b b2 c x + = 2 − a 2a 4a 2 b b2 − 4ac x + = 4a 2 2a 2 © E¹ E DF [ ª º"t D D « » E E DF [" t D D E E DF [" t D D E t E DF [" D Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 ACADEMIC MATHEMATICS RESOURCEDECATHLON GUIDE ® MATHEMATICS RESOURCE GUIDE 13 13 We have just developed a formula for solving quadratic equationsD[ E[ F " . This formula is called the quadratic formula: [ " E t E DF . D 2.2.4 2.2.4THE THEDISCRIMINANT DISCRIMINANT The expression E DF tells us when a quadratic equation has a real solution, and if it does, whether it has expression is called the discriminant. It is denoted by the Greek letter 14 one or two real solutions. This USAD Mathematics Resource Guide • 2016–2017 ) (delta). 2a 2 .2 .4 2 .2 .4The theDIScrImInAnT discriminant 2.2.4 The Discriminant The expression b2 − 4ac tells us when a quadratic equation has a real solution, and if it does, whether it has one or two real solutions. This expression is called the discriminant. It is denoted by the Greek letter ∆ (delta). From the quadratic formula we developed previously, we get: −b + b2 − 4ac −b − b2 − 4ac or x = . 2a 2a −b 2. When ∆ = b2 − 4ac = 0 , the equation has only one real solution: x = . 2a 1. When ∆ = b2 − 4ac > 0, the equation has two real solutions: x = It is not always necessary to use the quadratic formula to solve a quadratic equation. Quadratic equations of the form ax 2 + bx = 0 are easy to solve by factoring, as is shown below: ax 2 + bx = 0 x ( ax + b) = 0 x = 0 or ax + b = 0 x = 0 or x = − b a Here we utilized the fact that the product of two numbers is zero if and only if at least one of the numbers is zero. When using the quadratic formula, we naturally get the same solution: −b ± b2 − 4ac −b ± b2 − 4a0 −b ± b2 − 0 −b ± b2 −b ± b = = = = 2a 2a 2a 2a 2a −2b b Thus, x = 0 and x = = − is the solution set of the equation ax 2 + bx = 0 . 2a a x= Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 14 14 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MATheMATics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 –2014 2013 2010–2011 15 Rancho Cucamonga High School - Ontario, CA 3. When ∆ = b2 − 4ac < 0 , the equation has no real solution. kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk Section 2.2 Section 2 .2: section 2 .2: exercises Exercises Exercises kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 1. Solve the equations: a. x 2 − x − 90 = 0 b. 25x 2 + 90x + 81 = 0 c. (3x − 2)(x − 3) = 20 d. 3(5x + 3)(4x 2 −1) = 8(4x 2 −1)2 e. x 2 − 13x + 4 = 0 Rancho Cucamonga High School - Ontario, CA 3x 2 −14x +11 x + 9 x 2 + x +1 = − 2 14 5 2 g. x + 2(1+ 8)x + 8 2 = 0 (Do not round off the coefficients.) f. 2. One of the digits of a two-digit number is 3 less than the other digit. When the digits are interchanged, we get a new number. The sum of the squares of the new number and the initial number is 1877. Find the two numbers. 3. Make up a word problem whose solution can be found by solving the equation x ( x + 36) = 180 . 4. Two planes start flying from the same airport at the same time. One plane flies directly north, and the other flies directly east. After 2 hours from the time they started, the distance between the two planes is 2000 miles. Find the speeds of each plane if the speed of one is 75% of the speed of the other. 5. A company has decided to double its production within two years. By how many percents should the company increase its production each year if it wants the percentage increase in the first year to be equal to the percentage increase in the second year? kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk secTion 3: Polynomial equations SecTIon 3: Polynomial equations So far we have learned how to solve linear and quadratic equations. Now, you may be wondering, what about equations of similar form but with higher degrees? For example, how do we solve equations like 2 x 3 + 4 x 2 − x + 9 = 0 and −3x 4 + 5 x 3 − 2 x + 12 = 0 ? Equations of this kind are called polynomial equations. 16 2013 –2014 USAD Mathematics Resource Guide • 2016–2017 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement AcAdemic decAthlon ® mAthemAtics ResouRce Guide 15 15 and the other flies directly east. After 2 hours from the time they started, the distance between the two planes is 2000 miles. Find the speeds of each plane if the speed of one is 75% of the speed of the other. 5. A company has decided to double its production within two years. By how many percents should the company increase its production each year if it wants the percentage increase in the first year to be equal to the percentage increase in the second year? Section kkk3 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk Equations secTion 3:Polynomial Polynomial equations SecTIon 3: Polynomial equations about equations of similar form but with higher degrees? For example, how do we solve equations like A polynomial is an expression of the form an x n + an −1 x n −1 + + a3 x 3 + a2 x 2 + a1 x1 + a0 , where n is a non3 2 4 3 2 x + 4 x − x + 9 = 0 and −3x + 5 x − 2 x + 12 = 0 ? Equations of this kind are called polynomial equations. with a 0 . a , a , a , , a are called the coefficients negative integer, and a0 , a1 , a2 , , an are real numbers A polynomial is an expression of the form an x n + an −1 x n −1 +n + a3 x03 +1 a2 2x 2 + a1nx1 + a0 , where n is a nonDuplication is Prohibited Outside the Terms of Yourthe License Agreement n , the highest power in the polynomial, is called degree of the polynomial. The of the polynomial.Unauthorized a 0 a , a , a , , a are real numbers with . a , a , a , , an are called the coefficients negative integer, and ® 0 1 2 n 0 1 2 n 2013 –2014 MATheMATics 15 AcAdemic decAthlon mAthemAtics ResouRce Guide 15 2010–2011 ResouRce Guide expressions, a0 , a1 x1 , a2 x 2 , a3 x 3 , , an −1 x n −1 , an x n , are called the terms of the polynomial. Expressions of of the polynomial. n , the highest power in the polynomial, is called the degree of the polynomial. The integer, are called monomials. the form x k , where1 k is 2a non-negative expressions, a0 , a1 x , a2 x , a3 x 3 , , an −1 x n −1 , an x n , are called the terms of the polynomial. Expressions of the form x k , where k is a non-negative integer, are called monomials. ExamplE 3a: ExamplE 3a: 2 The coefficients of the polynomial 4 x 5 − 3x 4 + x 3 + 3x 2 − 6 x + 1 are: 5 2 2 5 4 3 a5 = coefficients 4 , a4 = −3 ,ofa3the , a2 = 3 , a41x= − −63,x a+ this polynomial is n = 5 . = polynomial The 0 = 1xand + the 3x 2degree − 6 x + of 1 are: 5 5 2 2 the 5 polynomial is n = 5 . a5 = terms 4 , a4 of = −this 3 , apolynomial , a2 = are: −6 ,, a03x = 21,and of 4x this 3 , 1a1, =−6x 3 = . The 3x 4 , and x 3 , −degree 5 5 2 The terms of this polynomial are: 1 , −6x , 3x 2 , x 3 , −3x 4 , and 4x 5. 5 ExamplE 3b: ExamplE 3b: 3 The coefficients of the polynomial 17 x 4 + 0 ⋅ x 3 − 12 x 2 + 0 ⋅ x − are: 4 3 3 3 3 4 2 a4 = coefficients 17 , a3 = 0 , of a2 the = −12 , a1 = 0 , a17 The polynomial x —+ 04 ⋅ x − 12 x + 0 ⋅ x − are: 0 = 4 4 3 This polynomial is of degree . 4 a4 = 17 , a3 = 0 , a2 = −12 , a1 = 0 , a0 = 4 3 This terms polynomial of degree 4are: . − , 0 ⋅ x , −12x 2 , 0 ⋅ x 3, and 17x 4. The of thisispolynomial 4 3 The terms of this polynomial are: − , 0 ⋅ x , −12x 2 , 0 ⋅ x 3, and 17x 4. 4 17 USAD Mathematics Resource Guide • 2016–2017 When a coefficient is zero, there is no need to write the corresponding term. For example, instead of Rancho Cucamonga High School - Ontario, CA So far we have learned how to solve linear and quadratic equations. Now, you may be wondering, what a4 = 17 , a3 = 0 , a2 = −12 , a1 = 0 , a0 = 3 4 This polynomial is of degree 4 . 3 The terms of this polynomial are: − , 0 ⋅ x , −12x 2 , 0 ⋅ x 3, and 17x 4. 4 When a coefficient is zero, there is no need to write the corresponding term. For example, instead of writing −4x 5 + 0x 4 + 5x 3 + 0x 2 − 4x + 7 , we just write −4x 5 + 5x 3 − 4x + 7 . When all the coefficients of a polynomial are zeros, the polynomial is called the zero polynomial. For convenience, we denote polynomials by Q(x), R(x), etc. 3 .11When Areare TWo PoLynomIALS eQuAL? 3 . When two Polynomials Polynomials equal? 3.1 When Are Two Equal? ExamplE 3.1a: Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 16 16 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MATheMATics ResouRce Guide –2014 2013 2010–2011 The polynomials P ( x ) = x 3 + x 2 + x and Q ( x ) = 0 ⋅ x 4 + x 3 + x 2 + x + 0 are equal. The polynomials R ( x ) = x 3 + x 2 + x and S ( x ) = x 4 + x 3 + x 2 + x , on the other hand, are not equal because the S(x). term x 4 is in R(x) S(x) but not in R(x). 3 .2 hoW Do We SuBTrAcT PoLynomIALS? 3 .2 how doADD WeAnD add and Polynomials? 3.2 How Do We Add and subtract Subtract Polynomials? Naturally, the sum and difference of two polynomials are defined. We simply add or subtract the coefficients of the like terms. ExamplE 3.2a: Q(x) = −4x 5 + 3x 3 + 2x 2 − 6x + 9 P(x) = 2x 5 + 7x 3 − 2x 2 + 6x +19 Q(x) + P(x) = (−4 + 2)x 5 + (3+ 7)x 3 + (2 + (−2))x 2 + (−6 + 6)x + (9 +19) = = −2x 5 + 10x 3 + 0x 2 + 0x + 28 = −2x 5 + 10x 3 + 28 Q(x) − P(x) = (−4 − 2)x 5 + (3 − 7)x 3 + (2 − (−2))x 2 + (−6 − 6)x + (9 −19) = = −6x 5 − 4x 3 + 4x 2 − 12x − 10 = −6x 5 − 4x 3 + 4x 2 − 12x − 10 18 USAD Mathematics Resource Guide • 2016–2017 Rancho Cucamonga High School - Ontario, CA Two polynomials Q(x) and R(x) are equal if they have the same terms. Q(x) + P(x) = (−4 + 2)x + (3+ 7)x + (2 + (−2))x + (−6 + 6)x + (9 +19) = 5 2 Q(x) + P(x) = (−4 + (3+ 7)x 3 + (25 + (−2))x + (−6 + 6)x + (9 +19) = 5 3 + 2)x 2 3 = −2x + 10x + 0x + 0x + 28 = −2x + 10x + 28 = −2x 5 + 10x 3 + 0x 2 + 0x + 28 = −2x 5 + 10x 3 + 28 Q(x) − P(x) = (−4 − 2)x 5 + (3 − 7)x 3 + (2 − (−2))x 2 + (−6 − 6)x + (9 −19) = 5 2 Q(x) − P(x) = (−4 − 2)x + (3 − 7)x 3 + (25− (−2))x +2(−6 − 6)x + (9 −19) = 5 3 2 3 = −6x − 4x + 4x − 12x − 10 = −6x − 4x + 4x − 12x − 10 = −6x 5 − 4x 3 + 4x 2 − 12x − 10 = −6x 5 − 4x 3 + 4x 2 − 12x − 10 ExamplE 3.2b: ExamplE 3.2b: ExamplE 3.2c: Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement ® Outside the Terms of Your License Agreement 2013Q –2014 mAthemAtics ResouRce Guide 2010–2011 ( x) = 6 x MATheMATics + 23Unauthorized x 3 + AcAdemic 2ResouRce x 2 −Duplication 3 x −decAthlon 8Guideis Prohibited ® 2013 –2014 MATheMATics AcAdemic mAthemAtics ResouRce Guide 2010–2011 ResouRcedecAthlon Guide 4 17 17 17 17 P ( x) = 32 x 5 − 2 x3 − 18 x + 19 Q(x) − P(x) = (0 − 32)x 5 + (6 − 0)x 4 + (23 − (−2))x 3 + ( 2 − 0)x 2 + (−3 − (−18))x + (−8 −19) = = −32x 5 + 6x 4 + 25x 3 + 2x 2 + 15x − 27 . 3.3 How We Polynomials? 3 .3How how do We Multiply multiply Polynomials? 3 .3 DoDo We Multiply Polynomials? To multiply two polynomials, we multiply each term of one polynomial by each term of the second polynomial and add like terms. ExamplE 3.3a: Q(x) = 5x 3 − 3x 2 + 2x P(x) = −4x 4 −12x 3 +15 ( )( ) = (5x ) ⋅ (−4x − 12x + 15) + (−3x ) ⋅ (−4x − 12x + 15) + ( 2x ) ⋅ (−4x − 12x + 15) = = (5x ) ⋅ (−4x ) + (5x ) ⋅ (−12x ) + (5x ) ⋅ (15) + (−3x ) ⋅ (−4x ) + (−3x ) ⋅ (−12x ) + (−3x ) ⋅ (15)+ + ( 2x ) ⋅ (−4x ) + ( 2x ) ⋅ (−12x ) + ( 2x ) ⋅ (15) = Q(x) ⋅ P(x) = 5x 3 − 3x 2 + 2x ⋅ −4x 4 − 12x 3 + 15 = 3 4 3 3 4 3 2 4 3 4 3 3 4 2 4 3 2 3 6 3 5 4 = −20x 7 − 60xUSAD + 75xMathematics + 12x 6 + 36x 5Resource − 45x 2 − 8x − 24x + 30x = Guide • 2016–2017 = −20x − 48x + 28x − 24x + 75x − 45x + 30x 7 2 3 6 5 4 3 2 19 Rancho Cucamonga High School - Ontario, CA Q(x) = 65x 7 +13x 3 + 2x 2 − 8 2 Q(x) = 65x 75 +13x4 3 + 2x −8 3 P(x) = 32x + 4x − 2x + 6x +19 P(x) = 32x 5 + 4x 4 − 2x 3 + 6x +19 Q(x) + P(x) = (65 + 0)x 7 + (0 + 32)x 5 + (0 + 4)x 4 + (13 − 2)x 3 + ( 2 + 0)x 2 + (0 + 6)x + (−8 +19) = Q(x) + P(x) = (65 + 0)x 7 + (0 + 32)x 5 + (0 + 4)x 4 + (13 − 2)x 3 + ( 2 + 0)x 2 + (0 + 6)x + (−8 +19) = = 65x 7 + 32x 5 + 4x 4 +11x 3 + 2x 2 + 6x +11 = 65x 7 + 32x 5 + 4x 4 +11x 3 + 2x 2 + 6x +11 ( )( ) = (5x ) ⋅ (−4x − 12x + 15) + (−3x ) ⋅ (−4x − 12x + 15) + ( 2x ) ⋅ (−4x − 12x + 15) = = (5x ) ⋅ (−4x ) + (5x ) ⋅ (−12x ) + (5x ) ⋅ (15) + (−3x ) ⋅ (−4x ) + (−3x ) ⋅ (−12x ) + (−3x ) ⋅ (15)+ + ( 2x ) ⋅ (−4x ) + ( 2x ) ⋅ (−12x ) + ( 2x ) ⋅ (15) = Q(x) ⋅ P(x) = 5x 3 − 3x 2 + 2x ⋅ −4x 4 − 12x 3 + 15 = 3 4 3 3 4 3 2 3 4 4 3 3 4 2 4 2 3 3 2 3 = −20x 7 − 60x 6 + 75x 3 + 12x 6 + 36x 5 − 45x 2 − 8x 5 − 24x 4 + 30x = = −20x 7 − 48x 6 + 28x 5 − 24x 4 + 75x 3 − 45x 2 + 30x ( )( ) Q(x) ⋅ P(x) = 5x 3 − 3x 2 + 2x ⋅ −4x 4 − 12x 3 + 15 = −20x 7 − 48x 6 + 28x 5 − 24x 4 + 75x 3 − 45x 2 + 30x 3 .4 how do We Divide divide Polynomials? 3 . 4 How DoDo We Divide Polynomials? 3.4 How We Polynomials? Division of polynomials is similar to the long division algorithm you learned in elementary school. Before we show how to divide one polynomial by another, it is worth discussing a few properties of polynomials Any integer can be factored into the product of other, smaller integers, unless the integer is a prime number. For example, 12 = 2 2 ⋅ 3 , 60 = 2 2 ⋅ 3 ⋅ 5 , 525 = 3 ⋅ 52 ⋅ 7 , etc. 3 divides 12 because 12 ÷ 3 = 4 , or Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 12 18 18 = 3 ⋅ 4 . On the other hand, 5 does not divide® 12, because when we divide 12 by 5, Guide we get a remainder 2013 –2014 mAthemAtics ResouRce Guide AcAdemic decAthlon MATheMATics ResouRce 2010–2011 of 2. We can say the same thing using different words: 5 does not divide 12 because there is no integer whose product with 5 equals 12. In general, we say an integer a ≠ 0 divides an integer b , if there exists an integer c such that b = ac . A similar situation exists with polynomials. Polynomials can also be factored into the product of other polynomials with smaller degrees. For example, 3x 2 + 5 x − 2 = (3x − 1)( x + 2) . Therefore, we say that the polynomial 3x − 1 divides the polynomial 3x 2 + 5 x − 2 . Also, the polynomial x + 2 divides the polynomial 3x 2 + 5 x − 2 . We note that 2(x + 2) also divides the polynomial 3x 2 + 5 x − 2 because 3 1 x 2x + 4 . The coefficients of the polynomials can be real numbers and even 2 2 complex numbers in general. But, for computational simplicity, most of our polynomials will have integer 3x 2 + 5x 2 = ( ) coefficients. Similarly, each of the polynomials x − 2 and x + 2 divides the polynomial x 2 − 4 because x 2 − 4 = ( x − 2)( x + 2) . On the other hand, the polynomial x does not divide x − 2 because there is no polynomial whose product with x results in x − 2 . definition 20 A polynomial K ( x) divides a polynomial P ( x) if there exists USAD Mathematics Resource Guide • 2016–2017 a polynomial Q( x) such that P ( x) = K ( x)Q( x) . Rancho Cucamonga High School - Ontario, CA that are analogous to those of whole numbers. polynomial whose product with x results in x − 2 . definition A polynomial K ( x) divides a polynomial P ( x) if there exists a polynomial Q( x) such that P ( x) = K ( x)Q( x) . We will now learn how to carry out the division of polynomials. ExamplE 3.4a: Let P ( x) = 2 x 4 + 5 x 3 + 3 x 2 − 2 x − 8 and K ( x) = x − 1 . Divide P ( x) by K ( x) . solution: As we said before, the division of polynomials is similar to the division of multithe right. 2x 3 + 7x 2 +10x + 8 3 1. Divide 2x by x . You get 2x . x −1 2x 4 + 5xUnauthorized + 3x 2 − 2xDuplication −8 is Prohibited Outside the Terms of Your License Agreement 4 3 2x − 2x ® 3 2013 –2014 4MATheMATics AcAdemic mAthemAtics 19 2010–2011 ResouRcedecAthlon Guide 2. Multiply divisor x − 1Guide and subtract the result from the 19 2x 3 by theResouRce 3 2 polynomial. You get 7x + 3x − 2x − 8. 7x 3 + 3x 2 − 2x − 8 3. Divide 7x 3 − 7x 2 4. Multiply 7x by the divisor x − 1 and subtract the result from the 2 polynomial. You get 10x − 2x − 8. 7x 3 by x . You get 7x 2. 2 10x 2 − 2x − 8 10x 2 −10x 8x − 8 8x − 8 0 5. Divide 10x 2 by x . You get 10x . 6. Multiply 10x by the divisor polynomial. You get 8x − 8. 7. Divide 8x by x − 1 and subtract the result from the x . You get 8 . 8. Multiply 8 by the divisor polynomial. You get 0 . x −1 and subtract the result from the Thus, P(x) divided by K ( x) is the polynomial Q(x) = 2x 3 + 7x 2 +10x + 8. We write: 2x 4 + 5x 3 + 3x 2 − 2x − 8 = 2x 3 + 7x 2 +10x + 8 x −1 or ( )( ) 2x 4 + 5x 3 + 3x 2 − 2x − 8 = 2x 3 + 7x 2 + 10x + 8 x − 1 ExamplE 3.4b: Let P ( x) = 2 x 3 + 7 x 2 + 10 x + 8 and K ( x) = x + 2 . Divide P ( x) by K ( x) . solution: x+2 2 Resource Guide • 2016–2017 2xUSAD + 3xMathematics +4 2x 3 + 7x 2 +10x + 8 1. Divide 2x by x . You get 2x . 3 2 21 Rancho Cucamonga High School - Ontario, CA digit numbers. Each step in the following division process is accompanied by an explanation on ExamplE 3.4b: Let P ( x) = 2 x 3 + 7 x 2 + 10 x + 8 and K ( x) = x + 2 . Divide P ( x) by K ( x) . solution: 1. Divide −(2x 3 + 4x 2 ) 2x 3 by x . You get 2x 2 . 2. Multiply 2x by the divisor x + 2 and subtract the result from the 2 polynomial. You get 3x + 10x + 8. 2 3x 2 +10x + 8 3. Divide − (3x 2 + 6x) 3x 2 by x . You get 3x . 4. Multiply 3x by the divisor x + 2 and subtract the result from the polynomial. You get 4x + 8. 4x + 8 − (4x + 8) 5. Divide 4x by x . You get 4. 6. Multiply 4 by the divisor polynomial. You get 0. x+2 and subtract the result from the 0 Therefore, we get: 2x 3 + 7x 2 +10x + 8 = (x + 2)(2x 2 + 3x + 4) . In both of these examples, theDuplication remainder dividing is zero. This is not always the case, as P ( xthe ) by K (ofx)Your Unauthorized is of Prohibited Outside Terms License Agreement 20 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MATheMATics ResouRce Guide is shown in the following example. –2014 2013 2010–2011 ExamplE 3.4c: Let P(x) = −4x 4 + 9x 3 −12x + 5 and K(x) = x 2 + 2 . Divide P ( x) by K ( x) . solution: Note, the coefficient of x 2 in polynomial K ( x) is zero. −4x 2 + 9x + 8 x 2 + 2 − 4x 4 + 9x 3 + 0 ⋅ x 2 −12x + 5 −(−4x 9x + 8x 3 − (9x −4x 4 by x 2 . You get 4x 2. 2 2. Multiply 4x by the divisor x + 2 and subtract the 3 2 result from the polynomial. You get 9x + 8x . 2 3. Divide +18x) 3 8x 2 − 30x − (8x 2 2 2 − 8x ) 4 1. Divide +16) 9x 3 by x 2 . You get 9x . 4. Multiply 9x by the divisor x + 2 and subtract the 2 result from the polynomial. You get 8x − 30x . 2 5. Divide 8x 2 by x 2 . You get 8. 6. Multiply 8 by the divisor x + 2 and subtract the result from the polynomial. You get − 30x − 11. 2 − 30x −11 Since −30x cannot be divided by x 2 , we have the remainder −30x −11 . Therefore, we get: 22 −4x 4 + 9x 3 + 0 ⋅ x 2 −12x + 5 USAD2 Mathematics Guide • 2016–2017 −30xResource −11 x2 + 2 = −4x + 9x + 8 + x2 + 2 Rancho Cucamonga High School - Ontario, CA x+2 2x 2 + 3x + 4 2x 3 + 7x 2 +10x + 8 8x − 30x − (8x 2 6. Multiply 8 by the divisor x + 2 and subtract the result from the polynomial. You get − 30x − 11. 2 +16) − 30x −11 Since −30x cannot be divided by x 2 , we have the remainder −30x −11 . Therefore, we get: −30x −11 −4x 4 + 9x 3 + 0 ⋅ x 2 −12x + 5 = −4x 2 + 9x + 8 + 2 x2 + 2 x +2 or 3 −4x 4 + 9x 2 −12x + 5 = (x 2 + 2)(−4x 2 + 9x + 8) + (−30x −11) The division of polynomials just shown is based on the following theorem, which we accept without division algorithm Let P ( x) and K ( x) be two polynomials, where K ( x) is not the zero polynomial. Then, there exist two unique polynomials Q( x) and R ( x ) such that P(x) = K (x)Q(x) + R(x) ; the polynomial Q( x) is the quotient, and the remainder R ( x ) is either the zero polynomial or of a degree that is less than the degree of K ( x) . Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 AcAdemic MATheMATics ResouRcedecAthlon Guide ExamplE 3.4d: ® mAthemAtics ResouRce Guide 21 21 In the last example, we saw that −4 x 4 + 9 x 23 − 12 x + 5 = ( x 2 + 2)(−4 x 2 + 9 x + 8) + (−30 x − 11). Here 3 P( x) = −4 x 4 + 9 x3 − 12 x + 5 , K ( x) = x 2 + 2 , Q( x) = −4 x 2 + 9 x + 8 , and R( x) = −30 x − 11. The degree of the remainder R ( x) is less than the degree of the divisor K ( x) . ExamplE 3.4e: We have seen in previous examples that 2 x 3 + 7 x 2 + 10 x + 8 = ( x + 2)(2 x 2 + 3 x + 4) Here Q( x) = 2 x 2 + 3 x + 4 , and the remainder R ( x ) is the zero-polynomial. The following theorem relates evaluating a polynomial to the division of polynomials. USAD Mathematics Resource Guide • 2016–2017 23 Rancho Cucamonga High School - Ontario, CA proof. The following theorem relates evaluating a polynomial to the division of polynomials. remainder theorem If R is the remainder when a polynomial P ( x) is divided by x − a , then R = P (a ) . Proof We use the Division Algorithm. Let Q( x) be the quotient and R be the remainder when P ( x) is divided by x − a . Thus, P(x) = Q(x)(x − a) + R . ExamplE 3.4f: When dividing the polynomial P(x) = 3x 3 − 4x 2 + 5 by x + 2 using the Division Algorithm, we get the remainder −35 . Using the Remainder Theorem, we obtain the same result: P(−2) = 3⋅ (−2)3 − 4 ⋅ (−2)2 + 5 = −35 3.5 H ow Does the of 3 .5 how does the Division division of Polynomials Polynomials Help 3 .5 How Does the Division of Polynomials help with Polynomial equations? Help Us Usus with Polynomial Equations? with Polynomial Equations? Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 22 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MATheMATics ResouRce Guide –2014 2013 2010–2011 Division of polynomials can be very helpful in solving polynomial equations. We have no general method of solving polynomial equations of any degree. However, sometimes we can solve a polynomial equation of a high degree by factoring the polynomial into the product of lower-degree polynomials. ExamplE 3.5a: Solve the equation x 3 + 3x 2 − 18x − 40 = 0 . solution: We first factor the polynomial x 3 + 3x 2 − 18x − 40 (we will learn about factoring polynomials later): x 3 + 3x 2 − 18 x − 40 = ( x + 2)( x 2 + x − 20) . With this factorization, we reduce the task of solving an unfamiliar equation x 3 + 3x 2 − 18 x − 40 = 0 into solving two familiar equations: x + 2 = 0 or x 2 + x − 20 = 0 . The solution of x + 2 = 0 is x = −2 . 24 The solutions of 4. x = −5 , x = Resource x 2 + x − 20 USAD = 0 are Mathematics Guide • 2016–2017 Hence, the given equation has three solutions: x = −2 , x = −5, and x = 4 . Rancho Cucamonga High School - Ontario, CA Setting x = a in the above equation, we get P(a) = Q(a)(a − a) + R . That is, R = P (a ) . the task of solving an unfamiliar equation x 3 + 3x 2 − 18 x − 40 = 0 into solving two familiar equations: x + 2 = 0 or x 2 + x − 20 = 0 . The solution of x + 2 = 0 is x = −2 . The solutions of x 2 + x − 20 = 0 are x = −5 , x = 4 . Hence, the given equation has three solutions: x = −2 , x = −5, and x = 4 . The question now is: How would one know how to factor the polynomial x 3 + 3x 2 − 18 x − 40 or any other high-degree polynomial? The following two theorems, together with the Division Algorithm, can be very helpful in factoring certain polynomials. Let P(x) = an x n + an −1 x n −1 ++ a1 x + a0 be a polynomial with integer coefficients. If a reduced m is a solution to the equation P ( x) = 0 , then m divides a0 and k divides an . fraction k factor theorem The polynomial x − a is a factor of P ( x) if and only if P (a ) = 0 . The Factor Theorem follows from the Remainder Theorem, and a direct proof of it by using the Division Algorithm is presented in Section 3.7. But, before we present the proofs of these theorems, we will first consider how to apply them to solve polynomial equations. Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 AcAdemic decAthlon ® mAthemAtics ResouRce Guide ResouRce x 3 − 2 xGuide −1 = 0. ExamplEMATheMATics 3.5b: Solve 23 solution: Let P ( x) = x 3 − 2 x − 1 . In this polynomial, a0 = −1 and an = 1 . By the Rational m is a solution to P ( x) = 0 , then m divides −1 and k Root Theorem, if a reduced fraction k divides 1 . The integers that divide −1 are m = ± 1 . And, the integers that divide 1 are k = ±1 . m , with these values of m and k , is a candidate to solve P ( x) = 0 . k m m +1 −1 + 1 − 1 The following are all the possible fractions : = , , , k k +1 −1 −1 +1 Therefore, any fraction After removing repetitions, we get two candidates for rational solutions to P ( x) = 0 . These are: +1 +1 = 1 and = −1 +1 −1 USAD Mathematics Resource Guide • 2016–2017 Now we only need to find out which of these two numbers is a solution to P ( x) = 0 . 25 Rancho Cucamonga High School - Ontario, CA the rational root theorem The following are all the possible fractions m m +1 −1 + 1 − 1 : = , , , k k +1 −1 −1 +1 After removing repetitions, we get two candidates for rational solutions to P ( x) = 0 . These are: +1 +1 = 1 and = −1 +1 −1 Now we only need to find out which of these two numbers is a solution to P ( x) = 0 . We substitute x = 1 in P ( x) = 0 , and get 1⋅13 − 2 ⋅1− 1 = −2 ≠ 0 Thus, x = 1 is not a solution. ( ) ( ) We substitute x = −1 in the same equation, and get 1⋅ −1 − 2 ⋅ −1 − 1 = 0 Thus, x = −1 is a 3 solution. By the Factor Theorem, x − (−1) = x + 1 divides x 3 − 2 x − 1 . Let’s do the division: x 2 − x −1 x +1 x 3 + 0 ⋅ x 2 − 2x − 1 Rancho Cucamonga High School - Ontario, CA −(x 3 + x 2 ) − x 2 − 2x −1 − (−x 2 − x) − x −1 − (−x −1) 0 ( )( ) This shows that x 3 − 2x − 1 = x 2 − x − 1 x + 1 ( )( ) Solving x 2 − x − 1 x + 1 = 0 , we get 24 x +1 = 0 x2 − x − 1 = 0 x = −1 x= 1± 1+ 4 1± 5 = 2 2 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 1+ 5 1− 5 ® x = −1 ResouRce Thus, our polynomial equationdecAthlon has three solutions: , MATheMATics x= ,ResouRce x= . mAthemAtics Guide AcAdemic 2 2 Guide –2014 2013 2010–2011 Thus, we can see how the Division Algorithm, the Rational Root Theorem, and the Factor Theorem help us to solve a polynomial equation of degree three. ExamplE 3.5c: Solve 2 x 3 − 5 x + 3 = 0 . solution: Let P ( x) = 2 x 3 − 5 x + 3 . In this polynomial, a0 = 3 and a3 = 2 . By the Rational m is a solution to P ( x) = 0 , then m divides 3 and k Root Theorem, if a reduced fraction k 26 divides 2 . USAD Mathematics Resource Guide • 2016–2017 The integers that divide 3 are m = ±3, ± 1 . us us to to solve solve a a polynomial polynomial equation equation of of degree degree three. three. 3 ExamplE 2 xx 3 − −5 5 xx + +3 3= =0 0 .. ExamplE 3.5c: 3.5c: Solve Solve 2 3 solution: solution: Let Let P P(( xx)) = = 22 xx3 − − 55 xx + + 33 .. In In this this polynomial, polynomial, m m is Root is aa solution solution to to Root Theorem, Theorem, if if aa reduced reduced fraction fraction kk divides divides 22 .. aa00 = = 33 and and a a33 = =2 2 .. By By the the Rational Rational P then m m divides divides 3 and and k k P (( xx)) = =0 0 ,, then The are m The integers integers that that divide divide 3 are m= =± ±3, 3, ± ±1 1 .. And =± ±2, 2, ± ±1 1 .. are kk = And the the integers integers that that divide divide 22 are m m: The : The following following are are the the possible possible fractions fractions kk m m 3, m=+ +3 +3 +3 +3 , −3 −3 , −3 −3 m = +3 = , = ,, , , kk + kk +2 +1 1 +2 −2 −2 +2 +2 −2 −2 m m m=+ 1, + 1, − 1, − 1 m=+ 1, +1 +1 −1 −1 +1 = , , , = , kk + kk + +2 2 − −2 2 + +2 2 − −2 2 +1 1 After After removing removing all all the the 33 11 P ± ,, ± ± ,, P (( xx)) = =0 0 :: ± ±3 3 ,, ± 22 22 3 +3 + ,, −1 1 − + 1, +1 , − −1 1 3 −3 − ,, +1 1 + − 1, −1 , + +1 1 Rancho Cucamonga High School - Ontario, CA m m with these values of m and k is a candidate to solve the equation Therefore, with these values of m and is a candidate to solve the equation Therefore, any any fraction fraction kk P P (( xx)) = =0 0 .. 3 −3 − −1 1 − 1 − −1 − −1 1 repetitions, repetitions, we we get get the the following following eight eight candidates candidates for for solutions solutions to to ± ±1 1 Now Now we we only only need need to to examine examine which which of of these these 88 numbers numbers are are solutions solutions to to P(x) We start start with with P(x) = =0 0 .. We the and ±3 .. the easy easy ones: ones: ± ± 11 and 3 22 ⋅1 ⋅13 − − 55 ⋅1+ ⋅1+ 3⋅1 3⋅1 = = 00 3 22 ⋅⋅ (−1) (−1)3 − − 55 ⋅⋅ (−1) (−1) + + 33 = = 66 x = 1 is is aa solution solution x = −1 is is not not aa solution solution 3 2 2 ⋅⋅ 3 33 − −5 5 ⋅⋅ 3+ 3+ 3 3= = 42 42 x x= is not not aa solution solution = 33 is 3 22 ⋅⋅ (−3) (−3)3 − − 55 ⋅⋅ (−3) (−3) + + 33 = = −36 −36 x x = is not not aa solution solution =33 is You can check on your own that none of the other candidates is a solution to P(x) = 0 . 1 Your mustLicense divide , x −of Now we can apply the Factor Theorem. Since,Outside P(1) = P(x) . After dividing Unauthorized Duplication is the Agreement Unauthorized Duplication is Prohibited Prohibited Outside the0Terms Terms of Your License Agreement ® 2 ® mAthemAtics 2013 –2014 decAthlon 2010–2011 xMATheMATics ResouRce Guide 2013 –2014 by AcAdemic decAthlon mAthemAtics ResouRce Guide 2010–2011 MATheMATics ResouRce Guide −1 , we AcAdemic get 2x 3 − 5x + 3 = (x −1)(2x + 2x − 3) .ResouRce Therefore,Guide the problem Q(x) P(x) of solving 25 25 the equation 2x 3 − 5x + 3 = 0 is reduced to the problem of solving the equations x − 1 = 0 and 2x 2 + 2x − 3 = 0 . Solving, we get: x = 1 , x = −1− 7 −1+ 7 , x= 2 2 Hence, these three values are solutions of the equation 2x 3 − 5x + 3 = 0 . USAD Mathematics Resource Guide • 2016–2017 ExamplE 3.5d: Solve 2x 4 + 5x 3 + 3x 2 − 2x − 8 = 0 . 27 ExamplE 3.5d: Solve 2x 44 + 5x 33 + 3x 22 − 2x − 8 = 0 . ExamplE 3.5d: Solve 2x + 5x + 3x − 2x − 8 = 0 . solution: Let P(x) = 2x 44 + 5x 33 + 3x 22 − 2x − 8 . In this polynomial, a0 = −8 and an = 2 . By the solution: Let P(x) = 2x + 5x + 3x − 2x − 8 . In this polynomial, a0 = −8 and an = 2 . By the m Rational Root Theorem, if a reduced fraction m is a solution to P ( x) = 0 , then m divides −8 Rational Root Theorem, if a reduced fraction k is a solution to P ( x) = 0 , then m divides −8 k and k divides 2 . and k divides 2 . m Therefore, any fraction m , with these values of m and k , is a solution candidate for P ( x) = 0 . Therefore, any fraction k , with these values of m and k , is a solution candidate for P ( x) = 0 . m The following are all the kpossible fractions m : The following are all the possible fractions k : k m +8 +8 −8 −8 m +8 +8 −8 −8 m = +8, +8 , −8, −8 m = +8, +8, −8, −8 k = +2 , −2 , +2 , −2 k = +1 , −1 , +1 , −1 k +2 −2 +2 −2 k +1 −1 +1 −1 m +4 +4 −4 −4 m +4 +4 −4 −4 m = +4 , +4 , −4 , −4 m = +4 , +4 , −4 , −4 k = +2 , −2 , +2 , −2 k = +1 , −1 , +1 , −1 k +2 −2 +2 −2 k +1 −1 +1 −1 m +1 +1 −1 −1 m +1 +1 −1 −1 m = +1, +1, −1, −1 m = +1, +1, −1, −1 k = +2 , −2 , +2 , −2 k = +1 , −1 , +1 , −1 k +2 −2 +2 −2 k +1 −1 +1 −1 After removing all the repetitions, we get the following candidates as possible solutions for After removing all the repetitions, we get the following candidates as possible solutions for 1 P ( x) = 0 : ±8, ± 4, ± 2, ± 1, ± 1. P ( x) = 0 : ±8, ± 4, ± 2, ± 1, ± 2 . 2 Now we only need to examine which of these 10 numbers are solutions to P ( x) = 0 . Now we only need to examine which of these 10 numbers are solutions to P ( x) = 0 . We start with the easy ones: ± 1 and ±2. We start with the easy ones: ± 1 and ±2. 2 ⋅144 + 5 ⋅133 + 3⋅122 − 2 ⋅1− 8 = 0 2 ⋅1 + 5 ⋅1 + 3⋅1 − 2 ⋅1− 8 = 0 26 26 26 26 xx ==11 isis aa solution solution 2 ⋅ (−1)4 + 5 ⋅ (−1)3 + 3⋅ (−1)2 − 2 ⋅ (−1) − 8 = −6 x = −1 is not a solution 2 ⋅ 24 + 5 ⋅ 23 + 3⋅ 22 − 2 ⋅ 2 − 8 = 72 x = 2 is not a solution 2 ⋅ (−2)4 x = −2 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement Unauthorized Outside the Terms of Your License Agreement 3 2 + 5 ⋅ (−2) + 3⋅ (−2)Duplication − 2 decAthlon ⋅ (−2)is−Prohibited 8 = 0® mAthemAtics is a solution ResouRce Guide AcAdemic MATheMATics ResouRce Guide ® Guide AcAdemic decAthlon mAthemAtics ResouRce MATheMATics ResouRce Guide –2014 2013 2010–2011 –2014 2013 2010–2011 You can check on your own that none of the other candidates is a solution to P(x) = 0 . After dividing 2x 4 + 5x 3 + 3x 2 − 2x − 8 by x − 1 , we get 2x 3 + 7x 2 + 10x + 8 . ( )( ) Therefore, we have 2x 4 + 5x 3 + 3x 2 − 2x − 8 = 2x 3 + 7x 2 + 10x + 8 x − 1 . Now we solve two equations: 1. K (x) = x − 1 = 0 2. Q( x) = 2 x 3 + 7 x 2 + 10 x + 8 = 0 The first will give us x = 1 . To solve the second equation, we need to factor Q( x) . We have: P( x) = Q( x) K ( x) . 28 Recall x = −2 solves the equation P ( x) = 0 . Hence, P (2) = Q(−2) K (−2) = 0 . USAD Mathematics Resource Guide • 2016–2017 Since K (−2) ≠ 0 , Q(−2) = 0 . Rancho Cucamonga High School - Ontario, CA The integers that divide a0 = −8 are m = ±8, ± 4, ± 2, ± 1 . And, the integers that divide an = 2 The integers that divide a0 = −8 are m = ±8, ± 4, ± 2, ± 1 . And, the integers that divide an = 2 are k = ±2, ± 1. are k = ±2, ± 1. 2 ( x) will = 2 x3give + 7 xus +x10 2. Q The first = x1 +. 8To= 0solve the second equation, we need to factor Q( x) . We have: P ( x) =first Q( xwill ) K ( xgive ) . us x = 1 . To solve the second equation, we need to factor The Q( x) . We have: P ( x) =xQ=( x−)2Ksolves ( x) . the equation P ( x) = 0 . Hence, P (2) = Q(−2) K (−2) = 0 . Recall the=equation RecallK (x−=2)−≠2 0solves P ( x) = 0 . Hence, P (2) = Q(−2) K (−2) = 0 . , Q(−2) Since 0. Sincewe K (use −2) the ≠ 0 ,Factor Q (−2)Theorem = 0. Now, again and divide Q( x) by x + 2 to get: 2Now, x 3 + 7we x 2 use + 10the x + 8Factor = (2 x 2Theorem + 3 x + 4)(again x + 2)and divide Q( x) by x + 2 to get: 2 x 3 solution + 7 x 2 + 10 + 8 = (2 x 2 2x + 33 x++7x4)( 2 x + 2) The ofx equation + 10x + 8 = 0 is reduced to the solution of two equations: 1.Thexsolution + 2 = 0 of equation 2x 3 + 7x 2 + 10x + 8 = 0 is reduced to the solution of two equations: 2.1. 2 x 2 + 3 x + 4 = 0 x+2=0 2. 2 x 2the + 3 xtwo + 4equations, =0 Solving we get: x = −2 in the first one and no real solutions in the second one. Thus, the given equation has two real solutions: x = −2 and x = 1 . 3 .6 Proof theRational rational root theorem 3 . 6 Proof of the Root Theorem 3.6 Proof ofofRational the Root Theorem 3 .6 Proof the rational root theorem 3 . 6 Proof of theof Rational Root Theorem the rational root theorem 2013 –2014 2010–2011 2013 –2014 2010–2011 1 x) = an x nrational + an −1 x n −1 + + a1 xroot + a0 , wheretheorem Let P ( the all the coefficients are m m If allis the a solution to the integers, n na −1 reduced fraction. P( x) and = an xlet + kan −be + + a1 x1 + a0 , where Let coefficients are 1x k m k divides an . m divides a0 Ifand equation P(m x) =be0 a, then reduced fraction. is a solution to the integers, and let k k Proof m divides a0 and k divides an . equation P ( x) = 0 , then m Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement Let a reduced fraction be a solution to the equation P ( x) = 0 . ® AcAdemic mAthemAtics ResouRce Guide MATheMATics ResouRcedecAthlon Guide k n n −1 1 Unauthorized Duplication the Terms m is Prohibited mOutside mof Your License Agreement a Guide 0. Then, + an −1®mAthemAtics ++ a1 ResouRce + a0 = Guide AcAdemic MATheMATics ResouRcendecAthlon k k k 27 27 Multiplying both sides of this equation by k n −1 we get n m an + an −1 mn −1 + an − 2 mn − 2 k ++ a1 m1 k n − 2 + a0 k n −1 = 0 . k Since a0 , a1 , a2 ,, an −1 , k and m are all integers, n −1 n−2 1 n−2 n −1 an −1 m + an − 2 m k ++ a1 m k + a0 k is an integer. mn + Integer = 0 . We have: an k Hence, an mn must be an integer as well. k USAD Mathematics Resource Guide • 2016–2017 This implies that k divides an m n . 29 Rancho Cucamonga High School - Ontario, CA = −one 2 and x =1 Thus, thethe given haswe two x =solutions: −2 in the xfirst Solving twoequation equations, get:real and no.real solutions in the second one. We have: an mn + Integer = 0 . k mn Hence, an must be an integer as well. k This implies that k divides an m n . m is a reduced fraction, k and m have no common factor k n (other than 1), therefore k and m have no common factor too. Since In a similar manner, we can prove that m divides a0 . To do this, we multiply both sides n n −1 1 m m m kn . of the equation an + an −1 ++ a1 + a0 = 0 by m k k k kn = 0. We get an mn −1 + an −1 mn − 2 k ++ a1 k n −1 + a0 m As before, an mn −1 + an −1 mn − 2 k ++ a1 k n −1 is an integer. kn kn = 0 , a0 must be an integer as well. Again, since Integer + a0 m m Since m and k have no common factor, m and kn have no common factor either. Hence m divides a0 . This completes the proof. 3 . Proof of the factor theorem 3 . 77Proof of Unauthorized the Factor Theorem Duplication is Prohibited Outside the Terms of Your License Agreement 3.7 Proof of the Factor Theorem 28 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MATheMATics ResouRce Guide –2014 2013 2010–2011 factor theorem The polynomial x − a divides a polynomial P ( x) if and only if P (a ) = 0 . Proof We have two assertions in this theorem: 1 . 2 . If x − a divides P ( x) , then P (a ) = 0 . If P (a ) = 0 , then x − a divides P ( x) . Proof of 1 Since x − a divides P ( x) , there exists a polynomial Q( x) such that P ( x) = Q( x)( x − a ) . Substituting a for x , we get: P(a) = Q(a )(a − a) = Q(a) ⋅ 0 = 0 . Proof of 2 30 By the Division Algorithm for polynomials, there exist two unique polynomials USAD Mathematics Resource Guide • 2016–2017 Q ( x) and R ( x ) such that: P(x) = Q(x)(x − a) + R(x) , where either R ( x ) is the Rancho Cucamonga High School - Ontario, CA Hence, k divides an . 2 . If P (Proof − a divides a ) = 0 , then xof 1 P( x) . Proof of Q1( x) such that P( x) = Q( x)( x − a) . Since x − a divides P ( x) , there exists a polynomial that P divides P ( x) ,athere Since x − a Substituting a polynomial ( x)−such for xexists , we get: P(a) = Q(aQ)(a a) = Q(a) ⋅ 0( x=) 0=.Q( x)( x − a ) . , we get: P(a)of Substituting a for x Proof = Q(a2 )(a − a) = Q(a) ⋅ 0 = 0 . Proof of By the Division Algorithm for polynomials, there2exist two unique polynomials By Division polynomials, there exist two unique such that: for , where either polynomials Q (the x) and R ( x )Algorithm P(x) = Q(x)(x − a) + R(x) R ( x ) is the that:degree , where either is athe Qzero ( x) and R ( x ) such P(x) of = Q(x)(x − a) R(x) the R ( x )x − less+ than degree of the . polynomial or the R ( x ) is zero polynomial or the degree of R ( x ) is less than the degree of the x − a . We will show that if P (a ) = 0 , then R ( x ) is the zero polynomial, and so x − a divides P ( x) . and so x − We will show that if P (xa−) =a 0is, then ) is the x) .r . 1, R ( R be zero eitherpolynomial, the zero polynomial oraa divides number,P (say Since the degree of the x )( xmust Since the degree of the x − a is 1, R ( x ) must be either the zero polynomial or a number, say r . Let’s examine these two cases: Let’s examine these= two cases: Then P(x) 1. Let R ( x ) be the zero polynomial. Q(x)(x − a) , and so x − a divides P ( x) . 2. Let R ( x ) = r . Substituting x = a in P(x) = Q(x)(x − a) + r , we get P(a) = (a − a)Q(a) + r . Since P (a ) = 0 , we get 0 = (a − a)Q(a) + r . This implies that r = 0 that r = 0 Since Pand (a ) so = 0x, we 0 = (aP− r . This implies − a get divides the proof. ( xa)Q(a) ) . This+completes and so x − a divides P ( x) . This completes the proof. Before finishing our discussion of polynomial equations, we should note that polynomials of degree three Before finishing discussion polynomial equations, we in should that polynomials of degree three or higher usuallyour have rational of roots when they are studied high note school. or higher usually have rational roots when they are studied in high school. kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 3 .1–3 .7: sections 3 .1–3 . 7Section : exercises Sections 3.1–3.7 Exercises Section 3 . 1–3 .7: 1k–3 . k sections k k k k k k k k k 3 . kk k k7 k:kexercises kkkkkkkkkkkkkkkkkkkkkk kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk Exercises Exercises kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 1. Solve the following equations 1. Solve the following equations Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2 a. x 3+ xMATheMATics − 4 x − 4AcAdemic = ResouRce 0 ® 2013 –2014 decAthlon 2010–2011 Guide mAthemAtics ResouRce Guide 29 ® 2013 –2014 mAthemAtics ResouRce Guide 2010–2011 Guide b. 24x 3 +MATheMATics 16 x 2 − 3AcAdemic x −ResouRce 2 = 0 decAthlon 29 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement c. x 3 + 1991x + 1992 = 0 2. Find the quotient and the remainder of the division of P ( x) by K ( x) for P(x) = 30x 4 − 31x 3 −180x 2 + 7x + 6 , K ( x) = x + 1 . 3. The remainder of the division of 2x 5 − 3x 3 +11x 2 − x + a by x + 2 is 3. Find the value of a . 4. Solve the equation ax 3 − 2 x 2 − 5 x + 6 = 0 , knowing 2 is one of its solutions. 5. Find all rational solutions to the following equations: USAD Mathematics Resource Guide • 2016–2017 2 a. 6x 4 + 7x 3 − 22x − 28x − 8 = 0 31 Rancho Cucamonga High School - Ontario, CA divides ber the zero polynomial. andget soP(a) x−a 1. = Q(x)(x P( x+ ) .r . . Substituting x = a inThen 2. Let R ( x ) = P(x)P(x) = Q(x)(x − a)−+ a) r ,,we = (a − a)Q(a) 3 2 4. , knowing 2 is one of its solutions. 4. Solve Solve the the equation equation ax ax 3 −− 22xx 2 −− 55xx ++ 66 == 00 , knowing 2 is one of its solutions. 5. 5. Find Find all all rational rational solutions solutions to to the the following following equations: equations: 4 3 2 a. a. 6x 6x 4 ++ 7x 7x 3 −− 22x 22x 2 −− 28x 28x −− 88 == 00 3 2 b. b. xx 3 ++ 3x 3x 2 −−18x 18x −− 40 40 == 00 6 4 2 c. c. xx 6 −− xx 4 −− xx 2 ++11 == 00 6 5 4 x = −3 and xx == −−44 solutions 6. ? solutions to to the the equation equation xx 6 ++ bx 6. For For what what values values of of bb and and cc are are x = −3 and bx 5 + + cx cx 4 = = 00 ? 7. 7. Solve Solve the the following following equations: equations: 2 2 a. a. (x (x 2 ++ xx +1)(x +1)(x 2 ++ xx ++ 2) 2) == 12 12 8. does not divide ( x) = x − 2 (Hint: Assume that there exists a polynomial 8. Prove Prove that that K K ((xx)) == xx does not divide P P ( x) = x − 2 (Hint: Assume that there exists a polynomial such that , and conclude that ( x) must be a number. Proceed from Q Q((xx)) such that P(x) P(x) = = Q(x)K Q(x)K(x) (x) , and conclude that Q Q ( x) must be a number. Proceed from there does not exist.) there to to conclude conclude that that such such Q Q((xx)) does not exist.) kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 3 .8 complex numbers 3 . Numbers 3.8 Complex Numbers 3 .8 complex numbers 3 .88 Complex Complex Numbers Earlier Earlier we we concluded concluded that that there there are are no no real real solutions solutions to to aa quadratic quadratic equation equation when when its its discriminant discriminant is is less less than than zero. zero. What What is is meant meant by by this this is is that that there there are are no no real real solutions, solutions, numbers numbers of of the the set set of of real real numbers numbers .. For For example, example, if if the the discriminant discriminant is is −−11,, there there are are no no real real solutions solutions because because there there is is no no real real number number xx for for 2 x = − 1 2 which which x = −1.. That That is, is, the the expression expression −−11 does does not not represent represent aa real real number. number. Mathematicians Mathematicians encountered encountered certain certain problems problems in in science science and and mathematics mathematics that that necessitated necessitated extending extending the the real real numbers numbers into into aa new new set set of of numbers numbers called called complex complex numbers. numbers. This This was was done done by by defining defining the the expression Duplication Prohibited Outside of Your License Agreement a number that solvesis the equation denoted this number by the letter i. −1 asUnauthorized x 2 the = −Terms 1 . They Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 30 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MATheMATics ResouRce Guide 30 new number, in combination mAthemAtics ResouRce Guide AcAdemicwith decAthlon MATheMATics ResouRce Guide This the real® numbers, generates the complex numbers. –2014 2013 2010–2011 –2014 2013 2010–2011 Let’s consider the equation x 2 − 2 x + 5 = 0 . Solving, we get: x = 2 ± 4 − 4 ⋅ 5 2 ± −16 2 ± 4 −1 2 ± 4i = = = = 1± 2i 2 2 2 2 The expressions 1 + 2i and 1 − 2i are complex numbers; they are solutions to the equation x 2 − 2 x + 5 = 0. A complex number is an expression of the form a + bi , where a and b are real numbers. The set of all complex numbers is denoted by . Clearly, any real number r is a complex number since it can be written as r + 0i . Hence, is a subset of . With this relation, we can extend Figure 1–1 from Section 1 of the 32 USAD Mathematics Resource Guide • 2016–2017 guide, as shown in Figure 3–1: Rancho Cucamonga High School - Ontario, CA b. (x +1)(x +1)(x ++ 3)(x 3)(x ++ 5)(x 5)(x ++ 7) 7) +15 +15 == 00 b. (x A complex number is an expression of the form a + bi , where a and b are real numbers. The set of all complex numbers is denoted by . Clearly, any real number r is a complex number since it can be written as r + 0i . Hence, is a subset of . With this relation, we can extend Figure 1–1 from Section 1 of the guide, as shown in Figure 3–1: Figure 3–1 Every complex number z = a + ib corresponds to the point (a, b) in the coordinate plane, and, conversely, every point (a, b) in the coordinate plane corresponds to the complex number z = a + ib . (See Figure 3–2 below.) y z b a 0 x Figure 3–2 3 .8 . hoW comPlex numBers? 3 .8 .1 hoW Do We ADD comPLex numBerS? 3.8.11How Dodo WeWe Addadd Complex Numbers? Addition and subtraction of complex numbers are defined as follows: Let z1 = a1 + b1i and z2 = a2 + b2i be two complex numbers; then z1 ± z2 = (a1 ± a2 ) + (b1 ± b2 )i . Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 AcAdemic MATheMATics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide 31 31 ExamplE 3.8a: 1. (5 + 3i) + (4 − 5i) = (5 + 4) + (3 − 5)i = 9 − 2i 2. (3+ 15i) − (4 + 5i) = (3 − 4) + (15 − 5)i = −1+ 10i 1 3 2 1 1 2 3 1 1 3. + i − − 2i + i = − + − −2 + i = − + 3i 4 6 2 4 3 4 2 3 4 ( ) USAD Mathematics Resource Guide • 2016–2017 33 Rancho Cucamonga High School - Ontario, CA Properties 1–8 from Section 1 hold for the complex numbers as well. 1. (5 + 3i) + (4 − 5i) = (5 + 4) + (3 − 5)i = 9 − 2i 2. (3+ 15i) − (4 + 5i) = (3 − 4) + (15 − 5)i = −1+ 10i 1 3 2 1 1 2 3 1 1 3. + i − − 2i + i = − + − −2 + i = − + 3i 4 6 2 4 3 4 2 3 4 ( ) 3 .8 .2 hoW multiPly comPlex numBers? 3 .8 .2 hoW Do We muLTIPLy comPLex numBerS? 3.8.2 How Dodo WeWe Multiply Complex Numbers? Multiplication of complex numbers is defined as follows: Let z1 = a1 + b1i and z2 = a2 + b2i be two complex numbers. Then z1 ⋅ z2 = a1 ⋅ (a2 + b2 i) + b1i ⋅ (a2 + b2 i) = a1 a2 + a1b2 i + a2 b1i + b1b2 i 2 = ( ) Notice that i = ( −1) = 2 −1 + (a1b2 + a2 b1 )i = a1 a2 − b1b2 + (a1b2 + a2 b1 )i . 2 2 −1 ⋅ −1 = −1 . Rancho Cucamonga High School - Ontario, CA = a1 a2 + b1b2 ExamplE 3.8b: 1. (4 +15i) ⋅ (17 − 2i) = 4 ⋅17 + 4 ⋅ (−2)i +15i ⋅17 +15 ⋅ (−2)i 2 = 68 − 8i + 255i − 30 ⋅ (−1) = = 68 + 30 + (255 − 8)i = 98 + 247i 1 1 1 1 1 1 1 4 1 2. 1+ i ⋅ 4 − i = 1⋅ 4 +1⋅ − i + i ⋅ 4 + − i 2 = 4 − i + i − ⋅ (−1) = 3 5 5 3 15 3 5 5 3 1 4 1 2 1 = 4 + + − i = 4 +1 i 15 3 5 15 15 3. (3 − 5i)2 = 32 − 2 ⋅ 3⋅ 5i + (5i)2 = 9 − 30i + 25 ⋅ (−1) = 9 − 25 − 30i = −16 − 30i 4. (12 + 3i) ⋅ (12 − 3i) = 122 − (3i)2 = 144 + 9 = 153 In the last example, we multiplied two complex numbers that differ by the sign in front of 3i . One is referred to as the conjugate of the other. In general, we say that a − bi is the conjugate of a + bi , and vice versa: a + bi is the conjugate of a − bi . The product of a complex number and isitsProhibited conjugate is a the realTerms number: Unauthorized Duplication Outside of Your License Agreement 32 (a + bi) ⋅ (a − bi) = 2 2 2 AcAdemic a 2 − (bi) = a 2 − bdecAthlon i = a 2 + b2 . mAthemAtics ® ResouRce Guide MATheMATics ResouRce Guide –2014 2013 2010–2011 The conjugate of a complex number z is denoted by z . Therefore, we have: z ⋅ z = a 2 + b 2 . z zz a 2 + b2 = = =1 And, z ⋅ 2 a + b2 a 2 + b2 a 2 + b2 34 the last equality, we see that USADzMathematics Resource • 2016–2017 From is the reciprocal of zGuide because their product is 1. As with real 2 2 a +b −1 The conjugate of a complex number ] is denoted by ] . Therefore, we have: ] ] " D E . zz a 2 + b2 z = = =1 And, z ⋅ 2 a + b2 a 2 + b2 a 2 + b2 ] is the reciprocal of ] because their product is 1. As with real D E numbers, the reciprocal of ] is denoted by ] . From the last equality, we see that D EL D E ] If] " D EL , then ] " " " L. D E D E D E D E As with real numbers, the absolute value of a complex number ] , also denoted by ] , is defined as the distance of ] from the origin. Therefore: ] " D E . Let ] " D EL and ] " D EL be two complex numbers, and ] | . And, let ] be a complex number ] such that: ] " . ] We rewrite the last expression as ] " ] ] . Since ] " ] D E " D D E E D E L , we have: ] D E L D E L D E L D E L ] ] ] " ] ] " D EL " " " . D E L D E L ] ] D E D E or ] ] ] ]" " ] ] ] Substituting in the last equation ] " D EL and ] " D EL and multiplying, we get: D E L D E L D D EE D E D E ]" " L . D E L D E L D E D E This shows that the result of dividing two complex numbers is a complex number. EXAMPLE 3.8c: 1. L L L L L L L L " " " " " L L L L L 2. " L L L L L L L ® LMATHEMATICS 2013 –2014 ACADEMIC MATHEMATICS RESOURCE GUIDE 2010–2011 RESOURCEDECATHLON GUIDE L " L L Unauthorized Duplication " is Prohibited Outside the Terms of Your " License Agreement USAD Mathematics Resource Guide • 2016–2017 " 33 35 Rancho Cucamonga High School - Ontario, CA 3.8.3 HOW DIVIDE COMPLEX NUMBERS? 3.8.3 HOW DO WE DIVIDE COMPLEX NUMBERS? 3.8.3 How doDO WeWE Divide Complex Numbers 2. 2 −i 3 − 4i = = ( )( ( 3 − 4i) ⋅ ( )= 3 + 4i) 2 ⋅ 3 + 2 ⋅ 4i − i 3 − 4i ( 3) − (4i) 2 6+4 4 2 − 3 + i 7 7 2 = ( ) 3+ 4 = ExamplE 3.8d: Solve the equation 5x 2 − 6x + 5 = 0 . solution: 5 x 2 − 6 x + 5 = 0 ∆ = 36 − 5 ⋅ 5 ⋅ 4 = 36 − 100 = −64 The equation does not have real solutions. 6 ± −64 6 ± 64 ⋅ −1 6 ± 8 ⋅ i 3 4 = = = ± i 10 5 5 10 10 3 4 Thus, x = ± i . 5 5 ExamplE 3.8e: Solve the equation x 3 + 3x 2 + 7x + 10 = 0 . 3 2 solution: x + 3x + 7x + 10 = 0 By the Rational Root Theorem, a rational solution to this equation must be one of the following: m +10 +10 −10 −10 , = , , , +1 −1 +1 −1 k m +2 +2 −2 −2 , = , , , k +1 −1 +1 −1 m +5 +5 −5 −5 , = , , , k +1 −1 +1 −1 m +1 +1 −1 −1 , = , , , k +1 −1 +1 −1 After removing all the repetitions, we get the following candidates for rational solutions: ±1, ± 2, ± 5, ± 10 . Of all these, only x = −2 is a rational solution of the equation. By the Factor Theorem, the polynomial x 3 + 3x 2 + 7x + 10 is divisible by x + 2 . We get: x 3 + 3x 2 + 7x +10 = (x + 2)(x 2 + x + 5) (x + 2)(x 2 + x + 5) = 0 x+2=0 or 34 x2 + x + 5 = 0 . Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement Guide AcAdemic decAthlon ® mAthemAtics ResouRce MATheMATics ResouRce Guide Solving the first equation, we get x = −2 . 36 USAD Mathematics Resource Guide • 2016–2017 −1± −19 −1± −1 ⋅ 19 −1± i 19 . Solving the second equation, we get x = = = –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA x= x2 + x + 5 = 0 . Solving the first equation, we get x = −2 . Solving the second equation, we get x = −1± −19 −1± −1 ⋅ 19 −1± i 19 = = . 2 2 2 Thus, the solutions to the given equation are: x = −2 , x = −1 ± i 19 . 2 ExamplE 3.8f: Solve the equation ( 2 + i)x 2 + 2x − ( 2 − i) = 0 . solution: ( 2 + i)x 2 + 2x − ( 2 − i) = 0 x= x= −2 ± 4 2( 2 + i) 1 = Rancho Cucamonga High School - Ontario, CA ∆ = 4 + 4( 2 − i)( 2 + i) = 4 + 4 ⋅ (2 +1) = 16 −1± 2 2+i 2 +i or x= −3 . 2 +i Each of these solutions can be expressed in the form of a + bi . For this, we utilize the division of complex numbers as we learned earlier: 1 = 1⋅ 2 −i ) = 2 −i 2 1 = − i 2 +1 3 3 ) ( 2 − i) −3⋅ ( 2 − i) −3 2 + 3i −3 = = =− 2 +1 2 + i ( 2 + i) ⋅ ( 2 − i) 2+i ( ( 2+i ⋅ 2+i Thus, the given equation has the solutions: x = 2 1 − i and x = − 2 + i . 3 3 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 AcAdemic MATheMATics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 35 37 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk Section Section 3 .3.8 8: Section 3.8: exercises Exercises Exercises kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 1. Find the conjugate z and the absolute value z of the given complex numbers z : a. 4 − 13i 2 b. (1− i) c. (3+ i)2 2 − 5i a. z1 + z2 b. z1 ⋅ z2 c. z1 + z2 i z2 3. Present the following complex number in the standard form a + bi : (2 + i)3 − (2 − i)3 (2 + i)2 − (2 − i)2 4. Solve the following equation in the set � : x 5 + 2x 4 + 2x 3 + 2x 2 + x = 0 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk section 4: Functions SecTIon 4: functions 4.1 Preliminaries 4 .1 Preliminaries The term function is often used in everyday language to indicate dependencies between quantities. For example, we say “the commuting time from work to home is a function of traffic density”; “the price of gasoline is a function of supply from oil-producing countries”; “population growth is a function of time”; etc. In order to understand the properties of such dependencies, it is necessary to express them mathematically. For example, to study the growth rate of a population, it is necessary to model the size of the population at a given time. The equation P (t ) = 67.38(1.026)t is an example of such a model; it models the population growth in a particular country at any given time. The equation says that the size of the population (in millions) at a given time t is the number P(t) = 67.38(1.026)t . For example, the initial population (when the measurement started) is 67.38 million ( P(0) = 67.38(1.026)0 = 67.38 ), and after 5 38 36 36 USAD Mathematics Resource Guide 2016–2017 Unauthorized Duplication is Prohibited Outside the Terms of Your• License Agreement Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA 2. Given z1 = −i + 3 , z2 = 3i −1 , find: 1 2 b. z1 ⋅ z2 c. z1 + z2 i z2 3. Present the following complex number in the standard form a + bi : (2 + i)3 − (2 − i)3 (2 + i)2 − (2 − i)2 4. Solve the following equation in the set � : x 5 + 2x 4 + 2x 3 + 2x 2 + x = 0 Section kkk4 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk FUNCTIONS section 4: Functions SecTIon 4: functions The term function function is often used in everyday language to indicate dependencies between quantities. For example, we say “the commuting time from work to home is a function of traffic density”; “the price of gasoline is a function of supply from oil-producing countries”; “population growth is a function of time”; etc. In order to understand the properties of such dependencies, it is necessary to express them mathematically. For example, to study the growth rate of a population, it is necessary to model the size of the population at a given time. The equation P (t ) = 67.38(1.026)t is an example of such a model; it models the population growth in a particular country at any given time. The equation says that the size of the population (in millions) at a given time t is the number P(t) = 67.38(1.026)t . For example, the initial population (when the measurement started) is 67.38 million ( P(0) = 67.38(1.026)0 = 67.38 ), and after 5 years, it is 76.607 ( P(5) = 67.38(1.026)5 = 76.607 ). Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 36 36 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide –2014 2013 2010–2011 linear equations—equations equations—equations of of the the form form ax ax++bb = =0 — —and and quadratic equations— Earlier we dealt with linear equations of the form ax 2 + bx + c = 0 . The expressions ax + b and ax 2 + bx + c are functions. We can think of the expression 2 x − 7 in terms of a dependency between quantities: the value of 2 x − 7 depends on the value of x , or we can say 2 x − 7 is a function of x . For x = 3 , 2 x − 7 = −1 ; for x = 0 , 2 x − 7 = −7 ; for x = 2 , 2x − 7 = 2 2 − 7 −4.1716 . Functions of the form y = ax + b are called linear functions, and functions of the form y = ax 2 + bx + c , where a ≠ 0 , are called quadratic functions. When thinking of an expression as a function, we use symbols such as f ( x ) , g ( x ) , k ( x ) , etc. For example, the function f ( x ) = 2 x − 7 indicates that for any value of x , there corresponds a value f ( x ) (read f of x ). For example, x = 3 corresponds to f (3) = −1 ; x = 0 corresponds to f (0) = −7 ; x = 2 corresponds to f ( 2) = 2 2 − 7 , etc. Often, instead of f ( x ) = 2 x − 7 we write y = 2 x − 7 , meaning the value of x corresponds to the value y = 2 x − 7 . 4.2 Definition of a Function USAD Resource Guide • 2016–2017 4 .2 Definition of aMathematics Function 39 Rancho Cucamonga High School - Ontario, CA 4.1 Preliminaries 4 .1 Preliminaries 4.1 Preliminaries corresponds to the value y = 2 x − 7 . 4.2 Definition of a Function 4 .2 Definition of a Function 4.2 Definition of a Function The function f ( x ) = 2 x − 7 admits any value of x . That is to say, for any real number x , there is a corresponding value f ( x ) = 2 x − 7 . Likewise, the function u( x ) = 3x 2 − 2 x + 4 admits any value of x . 1 Not all functions admit all real values. For example, the function g ( x ) = does not admit x = 0 , because x 1 f (0) = is not a number. Likewise, the function h( x ) = x admits only non-negative numbers. 0 A number that can be admitted into a function is called an input, and its corresponding number under the function is called an output. Thus, for example, for the function h ( x ) = x 2, any real number can be an input. Can any real number be an output for this function? The answer is obviously no. For example, –1 cannot be an output, because there is no real number x for which x 2 = −1 . outputs of a function is called the range of the function. Thus, the domain of h ( x ) = x 2 is the set of all real numbers. The range of this function is the set of all non-negative numbers. ExamplE 4.2a: What are the domain and range of the function h( x ) = x ? answEr: Both the domain and range of this function are the set of all non-negative numbers. Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement Sometimes determining the domain and range of a function requires a little more work. 2013 –2014 2010–2011 AcAdemic MatheMatics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide ExamplE 4.2b: What are the domain and range of the function k(x) = 2 ? x−3 answEr: Clearly, any real number different from 3 is in the domain of this function. So, the domain consists of real numbers x ≠ 3 . What about the range of this function? By definition, a real number a is in the range if there exists x such that 2 =a. x−3 Multiplying both sides of this equation by x − 3 , we get 2 = a ( x − 3) . Clearly a = 0 is not in the range of the function because otherwise we get 2 = 0( x − 3) = 0 , which is not possible. 40 On the other hand, any number USAD Resource Guide a ≠Mathematics 0 is in the range. To show this,• 2016–2017 we need to show that for any a ≠ 0 , the equation 2 = a is solvable for x . Indeed: 37 Rancho Cucamonga High School - Ontario, CA The set of all possible inputs of a function is called the domain of the function; and the set of all the With this background, we can now introduce the formal definition of a function. With this background, we can now introduce the formal definition of a function. DeFinition DeFinition A function f is a correspondence between two sets, A and B , that assigns DeFinition A function f is a correspondence between two sets, A and B , that assigns to each element x in A one and only one element f ( x ) in B . DeFinition A function a correspondence that A andf ( B f iselement x in A onebetween B assigns and onlytwo onesets, element . to each x ), in A function a correspondence that A andf ( B f iselement x in A onebetween B assigns and onlytwo onesets, element . to each x ), in in A one notation: and only one each element This state of affairs istoindicated by thex following f : Aelement → B f ( x ) in B . This affairs is indicated by the following : A →ofB the function. You might think of a calledofthe domain of the function, and B isnotation: called thef range A is state Unauthorized Duplication is Prohibitednotation: Outside the Terms of Your License Agreement This statefofasaffairs is indicated the following f : A→ a machine with anbyinput and output. The domain ofBf is all the numbers that can be fed function ® Unauthorized Duplication is Prohibited Outside the Terms of Your AgreementGuide 2013 38 –2014 ResouRce Guide AcAdemic 38 MatheMatics ResouRce 2010–2011 This state of affairs is indicated by decAthlon the followingmAthemAtics notation: f : A → B License ® igure 4–1.) The into the machine, and the range is all the numbers that come out of the machine. (See F 38 2013 –2014 mAthemAtics ResouRce Guide AcAdemic decAthlon 38 MatheMatics ResouRce Guide 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 38 machine 38 38 38 ® mAthemAtics ResouRce Guide AcAdemic decAthlon never produces two outputs foris the same input.the Terms Unauthorized Duplication Prohibited Outside of Your License AgreementGuide MatheMatics ResouRce ® Guide AcAdemic decAthlon mAthemAtics ResouRce MatheMatics ResouRce Guide x –2014 2013 2010–2011 –2014 2013 2010–2011 ff ff(x) (x) Figure 4–1 Thinking of a function as a machine helps us see how functions are composed of more elementary 41 USAD Mathematics Resource Guide • 2016–2017 2 functions. For example, the function f (x) = (x + 2) is composed of two functions: g(x) = x + 2 and Rancho Cucamonga High School - Ontario, CA is not in the range of the function Clearly a = 0both we get 2 otherwise Multiplying sides of this equation by x − 3 , because = a ( x − 3) .we get 2 = 0( x − 3) = 0 , which is not in the range of the function Clearly a = 0both we get 2 otherwise Multiplying sides of this equation by x − 3 , because = a ( x − 3) .we get 2 = 0( x − 3) = 0 , which is not possible. a = 0 is not in the range of the function because otherwise we get 2 = 0( x − 3) = 0 , which Clearly is not possible. a = 0 hand, is notany in the range aof≠the because which Clearly 2 =to0(show x − 3)that = 0 ,for 0 isfunction in the range. To otherwise show this, we we get need any On thepossible. other number is not On the other hand, any2 number a ≠ 0 is in the range. To show this, we need to show that for any isa not ≠ 0 ,possible. the equation = a is solvable for x . Indeed: 2− number x 3 a ≠ 0 is in On the other hand, any x . range. a ≠ 0 , the equation = a is solvable forthe Indeed:To show this, we need to show that for any 2 x − 3 a ≠ 0 is in On the= other hand, any2 number a x . range. ax ≠ = a is solvable forthe Indeed:To show this, we need to show that for any 2− 03 , the equation 2 =a a ≠ 03 , the equation x − 3 = a is solvable for x . Indeed: 2x =2− a(x − 3) x−3 =a 2x =2− a(x 3 − 3) 2 =a x =−=a(x 3x − −3 3) 22 a = x−3 2 − 3) 2a = a(x x2 == x −+ 33 2 x2a = a + 3 = a2x − 3 2 xa =summarize, +3 To the domain of the function k ( x ) = consists of all the real numbers different 2 2 a x =summarize, +3 To the domain of the function k ( x ) = x − 3 consists of all the real numbers different from 0 . froma3 , and its range consists of all the real numbers x 2−different 3 To summarize, domain of of theall function ( x) = consists of all 3 , and its the 0 . the real numbers different range consists the realknumbers different from from x 2− 3 To summarize, domain of of theall function ( x) = consists of all 3 , and its the 0 . the real numbers different range consists the realknumbers from from x −different 3 from 3 , and its range consists of all the real numbers different from 0 . With this background, we can now introduce the formal definition of a function. With this background, we can now introduce the formal definition of a function. Figure 4–1 Thinking of a function as a machine helps us see how functions are composed of more elementary functions. For example, the function f (x) = (x + 2)2 is composed of two functions: g(x) = x + 2 and h ( x ) = x 2 . Any given input x corresponds to the output g ( x ) = x + 2 . This output, in turn, is an input for the function h . We get: h(g(x)) = h(x + 2) = (x + 2)2 = f (x) . Thus, f is a composition of h and g : f (x) = h(g(x)) . Figure 4–2 below illustrates this composition. xx g(x) h ff(x)=h(g(x)) (x)=h (g(x)) Figure 4–2 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 MatheMatics AcAdemic decAthlon ResouRce 2010–2011 ResouRce Guide ExamplE 4.2c: Consider the functions® mAthemAtics . These h(x) = xGuide g ( x) = 2 x − 3 and functions can be composed in two ways: h( g ( x)) and g (h( x)) . 1. For a given input x, we get the output g ( x) = 2 x − 3 , which, in turn, is an input for the function h . Thus, we have h(g(x)) = g(x) = 2x − 3 . 2. For a given input x, we get the output h( x) = x , which, in turn, is an input for the function ( ) g . We have g h(x) = 2h(x) − 3 = 2 x − 3 . ExamplE 4.2d: The function f (x) = ( ) 2 x 2 − 9 + 3 can be looked at as a composition of three functions: g(x) = x − 9 , h(x) = x and k(x) = (x + 3)2 . 2 42 1. Every non-negative output of the function g is an input of the function h : Mathematics Resource Guide • 2016–2017 h(g(x)) = g(x) = x 2 − USAD 9; 39 Rancho Cucamonga High School - Ontario, CA g ( x − 9 + 3) can be looked at as a composition of three = x The − 9 ,function and k(x) = (x + 3) .can be looked at as a composition of three functions: g(x) h(x) = xf (x) ExamplE 4.2d: f (x) ==(( xx −−99++33)) can be looked at as a composition of three ExamplE 4.2d: The function functions: g(x) = x − 9 , h(x) = x and k(x) = (x + 3) . ExamplE 4.2d: The function f (x) = 2 2 2 2 2 2 2 22 2 functions: g(x) = x 2 − 9 , h(x) = x and k(x) = (x + 3)2 . 1. Every non-negative output of the function g is an input of the function h : 2.2. 3. 3.3. h(g(x)) = g(x) = output x 2 − 9 ; of the function g is an input of the function h : Every non-negative Every non-negative output of the function g is an input of the function h : 2 h is an input of the function k : Every output of the function h(g(x)) g(x) h(g(x)) == g(x) == xx2 −−99;; 2 2 2 k h =output h + 3 of= thexfunction − 9 + 3 h; is an input of the function k : Every is an input of the function k : Every output of the function h 2 2 2 2 2 Finally, every output of the function k determines the function f : kk hh == hh++33 == xx2 −−99++33 ;; 2 2 f (x) = kevery h g(x) = ofxthe − 9function + 3 . k determines the function f : Finally, output k determines the function f : Finally, every output of the function 2 2 2 (x)==kk hh g(x) g(x) == xx2 −−99++33 .. ff(x) () ( ) ( (( )) (( )) (( ( ( )) (( (( )))) ( (( ) )) ) )) 4.3 Functions 4.3 M Many-to-one any-to-One Functions versus Functions One-to-One 4 .3 Many-to-One Functions versus One-to-One versus one-to-one Functions 4.3 Many-to-one Many-to-one Functions Functions 4.3 Functions 4 .3 Many-to-One Functions versus One-to-One One-to-One Functions Functions 4 .3 Many-to-One FunctionsFunctions versus versus one-to-one Functions versus one-to-one Is it possible that different inputs of a function correspond to one output of the same function? The = −same 2same = 2 andof answer is yes. that For different example, inputs for theoffunction correspond toThe the h ( x )correspond = x 2 , the inputs possible function correspond to one onexoutput output ofxthe the function? IsIs itit possible that different inputs of aa function to function? The 2 same output . Such functions arefunction often called many-to-one functions, meaning more than one 4For answer yes. For example, for the the the inputs inputs and correspond to the the xx==−−22that xx==22 and answer isis yes. example, for function hh((xx))==xx2 ,, the correspond to input corresponds to one output. are often called many-to-one functions, meaning that more than one same output 44.. Such Such functions same output functions are often called many-to-one functions, meaning that more than one inputcorresponds corresponds toone one output. input We have already to learnedoutput. that any quadratic equation ax 2 + bx + c = 0 for which the discriminant −b − b2 − 4ac −b + b2 − 4ac 2 2 has two distinct solutions: and .discriminant This shows ∆ = b − 4 ac > 0 x = x = 2 for which the the discriminant We have already learned that any quadratic equation ax + bx + c = 0 We have already learned that any quadratic 1 equation2aax + bx + c = 02 for which 2a2 2 −b−− bb2 −−4ac −b++ bb2 −−4ac 4ac 4ac 2 −b −b that quadratic functions (for which the discriminant is positive) are many-to-one functions. 2 − 4ac that has two distinct solutions: and This shows ∆ = b > 0 x = x = Is it possible functions are one-to-many: that is, one input of a function corresponds to more than one 1 2 and .. This shows ∆ = b − 4ac > 0 has two distinct solutions: x1 = x = 2 2a 2a 2a 2a Is it possible functions arewhich one-to-many: thatThis is, one input ofisaare function corresponds to more than one output of thethat same function? The answer no. possibility excluded in order to avoid confusion. If that quadratic functions (for thediscriminant discriminant positive) many-to-one functions. Unauthorized Duplication is is Prohibited Outside the Termsare of Your License Agreement that quadratic functions (for which the isispositive) many-to-one functions. ® 40 allowed, mAthemAtics ResouRce Guide AcAdemic decAthlon Guide f 2013 output of the function? is ano. This possibility isMatheMatics excluded inResouRce order tosuch avoidasconfusion. we forsame example, function, upon evaluating an expression f ( x )The = ± answer x to be ( x2010–2011 ) +–2014 x +If1 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement ® we example, to outcome be a function, expression such f ( xwhether ) = ± decAthlon xthe ( x2010–2011 ) +–2014 x +1 40 2013 mAthemAtics Guide AcAdemic MatheMatics Guide asf2013 = 4 , we for 4 + 1evaluating = ResouRce 7ResouRce +ResouRce 1ResouRce = 3 . Guide at xallowed, wouldn’t know is 2 +upon or −2 +an4Guide ® 40 –2014 mAthemAtics AcAdemic decAthlon MatheMatics 2010–2011 at x = 4 , we wouldn’t know whether the outcome is 2 + 4 + 1 = 7 or −2 + 4 + 1 = 3 . Finally, there are functions that are one-to-one, meaning that two distinct inputs must correspond to two Finally, there are functions that are one-to-one, meaning that two distinct inputs must correspond to two distinct outputs. distinct outputs. ExamplE 4.3a: Any linear function y = ax + b , for which a ≠ 0 , is one-to-one. ExamplE 4.3a: Any linear function y = ax + b , for which a ≠ 0 , is one-to-one. proof: Take any two distinct inputs x1 and x2 ; their corresponding outputs y1 = ax1 + b and proof: any distinct inputsthis their corresponding must be two different. Assume is not namely that y1 = y2outputs . Then: y1 = ax1 + b and x1 and x2 ;so, y2 = ax2 + bTake y2 1=+ax ax b= axb2 must + b be different. Assume this is not so, namely that y1 = y2 . Then: 2 + ax1 + = bax=2 ax2 + b USAD Mathematics Resource Guide • 2016–2017 = ax ax22 = 0 ax1 − 43 Rancho Cucamonga High School - Ontario, CA 1.1. 2. proof: Take any two distinct inputs x1 and x2 ; their corresponding outputs y1 = ax1 + b and proof: any distinct inputsthis their corresponding x1 and x2 ;so, must be two different. Assume is not namely that y1 = y2outputs . Then: y1 = ax1 + b and y2 = ax2 + bTake y2 =+ax ax b= axb2 must + b be different. Assume this is not so, namely that y1 = y2 . Then: 2 + 1 ax1 + b = ax2 + b ax 1 = ax2 ax1 − = ax ax2 = 0 ax 1 2 ax a (1x1−−axx22 )==00 a ( x1 −ax≠2 )0=, x0 = x . Since 1 2 Since a ≠ 0 , x1 = x2 . This, of course, is not possible since we started with two distinct inputs x1 and x2 . Hence, y1 ≠ y2 . This, of course, is not possible since we started with two distinct inputs x1 and x2 . Hence, y1 ≠ y2 . ExamplE 4.3b: y = x is one-to-one proof: As before, take any two distinct inputs x1 and x2 . We are to show that their correspondproof: inputs Assume . Weis are thatthat their x1 and x2this and any are different. notto so,show namely ing outputsAsybefore, = x take y =twoxdistinct y correspond= y . Then: 1 1 ingx outputs x2 y1 = 1 = ( xx =) =x( x ) (x =x x) = ( x ) 2 1 1 1 1 2 2 2 2 2 2 x1 and y2 = 1 2 x2 are different. Assume this is not so, namely that y1 = y2 . Then: 2 2 2 x1 = x2 This of course is not possible since we started with two distinct inputs x1 and x2 . Hence, it is not This that of course true y1 = y2is. not possible since we started with two distinct inputs x1 and x2 . Hence, it is not true that y1 = y2 . 4.4 inverse Functions 4.4 Inverse Functions 4 .4 Inverse Functions Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 AcAdemic MatheMatics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide 41 41 2013 –2014 2010–2011 AcAdemic MatheMatics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide 41 41 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement Let f be a function. If there exists a function g such that y = f ( x ) if and only if x = g ( y ) , then we say that f is invertible and g is the inverse of f . The function g is unique (see the next page for a proof) and is denoted by f −1 . ExamplE 4.4a: The inverse of the function f ( x ) = x 3 is g ( x ) = proof: If y = x 3 then 3 y = x , and conversely, if 3 3 x. y = x , then y = x 3 . Thus, g(x) = f −1 (x) . 4 4 y=x 3 3 3 1 44 –6 –4 –2 y=3: x x 2 2 1 USAD Mathematics Resource Guide • 2016–2017 2 4 –66 –4 –2 2 4 6 Rancho Cucamonga High School - Ontario, CA ExamplE 4.3b: y = x is one-to-one proof: If y = x 3 then 3 y = x , and conversely, if proof: If y = x 3 then 3 y = x , and conversely, if 3 3 y = x , then y = x 3 . Thus, g(x) = f −1 (x) . y = x , then y = x 3 . Thus, g(x) = f −1 (x) . 4 4 3 2 y=x3 y=x3 4 3 2 1 2 1 1 1 –6 –4 –2 –6 –4 –2 –1 2 4 –66 –4 –2 2 4 –66 –4 –2 –1 –2 –2 –3 –2 –3 –3 –4 –3 –4 –4 –4 x −1 ExamplE 4.4b: The inverse of f ( x ) = 2 x + 1 is g ( x ) = . x 2− 1 ExamplE 4.4b: The inverse of f ( x ) = 2 x + 1 is g ( x ) = . 2 y −1 proof: y = 2 x + 1 if and only if x = . Thus g ( x ) = f −1 ( x ) y 2− 1 proof: y = 2 x + 1 if and only if x = . Thus g ( x ) = f −1 ( x ) 4 2 3 2 –6 –4 –2 –1 –2 4 6 2 4 6 4 3 2 2 1 1 –4 2 4 3 yy=2x x+1 yy=2x x+1 2 1 –6 –1 –1 –2 4 3 y=3: x x 3 y= : x x 3 2 Rancho Cucamonga High School - Ontario, CA 4 3 1 2 4 –6 6 2 4 –6 6 –4 –4 –2 –2 –1 –1 –2 –1 –2 –2 –3 –2 –3 –3 –4 –3 –4 –4 –4 y= x – 1 2 y= x – 1 2 2 4 6 2 4 6 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 42 42 ® x + 5 the Terms 2Outside Unauthorized Duplication is Prohibited of Your License AgreementGuide ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce ExamplE 4.4c: Find the inverse of . f ( x) =mAthemAtics ® −3 x + 7 ResouRce Guide AcAdemic decAthlon mAthemAtics MatheMatics ResouRce Guide –2014 2013 2010–2011 –2014 2013 2010–2011 7y −5 2x + 5 7x − 5 if and only if x = (check). Hence the function g ( x) = is 3 y 2 + 3 x + 2 −3 x + 7 2x + 5 . the inverse of the function f ( x) = −3 x + 7 7 The domain of f is the set all real numbers different from , and the domain of g is the set of all 3 2 real numbers different from − . 3 solution: y = USAD Mathematics Resource Guide • 2016–2017 theoreM 45 real numbers different from − 3 .3 theoreM theoreM An invertible function f can have only one inverse function. An invertible function f can have only one inverse function. ProoF ProoF Let f1 and f 2 be two different inverse functions of f . Let f1 and f 2 be two different inverse functions of f . Then y = f ( x ) if and only if x = f1 ( y ) Then y = f ( x ) if and only if x = f1 ( y ) and and y = f ( x ) if and only if x = f ( y ) y = f ( x ) if and only if x = f 2 ( y2 ) Rancho Cucamonga High School - Ontario, CA This implies that f ( y ) if and only if x = f ( y ) . We see then that if This implies that x =x = f1 ( y1 ) if and only if x = f 2 ( y2 ) . We see then that if f1 and f 2 are both inverse functions of f , then f1 = f 2 . f1 and f 2 are both inverse functions of f , then f1 = f 2 . theoreM theoreM The graph of −1 is the reflection of f in the line y = x . The graph of f −1f is the reflection of f in the line y = x . ProoF ProoF Implied from the definition of an inverse function is that ( x0 , y0 ) is on the Implied from the definition of an inverse function is that A( A x0 , y0 ) is on the graph of f if and only if A '( y0 , x0 ) is on the graph of−1f −1. Notice that A and graph of f if and only if A '( y0 , x0 ) is on the graph of f . Notice that A and A ' are symmetric with respect to the line y = x . This is so, because the midpoint A ' are symmetric with respect to the line y = x . This is so, because the midpoint x +y y +x between A and A ' is M x +0 y 0y, +0 x 0 , and M is on the line y = x . between A and A ' is M 0 20 , 0 20 , and M is on the line y = x . 2 2 ExamplE 4.4d: Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement ® 2013 –2014 MatheMatics AcAdemic mAthemAtics ResouRce Guide 2010–2011 ResouRcedecAthlon Guide ® 2013 –2014 MatheMatics AcAdemic mAthemAtics ResouRce Guide 2010–2011 ResouRcedecAthlon Guide 43 3 The property we have just proven can clearly be seen in the graphs of f ( x ) = x and f −1 ( x ) = 3 x which we discussed earlier. This property is useful because if we know the graph of an invertible function f , then we can graph its inverse f −1 . The two graphs are symmetric with respect to y= x. y=x3 4 y=x y=x 3 y=3: x 2 1 46 USAD Mathematics Resource Guide • 2016–2017 –6 –4 –2 2 4 6 43 which we discussed earlier. This property is useful because if we know the graph of an invertible function f , then we can graph its inverse f −1 . The two graphs are symmetric with respect to y= x. y=x3 4 y=x y=x 3 y=3: x 2 1 –6 –4 –2 2 4 6 –1 –2 –3 Not all functions are invertible of course. For example, y = x 2 is not invertible. To see why, we graph y = x 2 , and then graph its symmetric image with respect to the line y = x . (See Figure 4–3 below.) We see that the resulting graph is not a function because for one input, x , there are two outputs, y1 and y2 . 5 y=x y=x 4 3 • y2 • 2 1 –6 –4 •& –2 2 4 • xx 6 8 –1 -2 y1 • • –3 –4 –5 Figure 4–3 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 44 44 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 –2014 2013 2010–2011 47 Rancho Cucamonga High School - Ontario, CA –4 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk Section Section 4:4 Section 4: exercises Exercises Exercises kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 1. Find the domain and the range of the following functions: 3 2− x b. f ( x) = x − 5 a. f ( x) = 2 c. f ( x) = 3 + ( x − 2) a. f ( x) = x − 3 5 b. f ( x) = x − 3 2 3. a. h( x) = x , g ( x) = 2 − x . Find h( g ( x)) and g (h( x)) . b. h( x) = 6 x − 1 , g ( x) = x + 1 , k ( x) = 1 . Find h( g ( x)) , h( g (k ( x))) and k ( g (h( x))) . x 2 − 5.7 x . 4. Find the range and domain of f ( x) = 7 1 23 + x 6 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk SecTIon 5: Graphing section 5: Graphing The graph of a function f is the set of all ordered pairs ( x, f ( x )) , where x belongs to the domain of f . In this section, we will learn what the graphs of different functions look like. 5.1 What theofGraph a 5 .1 What DoesDoes the Graph a Linear of Function yaxb Look Like? Linear Function yaxb Look Like? In Section 8 we will prove that the graph of a linear function y = ax + b is a line. The coefficients a and b determine the form of the graph. For x = 0 , y = b . Therefore, (o, b) is the point where the line intersects the y-axis. This number b is called the y-intercept. The graph is shown in Figure 5–1. 48 2013 –2014 2010–2011 USAD Mathematics Resource Guide 2016–2017 Unauthorized Duplication is Prohibited Outside the Terms of Your• License Agreement AcAdemic MatheMatics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide 45 Rancho Cucamonga High School - Ontario, CA 2. Determine which of the functions have an inverse. If the function is invertible, find its inverse. 5 b. f ( x) = x − 3 2 3. a. h( x) = x , g ( x) = 2 − x . Find h( g ( x)) and g (h( x)) . b. h( x) = 6 x − 1 , g ( x) = x + 1 , k ( x) = 1 . Find h( g ( x)) , h( g (k ( x))) and k ( g (h( x))) . x 2 − 5.7 x . 4. Find the range and domain of f ( x) = 7 1 23 + x Section 6 k k k5 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk Graphing SecTIon 5: Graphing section 5: Graphing The graph of a function f is the set of all ordered pairs ( x, f ( x )) , where x belongs to the domain of f . 5.1 W hat Does the Graph of aaLinear Function 5.1 What theof Graph 5 .1 What DoesDoes the Graph a Linear of Function yaxb Look Like? Linear Function yaxb Look Like? y = ax + b Look Like? In Section 8 we will prove that the graph of a linear function y = ax + b is a line. The coefficients a and b determine the form of the graph. For x = 0 , y = b . Therefore, (0, (o, bb)) is the point where the line intersects the y-axis. This number b is called the y-intercept. The graph is shown in Figure 5–1. Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 AcAdemic MatheMatics ResouRcedecAthlon Guide yy ® mAthemAtics ResouRce Guide 45 (0, b) b) y=ax+b y = ax + b x ○ Figure 5–1 When a = 0 , we have y = 0 ⋅ x + b = b . We see that the output value for any x is b . Hence, the graph is simply a line parallel to the x-axis. Its y-intercept is b . (See Figure 5–2.) yy USAD Mathematics Resource Guide • 2016–2017 49 Rancho Cucamonga High School - Ontario, CA In this section, we will learn what the graphs of different functions look like. When a = 0 , we have y = 0 ⋅ x + b = b . We see that the output value for any x is b . Hence, the graph is simply a line parallel to the x-axis. Its y-intercept is b . (See Figure 5–2.) yy yy=b =b (0, b) x ○ Figure 5–2 Figure 5–3 below.) b (why?). This point is called the x-intercept. (See a yy y =y=ax+b ax + b – –b , 0 a xx ○ Figure 5–3 Unauthorized Duplication is Prohibited Outside the Terms Your LicenseFor Agreement When a > 0 , the function This means theoffollowing: any numbers x1 and x2 , y = ax + b is increasing. 46 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide if x1 ≥ x2 , then ax1 + b ≥ ax2 + b (See Figure 5–4.). y y =y=ax+b ax + b axax ++b b 2 2 ax1 1++b b ax ○ xx1 1 xx22 x Figure 5–4 50 USAD Mathematics Resource Guide • 2016–2017 ProoF –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA When a ≠ 0 , the graph intersects the x-axis at x = − Figure 5–4 ProoF Since a > 0 and x1 ≥ x2 , a ( x1 − x2 ) ≥ 0 . Therefore, ax1 − ax2 ≥ 0 . Therefore, ax1 ≥ ax2 . Adding b to both sides of this inequality, we get: ax1 + b ≥ ax2 + b . When a < 0 , the function y = ax + b is decreasing. This means the following: For any numbers x1 and x2 , if x1 ≥ x2 , then ax1 + b ≤ ax2 + b . ProoF Since a < 0 and x1 ≥ x2 , a ( x1 − x2 ) ≤ 0 . Therefore, ax1 − ax2 ≤ 0 . Therefore, ax1 ≤ ax2 . y y=ax+b y = ax + b ax1 + +b ax b 1 ax b ax2 2++b ○ xx11 xx22 x Figure 5–5 5.2 W hat Does the Graph of a Quadratic 5.2 What the of of a 5 .2 What DoesDoes the Graph a Quadratic 2Graph 2 y = ax + bx 2+ c Look Like? Function Function yaxFunction bxc Look Like? Quadratic yax bxc Look Like? Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 AcAdemic MatheMatics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide 47 47 22 2 5.2.1 Case y yx = xyx 5 .2 .1 The The cASe 5.2.1 the caSe We can observe several properties of the graph of the function y = x 2 . ProPerty 1: The graph has a minimum at x = 0 . It easy to see that (0,0) is the lowest point in the graph of this function. It is so because for any real number x , x 2 ≥ 0 . ProPerty 2: The graph is symmetric. The graph of y = x 2 is symmetric with respect to the y-axis. This means that any line parallel to the 51 USAD Mathematics Resource Guide • 2016–2017 x-axis that intersects the graph will intersect it in two points that are equidistant from the y-axis. Rancho Cucamonga High School - Ontario, CA Adding b to both sides of the inequality, we get: ax1 + b ≤ ax2 + b . ProPerty 2: The graph is symmetric. The graph of y = x 2 is symmetric with respect to the y-axis. This means that any line parallel to the x-axis that intersects the graph will intersect it in two points that are equidistant from the y-axis. Let y = d be such a line, and let P and Q be the intersection points of the line with the graph. Also, let Y be the intersection of the line with the y-axis. We are to show that PY = QY . Clearly the coordinates of Y are (0, d ) . To find the coordinates of P and Q , we solve the system y = x2 y = d Solving the system, we get x2 = d x=± d Thus, P = ( d , d ) and Q = (− d , d ) . From this, we get that PY = QY = d . Hence, the graph y = x 2 is symmetric. ProPerty 3: The function y = x 2 is increasing on the positive side of the x-axis and decreasing on the negative side of the x-axis. This means: A. If x1 > x2 ≥ 0 , then x12 > x2 2 (the function is increasing) And B. If 0 ≥ x1 > x2 , then x12 < x2 2 (the function is decreasing) Proof of A: x12 > x2 2 if and only if x12 − x2 2 > 0 if and only if ( x1 − x2 )( x1 + x2 ) > 0 . Since x1 and x2 areUnauthorized non-negative and different, Duplication is Prohibited Outside the Terms of Your License Agreement 48 (1) x1 + x2 > 0 . Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide –2014 2013 2010–2011 Since x1 > x2 ≥ 0 , (2) x1 − x2 > 0 From (1) and (2), ( x1 − x2 )( x1 + x2 ) > 0 . Therefore, x12 − x2 2 > 0 . Hence, x12 > x2 2 . Proof of B: The proof for B is similar to the Proof of A, and you can complete the proof for B on your own as an exercise. The three properties just discussed determine the graph of the function y = x 2 . The graph is shown in Figure 5–6: yy yy=x = x22 52 USAD Mathematics Resource Guide • 2016–2017 y=d y=d Rancho Cucamonga High School - Ontario, CA y=d The three properties just discussed determine the graph of the function y = x 2 . The graph is shown in Figure 5–6: yy yy=x = x22 yy=d =d ( d, 0) ○ (0, 0) x (: d, 0) Figure 5–6 Rancho Cucamonga High School - Ontario, CA 2 5 .2 .2 The The GenerAL cASe bxc 5.2.2 the GeneraL yax 5.2.2 General Case ycaSe =yax ax2 + bx 2+bxc c 2 b b 2 − 4ac In Section 2, we showed that y = ax 2 + bx + c = a x + − 2a 4a 2 2 b − 4ac b Let k = − . Then, y = a x + +k . 4a 2a 2 b As before, we can observe several properties of the graph of y = a x + +k . 2a ProPerty 1: When a > 0 , the graph has a minimum at x = − b . 2a b Proof: We are to show that for any x , y − ≤ y ( x ) . 2a 2 b b b (1) y − = a − + + k = a⋅0 + k = k 2a 2a 2a 2 b (2) y ( x) = a x + +k 2a 2 2 b b Since a > 0 and x + ≥ 0 , a x + ≥ 0. 2a 2a Therefore: 2Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement b ® 2013 –2014 mAthemAtics ResouRce Guide (3) k a x ≤ + 2010–2011 MatheMatics ResouRcedecAthlon Guide + k AcAdemic 2a b From (1), (2), and (3), we get y − ≤ y ( x ) . 2a ProPerty 2: When a < 0 , the graph has a maximum at x = − 49 b . 2a Proof: The proof is very similar to the proof of Property 1, and you can complete it on your own as an exercise. ProPerty 3: The graph is symmetric with respect to the line x = − b . 2a Proof: We are to show that any line parallel to the x-axis that intersects the graph will intersect it53 at USAD Mathematics Resource Guide • 2016–2017 two points that are equidistant from the line x = − b . an exercise. ProPerty 3: The graph is symmetric with respect to the line x = − b . 2a Proof: We are to show that any line parallel to the x-axis that intersects the graph will intersect it at two points that are equidistant from the line x = − b . 2a Let y = d be such a line, and let P and Q be the intersection points of the line with the graph. Also, let Y be the intersection of the line y = d with the line x = − b . We will show that PY = QY . 2a b ,d . 2a Clearly, the coordinates of Y are − Rancho Cucamonga High School - Ontario, CA To find the coordinates of P and Q , we solve the system 2 b = + y a x +k 2a y = d Solving the system, we get 2 b a x + +k =d 2a b d −k ± x = − 2a a y = d b b d −k d −k Thus, P = − and Q = − + , d − , d . 2a 2a a a Now we calculate PY and QY , using the distance formula (which we will prove in Section 8): 2 b d −k PY = − + 2a a b d −k 2 + + (d − d ) = a 2a b d −k QY = − − a 2a b d −k 2 + − (d − d ) = a 2a 2 y Agreement = ax 2 + bx + c is symmetric Hence: PY = QY . This completes the proof that the graph theoffunction Unauthorized Duplication is Prohibited Outside the of Terms Your License b 2a 50 =− . with respect to the line xAcAdemic Guide decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide –2014 2013 2010–2011 Note that Property 3 implies that the quadratic function has no inverse since it is not a one-to-one function. ProPerty 4: When a > 0 , the function y = ax 2 + bx + c is increasing on the right of x = − decreasing on the left of x = − yy=ax = ax2+bx+c + bx + c 54 b . 2a y b xx= =– 2a Resource Guide • 2016–2017 USAD Mathematics b and 2a ProPerty 4: When a > 0 , the function y = ax + bx + c is increasing on the right of x = − b . decreasing on the left of x = − 2a b xx= =– 2a yy=ax = ax2+bx+c + bx + c 2a and y – b ,0 2a 2a x x ○ (0, 0) 22 4ac 0, – bb –– 4ac 4a Figure 5–7 Proof: We are to prove: b , ax12 + bx1 + c ≥ ax2 2 + bx2 + c 2a b B. For any x1 ≥ x2 ≥ − , ax12 + bx1 + c ≤ ax2 2 + bx2 + c 2a We will prove A here. The proof for B is similar, and you can complete it on your own as an exercise. As before, we write: 2 b (1) ax + bx1 + c = a x1 + +k 2a 2 b (2) ax2 2 + bx2 + c = a x2 + +k 2a b Since x1 > x2 > − , 2a 2 2 b b (3) x1 + > x2 + 2a 2a 2 1 Since a > 0 2 b b (4) a x1 + > a x2 + 2a 2a 2 Adding k to both sides of this inequality, we get 22 22 bb Unauthorized bDuplication b is Prohibited Outside the Terms of Your License Agreement (5) (5) aaxx11++ ++kk >>aaxx22 ++ ++kk ® 2013 –2014 AcAdemic decAthlon mAthemAtics ResouRce Guide 2 22aa Guide 2010–2011 2aaMatheMatics ResouRce 51 51 From From(1), (1),(2), (2),and and(5), (5),we weget getax asdesired. desired. ax1122 ++bx bx11++cc≥≥ax ax2222 ++bx bx22 ++cc,,as The Thegraph graphof of aaquadratic quadraticfunction functionisiscalled calledaaparabola, parabola,and andthe theminimum minimumor ormaximum maximumpoint pointisiscalled calledthe the vertex vertexof of the theparabola. parabola. ProPerty theset setof ofall allreal realnumbers. numbers.IfIfaa>>00,, ProPerty5: 5:The Thedomain domainof ofthe thequadratic quadraticfunction function yy==ax ax22 ++bx bx++cc isisthe bb bb 55 USAD Mathematics Resource Guide • 2016–2017 the the range range of of the the quadratic quadratic function function isis the the set set of of all all numbers numbers yy for for which whichyy≥≥aa−− ++bb−− ++cc.. IfIf 22aa 22aa 22 Rancho Cucamonga High School - Ontario, CA A. For any x1 ≥ x2 ≥ − vertex of the parabola. ProPerty 5: The domain of the quadratic function y = ax 2 + bx + c is the set of all real numbers. If a > 0 , 2 b b the range of the quadratic function is the set of all numbers y for which y ≥ a − + b − + c . If 2a 2a 2 b b a < 0 , the range of the quadratic function is the set of all numbers y for which y ≤ a − + b − + c . 2a 2a You can complete the proof of this property on your own as an exercise. 5.3 hat Do 5.3 W What Do the the Graphs Graphsof ofSome Some Polynomials 5 .3 What Do the Graphs Some Polynomials Look Like? Polynomials Look of Like? Look Like? solution: We first find where the polynomial intersects the x-axis. To do this, we need to solve the equation x 3 − 2 x 2 − 5 x + 6 = 0 . We use the Rational Root Theorem to find that this equation has three solutions: x = 1 , x = −2 , and x = 3 . Also, substituting x = 0 in the polynomial, we find that the polynomial intersects the y-axis at y = 6. We also note that as x gets larger moving toward infinity, y gets larger moving toward infinity; and as x gets smaller moving toward negative infinity, y gets smaller moving toward negative infinity. We can use this information to sketch the graph of the polynomial: y = x3 – 2x2 – 5x + 6 lled the fa > 0, Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 52 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide + c . If b +c. a 56 3 ExamplE 5.3b:USAD Graph Mathematics the polynomial y = xResource +1. Guide • 2016–2017 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA ExamplE 5.3a: Graph the polynomial y = x 3 − 2 x 2 − 5 x + 6 . ExamplE 5.3b: Graph the polynomial y = x 3 + 1 . solution: We first find where the polynomial intersects the x-axis. To do this, we need to solve the equation x 3 + 1 = 0 . This equation has one solution: x = −1 . The polynomial intersects the y-axis at y = 1 . We also note that as x gets larger toward infinity, y gets larger toward infinity; and as x gets smaller toward negative infinity, y gets smaller toward negative infinity. We can use this information to sketch the graph of the polynomial: 5.4 W hat Does the of 5.4 What Does the Graph Graph ofthe theExponential 5 .4 What Does the Graph of the Exponential x exponential Function ya Look Like? Function ya x Look Function y =Like? ax Look Like? Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 AcAdemic MatheMatics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide 53 The function y = a x is called the exponential function. The exponential function is only defined for a > 0 since if a > 0 , y = a x is positive for any real number x , and so its graph is above the x-axis. Another feature of this function is that when a > 1 , the function is increasing, and when 0 < a < 1 , the function is decreasing. ExamplE 5.4a: x 1 Consider the functions y = 2 and y = : 2 x In y = 2 x , as x gets larger, 2 x gets larger. 7 USAD Mathematics Resource y = 2x 6 Guide • 2016–2017 x 2 2 57 Rancho Cucamonga High School - Ontario, CA y = x3 + 1 1 2 x Consider the functions y = 2 x and y = : In y = 2 x , as x gets larger, 2 x gets larger. 7 x 2 2 y = 2x 6 5 4 3 x 2 1 2 –8 –6 –4 x1 –2 2 x2 4 6 8 –1 –2 That is: if x1 > x2 , then 2 x1 > 2 x2 . For example, 2 4 > 23 . 11 1 1 In yy==( )x = = x , on the other hand, as x gets larger, gets smaller. 22 2 2 x1 x2 3 2 1 1 1 1 That is: if x1 > x2 , then < . For example, < . 2 2 2 2 x x x y= x 1 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 54 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide –2014 2013 2010–2011 x 2 x1 x2 58 USAD Mathematics Resource Guide • 2016–2017 The domain of the exponential function is the set of all real numbers, whereas its range is the set of 10 Rancho Cucamonga High School - Ontario, CA 1 The domain of the exponential function is the set of all real numbers, whereas its range is the set of all positive real numbers. Since the exponential function is increasing, it is one-to-one, and thus has an inverse. The inverse of the exponential function is called the logarithmic function, which we will study next. 14 16 The properties of the logarithmic function can be derived from the properties of the exponential function because the logarithmic function is the inverse of the exponential function, as the following definition indicates. DeFinition c c = log a b if and only if a = b . The function y = log a x is called the logarithmic function. a is called the base of the logarithm. ProPerty 1: The expression y = log a x is equivalent to a y = x . Since a y is defined only for a > 0 , and Unauthorized Duplication is Prohibited Outside Terms of Your License Agreement of the logarithmic x must (as we showed earlier), bethepositive. Thus, the domain a y is always positive 2013 –2014 2010–2011 AcAdemic MatheMatics ResouRcedecAthlon Guide function consists of the positive real numbers. ® mAthemAtics ResouRce Guide 55 ProPerty 2: Another important condition to observe is that a , the base of the logarithm, must be different from 1. This is so because for a = 1 , log a x is not a function. Recall, a function cannot be one-tomany. If a = 1 , then the input x = 1 would have infinitely many outputs. For example, log1 1 = 4 because 14 = 1 ; log1 1 = −3 because 1−3 = 1 ; log1 1 = 6.9 because 16.9 = 1 ; log1 1 = 5 because 1 5 = 1 , etc. Thus, the logarithmic function y = log a x is meaningful only for x > 0 and 1 ≠ a > 0 . ProPerty 3: Since the function admits only positive inputs, its graph is located in the half plane to the right of the y-axis. ProPerty 4: Next, we can observe that the x-intercept of the logarithmic function is x = 1 , because log a 1 = 0 . Recall that y = log a x is defined only for x > 0 . Hence, there is no y-intercept—the function does not intersect the y-axis. ProPerty 5: Finally, y = log a x is increasing when a > 1 and decreasing when 0 < a < 1 . Putting all these properties together, we can see that the graph of y = log a x when a > 1 is of the following form: USAD Mathematics Resource Guide • 2016–2017 59 Rancho Cucamonga High School - Ontario, CA 12 5.5 W hat Does the Graph of 5.5 What Does the of Graph ofthe theLogarithmic 5 .5 What Does the Graph the Logarithmic Logarithmic ylog x Look Like? Function ylog aFunction xy Look Like? a Like? Function = log x Look a ProPerty 5: Finally, y = log a x is increasing when a > 1 and decreasing when 0 < a < 1 . Putting all these properties together, we can see that the graph of y = log a x when a > 1 is of the following form: y = log2x log2x2 log2x1 x2 Figure 5–8 Duplication Outside the Termsform: of Your License Agreement 0 < ais <Prohibited 1 is of the And, the graph of Unauthorized following y = log a x when 56 56 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide –2014 2013 2010–2011 y = log 1 x 2 x1 x2 log1x1 2 log1x2 2 Figure 5–9 x 1 The graph of y = log 1 x is symmetric to the graph of y = as is shown in Figure 5–10. This is 2 2 because the two functions are the inverse of one another. 60 USAD Mathematics Resource Guide • 2016–2017 Rancho Cucamonga High School - Ontario, CA x1 Figure 5–9 x 1 The graph of y = log 1 x is symmetric to the graph of y = as is shown in Figure 5–10. This is 2 2 because the two functions are the inverse of one another. x y y = log 1 x Figure 5–10 The domain of the logarithmic function is the set of all positive real numbers, whereas its range is the set of all real numbers. In science, the most frequently used base is the number e = 2.718281828459045 23536 …. . Logarithms with base e are called natural logarithms. The number e is irrational since it cannot be expressed as a repeating decimal. The natural logarithm is written ln x and is defined to be the inverse function of e x . Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 AcAdemic MatheMatics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide 57 Sections 5.1-5.5 Section 5 .1–5 .5: kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk SectionS 5.1–5.5: exercises Exercises Exercises kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 1. Determine the maximum or minimum and axis of symmetry of the following quadratic functions. Graph each parabola including the vertex and the axis of symmetry. a. f ( x) = x 2 − 6 x + 8 b. f ( x) = − x 2 + 6 x − 9 c. f ( x) = 2 x 2 − 3 x − 3 2. Graph the following functions: x 1 a. f ( x) = 3 x 3 b. f ( x) = 2 USAD Mathematics Resource Guide • 2016–2017 3. Sketch the graphs of the following functions: a. f ( x) = 4 x 3 − 8 x 2 − 15 x + 9 61 Rancho Cucamonga High School - Ontario, CA 2 2. Graph the following functions: x 1 2. Graph functions: a. f ( xthe ) = following x 13 a. f ( x) = x 3 b. f ( x) = x 3 b. f ( x) = 2 2 3. Sketch thegraphs of the following functions: 3. Sketch 2 the following functions: a. f ( x)the = 4graphs x3 − 8 xof − 15 x + 9 4. 4. 5. 5. 3 2 a. 9 x − 45 b. ff (( xx)) = = 4x x4 −−68xx3 −−415 x 2x++54 4 3 2 b. f ( x) = x − 6 x − 4 x + 54 x − 45 Let f ( x) = x 2 + px + q . Find the values of p and q if the minimum of this function is (1, −2) . Let f ( x) = x 2 + px + q . Find the values of p and q if the minimum of this function is (1, −2) . Find the inverse of the following functions.: Find x the following functions. a. f (the x) =inverse 3 − 22−of kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 5.6 transformations ofGraphs Graphs 5 .6 Transformations of Graphs 5.6 Transformations of 5.6 transformations of 5 .6 Transformations of Graphs Graphs It is recommended that you review the section on the composition of functions (Section 4) before It is recommended studying this section.that you review the section on the composition of functions (Section 4) before studying this section. If we know the graph of a function y = f ( x) , we are able to construct the graphs of the functions If we know ,the graph of ,a function y, and , we are able to construct the graphs of the functions y = f ( x + c) y = f ( x) + C y = f (ax) = f (yx)= Af ( x) , where a , c , A , and C are constants. y = f ( x + c) , y = f ( x) + C , y = f (ax) , and y = Af ( x) , where a , c , A , and C are constants. 5 .6 .1 GrAPhInG yfyf (xc) from The the GrAPh of yfoF (x)yf (x) 5.6.1 GraPhinG (xc) FroM GraPh 5 .6 .1 Graphing GrAPhInG from TheGraph GrAPh (x)yf (x) 5.6.1 GraPhinG FroM the GraPh oF 5.6.1 y yf = fyf (x(xc) +(xc) c) from the ofof y =yf f (x) The function y = f ( x + c) is the composition of two functions: y = x + c and y = f ( x) . For any input x , the composition of two functions: The function and any input x , y = xfor + cthe = f ( x +x c+)cis. This y = f ( xy) =. For we first get theyoutput output, in turn, becomes an input function f ( x) , giving the we first get the output x + c . This output, in turn, becomes an input for the function y = f ( x) , giving the output f ( x + c). 58 58 exAmPles: Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement ® Unauthorized Duplication is Prohibited Outside the Terms of Your License AgreementGuide mAthemAtics ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce ® Guide AcAdemic decAthlon mAthemAtics ResouRce MatheMatics ResouRce Guide –2014 2013 2010–2011 –2014 2013 2010–2011 (1) y = ( x + 2) 2 is the composition y = x + 2 and y = x 2 : to any given x , we first add the 2, and then we square the resulting sum. (2) y = 2 x − 4 is the composition of y = x − 4 and y = 2 x : to any given x , we first subtract 4, and then we raise 2 to the power of x − 4 . (3) y = log 2 ( x − 5) is the composition of y = x − 5 and y = log x : to any given x , we first subtract 5, and then we take the log of x − 5 . Notice that the two functions y = f ( x) and y = f ( x + c) have the same output for any two inputs that differ by c . For the inputs x and x − c , we have f ( x) = f (( x − c) + c) . On the basis of this fact, we can 62 USAD Mathematics Resource Guide • 2016–2017 now construct the graph of y = f ( x + c) from the graph of y = f ( x) . Rancho Cucamonga High School - Ontario, CA 2− x 3+ x a. ff (( xx)) = = 32 − b. log 3 −2 3 + x b. f ( x) = 2 − log 3 x x then we take the log of x − 5 . Notice that the two functions y = f ( x) and y = f ( x + c) have the same output for any two inputs that differ by c . For the inputs x and x − c , we have f ( x) = f (( x − c) + c) . On the basis of this fact, we can now construct the graph of y = f ( x + c) from the graph of y = f ( x) . For any point ( x0 , y0 ) on the graph of y = f ( x) , we have y0 = f ( x0 ) . What input x of the function ( ) y = f ( x + c) gives the output y0 ? The answer is x = x0 − c because f (x0 + c) = f (x0 − c) + c = f (x0 ) = y0 . Conversely, if ( x0 − c, y0 ) is on the graph of y = f ( x + c) , then ( x0 , y0 ) is on the graph of y = f ( x) . The conclusion that we can draw from this is that to construct the graph of the function y = f ( x + c) from the graph of y = f ( x) , we simply shift each point of the graph y = f ( x) by c units along the x-axis. We move the graph by c units to the left if c > 0 and by c units to the right if c < 0 . Rancho Cucamonga High School - Ontario, CA ExamplE 5.6a: Construct y = 2 x − 4 . The function y = 2 x − 4 is of the form y = f ( x + c) where f ( x) = 2 x and c = −4 < 0 . Therefore, the graph of y = 2 x − 4 is the result of shifting the graph of y = 2 x along the x-axis to the right by 4 units. x–4 5.6.2 y yf = fyf (x) +(x)C Cisfrom the Graph of yof = yf f (x) Unauthorized Duplication Prohibited Outside the Terms of Your License Agreement 5.6.2Graphing GraPhinG FroM the GraPh oF 5 .6 .2 GrAPhInG (x)C from The GrAPh (x)yf (x) 2013 –2014 2010–2011 AcAdemic MatheMatics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide 59 The function y = f ( x) + C is the composition of two functions: y = f ( x) and y = x + C . For any input x , we get the output f ( x) . This output, in turn, becomes an input for the function y = x + C , giving the output f ( x) + C . exAmPles: (1) y = x 2 − 3 is the composition of y = x 2 and y = x − 3 : We first square an input x , and then add −3 to the result. x x 1 1 1 (2) y = + 2 is the composition of y = and y = x + 2 : We first raise to the power x and then 3 3 3 63 USAD Mathematics Resource Guide • 2016–2017 add 2 to the result. (1) y = x − 3 is the composition of y = x and y = x − 3 : We first square an input x , and then add −3 to the result. x x 1 1 1 (2) y = + 2 is the composition of y = and y = x + 2 : We first raise to the power x and then 3 3 3 add 2 to the result. 1 3 1 and then add − to the result. 3 1 3 (3) y = log 2 x − is the composition of y = log 2 x and y = x − : We first compute the log of an input x Notice that for any input x , the outputs y = f ( x) + C and y = f ( x) differ by C units. This fact helps us construct the graph of y = f ( x) + C from the graph of y = f ( x) . For any point ( x0 , y0 ) on the graph of y = f ( x) , we have y0 = f ( x0 ) . Clearly, the point ( x0 , y0 + C ) belongs to the graph of y = f ( x) + C (because y0 + C = f ( x0 ) + C ). graph of y = f ( x) . The conclusion that we can draw from this is that to construct the graph of the function y = f ( x) + C , we simply shift the graph y = f ( x) by C units along the y-axis. The graph moves up by C units if C > 0 , and it moves down by C units if C < 0 . ExamplE 5.6b: Construct the graph of y = log 2 x − 3 . The function y = log 2 x − 3 is of the form y = f ( x) + C where f ( x) = log 2 x and C = −3 < 0 . Therefore, moving f ( x) = log 2 x down the y-axis by 3 units results in the graph of y = log 2 x − 3 . Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 60 60 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide 5.6.3 GraPhinG (ax) FroM the GraPh oF 5 .6 .3 GrAPhInG yfyf (ax) TheResource GrAPh of •yf (x)yf (x) 64 USAD from Mathematics Guide 2016–2017 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA Conversely, if the point ( x0 , y0 + C ) belongs to the graph of y = f ( x) + C , then the point ( x0 , y0 ) is on the 5.6.3 GraPhinG yf FroM the 5 .6 .3 Graphing GrAPhInG (ax)(ax) from GrAPh yfoF (x)yf (x) 5.6.3 y yf = f(ax) from theThe Graph ofGraPh yof = f(x) The function y = f (ax) , a ≠ 0 , is the composition of y = ax and y = f ( x) . For any input x , we first get the output ax . This output, in turn, becomes an input for the function y = f ( x) , giving the output y = f (ax) . exAmPles: 2 x x (1) y = is the composition of y = and y = x 2 (How?) 2 2 (2) y = 23 x is the composition of y = 3 x and y = 2 x (How?) 1 1 x is the composition of y = x and y = log 2 x (How?) 7 7 Let ( x0 , y0 ) be a point on the graph of y = f ( x) , then y0 = f ( x0 ) . What input x for the function y = f (ax) x x gives the output y0 ? Clearly, the answer is x = 0 . Indeed: f (ax0 ) = f a 0 = f ( x0 ) = y0 . a a x This shows that if the point ( x0 , y0 ) belongs to the graph of y = f ( x) , then point 0 , y0 belongs to the a x0 graph of y = f (ax) . And, conversely, if the point , y0 belongs to the graph of y = f (ax) , then the a point ( x0 , y0 ) belongs to the graph of y = f ( x) . The relation between the graph of y = f (ax) and the graph of y = f ( x) depends on the value of a : k If a > 1 , then we get the graph of y = f (ax) by “shrinking” the graph of y = f ( x) along the x-axis by a factor of k 1 . a Unauthorized the Terms of the Yourgraph Licenseof Agreement If 0 < a < 1 , then we get theDuplication graph of isy Prohibited by “expanding” = f (ax) Outside y = f ( x) along the x-axis 2013 –2014 2010–2011 AcAdemic 1 MatheMatics ResouRcedecAthlon Guide by a factor of a . ® mAthemAtics ResouRce Guide 61 61 You can investigate the cases a < −1 and −1 < a < 0 on your own as an exercise. (Hint: note that for any function y = g ( x ) the graphs of y = g ( x ) and y = g ( − x ) are symmetric with respect to the y-axis.) ExamplE 5.6c: Construct the graph of y = 23 x . The function y = 23 x is of the form y = f (ax) , a ≠ 0 , where f ( x) = 2 x and a = 3 > 1 . To get the 1 along the x-axis toward the graph of y = 23 x , we “shrink” the graph of f ( x) = 2 x by a factor of 3 y-axis. USAD Mathematics Resource Guide • 2016–2017 65 Rancho Cucamonga High School - Ontario, CA (3) y = log 2 ExamplE 5.6d: Construct the graph of y = 2−3 x . The function y = 2−3 x is of the form y = f (ax) , a ≠ 0 , where f ( x) = 2 x and a = −3 < −1 . We first 1 along the x-axis obtain the graph of y = 23 x by expanding the graph f ( x) = 2 x by a factor of 3 3x toward the y-axis. Following this, we reflect the graph of y = 2 with respect to the y-axis to obtain the graph of y = 2 −3 x . Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 62 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide –2014 2013 2010–2011 66 USAD Mathematics Guide 5.6.4 GraPhinG yAf FroM the GraPh 5 .6 .4 GrAPhInG yAf (x) (x) from TheResource GrAPh of• 2016–2017 yfoF (x)yf (x) Rancho Cucamonga High School - Ontario, CA The function y = 23 x is of the form y = f (ax) , a ≠ 0 , where f ( x) = 2 x and a = 3 > 1 . To get the 1 along the x-axis toward the graph of y = 23 x , we “shrink” the graph of f ( x) = 2 x by a factor of 3 y-axis. 5.6.4 y yAf = Af(x) from theThe Graph ofGraPh y of = f(x) 5.6.4 GraPhinG yAf (x) FroM the 5 .6 .4 Graphing GrAPhInG (x) from GrAPh yfoF (x)yf (x) To obtain the graph of the function y = Af ( x) , A ≠ 0 , from the graph of the function y = f ( x) , we reason as before: 1. ( x0 , y0 ) is on the graph of y = f ( x) if and only if ( x0 , Ay0 ) is on the graph of y = Af ( x) (Why?) 2. Therefore, to plot the graph of y = Af ( x) , we simply transform each point ( x0 , y0 ) on the graph ( ) y = f ( x) to the point x0 , Ay0 . From there: 3. If A > 1 , then we get the graph of y = Af ( x ) by “expanding” the graph of y = f ( x) along the y-axis 4. If 0 < A < 1 , then we get the graph of y = Af ( x ) by “shrinking” the graph of y = f ( x) along the y-axis by a factor of A . 5. If A = −1 , we get the graph of y = Af ( x ) by simply reflecting the graph of y = f ( x) with respect to the x-axis. You can work out the cases −1 < A < 0 and A < −1 as an exercise on your own. (Hint: note that for any function y = g ( x ) , the graphs of y = g ( x ) and y = − g ( x ) are symmetric with respect to the x-axis. Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement ® 2013 –2014 5.6e: AcAdemic ResouRce Guide 2010–2011 MatheMatics ResouRce Guide of y =mAthemAtics ExamplE Construct the decAthlon graph − log 1 x . 63 63 3 The function y = − log 1 x is of the form y = Af ( x) where f ( x) = log 1 x and A = −1 . Therefore, 3 3 the graph of y = − log 1 x is symmetrical to the graph of f ( x) = log 1 x with respect to the x-axis. 3 3 USAD Mathematics Resource Guide • 2016–2017 67 Rancho Cucamonga High School - Ontario, CA by a factor of A . 1 4 ExamplE 5.6f: Construct the graph of y = − x 2 . 1 1 . Therefore, the 4 4 1 graph of y = − x 2 is obtained in two steps: first we shrink the graph of f ( x) = x 2 along the 4 1 The function y = − x 2 is of the form y = Af ( x) where f ( x) = x 2 and A = − x-axis. 4 , and then we take the reflection of the latter graph with respect to the Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 64 ® 2013 decAthlon mAthemAtics ResouRce ResouRce k 2010–2011 k k k k k k k k k AcAdemic kkkkk k k k k k k k k k kMatheMatics k k k kGuide k k k kGuide kkk k–2014 kk Section 5 .5.6 6: Section Section 5.6: exercises Exercises Exercises kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 1. Construct the graphs of the following functions: a. y = −( x + 3) 2 − 6 1 2 c. y = − log 3 x + 4 b. y = log 2 ( x + ) 2. Find the inverse of the following functions and sketch both the function and its inverse in the same graph: 1 a. f (x) = ln 2x 3 b. f ( x) = e −2 x + 2 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 68 USAD Mathematics Resource Guide • 2016–2017 section 6: non-polynomial equations SecTIon 6: non-Polynomial equations Rancho Cucamonga High School - Ontario, CA y-axis by a factor of b. y = log 2 ( x + ) 2 c. y = − log 3 x + 4 2. Find the inverse of the following functions and sketch both the function and its inverse in the same graph: 1 a. f (x) = ln 2x 3 b. f ( x) = e −2 x + 2 Section k k k6 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk non-polynomial equations section 6: non-polynomial equations SecTIon 6: non-Polynomial equations In the previous sections, we learned how to solve linear equations, quadratic equations, and some new type of equation that is not a polynomial equation. 6.1 rational equations 6 .1 Rational Equations 6.1 Rational Equations The first non-polynomial equations we will learn about in this section are called rational equations. A rational equation is an equation that involves only polynomials or rational expressions. By “rational expression,” we mean a fraction whose numerator and denominator are polynomials. exAmPles: 1. 2. 3. 1 1 − x = 0 is a rational equation. It involves the polynomials x and 0 and the rational expression . x x x + x 3 − 2 = x is not a rational equation since x is not a rational expression. x−3 + 7 x 2 + 10 = 5 x is a rational equation since it involves only polynomials ( 7 x 2 + 10 and 5x) 2 ( x − 4) x−3 and a rational expression ( x − 4) 2 . Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 4. 2 x ( x − 4) 2 = 0 is a polynomial equation, but we can view it as a rational equation since we can write ® 2013 –2014 AcAdemic mAthemAtics ResouRce Guide 2010–2011 ResouRcedecAthlon Guide 2 x( x −MatheMatics 4) 2 it as 1 = 0. 5. ( x − 5)( x − 4) − ( x − 2)(2 x + 9) = 6. 65 65 17 is a rational equation for the same reason given in 1 and 3. 2 x + 1.9 ( x + 3)( x + 5) 11x + 13 x−2 is not a rational equation (why?) − =4+ 3 10 7 A polynomial admits any real number x ; that is, its domain is the set of all real numbers. A rational expression, on the other hand, may not admit all real numbers. For example, the rational expression 17 + 1 does not admit the values x = ±2 2 because the denominator x 2 − 8 = 0 is zero for these 2 69 x −8 USAD Mathematics Resource Guide • 2016–2017 values. We say that the rational expression is not defined for x = ±2 2 or the domain of the rational Rancho Cucamonga High School - Ontario, CA techniques for solving higher-degree polynomial equations. In this section, we will learn how to solve a 3 10 7 A polynomial admits any real number x ; that is, its domain is the set of all real numbers. A rational expression, on the other hand, may not admit all real numbers. For example, the rational expression 17 + 1 does not admit the values x = ±2 2 because the denominator x 2 − 8 = 0 is zero for these 2 x −8 values. We say that the rational expression is not defined for x = ±2 2 or the domain of the rational 17 function f (x) = 2 +1 is the set of all real numbers x ≠ ±2 2 . On the other hand, the expression x − 8 1 − 4 is defined for all real numbers because there is no real number x for which the denominator 2 5x + 1 1 − 4 is the set of all real numbers. 5 x 2 + 1 is zero. So, the domain of the rational function f ( x) = 2 5x + 1 6.1.1 hoW SoLve rationaL eQuationS? 6 .1 .1 hoW DoDo WeWe SoLve rATIonAL eQuATIonS? 6.1.1 How Do We Solve Rational Equations? dealt with like fractions. However, since they involve variables, we must pay attention to their domains. x2 − 1 Consider, for example, the rational expression A( x ) = . ( x − 1)( x + 3) x +1 One might hastily simplify this expression into B ( x ) = by factoring the numerator into ( x − 1)( x + 1) x+3 and canceling the expression expressing x − 1 , which appears in both the numerator and denominator. 1 Is A( x ) = B ( x ) for all x ? The answer is no because while for x = 1 , B ( x ) = , A( x ) is not defined for 2 x = 1 . The lesson we learn from this is that before we simplify an expression, we must pay attention to the domain of the expression. Thus, for example, we can say A( x ) = B ( x ) for all x ≠ 1 . Likewise, x 2 − 4 x + 4 ( x − 2) 2 = = x − 2 for all x ≠ 2 . x−2 x−2 Let us now turn to solving rational equations. ExamplE 6.1a: Solve the equation 5 − 4x = 0 . x−2 solution: 1. We first exclude the numbers that cannot be solutions. Since x − 2 appears in the denominator, x= 2 cannot be a solution. Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 66 66 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide 4. Since a fraction is zero if and only if its numerator is zero, we get 5 − 4 x ( x − 2) = 0 . 70 –2014 2013 2010–2011 2. Next, as we do with fractions, we write all expressions in equivalent forms with the same 5 4 x ( x − 2) − =0. denominator. In this case, x − 2 is a common denominator. We write: x−2 x−2 5 − 4 x ( x − 2) =0. 3. Now we add the two rational expressions as we do with fractions: x−2 5. Now we proceed to solve this familiar equation: USAD Mathematics Resource Guide • 2016–2017 2 5 − 4 x + 8x = 0 Rancho Cucamonga High School - Ontario, CA When solving rational equations often we need to simplify rational expressions. Rational expressions are 3. Now we add the two rational expressions as we do with fractions: 5 − 4 x ( x − 2) =0. x−2 4. Since a fraction is zero if and only if its numerator is zero, we get 5 − 4 x ( x − 2) = 0 . 5. Now we proceed to solve this familiar equation: 5 − 4 x2 + 8x = 0 4 x2 − 8x − 5 = 0 8 ± 64 + 80 x= 8 8 ± 12 x= 8 6. The solutions to the equation are: x = 6x = 4. x−9 Rancho Cucamonga High School - Ontario, CA ExamplE 6.1b: Solve the equation 5 1 , x =− . 2 2 solution: 1. x ≠ 9 since the expression 6x is not defined for x = 9 . x−9 6x =4 x−9 6x − 4 = 0 Subtract 4 from both sides. 3. x−9 6x x−9 −4 = 0 Write all expressions in an equivalent form with the same denominator. 4. x−9 x−9 2 x + 36 = 0 Add the rational expressions. 5. x−9 2. 6. 2 x + 36 = 0 A rational expression is zero if and only if the numerator is zero. 7. 2 x = −36 Solve 8. x = −18 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement x = −18®. mAthemAtics 9.–2014 Theequation hasAcAdemic one solution: 2013 decAthlon 2010–2011 MatheMatics ResouRce Guide ResouRce Guide 67 In the next set of examples, you are expected to explain most of the steps on your own. x2 + 1 − 2 x − x = 0. ExamplE 6.1c: Solve the equation x −1 USAD Mathematics Resource Guide • 2016–2017 71 In the next set of examples, you are expected to explain most of the steps on your own. ExamplE 6.1c: Solve the equation x2 + 1 − 2 x − x = 0. x −1 solution: Condition: x ≠ 1 . 1. 2. x2 + 1 − 2 x −x=0 x −1 x 2 + 1 − 2 x − x ( x − 1) =0 x −1 3. x 2 + 1 − 2 x − x ( x − 1) = 0 4. x2 + 1 − 2 x − x2 + x = 0 6. x =1. Since x = 1 was excluded as a solution, there is no solution to this equation. You can see how important it is to specify the domains of the expressions involved beforehand. ExamplE 6.1d: Solve the equation x +1 x − 3 = . x −5 x +6 solution: Condition: x ≠ 5 and x ≠ −6 . 1. 2. x +1 x − 3 = x −5 x +6 x +1 x − 3 − =0 x −5 x +6 3. A common denominator here is ( x − 5)( x + 6) . Hence: 4. 68 68 ( x + 1)( x + 6) ( x − 3)( x − 5) − =0 Unauthorized x − 5) is Prohibited Outside the Terms of Your License Agreement ( x − 5)( x + 6) ( x + 6)(Duplication Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide 5. ( x + 1)( x + 6) − ( x − 3)( x − 5) = 0 ( x − 5)( x + 6) 6. ( x 2 + 7 x + 6) − ( x 2 − 8 x + 15) = 0 7. x 2 + 7 x + 6 − x 2 + 8 x − 15 = 0 8. 15 x − 9 = 0 72 9. x = 9 15 USAD Mathematics Resource Guide • 2016–2017 9 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA 5. 1 − x = 0 7. x 2 + 7 x + 6 − x 2 + 8 x − 15 = 0 8. 15 x − 9 = 0 9. x= 9 15 10. The solution to the given equation is x = ExamplE 6.1e: Solve the equation 9 . 15 2x − 9 2x + 1 1 + = . 1− x x + 1 1 − x2 solution: 2x − 9 2x + 1 1 + − =0 1− x x + 1 1 − x2 Rancho Cucamonga High School - Ontario, CA 1. 2. First, we factor 1 − x 2 . 3. 1 − x 2 = (1 − x )(1 + x ) 4. Thus, (1 − x )(1 + x ) is a common denominator for the three rational expressions in the given equation. 5. Condition: x ≠ 1 and x ≠ −1 . 6. Changing rational expressions into equivalent forms that have the same common denominator, we get 7. (2 x − 9)( x + 1) (2 x + 1)(1 − x ) 1 + − =0 (1 − x )( x + 1) ( x + 1)(1 − x ) (1 − x )( x + 1) (2 x − 9)( x + 1) + (2 x + 1)(1 − x ) − 1 =0 (1 − x )( x + 1) 8. (2 x 2 + 2 x − 9 x − 9) + (2 x − 2 x 2 + 1 − x) − 1 = 0 9. −6 x − 9 = 0 99 = –1.5 10. xx == – = 1.5 66 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement ® 2013 –2014 MatheMatics AcAdemic mAthemAtics ResouRce Guide 2010–2011 ResouRcedecAthlon xx == 1.5 11. The given equation has only Guide one solution: . –1.5. ExamplE 6.1f: Solve the equation 69 69 2x 1 1 − 2 = . x − 9 x + 14 x − 3x + 2 x − 1 2 USAD Mathematics Resource Guide • to 2016–2017 solution: As before, we find a common denominator the rational fractions in this equation. For this, as we did in the previous examples, we factor the given denominators into unfactorable 73 ExamplE 6.1f: Solve the equation 2x 1 1 − 2 = . x − 9 x + 14 x − 3x + 2 x − 1 2 solution: As before, we find a common denominator to the rational fractions in this equation. For this, as we did in the previous examples, we factor the given denominators into unfactorable expressions. 1. x − 1 is unfactorable. 3. x 2 − 9 x + 14 = ( x − 2)( x − 7) 4. x 2 − 3x + 2 = ( x − 1)( x − 2) 5. Thus, our equation can be written as 2x 1 1 − − =0 ( x − 2)( x − 7) ( x − 1)( x − 2) x − 1 6. ( x − 1)( x − 2)( x − 7) is a common denominator of the three rational expressions in our equation, because each denominator of the rational expressions divides it. 7. Thus, we have the condition: x ≠ 1 , x ≠ 2 , x ≠ 7 . 8. Writing these rational expressions in equivalent forms that have this common denominator, 2 x ( x − 1) x−7 ( x − 2)( x − 7) − − =0 ( x − 1)( x − 2)( x − 7) ( x − 1)( x − 2)( x − 7) ( x − 1)( x − 2)( x − 7) − x– 7 x2 x+2 6x 9. Adding, we get =0 2)(xx − – 7)( 7)(xx –− 1) 1) ((x x −– 2)( we get 10. Solving, we get (x x 2 +−7)(x x = 0– 1) = 0 0 or 11. xx == –7 or xx==11 = –7 0. 12. Since x = 1 cannot be a solution, the only solution to the equation is xx = Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 70 74 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA 2. Next we factor the other two denominators: x−3 =0. x − 5x + 6 xx −− 33 ExamplE ExamplE 6.1g: 6.1g: Solve Solve the the equation equation x 22 − 5 x + 6 == 00 .. + 6which the denominator x 2 − 5 x + 6 is 0 , and we solution 1: First we find the values xof −x5 xfor ExamplE 6.1g: Solve the equation 2 exclude these values. 2 solution solution 1: 1: First First we we find find the the values values of of xx for for which which the the denominator denominator xx 2 −− 55xx ++ 66 is is 00,, and and we we 2 exclude these 1. Solving x values. − 5 x + 6 = 0 , we get x = 2 , x = 3 . exclude these values. 2 1. Solving == 00 ,,xwe ≠ 3get . xx == 22 ,, xx == 33 .. 2. we get 1. Condition: Solving xx 2 −−x55x≠ x ++266and xx ≠≠ 22to xx ≠≠the 33 .. equation: and 2. Condition: 3. we proceed solve and 2. Now Condition: x−3 3. we = 0 to 4. 3. Now Now we proceed proceed to solve solve the the equation: equation: 2 x − 5x + 6 xx −− 33 4. 3 = 0 == 00 5. 4. xx 2− x 2 −− 55xx ++ 66 xx == 33 x = 3 was excluded as a solution, our equation has no solution. 6. 7. 6. Since x−3 33 was 7. excluded aa solution, our =0 solution Factoring outas denominator, we get 2has 7. Since Since xx ==2: was excluded asthe solution, our equation equation has no no solution. solution. x − 5x + 6 xx −− 33 x−3 solution out =0 ≠ 3denominator, 1. solution 2: 2: Factoring Factoring outxthe the denominator, we we get get x 22 − 5 x + 6 == 00 ( x − 3)( x − 2) x − 5x + 6 1x−3 x −=30 = 0 1. 2. =0 1. 3)( ( xx −− 2) 3)(xx −− 2) 2) xx ≠≠ 33 1 1 the= denominator 0 2. 3. is different from zero for all x , the equation has no solution. 2. Since ( x − 2) = 0 ( x − 2) 3. 3. Since Since the the denominator denominator is is different different from from zero zero for for all all xx ,, the the equation equation has has no no solution. solution. kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk Section 6 .1 .1: Section 6.1.1: exercises Section 6.1.1 k kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk k Exercises Section 6 .1 .1: Exercises k Section k k k k k k k k6.1.1: k k k k kexercises k kkkkkkkkkkkkkkkkkkkkkkkk Section 6.1.1: exercises Exercises k kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk k 1. Solve the following equations: 1. 3 x the − 2 following x + 2 equations: 1. Solve Solve the =following equations: a. x −3 x +3 33xx −− 22 xx ++222 == 8 x + 3 x + 2 a. a. xx+−0.5 b. x − 33 + xx ++233 = 9 x + 3 9 x − 1 3x − 1 xx ++ 0.5 88xx22 ++ 33 xx ++ 22 0.5 = b. b. x92x−+23x+++ 19Unauthorized is Prohibited Outside the Terms of Your License Agreement xx22 x−−+ 111==33x4xDuplication −−11 c. 9 x + 3 9 + ® 2013 –2014 x −MatheMatics mAthemAtics ResouRce Guide 3 3 −AcAdemic x ResouRcedecAthlon 2010–2011 Guide Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement Unauthorized 2 x + 1 Duplication 8 is Prohibited Outside the Terms of Your License Agreement 1− 2x ® + = decAthlon AcAdemic ® mAthemAtics ResouRce MatheMatics ResouRce 2Guide mAthemAtics ResouRce Guide 6 x + MatheMatics 3 x 14 x 2AcAdemic − 7ResouRce x 12 xdecAthlon −3 USAD Mathematics Resource Guide • 2016–2017 x+3 3− x 2 − = e. d. 2013 2010–2011 2013–2014 –2014 2010–2011 2 Guide Guide 71 71 71 71 71 71 75 Rancho Cucamonga High School - Ontario, CA 5. 6. − 33 == 00 5. xx −= x2 − 2x + 1 x +1 c. x 2 − 2 x + 1 + x + 1 = 4 c. x −3 + 3− x = 4 d. d. e. e. f. f. x −3 3− x 1− 2x 2x +1 8 1 −2 2 x + 2 x2 + 1 = 82 6 x + 3 x + 14 x − 7 x = 12 x − 3 6 x 2 + 3 x 14 x 2 − 7 x 12 x 2 − 3 x+3 3− x 2 = 2 x 2+ 3 − 2 3 − x 4 x − 9 − 4 x + 12 x + 9 = 2 x − 3 4 x 2 − 9 4 x 2 + 12 x + 9 2 x − 3 2x + 7 3 1 + 2 3 = 1 22 x + 7 x + 5 x − 6 + x + 9 x + 18 = x + 3 x 2 + 5 x − 6 x 2 + 9 x + 18 x + 3 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 6.1.2 What the GraPhS oF SoMe 6 .1 .2 WhAT DoDo The GrAPhS of Some 6.1.2 What Do the GraPhS oF SoMe 6 .1 .2 WhAT Do The GrAPhS of Some rationaL FunctionS Look Like? rATIonAL funcTIonS Look LIke? 6.1.2 What Do the Graphs of Some Rational Functions Look Like? rationaL FunctionS Look rATIonAL funcTIonS Look LIke?Like? y y x x Figure 6–1 Figure 6–1 1 The graph of the function f ( x) = 1 is shown in Figure 6–1. The graph of the function f ( x) = x is shown in Figure 6–1. x Note that the function is not defined for x = 0 . As x gets closer to zero along the positive x-axis, y gets Note that the function is not defined for x = 0 . As x gets closer to zero along the positive x-axis, y gets larger; and as x gets closer to zero along the negative x-axis, y gets smaller, and in both cases, the graph larger; and as x gets closer to zero along the negative x-axis, y gets smaller, and in both cases, the graph approaches the y-axis. approaches the y-axis. 72 72 76 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide ® Guide AcAdemic decAthlon mAthemAtics ResouRce MatheMatics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 –2014 2013 2010–2011 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA 2. Find the domain of the following functions: 2. Find the domain of the following functions: 2 a. y = 22 a. y = x2 + 2 x + 2 x −3 b. y = x −3 b. y = 2 x33 + x22 − 13 x + 6 2 x + x − 13 x + 6 Also, as x gets farther from zero along the x-axis, y gets closer to zero, and the graph approaches the x-axis. The graph never intersects the x-axis or the y-axis. In general, rational functions can approach, without intersecting, certain lines. These lines are called asymptotes. The lines x = 0 and y = 0 are 1 asymptotes of the function f ( x) = . x ExamplE 6.1h: Graph the function f ( x) = x . x −3 solution: 3 . x −3 2. You can get this by the Division Algorithm or by noting that x ( x − 3) + 3 3 f ( x) = = = 1+ x −3 x −3 x −3 3. From this, we see that f ( x) is of the form f ( x) = Ag ( x) + C where g ( x) = C = 1. Rancho Cucamonga High School - Ontario, CA 1. We first note that f ( x) = 1 + 1 , A = 3 , and x −3 4. We can graph this function in three steps: Step 1: Graph the function g ( x) = along the x-axis by 3 units. 1 1 . We can graph this function by shifting the graph of x −3 x y x 3 1 . We can graph this function by expanding the graph of x −3 x −3 by a factor of 3 along the y-axis to the right. Step 2: Graph the function Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 USAD Mathematics Resource Guide • 2016–2017 ® AcAdemic MatheMatics ResouRcedecAthlon Guide mAthemAtics ResouRce Guide 77 73 73 Step 3: Graph the function 1 + by 1 unit along the y-axis. x 3 3 . We graph this function by moving the graph of up x −3 x −3 y ExamplE 6.1i: Graph the function f ( x) = solution: We first note that f ( x) = 2 − x 2x + 5 . x+4 3 . x+4 1. You can get this by the Division Algorithm or by noting that 2 x + 5 2( x + 4) − 3 3 f ( x) = = = 2− x+4 x+4 x+4 78 74 74 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement ® USAD Mathematics Resource ResouRce Guide • 2016–2017 mAthemAtics Guide AcAdemic decAthlon MatheMatics ResouRce Guide –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA y . 2. From here we see that f ( x) is of the form f ( x) = Ag ( x) + C where g ( x) = C = 2. 1 , A = −3 , and x+4 We can graph this function in four steps: 1 1 . We graph this function by shifting the graph of x+4 x along the x-axis by 4 units to the left. Step 1: Graph the function g ( x) = 6 yy = 1 x+4 4 Rancho Cucamonga High School - Ontario, CA 2 -4 -5 -2 -4 Step 2: Graph the function a factor of 3 along the y-axis. 3 1 . We graph this function by expanding the graph of by x+4 x+4 y x Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 AcAdemic MatheMatics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 75 75 79 with respect to the y-axis. 3 3 . We graph this function by reflecting the graph of x+4 x+4 y Step 4: Graph the function 2 − up by 2 units along the y-axis. x 3 3 . We graph this function by moving the graph of − x+4 x+4 y x kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk Section 6 .1 .2: SectionS 6.1.2: exercises Exercises kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 1. Graph the following functions: 8076 76 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource Guide • 2016–2017 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA Step 3: Graph the function − kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk Section 6 .6.1.2 1 .2: SectionS 6.1.2: Section exercises Exercises Exercises kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 1 )=− a. f ( xthe 1. Graph following x − 2 functions: 1 13 b. f ( x) = − − a. Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2 x −x 2+ 1 76 –2014 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide 2013 2010–2011 1x + 1 3 c. f ( x) = − b. 2x − 3x + 1 x +1 ( x)k= k k k k k k k k k k k k k k k k k k k k k k k k k k k k k k k k k c. kf k kkk x −3 6.2 exponential equations 6 .2 Exponential Equations 6.2 Exponential Equations 6.2 exponential equations 6.2.1 BaSicProPerTIeS ProPertieS 6 .2 Exponential Equations 6 .2 .1 BASIc 6.2.1 Basic Properties 6.2.1 BaSic ProPertieS Recall meanings of expressions of the form a n . For example: 6 .2 .1theBASIc ProPerTIeS 25 = 2 ⋅the 2 ⋅ 2meanings ⋅2⋅2 Recall of expressions of the form a n . For example: 1 9252 = 2 ⋅92 ⋅ 2 ⋅ 2 ⋅ 2 1 1 9272 3== 93 27 1 −23 1 627 == 32 27 6 1 The a n is called a power. a is called the base of the power, and n is called the exponent of the 6−2 =expression 2 6 power. Recall the following important properties of powers: For any real numbers a > 0, b > 0 a ≠ 0 , and The expression a n is called a power. a is called the base of the power, and n is called the exponent of the real numbers m, and n, the following properties, 1 through 6, hold true: power. Recall the following important properties of powers: For any real numbers a > 0, b > 0 a ≠ 0 , and 0 1. anumbers = 1 ; m, and n, the following properties, 1 through 6, hold true: real 2. ⋅ a1n; = a m + n 1. a 0m = m n m−n 3. a m ÷ 2. ⋅: aan == aam + n m n n n m −nn 4. 3. (aab÷:) a= =a a⋅ b n n a n an n 5. (ab) = a n ⋅ b 4. b b n n a a 6. ( a n ) m == ann⋅m 5. b b 81 USAD Mathematics Resource Guide • 2016–2017 When 6. ( a nm) mand = ann⋅mare integers, the above properties, 1 through 6, hold true for all real numbers a ≠ 0,b ≠ 0. Rancho Cucamonga High School - Ontario, CA kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk n n a a 5. = n b b 6. ( a n ) m = a n⋅m When m and n are integers, the above properties, 1 through 6, hold true for all real numbers a ≠ 0,b ≠ 0. And, for a > 0 and n a natural number, 1 n 7. a = n a . 2013 –2014 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement AcAdemic MatheMatics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide 77 77 x We note that (−2) = −8 has the solution x = 3. 6 .2 .2 hoW DoDo We Solve SoLve exPonenTIAL eQuATIonS? 6.2.2 hoW We SoLve exPonentiaL eQuationS? 6.2.2 How Do We Exponential Equations? The basic idea for solving exponential equations of the form a x = b is to write b as the power with the becomes: 3x−4 = 34 . Since a b = a c if and only if b = c , we get x − 4 = 4 . Hence, the solution to the equation is x = 8 . Let’s now turn to solve more complicated exponential equations. . ExamplE 6.2a: Solve 32 x +1 − 2 ⋅ 9 x = 5 − 32 x −1 solution: All the powers involving the variable x can be written as powers with base 3: 32 x +1 − 2 ⋅ (32 ) x = 5 − 32 x −1 32 x +1 − 2 ⋅ 32 x = 5 − 32 x −1 Notice that each exponent in the last equation involves 2x: 32 x +1 = 32 x ⋅ 31 = 3 ⋅ 32 x 32 x 3 = 3 ⋅3 = 3 2x 3 2x 2x 3⋅ 3 − 2 ⋅ 3 = 5 − 3 2 x −1 2x −1 Writing the given equation in terms of 32 x , we get 32 x 3⋅ 3 − 2 ⋅ 3 = 5 − 3 2x 2x Let 32 x = t . Now, our equation turns into a linear equation: t 3⋅ t − 2 ⋅ t = 5 − . 3 82 Solving for t, we get t= 15 = 3.75 USAD Mathematics Resource Guide • 2016–2017 Rancho Cucamonga High School - Ontario, CA base a. Consider, for example, the equation 3x−4 = 81.. We write 81 = 34 , and accordingly our equation [ [ " Let [ " W . Now, our equation turns into a linear equation: W W W " . Solving for t, we get W " " Thus, [ " . 78 Recall that D E " F if and only if ORJ D F " E . Therefore: [ " if and only if [ " ORJ , Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement if and only if [ " ORJ –2014 GUIDE ACADEMIC DECATHLON ® MATHEMATICS RESOURCE MATHEMATICS RESOURCE GUIDE 2013 2010–2011 . Thus: [ " ORJ is the only solution to the given equation. EXAMPLE 6.2b: Solve the equation [ [ " [ [ SOLUTION: Notice [ " [ Hence, our equation can be written as [ [ " [ [ [ We separate between the powers of 4 and the powers of 5: " [ Factoring out , we get: [ – " [ [ [ [ [ " [ We will compute the part on the right of the equation separately: 1 1 1 3 2 2 5 + 3 1 1 1 5 5 52 ⋅ 52 + 1 4 3 − 2+ 2 2 2 2 2 2 2 2 2 5 +5 (25 + 1)4 26 ⋅ 4 2⋅4 4 ⋅4 4 4 4 2 52 = = = = = = = = = 3 3 3 3 3 3 5 1+ 1 4 + 4−2 5 2 2 2 2 2 2 2 2 4+ 2 5 4 ⋅ 4 +1 (64 + 1)5 65 ⋅ 5 5⋅5 5⋅5 5 5 4 ( [ ©¹ ©¹ We have ª º " ª º «» «» Hence: [ " . ) The solution to the given equation is [ " . kkkkkkkkk k kMathematics k k k k kResource k k k kGuide kkk k k k k k k k k k k k k k k k83 k USAD • 2016–2017 Section 6.2: SECTION 6.2: Exercises Rancho Cucamonga High School - Ontario, CA Using a calculator, we can find that solution [ is approximately 0.60155. 2 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk Section 6 .2: Section 6.2 Section 6.2: exercises Exercises Exercises kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk x following −5 1. Solve the equations: 3 7 a. = 73 x 73Unauthorized −5 Duplication is Prohibited Outside the Terms of Your License Agreement a. x2 +x −0.5 = ® 2013 –2014 AcAdemic mAthemAtics ResouRce Guide b. 2 = 4 2 2010–2011 7 MatheMatics ResouRcedecAthlon Guide 3 79 2 b. 2 x 1++ xx−2 0.5 = 4 22 c. 10 − 101− x = 99 2 2 c. 10x211+− xx − 101x1− x 3= 99 d. 5 ⋅ 0.2 = 25 1 1 x d. 5 xx − x ⋅x0.2 = 3 25 e. 4 − 3 −0.5 = 3x + 0.5 − 22 x −1 e. 4 x x− 3x −0.5 =x 3x + 0.5 −x+212 x −1 f. 27 − 13 ⋅ 9 + 13 ⋅ 3 − 27 = 0 f. 27 x −x 13 ⋅ 9xx + 13 ⋅ 3x+x 1 − 27 = 0 g. 3 ⋅16 + 36 = 2 ⋅ 81 g. 3 ⋅16 x + 36 x = 2 ⋅ 81x kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 6.3 Logarithmic equations 6 .3 Logarithmic Equations 6.3 Logarithmic Equations 6.3 Logarithmic equations 6 .3 Logarithmic Equations 6.3.1 BaSic ProPertieS 6 .3 .1 BASIc ProPerTIeS 6.3.1 Basic Properties 6.3.1 BaSic ProPertieS 6 .3 .1 BASIc ProPerTIeS The simplest logarithmic equation is of the form log a x = b . Its solution is x = a b . For example, the solution to the equation log 2 x = 4 is x = 2 4 = 16 . Logarithmic equations can, however,b be more complicated. For The simplest logarithmic equation is of the form log a x = b . Its solution is x = a . For example, the solution example, to solve the equation log4 2 x + log 4 x 2 = 17 we need to apply some properties of logarithms. to the equation log 2 x = 4 is x = 2 = 16 . Logarithmic equations can, however, be more complicated. For 2 example, the equation we need to only applyfor some 2 x + loglog 4 xa x= 17 a >properties 0 and a ≠ of 1 , logarithms. is meaningful and its domain Before wetodosolve so, recall that thelog function is the positive real numbers. Also, before proceeding, you should make sure that you understand the Before we do so, recall that the function log a x is meaningful only for a > 0 and a ≠ 1 , and its domain meaning of log c = b . Here are a few identities that will help you review this meaning: is the positive areal numbers. Also, before proceeding, you should make sure that you understand the meaning = bsame . Here a few that will help you review this meaning: log 2 32of=log 5 isa cthe as are saying 25identities = 32 . 1 = 5 is the same as saying 25 =−332 . 1 log 24 32 = −3 is the same as saying 4 = . 64 64 1 1 1 −3 log84 2 2==−3 isis the the same same as as saying saying 4 8 == 2 .2 . 64 64 2 1 2 =is not is the same as saying 8 = 2 is2no . real number b such that 6b = −36 . log 68 (2−36) meaningful because there 2 b such that −36)now is not because there is no real In number 6b = −36 We log will6 (turn tomeaningful the basic properties of logarithms. the following, 1) through 8),.let a be a positive 84 USAD Mathematics Resource Guide • 2016–2017 real number different from 1. Then, We will turn now to the basic properties of logarithms. In the following, 1) through 8), let a be a positive Rancho Cucamonga High School - Ontario, CA 2 log 6 ( −36) is not meaningful because there is no real number b such that 6b = −36 . We will turn now to the basic properties of logarithms. In the following, 1) through 8), let a be a positive real number different from 1. Then, 1) blogbea 1a=positive 0 Let real number. Then log Let beaa baa==positive real number. Then log 2) ba 3) b1 log Let bea b a=positive real number. Then 3) b b ba be Let a positive real number and pis be any real number. Then, Unauthorized Duplication Prohibited Outside the Terms of Your License Agreement log a b ® a be b ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce Guide a=ppositive real number and p be any mAthemAtics real number. Then, 4) blog a b = p log a b 3) 80 80 Let –2014 2013 2010–2011 Let be ab ppositive and p be any real number. Then, 4) b = p log areal b number blog Let bea a positive real number and q be any real number different from zero. Then, 4) blogbea bap positive = 1p log bb number and q be any real number different from zero. Then, Let 5) log a q b = log aareal q 1 Let ab positive and q be any real number different from zero. Then, b number = log a real 5) b blogbe Let and real numbers. Then, a q d be positive 1q Rancho Cucamonga High School - Ontario, CA b = log 5) blogand ab Let positive real numbers. Then, 6) log aa qbdd =be qlog a b + log a d b d =belog Let positive 6) blog logand bd logreal d numbers. Then, 7) = log a abb−+log aa a ad db 6) log log a bd log abb−+log log add 7) ==log Let b bea a positive different from 1 , and let c be any positive real number. Then, a real number a db 7) log =log logbacbreal − lognumber a d Let bea ac positive different from 1 , and let c be any positive real number. Then, 8) blog a d= log a logbb c real number different from 1 , and let c be any positive real number. Then, Let be ac positive 8) blog Property (8) a is=called the Change of Base formula. log logbb ac 8) log Property (8) a cis=called the Change of Base formula. log byou a will be guided as to how to prove these properties. In the next section, Property (8) is called the Change of Base formula. In the next section, you will be guided as to how to prove these properties. 6.3.2 hoW We SoLve eQuationS? 6 .3 .2 hoW DoDo Wewill SoLve LoGArIThmIc eQuATIonS? In the next section, you be guided as to LoGarithMic how to prove these properties. 6.3.2 hoW SoLve LoGarithMic eQuationS? 6 .3 .2 hoW DoDo We We SoLve LoGArIThmIc eQuATIonS? 6.3.2 How Do We WeWe Solve Logarithmic Equations? 6.3.2 hoW SoLve LoGarithMic eQuationS? 6 .3 .2 hoW DoDo SoLve LoGArIThmIc eQuATIonS? ExamplE 6.3a: Solve the equation log5 x = 3 . ExamplE 6.3a: Solve the equation log5 x = 3 . ExamplE Solve the equation log5 x = 3 log . 5 x = 3 if and only 53 = x . Hence, x = 53 = 125 solution:6.3a: By the definition of a logarithm, is the solution By to this equation. of a logarithm, log5 x = 3 if and only 53 = x . Hence, x = 53 = 125 solution: the definition is the solution By to this equation. of a logarithm, log5 x = 3 if and only 53 = x . Hence, x = 53 = 125 solution: the definition is the solution to this equation. . ExamplE 6.3b: Solve the equation log10 ( x 2 − 9 x + 10) = log10 ( x − 6) . ExamplE 6.3b: Solve the equation log10 ( x 2 − 9 x + 10) = log10 ( x − 6) . ExamplE 6.3b: Solve . set of the positive real ( x 2logarithmic − 9 x + 10) =function log10 ( x −is6)the solution: Recall thatthe theequation domainlog of 10the numbers. Hence, we must the following conditions: solution: Recall that state the domain of thetwo logarithmic function is the set of the positive real 1. x 2 − 9Hence, x + 10Recall >we 0 must numbers. the following conditions: solution: that state the domain of thetwo logarithmic function is the set of the positive real 2. 1. x 2−−69>Hence, x 0+ 10 >we 0 must state the following two conditions: numbers. Since function is one-to-one, log10 ( x 2 − 9 x + 10) = log10 ( x − 6) if and only 0+ 10 > 0 2. 1. x 2−the −69>xlogarithmic 2 Mathematics Resource Guide x 2 −x9−xthe 6. Since function is one-to-one, log10 (•x2016–2017 − 9 x + 10) = log10 ( x − 6) if and only 6+ >10logarithmic 0= x − USAD 2. 2 2 85 numbers. Hence, we must state the following two conditions: 1. x 2 − 9 x + 10 > 0 2. x−6>0 Since the logarithmic function is one-to-one, log10 ( x 2 − 9 x + 10) = log10 ( x − 6) if and only Solving the last equation, we get two solutions: x = 8 or x = 2 . x 2 − 9 x + 10 = x − 6 . x = 8 or1xand = 2 .2 above. Solving last whether equation, we get twosatisfy solutions: Now, wethe check these values inequalities Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement ® 2013 –2014we AcAdemic decAthlon mAthemAtics ResouRce Guide 2010–2011 check MatheMatics ResouRce Guide satisfy inequalities Now, whether these values 1 and 2 above. Substituting x = 8 in these inequalities, we get: 81 81 82 − 9 ⋅ 8 + 10 = 2 > 0 Substituting x = 8 in these inequalities, we get: 82− 6 = 2 > 0 8 − 9 ⋅ 8 + 10 = 2 > 0 x = 8 satisfies conditions 1 and 2, and hence it is a solution. 8−6 = 2 > 0 x = 8 satisfiesx conditions 1 and 2, hence = 2 in inequality 1, and we get Substituting 22 −it9is⋅ 2a+solution. 10 = −4 < 0 Since x = 2 does not satisfy at least one of conditions 1 and 2 , it is not a solution. ExamplE 6.3c: Solve the equation log 3 x + log 3 ( x + 8) = 2 . ExamplE 6.3c: Solve the equation log 3 x + log 3 ( x + 8) = 2 . solution: log 3 x + log 3 ( x + 8) = 2 Condition: solution: log 3 x + log 3 ( x + 8) = 2 1. x > 0 Condition: 2. x + 8 > 0 1. x > 0 log 3xx+( x8+>8) By Property 6 0=2 2. log x ( x3 +x (8)x += 8) 32 = 2 By Property 6 2 xx (2 x++88) x ==93 2 + 88xx −= 99 = 0 xx 2 + xx 2=+−88x ±− 9 64 = 0+ 36 2 − 8 ± 64 + 36 = 8 ± 10 xx = 2 2 8 ± 10 xx = = 9 , 2x = −1 x = 9 ,xx==−−11violates condition 1, the only solution our equation has is x = 9 . Since Since x = −1 violates condition 1, the only solution our equation has is x = 9 . 86 USAD Mathematics Resource Guide • 2016–2017 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement Rancho Cucamonga High School - Ontario, CA x = 2not in satisfy inequality 1, we getof22conditions Substituting − 9 ⋅ 2 + 10 1= and −4 <2 0, it is not a solution. at least one Since x = 2 does ExamplE 6.3d: Solve the equation log 3 x + 2 log 3 ( x + 8) = 4 solution: Condition: 1. x>0 2. x +8> 0 log 3 x + log 3 ( x + 8) 2 = 4 By Property 4 log 3 x ( x + 8) 2 = 4 By Property 6 Rancho Cucamonga High School - Ontario, CA log 3 x + 2 log 3 ( x + 8) = 4 x ( x + 8) 2 = 43 x 3 + 16 x 2 + 64 x = 81 x 3 + 16 x 2 + 64 x − 81 = 0 By the Rational Root Theorem, since x = 1 is a solution to the last equation, the polynomial x 3 + 16 x 2 + 64 x − 81 is divisible by x − 1 . When dividing x 3 + 16 x 2 + 64 x − 81 by x − 1 , we get x 2 + 17 x + 81 . Hence, the last equation can be written as ( x − 1)( x 2 + 17 x + 81) = 0 . Equation x 2 + 17 x + 81 = 0 has no real solution (verify). Therefore x = 1 is the only solution to the last equation. Since x = 1 satisfies conditions 1 and 2, it is the (only) solution to the given equation. . ExamplE 6.3e: Solve the equation log 32 ( x − 1) − log 3 ( x − 1) − 2 = 0 solution: Condition: x − 1 > 0 . Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 USAD Mathematics Resource Guide • 2016–2017 ® AcAdemic MatheMatics ResouRcedecAthlon Guide mAthemAtics ResouRce Guide 87 83 83 ( ) 22 log3232(x (x 1) 1) == log log33(x (x 1) 1) . We note that log Let log 3 ( x − 1) = t . We have the quadratic equation t 2 − t − 2 = 0 . Solving for t we get t = −1 or t = 2 . Hence: log 3 ( x − 1) = −1 or log 3 ( x − 1) = 2 . Solving each of these two equations, we get: 3−1 = x − 1 or 32 = x − 1 x= 4 or x = 10 . 3 Rancho Cucamonga High School - Ontario, CA Since these two values satisfy the condition x − 1 > 0 , they are both solutions to the given equation. ExamplE 6.3f: Solve the equation ln x + ln 2 x = eln1 solution: Condition: x > 0 . First we note that eln1 = 1 (why?) ln x + ln 2 x = eln1 ln x ⋅ 2 x = 1 (why?) 2x 2 = e x2 = e 2 x=± e 2 Since x > 0 , the equation has only one solution: x = 84 88 e . 2 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement ® mAthemAtics Guide AcAdemic decAthlon USAD Mathematics Resource ResouRce Guide • 2016–2017 MatheMatics ResouRce Guide –2014 2013 2010–2011 Section 6.3 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk Section 6 .3: Section 6.3: exercises Exercises Exercises kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 1. Simplify the following expressions: a. log 3 b. log c. 3 3 3 3 25 − log 3 7 58 81 log10 8 + log10 18 2 log10 2 + log10 3 Rancho Cucamonga High School - Ontario, CA d. log 3 2 ⋅ log 4 3 ⋅ log 5 4 ⋅ log 6 5 ⋅ log 7 6 ⋅ log8 7 (Hint: Use the Change of Base Formula.) 2. Use the Change of Base Formula and a calculator to evaluate the following logarithms, correct to four decimal places: a. log 3 19 b. log 2 3 3. Solve the following equations: a. log 0.5 x = −1 b. log 22 ( x + 1) − log 1 ( x + 1) = 5 4 c. log 2 x + log 4 x + log8 x = 11 d. log 2 x + 4 =5 log x 2 e. 2 log10 x − log10 4 = − log10 (5 − x 2 ) f. ln x + log x e = 2 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 6.4 radical equations 6 .4 Radical Equations The following are examples of radical equations: k x+5 = 2 USAD Mathematics Resource Guide • 2016–2017 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 AcAdemic decAthlon ® mAthemAtics ResouRce Guide 89 85 f. ln x + log x e = 2 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 6.4 radical equations 6 .4 Radical Equations 6.4 Radical Equations The following are examples of radical equations: k x+5 = 2 k x −4 + 2− x =1 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement ® 2013 –2014 AcAdemic mAthemAtics ResouRce Guide 2010–2011 ResouRcedecAthlon Guide k 5 x − 1− xMatheMatics =1 k 85 2 3 x + 3 x =7 In all these equations, the variable x appears under the radical sign, and none of them is of the types we The fact that an equation includes a radical does not necessarily make it a radical equation. For example, the following equations are not radical equations: k x− 2 = 5 k x2 + 3 3 = 5 k x2 − 2 x = 3 The first two are not radical because the variable x does not appear under the radical sign. The third equation is not radical because it is equivalent to x − 2 x = 3 , which is not radical. 6.4.1Method MethoD 6 .4 .1 meThoD 6.4.1 11 1 One useful method for solving a radical equation of the form n u = n v is to raise both sides of the equation to the power n . In this way, we get a simpler equation of the form u = v . We must be careful, however, for the two equations may not be equivalent. This is the case because not every solution to the second equation is necessarily a solution to the first equation. For example, consider the following equation: x+2 = x. 1. Squaring both sides, we get x + 2 = x 2 2. Solving this quadratic equation, we get two solutions: x = −1 and x = 2 . 3. Note that while x = 2 is a solution to our equation, x = −1 is not. Let’s demonstrate this method further by working through several examples: 90 ExamplE 6.4a: Solve theUSAD Mathematics 5 x − 1 − Resource 3x − 2 = Guide 2 x + 1• 2016–2017 equation Rancho Cucamonga High School - Ontario, CA have learned about so far. In this section of the resource guide, we will learn how solve such equations. ExamplE ExamplE 6.4a: 6.4a: Solve Solve the the equation equation 55xx −−11 −− 33xx −− 22 == 22xx ++11 solution: solution: ( ) 55xx −−11 −− 33xx −− 22 == 22xx ++11 .. 2 5x −1 − 2 5x −1 ⋅ 3x − 2 + ( 3x − 2 ) =( 2 2x +1 ) 2 Square both sides the Terms of Your License Agreement 5 x − 1 − 2 (5 x −Unauthorized 1)(3x − 2) Duplication + 3x − 2 =isis2Prohibited x + 1 Outside Unauthorized Duplication Prohibited Outside the Terms of Your License Agreement ® –2014 Guide AcAdemic MatheMatics ResouRce Guide 2013 2010–2011 –2014 mAthemAtics ResouRce ResouRce Guide AcAdemic decAthlon decAthlon ® mAthemAtics MatheMatics ResouRce Guide 2013 2010–2011 8 x − 3 − 2 (5 x − 1)(3x − 2) = 2 x + 1 −2 (5 x − 1)(3x − 2) = −6 x + 4 2 (5 x − 1)(3x − 2) = 6 x − 4 15 x 2 − 13x + 2 = 3x − 2 15 x 2 − 13x + 2 = 9 x 2 − 12 x + 4 Rancho Cucamonga High School - Ontario, CA 86 86 86 86 Square both sides 6x2 − x − 2 = 0 2 1 , x=− 3 2 1 x = − cannot be a solution to our original equation because some of the expressions in the 2 1 original equation are not defined for this value. (For example, when x = − is substituted in 2 the expression 5 x − 1 , we get a square root of a negative number.) On the other hand, all the 2 expressions in our original equation are defined for x = . 3 2 Hence x = is the only solution to our original equation. 3 x= ExamplE 6.4b: Solve the equation x − 5 + 3 − x = 1 . solution: ( ) 2 x − 5 + 2 x − 5 ⋅ 3− x + ( 3− x ) =1 2 Square both sides x − 5 + 2 x − 5 ⋅ 3− x + 3− x = 1. 2 − x 2 + 8 x − 15 = 3 − x 2 + 8 x − 15 = 3 2 USAD Mathematics Resource Guide • 2016–2017 91 2 − x 2 + 8 x − 15 = 3 − x 2 + 8 x − 15 = 3 2 99 2 − − xx 2 + + 88 xx − − 15 15 = = 44 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement ® 4 69 x ResouRcedecAthlon Guide 4 xx − − 32 32 x+ +MatheMatics 69 = =0 0 AcAdemic 2 2 2013 –2014 2010–2011 mAthemAtics ResouRce Guide 87 This This equation equation has has no no solution. solution. Hence, Hence, our our original original equation equation too too has has no no solution. solution. 6.4.2 MethoD 6 .4 .2 meThoD 6.4.2 MethoD 2 6 .4 .2 meThoD 6.4.2 Method 222 2 ExamplE ExamplE 6.4c: 6.4c: Solve Solve the the equation equation solution: solution: Note Note that that if if we we let let 33 33 2 3 (( xx − − 1) 1) 2 − − 3 xx − − 11 = = 12 12 .. 2 xx − − 11 = = tt ,, we we will will get get the the quadratic quadratic equation: equation: tt 2 − − tt = = 12 12 .. Solving = 44 or =− −33 .. Solving for for tt ,, we we get get that that tt = or tt = Therefore, Therefore, 33 xx − − 11 = = 44 or or 33 xx − − 11 = =− −33 .. Raising − 11 = = 64 64 or − 11 = =− −27 27 or xx − Raising both both sides sides of of each each equation equation to to the the third third power, power, we we get get xx − xx = = 65 65 or =− −26 26 .. or xx = By By checking checking each each of of these these values values in in the the original original equation, equation, we we can can conclude conclude that that both both x x= = 65 65 and and xx = =− −26 26 are are solutions solutions to to our our equation. equation. k kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk k Section Section 6 .6.4 4: Section 6.4: exercises Exercises Exercises k kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk k 1. 1. Solve Solve the the following following equations: equations: 92 a. a. 2 33 − − 22 xx 2 = = 11 b. b. 4 55 + + 22 xx = = 10 10 − − 33 4 55 + + 22 xx 2 c. + 1) 1) xx 2 − − xx − − 66 = = 66 xx + + 66 c. (( xx + USAD Mathematics Resource Guide • 2016–2017 Unauthorized Unauthorized Duplication Duplication is is Prohibited Prohibited Outside Outside the the Terms Terms of of Your Your License License Agreement Agreement Rancho Cucamonga High School - Ontario, CA Now Now we we will will learn learn aa second second method method for for solving solving radical radical equations. equations. It It is is called called solution solution by by substitution. substitution. a. 3 − 2x = 1 b. 5 + 2 x = 10 − 3 4 5 + 2 x c. ( x + 1) x 2 − x − 6 = 6 x + 6 d. x − 3 + Unauthorized 6 − x = 3 Duplication is Prohibited Outside the Terms of Your License Agreement 88 88 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide e. x + x + 11 + x − x + 11 = 4 f. 3 x − 15 − 2 x = 3 − 5 − x –2014 2013 2010–2011 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 7 .1 Linear Inequalities 7.1 Linear Inequalities In Section 1, we listed the important set of properties of the real numbers concerning inequalities. These are: 1. For any two real numbers x and y , one and only one of the following is true: x < y , x = y , or y < x . 2. For any three real numbers x , y , and z , if x < y , y < z , then x < z . 3. For any three real numbers x , y , and z , if x < y then x + z < y + z . 4. For any three real numbers x , y , and z , if x < y and z > 0 , then zx < zy . 5. For any three real numbers x , y , and z , if x < y and z < 0 , then zx > zy . These properties are needed to solve inequalities. Let’s consider a few examples: ExamplE 7.1a: Solve the inequality 3x − 4 > 5 . We are asked to find the set of all numbers x that satisfy this inequality. Solution: 3x − 4 > 5 Add 4 to both sides of this inequality to get the following inequality: 3x > 9 Divide both sides of the inequality by 3 to get the following inequality: x>3 The solution set consists of all the numbers that are greater than 3. For example, x = 4 satisfies the inequality because 3 ⋅ 4 − 4 = 8 > 5 . On the other hand, 2 does not satisfy the inequality because 3⋅ 2 − 4 = 2 < 5 . USAD Mathematics Resource Guide • 2016–2017 93 Rancho Cucamonga High School - Ontario, CA section 7: inequalities SecTIon 7: Inequalities d. x −3 + 6− x = 3 e. x + x + 11 + x − x + 11 = 4 f. 3 x − 15 − 2 x = 3 − 5 − x Section k k k7 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk section 7: inequalities SecTIon 7: Inequalities Inequalities In Section 1, we listed the important set of properties of the real numbers concerning inequalities. These are: 1. For any two real numbers x and y , one and only one of the following is true: x < y , x = y , or xy >< y.x . 2. For any three real numbers x , y , and z , if x < y , y < z , then x < z . 3. For any three real numbers x , y , and z , if x < y then x + z < y + z . 4. For any three real numbers x , y , and z , if x < y and z > 0 , then zx < zy . 5. For any three real numbers x , y , and z , if x < y and z < 0 , then zx > zy . These properties are needed to solve inequalities. Let’s consider a few examples: ExamplE 7.1a: Solve the inequality 3x − 4 > 5 . We are asked to find the set of all numbers x that satisfy this inequality. Solution: 3x − 4 > 5 Add 4 to both sides of this inequality to get the following inequality: 3x > 9 Divide both sides of the inequality by 3 to get the following inequality: x>3 The solution set consists of all the numbers that are greater than 3. For example, x = 4 satisfies the inequality because 3 ⋅ 4 − 4 = 8 > 5 . On the other hand, 2 does not satisfy the inequality because 3⋅ 2 − 4 = 2 < 5 . 94 2013 –2014 2010–2011 USAD Mathematics Resource Guide 2016–2017 Unauthorized Duplication is Prohibited Outside the Terms of Your• License Agreement AcAdemic MatheMatics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide 89 89 Rancho Cucamonga High School - Ontario, CA 7 .1 Linear Inequalities 7.1 Linear Inequalities 7.1 Linear Inequalities ExamplE 7.1b: Solve the inequality 8 x − 2 > 2 x − 3 . Solution: 8x − 2 > 2 x − 3 Add 2 to both sides of this inequality to get the following inequality: 8x > 2 x − 1 Add −2x to both sides of this inequality to get the following inequality: 6 x > −1 Divide both sides of this inequality by 6 to get the following inequality: 1 6 Thus, the solution set of the given inequality consists of all the numbers that are greater than − 1 . 6 ExamplE 7.1c: Solve the inequality −3x + 12 > 45 . Solution: −3x + 12 > 45 Add −12 to both sides of this inequality to get the following inequality: −3x > 33 Divide both sides of the inequality by −3 to get the following inequality: x < −11 If you feel you understand the process of solving an inequality, there is no need to explain each step. Nor is there a need to say explicitly the meaning of the last expression. The answer x < −11 is understood to mean “the solution set of the given inequality consists of all the numbers that are smaller than −11 .” 1 8 ExamplE 7.1d: Solve the inequality 4( x − 12.4) ≤ 8 x − . Solution: You should justify each step in the following process on your own: 4( x − 12.5) ≤ 8 x − 1 8 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 90 90 mAthemAtics ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 ® –2014 2013 2010–2011 95 Rancho Cucamonga High School - Ontario, CA x>− 4 x − 50 ≤ 8 x − 1 8 32 x − 400 ≤ 64 x − 1 −32 x − 400 ≤ −1 −32 x ≤ 399 399 15 = −12 32 32 Rancho Cucamonga High School - Ontario, CA x≥− ExamplE 7.1e: Solve the inequality 2( − x + 3) > −2 x + 19 . Solution: 2( − x + 3) > −2 x + 19 −2 x + 6 > −2 x + 19 Look carefully at this inequality. Clearly, there is no number that satisfies it because if there were such a number, then we would get 6 > 19 , which is not true. Thus, the given inequality has no solution. ExamplE 7.1f: Solve the inequality −22(3x − 3) > −6(11x + 1) − 35 . Solution: −22(3x − 3) > −6(11x + 1) − 35 −66 x + 66 > −66 x − 6 − 35 −66 x + 66 > −66 x − 41 Since 66 > −41 , the given inequality has infinitely many solutions: any number x satisfies it. 96 –2014 2013 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource ResouRce Guide • 2016–2017 AcAdemic decAthlon mAthemAtics Guide MatheMatics ResouRce Guide ® 91 91 5( x − 3) > 3x − 7 ExamplE 7.1g: Solve the system of inequalities 2 x + 3 > 36 − x Solution: This time, we are looking for the set of all the numbers that satisfy two inequalities. We maintain the two inequalities together while solving each separately: 5 x − 15 > 3x − 7 2 x + 3 > 36 − x 5 x − 3x > −7 + 15 2 x + x > 36 − 3 x > 4.5 x > 11 The solution set of the given system consists of all the numbers that are greater than 4.5 and greater than 11. Clearly, any number that is greater than 11 is also greater than 4.5. Thus, the solution set is the set all the numbers x that are greater than 11. Or we simply write: x > 11 . Some linear inequalities involve absolute values. The absolute value of a number is the distance of the number from zero on the number line. For example, the absolute value of 5 , is 5 , and the absolute value of −5 is also 5. The absolute value of zero is obviously zero. The absolute value of a number x is denoted by x . The absolute value of a number x depends on where x is located on the number line: if x is located to right of 0 , then x = x ; if x is located to left of 0 , then x = − x ; and if x is zero, then x = 0 . The above paragraph can be written in more compact form: if x > 0 x x = 0 if x = 0 − x if x < 0 The outputs of the absolute-value function are always non-negative. Its graph consists of two linear functions: y = x for x ≥ 0 , and y = − x for x < 0 . (See Figure 7–1.) Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 92 ® USAD Mathematics Resource Guide • 2016–2017 mAthemAtics ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce Guide 97 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA 2 x > 9 3x > 33 Rancho Cucamonga High School - Ontario, CA y = | x| Figure 7–1 Let’s now solve a few linear inequalities involving absolute values. ExamplE 7.1h: Solve x < 3 . Solution: We are looking for all the numbers x whose distance from zero is smaller than 3. The only numbers x that satisfy this inequality are −3 < x < 3 . –3 3 0 ExamplE 7.1i: Solve x ≥ 6 . Solution: We are looking for all the numbers x whose distance from zero is greater than or equal to 6. The only numbers x that satisfy this inequality are: x ≥ 6 or x ≤ −6 . –6 98 2013 –2014 2010–2011 0 6 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource Guide • 2016–2017 ® AcAdemic MatheMatics ResouRcedecAthlon Guide mAthemAtics ResouRce Guide 93 ExamplE 7.1j: Solve x + x < 1 . 2 Solution: This inequality is a bit harder. The algebraic definition of absolute value can be very helpful in solving this kind of inequality. We will deal with two cases: CaSE 1: x ≥ 0 . In this case x = x , and so the given inequality is: x + x < 1 2 Solving: 2x < 1 4 So, when x ≥ 0 the solution of the given inequality is x < Hence, any x where 0 ≤ x < Rancho Cucamonga High School - Ontario, CA x< 1 2 1 . 4 1 is a solution to the inequality. 4 CaSE 2: x < 0 . In this case, x = − x , and so the given inequality is: x − x < 1 . 2 Solving: 0⋅ x < 1 2 1 We are looking for all the numbers x for which 0 ⋅ x < . Obviously, any multiple of zero is zero, 2 1 1 which is less than . Therefore, any number x satisfies the inequality 0 ⋅ x < . 2 2 But, remember that we are dealing with the case x < 0 . Hence the solution in this case is x < 0 . Summarizing: The solution to the given inequality consists of two sets of numbers: 1 The set of all the numbers x for which 0 ≤ x < 4 The set of all the numbers x for which x < 0 . Combining these two sets, we get: The solution to the given inequality is x < 1 . 4 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 94 USAD Mathematics Resource Guide • 2016–2017 ® Guide AcAdemic decAthlon mAthemAtics ResouRce MatheMatics ResouRce Guide 99 –2014 2013 2010–2011 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk Section 7 .1: SectIon 7.1: exercises Section 7 .7.1 1: Exercises Section SectIon 7.1: exercises kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk Exercises Exercises kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 1. Solve the inequalities: 1. Solve the inequalities: a. −3 x + 21 > 0 a. −3 x + 21 > 0 b. 2( x − 2) − 5(1 − 3 x) < 2 c. 6(2 − x) ≥ 5 − 2 x 2x − 1 < 3 − 2(1 − 2 x) d. 23x − 1 < 3 − 2(1 − 2 x) d. 3 − 3 − 7 x + x + 1 < 14 − 7 − 8 x e. 13 107 x x 2+ 1 3− 7 −28 x + < 14 − e. 13 − 10 2 2 x 7 5x 7 2 − 4 > 2 − 8 x 7 5x 7 f. − > 22 x +41 2 −1 − 2x < 5− 8 f. 2 x4+ 1 1 −32 x < 5− 4 3 g. x − 2 ≤ 5 − x g. x − 2 ≤ 5 − x kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 7.2 Quadratic Inequalities 7 .2 Inequalities 7.2Quadratic Quadratic Inequalities 7.2Quadratic Quadratic Inequalities 7 .2 Inequalities The following are examples of quadratic inequalities: The following are examples of quadratic inequalities: k x 2 − 3x + 6 < 0 k k x 2 − 3x + 6 < 0 5 x 2 < 125 k k 5 x 2 < 125 ( x − 3)( x + 5) > 0 k k ( x −2 3)( x + 5) > 0 − x − 3x ≤ 0 k k − x 2 − 3x ≤ 0 (− x + 5)(3 − x) ≤ 6 (− x + 5)(3 − x) ≤ 6 Any quadratic inequality can be brought by algebraic operations into one of two forms: k Any quadratic inequality can be brought by algebraic operations into one of two forms: 2 1. ax + bx + c > 0 and a > 0 ax 22 + bx + c > 0 and a > 0 1. ax + bx + c < 0 and a > 0 2. 2 2. ax + bx + c < 0 and a > 0 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 2010–2011 100 –2014 AcAdemic decAthlon mAthemAtics ResouRce Guide MatheMatics ResouRce Guide USAD Mathematics Resource Guide 2016–2017 Unauthorized Duplication is Prohibited Outside the Terms of Your• License Agreement 2013 –2014 2010–2011 AcAdemic MatheMatics ResouRcedecAthlon Guide ® ® mAthemAtics ResouRce Guide 95 95 Rancho Cucamonga High School - Ontario, CA b. 2( x − 2) − 5(1 − 3 x) < 2 c. 6(2 − x) ≥ 5 − 2 x For example, the inequality −2 x 2 + 3 x − 9 > 0 can be changed into the second form by multiplying both sides of the inequality by −1 . It becomes: 2 x 2 − 3 x − 9 < 0 . (Note that a , the coefficient of x 2 , is positive). 7.2.1 IneQuaLItIeS the a0 7 .2 .1 Inequalities IneQuALITIeS ofof The form a0 7.2.1 of the Form ax2form + ax bx2bxc0 +ax c >2bxc0 0 andAnD a > and 0 We first observe that when a > 0 , the function ax 2 + bx + c has a minimum. Next we ask: Does the parabola intersect the x-axis? And, if it does, at how many points does it intersect the x-axis? Recall that the answer to these questions depends on the value of the discriminant ∆ = b2 − 4ac . Case 1: When ∆ = b2 − 4ac < 0 and a > 0 , the parabola does not intersect the x-axis, and it looks like Figure 7–2 This implies that for any value of x , ax 2 + bx + c > 0 . Hence: When ∆ = b2 − 4ac < 0 and a > 0 , the solution to the inequality ax 2 + bx + c > 0 is −∞ < x < ∞ . Case 2: When ∆ = b2 − 4ac = 0 , the parabola touches the x-axis at one point: x = − b 2a bb 2a 2a Figure 7–3 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 96 96 USAD Mathematics Resource Guide • 2016–2017 ® mAthemAtics ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce Guide 101 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA this: b , ax 2 + bx + c > 0 . Hence: When 2a b . ∆ = b2 − 4ac = 0 and a > 0 , the solution to the inequality ax 2 + bx + c > 0 is any number x ≠ − 2a This implies that for any value of x that is different from − Case 3: When ∆ = b2 − 4ac > 0 and a > 0 , the parabola intersects the x-axis at two points: −b − b2 − 4ac −b + b2 − 4ac and x2 = . 2a 2a x < x1 x > x2 x1 x2 Figure 7–4 This implies that for any value of x for which x > x2 or x < x1 , ax 2 + bx + c > 0 . Hence: When ∆ = b2 − 4ac > 0 and a > 0 , the solution to the inequality ax 2 + bx + c > 0 is x < x1 or x > x2 . ExamplE 7.2a: Solve the inequality x 2 + 6 x > −10 . Solution: First we bring this inequality into standard form ax 2 + bx + c > 0 by adding 10 to each side: x 2 + 6 x + 10 > 0 We find the value ∆ for x 2 + 6 x + 10 : ∆ = 62 − 10 ⋅ 4 = 36 − 40 = −4 < 0 . Since a = 1 , the quadratic function has a minimum, and its graph is above the x-axis. Therefore, any real number x satisfies the given inequality. To show this, we write: −∞ < x < ∞ . 102 2013 –2014 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource Guide • 2016–2017 ® AcAdemic MatheMatics ResouRcedecAthlon Guide mAthemAtics ResouRce Guide 97 Rancho Cucamonga High School - Ontario, CA x1 = ExamplE 7.2b: Solve x 2 + 6 x + 9 > 0 . Solution: We find the value of ∆ for x 2 + 6 x + 9 ∆ = 62 − 9 ⋅ 4 = 36 − 36 = 0 . Since a = 1 , the graph of the quadratic function touches the x-axis from above at one point. To find this point, we solve: x 2 + 6 x + 9 = 0 . Solving, we get x = −3. ExamplE 7.2c: Solve x 2 + 6 x + 5 > 0 . Solution: The value of ∆ for x 2 + 6 x + 5 is ∆ = 62 − 5 ⋅ 4 = 36 − 20 = 16 > 0 Since a = 1 , the function has a minimum, and it intersects the x-axis at two points. These points are the solutions to the equation x 2 + 6 x + 5 = 0 . Solving, we get: x= −6 ± 4 2 x = −5 or x = −1 . –5 –1 For x < −5 or x > −1 , y = x 2 + 6 x + 5 is positive. Thus, the solution to the given inequality is the set: x < −5 or x > −1 . Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 98 98 USAD Mathematics Resource Guide • 2016–2017 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide 103 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA This implies that any x ≠ –3 satisfies the given inequality. 2 2 7.2.2 IneQuaLItIeS a0 7 .2 .2 IneQuALITIeS ofof Thethe form a0 7.2.2 Inequalities of the Form ax2form +ax bxbxc0 +ax c <bxc0 0 andAnD a > and 0 Recall that when a > 0 the function ax 2 + bx + c has a minimum. As before we ask: Does the parabola intersect the x-axis, and if it does, at how many points does it intersect the x-axis? The answer to this question depends on the value of the discriminant ∆ = b2 − 4ac . As before, we have three cases: Figure 7–5 This implies that there is no x for which ax 2 + bx + c < 0 . Hence: When ∆ = b2 − 4ac < 0 and a > 0 , the inequality ax 2 + bx + c < 0 has no solution. Case 2: When ∆ = b2 − 4ac = 0 , the parabola touches the x-axis at one point: x = − – b 2a b 2a Figure 7–6 This implies that there is no x for which ax 2 + bx + c < 0 . Hence: When ∆ = b2 − 4ac = 0 and a > 0 , the inequality ax 2 + bx + c < 0 has no solution. 104 –2014 2013 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource Guide • 2016–2017 ® mAthemAtics ResouRce Guide AcAdemic MatheMatics ResouRcedecAthlon Guide 99 99 Rancho Cucamonga High School - Ontario, CA Case 1: When ∆ = b2 − 4ac < 0 , the parabola does not intersect the x-axis, as is shown in Figure 7–5. Case 3: When ∆ = b2 − 4ac > 0 , the parabola intersects the x-axis at two points: x1 = −b − b2 − 4ac −b + b2 − 4ac and x2 = . 2a 2a x1 < x < x2 x1 x2 This implies that for any value of x between x1 and x2 , ax 2 + bx + c < 0 . Hence: When ∆ = b2 − 4ac > 0 and a > 0 , the solution to the inequality ax 2 + bx + c < 0 is x1 < x < x2 . ExamplE 7.2d: Solve the inequality x 2 + 8 x + 18 < 0 . Solution: ∆ = 82 − 18 ⋅ 4 = 64 − 72 = −8 < 0 Therefore, the graph of the function y = x 2 + 8 x + 18 does not intersect the x-axis. Since a = 1 , the graph is above the x-axis. Therefore, the given inequality has no solution. ExamplE 7.2e: Solve the inequality x 2 + 8 x + 16 < 0 . Solution: ∆ = 82 − 16 ⋅ 4 = 64 − 64 = 0 Therefore, the graph of the function y = x 2 + 8 x + 16 touches the x-axis at one point. Since a = 1 , the graph is above the x-axis. Hence, the given inequality has no solution. Unauthorized Duplication is Prohibited Outside the •Terms of Your License Agreement USAD Mathematics Resource Guide 2016–2017 100 100 ® Guide AcAdemic decAthlon mAthemAtics ResouRce MatheMatics ResouRce Guide 105 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA Figure 7–7 ExamplE 7.2f: Solve the inequality x 2 + 8 x + 7 < 0 . Solution: ∆ = 82 − 7 ⋅ 4 = 64 − 28 = 36 > 0 Therefore, the graph of the function y = x 2 + 8 x + 16 intersects the x-axis at two points. These points are the solutions to the equation x 2 + 8 x + 7 = 0 . Solving, we get x= −8 ± 6 2 x = −7 x = −1 Since a = 1 , the function y = x 2 + 8 x + 16 has a minimum. Hence, for −7 < x < −1 the inequality holds. Section 7.2 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk Section 7 .2: SectIon 7.2: exercises Exercises Exercises kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 1. Solve the inequalities: a. x 2 − 5 x + 6 < 0 b. 4 x 2 − 4 x + 1 ≤ 0 c. 81 − (3 + 2 x) 2 < 0 d. 3 + 2x ≤ x 2 e. 5 − x 2 ≥ −4 x f. 3 >0 2x −1 106 2013 –2014 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource ResouRce Guide • 2016–2017 AcAdemic decAthlon mAthemAtics Guide MatheMatics ResouRce Guide ® 101 101 Rancho Cucamonga High School - Ontario, CA or g. 2x − 9 <1 x−6 2. Find the domain of the following functions: a. y = (3 x − 2)( x − 5) b. y = 1 112 x + 64 + 49 x 2 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk section 8: coordinate Geometry SecTIon 8: coordinate Geometry through equations (and sometimes through inequalities). In this way, an investigation of an equation representing a geometric object can reveal important properties of the geometric object. The most basic geometric object is the point. A basic idea of coordinate geometry is the creation of a correspondence between the set of all ordered pairs of real numbers and the set of all points in the plane: For any point in the plane, there corresponds exactly one ordered pair of real numbers ( x, y ) , and, conversely, for any ordered pair of real numbers ( x, y ) , there corresponds exactly one point in the plane. 8.1 The Pythagorean 8 .1 The Pythagorean Theorem Theorem Before we start with coordinate geometry, we will first prove an ancient theorem, called the Pythagorean Theorem. We will need this theorem to prove the distance formula. Pythagorean Theorem: In a right triangle, the square of the hypotenuse is the sum of the squares of the other two sides of the triangle. Proof: Consider a right triangle with sides a and b and hypotenuse c as shown in Figure 8–1 below. We are to prove that a 2 + b 2 = c 2 . Figure 8–1 Unauthorized Duplication is Prohibited Outside the •Terms of Your License Agreement USAD Mathematics Resource Guide 2016–2017 102 102 ® Guide AcAdemic decAthlon mAthemAtics ResouRce MatheMatics ResouRce Guide 107 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA The goal of coordinate geometry is to describe geometric objects, such as lines and circles, algebraically g. 2x − 9 <1 x−6 2. Find the domain of the following functions: a. y = (3 x − 2)( x − 5) b. y = 1 Section 112 x + 64 + 49 x k k k8 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 2 section 8: coordinate Geometry Geometry SecTIonCoordinate 8: coordinate Geometry The goal of coordinate geometry is to describe geometric objects, such as lines and circles, algebraically representing a geometric object can reveal important properties of the geometric object. The most basic geometric object is the point. A basic idea of coordinate geometry is the creation of a correspondence between the set of all ordered pairs of real numbers and the set of all points in the plane: For any point in the plane, there corresponds exactly one ordered pair of real numbers ( x, y ) , and, conversely, for any ordered pair of real numbers ( x, y ) , there corresponds exactly one point in the plane. 8.1 The Pythagorean Theorem 8 .1 The Pythagorean TheoremTheorem 8.1 The Pythagorean Before we start with coordinate geometry, we will first prove an ancient theorem, called the Pythagorean Theorem. We will need this theorem to prove the distance formula. Pythagorean Theorem: In a right triangle, the square of the hypotenuse is the sum of the squares of the other two sides of the triangle. Proof: Consider a right triangle with sides a and b and hypotenuse c as shown in Figure 8–1 below. We are to prove that a 2 + b 2 = c 2 . c a b Figure 8–1 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 102 102 108 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA through equations (and sometimes through inequalities). In this way, an investigation of an equation Four copies of the right triangle are assembled to form Figure 8–2 below. In this figure we have two squares: the outer square with side a + b , and the inner square with side c . b a b c a c c a c b b a Let A be the area of the outer square: On the one hand, A = (a + b) 2 . On the other hand, A is the sum 1 of the areas of the four triangles and the area of the inner square. That is: A = 4 ab + c 2 2 1 Therefore, (a + b) 2 = 4 ab + c 2 . 2 This leads to: a 2 + 2ab + b 2 = 2ab + c 2 Hence: a 2 + b 2 = c 2 , as was required. The Converse of the Pythagorean Theorem: If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle. Proof: Let ABC be a triangle such that (1) AC 2 + BC 2 = AB 2 . We are to prove that C = 90° . Construct a triangle AB′C such that B′C is perpendicular to AC and B′C = BC , as is shown in Figure 8–3: A B' C B Figure 8–3 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 USAD Mathematics Resource Guide • 2016–2017 ® AcAdemic MatheMatics ResouRcedecAthlon Guide mAthemAtics ResouRce Guide 109 103 103 Rancho Cucamonga High School - Ontario, CA Figure 8–2 Since triangle AB′C is a right triangle, according to the Pythagorean Theorem, (2) AC 2 + B ' C 2 = AB '2 Since B′C = BC , we get (3) AC 2 + BC 2 = AB '2 From (1) and (3), we get AB′ = AB . The triangles ABC and AB′C are congruent by the side-side-side triangle congruence theorem (SSS). Hence C ' = C = 90 . 8.2 Points 8 .2 Points 8.2 Points Distance between two points: Given two points P ( x1 , y1 ) and Q( x2 , y2 ) , find the distance d between them. SoluTion: In Figure 8–4, QL ⊥ PR 2 By the Pythagorean Theorem, PQ = QL + PL 2 QL = x1 − x2 PL = y1 − y2 Therefore d = ( x1 − x2 ) 2 + ( y1 − y2 ) 2 This is called the distance formula. yy xx 1 P 1 dd yy11 xx22 L Q yy22 xx 0 R Figure 8–4 110 104 104 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource Guide • 2016–2017 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA This completes the proof. Midpoint: Given two points P ( x1 , y1 ) and Q( x2 , y2 ) , find the midpoint M ( x, y ) between them. SoluTion: We solve this problem for the case where the segment PQ is not parallel to the y-axis (see Figure 8–5). You can pursue the case where PQ is parallel to the y-axis on your own as an independent yy exercise. Q M P xx O 1 M (x 0) , 0) M11 (x, 1 Q Q1 (x(x2, 0) , 0) 1 2 Figure 8–5 Since PQ is not parallel to the y-axis, x1 ≠ x2 . From the points P, Q, and M, drop lines parallel to the y-axis, and let P1 ( x1 , 0) , Q1 ( x2 , 0) , and M 1 ( x, 0) be the intersections of these lines with the x-axis, respectively. From geometry, we know that the point M 1 is the midpoint of PQ 1 1 . This implies that PM 1 1 = M 1Q1 . Therefore, x − x1 = x − x2 . Solving for x , we get either x − x1 = x − x2 , which is not possible since x1 ≠ x2 , or x − x1 = −( x − x2 ) , which x +x implies that x = 1 2 . 2 y + y2 x + x2 y1 + y2 . Thus, M 1 In a similar way, we find that y = 1 ; is the midpoint between P ( x1 , y1 ) 2 2 2 and Q( x2 , y2 ) . Dividing a Segment into a Given ratio: Given two points P ( x1 , y1 ) and Q( x2 , y2 ) , find the point C on the PC m = , where m and n are two positive integers. CQ n PC m = (see Figure 8–6). SoluTion: Let C ( x, y ) be the point on the segment PQ such that CQ n segment PQ such that From the points P, C, and Q, drop lines parallel to the y-axis, and let P1 ( x1 , 0), Q1 ( x2 , 0), C1 ( x, 0) be the intersections of these lines with the x-axis. From geometry, we know that PC PC 1 1 . = C1Q1 CQ Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 AcAdemic MatheMatics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 105 105 111 Rancho Cucamonga High School - Ontario, CA P 0) P1 (x (x 1,, 0) Q C P PP1 (x1, 0) 1 (x 1 , 0) C1 (x, (x 0) , 0) 1 Q Q1 (x2, 0) 1 (x , 0) xx 2 Figure 8–6 Therefore, Rancho Cucamonga High School - Ontario, CA x − x1 m = x2 − x n n( x − x1 ) = m( x2 − x) nx − nx1 = mx2 − mx x(n + m) = nx1 + mx2 x= nx1 + mx2 n+m In a similar way, we find that y = ny1 + my2 . n+m nx + mx2 ny1 + my2 Thus, C 1 , . n+m n+m 8.3 Lines 8 .3 Lines 8.3 Lines In the previous section, we dealt with points in the plane. In this section, we will deal with straight lines. We will begin with the following problem: Perpendicular Bisector: Let P ( x1 , y1 ) and Q( x2 , y2 ) be two points in the plane, and let l be the perpendicular bisector of the segment PQ .1 Find the equation of l . 1 The perpendicular bisector of the segment PQ is the line that goes through the midpoint of PQ and is perpendicular to PQ. Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 106 112 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 –2014 2013 2010–2011 yy l A (xx,, yy) P (xx11,, yy11 ) (x2, yy22)) Q (x O xx Figure 8–7 P and Q . By the distance formula, AP = ( x − x1 ) 2 + ( y − y1 ) 2 AQ = ( x − x2 ) 2 + ( y − y2 ) 2 Thus, A( x, y ) is on l if and only if ( x − x1 ) 2 + ( y − y1 ) 2 = ( x − x2 ) 2 + ( y − y2 ) 2 . Squaring both sides and expanding, we get x 2 − 2 x1 x + x12 + y 2 − 2 y1 y + y12 = x 2 − 2 x2 x + x2 2 + y 2 − 2 y2 y + y2 2. The last equation is equivalent to 2( x2 − x1 ) x + 2( y2 − y1 ) y = x2 2 + y2 2 − x12 − y12 . Thus, we have obtained a formula for the equation of the perpendicular bisector of a segment given by its endpoints P ( x1 , y1 ) and Q( x2 , y2 ) . The formula is 2( x2 − x1 ) x + 2( y2 − y1 ) y = x2 2 + y2 2 − x12 − y12 . 1 ExamplE 8.3a: Find the equation of the perpendicular bisector of P (2, − 3) and Q(− ,5) . 2 Solution: The equation of the perpendicular bisector of the given segment PQ is: 1 1 2(− − 2) x + 2(5 − (−3)) y = (− ) 2 + 52 − 22 − (−3) 2 . 2 2 49 . Simplifying, we get −5 x + 16 y = 4 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 USAD Mathematics Resource Guide • 2016–2017 ® AcAdemic MatheMatics ResouRcedecAthlon Guide mAthemAtics ResouRce Guide 113 107 107 Rancho Cucamonga High School - Ontario, CA SoluTion: Any point A( x, y ) is on the perpendicular bisector l if and only if it is equidistant from Multiplying both sides by 4 , we get that the equation of the perpendicular bisector of PQ is −20 x + 64 y = 49 . The formula we have obtained for a perpendicular bisector will help prove several important theorems. The first of these theorems concerns the equation of a line in the plane: TheoreM 1: A line in the plane has the equation ax + by + c = 0 , where a, b, c are real numbers, and a and b are not both zero. Proof: Let l be a line in the plane, and regard l as the perpendicular bisector of a certain segment, say, By the Perpendicular Bisector Formula, l has the equation 2( x2 − x1 ) x + 2( y2 − y1 ) y = x2 2 + y2 2 − x12 − y12 . Let a = 2( x2 − x1 ) , b = 2( y2 − y1 ) , and c = −( x2 2 + y2 2 − x12 − y12 ) . We get that l has the equation ax + by + c = 0 . Now, we must show that a and b are not both zero. This is easy: notice that since P and Q are different points, x2 − x1 ≠ 0 or y2 − y1 ≠ 0 . Hence, a and b cannot both be zero. ExamplE 8.3b: Find the equation of the line that goes through the points A (−2,3) and B (0, 4) . Solution: Let ax + by + c = 0 be the equation of the line through A and B. Recall that a and b are not both zero. Since A and B are on the line, their coordinates satisfy this equation. Namely: (1) − 2 a + 3b + c = 0 (2) 0 + 4b + c = 0 Solving for a and b in terms of c , we get the following equation: c c (3) x − y + c = 0 . 8 4 Multiplying both sides of the equation (3) by 8 and then dividing both sides by c, we get that the equation of the line that goes through the points A (−2,3) and B (0, 4) is x − 2 y + 8 = 0 . 114 108 108 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource Guide • 2016–2017 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA PQ . Let ( x1 , y1 ) and ( x2 , y2 ) be the coordinates of P and Q , respectively. 8 4 Multiplying both sides of the equation (3) by 8 and then dividing both sides by c, we get that the equation of the line that goes through the points A (−2,3) and B (0, 4) is x − 2 y + 8 = 0 . Since we divided both sides by c , we must show that that c ≠ 0 . Indeed, if c = 0 then from equation Duplication is Prohibited License Agreement , and so from equation (1) a = 0Outside . Butthe and bof Your are not both zero. Hence, c ≠ 0 . (2) we get b = 0Unauthorized a Terms 108 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide –2014 2013 2010–2011 8.3.1 Form 8 .3 .1 SLoPe form 8.3.1Slope SLoPe Form When b ≠ 0 , we can write the equation of a line ax + by + c = 0 as y = − If we denote − a c x− . b b a c = m and − = d , we get the equation y = mx + d . b b A line whose equation is of the form y = mx + d cannot be vertical; that is, it cannot be parallel to the equation. Take two distinct points with the same x -coordinate: A( x1 , y1 ) and B ( x1 , y2 ) . Since y1 ≠ y2 , it is not possible that y1 = mx1 + d and y2 = mx1 + d . The equation y = mx + d is called the slope-intercept form of a line. The reason for this name has to do with the geometric meanings of the coefficients m and d. Take any two distinct points A( x1 , y1 ) and B ( x2 , y2 ) on the line whose equation is y = mx + d . Clearly, (1) y1 = mx1 + d (2) y2 = mx2 + d First note that x1 ≠ x2 . It is so because if x1 = x2 , then from equations (1) and (2), y1 = y2 , which implies that A( x1 , y1 ) and B ( x2 , y2 ) are the same points. But, we are given that they are distinct points. Now, subtracting equation (1) from equation (2), we get (3) y2 − y1 = m( x2 − x1 ) . Since x1 ≠ x2 , we divide both sides of equation (3) by x2 − x1 to get m = y2 − y1 . x2 − x1 Thus, m is independent of the choice of the points A( x1, y1) and B ( x2 , y2) on the line. m is called the slope of the line. It is also called the rise over run. Geometrically, m represents the “steepness” of the line: the larger | m | is, the steeper the line is. Now, in the equation y = mx + d what does d represent? Note that for x = 0 , y = d . Thus, d is the y-intercept—it is the point where the line intersects the y-axis. The equation y = mx + d has an advantage over the equation ax + by + c = 0 : the values m and d tell us USAD Mathematics Resource Guide • 2016–2017 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 AcAdemic MatheMatics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide 115 109 109 Rancho Cucamonga High School - Ontario, CA y-axis. The reason is simple: there are no two distinct points with the same x -coordinate that satisfy this Note that for x = 0 , y = d . Thus, d is the y-intercept—it is the point where the line intersects the y-axis. The equation y = mx + d has an advantage over the equation ax + by + c = 0 : the values m and d tell us how the line is positioned in the coordinate system. The equation y = mx + d is called the slope form of a line. 2013 –2014 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement AcAdemic MatheMatics ResouRcedecAthlon Guide ® 109 109 mAthemAtics ResouRce Guide 8.3.2 PoinT-PoinT Form 8 .3 .2 Point–Point PoInT-PoInT form 8.3.2 Form Determining a line by two points. The equation of a non-vertical line that goes through two distinct points y −y P( x1 , y1 ) and Q( x2 , y2 ) is y − y1 = 2 1 ( x − x1 ) . x2 − x1 Since P ( x1 , y1 ) is on l , its coordinates satisfy equation (1). Namely: y2 − y1 x1 + d x2 − x1 y − y1 (3) d = y1 − 2 x1 x2 − x1 (2) y1 = Substituting for d in equation (1), we get (4) y= y2 − y1 y − y1 x + y1 − 2 x1 x2 − x1 x2 − x1 Moving y1 to the left-hand side and factoring out (5) y − y1 = y2 − y1 ( x − x1 ) . x2 − x1 y2 − y1 , we get x2 − x1 Theorem 1 asserts that any line in the plane has an equation of the form ax + by + c = 0 . What about the converse of this theorem? Namely: Does every equation of the form ax + by + c = 0 represent a line? The following theorem answers this question affirmatively. TheoreM 2: The graph of the equation ax + by + c = 0 , where a and b are not both zero, is a line. Proof: Consider first the case where b ≠ 0 . Recall that in this case the equation ax + by + c = 0 had the form y = mx + d , where m = − a c and d = − . b b A graph of an equation is the set of all points ( x, y ) that satisfy the equation. Let γ be the graph of the equation y = mx + d . We will show that there exists a line l that has this equation. This will prove that γ and l are identical. Hence, γ is a line. 116 USAD Mathematics Resource Guide • 2016–2017 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement ® Rancho Cucamonga High School - Ontario, CA Proof: Let l be the line through P ( x1 , y1 ) and Q( x2 , y2 ) . We have seen that l has the equation y −y y = mx + d , where m = 2 1 . Hence l has the equation: x2 − x1 y2 − y1 (1) y = x+d . x2 − x1 We will show that there exists a line l that has this equation. This will prove that γ and l are identical. Hence, γ Take any a B 1, 110 110 b is a line. c two distinct points on γ . For simplicity’s sake, we will take the points A 0, and b Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement c ® . (Check that these points are indeed on γ .) –2014 mAthemAtics ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce Guide 2013 2010–2011 b Let l be the line between A and B. Using the Point-Point Form, the equation of l is a c c − − −− c b b b y= x− . 1− 0 b a c Simplifying, we get y = − x − . b b So, the graph γ and the line l have exactly the same equation. Position of a line. How does the position of the line whose equation is ax + by + c = 0 depend on the coefficients a , b , and c ? SoluTion: We will distinguish among several cases: CaSe 1: a = 0 and b ≠ 0 . c In this case, the equation of the line is y = − . (See Figure 8–8.) This line is parallel to the x-axis and b c passes through the point (0, − ) (Why?) b y O xx B y = –c/b (0, –c/b) –c/b) Figure 8–8 CaSe 2: a ≠ 0 and b = 0 . c In this case, the equation of the line is x = − . (See Figure 8–9.) This line is parallel to the y-axis and a c passes through the point (− , 0) (Why?) a Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource Guide • 2016–2017 2013 –2014 2010–2011 AcAdemic MatheMatics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide 117 111 111 Rancho Cucamonga High School - Ontario, CA You can work through the case where b = 0 as an exercise on your own. CaSe 2: a ≠ 0 and b = 0 . c In this case, the equation of the line is x = − . (See Figure 8–9.) This line is parallel to the y-axis and a c passes through the point (− , 0) (Why?) a yy x = – c/a Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement AcAdemic MatheMatics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide O A 111 111 x (–c/a, 0) (–c/a, Figure 8–9 CaSe 3: c = 0 . In this case, the equation of the line is ax + by = 0 . (See Figure 8–10.) This line goes through the origin yy (Why?) ax + by = 0 O x Figure 8–10 8.3.3 Form 8 .3 .3 SLoPe-PoInT form 8.3.3Slope–Point SLoPe-PoinT Form Determining a line by slope and one point. The equation of a line with slope m that passes through the point P1 ( x1 , y1 ) is y − y1 = m( x − x1 ) . Proof: Recall that the slope of a line is independent of the two distinct points we choose on the line to calculate it. y − y1 other than P1 ( x1 , yResource . Let 1 ) . Then Guide=• m 118 P( x, y ) be any point on the line USAD Mathematics x − x1 2016–2017 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement Rancho Cucamonga High School - Ontario, CA 2013 –2014 2010–2011 to calculate it. Let P ( x, y ) be any point on the line other than P1 ( x1 , y1 ) . Then y − y1 =m. x − x1 x1 we get: Duplication y − y1 = mis( xProhibited − x1 ) . Outside the Terms of Your License Agreement Multiplying by x − Unauthorized 112 112 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide 8.3.4 Positions of Lines 8 .3 .4 muTuAL PoSITIonS of LIneS 8.3.4Mutual muTuaL PoSiTionS oF LineS –2014 2013 2010–2011 If we are given two lines by their equations, can we determine whether the lines are parallel to each other? Can we determine whether they are perpendicular to each other? The following two theorems answer these questions. TheoreM 4: Let l1 and l2 be two distinct lines, whose equations are y = m1 x + d1 and y = m2 x + d 2 , respectively. l1 and l2 are parallel if and only if m1 = m2 . Since the two lines are different, d1 ≠ d 2 . Clearly, there is no ordered pair ( x, y ) that satisfies both equations (why?). Hence, l1 and l2 have no point in common. This proves that l1 and l2 are parallel. Now let l1 and l2 be two distinct parallel lines. Therefore, the equations of these lines do not have any common solution. Therefore, when subtracting the two equations, we get that the equation 0 = (m2 − m1 ) x + (d 2 − d1 ) has no solution. Since this equation has no solution, m1 = m2 . This is so because if m1 ≠ m2 , then this equation does have a solution. It is x = d1 − d 2 . m2 − m1 TheoreM 5: Let l1 and l2 be two lines with non-zero slopes whose equations are y = m1 x + d1 and 1 y = m2 x + d 2 , respectively. l1 and l2 are perpendicular to each other if and only if m2 = − . m1 Proof: Let P1 ( x1 , y1 ) and P2 ( x2 , y2 ) be two distinct points on l1 . Then, the slope of l1 is m1 = y2 − y1 (Note that x2 − x1 ≠ 0 .) x2 − x1 We have seen that the perpendicular bisector l ' of P1 P2 has the equation 2( x2 − x1 ) x + 2( y2 − y1 ) y = x22 + y22 − x12 − y12 . 2( x2 − x1 ) x −x 1 From here, the slope of l ' is m ' = − =− 2 1 =− . 2( y2 − Resource y1 ) y2 Guide m1 − y1 • 2016–2017 USAD Mathematics Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 119 Rancho Cucamonga High School - Ontario, CA Proof: Let m1 = m2 = m . Then, the equations of l1 and l2 are, respectively, y = mx + d1 and y = mx + d 2 . We have seen that the perpendicular bisector l ' of P1 P2 has the equation 2( x2 − x1 ) x + 2( y2 − y1 ) y = x22 + y22 − x12 − y12 . 2( x2 − x1 ) x −x 1 =− 2 1 =− . 2( y2 − y1 ) y2 − y1 m1 l l From geometry, weUnauthorized can conclude that is perpendicular to and only if l ' is parallel to l2 . 2 if of Duplication1is Prohibited Outside the Terms Your License Agreement From here, the slope of l ' is m ' = − 2013 –2014 2010–2011 AcAdemic MatheMatics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide On the other hand, l ' is parallel to l2 if and only if m2 = m2′ = − 1 . m1 113 113 8.4 Circles 8 .4 Circles 8.4 Circles A circle is the set of all points in the plane that are equidistant from a given point in the plane. What is the equation of the circle? Let’s begin with a simple case where a circle with radius r is centered at the origin. By the Pythagorean Theorem (see Figure 8–11), for any point A( x, y ) on the circle, x 2 + y 2 = r 2 . A ((x, x, yy)) ( 0, yy ) rr O x x ((x, x, 00)) Figure 8–11 Thus, x 2 + y 2 = r 2 is the equation of a circle with radius r centered at the origin. Now let’s take a circle with a center at any arbitrary point C (m, n) in the plane, and let r be the radius of the circle. (See Figure 8–12.) By the distance formula, ( x − m) 2 + ( y − n) 2 = r 2 yy A (x, ( x y) ,y) rr C (m, ( m n) ,n) O x Figure 8–12 Thus, ( x − m) 2 + ( y − n) 2 = r 2 is the equation of a circle with radius r centered at C (m, n) . 120 USAD Mathematics Resource Guide • 2016–2017 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement Rancho Cucamonga High School - Ontario, CA yy Figure 8–12 Thus, ( x − m) 2 + ( y − n) 2 = r 2 is the equation of a circle with radius r centered at C (m, n) . Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 8.5 Solving Geometry Problems Using 8.5 Solving Geometry Problems 8 .5 Solving GeometryGeometry Problems Using Coordinate Geometry Coordinate using Coordinate Geometry 114 114 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide –2014 2013 2010–2011 TheoreM: The diagonals of a parallelogram bisect each other. Proof: We are given a parallelogram ABCD . To prove this theorem using coordinate geometry, we must first introduce a coordinate system. We choose the coordinate axes so that AB is on the x-axis and the point A is at the origin (see Figure 8–13). D(c, d) Rancho Cucamonga High School - Ontario, CA y C(x, d) B(b, 0) x Figure 8–13 Therefore, B = (b, 0) . Let D = (c, d ) . Since CD is parallel to AB, the y-coordinate of C is d. Let C = ( x, d ) . To find x, we use the fact that BC is parallel to AD. We will consider two cases: CaSe 1. AD and BC are not parallel to the y-axis. Since they are parallel to each other, their slopes must be equal. Hence, d d = . c x −b From this we get x = b + c . CaSe 2. AD and BC are parallel to the y-axis. In this case, c = 0 and x = b = b + c . USAD Mathematics Resource Guide • 2016–2017 So, in both cases we get . 121 CaSe 2. AD and BC are parallel to the y-axis. In this case, c = 0 and x = b = b + c . So, in both cases we get C = (b + c, d ) . To show that the diagonals bisect each other, it is sufficient to show that the midpoint of AC is the same Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement ® as the–2014 midpoint of BD. AcAdemic 2013 2010–2011 MatheMatics ResouRcedecAthlon Guide mAthemAtics ResouRce Guide The midpoint M 1 of AC has the coordinates: x = 115 115 1 1 1 1 ⋅ (0 + (b + c)) = (b + c) , y = (0 + d ) = d . 2 2 2 2 1 1 Therefore, M 1 = (b + c), d . 2 2 The midpoint M 2 of BD has the coordinates: x = 1 1 1 1 ⋅ (b + c) = (b + c) , y = (0 + d ) = d . 2 2 2 2 Hence: M 1 = M 2 . This implies that the diagonals of a parallelogram bisect each other. TheoreM: The three perpendicular bisectors of the sides of a triangle are concurrent. Proof: Let ABC be any triangle positioned in a coordinate system as shown in Figure 8–14: y C(c, 0) K(x, y) A(a, 0) B(b, 0) x Figure 8–14 A = (a, 0) , B = (b, 0) , C = (0, c) . Let us find the equation of the perpendicular bisector of AB. Take any point K = ( x, y ) on the perpendicular bisector of AB. Then, since it is equidistant from A and B, we have: ( x − a ) 2 + ( y − 0) 2 = ( x − b) 2 + ( y − 0) 2 Opening parentheses and simplifying, we get: 122 USAD Mathematics Resource Guide • 2016–2017 1 (1) x = (a + b) . Rancho Cucamonga High School - Ontario, CA 1 1 Therefore, M 2 = (b + c), d . 2 2 B, we have: ( x − a ) 2 + ( y − 0) 2 = ( x − b) 2 + ( y − 0) 2 Opening parentheses and simplifying, we get: (1) x = 1 ( a + b) . 2 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement The equations of the other two perpendicular bisectors can be found in a similar way. They are 116 116 Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide 2ax − 2cy = a − c and 2bx − 2cy = b 2 − c 2 . 2 2 –2014 2013 2010–2011 To find the point of intersection of the last two bisectors, we solve their system of equations by subtracting the second from the first. We get: 2ax − 2bx = a 2 − b 2 , From (1) and (2) we get that the point of intersection lies on the first perpendicular bisector. Hence, the three perpendicular bisectors are concurrent. TheoreM: The three medians of a triangle are concurrent, and the distance from any vertex to the intersection point is two-thirds the length of the median drawn from that vertex. Proof: Let P1 P2 P3 be a triangle in a coordinate system as shown in Figure 8–15: P1 = ( x1 , y1 ) , P2 = ( x2 , y2 ) , P3 = ( x3 , y3 ) . y x1+x2+x3 y1+y2+y3 x Figure 8–15 x + x2 y1 + y2 The midpoint of P1 P2 has the coordinates 1 , 2 2 . Thus, the point that is two-thirds the way down the median from P3 has the coordinates 1 1 1 ⋅ x3 + 2 ⋅ ( x1 + x2 ) x + x + x 1⋅ y3 + 2 ⋅ ( y1 + y2 ) y + y + y 3 2 2 . x= = 1 2 3 , y= = 1 2 3 3 3 3 1232 USADpoint Mathematics • 2016–2017 can simplyGuide interchange indices. (Replace 3 by 1, 1 by 2, and To find the corresponding for P1 , we Resource by 3.) We do the same for P . Rancho Cucamonga High School - Ontario, CA 2 x ( a − b) = a 2 − b 2 , 1 (2) x = (a + b) . 2 1 2 2 2 Thus, the point that is two-thirds the way down the median from P3 has the coordinates 1 1 1 ⋅ x3 + 2 ⋅ ( x1 + x2 ) x + x + x 1⋅ y3 + 2 ⋅ ( y1 + y2 ) y y y + + 3 2 2 . x= = 1 2 3 , y= = 1 2 3 3 3 3 To find the corresponding point for P1 , we can simply interchange indices. (Replace 3 by 1, 1 by 2, and 2 by 3.) We do the same for P2 . But the above expressions for x and y are invariant under a permutation of the indices. This proves the Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement ® theorem. 2013 –2014 AcAdemic mAthemAtics ResouRce of Guide 117 2010–2011 expressions MatheMatics ResouRce Guide But the above for x anddecAthlon y are invariant under a permutation the indices. This proves 117 the theorem. kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk Section 8: SeCTion 8: exercises Section 8 Exercises Section 8: k SeCTion k k k k k k k k8: k k exercises kkkkkkkkkkkkkkkkkkkkkkkkkkkk Exercises kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 1. Which of the given points A(3, − 4) , B (1, 0) , C (0,5) , D(0, 0) , and E (0,1) belong to the following circles?of the given points A(3, − 4) , B (1, 0) , C (0,5) , D(0, 0) , and E (0,1) belong to the following 1. Which a. ( x − 1) 2 + ( y + 3) 2 = 9 circles? 2 +1)y 2 += (25 b. y + 3) 2 = 9 a. (xx − b. x 2 + y 2 = 25 2. Find which of the following equations represent a circle. If it is a circle, identify the radius and the center 2. Find whichofofthe thecircle. following equations represent a circle. If it is a circle, identify the radius and 2 + y 2 of + 8the x − 4circle. y + 40 = 0 a. xcenter the xx22 + −=1=−10 x 2y+22 − y+248−x 4−x42−y2+=y40 b. a. b. x2 + x 2y+2 − y 24−x 4−x2−y2=y−=1 −1 b. b. 3. The vertices of a triangle ABC are A(4, 6) , B (−4, 0) , C (−1, − 4) . Find the equation of the median CM.vertices of a triangle ABC are A(4, 6) , B (−4, 0) , C (−1, − 4) . Find the equation of the median 3. The CM. 4. What are the values of a and b in the equation of the line ax + by = 1 going through the points ? (1, 2) and 4. What are (2,1) the values of a and b in the equation of the line ax + by = 1 going through the points (1, 2) and (2,1) ? 5. Three vertices of a parallelogram ABCD are A(1, 0) , B (2,3) , C (3, 2) . Find the coordinates of fourth vertex, find the coordinates of A the point of intersection thethe diagonals. 5. the Three vertices of aand parallelogram ABCD are , B (2,3) , C (3, 2) . of Find coordinates of (1, 0) the fourth vertex, and find the coordinates of the point of intersection of the diagonals. 6. The distance of the point A from the x-axis is q1 , and its distance from the y-axis is q2 , and its = q2 the = q3y-axis fromof thethe point theiscoordinates of A if q1from . B (3, 6) is qthe 3 . Find q1 , and its distance 6. distance The distance point A from x-axis is q2 , and its distance from the point B (3, 6) is q3 . Find the coordinates of A if q1 = q2 = q3 . 7. Find the point that is one-fourth of the distance from the point A(–1, 2) to the point B(6, 7) alongthe the point segment 7. Find thatAB. is one-fourth of the distance from the point A(–1, 2) to the point B(6, 7) 3 124 along the segment AB. USAD Mathematics Resource Guide • 2016–2017 8. Find the point that is five-sevenths of the distance from the point A(− , 2) to the point B (2, 2) 2 Rancho Cucamonga High School - Ontario, CA Exercises kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 1 2 distance from the point B (3, 6) is q3 . Find the coordinates of A if q1 = q2 = q3 . 7. Find the point that is one-fourth of the distance from the point A(–1, 2) to the point B(6, 7) along the segment AB. 3 8. Find the point that is five-sevenths of the distance from the point A(− , 2) to the point B (2, 2) 2 along the segment AB. kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement Guide AcAdemic decAthlon ® mAthemAtics ResouRce MatheMatics ResouRce Guide –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA 118 118 USAD Mathematics Resource Guide • 2016–2017 125 part II T R PA 2 TRIGONOMETRY Part II: TrigonomeTry Trigonometry Section 9 TRIGonometric Functions Ssection ecTIon 9: trigonometric Trigonometric functions Functions Trigonometry means triangle measurement. Trigonometry is used to compute the lengths of segments astronomy, geography, and navigation. To illustrate how trigonometry is used in measurement, let’s consider the following example. In Figure 9–1, AD is the height of a mountain sitting on a plane. There is no direct way to measure the height of the mountain. However, with a special instrument, we can measure the angle in which we see the summit A of the mountain from a point B on the plane. Let’s denote the measure of this angle by β . Now, along the segment BD we move a meters toward the mountain, and again measure the angle in which we see the summit A . Let’s denote the measure of this angle by γ . The numbers β , γ , and a are sufficient to compute the height AD . In this section of the resource guide, you will learn trigonometric tools for finding measurements of sides and angles of a triangle. A β B D Figure 9–1 We will begin with the trigonometric functions of an acute angle. Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 2010–2011 126 –2014 ® AcAdemic decAthlon mAthemAtics Guide MatheMatics ResouRce Guide USAD Mathematics Resource ResouRce Guide • 2016–2017 119 119 Rancho Cucamonga High School - Ontario, CA and measures of angles, and therefore it is an important tool in many branches of science, such as 9.1 the Sine Function for acuteAngles angles 9 . 1 The SineSine Function for Acutefor Angles 9.1 The Function Acute Take an acute angle α and position it in a coordinate system so that its vertex is at the origin O and one of its sides is on the positive x-axis. Let P ( x, y ) be a point on the second side of the angle, and let r be the distance between P and the origin O . (See Figure 9–2.) yy xx rr PP(x, (x , y) y) yy xx Figure 9–2 y y gets larger, and As α varies from 0° to 90° , the ratio also varies. Moreover, as α gets larger, r r y y as α gets smaller, gets smaller. The ratio is called the sine of α and is written as sin α . r r Since α is acute, and y is positive, the sin α is therefore positive. And, since y is smaller than r , sin α is smaller than 1 . We summarize this as follows: For 0° < α < 90° , 0 < sin α < 1 . At 90° , y = r . Hence: sin 90° = 1 . At 0° , y = 0 . Hence: sin 0° = 0 . Is sin α a function? As defined, there is no guarantee that sin α is a function, or in other words there is no guarantee that for any input α , there exists exactly one output sin α . Thus, to insure that sin α is a function, we must prove the following theorem, which asserts that the value sin α is independent of the choice of the point P( x, y ) on the second side of the angle. Theorem: Let angle α be an acute angle such that its vertex is at the origin and one of its sides is on the positive x-axis, as is shown in Figure 9–3. Let P( x, y ) and P '( x ', y ') be two distinct points on y y' the non-horizontal side of the angle. Call OP = r and OP ' = r ' . Then, = . r r' Proof: Drop two perpendiculars PC and PC ' as shown in Figure 9–3. Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 120 120 USAD Mathematics Resource Guide • 2016–2017 ® Guide AcAdemic decAthlon mAthemAtics ResouRce MatheMatics ResouRce Guide 127 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA O y x P(x, y) x' P'(x', y') r' y' y x Clearly, OPC and OP ' C ' are similar. Hence, y y' = . r r' One of the important applications of the sine function is expressed in the following theorem: Theorem: The area of any triangle is one-half the product of two sides of the triangle and the sine of the angle between them. That is, the area of ABC is given by the formula: 1 area ( ABC ) = AB ⋅ AC ⋅ sin α . 2 Proof: We will prove this theorem for the case where α is acute. The theorem, however, is also valid when α is not acute. The area of any triangle ABC (see Figure 9–4) is half the product of a side of the triangle and the 1 altitude to that side. Hence, area ( ABC ) = AB ⋅ CH . 2 Figure 9–4 Point C is at distance CH from the line AB. This distance can be seen from the angle CAB (let’s call CH = sin α . From here, CH = AC ⋅ sin α . Substituting the value of CH into the it α ). Therefore, ratio AC 1 1 initial formula, we get area ( ABC ) = AB ⋅ CH = AB ⋅ AC ⋅ sin α . 2 2 128 –2014 2013 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource Guide • 2016–2017 ® mAthemAtics ResouRce Guide AcAdemic MatheMatics ResouRcedecAthlon Guide 121 121 Rancho Cucamonga High School - Ontario, CA Figure 9–3 Therefore, area ( ABC ) = 1 AB ⋅ AC ⋅ sin α . 2 9 .2 Tangent Function for Acute for Angles 9.2The the tangent Function forAcute acuteAngles angles 9.2 The Tangent Function As with the sine function, take an acute angle α and position it in a coordinate system so that its vertex is at the origin O and one of its sides is on the positive x-axis. Let P ( x, y ) be a point on the second side x P(x, y) r y Figure 9–5 y also varies but always remains positive. Moreover, as α gets x y larger, y gets larger and x gets smaller, and so gets larger. As α gets smaller, y gets smaller and x y gets smaller. x gets larger, and so xr As α varies from 0° to 90° , the ratio y is called the tangent of α , and is written as tanα . Indeed tanα is the slope of the line x going through the origin O and the point P. For 0 < α < 90 , 0 < tanα < ∞ . The ratio As with the sine function, here too we must guarantee that tan α is a function. That is, we must show that the value of tan α is independent of the choice of the point P( x, y ) on the second side of the angle. Theorem: Let angle α be an acute angle such that its vertex is at the origin and one of its sides is on the positive x-axis, as is shown in Figure 9–6. Let P( x, y ) and P '( x ', y ') be two distinct points on the non-horizontal side of the angle. Then y y' = . x x' Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 122 122 ® mAthemAtics ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 –2014 2013 2010–2011 129 Rancho Cucamonga High School - Ontario, CA of the angle. (See Figure 9–5.) yy xx P(x, y) P(x ,y) P'(x',y') P'(x', y') r'r' x' x' O yy y' y' xx C' C Figure 9–6 are similar. Hence, y y' = . x x' 9.3 T the he Cosine and Functions 9.3 Cosine and Cotangent Cotangent 9 .3 The Cosine and Cotangent Functions for Acute Anglesfor Functions for acute angles Acute Angles In addition to the functions sine and tangent, we will define two additional trigonometric functions. The four functions are needed to write trigonometric formulas more effectively. The cosine of an angle 0° < α < 90° is denoted by cos α and is defined as follows: cos α = sin(90° − α ) . The cotangent of an angle 0° < α < 90° is denoted by cot α and is defined as follows: cot α = tan(90° − α ) From here, it is easy to see that sin α = cos(90° − α ) and that tan α = cot(90° − α ) Here is why: By the definition of cosine, cos(90° − α ) = sin(90° − (90° − α )) = sin α . By the definition of cotangent, cot(90° − α ) = tan(90° − (90° − α )) = tan α . Thus, in a right triangle ABC (see Figure 9–7), we have the following relations: a = cos B c b 2. cos A = = sin A c a 3. tan A = = cot B b 1. sin A = Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 130 AcAdemic decAthlon mAthemAtics Guide MatheMatics ResouRce Guide USAD Mathematics Resource ResouRce Guide • 2016–2017 ® 123 123 Rancho Cucamonga High School - Ontario, CA Proof: Drop two perpendiculars PC and PC ' as shown in Figure 9–6. Clearly, OPC and OP ' C ' 4. cot A = b = tan B a a b c Figure 9–7 There are six basic relations among trigonometric functions. They are referred to as basic identities: 1. tan α = sin α cos α 2. cot α = cos α sin α 3. tan α ⋅ cot α = 1 4. sin 2 α + cos 2 α = 1 5. tan 2 α + 1 = 1 , cos α ≠ 0 cos 2 α 6. 1 + cot 2 α = 1 , sin α ≠ 0 sin 2 α Often 1 1 is denoted by csc α and by sec α . sin α cos α Thus identities 5 and 6 can be written as: 7. tan 2 α + 1 = sec 2 α 8. 1 + cot 2 α = csc 2 α . To prove these formulas for an acute angle α , let ABC be a right triangle in C with the acute angle BAC = α . Let AC = b , CB = a , and AB = c , as shown in Figure 9–8. Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 124 124 ® USAD Mathematics Resource Guide • 2016–2017 mAthemAtics ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce Guide 131 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA 9 .4 Relations among Trigonometric Functions 9.4 relations among trigonometric Functions 9.4 Relations among Trigonometric Functions C 90° aa bb B cc A Figure 9–8 sin α cos α Proof: Since sin α = tan α = sin α . cos α 2. cot α = sin α a b a a b a = : = . By definition, = tan α . Therefore, and cos α = , the ratio cos α c c b c b c cos α sin α Rancho Cucamonga High School - Ontario, CA 1. tan α = cos α b a b b = : = , we have: sin α c c a c b cos α . Since = cot α , we get cot α = a sin α Proof: Since cos α = 3. tan α ⋅ cot α = 1 Proof: By definition, cot α = b a a b and tan α = . Hence: tan α ⋅ cot α = ⋅ = 1 . a b b a 4. sin 2 α + cos2 α = 1 Proof: By the Pythagorean Theorem, in a right triangle ABC, we have a 2 + b 2 = c 2 . Dividing both sides of the equality by the non-zero c 2 , we get Since a 2 b2 c 2 + = =1 c2 c2 c2 a b = sin α and = cos α , we get: sin 2 α + cos2 α = 1 . c c 5. tan 2 α + 1 = 1 cos2 α 2 sin 2 α sin 2 α + cos2 α 1 sin α Proof: tan α + 1 = + 1 = + 1 = = . 2 2 cos α cos α cos2 α cos α 2 6. cot 2 α + 1 = 1 sin 2 α 2 cos2 α cos2 α + sin 2 α 1 cos α Proof: cot α + 1 = + 1 = + 1 = = 2 . 2 2 sin α sin α sin α sin α 2 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 2010–2011 132 –2014 AcAdemic MatheMatics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 125 125 9.5 trigonometric Functions of Special SpecialAngles angles 9 .5 Functions of Special Angles 9.5Trigonometric Trigonometric Functions of We will now show how the identities we just proved can be used to compute the trigonometric functions for the angles 30°, 45°, 60°, and 90°. First, let’s determine the values of all trigonometric functions for 30º and 60º. For this, consider a right triangle ABC ( AC = b , CB = a , and AB = c ) with acute angles A = α = 30° and B = β = 60° . From geometry, we know that the hypotenuse c is twice the side opposite the 30º. Hence, 1 . 2 Using the identity sin 2 α + cos2 α = 1 , it is easy to get cos 30° = We have cos 30° = 1 − sin 2 30° = 1 − 3 . 2 1 3 3 . = = 4 4 2 1 . Also, sin 60° = cos 30° , cos 30° = 2 sin α sin 30° 1 = : , we compute tan 30° = Using the identity tan α = cos α cos 30° 2 Since sin 30° = cos 60° , cos 60° = 3 . 2 3 1 = . 2 3 Rancho Cucamonga High School - Ontario, CA sin 30° = a 1 = , and so: c 2 And sin 60° 3 1 : = 3. = cos 60° 2 2 cos 30° 3 1 cos α : = 3. = , we get cot 30° = Using the identity cot α = sin α sin 30° 2 2 tan 60° = And cot 60° = cos 60° 1 3 1 = : = . sin 60° 2 2 3 Now let’s compute the values of all trigonometric functions for the angle 45º. For this we take a right isosceles (i.e., having two equal sides) triangle ABC ( AC = a , CB = a , and AB = c ) with β = α = 45° . By the Pythagorean Theorem, a 2 + a 2 = c 2 . Hence: 2a 2 = c 2. From this, we get Since a 1 . = c 2 a 1 = sin 45° = cos 45° , we have: sin 45° = cos 45° = . c 2 And tan 45° = cot 45° = 1 1 : = 1. 2 2 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 126 126 mAthemAtics ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 ® –2014 2013 2010–2011 133 We will now conclude with the values of the trigonometric functions for angle 90º. We have already seen that: sin 90° = 1 and sin 0° = 0 . From these two facts, we can calculate the rest: cos 90° = 1 − sin 2 90° = 1 − 1 = 0 . tan 90° = sin 90° 1 = =∞. cos 90° 0 cot 90° = cos 90° 0 = =0. sin 90° 1 α 30º 45º 60º 90º sin α 1 2 1 2 3 2 1 cos α 3 2 1 2 1 2 0 tan α 1 3 1 3 − cot α 3 1 1 3 0 Rancho Cucamonga High School - Ontario, CA The following table summarizes the values of the special angles: 9.6 Functions of ofAngles of Any 9.6 Ttrigonometric rigonometric Functions 9 .6 Trigonometric Functions of Angles of Any Measure angles of any Measure Measure 9 .6 .1 DefInITIonS ProPerTIeS 9.6.1 DeFInItIonS anD ProPertIeS 9.6.1 Definitions andAnD Properties The definitions of the trigonometric functions for acute angles we learned in the previous sections are extendable to angles of any measure. As before, we take any angle α and position it in a coordinate system such that its vertex is at the origin O and one of its sides is on the positive x-axis. Let P( x, y ) be a point on the second side of the angle, and let PO = r . Here we think of the angle α as the angle generated by moving counterclockwise from the positive x-axis to the segment PO . See Figure 9–9 . Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 2010–2011 134 –2014 AcAdemic decAthlon mAthemAtics Guide MatheMatics ResouRce Guide USAD Mathematics Resource ResouRce Guide • 2016–2017 ® 127 127 yy PP(x, (x ,y)y) xx rr yy xx O Rancho Cucamonga High School - Ontario, CA Figure 9–9 The definitions of the four trigonometric functions for such an angle α are as follows: y r x 2. cos α = r y 3. tan α = , x ≠ 0 x x 4. cot α = , y ≠ 0 y 1. sin α = From these definitions, it is easy to derive the following identities: l l sin α cos α cos α cot α = sin α tan α = Since r is positive, the sign of the sine function (that is, whether the function is positive or negative) is determined by the sign of y , whereas the sign of cosine is determined by the sign of x . l For 0 < α < 90° , y > 0 and x > 0 . Hence: sin α > 0 and cos α > 0 l For 90° < α < 180° , y > 0 and x < 0 . Hence: sin α > 0 and cos α < 0 l For 180° < α < 270° , y < 0 and x < 0 . Hence: sin α < 0 and cos α < 0 l For 270° < α < 360° , y < 0 and x > 0 . Hence: sin α < 0 and cos α > 0 α == 00, °, 90, 90°, 180, 180°, 270, 270°,360 In addition, the following table gives the values of sine and cosine for 360°:: Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 128 128 mAthemAtics ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 ® –2014 2013 2010–2011 135 α 0º 90º 180º 270º 360º sin α 0 1 0 −1 0 cos α 1 0 −1 0 1 As for tan α and cot α , these functions are not defined for all angles. tan α is not defined for α = 90° and α = 270° because for these values x = 0 ; cot α is not defined for α = 0° , α = 180° , and α = 360° because for these values y = 0 . l For 0° < α < 90° , tan α > 0 and cot α > 0 l For 90° < α < 180° , tan α < 0 and cot α < 0 l For 180° < α < 270° , tan α > 0 and cot α > 0 l For 270° < α < 360° , tan α < 0 and cot α < 0 In addition, the following table gives the values of tangent and cotangent for == 00, 90, 180, 360°:: α °, 90 °, 180 °, 270, 270°,360 α 0º 90º 180º 270º 360º tan α 0 – 0 − 0 cot α − 0 − 0 − 9.6.2Negative Negative aNgles 9 .6 .2 neGATIve AnGLeS 9.6.2 Angles Using the above identities, we can determine the values of trigonometric functions for negative angles. As before, we take any angle α and position it in a coordinate system so that its vertex is at the origin O and one of its sides is on the positive x-axis. And, as before, we let P ( x, y ) be a point on the second side of the angle, and PO = r . Here, however, we think of the angle α as the angle generated by moving clockwise from the positive x-axis to the segment PO . In the exercises at the end of this section, you will prove the following relations using identities we will learn in this section: 1. sin(−α ) = − sin α 2. cos(−α ) = cos α 3. tan(−α ) = − tan α 4. cot(−α ) = − cot α 136 2013 –2014 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource Guide • 2016–2017 ® AcAdemic MatheMatics ResouRcedecAthlon Guide mAthemAtics ResouRce Guide 129 129 Rancho Cucamonga High School - Ontario, CA From the definitions of tangent and cotangent we get: 9.7 trigonometric 9 .7 Trigonometric Identitiesidentities 9.7 Trigonometric Identities 9.7.1 sum aNd differeNce ideNtities 9 .7 .1 Sum AnD DIfference IDenTITIeS 9.7.1 Sum and Difference Identities The following identities are helpful in computing the measure of many angles: 1. sin(α + β ) = sin α cos β + cos α sin β 2. cos(α + β ) = cos α cos β − sin α sin β 3. sin(α − β ) = sin α cos β − cos α sin β 4. cos(α − β ) = cos α cos β + sin α sin β Proofs of Identities 1 and 2: These identities hold for α and β of any measure. Here we will prove them only for the case where Rancho Cucamonga High School - Ontario, CA 0 < α + β < 90 . We first construct a figure like that shown in Figure 9–10. B α sinβ F 90° 1 90° D cosβ sin(α+β) β A α α+β cos(α+β) 90° 90° C E Figure 9–10 In this figure we have: BC = sin(α + β ) AC = cos(α + β ) BD = sin β AD = cos β sin(α + β ) = BC = BF + DE cos(α + β ) = AC = AE − FD BF = BD ⋅ cos α = sin β cos α Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 130 130 ® mAthemAtics ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 –2014 2013 2010–2011 137 DE = AD ⋅ sin α = cos β sin α AE = AD ⋅ cos α = cos β cos α FD = BD ⋅ sin α = sin β sin α Therefore, sin(α + β ) = sin α cos β + cos α sin β and cos(α + β ) = cos α cos β − sin α sin β . This completes the proofs for Identities 1 and 2. We will now prove Identities 3 and 4: Let α = (α − β) + β . Then, sinα = sin((α − β) + β) = sin(α − β)cos β + cos(α − β)sin β cosα = cos((α − β) + β) = cos(α − β)cos β − sin(α − β)sin β . We treat these two equations as the system of equations with two unknowns sin(α − β ) and cos(α − β ). Multiplying the first equation by cos β , and the second by − sin β and adding, we get: sinα cos β − cosα sin β = sin(α − β) ⋅ (cos 2 β + sin 2 β) = sin(α − β) ⋅1 = sin(α − β) . This completes the proof for Identity 3. To prove Identity 4, solve the system of equations 1 and 2 for cos(α − β ) . To do this, multiply the first equation by sin β , and the second equation by cos β , and then add them. You will get: sinα sin β + cosα cos β = cos(α − β) ⋅ (cos 2 β + sin 2 β) = cos(α − β) ⋅1 = cos(α − β) . The following identities are also useful: 1. tan(α + β ) = integer. 2. tan(α − β ) = integer. 138 2013 –2014 2010–2011 tan α + tan β , where α + β ≠ (1 + 2n )90°, α ≠ (1 + 2n )90°, β ≠ (1 + 2n)90°, n is an 1 − tan α ⋅ tan β tan α − tan β where α − β ≠ (1 + 2n )90°, α ≠ (1 + 2n )90°, β ≠ (1 + 2n )90°, n is an 1 + tan α ⋅ tan β Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource Guide • 2016–2017 ® AcAdemic MatheMatics ResouRcedecAthlon Guide mAthemAtics ResouRce Guide 131 131 Rancho Cucamonga High School - Ontario, CA Proofs of Identities 3 and 4: sinα cos β cosα sin β + sin(α + β) sinα cos β + cosα sin β cosα cos β cosα cos β tan(α + β) = = = = cos(α + β) cosα cos β − sinα sin β cosα cos β sinα sin β − cosα cos β cosα cos β ( tanα ) ⋅1+1⋅ ( tan β ) = ( tanα ) + ( tan β ) . = 1− tanα tan β 1− tanα tan β Proof: The proof for the second identity is similar. ExamplE 9.7a: Find cos135° Solution: 1 2 −1⋅ 1 2 =− 1 2 . Rancho Cucamonga High School - Ontario, CA cos135° = cos(90° + 45°) = cos90°cos 45° − sin 90°sin 45° = 0 ⋅ ExamplE 9.7b: Find tan 210° 1 tan180° + tan 30° 3 = 1 . tan 210° = tan(180° + 30°) = = 1 − tan180°⋅ tan 30° 1 − 0 ⋅ 1 3 3 Solution: 0+ ExamplE 9.7c: Find sin 300° sin 300° = sin(270° + 30°) = sin 270° cos 30° + co Solution: sin 300° = sin(270° + 30°) = sin 270° cos 30° + cos 270° sin 30° == (−1) ⋅ = (−1) ⋅ 3 1 3 + 0⋅ = − . 2 2 2 3 1 3 + 0⋅ = − . 2 2 2 ExamplE 9.7d: Find cot 75° 1 1 − tan 45° tan 30° 3 = 3 −1 = 4 − 2 3 = 2 − 3 . cot 75° = cot(45° + 30°) = = tan 45° + tan 30° 1 + 1 2 3 +1 3 Solution: 1− Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 132 132 USAD Mathematics Resource Guide • 2016–2017 ® Guide AcAdemic decAthlon mAthemAtics ResouRce MatheMatics ResouRce Guide 139 –2014 2013 2010–2011 Summarizing: we have the following Sum and Difference Identities: 1. sin(α + β ) = sin α cos β + cos α sin β 2. cos(α + β ) = cos α cos β − sin α sin β 3. sin(α − β ) = sin α cos β − cos α sin β 4. cos(α − β ) = cos α cos β + sin α sin β integer. 6. tan(α − β ) = integer. tan α + tan β , where α + β ≠ (1 + 2n )90°, α ≠ (1 + 2n )90°, β ≠ (1 + 2n)90°, n is an 1 − tan α ⋅ tan β tan α − tan β , where α − β ≠ (1 + 2n )90°, α ≠ (1 + 2n )90°, β ≠ (1 + 2n )90°, n is an 1 + tan α ⋅ tan β 9 .7 .2 IDenTITIeS 9.7.2DouBLe-AnGLe double-aNgle ideNtities 9.7.2. Double-Angle Identities From the above identities, it is easy to prove the following identities. You can complete the proofs of these identities as an exercise on your own: 1. sin 2α = 2 sin α cos α 2. cos 2α = cos 2 α − sin 2 α 3. tan 2α = 2 tan α 1 − tan 2 α ExamplE 9.7e: Given that sin α = cot 2α . 4 where 0 < α < 90°, find sin 2α , cos 2α , tan 2α and 5 Solution: 2 9 3 4 cos α = ± 1 − sin 2 α = ± 1 − = ± =± . 25 5 5 Since 0° < α < 90° , we take cos α positive, and so cos α = 3 . 5 4 3 24 = 5 5 25 From here, we have sin 2α = 2sin α cos α = 2 ⋅ ⋅ 140 2013 –2014 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource Guide • 2016–2017 ® AcAdemic MatheMatics ResouRcedecAthlon Guide mAthemAtics ResouRce Guide 133 133 Rancho Cucamonga High School - Ontario, CA 5. tan(α + β ) = 2 2 7 3 4 cos 2α = cos α − sin α = − = − 25 5 5 2 2 4 sin α 5 4 tan α = = = . cos α 3 3 5 1 1 3 = = cot α = tan α 4 4 3 4 8 2⋅ 2 tan α 3 = 3 = − 24 = tan 2α = 2 2 7 1 − tan α 7 4 − 1− 9 3 1 7 cot 2α = =− . tan 2α 24 Thus, if sin α = cot 2α = − 7 . 24 4 24 7 24 where 0 < α < 90° , then sin 2α = , cos 2α = − , tan 2α = − , 5 25 25 7 ExamplE 9.7f: Given that cos 2α = 4 and 90° < α < 180°, find tan α + cot α . 5 Solution: Recall that tan 2 2α + 1 = 1 . Hence: cos 2 2α 4 1− 1− cos 2 2α 9 3 5 tan 2α = ± =± =± =± . 2 2 16 4 cos 2α 4 5 2 Since 90° < α < 180° , so 180° < 2α < 360° , but cos 2α > 0 , therefore 270° < 2α < 360° and 3 tan 2α = − . 4 To find tan α , we use the identity tan 2α = We have: − 2 tan α . 1 − tan 2 α 3 2 tan α = , 1 − tan 2 α ≠ 0 . 4 1 − tan 2 α Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 134 134 ® USAD Mathematics Resource Guide • 2016–2017 mAthemAtics ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce Guide 141 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA To find tan 2α and cot 2α , we first find tan α and cot α : Solving for tan α , we get: 3 tan 2 α − 8 tan α − 3 = 0 1 3 tan α = 3 or tan α = − . Since 90° < α < 180° , tan α = − 1 and, therefore, cot α = −3 . 3 1 3 Now we are ready to compute tan α + cot α : tan α + cot α = − − 3 = −3 1 3 9 .7 .3 hALf-AnGLe IDenTITIeS 9.7.3 Half-aNgle ideNtities 9.7.3 Half-Angle Identities We can also obtain the following identities for half angles: 1 − cos α 2 2 α 1 + cos α 2. cos = ± 2 2 α sin α 3. tan = ± 2 1 + cos α α 1 + cos α 4. cot = ± 2 sin α α =± Rancho Cucamonga High School - Ontario, CA 1. sin Proof: Proof of (1): α α α α α α cosα = cos 2 = cos 2 − sin 2 = 1− sin 2 − sin 2 = 1− 2sin 2 2 2 2 2 2 2 sin α 2 =± 1 − cos α 2 Proof of (2): α α α α α α cosα = cos 2 = cos 2 − sin 2 = cos 2 − 1− cos 2 = 2cos 2 −1 2 2 2 2 2 2 cos α 2 =± 1 + cos α 2 Proof of (3): α sin 2 α 2 = ± 1− cosα = ± 1− cosα ⋅ 1+ cosα = ± 1− cos α = ± sinα tan = = α 2 1+ cosα 1+ cosα 1+ cosα ±(1+ cosα) (1+ cosα)2 cos sin α 2 , where α ≠ 180°n and n is an integer. = (1 + cos α ) 142 2013 –2014 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource Guide • 2016–2017 ® AcAdemic MatheMatics ResouRcedecAthlon Guide mAthemAtics ResouRce Guide 135 135 Proof of (4): You should complete the proof of (4) on your own as an exercise. ExamplE 9.7g: Find sin15°. Solution: 3 1− cos30° 2 =± 2− 3 =± 2− 3 . sin15° = ± =± 2 2 4 2 2− 3 . Since 0° < α < 90° , we take the positive value: sin15° = 2 ExamplE 9.7h: Simplify the expression A = 0.125cos 4α + sin 2 α cos 2 α . Solution: (Justify each step using the identities you have learned thus far.) A = 0.125cos 4α + (sin α cos α ) 2 (2sin α cos α ) 2 4 (sin 2α ) 2 A = 0.125cos 4α + 4 1 A = 0.125cos 4α + sin 2 2α . 4 A = 0.125cos 4α + 1 1 1− cos 4α A = 0.125cos 4α + sin 2 2α = 0.125cos 4α + ⋅ = 0.125cos 4α + 0.125(1− cos 4α) = 4 4 2 = 0.125cos 4α + 0.125 − 0.125cos 4α = 0.125. Thus, A = 0.125 . 9.7.4 Identities 9 .7 .4 Sum–to–Product Sum-To-ProDucT IDenTITIeS 9.7.4 Sum-to-Product IdentItIeS In the previous sections, we learned identities of trigonometric functions for the sum and difference of angles. In this section, we will learn identities that will help us deal with the sum and difference of trigonometric functions such as sin18° + cos 42° . Recall the following identities: 1. sin( x + y ) = sin x cos y + cos x sin y Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 136 136 ® USAD Mathematics Resource Guide • 2016–2017 mAthemAtics ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce Guide 143 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA 1− 2. sin( x − y ) = sin x cos y − cos x sin y Adding Equations 1 and 2, we get sin( x + y ) + sin( x − y ) = 2sin x cos y Subtracting Equations 1 and 2, we get sin( x + y ) − sin( x − y ) = 2 cos x sin y Let x + y = α and x − y = β Adding these two equations, we get: x = α +β 2 Subtracting these two equations, we get: y = α −β 2 Substituting the values of x and y in sin( x + y ) + sin( x − y ) = 2sin x cos y , we get: sin α + sin β = 2sin α +β 2 cos α −β 2 . sin α − sin β = 2 cos α +β 2 sin Rancho Cucamonga High School - Ontario, CA Substituting the values of x and y in sin( x + y ) − sin( x − y ) = 2 cos x sin y , we get: α −β 2 In a similar manner, using sum and difference identities for cosine, we can prove that: 1. cos α + cos β = 2 cos α+β cos α −β 2 2 α+β α −β sin 2. cos α − cos β = −2sin 2 2 You can complete this proof on your own as an exercise. ExamplE 9.7i: Compute sin18° + cos 42° . Solution: sin18° + sin 42° = 2sin 18° + 42° 18° − 42° 1 cos = 2sin 30°cos(−12°) = 2 ⋅ cos12° = cos12° . 2 2 2 We will now derive similar formulas for tangent and cotangent : tanα + tan β = sinα sin β sinα cos β + cosα sin β sin(α + β) + = = cosα cos β cosα cos β cosα cos β and sinα sin β sinα cos β − cosα sin β sin(α − β) , where − = = cosα cos β cosα cos β cosα cos β α ≠ 90°(1 + 2n ), β ≠ 90°(1 + 2n ) , and n is an integer. tanα − tan β = 144 –2014 2013 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement ® USAD Mathematics Resource ResouRce Guide • 2016–2017 mAthemAtics Guide AcAdemic MatheMatics ResouRcedecAthlon Guide 137 137 Similarly, cotα + cot β = cosα cos β sin β cosα + cos β sinα sin(β + α) + = = sinα sin β sinα sin β sinα sin β and cotα − cot β = integer. cosα cos β sin β cosα − cos β sinα sin(β − α) , α ≠ 180° ⋅ n, β ≠ 180° ⋅ n , and n is an − = = sinα sin β sinα sin β sinα sin β Summarizing: we have the following Sum-to-Product Identities: 1. sin α + sin β = 2sin 3. 4. 5. 6. cos α −β 2 2 α+β α −β sin α − sin β = 2 cos sin 2 2 α+β α −β cos α + cos β = 2 cos cos 2 2 α+β α −β cos α − cos β = −2sin sin 2 2 sin(α ± β ) , α ≠ 90°(1 + 2n ), β ≠ 90°(1 + 2n ) , and n is an integer. tan α ± tan β = cos α cos β sin( β ± α ) , α ≠ 180° ⋅ n, β ≠ 180° ⋅ n , and n is an integer. cot α ± cot β = sin α sin β ExamplE 9.7j: Compute (without calculator) Rancho Cucamonga High School - Ontario, CA 2. α+β 2sin 2 49° −1 . cos53° − cos37° Solution: (Justify each step in the solution.) 1− cos98° −1 1− cos98° −1 cos98° 2sin 49° −1 2 = = = cos53° − cos37° −2sin 45°sin8° − 2 sin8° 2 sin8° 2 2 Notice, cos 98° = cos(90° + 8°) = − sin 8° . Thus, − sin 8° 1 . =− 2 sin 8° 2 ExamplE 9.7k: Simplify the expression sin 7 β + sin11β and find its value for β = 30° . cos10 β − cos8β Solution: Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 138 138 USAD Mathematics Resource Guide • 2016–2017 ® Guide AcAdemic decAthlon mAthemAtics ResouRce MatheMatics ResouRce Guide 145 –2014 2013 2010–2011 sin 7 β + sin11β 2sin 9 β cos(−2 β ) . = −2sin 9 β sin β cos10 β − cos8β sin 7 β + sin11β 2sin 9 β cos(−2 β ) cos 2 β . = =− cos10 β − cos8β sin β −2sin 9 β sin β cos 2 β cos 60° 1 1 For β = 30° , we have − =− = − : = −1 sin β sin 30° 2 2 Recall that cos(−2 β ) = cos 2 β . Hence: 9 .7 .5 ProDucT-To-Sum IDenTITIeS 9.7.5 Identities 9.7.5Product–to–Sum Product-to-Sum IdentItIeS The following are three more useful identities; they allow us to transfer the product of trigonometric functions into the sum or difference of trigonometric functions. 1 cos(α − β) − cos(α + β) 2 1 2. cosα cos β = cos(α + β) + cos(α − β) 2 1 3. sinα cos β = sin(α + β) + sin(α − β) 2 ( ) ( ( ) Rancho Cucamonga High School - Ontario, CA 1. sinα sin β = ) Use the sum and difference identities to prove these identities. ExamplE 9.7l: Compute (without a calculator) A = sin15°cos7° − cos11°cos79° − sin 4°sin86° . Solution: 1 1 1 sin 22° + sin8° − cos90° + cos68° − cos(−82°) − cos90° . 2 2 2 1 Opening the parentheses, we get A = sin 22° + sin8° − cos68° − cos82° . 2 A= ( ) ( ) ( ( ) ) (Recall that cos(−82°) = cos82° .) Note, cos 68° = cos(90° − 22°) = sin 22° and cos82° = cos(90° − 8°) = sin 8° . We have: A = 1 sin 22° + sin8° − sin 22° − sin8° = 0 . 2 ( ) ExamplE 9.7m: Show that tan 30° + tan 40° + tan50° + tan 60° = Solution: Recall that tan α + tan β = 8cos 20° 3 . sin(α + β ) . cos α cos β Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 146 AcAdemic MatheMatics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 139 139 Hence: tan 30° + tan 40° + tan50° + tan 60° = (tan 30° + tan 60°) + (tan 40° + tan50°) = sin(30° + 60°) sin(40° + 50°) sin 90° sin 90° 4 1 + = + = + = cos30°cos60° cos 40°cos50° 3 1 cos 40°cos50° 3 cos 40°cos50° ⋅ 2 2 4 cos 40° cos 50° + 3 = . 3 cos 40° cos 50° ( ) ( ) ( ) Now we can use the identity cosα + cos β = 2cos α + β cos α − β . 2 2 4(cos10° + cos30°) 4 ⋅ 2cos 20°cos(−10°) 8cos 20° cos10° 8cos 20° = = = . We get 3cos10° 3 cos10° 3 3 cos10° kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk k k Sections 9.1 – 9.7 Section 9 .1–9 .7: SectIonS 9.1–9.7: exercises Exercises Exercises kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk k k 7 270° < α < 360° 1. Find sin 2α and cos 2α if cos α = , . 25 3 1 2. Find sin(α + β ) if sin α = , cos β = − , 90° < α < 180° , 180° < β < 270° . 4 5 3. An 18-ft ladder leans against a building so that the angle between the ground and the ladder is 70 degrees. At what height does the ladder reach the building? 4. In a triangle ABC, the length of BC is 1 inch, and angles A and B are 30° and 45° , respectively. Find the other two sides of the triangle. 5. Simplify the following expressions: ( ) 2222 a. (cos cos222218° 18–cos a. − cos72 72°) • :2cos27 2cos 27° Unauthorized Duplication Duplication is is Prohibited Prohibited Outside Outside the the Terms Terms of of Your Your License License Agreement Agreement Unauthorized 140 140 140 140 ®® mAthemAtics USAD Mathematics Resource Guide • 2016–2017 ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce Guide Guide mAthemAtics ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce 147 2013 –2014 2013 2010–2011 –2014 2010–2011 Rancho Cucamonga High School - Ontario, CA 1 To proceed, we use cosα cos β = cos(α + β) + cos(α − β) : 2 1 4 ⋅ cos90° + cos(−10°) + 3 4cos 40°cos50° + 3 = 2 = 2(0 + cos10°) + 3 = 4cos10° + 2 3 3 3 cos 40°cos50° 3 cos10° 3 ⋅ 1 cos90° + cos(−10°) (0 + cos10°) 2 2 3 4 cos10° + 2 4(cos10° + cos30°) 3 4cos10° + 2 3 = cos 30° . Hence Recall that = = 2 3cos10° 3cos10° 3cos10° ( ) ( b. cos 2 α + cos 2 β − cos α + β cos α − β c. 1 + cos α + cos 2α + cos 3α cos α + 2 cos 2 α − 1 d. sin 58° cos 52° + sin 52° cos 58° cos 72° cos 37° + sin 72° sin 37° ) 6. Prove: a. sin(−α ) = − sin α b. cos(−α ) = cos α d. cot(−α ) = − cot α kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 9.8 Graphs ofTrigonometric trigonometric Functions 9.8 Graphs of 9 .8 Graphs of Trigonometric Functions Functions You to to verify thatthat the the graphs of the functions, sine, cosine, You can canuse usea acalculator calculator verify graphs of six thetrigonometric six trigonometric functions, sine,tangent, cosine, cotangent, secant, andsec, cosecant like the shownshown here for each: tangent, cotangent, andlook csc look likegraphs the graphs here for each: y = sin x Figure 9–11 It is important to note in this graph (Figure 9–11) that for any integer k , l sin(α + 360°k ) = sin α l sin(180°k − α ) = sin α l sin( −α ) = − sin α 148 2013 –2014 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource Guide • 2016–2017 ® AcAdemic MatheMatics ResouRcedecAthlon Guide mAthemAtics ResouRce Guide 141 141 Rancho Cucamonga High School - Ontario, CA c. tan(−α ) = − tan α You can use the identities you learned in the previous sections to verify these identities. yy==cos x cosx 1 –540° –360° –180° –α 180° 0° α 540° 360° Rancho Cucamonga High School - Ontario, CA –1 Figure 9–12 It is important to note that in this graph (Figure 9–12), for any integer k , l cos(α + 360°k ) = cos α l cos( −α ) = cos α You can use the identities you learned in the previous sections to verify these identities. tanxx yy== tan –540° –360° –180° 0° α 180° 180°+α 360° 540° Figure 9–13 It is important to note in this graph (Figure 9–13) that for any integer k , l tan(α + 180°k ) = tan α l tan( −α ) = − tan α Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 142 142 USAD Mathematics Resource Guide • 2016–2017 ® Guide AcAdemic decAthlon mAthemAtics ResouRce MatheMatics ResouRce Guide 149 –2014 2013 2010–2011 You can use the identities you learned in the previous sections to verify these identities. y = cot x Rancho Cucamonga High School - Ontario, CA Figure 9–14 It is important to note in this graph (Figure 9–14) that for any integer k , l cot(α + 180°k ) = cot α l cot( −α ) = − cot α You can use the identities you learned in the previous sections to verify these identities. y = sec x 1 –540° –360° 180° –180° 360° 540° 0° –1 Figure 9–15 150 2013 –2014 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource Guide • 2016–2017 ® AcAdemic MatheMatics ResouRcedecAthlon Guide mAthemAtics ResouRce Guide 143 143 y = csc x 1 –540° –360° 180° –180° 360° 540° 0° Figure 9–16 Before proceeding, it is recommended that you revisit Section 5.6. ExamplE 9.8a: Graph the function y = 4 − 2 cos x . Solution: This function is of the form y = Af ( x) + C , where f ( x) = cos x , A = −2 , C = 4 . The graph of this function can be obtained from the graph of y = cos x in the following steps: Step 1: Construct the graph of y = 2 cos x by expanding the graph of y = cos x along the y-axis by a factor of 2. 2cos x yy == 2cosx 1 –360° –180° 180° 360° 540° 0° Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 144 144 USAD Mathematics Resource Guide • 2016–2017 ® Guide AcAdemic decAthlon mAthemAtics ResouRce MatheMatics ResouRce Guide 151 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA –1 STEP 2: Reflect the graph of \ " FRV [ with respect to the x-axis to get the graph of \ " FRV [ . y = – 2cos x y = –2cosx 1 –360° –180° 180° 360° 0° STEP 3: Move the graph of \ " FRV [ along the y-axis by 4 units to get the graph of Rancho Cucamonga High School - Ontario, CA \ " FRV [ . 2cosx yy==44––2cos x 1 –360° –180° 180° 360° 0° EXAMPLE 9.8b: Graph the function \ " FRV [ s . SOLUTION: First we rewrite this function as \ " FRV [ s . This function is of the form \ " $I D[ F & , where I [ " FRV [ , $ " , & " , D " , F " s . The graph of this function can be obtained from the graph of \ " FRV [ in the following steps: STEP 1: Move the graph of \ " FRV [ along the x-axis by s to the left to get the graph of \ " FRV [ s . 152 –2014 2013 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource RESOURCE Guide • 2016–2017 ACADEMIC DECATHLON MATHEMATICS GUIDE MATHEMATICS RESOURCE GUIDE ® 145 145 yy== cos (x+30˚) cos(x+30°) 1 –360° –180° 180° 360° 0° y y==cos 2(x+30˚) cos2(x+30°) 1 –360° –180° 180° 360° 0° Step 3: Construct the graph of y = 5cos 2( x + 30°) by expanding the graph of y = cos 2( x + 30°) along the y-axis by a factor of 5. y y==55cos2(x+30°) cos 2(x+30˚) 1 –360° –180° 180° 360° 0° Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 146 146 ® Guide AcAdemic decAthlon mAthemAtics ResouRce MatheMatics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 –2014 2013 2010–2011 153 Rancho Cucamonga High School - Ontario, CA Step 2: Construct the graph of y = cos 2( x + 30°) by shrinking the graph of y = cos( x + 30°) 1 along the x-axis by a factor of . 2 Step 4: Reflect the graph of y = 5cos 2( x + 30°) with respect to the y-axis to get the graph of y = −5cos 2( x + 30°) . yy==5–5cos2(x+30 cos 2(x+30˚) °) 1 –360° –180° 180° 360° Rancho Cucamonga High School - Ontario, CA 0° Step 5: Move the graph of y = −5cos 2( x + 30°) along the y-axis by 1 unit to get the graph of yy== 5–5cos2(x+30°)+1 cos 2(x+30˚)+1 y = −5cos 2( x + 30°) + 1 . 1 –360° –180° 180° 360° 0° Graphs of the trigonometric functions are periodic. This means that there exists a number T for which f ( x + T ) = f ( x) for any x in the domain of the function. T is called the period of the function. For the cosine and sine functions, the period is 360° because cos(α + 360°k ) = cos α and sin(α + 360°k ) = sin α . For the tangent and cotangent functions, the period is 180° because tan(α + 180°k ) = tan α and cot(α + 180°k ) = cot α . The period of a function can change if the function is of the form y = f (ax) a ≠ 0 . Recall from Section 5.6 that the graph of the function y = f (ax) shrinks or expands along the x-axis by a factor 154 –2014 2013 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource ResouRce Guide • 2016–2017 AcAdemic decAthlon mAthemAtics Guide MatheMatics ResouRce Guide ® 147 147 1 . From here we conclude that the period of y = f (ax) is determined from the period of y = f ( x) a T by the following formula: T1 = , where T1 is the period of y = f (ax) and T is the period of y = f ( x) . a of ExamplE 9.8c: Find the period of the function y = −5cos(2 x + 60°) + 1. Solution: By the formula just discussed, the period is: T1 = 360° = 180°. 2 When a periodic function is of the form y = Af ( x) , its graph expands or shrinks along the y-axis by the l The amplitude of the function y = −5cos(2 x + 60°) + 1 is −5 = 5 . l The amplitude of the function y = 4 − 2 cos x is −2 = 2 . 9.9 Inverse Trigonometric Functions 9 . 9 Inverse Trigonometric Functions Functions 9.9 Inverse Trigonometric Clearly, the sine function is a many-to-one function because for a value of y between −1 and 1, there are infinitely many values of x such that sin x = y , as is shown in Figure 9–17: = sin sinxx yy = 1 α1 α2 α3 –90 ° 0° α 4 90° α5 α6 α7 –1 Figure 9–17 Recall from Section 4 that this implies that the function y = sin x is not invertible. However, if we restrict the domain of the function to −90° ≤ x ≤ 90° , the function y = sin x over this interval is oneto-one, and, hence, it is invertible. The inverse of the function y = sin x over the interval −90° ≤ x ≤ 90° is called arcsine and is denoted by y = arcsin x . Recall, that the domain of function f will be the range of function f −1 (see Section Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 148 148 USAD Mathematics Resource Guide • 2016–2017 ® Guide AcAdemic decAthlon mAthemAtics ResouRce MatheMatics ResouRce Guide 155 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA factor A . The absolute value A is called the amplitude of the function. 4). Since we restrict the domain of y = sin x to be −90° ≤ x ≤ 90° , the range of y = arcsin x is −90° ≤ y ≤ 90° . Likewise, the range of y = sin x is the domain of y = arcsin x . Hence the domain 4). Since we restrict the domain of y = sin x to be −90° ≤ x ≤ 90° , the range of y = arcsin x is ). Sinceofwe of y = sin x to be −90° ≤ x ≤ 90° , the range of y = arcsin x is y =restrict arcsin xthe is: domain −1 ≤ x ≤ 1 . In summary: y = arcsin x if only if sin y = x and −90° ≤ y ≤ 90° . The −90° ≤ y ≤ 90° . Likewise, the range of y = sin x is the domain of y = arcsin x . Hence the domain 90° ≤ y graph ≤High 90° .school yoften = sin use x is9–18. Likewise, thex range of in the –1 domain of y = arcsin x . Hence the domain books sin instead of arcsin. Consider the following example: sin–1 y =mathematics arcsin is shown Figure of if sin y = x and −90° ≤ y ≤ 90° . The of y = arcsin x is: −1 ≤ x ≤ 1 . In summary: y = arcsin x if only 1 x is: y = arcsin x if only x and ° ≤1y=≤ sin 90°90. if sin y = 90, . The f y = arcsin −1 ≤can x ≤use 1 . In = 90, so we thissummary: easy (though semi invalid) step: and−90 thus The graph of sin y = arcsin x graph of y = arcsin x is shown in Figure 9–18. raph of y = arcsin x is shown in Figure 9–18. Rancho Cucamonga High School - Ontario, CA y = arcsin x y = arcsin x Figure 9–18 exampleS: Figure 9–18 Figure 9–18 1 1 xampleS: 2 2 1 1 arcsin 0 ==90 30°°because 30°° == 0 becausesin l1 arcsin 1sin90 arcsin = 30° because sin 30° = 2 2 2 2 exampleS: l arcsin = 30° because sin 30° = because l arcsin 0 90°not 90°there = 0 is no angle α for which sin α = 2 . does existsin since 2= arcsin 0 = 90° because sin 90° = 0 3 3 since there no°)angle arcsin 2−does not = −exist 60° because sin(is−60 = − α for which sin α = 2 . does not exist since there is no angle for which α arcsin 2 2 sin α = 2 . 2 3 3 arcsin l 3 − 3 °) = − = −60° because sin(−60 arcsin − sin(−60°) = − 2 =−602° because 2 2 ExamplE 9.9a: For any a ≤ 1 , arcsin(− a ) = − arcsin a . l ExamplE 9.9a: For any a ≤ 1 , arcsin(− a ) = − arcsin a . ExamplE 9.9a: Let: For any a ≤ 1 , arcsin(− a ) = − arcsin a . proof: (1) arcsin(− a ) = α , −90° ≤ α ≤ 90° . proof: Let: proof: Let: Therefore, (1) arcsin(sin − a )α==α−,a−. 90° ≤ α ≤ 90° . (1) arcsin(− a ) = α , −90° ≤ α ≤ 90° . Dividing both Therefore, sin αsides = − aof. this equation by −1 , we get − sin α = a . Therefore, sin α = − a . Dividing both sides of this equation by −1 , we get − sin α = a . Dividing both sides of this equation by −1 , we get − sin α = a . Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 156 USAD Mathematics Resource Guide • 2016–2017 ® 2013 –2014 2010–2011 AcAdemic MatheMatics ResouRcedecAthlon Guide mAthemAtics ResouRce Guide Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 149 149 Recall that sin(−α ) = − sin α . So, sin(−α ) = a . Note that −90° ≤ −α ≤ 90° . Thus, −α = arcsin a . Therefore: (2) α = − arcsin a From (1) and (2), we get arcsin(− a ) = − arcsin a . The cosine function, too, is not one-to-one because for a value of y between −1 and 1, there are infinitely many values of x such that cos x = y , as is shown in Figure 9–19: yy==cosx cos x 1 α1 α2 α3 α4 0° α5 180° α6 α7 α8 –1 Figure 9–19 This implies that the function y = cos x is not invertible. However, if we restrict the domain of the function to 0° ≤ x ≤ 180° , the function y = cos x over this interval is one-to-one, and, hence, it is invertible. Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 150 150 mAthemAtics ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 ® –2014 2013 2010–2011 157 Rancho Cucamonga High School - Ontario, CA 1 1 ExamplE 9.9b: arcsin − = −arcsin = −30° . 2 2 The inverse of the function y = cos x over the interval 0° ≤ x ≤ 180° is called arccosine and is denoted by y = arccos x . Again, since we take the interval 0° ≤ x ≤ 180° as the domain of the function y = cos x , the range of the function y = arccos x is 0° ≤ y ≤ 180° . Likewise, the range of y = cos x is the domain of y = arccos x . Hence the domain of y = arccos x is: −1 ≤ x ≤ 1 . In summary: y = arccos x if and only if cos y = x and 0° ≤ y ≤ 180° . The graph of y = arcsin x is shown in Figure 9–20. f(x) = cos–1(x) Rancho Cucamonga High School - Ontario, CA y = arccos x Figure 9–20 exampleS: 2 2 = 45° because cos 45° = 2 2 l arccos l arccos1 = 0° as cos 0° = 1 l arccos(−3) does not exist because there is no angle α for which cos α = −3 . l 1 1 arccos − = −120° because cos120° = − . 2 2 ExamplE 9.9c: For any a ≤ 1 , arccos(− a ) = 180° − arccos a . proof: Let: (1) arccos(− a ) = α , 0° ≤ α ≤ 180° . 158 2013 –2014 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource Guide • 2016–2017 ® AcAdemic MatheMatics ResouRcedecAthlon Guide mAthemAtics ResouRce Guide 151 151 Therefore, cos α = − a . Using the fact that cos α = − cos(180° − α ) , we get −a = cos α = − cos(180° − α ) or just a = cos(180° − α ) . Note that 0° ≤ 180° − α ≤ 180°. Thus, 180° − α = arccos a . Therefore, (2) α = 180° − arccos a . 2 2 = 180° − arccos = 180° − 45° = 135° . ExamplE 9.9d: arccos − 2 2 As with the sine and cosine functions, the tangent and cotangent functions too are not one-to-one because for any value of y there are infinitely many values of x such that tan x = y and cot x = y , as is shown in the following figures: yy = tanx = tan x α1 α 2 –90 ° 0° α 3 90° α4 α5 Figure 9–21 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 152 152 ® USAD Mathematics Resource Guide • 2016–2017 mAthemAtics ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce Guide 159 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA From (1) and (2), we get arccos(− a ) = 180° − arccos a . Figure 9–22 This implies that the functions y = tan x and y = cot x are not invertible. However, if we appropriately restrict their domains, we can make them invertible functions. For the tangent function, we only consider the interval −90° < x < 90° , and for the cotangent function, we only consider the interval 0° < x < 180° . The inverse of the function y = tan x over the interval −90° < x < 90° is called the arctangent, and is denoted by y = arctan x . The domain of this function is −∞ < x < ∞ because the range of y = tan x is any number. The range of y = arctan x is −90° < y < 90° . Note that since y = tan x is undefined for all α = ±90°(1 + 2k ), k ∈ , the angles 90° and –90° will be excluded from the range of y = arctan x . The inverse of the function y = cot x over the interval −90° < x < 90° is called the arccotangent, and is denoted by y = arc cot x . The domain of this function is also −∞ < x < ∞ and the range is 0° < y < 180° . Note, angles 0° and 180° will be excluded from the range of y = arc cot x for y = cot x is undefined for all α = 180°k , k ∈ . Thus: y = arctan x if only if tan y = x and −90° < y < 90° . y = arccot x if and only if cot y = x and 0° < x < 180° . 160 2013 –2014 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource Guide • 2016–2017 ® AcAdemic MatheMatics ResouRcedecAthlon Guide mAthemAtics ResouRce Guide 153 153 Rancho Cucamonga High School - Ontario, CA y = cot x y = arctan x Figure 9–23 y = arccot x Figure 9–24 l arctan1 = 45° because tan 45° = 1 l 3 3 arccot = 60° because cot 60° = 3 3 l arctan − 3 = −60° because tan(−60°) = − 3 l arc cot(−1) = 135° because cot135° = −1 Rancho Cucamonga High School - Ontario, CA ExamplEs: ( ) ExamplE 9.9e: Prove that for all −1 ≤ y ≤ 1 cos(arccos y ) = y . proof: For any y and x , −1 ≤ y ≤ 1 and 0° ≤ x ≤ 180° , arccos y = x if and only if cos x = y . Substituting x in the last equation, we get cos x = cos(arccos y ) = y . In the same way, we can prove that for any y , −1 ≤ y ≤ 1 , sin(arcsin y ) = y ; and for any y ∈ l tan(arctan y ) = y l cot(arc cot y ) = y (You can complete the proof of this on your own as an exercise.) ExamplE 9.9f: Prove that for all −90° ≤ x ≤ 90° arcsin(sin x) = x . proof: For any y , −1 ≤ y ≤ 1 and any x , −90° ≤ x ≤ 90° , arcsin y = x if and only if sin x = y . Substituting y in the first equation, we get arcsin y = arcsin(sin x) = x . Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 154 154 USAD Mathematics Resource Guide • 2016–2017 ® Guide AcAdemic decAthlon mAthemAtics ResouRce MatheMatics ResouRce Guide 161 –2014 2013 2010–2011 In the same way, we can prove that for any x , 0° ≤ x ≤ 180° , arccos(cos x) = x ; And for any x , −90° < x < 90° , arctan(tan x) = x ; And for any x , 0° < x < 180° , arc cot(cot x) = x . (You can complete the proof of this on your own as an exercise.) ExamplEs: l 3 3 )= 2 2 arcsin(sin 30°) = 30° l arctan(tan(−60°)) = −60° l arc cot(cot 270°) is not defined (why?) l cos(arccos Solution: Let arcsin Then, sin α = Rancho Cucamonga High School - Ontario, CA 2 ExamplE 9.9g: Compute cos(arcsin ) . 5 2 = α where −90° ≤ α ≤ 90° . 5 2 , 0° ≤ α ≤ 90° (why?) 5 2 21 2 . Therefore, cos α = 1 − sin 2 α = 1 − = ± 5 5 21 . Since 0° ≤ α ≤ 90° , cos α = 5 21 =α . From here we have arccos 5 21 2 = arcsin . Therefore, arccos 5 5 2 21 21 )= . Hence, cos(arcsin ) = cos(arccos 5 5 5 4 ExamplE 9.9h: Compute sin(arctan(− )) . 5 4 Solution: Let arctan(− ) = α , where −90° < α < 90° . 5 4 Then tan α = − , −90° < α < 0° (why?) 5 162 2013 –2014 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource Guide • 2016–2017 ® AcAdemic MatheMatics ResouRcedecAthlon Guide mAthemAtics ResouRce Guide 155 155 To find sin α , we first find cot α using the identity cot α ⋅ tan α = 1 and then we find sin α using 1 . the identity cot 2 α + 1 = sin 2 α cot α = 1 5 =− . tan α 4 1 16 4 . =± =± 25 cot α + 1 41 41 +1 16 4 . Since −90° < α < 0° , sin α = − 41 sin α = ± 1 =± 2 From here, arcsin(− 4 ) =α . 41 4 4 4 . Hence, sin(arctan( − )) = sin(arcsin( − )) = − 5 41 41 9.10 Trigonometric Equations 9 .10 Trigonometric Equations 9.10 Trigonometric Equations The simplest trigonometric equations are equations such as sin x = a , cos x = a , tan x = a and cot x = a . Since the range of the function y = sin x is −1 ≤ y ≤ 1 , the equation sin x = a has a solution if and only if −1 ≤ a ≤ 1 . 1 2 ExamplE 9.10a: Solve the equation sin 2 x = − . Solution: sin 2 x = − 1 2 1 2 x = arcsin − + 360°k 2 or 1 2 x = 180° − arcsin − + 360°k 2 1 1 Since arcsin − = − arcsin = −30° , we get 2 x = −30° + 360°k 2 2 or 2 x = (180° − (−30°)) + 360°k = 210° + 360°k . Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 156 156 ® Guide AcAdemic decAthlon mAthemAtics ResouRce MatheMatics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 –2014 2013 2010–2011 163 Rancho Cucamonga High School - Ontario, CA 4 4 Therefore, arctan(− ) = arcsin( − ). 5 41 Thus: x = −15° + 90°k or x = 105° + 180°k ExamplE 9.10b: Solve the equation Solution: 1 sin 2 x = 3 . 2 1 sin 2 x = 3 2 sin 2 x = 2 3 . ExamplE 9.10c: Solve the equation cos( x − 45°) = Solution: cos( x − 45°) = x − 45° = arccos Rancho Cucamonga High School - Ontario, CA Since 2 3 > 1 , the equation has no solution. 3 . 2 3 2 3 + 360°k 2 or x − 45° = − arccos Since arccos 3 + 360°k 2 3 = 30° , we get x − 45° = 30° + 360°k 2 or x − 45° = −30° + 360°k Thus: x = 75° + 180°k or x = 15° + 180°k Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 2010–2011 164 –2014 AcAdemic decAthlon mAthemAtics Guide MatheMatics ResouRce Guide USAD Mathematics Resource ResouRce Guide • 2016–2017 ® 157 157 ExamplE 9.10d: Solve the equation cos(−2 x) = 1 . Solution: cos(−2 x) = 1 −2 x = arccos1 + 360°k or −2 x = − arccos1 + 360°k Since arccos1 = 0° , we get −2 x = 0 + 360°k x = −180°k 2 3 Rancho Cucamonga High School - Ontario, CA Since k varies over all the integers, we can write the solution as: x = 180°k . ExamplE 9.10e: Solve the equation tan x − 30° = 3 . 2 3 Solution: tan x − 30° = 3 2 x − 30° = arctan 3 + 180°k 3 2 x − 30° = 60° + 180°k 3 x = 135° + 270°k . ExamplE 9.10f: Solve the equation cot 2 x = − Solution: cot 2 x = − 1 . 3 1 3 1 2 x = arc cot − + 180°k 3 2 x = 120° + 180°k x = 60° + 90°k . Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 158 158 USAD Mathematics Resource Guide • 2016–2017 ® Guide AcAdemic decAthlon mAthemAtics ResouRce MatheMatics ResouRce Guide 165 –2014 2013 2010–2011 ExamplE 9.10g: Solve the equation sin 4 x x 1 − cos 4 = . 2 2 2 x x 1 − cos 4 = 2 2 2 1 2x 2 x 2 x 2 x sin − cos sin + cos = 2 2 2 2 2 1 x x x x Since sin 2 + cos 2 = 1 and sin 2 − cos 2 = − cos x , we get − cos x = 2 2 2 2 2 1 Multiplying both sides of the equation by −1 , we get cos x = − 2 1 x = ± arccos − + 360°k 2 1 1 Since arccos − = 180° − arccos = 180° − 60° = 120° , we get x = ±120° + 360°k . 2 2 Rancho Cucamonga High School - Ontario, CA Solution: sin 4 ExamplE 9.10h: Solve the equation cos x + 3 sin x = 2 . Solution: cos x + 3 sin x = 2 1 3 cos x + sin x = 1 . 2 2 1 3 . Notice that sin 30° = and cos 30° = 2 2 Hence, the equation can be written as sin 30° cos x + cos 30° sin x = 1 . Using one of the sum identities we learned earlier, we get sin(30° + x) = 1 . 30° + x = 90° + 360°k x = 60° + 360°k . ExamplE 9.10i: Solve the equation cos 2 x − cos 6 x = 0 . Solution: −2sin 2x − 6x 2x + 6x sin =0 2 2 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 –2014 2010–2011 166 AcAdemic MatheMatics ResouRcedecAthlon Guide ® mAthemAtics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 159 159 −2sin(−2 x) sin 4 x = 0 2 sin 2 x sin 4 x = 0 In the last equation the product is equal to zero. Therefore, sin 2 x = 0 or sin 4 x = 0 Solving, we get 2 x = 180°k or 4 x = 180°k That is, x = 90°k Rancho Cucamonga High School - Ontario, CA or x = 45°k It is not difficult to see that the last set of solutions comprises the first one. Thus: x = 45°k . ExamplE 9.10j: Solve the equation 6sin 2 x − 5sin x + 1 = 0 . Solution: Let sin x = t . Then, we get 6t 2 − 5t + 1 = 0 . 5 ±1 12 1 t= 2 1 or t = . 3 t= Hence, sin x = 1 1 or sin x = . 3 2 From the first equation we get x = 30° + 360°k or x = 150° + 360°k 1 3 From the second equation we get x = arcsin + 360°k 1 3 or x = 180° − arcsin + 360°k Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 160 USAD Mathematics Resource Guide • 2016–2017 ® mAthemAtics ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce Guide 167 –2014 2013 2010–2011 x x cos x = 2 . ExamplE + cos x = 2 . sin 2 + ExamplE 9.10k: 9.10k: Solve Solve the the equation equation sin 2 Since = sin sin xx and Since the the functions functions yy = and yy = = cos cos xx xx xx sin + cos = 22 if sin 2 + cos xx = sin 2 and if and and only only if if each each of of the the addends addends sin and 2 2 Solution: Solution: cannot cannot exceed exceed the the value value 1, 1, cos is 1. 1. Hence, Hence, we we have have the the cos xx is following following system system of of equations: equations: xx sin sin = =1 1 2 2 cos x = 1 cos x = 1 xx Solving = 90 + 360 90°° + 360°°kk Solving the the first first equation, equation, we we get get 2 = 2 Solving Solving the the second second equation, equation, we we get get xx = = 360 360°°kk .. The 180°° + + 720 720°°kk11 = = 360 360kk22 ?? and kk22 such such that that 180 The question question now now is: is: Are Are there there integers integers kk11 and This 180°° = = 360 360°°((kk22 − − 22kk11 )) This is is equivalent equivalent to to 180 1 1 k − 2k . The = k22 − 2k11 . The last last equation, equation, in in turn, turn, is is equivalent equivalent to to 2 = 2 This This obviously obviously is is not not possible, possible, since since the the difference difference between between two two integers integers cannot cannot be be aa fraction. fraction. Hence, Hence, our our equation equation has has no no solution. solution. k kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk k Section 9 .9.8 8–9 .–10: Sections 9.10 SEcTionS Exercises SEcTionS 9.8–9.10: 9.8–9.10:Exercises Exercises Exercises Exercises k kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk k 1. 1. Which Which of of the the following following expressions expressions are are meaningful? meaningful? a. a. arccos( arccos(− −1) 1) b. b. tan(arctan( tan(arctan(− −3)) 3)) 22 c. sin(arcsin(− − 2 )) )) c. sin(arcsin( 2 2. 2. Compute Compute the the following: following: 24 24 a. tan(arcsin 25 )) a. tan(arcsin 25 168 –2014 2013 2010–2011 2013 –2014 2010–2011 Unauthorized Unauthorized Duplication Duplication is is Prohibited Prohibited Outside Outside the the Terms Terms of of Your Your License License Agreement Agreement USAD Mathematics Resource Guide • 2016–2017 AcAdemic decAthlon ® mAthemAtics ResouRce Guide MatheMatics ResouRce Guide AcAdemic mAthemAtics ResouRce Guide MatheMatics ResouRcedecAthlon Guide ® 161 161 161 161 Rancho Cucamonga High School - Ontario, CA xx = = 180 180°° + + 720 720°°kk .. 1 b. cos(arctan( − )) 3 3. Solve the following equations: ( ) a. sin −4x + 30° = − b. 1 − 2sin 2 x = ( 1 2 7 sin x 3 ) ( ) c. sin 15° + x + sin 45° − x = 1 d. 3 sin x − cos x = 3 f. 1 + cos x + cos 2 x = 0 x x − sin 4 2 2 1 = 25 h. cot 4 2 x + sin 4 2 x g. sin 2 x = cos 4 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 9.11 The Law of Sines and Cosines 9.11 The Law of SineS and CoSineS 9 .11 The Law of Sines and Cosines So far, we have learned how the trigonometric functions can help us solve geometric problems involving right triangles. In this section, you will learn two theorems that will help us solve geometric problems involving any triangle. The Law of Sines: Let ABC be any triangle, with sides a , b , c , and angles α , β , and γ , as shown in a b c Figure 9–25. Let r be the radius of the circle circumscribing the triangle. Then = = = 2r . sin α sin β sin γ Proof: Let ABC be any triangle, with sides a , b , c , and angles α , β , and γ , as shown in Figure 9–25. We are to prove: a b c . = = sin α sin β sin γ a b c 2) = = = 2r . sin α sin β sin γ 1) Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 162 162 USAD Mathematics Resource Guide • 2016–2017 ® Guide AcAdemic decAthlon mAthemAtics ResouRce MatheMatics ResouRce Guide 169 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA e. cos 2 x − 3sin x cos x = −1 C γ aa bb α A β cc B Figure 9–25 Proof of 1): We know that l l 1 ab sin γ 2 1 area ( ∆ABC ) = bc sin α 2 1 area ( ∆ABC ) = ac sin β . 2 area ( ∆ABC ) = Rancho Cucamonga High School - Ontario, CA l Hence: 1 1 ab sin γ = bc sin α 2 2 1 1 bc sin α = ac sin β (2) 2 2 (1) From (1), we get (3) a c . = sin α sin γ From (2), we get (4) a b . = sin α sin β From (3) and (4), we get (5) a b c . = = sin α sin β sin γ Proof of 2): It is sufficient to prove that a = 2r . sin α Circumscribe the triangle ABC. 170 –2014 2013 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource ResouRce Guide • 2016–2017 AcAdemic decAthlon mAthemAtics Guide MatheMatics ResouRce Guide ® 163 163 CaSe 1: α = 90° The center of the circle is on the side BC as is shown in Figure 9–26. C In this case, BC = a = 2r . γ a 2r 2r = = = 2r . Hence: sin α sin 90° 1 bb a a O A α β B Figure 9–26 CaSe 2 α ≠ 90° Let CA1 be a diameter. Clearly triangle A1BC is a right triangle (why?) with angle CBA1 = 90° . C γ a a bb A O α β cc B A1 Figure 9–27 We have: sin A1 = Therefore, a a . = A1C 2r a = 2r . sin A1 Since angle A1 and angle α are inscribed within the same arc, they are equal, and, hence, sin α = sin A1 . This shows that a = 2r , as was required. sin α Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 164 164 ® Guide AcAdemic decAthlon mAthemAtics ResouRce MatheMatics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 –2014 2013 2010–2011 171 Rancho Cucamonga High School - Ontario, CA cc Note, however, that the points $ and % can be on the same side of the diameter &$ , as is shown in FIGURE 9–28. C γ b b aa A α O cc β A1 B Rancho Cucamonga High School - Ontario, CA FIGURE 9–28 In this case, F $ " , or $ " F . Still, VLQ $ " VLQ s F " VLQ F , and so the relation D " U still holds. VLQ F EXAMPLE 9.11a: In the triangle ABC $& " , *$ " s , *& " s . Find side AB. SOLUTION: To find side $% " F we will use the Law of Sines: Since *% " s s s " s , we have F " VLQ s " VLQ s F " . VLQ % VLQ s " " . C 60° aa 12 75° A cc B The Law of Cosines: Let $%& be any triangle, with sides D , E , F , and angles �, �, and �, as shown in FIGURE 9–29. Then: 172 –2014 2013 2010–2011 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement USAD Mathematics Resource RESOURCE Guide • 2016–2017 ® MATHEMATICS GUIDE ACADEMIC MATHEMATICS RESOURCEDECATHLON GUIDE 165 165 a 2 = b2 + c 2 − 2bc cos α C b = a + c − 2ac cos β γ 2 2 2 c = a + b − 2ab cos γ 2 2 2 aa bb α β cc A B Figure 9–29 Proof: We will prove the first equality a 2 = b2 + c 2 − 2bc cos α . The proof for the other two are analogous. Situate the triangle in a coordinate system as is shown in Figure 9–30. C (b cos , b αsin bsin ) ) C (bcos α, β a a bb γ α A cc B (c, c 0) xx Figure 9–30 We have A = (0, 0) , B = (c, 0) and C = (b cos α , b sin α ) (Why? Hint: drop the altitude CH from the vertex C to the segment AB and consider the right triangle ACH). 2 2 2 2 2 2 2 2 2 By the distance formula: BC = a = (b cos α − c ) + (b sin α − 0) = b cos α + b sin α − 2bc cos α + c = b2 (sin 2 α + cos2 α ) − 2bc cos α + c 2 = b2 + c 2 − 2bc cos α . Thus, a 2 = b2 + c 2 − 2bc cos α . ExamplE 9.11b: In the triangle ABC A = 60° and AB = 6 6 , AC = 12 . Find side BC. Solution: By the Law of Cosines Theorem, we have: Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 166 ® mAthemAtics ResouRce Guide AcAdemic decAthlon MatheMatics ResouRce Guide USAD Mathematics Resource Guide • 2016–2017 –2014 173 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA yy ( ) 22 1 BC 22 = 6 6 +1222 − 2 ⋅ 6 6 ⋅12 ⋅ cos60° = 216 +144 −144 6 ⋅ = 360 − 72 6 . 2 Then, BC = 360 − 72 6 = 36(10 − 2 6) = 6 10 − 2 6 . Section 9.11 kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk Section 9 .11: Exercises SeCTion 9.11: exercises Exercises kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 1. Find all the sides and angles of the triangle ABC if C γ a a bb b. A = 80°, a = 16, b = 10 c. a = 14, b = 18, c = 20 . α 2. Determine the angles of a triangle ABC if its sides are A cc β B a. 5, 4, and 4 b. 17, 8, and 15 c. 9, 5, and 6. 3. Find the bisectors in the triangle with sides 5, 6, and 7. 4. A group of students was assigned to determine the width of a river. They marked two trees on one bank of the river, A and B, and measured the distance between them. The distance was 70 yards. On the other bank of the river, adjacent to the water, they marked another tree C. They also found that CAB = 12°30′ and ABC = 72°42′ . The students determined the width of the river from this information. What is the river’s width? 5. Use the Law of Cosines to prove the Pythagorean Theorem. 6. State the converse of the Pythagorean Theorem and prove it using the Law of Cosines. kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk 9.12 Radians 9 .12 Radians 9.12 Radians So far we have measured angles by degree units. In this section, we will learn how to measure angles by length units, called radians. To define the radian measure of an angle α , take any circle with radius r Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement 2013 174 –2014 2010–2011 ® AcAdemic decAthlon mAthemAtics Guide USAD Mathematics Resource ResouRce Guide • 2016–2017 MatheMatics ResouRce Guide ® 167 167 Rancho Cucamonga High School - Ontario, CA a. A = 60°, B = 40°, c = 14 centered at the vertex of F , as shown in FIGURE 9–31, and let O be the length of the arc intercepted by F . centered at the vertex of F , as shown in FIGURE 9–31, and let O be the length of the arc intercepted by F . O is considered to be positive if F is a positive angle and negative if F is a negative angle. O is considered to be positive if F is a positive angle and negative if F is a negative angle. l l r r FIGURE 9–31 FIGURE 9–31 O The radian measure of the angle F is defined to be the ratio, O . The radian measure of the angle F is defined to be the ratio, U . U For this definition to be meaningful, we must show that the radian measure of the angle F is independent For this definition to be meaningful, we must show that the radian measure of the angle F is independent of the choice of the radius U . In the exercises at the end of this section, you will be guided to justify this of the choice of the radius U . In the exercises at the end of this section, you will be guided to justify this fact. fact. O F ininradians radian is Since the ratio, O , is independent of the radius U , we may choose U " . So, the measure of � F ininradians radian is Since the ratio, U , is independent of the radius U , we may choose U " . So, the measure of � the length of theUarc intercepted by F in a circle of radius 1. When we say the radian measure of an angle the length of the arc intercepted by F in a circle of radius 1. When we say the radian measure of an angle F is, for example, 1.5, we mean that the length of the arc intercepted by F in a circle of radius 1 is . F is, for example, 1.5, we mean that the length of the arc intercepted by F in a circle of radius 1 is . Thus, it easy to see that: Thus, it easy to see that: 1. If the degree measure of an angle F is s , then the radian measure of F is U . 1. If the degree measure of an angle F is s , then the radian measure of F is U . 2. If the degree measure of an angle F is s , then the radian measure of F is U . 2. If the degree measure of an angle F is s , then the radian measure of F is U . U 3. If the degree measure of an angle F is s , then the radian measure of F is U . 3. If the degree measure of an angle F is s , then the radian measure of F is . And in general: And in general: U 4. If the degree measure of an angle F is Qs , then the radian measure of F is U Q . 4. If the degree measure of an angle F is Qs , then the radian measure of F is Q . We can also convert radians to degrees: We can also convert radians to degrees: © ¹ 5. If the radian measure of an angle F is 1, then the degree measure of F is ª© º¹s . 5. If the radian measure of an angle F is 1, then the degree measure of F is «ª U »º s . « U » 168 168 Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement Unauthorized Duplication is Prohibited Outside the Terms of Your License Agreement GUIDE ACADEMIC DECATHLON ® MATHEMATICS RESOURCE MATHEMATICS RESOURCE GUIDE ® USAD Mathematics Resource Guide • 2016–2017 RESOURCE GUIDE ACADEMIC DECATHLON MATHEMATICS MATHEMATICS RESOURCE GUIDE –2014 2013 2010–2011 175 –2014 2013 2010–2011 Rancho Cucamonga High School - Ontario, CA α α And in in general: general: And π π α is is ss ,, then then the the degree degree measure measure of of α α is 6. If If the the radian radian measure measure of of an an angle angle α ⋅⋅ ss °° .. is 6. 180 180 33π π , we mean that the radian measure of the angle whose degree measure is 135° When 135°° = = , we mean that the radian measure of the angle whose degree measure is 135° When we we write write 135 44 3π π 3 .. is is 44 examPLeS: examPLeS: l l l l l l π π ⋅ 360° = 2π 360 360°° = = ⋅ 360° = 2π 180 180°° π 3π 135°° = 135°° = = π ⋅⋅135 = 3π 135 180 44 180°° 180 180°° ⋅ 2.5 ≈ 143° 2.5 = ⋅ 2.5 ≈ 143° 2.5 = π π π 180 180°° π π π 30°° .. = = 30 = ⋅⋅ = 66 π π 66 k kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk k Section 9 .9.12 12: Section SeCTion 9.12: exercises Exercises Exercises Exercises k kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk k 1. 1. Find Find the the radian radian measure measure for for each each angle. angle. °° 15° 15° 30° 30° 45° 45° 60° 60° 90° 90° 120° 120° 150° 150° 180° 180° 225° 225° 270° 270° 315° 315° 360° 360° radians radians 2. 2. Determine Determine to to which which quarter quarter each each of of the the following following angles angles belongs: belongs: 22π 3 π π 5 π 7 π 2 π 15 2 π , 3π , π , 5π , 7π , 2π , 35π , 146π , 15 π , 1, 8, 2 . , , , , , , 35π , 146π , π , 1, 8, π . 33 44 66 18 22 18 12 12 55 π 3. 3. Find Find the the radian radian measures measures of of the the angles angles whose whose degree degree measures measures are: are: a. a. 45 45°°30 30′′ b. b. 12 12′′ 4. 4. Solve Solve the the following following equations: equations: a. tan(3sin π π xx)) = = 33 a. tan(3sin b. sin(cos 33 xx)) = = 11 b. sin(cos k kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk kk k 176 2013 –2014 2010–2011 2013 –2014 2010–2011 Unauthorized Unauthorized Duplication Duplication is is Prohibited Prohibited Outside Outside the the Terms Terms of of Your Your License License Agreement Agreement USAD Mathematics Resource Guide • 2016–2017 ® ® mAthemAtics ResouRce Guide AcAdemic MatheMatics ResouRce Guide AcAdemic decAthlon mAthemAtics ResouRce Guide MatheMatics ResouRcedecAthlon Guide 169 169 Rancho Cucamonga High School - Ontario, CA l l