Gr12 Mathematics Learner Eng PDF

EVERYTHING MATHS BY VERSION 1 CAPS VERSION 1 CAPS GRAdE 12 MATHEMATICS GRAdE 12 MATHEMATICS WRITTEN BY VOLUNTEERS THIS TEXTBOOK IS AVAILABLE ON YOUR MOBILE WRITTEN BY VOLUNTEERS WRITTEN BY VOLUNTEERS 21 6 89 4 27 000 39 MATHEMATICS GRadE 12 This book is available on web, mobi and Mxit. Read, check solutions and practise intelligently at www.everythingmaths.co.za 2.5 29 4 100 Trigonometry exercises in this book Geometry exercises in this book Algebra exercises in this book Everything Maths 179 Litres of ink used in the production of all the grade 10, 11 and 12 textbooks Breadth of this book (cm) Litres of glue used in the production of all the grade 10, 11 and 12 textbooks Height of this book (cm) Depth of this book (cm) 39 333 Number of words used in this book Hours spent being taught this book Number of pages Hours spent doing homework from this book 87 3 759 61 3.75 52 913 249 5 191 5 690 Length of pages side by side (cm) Length of pages top to bottom (cm) 30 81 3 800 How many times student scratches head while reading this book How many times student picks nose while reading this book Number of females who helped write this book Masters students who contributed to this book Hours spent getting book to school per week Number of males who helped write this book Honours students who contributed to this book Hours spent getting book home per week Undergraduate students who contributed to this book Hours spent with book in class per week How many times student clicks pen while reading this book 50 3.75 19 179 46 7 40 206 2 5 0.083 342 3 290 63 Average size of class being taught from this book Number of pages in Grade 12 Maths textbook Average age of a maths teacher teaching from this book Number of pages in Grade 11 Maths textbook Number of pages in Grade 10 Maths textbook Number of Afrikaans volunteers who helped write this book Number of English volunteers who helped write this book 36 6 1 Number of hours spent conceptualising this cover Weekly UCT hackathons that contributed to this book Number of hours it takes to manufacture this book Small office hackathons that contributed to this book Number of hours spent designing this cover Afrikaans hackathons that contributed to this book Virtual hackathons that contributed to this book EVERYTHING MATHS GRADE 12 MATHEMATICS VERSION 1 CAPS WRITTEN BY VOLUNTEERS COPYRIGHT NOTICE Your freedom to legally copy this book You are allowed and encouraged to copy any of the Everything Maths and Everything Science textbooks. 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Mathematics is the language that nature speaks to us in. As we learn to understand and speak this language, we can discover many of nature’s secrets. Just as understanding someone’s language is necessary to learn more about them, mathematics is required to learn about all aspects of the world – whether it is physical sciences, life sciences or even finance and economics. The great writers and poets of the world have the ability to draw on words and put them together in ways that can tell beautiful or inspiring stories. In a similar way, one can draw on mathematics to explain and create new things. Many of the modern technologies that have enriched our lives are greatly dependent on mathematics. DVDs, Google searches, bank cards with PIN numbers are just some examples. And just as words were not created specifically to tell a story but their existence enabled stories to be told, so the mathematics used to create these technologies was not developed for its own sake, but was available to be drawn on when the time for its application was right. There is in fact not an area of life that is not affected by mathematics. Many of the most sought after careers depend on the use of mathematics. Civil engineers use mathematics to determine how to best design new structures; economists use mathematics to describe and predict how the economy will react to certain changes; investors use mathematics to price certain types of shares or calculate how risky particular investments are; software developers use mathematics for many of the algorithms (such as Google searches and data security) that make programmes useful. But, even in our daily lives mathematics is everywhere – in our use of distance, time and money. Mathematics is even present in art, design and music as it informs proportions and musical tones. The greater our ability to understand mathematics, the greater our ability to appreciate beauty and everything in nature. Far from being just a cold and abstract discipline, mathematics embodies logic, symmetry, harmony and technological progress. More than any other language, mathematics is everywhere and universal in its application. SPONSOR This textbook was developed with corporate social investment funding from MMI Holdings. Well structured, impactful Corporate Social Investment (CSI) has the ability to contribute positively to nation building and drive positive change in the communities. MMI’s commitment to social investment means that we are constantly looking for ways in which we can assist some of South Africa’s most vulnerable citizens to expand their horizons and gain greater access to life’s opportunities. This means that we do not view social investment as a nice to have or as an exercise in marketing or sponsorship but rather as a critical part of our contribution to society. The merger between Metropolitan and Momentum was lauded for the complementary fit between two companies. This complementary fit is also evident in the focus areas of CSI programmes where Metropolitan and Momentum together cover and support the most important sectors and where the greatest need is in terms of social participation. HIV/AIDS is becoming a manageable disease in many developed countries but in a country such as ours, it remains a disease where people are still dying of this scourge unnecessarily. Metropolitan continues to make a difference in making sure that HIV AIDS moves away from being a death sentence to a manageable disease. Metropolitan’s other focus area is education which remains the key to economic prosperity for our country. Momentum’s focus on persons with disabilities ensures that this community is included and allowed to make their contribution to society. Orphaned and vulnerable children are another focus area for Momentum and projects supported ensure that children are allowed to grow up safely, to assume their role along with other children in inheriting a prosperous future. EVERYTHING MATHS & SCIENCE The Everything Mathematics and Science series covers Mathematics, Physical Sciences, Life Sciences and Mathematical Literacy. The Siyavula Everything Science textbooks The Siyavula Everything Maths textbooks DIGITAL TEXTBOOKS READ ONLINE Watch this textbook come alive on the web. In addition to all the content in this printed copy, the online version is also full of videos, presentations and simulations to give you a more comprehensive learning experience. www.everythingmaths.co.za and www.everythingscience.co.za CHECK YOUR ANSWERS ONLINE OR ON YOUR PHONE Want the answers? 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Contents 1 Sequences and series 1.1 Arithmetic sequences. 1.2 Geometric sequences. 1.3 Series......... 1.4 Finite arithmetic series 1.5 Finite geometric series 1.6 Infinite series..... 1.7 Summary................................................................................................................................................................. 4 4 13 19 24 30 36 43 2 Functions 2.1 Revision............. 2.2 Functions and relations..... 2.3 Inverse functions........ 2.4 Linear functions......... 2.5 Quadratic functions....... 2.6 Exponential functions...... 2.7 Summary............ 2.8 Enrichment: more on logarithms........................................................................................................................................................................ 48 48 53 56 58 61 70 90 96 3 Finance 3.1 Calculating the period of an investment 3.2 Annuities................ 3.3 Future value annuities.......... 3.4 Present value annuities......... 3.5 Analysing investment and loan options 3.6 Summary...................................................................................................................... 106 106 109 109 118 125 133 4 Trigonometry 4.1 Revision................. 4.2 Compound angle identities....... 4.3 Double angle identities......... 4.4 Solving equations............ 4.5 Applications of trigonometric functions 4.6 Summary...................................................................................................................... 138 138 144 151 154 161 171 5 Polynomials 5.1 Revision......... 5.2 Cubic polynomials... 5.3 Remainder theorem... 5.4 Factor theorem..... 5.5 Solving cubic equations. 5.6 Summary.............................................................................................................. 178 178 184 191 195 199 201............................................................................ 6 Differential calculus 204 6.1 Limits................................... 204 6.2 Differentiation from first principles.................... 216 6.3 6.4 6.5 6.6 6.7 6.8 Rules for differentiation....... Equation of a tangent to a curve.. Second derivative.......... Sketching graphs.......... Applications of differential calculus Summary.......................................................................................................................... 221 224 229 230 245 256 7 Analytical geometry 7.1 Revision............. 7.2 Equation of a circle....... 7.3 Equation of a tangent to a circle 7.4 Summary................................................................................................ 264 264 275 294 305 8 Euclidean geometry 8.1 Revision....... 8.2 Ratio and proportion 8.3 Polygons....... 8.4 Triangles....... 8.5 Similarity...... 8.6 Pythagorean theorem 8.7 Summary................................................................................................................................................................................................... 312 312 319 323 327 334 348 353 9 Statistics 9.1 Revision.. 9.2 Curve fitting 9.3 Correlation 9.4 Summary............................................................................................................. 360 360 372 387 394 10 Probability 10.1 Revision............... 10.2 Identities............... 10.3 Tools and Techniques........ 10.4 The fundamental counting principle 10.5 Factorial notation.......... 10.6 Application to counting problems. 10.7 Application to probability problems 10.8 Summary...................................................................................................................................................................... 402 402 403 413 425 429 431 436 440.................... Solutions to exercises 2...... 445 Contents CHAPTER 1 Sequences and series 1.1 Arithmetic sequences 4 1.2 Geometric sequences 13 1.3 Series 19 1.4 Finite arithmetic series 24 1.5 Finite geometric series 30 1.6 Infinite series 36 1.7 Summary 43 1 Sequences and series In earlier grades we learnt about number patterns, which included linear sequences with a common difference and quadratic sequences with a common second difference. We also looked at completing a sequence and how to determine the general term of a sequence. In this chapter we also look at geometric sequences, which have a constant ratio between consecutive terms. We will learn about arithmetic and geometric series, which are the summing of the terms in sequences. 1.1 Arithmetic sequences EMCDP An arithmetic sequence is a sequence where consecutive terms are calculated by adding a constant value (positive or negative) to the previous term. We call this constant value the common difference (d). For example, 3; 0; −3; −6; −9;... This is an arithmetic sequence because we add −3 to each term to get the next term: First term Second term Third term Fourth term Fifth term... T1 T2 T3 T4 T5... 3 + (−3) = 0 + (−3) = −3 + (−3) = −6 + (−3) =... 3 0 −3 −6 −9... See video: 284G at www.everythingmaths.co.za Exercise 1 – 1: Arithmetic sequences Find the common difference and write down the next 3 terms of the sequence. 1. 2. 3. 4. 5. 6. 7. 2; 6; 10; 14; 18; 22;... −1; −4; −7; −10; −13; −16;... −5; −3; −1; 1; 3;... −1; 10; 21; 32; 43; 54;... a − 3b; a − b; a + b; a + 3b;... −2; − 32 ; −1; − 12 ; 0; 12 ; 1;... More questions. Sign in at Everything Maths online and click ’Practise Maths’. Check answers online with the exercise code below or click on ’show me the answer’. 1. 284H 2. 284J 3. 284K 4. 284M www.everythingmaths.co.za 4 5. 284N 6. 284P m.everythingmaths.co.za 1.1. Arithmetic sequences The general term for an arithmetic sequence EMCDQ For a general arithmetic sequence with first term a and a common difference d, we can generate the following terms: T1 = a T2 = T1 + d = a + d T3 = T2 + d = (a + d) + d = a + 2d T4 = T3 + d = (a + 2d) + d = a + 3d............ Tn = Tn−1 + d = (a + (n − 2)d) + d = a + (n − 1) d Therefore, the general formula for the nth term of an arithmetic sequence is: Tn = a + (n − 1) d DEFINITION: Arithmetic sequence An arithmetic (or linear) sequence is an ordered set of numbers (called terms) in which each new term is calculated by adding a constant value to the previous term: Tn = a + (n − 1)d where Tn is the nth term; n is the position of the term in the sequence; a is the first term; d is the common difference. Test for an arithmetic sequence To test whether a sequence is an arithmetic sequence or not, check if the difference between any two consecutive terms is constant: d = T2 − T1 = T3 − T2 =... = Tn − Tn−1 If this is not true, then the sequence is not an arithmetic sequence. Worked example 1: Arithmetic sequence QUESTION Given the sequence −15; −11; −7;... 173. 1. Is this an arithmetic sequence? 2. Find the formula of the general term. 3. Determine the number of terms in the sequence. Chapter 1. Sequences and series 5 SOLUTION Step 1: Check if there is a common difference between successive terms T2 − T1 = −11 − (−15) = 4 T3 − T2 = −7 − (−11) = 4 ∴ This is an arithmetic sequence with d = 4 Step 2: Determine the formula for the general term Write down the formula and the known values: Tn b 8 6 Tn = a + (n − 1)d b 4 a = −15; d=4 2 b Tn = a + (n − 1)d = −15 + (n − 1)(4) = −15 + 4n − 4 = 4n − 19 0 n 1 2 3 −2 4 5 6 7 8 b −4 −6 b −8 −10 Tn = 4n − 19 b −12 −14 b −16 −18 A graph was not required for this question but it has been included to show that the points of the arithmetic sequence lie in a straight line. Note: The numbers of the sequence are natural numbers (n ∈ {1; 2; 3;...}) and therefore we should not connect the plotted points. In the diagram above, a dotted line has been used to show that the graph of the sequence lies on a straight line. Step 3: Determine the number of terms in the sequence Tn = a + (n − 1)d 173 = 4n − 19 192 = 4n 192 ∴n= 4 = 48 ∴ T48 = 173 Step 4: Write the final answer Therefore, there are 48 terms in the sequence. 6 1.1. Arithmetic sequences Arithmetic mean The arithmetic mean between two numbers is the number half-way between the two numbers. In other words, it is the average of the two numbers. The arithmetic mean and the two terms form an arithmetic sequence. For example, the arithmetic mean between 7 and 17 is calculated: 7 + 17 2 = 12 ∴ 7; 12; 17 is an arithmetic sequence T2 − T1 = 12 − 7 = 5 T3 − T2 = 17 − 12 = 5 Arithmetic mean = Plotting a graph of the terms of a sequence sometimes helps in determining the type of sequence involved. For an arithmetic sequence, plotting Tn vs. n results in the following graph: T9 Tn = a + (n − 1)d T8 b Term: Tn T7 b T6 b T5 gradient d b T4 b T3 b T2 b T1 b 0 1 2 3 4 5 6 Index: n 7 8 9 If the sequence is arithmetic, the plotted points will lie in a straight line. Arithmetic sequences are also called linear sequences, where the common difference (d) is the gradient of the straight line. Tn = a + (n − 1)d can be written as Tn = d(n − 1) + a which is of the same form as y = mx + c Chapter 1. Sequences and series 7 Exercise 1 – 2: Arithmetic Sequences 1. Given the sequence 7; 5,5; 4; 2,5;... a) Find the next term in the sequence. b) Determine the general term of the sequence. c) Which term has a value of −23? 2. Given the sequence 2; 6; 10; 14;... a) b) c) d) Is this an arithmetic sequence? Justify your answer by calculation. Calculate T55. Which term has a value of 322? Determine by calculation whether or not 1204 is a term in the sequence? 3. An arithmetic sequence has the general term Tn = −2n + 7. a) Calculate the second, third and tenth terms of the sequence. b) Draw a diagram of the sequence for 0 < n ≤ 10. 4. The first term of an arithmetic sequence is − 12 and T22 = 10. Find Tn. 5. What are the important characteristics of an arithmetic sequence? 6. You are given the first four terms of an arithmetic sequence. Describe the method you would use to find the formula for the nth term of the sequence. 7. A single square is made from 4 matchsticks. To make two squares in a row takes 7 matchsticks, while three squares in a row takes 10 matchsticks. a) b) c) d) e) Write down the first four terms of the sequence. What is the common difference? Determine the formula for the general term. How many matchsticks are in a row of 25 squares? If there are 109 matchsticks, calculate the number of squares in the row. 8. A pattern of equilateral triangles decorates the border of a girl’s skirt. Each triangle is made by three stitches, each having a length of 1 cm. 1 3 2 a) Complete the table: Figure no. No. of stitches 1 3 2 5 3 p q 15 r 71 n s b) The border of the skirt is 2 m in length. If the entire length of the border is decorated with the triangular pattern, how many stitches will there be? 8 1.1. Arithmetic sequences 9. The terms p; (2p + 2); (5p + 3) form an arithmetic sequence. Find p and the 15th term of the sequence. [IEB, Nov 2011] 10. The arithmetic mean of 3a − 2 and x is 4a − 4. Determine the value of x in terms of a. 11. Insert seven arithmetic means between the terms (3s − t) and (−13s + 7t). 12. More questions. Sign in at Everything Maths online and click ’Practise Maths’. Check answers online with the exercise code below or click on ’show me the answer’. 1. 284Q 2. 284R 3. 284S 4. 284T 5. 284V 6. 284W 7. 284X 8. 284Y 9. 284Z 10. 2852 11. 2853 www.everythingmaths.co.za m.everythingmaths.co.za DEFINITION: Quadratic sequence A quadratic sequence is a sequence of numbers in which the second difference between any two consecutive terms is constant. The general formula for the nth term of a quadratic sequence is: Tn = an2 + bn + c Tn 1st difference 2nd difference n=1 n=2 n=3 n=4 a+b+c 4a + 2b + c 9a + 3b + c 16a + 4b + c 3a + b 5a + b 2a 7a + b 2a It is important to note that the first differences of a quadratic sequence form an arithmetic sequence. This sequence has a common difference of 2a between consecutive terms. In other words, a linear sequence results from taking the first differences of a quadratic sequence. Worked example 2: Quadratic sequence QUESTION Consider the pattern of white and blue blocks in the diagram below. 1. Determine the sequence formed by the white blocks (w). 2. Find the sequence formed by the blue blocks (b). Chapter 1. Sequences and series 9 1 2 3 4 Pattern number (n) No. of white blocks (w) Common difference (d) 1 2 3 4 5 6 n Pattern number (n) No. of blue blocks (b) Common difference (d) 1 2 3 4 5 6 n SOLUTION Step 1: Use the diagram to complete the table for the white blocks Pattern number (n) No. of white blocks (w) Common difference (d) 1 4 2 8 4 3 12 4 4 16 4 5 20 4 6 24 4 n 4n We see that the next term in the sequence is obtained by adding 4 to the previous term, therefore the sequence is linear and the common difference (d) is 4. The general term is: Tn = a + (n − 1)d = 4 + (n − 1)(4) = 4 + 4n − 4 = 4n Step 2: Use the diagram to complete the table for the blue blocks Pattern number (n) No. of blue blocks (b) Difference 1 0 2 1 1 3 4 3 4 9 5 5 16 7 6 25 9 We notice that there is no common difference between successive terms. However, there is a pattern and on further investigation we see that this is in fact a quadratic sequence: 10 1.1. Arithmetic sequences Pattern number (n) No. of blue blocks (b) First difference Second difference Pattern 1 2 3 4 5 6 n 0 1 4 9 16 25 (n−1)2 − 1 3 5 7 9 − − − 2 2 2 2 − (1−1)2 (2−1)2 (3−1)2 (4−1)2 (5−1)2 (6−1)2 (n−1)2 Tn = (n − 1)2 Step 3: Draw a graph of Tn vs. n for each sequence Tn 26 b 24 b 22 20 b 18 Tn = 4n 16 White blocks: Tn = 4n b b 14 Blue blocks: Tn = (n − 1)2 = n2 − 2n + 1 12 b 10 b 8 Tn = (n − 1)2 b 6 4 b b 2 b n b 0 1 2 3 4 5 6 7 8 9 Since the numbers of the sequences are natural numbers (n ∈ {1; 2; 3;...}), we should not connect the plotted points. In the diagram above, a dotted line has been used to show that the graph of the sequence formed by the white blocks (w) is a straight line and the graph of the sequence formed by the blue blocks (b) is a parabola. Chapter 1. Sequences and series 11 Exercise 1 – 3: Quadratic sequences 1. Determine whether each of the following sequences is: a linear sequence; a quadratic sequence; or neither. a) b) c) d) 8; 17; 32; 53; 80;... 3p2 ; 6p2 ; 9p2 ; 12p2 ; 15p2 ;... 1; 2,5; 5; 8,5; 13;... 2; 6; 10; 14; 18;... e) f) g) h) 5; 19; 41; 71; 109;... 3; 9; 16; 21; 27;... 2k; 8k; 18k; 32k; 50k;... 2 21 ; 6; 10 12 ; 16; 22 12 ;... 2. A quadratic pattern is given by Tn = n2 + bn + c. Find the values of b and c if the sequence starts with the following terms: −1 ; 2 ; 7 ; 14 ;... 3. a2 ; −a2 ; −3a2 ; −5a2 ;... are the first 4 terms of a sequence. a) Is the sequence linear or quadratic? Motivate your answer. b) What is the next term in the sequence? c) Calculate T100. 4. Given Tn = n2 + bn + c, determine the values of b and c if the sequence starts with the terms: 2 ; 7 ; 14 ; 23 ;... 5. The first term of a quadratic sequence is 4, the third term is 34 and the common second difference is 10. Determine the first six terms in the sequence. 6. A quadratic sequence has a second term equal to 1, a third term equal to −6 and a fourth term equal to −14. a) Determine the second difference for this sequence. b) Hence, or otherwise, calculate the first term of the pattern. 7. More questions. Sign in at Everything Maths online and click ’Practise Maths’. Check answers online with the exercise code below or click on ’show me the answer’. 1a. 2854 1b. 2855 1c. 2856 1d. 2857 1e. 2858 1f. 2859 1g. 285B 1h. 285C 2. 285D 3. 285F 4. 285G 5. 285H 6. 285J www.everythingmaths.co.za 12 m.everythingmaths.co.za 1.1. Arithmetic sequences 1.2 Geometric sequences EMCDR DEFINITION: Geometric sequence A geometric sequence is a sequence of numbers in which each new term (except for the first term) is calculated by multiplying the previous term by a constant value called the constant ratio (r). See video: 285K at www.everythingmaths.co.za This means that the ratio between consecutive numbers in a geometric sequence is a constant (positive or negative). We will explain what we mean by ratio after looking at the following example. Example: A flu epidemic EMCDS Influenza (commonly called “flu”) is caused by the influenza virus, which infects the respiratory tract (nose, throat, lungs). It can cause mild to severe illness that most of us get during winter time. The influenza virus is spread from person to person in respiratory droplets of coughs and sneezes. This is called “droplet spread”. This can happen when droplets from a cough or sneeze of an infected person are propelled through the air and deposited on the mouth or nose of people nearby. It is good practice to cover your mouth when you cough or sneeze so as not to infect others around you when you have the flu. Regular hand washing is an effective way to prevent the spread of infection and illness. Assume that you have the flu virus, and you forgot to cover your mouth when two friends came to visit while you were sick in bed. They leave, and the next day they also have the flu. Let’s assume that each friend in turn spreads the virus to two of their friends by the same droplet spread the following day. Assuming this pattern continues and each sick person infects 2 other friends, we can represent these events in the following manner: Each person infects two more people with the flu virus. Chapter 1. Sequences and series 13 We can tabulate the events and formulate an equation for the general case: Day (n) 1 2 3 4 5... n No. of newly-infected people 2=2 4 = 2 × 2 = 2 × 21 8 = 2 × 4 = 2 × 2 × 2 = 2 × 22 16 = 2 × 8 = 2 × 2 × 2 × 2 = 2 × 23 32 = 2 × 16 = 2 × 2 × 2 × 2 × 2 = 2 × 24... 2 × 2 × 2 × 2 × · · · × 2 = 2 × 2n−1 The above table represents the number of newly-infected people after n days since you first infected your 2 friends. You sneeze and the virus is carried over to 2 people who start the chain (a = 2). The next day, each one then infects 2 of their friends. Now 4 people are newly-infected. Each of them infects 2 people the third day, and 8 new people are infected, and so on. These events can be written as a geometric sequence: 2; 4; 8; 16; 32;... Note the constant ratio (r = 2) between the events. Recall from the linear arithmetic sequence how the common difference between terms was established. In the geometric sequence we can determine the constant ratio (r) from: T2 T3 = =r T1 T2 More generally, Tn =r Tn−1 Exercise 1 – 4: Constant ratio of a geometric sequence Determine the constant ratios for the following geometric sequences and write down the next three terms in each sequence: 1. 2. 3. 4. 5. 6. 5; 10; 20;... 1 1 1 2; 4; 8;... 7; 0,7; 0,07;... p; 3p2 ; 9p3 ;... −3; 30; −300;... More questions. Sign in at Everything Maths online and click ’Practise Maths’. Check answers online with the exercise code below or click on ’show me the answer’. 1. 285M 2. 285N 3. 285P 4. 285Q www.everythingmaths.co.za 14 5. 285R m.everythingmaths.co.za 1.2. Geometric sequences The general term for a geometric sequence EMCDT From the flu example above we know that T1 = 2 and r = 2, and we have seen from the table that the nth term is given by Tn = 2 × 2n−1. The general geometric sequence can be expressed as: T1 T2 T3 T4 Tn =a =a×r =a×r×r =a×r×r×r = a × [r × r... (n − 1) times] = ar0 = ar1 = ar2 = ar3 = arn−1 Therefore the general formula for a geometric sequence is: Tn = arn−1 where a is the first term in the sequence; r is the constant ratio. Test for a geometric sequence To test whether a sequence is a geometric sequence or not, check if the ratio between any two consecutive terms is constant: T2 T3 Tn = = =r T1 T2 Tn−1 If this condition does not hold, then the sequence is not a geometric sequence. Exercise 1 – 5: General term of a geometric sequence Determine the general formula for the nth term of each of the following geometric sequences: 1. 2. 3. 4. 5. 6. 5; 10; 20;... 1 1 1 2; 4; 8;... 7; 0,7; 0,07;... p; 3p2 ; 9p3 ;... −3; 30; −300;... More questions. Sign in at Everything Maths online and click ’Practise Maths’. Check answers online with the exercise code below or click on ’show me the answer’. 1. 285S 2. 285T 3. 285V 4. 285W www.everythingmaths.co.za 5. 285X m.everythingmaths.co.za Chapter 1. Sequences and series 15 Worked example 3: Flu epidemic QUESTION We continue with the previous flu example, where Tn is the number of newly-infected people after n days: Tn = 2 × 2n−1 1. Calculate how many newly-infected people there are on the tenth day. 2. On which day will 16 384 people be newly-infected? SOLUTION Step 1: Write down the known values and the general formula a=2 r=2 Tn = 2 × 2n−1 Step 2: Use the general formula to calculate T10 Substitute n = 10 into the general formula: Tn = a × rn−1 ∴ T10 = 2 × 210−1 = 2 × 29 = 2 × 512 = 1024 On the tenth day, there are 1024 newly-infected people. Step 3: Use the general formula to calculate n We know that Tn = 16 384 and can use the general formula to calculate the corresponding value of n: Tn = arn−1 16 384 = 2 × 2n−1 16 384 = 2n−1 2 8192 = 2n−1 We can write 8192 as 213 So 213 = 2n−1 ∴ 13 = n − 1 ∴ n = 14 (same bases) There are 16 384 newly-infected people on the 14th day. 16 1.2. Geometric sequences For this geometric sequence, plotting the number of newly-infected people (Tn ) vs. the number of days (n) results in the following graph: Tn 16 b Tn = 2 × 2n−1 14 Day (n) 1 2 3 4 5 6 n No. of newly-infected people 2 4 8 16 32 64 2 × 2n−1 12 10 8 b 6 4 2 0 b b n 1 2 3 4 5 6 7 8 9 10 In this example we are only dealing with positive integers (n ∈ {1; 2; 3;...}, Tn ∈ {1; 2; 3;...}), therefore the graph is not continuous and we do not join the points with a curve (the dotted line has been drawn to indicate the shape of an exponential graph). Geometric mean The geometric mean between two numbers is the value that forms a geometric sequence together with the two numbers. For example, the geometric mean between 5 and 20 is the number that has to be inserted between 5 and 20 to form the geometric sequence: 5; x; 20 20 x = 5 x ∴ x2 = 20 × 5 Determine the constant ratio: x2 = 100 x = ±10 Important: remember to include both the positive and negative square root. The geometric mean generates two possible geometric sequences: 5; 10; 20;... 5; −10; 20;... In general, the geometric mean (x) between two numbers a and b forms a geometric sequence with a and b: For a geometric sequence: a; x; b b x = a x x2 = ab √ ∴ x = ± ab Determine the constant ratio: Chapter 1. Sequences and series 17 Exercise 1 – 6: Mixed exercises 1. The nth term of a sequence is given by the formula Tn = 6 1 n−1. 3 a) Write down the first three terms of the sequence. b) What type of sequence is this? 2. Consider the following terms: (k − 4); (k + 1); m; 5k The first three terms form an arithmetic sequence and the last three terms form a geometric sequence. Determine the values of k and m if both are positive integers. [IEB, Nov 2006] 3. Given a geometric sequence with second term 21 and ninth term 64. a) Determine the value of r. b) Find the value of a. c) Determine the general formula of the sequence. 4. The diagram shows four sets of values of consecutive terms of a geometric sequence with the general formula Tn = arn−1. Tn (4; 2) 2 b Tn = arn−1 (3; 1) 1 b (1; y) (2; x) b b 0 n 1 2 3 4 a) Determine a and r. b) Find x and y. c) Find the fifth term of the sequence. 5. Write down the next two terms for the following sequence: 1; sin θ; 1 − cos2 θ;... 6. 5; x; y is an arithmetic sequence and x; y; 81 is a geometric sequence. All terms in the sequences are integers. Calculate the values of x and y. 7. The two numbers 2x2 y 2 and 8x4 are given. a) Write down the geometric mean between the two numbers in terms of x and y. b) Determine the constant ratio of the resulting sequence. 1 8. Insert three geometric means between −1 and − 81. Give all possible answers. 9. More questions. Sign in at Everything Maths online and click ’Practise Maths’. Check answers online with the exercise code below or click on ’show me the answer’. 1. 285Y 2. 285Z 3. 2862 4. 2863 5. 2864 6. 2865 7. 2866 8. 2867 www.everythingmaths.co.za 18 m.everythingmaths.co.za 1.2. Geometric sequences 1.3 Series EMCDV It is often important and valuable to determine the sum of the terms of an arithmetic or geometric sequence. The sum of any sequence of numbers is called a series. Finite series We use the symbol Sn for the sum of the first n terms of a sequence {T1 ; T2 ; T3 ;... ; Tn }: Sn = T1 + T2 + T3 + · · · + Tn If we sum only a finite number of terms, we get a finite series. For example, consider the following sequence of numbers 1; 4; 9; 16; 25; 36; 49;... We can calculate the sum of the first four terms: S4 = 1 + 4 + 9 + 16 = 30 This is an example of a finite series since we are only summing four terms. Infinite series If we sum infinitely many terms of a sequence, we get an infinite series: S∞ = T1 + T2 + T3 + · · · Sigma notation EMCDW Sigma notation is a very useful and compact notation for writing the sum of a given number of terms of a sequence. P A sum may be written out using the summation symbol (Sigma), which is the capital letter “S” in the Greek alphabet. It indicates that you must sum the expression to the right of the summation symbol: For example, 5 X 2n = 2 + 4 + 6 + 8 + 10 = 30 n=1 In general, n X Ti = Tm + Tm+1 + · · · + Tn−1 + Tn i=m where i is the index of the sum; m is the lower bound (or start index), shown below the summation symbol; n is the upper bound (or end index), shown above the summation symbol; Ti is a term of a sequence; the number of terms in the series = end index − start index + 1. Chapter 1. Sequences and series 19 The index i increases from m to n by steps of 1. Note that this is also sometimes written as: n X ai = am + am+1 + · · · + an−1 + an i=m When we write out all the terms in a sum, it is referred to as the expanded form. If we are summing from i = 1 (whichPimplies summing from the first term in a sequence), then we can use either Sn or notation: Sn = n X ai = a1 + a2 + · · · + an (n terms) i=1 Worked example 4: Sigma notation QUESTION Expand the sequence and find the value of the series: 6 X 2n n=1 SOLUTION Step 1: Expand the formula and write down the first six terms of the sequence 6 X 2n = 21 + 22 + 23 + 24 + 25 + 26 (6 terms) n=1 = 2 + 4 + 8 + 16 + 32 + 64 This is a geometric sequence 2; 4; 8; 16; 32; 64 with a constant ratio of 2 between consecutive terms. Step 2: Determine the sum of the first six terms of the sequence S6 = 2 + 4 + 8 + 16 + 32 + 64 = 126 20 1.3. Series Worked example 5: Sigma notation QUESTION Find the value of the series: 7 X 2an n=3 SOLUTION Step 1: Expand the sequence and write down the five terms 7 X 2an = 2a(3) + 2a(4) + 2a(5) + 2a(6) + 2a(7) (5 terms) n=3 = 6a + 8a + 10a + 12a + 14a Step 2: Determine the sum of the five terms of the sequence S5 = 6a + 8a + 10a + 12a + 14a = 50a Worked example 6: Sigma notation QUESTION Write the following series in sigma notation: 31 + 24 + 17 + 10 + 3 SOLUTION Step 1: Consider the series and determine if it is an arithmetic or geometric series First test for an arithmetic series: is there a common difference? We let: We calculate: T1 = 31; T2 = 24; T3 = 17; T4 = 10; T5 = 3; d = T2 − T1 = 24 − 31 = −7 d = T3 − T2 = 17 − 24 = −7 There is a common difference of −7, therefore this is an arithmetic series. Chapter 1. Sequences and series 21 Step 2: Determine the general formula of the series Tn = a + (n − 1)d = 31 + (n − 1)(−7) = 31 − 7n + 7 = −7n + 38 Be careful: brackets must be used when substituting d = −7 into the general term. Otherwise the equation would be Tn = 31 + (n − 1) − 7, which would be incorrect. Step 3: Determine the sum of the series and write in sigma notation 31 + 24 + 17 + 10 + 3 = 85 ∴ 5 X (−7n + 38) = 85 n=1 Rules for sigma notation 1. Given two sequences, ai and bi : n X (ai + bi ) = i=1 n X ai + i=1 n X bi i=1 2. For any constant c that is not dependent on the index i: n X (c. ai ) = c. a1 + c. a2 + c. a3 + · · · + c. an i=1 = c (a1 + a2 + a3 + · · · + an ) n X =c ai i=1 3. Be accurate with the use of brackets: Example 1: 3 X (2n + 1) = 3 + 5 + 7 n=1 = 15 Example 2: 3 X (2n) + 1 = (2 + 4 + 6) + 1 n=1 = 13 Note: the series in the second example has the general term Tn = 2n and the +1 is added to the sum of the three terms. It is very important in sigma notation to use brackets correctly. 22 1.3. Series n X 4. The values of i: start at m (m is not always 1); increase in steps of 1; and end at n. ai i=m Exercise 1 – 7: Sigma notation 1. Determine the value of the following: 4 X a) 2 k=1 3 X b) i i=−1 5 X c) (3n − 2) n=2 2. Expand the series: 6 X a) 0k k=1 0 X b) 8 n=−3 5 X c) (ak) k=1 3. Calculate the value of a: 3 X a) a. 2k−1 = 28 k=1 4 X b) 2−j = a j=1 4. Write the following in sigma notation: 1 1 + +1+3 9 3 5. Write the sum of the first 25 terms of the series below in sigma notation: 11 + 4 − 3 − 10... 6. Write the sum of the first 1000 natural, odd numbers in sigma notation. 7. More questions. Sign in at Everything Maths online and click ’Practise Maths’. Check answers online with the exercise code below or click on ’show me the answer’. 1a. 2868 1b. 2869 1c. 286B 2a. 286C 2b. 286D 2c. 286F 3a. 286G 3b. 286H 4. 286J 5. 286K 6. 286M www.everythingmaths.co.za m.everythingmaths.co.za Chapter 1. Sequences and series 23 1.4 Finite arithmetic series EMCDX An arithmetic sequence is a sequence of numbers, such that the difference between any term and the previous term is a constant number called the common difference (d): Tn = a + (n − 1) d where Tn is the nth term of the sequence; a is the first term; d is the common difference. When we sum a finite number of terms in an arithmetic sequence, we get a finite arithmetic series. The sum of the first one hundred integers A simple arithmetic sequence is when a = 1 and d = 1, which is the sequence of positive integers: Tn = a + (n − 1) d = 1 + (n − 1) (1) =n ∴ {Tn } = 1; 2; 3; 4; 5;... If we wish to sum this sequence from n = 1 to any positive integer, for example 100, we would write 100 X n = 1 + 2 + 3 + · · · + 100 n=1 This gives the answer to the sum of the first 100 positive integers. The mathematician, Karl Friedrich Gauss, discovered the following proof when he was only 8 years old. His teacher had decided to give his class a problem which would distract them for the entire day by asking them to add all the numbers from 1 to 100. Young Karl quickly realised how to do this and shocked the teacher with the correct answer, 5050. This is the method that he used: 24 Write the numbers in ascending order. Write the numbers in descending order. Add the corresponding pairs of terms together. Simplify the equation by making Sn the subject of the equation. 1.4. Finite arithmetic series S100 + S100 ∴ 2S100 ∴ 2S100 ∴ S100 = 1 + 2 + 3 + · · · + 98 + 99 + 100 = 100 + 99 + 98 + · · · + 3 + 2 + 1 = 101 + 101 + 101 + · · · + 101 + 101 + 101 = 101 × 100 = 10 100 10 100 = 2 = 5050 General formula for a finite arithmetic series EMCDY If we sum an arithmetic sequence, it takes a long time to work it out term-by-term. We therefore derive the general formula for evaluating a finite arithmetic series. We start with the general formula for an arithmetic sequence of n terms and sum it from the first term (a) to the last term in the sequence (l): l X Tn = Sn n=1 Sn = a + (a + d) + (a + 2d) + · · · + (l − 2d) + (l − d) + l + Sn = l + (l − d) + (l − 2d) + · · · + (a + 2d) + (a + d) + a ∴ 2Sn = (a + l) + (a + l) + (a + l) + · · · + (a + l) + (a + l) + (a + l) ∴ 2Sn = n × (a + l) n ∴ Sn = (a + l) 2 This general formula is useful if the last term in the series is known. We substitute l = a + (n − 1)d into the above formula and simplify: n Sn = (a + [a + (n − 1)d]) 2 n ∴ Sn = [2a + (n − 1)d] 2 The general formula for determining the sum of an arithmetic series is given by: Sn = n [2a + (n − 1) d] 2 or Sn = n (a + l) 2 For example, we can calculate the sum S20 for the arithmetic sequence Tn = 3 + 7 (n − 1) by summing all the individual terms: S20 = 20 X [3 + 7 (n − 1)] n=1 = 3 + 10 + 17 + 24 + 31 + 38 + 45 + 52 + 59 + 66 + 73 + 80 + 87 + 94 + 101 + 108 + 115 + 122 + 129 + 136 = 1390 Chapter 1. Sequences and series 25 or, more sensibly, we could use the general formula for determining an arithmetic series by substituting a = 3, d = 7 and n = 20: n Sn = (2a + (n − 1)d) 2 20 S20 = [2(3) + 7 (20 − 1)] 2 = 1390 This example demonstrates how useful the general formula for determining an arithmetic series is, especially when the series has a large number of terms. See video: 286N at www.everythingmaths.co.za Worked example 7: General formula for the sum of an arithmetic sequence QUESTION Find the sum of the first 30 terms of an arithmetic series with Tn = 7n − 5 by using the formula. SOLUTION Step 1: Use the general formula to generate terms of the sequence and write down the known variables Tn = 7n − 5 ∴ T1 = 7(1) − 5 =2 T2 = 7(2) − 5 =9 T3 = 7(3) − 5 = 16 This gives the sequence: 2; 9; 16... a = 2; d = 7; n = 30 Step 2: Write down the general formula and substitute the known values n [2a + (n − 1)d] 2 30 = [2(2) + (30 − 1)(7)] 2 = 15(4 + 203) = 15(207) = 3105 Sn = S30 Step 3: Write the final answer S30 = 3105 26 1.4. Finite arithmetic series Worked example 8: Sum of an arithmetic sequence if first and last terms are known QUESTION Find the sum of the series −5 − 3 − 1 + · · · · · · + 123 SOLUTION Step 1: Identify the type of series and write down the known variables d = T2 − T1 = −3 − (−5) =2 d = T3 − T2 = −1 − (−3) =2 a = −5; Step 2: Determine the value of n d = 2; l = 123 Step 3: Use the general formula to find the sum of the series Tn = a + (n − 1)d ∴ 123 = −5 + (n − 1)(2) = −5 + 2n − 2 ∴ 130 = 2n ∴ n = 65 n (a + l) 2 65 = (−5 + 123) 2 65 = (118) 2 = 3835 Sn = S65 Step 4: Write the final answer S65 = 3835 Worked example 9: Finding n given the sum of an arithmetic sequence QUESTION Given an arithmetic sequence with T2 = 7 and d = 3, determine how many terms must be added together to give a sum of 2146. SOLUTION Step 1: Write down the known variables d = T2 − T1 ∴3=7−a ∴a=4 a = 4; d = 3; Sn = 2146 Chapter 1. Sequences and series 27 Step 2: Use the general formula to determine the value of n n (2a + (n − 1)d) 2 n 2146 = (2(4) + (n − 1)(3)) 2 4292 = n(8 + 3n − 3) Sn = ∴ 0 = 3n2 + 5n − 4292 = (3n + 116)(n − 37) 116 ∴n=− or n = 37 3 but n must be a positive integer, therefore n = 37. We could have solved for n using the quadratic formula but factorising by inspection is usually the quickest method. Step 3: Write the final answer S37 = 2146 Worked example 10: Finding n given the sum of an arithmetic sequence QUESTION The sum of the second and third terms of an arithmetic sequence is equal to zero and the sum of the first 36 terms of the series is equal to 1152. Find the first three terms in the series. SOLUTION Step 1: Write down the given information T2 + T3 = 0 So (a + d) + (a + 2d) = 0 ∴ 2a + 3d = 0...... (1) n (2a + (n − 1)d) 2 36 S36 = (2a + (36 − 1)d) 2 1152 = 18(2a + 35d) ∴ 64 = 2a + 35d...... (2) Sn = Step 2: Solve the two equations simultaneously 2a + 3d = 0...... (1) 2a + 35d = 64...... (2) Eqn (2) − (1) : 32d = 64 ∴d=2 And 2a + 3(2) = 0 2a = −6 ∴ a = −3 28 1.4. Finite arithmetic series Step 3: Write the final answer The first three terms of the series are: T1 = a = −3 T2 = a + d = −3 + 2 = −1 T3 = a + 2d = −3 + 2(2) = 1 −3 − 1 + 1 Calculating the value of a term given the sum of n terms: If the first term in a series is T1 , then S1 = T1. We also know the sum of the first two terms S2 = T1 +T2 , which we rearrange to make T2 the subject of the equation: T2 = S2 − T1 Substitute S1 = T1 ∴ T2 = S2 − S1 Similarly, we could determine the third and fourth term in a series: T3 = S3 − S2 And T4 = S4 − S3 Tn = Sn − Sn−1 , for n ∈ {2; 3; 4;...} and T1 = S1 Exercise 1 – 8: Sum of an arithmetic series 1. Determine the value of k: k X (−2n) = −20 n=1 2. The sum to n terms of an arithmetic series is Sn = n 2 (7n + 15). a) How many terms of the series must be added to give a sum of 425? b) Determine the sixth term of the series. 3. a) The common difference of an arithmetic series is 3. Calculate the values of n for which the nth term of the series is 93, and the sum of the first n terms is 975. b) Explain why there are two possible answers. 4. The third term of an arithmetic sequence is −7 and the seventh term is 9. Determine the sum of the first 51 terms of the sequence. 5. Calculate the sum of the arithmetic series 4 + 7 + 10 + · · · + 901. 4 + 8 + 12 + · · · + 100 6. Evaluate without using a calculator: 3 + 10 + 17 + · · · + 101 Chapter 1. Sequences and series 29 7. The second term of an arithmetic sequence is −4 and the sum of the first six terms of the series is 21. a) Find the first term and the common difference. b) Hence determine T100. [IEB, Nov 2004] 8. Determine the value of the following: 8 X a) (7w + 8) w=0 8 X b) 7j + 8 j=1 9. Determine the value of n. n X (2 − 3c) = −330 c=1 10. The sum of n terms of an arithmetic series is 5n2 − 11n for all values of n. Determine the common difference. 11. The sum of an arithmetic series is 100 times its first term, while the last term is 9 times the first term. Calculate the number of terms in the series if the first term is not equal to zero. 12. More questions. Sign in at Everything Maths online and click ’Practise Maths’. Check answers online with the exercise code below or click on ’show me the answer’. 1. 286P 2a. 286Q 2b. 286R 3. 286S 4. 286T 5. 286V 6. 286W 7. 286X 8a. 286Y 8b. 286Z 9. 2872 10. 2873 11. 2874 www.everythingmaths.co.za 1.5 m.everythingmaths.co.za Finite geometric series EMCDZ When we sum a known number of terms in a geometric sequence, we get a finite geometric series. We generate a geometric sequence using the general form: Tn = a. rn−1 where 30 n is the position of the sequence; Tn is the nth term of the sequence; a is the first term; r is the constant ratio. 1.5. Finite geometric series General formula for a finite geometric series EMCF2 Sn = a + ar + ar2 + · · · + arn−2 + arn−1... (1) r × Sn = ar + ar2 + · · · + arn−2 + arn−1 + arn...... (2) Subtract eqn. (2) from eqn. (1) ∴ Sn − rSn = a + 0 + 0 + · · · − arn Sn − rSn = a − arn Sn (1 − r) = a(1 − rn ) ∴ Sn = a(1 − rn ) 1−r (where r 6= 1) The general formula for determining the sum of a geometric series is given by: Sn = a(1 − rn ) 1−r where r 6= 1 This formula is easier to use when r < 1. Alternative formula: Sn = a + ar + ar2 + · · · + arn−2 + arn−1...... (1) r × Sn = ar + ar2 + · · · + arn−2 + arn−1 + arn...... (2) Subtract eqn. (1) from eqn. (2) ∴ rSn − Sn = arn − a Sn (r − 1) = a(rn − 1) ∴ Sn = a(rn − 1) r−1 ( where r 6= 1) The general formula for determining the sum of a geometric series is given by: Sn = a(rn − 1) r−1 where r 6= 1 This formula is easier to use when r > 1. Chapter 1. Sequences and series 31 Worked example 11: Sum of a geometric series QUESTION Calculate: 6 X 32 k=1 k−1 1 2 SOLUTION Step 1: Write down the first three terms of the series 0 1 = 32 T1 = 32 2 2−1 1 T2 = 32 = 16 2 3−1 1 T3 = 32 =8 2 k = 1; k = 2; k = 3; We have generated the series 32 + 16 + 8 + · · · Step 2: Determine the values of a and r a = T1 = 32 T2 T3 1 r= = = T1 T2 2 Step 3: Use the general formula to find the sum of the series Sn = a(1 − rn ) 1−r 32(1 − S6 = 1− = 32 1 − 1 6 2 ) 1 2 1 64 1 2 63 = 2 × 32 64 63 = 64 64 = 63 Step 4: Write the final answer 6 X k=1 32 32 k−1 1 = 63 2 1.5. Finite geometric series See video: 2875 at www.everythingmaths.co.za Worked example 12: Sum of a geometric series QUESTION Given a geometric series with T1 = −4 and T4 = 32. Determine the values of r and n if Sn = 84. SOLUTION Step 1: Determine the values of a and r a = T1 = −4 T4 = ar3 = 32 ∴ −4r3 = 32 r3 = −8 ∴ r = −2 Therefore the geometric series is −4 + 8 − 16 + 32... Notice that the signs of the terms alternate because r < 0. We write the general term for this series as Tn = −4(−2)n−1. Step 2: Use the general formula for the sum of a geometric series to determine the value of n a(1 − rn ) 1−r −4(1 − (−2)n ) ∴ 84 = 1 − (−2) −4(1 − (−2)n ) 84 = 3 Sn = 3 − × 84 = 1 − (−2)n 4 −63 = 1 − (−2)n (−2)n = 64 (−2)n = (−2)6 ∴n=6 Step 3: Write the final answer r = −2 and n = 6 Chapter 1. Sequences and series 33 Worked example 13: Sum of a geometric series QUESTION Use the general formula for the sum of a geometric series to determine k if n 8 X 1 255 k = 2 64 n=1 SOLUTION Step 1: Write down the first three terms of the series n = 1; n = 2; n = 3; 1 1 1 = k 2 2 2 1 1 = k T2 = k 2 4 3 1 1 T3 = k = k 2 8 T1 = k We have generated the series 21 k + 14 k + 18 k + · · · We can take out the common factor k and write the series as: k 8 n X 1 255 ∴k = 2 64 n=1 1 2 + 1 4 + Step 2: Determine the values of a and r 1 2 T3 1 T2 = = r= T1 T2 2 a = T1 = Step 3: Calculate the sum of the first eight terms of the geometric series a(1 − rn ) 1−r 8 1 (1 − 12 ) S8 = 2 1 − 12 1 1 8 2 (1 − 2 ) = 1 ∴ Sn = 2 1 =1− 256 255 = 256 ∴ 8 n X 1 n=1 34 2 = 255 256 1.5. Finite geometric series 1 8 + ··· So then we can write: k 8 n X 1 n=1 k 2 255 256 = 255 64 255 64 255 256 ∴k= × 64 255 256 = 64 =4 = Step 4: Write the final answer k=4 Exercise 1 – 9: Sum of a geometric series n −1) 1. Prove that a + ar + ar2 + · · · + arn−1 = a(rr−1 and state any restrictions. 2. Given the geometric sequence 1; −3; 9;... determine: a) The eighth term of the sequence. b) The sum of the first eight terms of the sequence. 4 X 3. Determine: 3. 2n−1 n=1 4. Find the sum of the first 11 terms of the geometric series 6 + 3 + 23 + 43 + · · · 5. Show that the sum of the first n terms of the geometric series 54 + 18 + 6 + · · · + n−1 5 13 is given by 81 − 34−n. 6. The eighth term of a geometric sequence is 640. The third term is 20. Find the sum of the first 7 terms. 7. Given: t n X 1 8 2 t=1 a) Find the first three terms in the series. b) Calculate the number of terms in the series if Sn = 7 63 64. 8. The ratio between the sum of the first three terms of a geometric series and the sum of the 4th , 5th and 6th terms of the same series is 8 : 27. Determine the constant ratio and the first 2 terms if the third term is 8. 9. More questions. Sign in at Everything Maths online and click ’Practise Maths’. Check answers online with the exercise code below or click on ’show me the answer’. 1. 2876 2a. 2877 2b. 2878 3. 2879 4. 287B 5. 287C 6. 287D 7a. 287F 7b. 287G 8. 287H www.everythingmaths.co.za m.everythingmaths.co.za Chapter 1. Sequences and series 35 1.6 Infinite series EMCF3 So far we have been working only with finite sums, meaning that whenever we determined the sum of a series, we only considered the sum of the first n terms. We now consider what happens when we add an infinite number of terms together. Surely if we sum infinitely many numbers, no matter how small they are, the answer goes to infinity? In some cases the answer does indeed go to infinity (like when we sum all the positive integers), but surprisingly there are some cases where the answer is a finite real number. Investigation: Sum of an infinite series 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Cut a piece of string 1 m in length. Now cut the piece of string in half and place one half on the desk. Cut the other half in half again and put one of the pieces on the desk. Repeat this process until the piece of string is too short to cut easily. Draw a diagram to illustrate the sequence of lengths of the pieces of string. Can this sequence be expressed mathematically? Hint: express the shorter lengths of string as a fraction of the original length of string. What is the sum of the lengths of all the pieces of string? Predict what would happen if these steps could be repeated infinitely many times. Will the sum of the lengths of string ever be greater than 1? What can you conclude? Worked example 14: Sum to infinity QUESTION Complete the table below for the geometric series Tn = that follow: 1 n 2 and answer the questions Terms Sn 1 − Sn 1 2 1 2 1 2 T1 T1 + T2 T1 + T2 + T3 T1 + T2 + T3 + T4 1. As more and more terms are added, what happens to the value of Sn ? 2. As more more and more terms are added, what happens to the value of 1 − Sn ? 3. Predict the maximum value of Sn for the sum of infinitely many terms in the series. 36 1.6. Infinite series SOLUTION Step 1: Complete the table Terms Sn 1 − Sn 1 2 1 2 3 4 7 8 15 16 1 2 1 4 1 8 1 16 T1 1 2 + 1 4 1 2 + 1 4 + + 1 4 + 1 8 T1 + T2 T1 + T2 + T3 1 2 T1 + T2 + T3 + T4 1 8 1 16 + Step 2: Consider the value of Sn and 1 − Sn As more terms in the series are added together, the value of Sn increases: 1 2 < 3 4 < 7 8 < ··· However, by considering 1 − Sn , we notice that the amount by which Sn increases gets smaller and smaller as more terms are added: 1 2 > 1 4 > 1 8 > ··· We can therefore conclude that the value of Sn is approaching a maximum value of 1; it is converging to 1. Step 3: Write conclusion mathematically We can conclude that the sum of the series 1 1 1 + + + ··· 2 4 8 gets closer to 1 (Sn → 1) as the number of terms approaches infinity (n → ∞), therefore the series converges. ∞ i X 1 =1 2 i=1 We express the sum of an infinite number of terms of a series as S∞ = ∞ X Ti i=1 Convergence and divergence If the sum of a series gets closer and closer to a certain value as we increase the number of terms in the sum, we say that the series converges. In other words, there is a limit to the sum of a converging series. If a series does not converge, we say that it diverges. The sum of an infinite series usually tends to infinity, but there are some special cases where it does not. Chapter 1. Sequences and series 37 Exercise 1 – 10: Convergent and divergent series For each of the general terms below: Determine if it forms an arithmetic or geometric series. Calculate S1 , S2 , S10 and S100. Determine if the series is convergent or divergent. 1. 2. 3. 4. 5. Tn = 2n Tn = (−n) Tn = ( 23 )n Tn = 2n More questions. Sign in at Everything Maths online and click ’Practise Maths’. Check answers online with the exercise code below or click on ’show me the answer’. 1. 287J 2. 287K 3. 287M 4. 287N www.everythingmaths.co.za m.everythingmaths.co.za Note the following: An arithmetic series never converges: as n tends to infinity, the series will always tend to positive or negative infinity. Some geometric series converge (have a limit) and some diverge (as n tends to infinity, the series does not tend to any limit or it tends to infinity). Infinite geometric series EMCF4 There is a simple test for determining whether a geometric series converges or diverges; if −1 < r < 1, then the infinite series will converge. If r lies outside this interval, then the infinite series will diverge. Test for convergence: If −1 < r < 1, then the infinite geometric series converges. If r < −1 or r > 1, then the infinite geometric series diverges. We derive the formula for calculating the value to which a geometric series converges as follows: n X a (1 − rn ) Sn = ari−1 = 1−r i=1 Now consider the behaviour of rn for −1 < r < 1 as n becomes larger. 38 1.6. Infinite series Let r = 12 : 1 1 1 n=1:r =r = = 2 2 2 1 1 1 1 1 n = 2 : rn = r2 = =. = < 2 2 2 4 2 3 1 1 1 1 1 1 n = 3 : rn = r3 = =.. = < 2 2 2 2 8 4 n 1 Since r is in the range −1 < r < 1, we see that rn gets closer to 0 as n gets larger. Therefore (1 − rn ) gets closer to 1. Therefore, a (1 − rn ) 1−r If − 1 < r < 1, then rn → 0 as n → ∞ Sn = a (1 − 0) 1−r a = 1−r ∴ S∞ = The sum of an infinite geometric series is given by the formula ∴ S∞ = ∞ X ari−1 = i=1 a 1−r (−1 < r < 1) where a is the first term of the series; r is the constant ratio. Alternative notation: a Sn → |{z} 1−r if − 1 < r < 1 n→∞ In words: as the number of terms (n) tends to infinity, the sum of a converging geoa metric series (Sn ) tends to the value 1−r. See video: 287P at www.everythingmaths.co.za Chapter 1. Sequences and series 39 Worked example 15: Sum to infinity of a geometric series QUESTION Given the series 18 + 6 + 2 + · · ·. Find the sum to infinity if it exists. SOLUTION Step 1: Determine the value of r We need to know the value of r to determine whether the series converges or diverges. Step 2: Determine the sum to infinity Write down the formula for the sum to infinity and substitute the known values: a = 18; T2 6 = T1 18 1 = 3 T3 2 = T2 6 1 = 3 1 ∴r= 3 r= 1 3 a 1−r 18 = 1 − 13 18 = 2 S∞ = 3 = 18 × Since −1 < r < 1, we can conclude that this is a convergent geometric series. 3 2 = 27 As n tends to infinity, the sum of this series tends to 27; no matter how many terms are added together, the value of the sum will never be greater than 27. Worked example 16: Using the sum to infinity to convert recurring decimals to fractions QUESTION Use two different methods to convert the recurring decimal 0, 5̇ to a proper fraction. SOLUTION Step 1: Convert the recurring decimal to a fraction using equations Let x = 0,5̇ ∴ x = 0,555...... (1) 10x = 5,55...... (2) (2) − (1) : 9x = 5 5 ∴x= 9 40 1.6. Infinite series Step 2: Convert the recurring decimal to a fraction using the sum to infinity or 0,5̇ = 0,5 + 0,05 + 0,005 +... 5 5 5 0,5̇ = + + +... 10 100 1000 This is a geometric series with r = 0,1 = that the series is convergent. 1 10. S∞ = = And since −1 < r < 1, we can conclude a 1−r 5 10 1− = 5 10 9 10 = 5 9 1 10 Worked example 17: Sum to infinity QUESTION Determine the possible values of a and r if ∞ X arn−1 = 5 n=1 SOLUTION Step 1: Write down the sum to infinity formula and substitute known values a 1−r a ∴5= 1−r a = 5(1 − r) ∴ a = 5 − 5r And 5r = 5 − a 5−a ∴r= 5 S∞ = Step 2: Apply the condition for convergence to determine possible values of a For a series to converge: −1 < r < 1 −1 < r < 1 5−a 1 Sum to infinity A convergent geometric series, with −1 < r < 1, tends to a certain fixed number as the number of terms in the sum tends to infinity. S∞ = ∞ X n=1 Tn = a 1−r Chapter 1. Sequences and series 43 Exercise 1 – 12: End of chapter exercises 1. Is 1 + 2 + 3 + 4 + · · · an example of a finite series or an infinite series? 2. A new soccer competition requires each of 8 teams to play every other team once. a) Calculate the total number of matches to be played in the competition. b) If each of n teams played each other once, determine a formula for the total number of matches in terms of n. 3. Calculate: k+2 6 X 1 3 3 k=2 4. The first three terms of a convergent geometric series are: x + 1; x − 1; 2x − 5. a) Calculate the value of x, (x 6= 1 or 1). b) Sum to infinity of the series. 5. Write the sum of the first twenty terms of the following series in 6+3+ 6. Determine: ∞ X k=1 P notation. 3 3 + + ··· 2 4 k−1 1 12 5 7. A man was injured in an accident at work. He receives a disability grant of R 4800 in the first year. This grant increases with a fixed amount each year. a) What is the annual increase if he received a total of R 143 500 over 20 years? b) His initial annual expenditure is R 2600, which increases at a rate of R 400 per year. After how many years will his expenses exceed his income? 8. The length of the side of a square is 4 units. This square is divided into 4 equal, smaller squares. One of the smaller squares is then divided into four equal, even smaller squares. One of the even smaller squares is divided into four, equal squares. This process is repeated indefinitely. Calculate the sum of the areas of all the squares. 9. Thembi worked part-time to buy a Mathematics book which costs R 29,50. On 1 February she saved R 1,60, and every day saves 30 cents more than she saved the previous day. So, on the second day, she saved R 1,90, and so on. After how many days did she have enough money to buy the book? 10. A plant reaches a height of 118 mm after one year under ideal conditions in a greenhouse. During the next year, the height increases by 12 mm. In each successive year, the height increases by 58 of the previous year’s growth. Show that the plant will never reach a height of more than 150 mm. 11. Calculate the value of n if: n X (20 − 4a) = −20 a=1 44 1.7. Summary 12. Michael saved R 400 during the first month of his working life. In each subsequent month, he saved 10% more than what he had saved in the previous month. a) How much did he save in the seventh working month? b) How much did he save all together in his first 12 working months? 13. The Cape Town High School wants to build a school hall and is busy with fundraising. Mr. Manuel, an ex-learner of the school and a successful politician, offers to donate money to the school. Having enjoyed mathematics at school, he decides to donate an amount of money on the following basis. He sets a mathematical quiz with 20 questions. For the correct answer to the first question (any learner may answer), the school will receive R 1, for a correct answer to the second question, the school will receive R 2, and so on. The donations 1; 2; 4;... form a geometric sequence. Calculate, to the nearest Rand: a) The amount of money that the school will receive for the correct answer to the 20th question. b) The total amount of money that the school will receive if all 20 questions are answered correctly. 14. The first term of a geometric sequence is 9, and the ratio of the sum of the first eight terms to the sum of the first four terms is 97 : 81. Find the first three terms of the sequence, if it is given that all the terms are positive. 15. Given the geometric sequence: 6 + p; 10 + p; 15 + p a) Determine p, (p 6= −6 or − 10). b) Show that the constant ratio is 45. c) Determine the tenth term of this sequence correct to one decimal place. 16. The second and fourth terms of a convergent geometric series are 36 and 16, respectively. Find the sum to infinity of this series, if all its terms are positive. 17. Evaluate: 5 X k (k + 1) 2 k=2 18. Sn = 4n2 + 1 represents the sum of the first n terms of a particular series. Find the second term. 19. Determine whether the following series converges for the given values of x. If it does converge, calculate the sum to infinity. ∞ X (x + 2) p p=1 a) x = − 25 b) x = −5 20. Calculate: ∞ X 5 4−i i=1 21. The sum of the first p terms of a sequence is p (p + 1). Find the tenth term. 22. The powers of 2 are removed from the following set of positive integers 1; 2; 3; 4; 5; 6;... ; 1998; 1999; 2000 Find the sum of remaining integers. Chapter 1. Sequences and series 45 23. Observe the pattern below: E A D E C D E B C D E B C D E B C D E C D E D E E a) If the pattern continues, find the number of letters in the column containing M’s. b) If the total number of letters in the pattern is 361, which letter will be in the last column. 24. Write 0,57̇ as a proper fraction. 25. Given: f (x) = ∞ X (1 + x)p p=1 1−x a) For which values of x will f (x) converge? b) Determine the value of f − 12. 26. From the definition of a geometric sequence, deduce a formula for calculating the sum of n terms of the series a2 + a4 + a6 + · · · 27. Calculate the tenth term of the series if Sn = 2n + 3n2. 28. A theatre is filling up at a rate of 4 people in the first minute, 6 people in the second minute, and 8 people in the third minute and so on. After 6 minutes the theatre is half full. After how many minutes will the theatre be full? [IEB, Nov 2001] 29. More questions. Sign in at Everything Maths online and click ’Practise Maths’. Check answers online with the exercise code below or click on ’show me the answer’. 1. 287Y 2a. 287Z 2b. 2882 3. 2883 4a. 2884 4b. 2885 5. 2886 6. 2887 7a. 2888 7b. 2889 8. 288B 9. 288C 10. 288D 11. 288F 12a. 288G 12b. 288H 13a. 288J 13b. 288K 14. 288M 15a. 288N 15b. 288P 15c. 288Q 16. 288R 17. 288S 18. 288T 19a. 288V 19b. 288W 20. 288X 21. 288Y 22. 288Z 23a. 2892 23b. 2893 24. 2894 25. 2895 26. 2896 27. 2897 28. 2898 www.everythingmaths.co.za 46 1.7. Summary m.everythingmaths.co.za 2 CHAPTER Functions 2.1 Revision 48 2.2 Functions and relations 53 2.3 Inverse functions 56 2.4 Linear functions 58 2.5 Quadratic functions 61 2.6 Exponential functions 70 2.7 Summary 90 2.8 Enrichment: more on logarithms 96 2 2.1 Functions Revision EMCF6 In previous grades we learned about the characteristics of linear, quadratic, hyperbolic and exponential functions. In this chapter we will demonstrate the ability to work with various types of functions and relations including inverses. In particular, we will look at the graphs of the inverses of: Linear functions: y = mx + c or y = ax + q Quadratic functions: y = ax2 Exponential functions: y = bx (b > 0, b 6= 1) Worked example 1: Linear function QUESTION Draw a graph of 2y + x − 8 = 0 and determine the significant characteristics of this linear function. SOLUTION Step 1: Write the equation in standard form y = mx + c 2y + x − 8 = 0 2y = −x + 8 1 ∴y =− x+4 2 1 ∴m=− 2 And c = 4 Step 2: Draw the straight line graph To draw the straight line graph we can use the gradient-intercept method: y − intercept : (0; 4) m=− 48 2.1. Revision 1 2 y 6 5 m = − 12 4 b 3 2 b y = − 21 x + 4 1 −2 −1 −1 x 0 1 2 3 4 5 6 7 8 9 10 11 −2 Alternative method: we can also determine and plot the x- and y-intercepts as follows: For the y-intercept, let x = 0: 1 y = − (0) + 4 2 ∴y =0+4 =4 This gives the point (0; 4). For the x-intercept, let y = 0: 1 0=− x+4 2 1 x=4 2 ∴x=8 This gives the point (8; 0). y 6 5 4 b 3 y = − 21 x + 4 2 1 x b −2 −1 −1 0 1 2 3 4 5 6 7 8 9 10 11 −2 Step 3: Determine the characteristics Gradient: − 12 Intercepts: (0; 4) and (8; 0) Domain: {x : x ∈ R} Range: {y : y ∈ R} Decreasing function: as x increases, y decreases. Chapter 2. Functions 49 Worked example 2: Quadratic function QUESTION Write the quadratic function 2y − x2 + 4 = 0 in standard form. Draw a graph of the function and state the significant characteristics. SOLUTION Step 1: Write the equation in standard form y = ax2 + bx + c 2y − x2 + 4 = 0 2y = x2 − 4 1 y = x2 − 2 2 Therefore, we see that: a= 1 ; 2 c = −2 b = 0; Step 2: Draw a graph of the parabola For the y-intercept, let x = 0: 1 2 (0) − 2 2 ∴y =0−2 = −2 y= This gives the point (0; −2). For the x-intercept, let y = 0: 1 2 x −2 2 0 = x2 − 4 0 = (x − 2)(x + 2) ∴ x = −2 or x = 2 0= This gives the points (−2; 0) and (2; 0). y 4 3 y = 12 x2 − 2 2 1 b −2 0 1 b −3 50 x b −4 −3 −2 −1 −1 2.1. Revision 2 3 4 Step 3: State the significant characteristics Shape: a > 0, therefore the graph is a “smile”. Intercepts: (−2; 0), (2; 0) and (0; −2) Turning point: (0; −2) b = − 2 01 = 0 Axes of symmetry: x = − 2a (2) Domain: {x : x ∈ R} Range: {y : y ≥ −2, y ∈ R} The function is decreasing for x < 0 and increasing for x > 0. Worked example 3: Exponential function QUESTION Draw the graphs of f (x) = 2x and g(x) = the two functions. 1 x 2 on the same set of axes and compare SOLUTION Step 1: Examine the functions and determine the information needed to draw the graphs Consider the function: f (x) = 2x If y = 0 : 2x = 0 But 2x 6= 0 ∴ no solution If x = 0 : 20 = 1 This gives the point (0; 1). Asymptotes: f (x) = 2x has a horizontal asymptote, the line y = 0, which is the x-axis. x f (x) −2 −1 1 4 1 2 Consider the function: g(x) = 0 1 1 2 2 4 1 x 2 x 1 If y = 0 : =0 2 x 1 But 6= 0 2 ∴ no solution 0 1 If x = 0 : =1 2 This gives the point (0; 1). Chapter 2. Functions 51 1 x 2 Asymptotes: g(x) = also has a horizontal asymptote at y = 0. −2 4 x g(x) −1 2 0 1 1 2 1 2 1 4 Step 2: Draw the graphs of the exponential functions y 4 g(x) = 1 x 2 f (x) = 2x 3 2 1 −4 −3 −2 −1 −1 b 0 x 1 2 3 4 Step 3: State the significant characteristics Symmetry: f and g are symmetrical about the y-axis. Domain of f and g: {x : x ∈ R} Range of f and g: {y : y > 0, y ∈ R} The function g decreases as x increases and function f increases as x increases. The two graphs intersect at the point (0; 1). Exercise 2 – 1: Revision 1. Sketch the graphs on the same set of axes and determine the following for each function: Intercepts Turning point Axes of symmetry Domain and range Maximum and minimum values a) f (x) = 3x2 and g(x) = −x2 b) j(x) = − 15 x2 and k(x) = −5x2 c) h(x) = 2x2 + 4 and l(x) = −2x2 − 4 2. Given f (x) = −3x − 6 and g(x) = mx + c. Determine the values of m and c if g k f and g passes through the point (1; 2). Sketch both functions on the same system of axes. 3. Given m : x2 − y3 = 1 and n : − y3 = 1. Determine the x- and y-intercepts and sketch both graphs on the same system of axes. 4. Given p(x) = 3x , q(x) = 3−x and r(x) = −3x. a) Sketch p, q and r on the same system of axes. b) For each of the functions, determine the intercepts, asymptotes, domain and range. 52 2.1. Revision 5. More questions. Sign in at Everything Maths online and click ’Practise Maths’. Check answers online with the exercise code below or click on ’show me the answer’. 1a. 2899 1b. 289B 1c. 289C 2. 289D 3. 289F 4a. 289G 4b. 289H www.everythingmaths.co.za 2.2 m.everythingmaths.co.za Functions and relations EMCF7 DEFINITION: Relation A rule which associates each element of set (A) with at least one element in set (B). DEFINITION: Function A rule which uniquely associates elements of one set (A) with the elements of another set (B); each element in set (A) maps to only one element in set (B). Functions can be one-to-one relations or many-to-one relations. A many-to-one relation associates two or more values of the independent (input) variable with a single value of the dependent (output) variable. The domain is the set of values to which the rule is applied (A) and the range is the set of values (also called the images or function values) determined by the rule. Example of a one-to-one function: y = x + 1 y 3 Input Output −2 −1 −1 0 0 1 1 2 −3 −2 −1 −1 2 3 −2 3 4 −3 b 2 b y =x+1 1 b x b 0 b 1 2 3 Example of a many-to-one function: y = x2 y Input Output 4 b b 3 −2 −1 0 0 4 y = x2 1 1 1 2 2 b b x b −3 −2 −1 −1 0 1 2 3 −2 Chapter 2. Functions 53 However, some very common mathematical constructions are not functions. For example, consider the relation x2 + y 2 = 4. This relation describes a circle of radius 2 centred at the origin. If we let x = 0, we see that y 2 = 4 and thus either y = 2 or y = −2. This is a one-to-many relation because a single x-value relates to two different y-values. Therefore x2 + y 2 = 4 is not a function. y b 2 1 b −2 x b −1 0 2 x2 + y 2 = 4 Vertical line test Given the graph of a relation, there is a simple test for whether or not the relation is a function. This test is called the vertical line test. If it is possible to draw any vertical line (a line of constant x) which crosses the graph of the relation more than once, then the relation is not a function. If more than one intersection point exists, then the intersections correspond to multiple values of y for a single value of x (one-to-many). If any vertical line cuts the graph only once, then the relation is a function (one-to-one or many-to-one). The red vertical line cuts the circle twice and therefore the circle is not a function. The red vertical line only cuts the parabola once and therefore the parabola is a function. y y 4 2 b 3 1 −2 −1 2 x 0 1 2 b −3 −2 −1 −1 −2 54 y = x2 2.2. Functions and relations 0 x 1 2 3 Exercise 2 – 2: Identifying functions 1. Consider the graphs given below and determine whether or not they are functions: a) e) b) f) c) g) d) h) 2. Sketch the following and determine whether or not they are functions: a) x + y = 4 d) x2 + y 2 = 9 x b) y = 4 e) y = tan x x c) y = 2 3. The table below gives the average per capita income, d, in a region of the country as a function of u, the percentage of unemployed people. Write down an equation to show that the average income is a function of the percentage of unemployed people. u d 1 22 500 2 22 000 3 21 500 4 21 000 Check answers online with the exercise code below or click on ’show me the answer’. 1a. 289J 1b. 289K 1c. 289M 1d. 289N 1e. 289P 1f. 289Q 1g. 289R 1h. 289S 2a. 289T 2b. 289V 2c. 289W 2d. 289X 2e. 289Y 3. 289Z www.everythingmaths.co.za m.everythingmaths.co.za Chapter 2. Functions 55 Function notation For the function y = f (x), y is the dependent variable, because the value of y (output) depends on the value of x (input). We say x is the independent variable, since we can choose x to be any number. Similarly, if g (t) = 2t + 1, then t is the independent variable and g is the function name. If h (x) = 3x − 5 and we need to determine when h (x) = 3, then we solve for the value of x such that: h (x) = 3x − 5 3 = 3x − 5 8 = 3x 8 ∴x= 3 If h (x) = 3x − 5 and we need to determine h (3), then we calculate the value for h (x) when x = 3: h (x) = 3x − 5 h (3) = 3 (3) − 5 =4 2.3 Inverse functions EMCF8 An inverse function is a function which does the “reverse” of a given function. More formally, if f is a function with domain X, then f −1 is its inverse function if and only if f −1 (f (x)) = x for every x ∈ X. y = f (x) : indicates a function y1 = f (x1 ) : indicates we must substitute a specific x1 value into the function to get the corresponding y1 value f −1 (y) = x : indicates the inverse function f −1 (y1 ) = x1 : indicates we must substitute a specific y1 value into the inverse to return the specific x1 value A function must be a one-to-one relation if its inverse is to be a function. If a function f has an inverse function f −1 , then f is said to be invertible. Given the function f (x), we determine the inverse f −1 (x) by: interchanging x and y in the equation; making y the subject of the equation; expressing the new equation in function notation. 56 2.3. Inverse functions Note: if the inverse is not a function then it cannot be written in function q notation. For example, the inverse of f (x) = 3x2 cannot be written as f −1 (x) = ± 13 x as it is not q a function. We write the inverse as y = ± 13 x and conclude that f is not invertible. If we represent the function f and the inverse function f −1 graphically, the two graphs are reflected about the line y = x. Any point on the line y = x has x- and y-coordinates with the same numerical value, for example (−3; −3) and 45 ; 45. Therefore interchanging the x- and y-values makes no difference. y 4 f 3 2 f −1 1 −4 −3 −2 −1 −1 0 x 1 2 3 4 −2 −3 −4 This diagram shows an exponential function (black graph) and its inverse (blue graph) reflected about the line y = x (grey line). Important: for f −1 , the superscript −1 is not an exponent. It is the notation for indi−1 cating the inverse of a function. Do not confuse this with exponents, such as 12 or 3 + x−1. Be careful not to confuse the inverse of a function and the reciprocal of a function: Inverse Reciprocal f −1 (x) f (x) and f −1 (x) symmetrical about y = x Example: g(x) = 5x ∴ g [f (x)]−1 = 1 f (x) 1 f (x) =1 f (x) × Example: −1 (x) = x 5 g(x) = 5x ∴ 1 g(x) = 1 5x See video: 28B2 at www.everythingmaths.co.za Chapter 2. Functions 57 2.4 Linear functions EMCF9 Inverse of the function y = ax + q EMCFB Worked example 4: Inverse of the function y = ax + q QUESTION Determine the inverse function of p(x) = −3x + 1 and sketch the graphs of p(x) and p−1 (x) on the same system of axes. SOLUTION Step 1: Determine the inverse of the given function Interchange x and y in the equation. Make y the subject of the new equation. Express the new equation in function notation. Let y = −3x + 1 Interchange x and y : x = −3y + 1 x − 1 = −3y 1 − (x − 1) = y 3 x 1 ∴y=− + 3 3 Therefore, p−1 (x) = − x3 + 13. Step 2: Sketch the graphs of the same system of axes y 5 4 p−1 (x) = − 31 x + 1 3 3 2 1 −5 −4 −3 −2 −1 −1 0 x 1 2 3 4 5 −2 −3 −4 −5 p(x) = −3x + 1 The graph of p−1 (x) is the reflection of p(x) about the line y = x. This means that every point on the graph of p(x) has a mirror image on the graph of p−1 (x). 58 2.4. Linear functions To determine the inverse function of y = ax + q: (1) Interchange x and y : x = ay + q (2) Make y the subject of the equation : x − q = ay x a − q a = ay a ∴ y = a1 x − 1 ax Therefore the inverse of y = ax + q is y = then its inverse will also be linear. q a − aq. If a linear function is invertible, Worked example 5: Inverses - domain, range and intercepts QUESTION Determine and sketch the inverse of the function f (x) = 2x − 3. State the domain, range and intercepts. SOLUTION Step 1: Determine the inverse of the given function Interchange x and y in the equation. Make y the subject of the new equation. Express the new equation in function notation. Let y = 2x − 3 Interchange x and y : x = 2y − 3 x + 3 = 2y 1 (x + 3) = y 2 x 3 ∴y= + 2 2 Therefore, f −1 (x) = x 2 + 32. Step 2: Sketch the graphs on the same system of axes y 3 2 f −1 b 1 b x b −4 −3 −2 −1 −1 0 1 2 3 f −2 −3 b The graph of f −1 (x) is the reflection of f (x) about the line y = x. Chapter 2. Functions 59 Step 3: Determine domain, range and intercepts Domain of f : {x : x ∈ R} Range of f : {y : y ∈ R} Intercepts of f : (0; −3) and 3 ;0 2 Domain of f −1 : {x : x ∈ R} Range of f −1 : {y : y ∈ R} 3 −1 Intercepts of f : 0; and (−3; 0) 2 Notice that the intercepts of f and f −1 are mirror images of each other. In other words, the x- and y-values have “swapped” positions. This is true of every point on the two graphs. Domain and range For a function of the form y = ax + q, the domain is {x : x ∈ R} and the range is {y : y ∈ R}. When a function is inverted the domain and range are interchanged. Therefore, the domain and range of the inverse of an invertible, linear function will be {x : x ∈ R} and {y : y ∈ R} respectively. Intercepts The general form of an invertible, linear function is y = ax + q (a 6= 0) and its inverse is y = a1 x − aq. The y-intercept is obtained by letting x = 0: 1 q (0) − a a q y=− a y= This gives the point 0; − aq. The x-intercept is obtained by letting y = 0: 1 q x− a a 1 q = x a a q=x 0= This gives the point (q; 0). It is interesting to note that if f (x) = ax + q (a 6= 0), then f −1 (x) = a1 x − aq and the y-intercept of f (x) is the x-intercept of f −1 (x) and the x-intercept of f (x) is the y-intercept of f −1 (x). See video: 28B3 at www.everythingmaths.co.za 60 2.4. Linear functions Exercise 2 – 3: Inverse of the function y = ax + q 1. Given f (x) = 5x + 4, find f −1 (x). 2. Consider the relation f (x) = −3x − 7. a) Is the relation a function? Explain your answer. b) Identify the domain and range. c) Determine f −1 (x). a) Sketch the graph of the function f (x) = 3x − 1 and its inverse on the same system of axes. Indicate the intercepts and the axis of symmetry of the two graphs. b) T 43 ; 3 is a point on f and R is a point on f −1. Determine the coordinates of R if R and T are symmetrical. 4. a) Explain why the line y = x is an axis of symmetry for a function and its inverse. b) Will the line y = −x be an axis of symmetry for a function and its inverse? 5. a) Given f −1 (x) = −2x + 4, determine f (x). b) Calculate the intercepts of f (x) and f −1 (x). c) Determine the coordinates of T , the point of intersection of f (x) and f −1 (x). d) Sketch the graphs of f and f −1 on the same system of axes. Indicate the intercepts and point T on the graph. e) Is f −1 an increasing or decreasing function? 6. More questions. Sign in at Everything Maths online and click ’Practise Maths’. 3. Check answers online with the exercise code below or click on ’show me the answer’. 1. 28B4 2a. 28B5 2b. 28B6 2c. 28B7 3. 28B8 4. 28B9 5. 28BB www.everythingmaths.co.za 2.5 m.everythingmaths.co.za Quadratic functions EMCFC Inverse of the function y = ax2 EMCFD Worked example 6: Inverse of the function y = ax2 QUESTION Determine the inverse of the quadratic function h(x) = 3x2 and sketch both graphs on the same system of axes. SOLUTION Step 1: Determine the inverse of the given function h(x) Interchange x and y in the equation. Make y the subject of the new equation. Chapter 2. Functions 61 Let y = 3x2 x = 3y 2 x = y2 3 r Interchange x and y : ∴y=± x 3 (x ≥ 0) Step 2: Sketch the graphs on the same system of axes y y = 3x2 4 3 2 1 −2 −1 −1 x 0 1 2 3 4 y=± −2 px 3 Notice that the inverse does not pass the vertical line test and therefore is not a function. y 2 1 −1 −1 0 x 1 2 3 −2 4 y=± px 3 To determine the inverse function of y = ax2 : (1) Interchange x and y : x = ay 2 (2) Make y the subject of the equation : x a = y2 p ∴ y = ± xa (x ≥ 0) The vertical line test shows that the inverse of a parabola is not a function. However, we can limit the domain of the parabola so that the inverse of the parabola is a function. 62 2.5. Quadratic functions y y = ax2 y=± px a x 0 Domain and range p Consider the previous worked example h(x) = 3x2 and its inverse y = ± x3 : px If we restrict the domain of h so that x ≥ 0, then h−1 (x) = 3 passes the vertical line test and is a function. y h(x) = 3x2 h−1 (x) = px 3 x 0 p If the restriction on the domain of h is x ≤ 0, then h−1 (x) = − x3 would also be a function. y h(x) = 3x2 x 0 h−1 (x) = − px 3 The domain of the function is equal to the range of the inverse. The range of the function is equal to the domain of the inverse. Similarly, a restriction on the domain of the function results in a restriction on the range of the inverse and vice versa. Function: Inverse: domain domain range range Chapter 2. Functions 63 Worked example 7: Inverses - domain, range and restrictions QUESTION Determine the inverse of q(x) = 7x2 and sketch both graphs on the same system of axes. Restrict the domain of q so that the inverse is a function. SOLUTION Step 1: Examine the function and determine the inverse Determine the inverse of the function: Let y = 7x2 x = 7y 2 x = y2 7 r Interchange x and y : ∴y=± x 7 (x ≥ 0) Step 2: Sketch both graphs on the same system of axes y q(x) = 7x2 4 3 2 y= 1 −2 −1 −1 0 px 7 x 1 2 3 4 5 y=− −2 6 px 7 Step 3: Determine the restriction on the domain Option 1: Restrict the domain of q to x ≥ 0 so that the inverse will also be a function q −1. The restriction x ≥ 0 on the domain of q will restrict the range of q −1 such that y ≥ 0. q: domain x ≥ 0 range y ≥ 0 q −1 : domain x ≥ 0 range y ≥ 0 y q(x) = 7x2 4 3 2 q −1 (x) = 1 −2 −1 −1 0 7 x 1 2 3 −2 OR 64 px 2.5. Quadratic functions 4 5 6 Option 2: Restrict the domain of q to x ≤ 0 so that the inverse will also be a function q −1. The restriction x ≤ 0 on the domain of q will restrict the range of q −1 such that y ≤ 0. q: domain x ≤ 0 range y ≥ 0 q −1 : domain x ≥ 0 range y ≤ 0 y q(x) = 7x2 4 3 2 1 −2 −1 −1 0 x 1 2 3 4 5 6 q −1 (x) = − −2 px 7 Worked example 8: Inverses - domain, range and restrictions QUESTION 1. Determine the inverse of f (x) = −x2. 2. Sketch both graphs on the same system of axes. 3. Restrict the domain of f so that its inverse is a function. SOLUTION Step 1: Determine the inverse of the function Let y = −x2 Interchange x and y : √ Note: −x is only defined if x ≤ 0. x = −y 2 −x = y 2 √ y = ± −x (x ≤ 0) Step 2: Sketch both graphs on the same system of axes y √ y = ± −x 3 2 1 −5 −4 −3 −2 −1 −1 0 x 1 2 3 −2 −3 −4 f (x) = −x2 Chapter 2. Functions 65 The inverse does not pass the vertical line test and is not a function. Step 3: Determine the restriction on the domain If f (x) = −x2 , for x ≤ 0: f: −1 domain x ≤ 0 range y ≤ 0 : domain x ≤ 0 range y ≤ 0 f: domain x ≥ 0 range y ≤ 0 f −1 : domain x ≤ 0 range y ≥ 0 f If f (x) = −x2 , for x ≥ 0: Exercise 2 – 4: Inverses - domain, range, intercepts, restrictions 1. Determine the inverse for each of the following functions: a) y = 34 x2 b) 4y − 8x2 = 0 c) x2 + 5y = 0 d) 4y − 9 = (x + 3)(x − 3) 2. Given the function g(x) = 21 x2 for x ≥ 0. a) b) c) d) e) Find the inverse of g. Draw g and g −1 on the same set of axes. Is g −1 a function? Explain your answer. State the domain and range for g and g −1. Determine the coordinates of the point(s) of intersection of the function and its inverse. 3. Given the graph of the parabola f (x) = ax2 with x ≥ 0 and passing through the point P (1; −3). y 1 −1 −1 0 x 1 2 3 −2 −3 −4 b P (1; −3) f (x) = ax2 a) Determine the equation of the parabola. b) State the domain and range of f. c) Give the coordinates of the point on f −1 that is symmetrical to the point P about the line y = x. 66 2.5. Quadratic functions d) Determine the equation of f −1. e) State the domain and range of f −1. f) Draw a graph of f −1. 2 a) Determine the inverse of h(x) = 11 5 x. b) Sketch both graphs on the same system of axes. c) Restrict the domain of h so that the inverse is a function. √ 5. The diagram shows the graph of g(x) = mx + c and f −1 (x) = a x, (x ≥ 0). Both graphs pass through the point P (4; −1). 4. y 2 1 g −2 −1 −1 x 0 1 2 3 b 5 f −1 P (4; −1) −2 a) b) c) d) e) 4 Determine the values of a, c and m. Give the domain and range of f −1 and g. For which values of x is g(x) < f (x)? Determine f. Determine the coordinates of the point(s) of intersection of g and f intersect. 6. More questions. Sign in at Everything Maths online and click ’Practise Maths’. Check answers online with the exercise code below or click on ’show me the answer’. 1a. 28BC 1b. 28BD 4. 28BK 5. 28BM 1c. 28BF 1d. 28BG www.everythingmaths.co.za 2. 28BH 3. 28BJ m.everythingmaths.co.za Worked example 9: Inverses - average gradient QUESTION Given: h(x) = 2x2 , 1. 2. 3. 4. 5. x≥0 Determine the inverse, h−1. Find the point where h and h−1 intersect. Sketch h and h−1 on the same set of axes. Use the sketch to determine if h and h−1 are increasing or decreasing functions. Calculate the average gradient of h between the two points of intersection. Chapter 2. Functions 67 SOLUTION Step 1: Determine the inverse of the function Let y = 2x2 (x ≥ 0) 2 (y ≥ 0) Interchange x and y : x = 2y x = y2 2 r y= −1 ∴h r (x) = x 2 (x ≥ 0, y ≥ 0) x 2 (x ≥ 0) Step 2: Determine the point of intersection r x 2 r 2 2 x 2x2 = 2 x 4 4x = 2 8x4 = x 2x2 = If x = 0, If 8x − 1 = 0 8x3 = 1 1 x3 = 8 1 ∴x= 2 1 1 If x = , y = 2 2 8x4 − x = 0 x(8x3 − 1) = 0 ∴ x = 0 or 8x3 − 1 = 0 Therefore, this gives the points A(0; 0) and B 1 1 2; 2 . Step 3: Sketch both graphs on the same system of axes y h(x) = 2x2 4 3 2 1 −2 −1 −1 h−1 (x) = 0 px 2 x 1 2 3 −2 68 y=0 3 2.5. Quadratic functions 4 Step 4: Examine the graphs From the graphs, we see that both h and h−1 pass the vertical line test and therefore are functions. h : as x increases, y also increases, therefore h is an increasing function. −1 h : as x increases, y also increases, therefore h−1 is an increasing function. Step 5: Calculate the average gradient Calculate the average gradient of h between the points A(0; 0) and B 1 1 2; 2 . yB − yA xB − xA 1 −0 = 21 2 −0 Average gradient: = =1 Note: this is also the average gradient of h−1 between the points A and B. Exercise 2 – 5: Inverses - average gradient, increasing and decreasing functions 1. a) Sketch the graph of y = x2 and label a point other than the origin on the graph. b) Find the equation of the inverse of y = x2. c) Sketch the graph of the inverse on the same system of axes. d) Is the inverse a function? Explain your answer. e) P (2; 4) is a point on y = x2. Determine the coordinates of Q, the point on the graph of the inverse which is symmetrical to P about the line y = x. f) Determine the average gradient between: i. the origin and P ; ii. the origin and Q. Interpret the answers. −1 2 2. Given the function f (x) = kx , x ≥ 0, which passes through the point 1 P 2 ; −1. y 1 −1 −1 x 0 b 1 P 2 3 1 2 ; −1 −2 −3 −4 f −1 (x) = kx2 Chapter 2. Functions 69 a) b) c) d) e) f) Find the value of k. State the domain and range of f −1. Find the equation of f. State the domain and range of f. Sketch the graphs of f and f −1 on the same system of axes. Is f an increasing or decreasing function? 3. Given: g(x) = 52 x2 , x ≥ 0. Find g −1 (x). Calculate the point(s) where g and g −1 intersect. Sketch g and g −1 on the same set of axes. Use the sketch to determine if g and g −1 are increasing or decreasing functions. e) Calculate the average gradient of g −1 between the two points of intersection. a) b) c) d) Check answers online with the exercise code below or click on ’show me the answer’. 1. 28BN 2. 28BP 3. 28BQ www.everythingmaths.co.za 2.6 m.everythingmaths.co.za Exponential functions EMCFF Revision of exponents base exponent/index bn An exponent indicates the number of times a certain number (the base) is multiplied by itself. The exponent, also called the index or power, indicates the number of times the multiplication is repeated. For example, 103 = 10 × 10 × 10 = 1000. Graphs of the exponential function f (x) = bx The value of b affects the direction of the graph: If b > 1, f (x) is an increasing function. If 0 < b < 1, f (x) is a decreasing function. If b ≤ 0, f (x) is not defined. y = bx b>1 y 0 70 b≤0 0 0. Note that the brackets around the number (x) are not compulsory, we use th

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