Precalculus. Thomas Tradler Holly Carley

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1 Precalculus Second Edition (.7) Thomas Tradler Holl Carle

2 ii Copright c 0 Thomas Tradler and Holl Carle This work is licensed under a Creative Commons Attribution- NonCommercial-ShareAlike.0 International License. (CC BY-NC-SA.0) To view a cop of the license, visit: Under this license, ou are free to: Share: cop and redistribute the material in an medium or format Adapt: remi, transform, and build upon the material The licensor cannot revoke these freedoms as long as ou follow the license terms. Under the following terms: Attribution: You must give appropriate credit, provide a link to the license, and indicate if changes were made. You ma do so in an reasonable manner, but not in an wa that suggests the licensor endorses ou or our use. NonCommercial: You ma not use the material for commercial purposes. ShareAlike: If ou remi, transform, or build upon the material, ou must distribute our contributions under the same license as the original. No additional restrictions: You ma not appl legal terms or technological measures that legall restrict others from doing anthing the license permits. Notices: You do not have to compl with the license for elements of the material in the public domain or where our use is permitted b an applicable eception or limitation. No warranties are given. The license ma not give ou all of the permissions necessar for our intended use. For eample, other rights such as publicit, privac, or moral rights ma limit how ou use the material. This document was created with L A TEX. The TI-8 images were created with the TI-SmartView software.

3 Preface These are notes for a course in precalculus, as it is taught at New York Cit College of Technolog - CUNY (where it is offered under the course number MAT 75). Our approach is calculator based. For this, we will use the currentl standard TI-8 calculator, and in particular, man of the eamples will be eplained and solved with it. However, we want to point out that there are also man other calculators that are suitable for the purpose of this course and man of these alternatives have similar functionalities as the calculator that we have chosen to use. An introduction to the TI-8 calculator together with the most common applications needed for this course is provided in appendi A. In the future we ma epand on this b providing introductions to other calculators or computer algebra sstems. This course in precalculus has the overarching theme of functions. This means that man of the often more algebraic topics studied in the previous courses are revisited under this new function theoretic point of view. However, in order to keep this tet as self contained as possible we alwas recall all results that are necessar to follow the core of the course even if we assume that the student has familiarit with the formula or topic at hand. After a first introduction to the abstract notion of a function, we stud polnomials, rational functions, eponential functions, logarithmic functions, and trigonometric functions with the function viewpoint. Throughout, we will alwas place particular importance to the corresponding graph of the discussed function which will be analzed with the help of the TI-8 calculator as mentioned above. These are in fact the topics of the first four (of the five) parts of this precalculus course. In the fifth and last part of the book, we deviate from the above theme and collect more algebraicall oriented topics that will be needed in calculus or other advanced mathematics courses or even other science courses. This part includes a discussion of the algebra of comple numbers (in particular comple numbers in polar form), the -dimensional real vector space R, seiii

4 iv quences and series with focus on the arithmetic and geometric series (which are again eamples of functions, though this is not emphasized), and finall the generalized binomial theorem. In short, here is an outline of the topics in this course and the five parts into which this course is divided: Part I: Functions and graphs Part II: Polnomials and rational functions Part III: Eponential and logarithmic functions Part IV: Trigonometric functions Part V: Comple numbers, sequences, and the binomial theorem The topics in this book are organized in 5 sessions, each session corresponding to one class meeting. Each session ends with a list of eercises that the student is epected to be able to solve. We cannot overstate the importance of completing these eercises for a successful completion of this course. These 5 sessions, together with scheduled eams and one review session give a total of 0 class sessions, which is the number of regularl scheduled class meetings in one semester. Each of the five parts also ends with a review of the topics discussed. This ma be used as a review for an of the eams during the semester. Finall, we point out that there is an overview of the important formulas used in this course at the end of the book. We would like to thank our colleagues and students for their support during the development of this project. In particular, we would like to thank Henr Africk, Johanna Ellner, Lin Zhou, Satanand Singh, Jean Camilien, Leo Chosid, Laurie Caban, Natan Ovshe, Johann Thiel, Wend Wang, Steven Karaszewski, Josue Enriquez, and Mohd Naum Parvez, Akindiji Fadei, Isabel Martinez, Erik Nowak, Sbil Shaver, Faran Hoosain, Kenia Rodriguez, Albert Jaradeh, Iftekher Hossain for man useful comments that helped to improve this tet. Thomas Tradler and Holl Carle New York Cit College of Technolog - CUNY September 05

5 Contents Preface Table of contents iii v I Functions and graphs The absolute value. Background regarding numbers The absolute value Inequalities and intervals Absolute value inequalities Eercises Lines and functions. Lines, slope and intercepts Introduction to functions Eercises Functions b formulas and graphs. Functions given b formulas Functions given b graphs Eercises Introduction to the TI-8 9. Graphing with the TI Finding zeros, maima, and minima Eercises v

6 vi CONTENTS 5 Basic functions and transformations 6 5. Graphing basic functions Transformation of graphs Eercises Operations on functions Operations on functions given b formulas Operations on functions given b tables Eercises The inverse of a function One-to-one functions Inverse function Eercises Review of functions and graphs 98 II Polnomials and rational functions 00 8 Dividing polnomials 0 8. Long division Dividing b ( c) Optional section: Snthetic division Eercises Graphing polnomials 9. Graphs of polnomials Finding roots of a polnomial with the TI Optional section: Graphing polnomials b hand Eercises Roots of polnomials 0 0. Optional section: The rational root theorem The fundamental theorem of algebra Eercises

7 CONTENTS vii Rational functions 7. Graphs of rational functions Optional section: Rational functions b hand Eercises Polnomial and rational inequalities 69. Polnomial inequalities Rational inequalities and absolute value inequalities Eercises Review of polnomials and rational functions 8 III Eponential and logarithmic functions 8 Eponential and logarithmic functions 8. Eponential functions and their graphs Logarithmic functions and their graphs Eercises Properties of ep and log 99. Algebraic properties of ep and log Solving eponential and logarithmic equations Eercises Applications of ep and log Applications of eponential functions Eercises Half-life and compound interest 6 6. Half-life Compound interest Eercises Review of eponential and logarithmic functions 9

8 viii CONTENTS IV Trigonometric functions 7 Trigonometric functions 7. Basic trigonometric definitions and facts sin, cos, and tan as functions Eercises Addition of angles and multiple angles 5 8. Addition and subtraction of angles Double and half angles Eercises Inverse trigonometric functions 6 9. The functions sin, cos, and tan Eercises Trigonometric equations Basic trigonometric equations Equations involving trigonometric functions Eercises Review of trigonometric functions 8 V Comple numbers, sequences, and the binomial theorem 86 Comple numbers 87. Polar form of comple numbers Multiplication and division of comple numbers Eercises Vectors in the plane 99. Introduction to vectors Operations on vectors Eercises

9 CONTENTS i Sequences and series. Introduction to sequences and series The arithmetic sequence Eercises The geometric series 5. Finite geometric series Infinite geometric series Eercises The binomial theorem 7 5. The binomial theorem Binomial epansion Eercises Review of comple numbers, sequences, and the binomial theorem 7 A Introduction to the TI-8 9 A. Basic calculator functions A. Graphing a function A.. Rescaling the graphing window A. Graphing more than one function A. Graphing a piecewise defined function A.5 Using the table A.6 Solving an equation using the solver A.7 Special functions (absolute value, n-th root, etc.) A.8 Programming the calculator A.9 Common errors A.0 Resetting the calculator to factor settings Answers to eercises 7 Session Session Session Session Session Session Session

10 CONTENTS Review part I Session Session Session Session Session Review part II Session Session Session Session Review part III Session Session Session Session Review part IV Session Session Session Session Session Review part V References 0 Important formulas used in precalculus 0 Inde

11 Part I Functions and graphs

12 Session Absolute value equations and inequalities. Background regarding numbers The natural numbers (denoted b N) are the numbers,,,,5,... The integers (denoted b Z) are the numbers...,,,,,0,,,,,5,... The rational numbers (denoted b Q) are the fractions a of integers a and b b with b 0. Here are some eamples of rational numbers: 5, 6,7,0, 8 The real numbers (denoted b R) are the numbers on the real number line 0 π Here are some eamples of real numbers:,π, 5,8,0,6.789

13 .. THE ABSOLUTE VALUE A real number that is not a rational number is called an irrational number. Here are some eamples of irrational numbers: π,,5,e. The absolute value The absolute value of a real numberc, denoted b c the non-negative number which is equal in magnitude (or size) to c, i.e., is the number resulting from disregarding the sign: { c, if c is positive or zero c = c, if c is negative Eample.. = Eample.. = Eample...5 =.5 Eample.. For which real numbers do ou have =? Solution. Since = and =, we see that there are two solutions, = or =. The solution set is S = {,}. Eample.5. Solve for : = 5 Solution. = 5 or = 5. The solution set is S = { 5,5}. Eample.6. Solve for : = 7. Solution. Note that 7 = 7 and 7 = 7 so that these cannot give an solutions. Indeed, there are no solutions, since the absolute value is alwas non-negative. The solution set is the empt set S = {}. Eample.7. Solve for : = 0. Solution. Since 0 = 0, there is onl one solution, = 0. Thus,S = {0}. Eample.8. Solve for : + = 6.

14 SESSION. THE ABSOLUTE VALUE Solution. Since the absolute value of + is 6, we see that + has to be either 6 or 6. We evaluate each case, The solution set is S = { 8,}. either + = 6, or + = 6, = = 6, = = 6, = = ; = = 8. Eample.9. Solve for : = 5 Solution. The solution set is S = {,}. Either = 5 or = 5 = = 9 = = = = = = Eample.0. Solve for : + = 8 Solution. Dividing both sides b gives + = 9. With this, we have the two cases The solution set is S = { 7, }. Either + = 9 or + = 9 = = = = = = = = 7. Inequalities and intervals There is an order relation on the set of real numbers: < 9 reads as is less than 9, reads as is less than or equal to, 7 7 is greater than, 6 6 reads as is greater than or equal to. Eample.. We have <, but >, which can be seen on the number line above.

15 .. INEQUALITIES AND INTERVALS 5 Eample.. We have 5 5 and 5 5. However the same is not true when using the smbol <. We write this as 5 5. The set of all real numbers greater than or equal to some number a and/or less than or equal to some number b is denoted in different was b the following chart: Inequalit notation Number line Interval notation a b a b [a,b] a < < b a b (a,b) a < b a b [a,b) a < b a b (a,b] a a [a, ) a < a (a, ) b b (,b] < b b (,b) Formall, we define the interval [a,b] to be the set of all real numbers such that a b: [a,b] = { a b } There are similar definitions for the other intervals shown in the above table. Warning.. Be sure to write the smaller number a < b first when writing an interval [a,b]. For eample, the interval [5,] = { 5 } would be the empt set! Eample.. Graph the the inequalit π < 5 on the number line and write it in interval notation. Solution. On the number line: π 5 Interval notation: (π, 5]

16 6 SESSION. THE ABSOLUTE VALUE Eample.5. Write the following interval as an inequalit and in interval notation: Solution. Inequalit notation: Interval notation: [, ) Eample.6. Write the following interval as an inequalit and in interval notation: 0 Solution. Inequalit notation: < Interval notation: (, ) Note.7. In some tets round and square brackets are also used on the number line to depict an interval. For eample the following displas the interval [, 5). 5. Absolute value inequalities Using the notation from the previous section, we now solve inequalities involving the absolute value. These inequalities ma be solved in three steps: Step : Solve the corresponding equalit. The solution of the equalit divides the real number line into several subintervals. Step : Using step, check the inequalit for a number in each of the subintervals. This check determines the intervals of the solution set. Step : Check the endpoints of the intervals.

17 .. ABSOLUTE VALUE INEQUALITIES 7 Here are some eamples for the above solution method. Eample.8. Solve for : a) +7 < b) 5 c) 5 Solution. a) We follow the three steps described above. In step, we solve the corresponding equalit, +7 =. +7 = +7 = = = 5 = = 9 The solutions = 5 and = 9 divide the number line into three subintervals: < 9 9 < < 5 5 < Now, in step, we check the inequalit for one number in each of these subintervals. Check: = 0 Check: = 7 Check: = 0 ( 0)+7? < ( 7)+7? < 0+7? <? < 0? < 7? < <? 0 <? 7 <? false true false Since = 7 in the subinterval given b 9 < < 5 solves the inequalit +7 <, it follows that all numbers in the subinterval given b 9 < < 5 solve the inequalit. Similarl, since = 0 and = 0 do not solve the inequalit, no number in these subintervals will solve the inequalit. For step, we note that the numbers = 9 and = 5 are not included as solutions since the inequalit is strict (that is we have < instead of ).The solution set is therefore the interval S = ( 9, 5). The solution on the number line is: b) We follow the steps as before. First, in step, we solve 5 =. 5 = 5 = = = 6 = = 6 = = 6 = =

18 8 SESSION. THE ABSOLUTE VALUE The two solutions = and = 6 = 5 subintervals displaed below. divide the number line into the < < < 5 5 < For step, we check a number in each subinterval. This gives: Check: = Check: = Check: = 6 ( ) 5? 5? 6 5? 9 5? 5? 8 5??????? true false true For step, note that we include and 5 in the solution set since the inequalit is greater than or equal to (that is, as opposed to >). Furthermore, the numbers and are not included, since ± are not real numbers. The solution set is therefore the union of the two intervals: ( S = ], [5 ), c) To solve 5, we first solve the equalit 5 =. 5 = 5 = = 5 = = 5 = = = =. = = 5 5 This divides the number line into three subintervals, and we check the original inequalit 5 for a number in each of these subintervals.

19 .. ABSOLUTE VALUE INEQUALITIES 9 Interval: <. Interval:. < <.6 Interval:.6 < Check: = Check: =. Check: = 5? 5.? 5? 5?? 5? 7? 0?? 7? 0?? false true false The solution set is the interval S = [.,.6], where we included =. and =.6 since the original inequalit less than or equal to ( ) includes the equalit. Note: Alternativel, whenever ou have an absolute value inequalit ou can turn it into two inequalities. Here are a couple of eamples. Eample.9. Solve for : 5 Solution. Note that 5 implies that so that 5 5 and b dividing b 5 (remembering to switch the direction of the inequalities when multipling or dividing b a negative number) we see that 5 5, or in interval notation, we have the solution set [ S = 5, ]. 5 Eample.0. If + 6 > then either + 6 > or + 6 < so that either > or < 8 so that in interval notation the solution is S = (, 8) (, ).

20 0 SESSION. THE ABSOLUTE VALUE Observation.. There is a geometric interpretation of the absolute value on the number line as the distance between two numbers: a b distance between a and b is b a which is also equal to a b This interpretation can also be used to solve absolute value equations and inequalities. Eample.. Solve for : a) 6 =, b) 6, c) 6. Solution. a) Consider the distance between and 6 to be on a number line: distance distance There are two solutions, = or = 0. That is, the distance between and 6 is and the distance between 0 and 6 is. b) Numbers inside the braces above have distance or less. The solution is given on the number line as: distance distance In interval notation, the solution set is the interval S = [,0]. One can also write that the solution set consists of all such that 0. c) Numbers outside the braces above have distance or more. The solution is given on the number line as: distance < distance < In interval notation, the solution set is the interval (,] and [0, ), or in short it is the union of the two intervals: S = (,] [0, ).

21 .5. EXERCISES One can also write that the solution set consists of all such that or 0..5 Eercises Eercise.. Give eamples of numbers that are a) natural numbers b) integers c) integers but not natural numbers d) rational numbers e) real numbers f) rational numbers but not integers Eercise.. Which of the following numbers are natural numbers, integers, rational numbers, or real numbers? Which of these numbers are irrational? a) 7 b) 5 c) 0 d) 7,000 e) f) 7 g) 5 Eercise.. Evaluate the following absolute value epressions: a) 8, b) 0, c) 99, d), e), f) 6, g) +, h) 9, i) 5., j), k) 5, l) 6. Eercise.. Solve for : a) = 8 b) = 0 c) = d) + = 0 e) +5 = 9 f) 5 = g) = 8 h) 7 = 0 i) = j) = k) = l) 8 = Eercise.5. Solve for using the geometric interpretation of the absolute value: a) = 8 b) = 0 c) = d) = e) +5 = 9 f) = 5

22 SESSION. THE ABSOLUTE VALUE Eercise.6. Complete the table. Inequalit notation Number line Interval notation < 5 7 [, 6] (, 0) 5 < 0.5 ( 7,π) Eercise.7. Solve for and write the solution in interval notation. a) 7 b) 7 c) > 7 d) +7 e) > 8 f) + < 7 g) 5 6 h) 5 > i) 8 j) + < 5 k) 5+5 l) 5+5 > 0

23 Session Lines and functions. Lines, slope and intercepts In this chapter we will introduce the notion of a function. Before we give the general definition, we recall one special kind of function which is alread familiar to the student. More precisel, we start b recalling functions that are given b straight lines. We have seen in the last section that each point on the number line is represented b a real number in R. Similarl, each point in the coordinate plane is represented b a pair of real numbers (,). The coordinate plane is denoted b R. Here is a picture of the coordinate plane: In this section, we will discuss the straight line in the coordinate plane. We will discuss its slope and intercepts. We will also discuss the line s

24 SESSION. LINES AND FUNCTIONS corresponding algebraic forms: the point-slope form and the slope-intercept form. B drawing a line on a coordinate plane we associate ever point on the line with a pair of numbers (a,b), where a is called the coordinate and b is called the coordinate. Consider, for eample, the picture L We see that (,0) lies on the line and so does (,). Of course, there are an infinite number of points on the line. This set of points can also be described b an equation relating the and coordinates of points on the line. For eample = is the equation of the line above. Notice that (,0) satisfies the equation since (0) ( ) = and (,) satisfies the equation since () () =. It is a fact that ever line has an equation of the form p+q = r, that is, given a line, there are numbers p,q and r so that a point (a,b) is on the graph of the line if and onl if pa+qb = r. Notice that, in the eample, if we put our finger on (, 0) and move it up and over to the point (,) we have moved units up and units to the right. If from there ((,)) we move another units up and units to the right then we land again on the line (at (6,)). In fact no matter where we start on the line if from there we move units up and units to the right we alwas land on the line again. The ratio rise = is called the slope of this line. The word run rise indicates vertical movement (down being negative) and run indicates horizontal movement (left being negative). Notice that = = and so on. So we could also see that if we start at (,0) and rise and run we land at (0,) which is on the line and if we start at (,) and rise (move down ) and run (move to the left ) we end up at (0,), which is again on the line. Generall, the slope describes how fast the line grows towards the right. For an two points P (, ) and P (, ) on the line L, the slope m is

25 .. LINES, SLOPE AND INTERCEPTS 5 given b the following formula (which is rise run ): P P Slope: m = (.) The slope determines how fast a line grows. When the slope m is negative the line declines towards the right. m = m = m = 0 m = m = From equation (.), we see that for a given slopem and a point P (, ) on the line, an other point (,) on the line satisfies m =. Multipling ( ) on both sides gives what is called the point-slope form of the line: = m ( ) (.) Other prominent features of a graph of a line (or other graphs) include where the graph crosses the -ais and the ais. For a given line L, let us recall what the - and -intercepts are. The -intercept is the point on the -ais where the line intersects the -ais. Similarl, the -intercept is the point on the -ais where the line intersects the -ais.

26 6 SESSION. LINES AND FUNCTIONS L -intercept -intercept One wa to describe a line using the slope and -intercept is the so-called slope-intercept form of the line. Definition.. The slope-intercept form of the line is the equation = m +b (.) Here, m is the slope and (0,b) is the -intercept of the line. Here is an eample of a line in slope-intercept form. Eample.. Graph the line = +. Solution. We calculate for various values of. For eample, when is,,0,,, or, we calculate In the above table each value is calculated b substituting the corresponding value into our equation = +: = = = ( )+ = + = = = = ( )+ = + = = 0 = = (0)+ = 0+ = = = = ()+ = + = 5 = = = ()+ = + = 7 = = = ()+ = 6+ = 9 In the above calculation, the values for were arbitraril chosen. Since a line is completel determined b knowing two points on it, an two values for would have worked for the purpose of graphing the line.

27 .. LINES, SLOPE AND INTERCEPTS 7 Drawing the above points in the coordinate plane and connecting them gives the graph of the line = +: Alternativel, note that the -intercept is (0, ) ( is the additive constant in our initial equation = +) and the slope m = determines the rate at which the line grows: for each step to the right, we have to move two steps up. To plot the graph, we first plot the -intercept (0,). Then from that point, rise and run so that ou find ourself at (,5) (which must be on the graph), and similarl rise and run to get to (,7) (which must be on the graph), etc. Plot these points on the graph and connect the dots to form a straight line. As noted above, an distinct points on the graph of a straight line are enough to plot the complete line. Before giving more eamples, we briefl want to justif wh, in the epression = m+b, the number (0,b) is the -intercept and m is the slope. This proof ma be skipped on a first reading.

28 8 SESSION. LINES AND FUNCTIONS Proof that (0,b) is the -intercept, and m is the slope. Given the line = m + b, we want to calculate its -intercept and its slope. The -intercept is the value of where = 0. Therefore, we have that the coordinate of the -intercept is = m 0+b = b. This shows the first claim. Net, to see wh m is the slope, note that P (, ) lies on the line eactl when = m +b. Similarl, P (, ) lies on the line eactl when = m +b. Subtracting = m +b from = m +b, we obtain: = (m +b) (m +b) = m +b m b = m ( ). Dividing b ( ) gives ( ) = m ( ) ( ) = m = m. B Equation (.) the fraction is the slope, establishing that our m is indeed the slope, as we claimed. Eample.. Find the equation of the line in slope-intercept form Solution. The -intercept can be read off the graph giving us that b =. As for the slope, we use formula (.) and the two points on the line P (0,) and P (,0). We obtain m = 0 0 = =. Thus, the line has the slope-intercept form = +.

29 .. LINES, SLOPE AND INTERCEPTS 9 Eample.. Find the equation of the line in slope-intercept form Solution. The -intercept is b =. To obtain the slope we can again use the -intercept P (0, ). To use (.), we need another point P on the line. We ma pick an second point on the line, for eample, P (, ). With this, we obtain m = ( ) ( ) 0 = + =. Thus, the line has the slope-intercept form =. Eample.5. Find the equation of the line in point-slope form (.) Solution. We need to identif one point (, ) on the line together with the slope m of the line so that we can write the line in point-slope form: = m( ). B direct inspection, we identif the two points P (5,) and P (8,) on the line, and with this we calculate the slope as m = 8 5 =.

30 0 SESSION. LINES AND FUNCTIONS Using the point (5,) we write the line in point-slope form as follows: = ( 5) Note that our answer depends on the chosen point (5,) on the line. Indeed, if we choose a different point on the line, such as (8,), we obtain a different equation, (which nevertheless represents the same line): = ( 8) Note, that we do not need to solve this for, since we are looking for an answer in point-slope form. Eample.6. Find the slope, find the -intercept, and graph the line + = 0. Solution. We first rewrite the equation in slope-intercept form. + = 0 ( +) = = + (divide ) = = + We see that the slope is and the -intercept is (0,). We can then plot the intercept (0,) and use the slope m = to find another point (, ). Plot that point, connect the two plotted points, and etend to see the graph below. We can also graph b plotting points. We can calculate the -values for some -values. For eample when =,,...,, we obtain: 0 5 5

31 .. LINES, SLOPE AND INTERCEPTS This gives the following graph: Eample.7. Find the slope, -intercept, and graph the line 5+ = 0. Solution. Again, we first rewrite the equation in slope-intercept form. 5 + = 0 (subtract ) = 5 = 0 (divide 5) = = 0 5 = = 5 Now, the slope is and the -intercept is (0, ). 5 We can plot the intercept and from there move units down and 5 units to the right to find another point on the line. The graph is given below. To graph it b plotting points we need to find points on the line. In fact, an two points will be enough to completel determine the graph of the line. For some smart choices of or the calculation of the corresponding value ma be easier than for others. We suggest that ou find the and intercepts, i.e., point of the form (?,0) ( = 0 in the equation and find ) and (0,?) (set = 0 in the equation and find ). Plugging = 0 into 5 + = 0, we obtain =0 = = 0 = 5 = 0 = =.

32 SESSION. LINES AND FUNCTIONS Similarl, substituting = 0 into 5 + = 0 gives =0 = 5 0+ = 0 = = 0 = = 5. We obtain the following table: This gives the following graph: Introduction to functions We now formall introduce the notion of a function. A first eample was provided b a straight line such as, for eample, = 5+. Note, that for each given we obtain an induced. (For eample, for =, we obtain = 5 + = 9.) Definition.8. A function f consists of two sets, a set D of inputs called the domain and a set C of possible outputs called the codomain, and an assignment that assigns to each input eactl one output. A function f with domain D and codomain C is denoted b f : D C. If is in the domain D (an input), then we denote b f() = the output that is assigned b f to. Sometimes it is of interest to know the set of all elements in the codomain that actuall occur as an output. This set is a subset of the codomain and is called the range. We have:

33 .. INTRODUCTION TO FUNCTIONS Definition.9. The range R of a function f is a subset of the codomain given b R = {f() in the domain of f}. That is, the range is the set of all outputs. Warning.0. Some authors use a slightl different convention b calling the range what we called the codomain above. Since we will be dealing with man functions it is convenient to name various functions (usuall with letters f, g, h, etc). Often we will implicitl assume that a domain and codomain are given without specifing these eplicitl. If the range can be determined and the codomain is not given eplicitl, then we take the codomain to be the range. If the range can not easil be determined and the codomain is not eplicitl given, then the codomain should be taken to be a simple set which clearl contains the range. There are several was to represent a particular function (all of which ma not appl to a specific function): via a table of values (listing the inputoutput pairs), via a formula (with the domain and range eplicitl or implicitl given), via a graph (representing input-output pairs on a coordinate plane), or in words, just to name a few. We have seen eamples of the first three of these in previous sections. Our discussion in this section goes into greater detail. Eample.. Define the assignment f b the following table The assignment f assigns to the input the output 6, which is also written as f() = 6. Similarl, f assigns to 5 the number 8, in short f(5) = 8, etc: f(5) = 8, f( ) = 6, f(0) =, f(7) =, f() = 8. The domain D is the set of all inputs. The domain is therefore D = {,0,,,5,7}.

34 SESSION. LINES AND FUNCTIONS The range R is the set of all outputs. The range is therefore R = {,,6,8}. The assignment f is indeed a function since each element of the domain gets assigned eactl one element in the range. Note that for an input number that is not in the domain, f does not assign an output to it. For eample, f() = undefined. Note also that f(5) = 8 and f() = 8, so that f assigns to the inputs 5 and the same output 8. Similarl, f also assigns the same output to the inputs and. Therefore we see that: A function ma assign the same output to two different inputs! Eample.. Consider the assignment f that is given b the following table This assignment does not define a function! What went wrong? Consider the input value 5. What does f assign to the input 5? The third column states that f assigns to 5 the output 8, whereas the sith column states that f assigns to 5 the output, f(5) = 8, f(5) =. However, b the definition of a function, to each input we have to assign eactl one output. So, here, to the input 5 we have assigned two outputs 8 and. Therefore, f is not a function. A function cannot assign two outputs to one input! We repeat the two bullet points from the last two eamples, which are crucial for the understanding of a function. Note.. A function ma assign the same output to two different inputs! f( ) = and f( ) = with is allowed!

35 .. INTRODUCTION TO FUNCTIONS 5 A function cannot assign two outputs to one input! f() = and f() = with is not allowed! Eample.. A universit creates a mentoring program, which matches each freshman student with a senior student as his or her mentor. Within this program it is guaranteed that each freshman gets precisel one mentor, however two freshmen ma receive the same mentor. Does the assignment of freshmen to mentor, or mentor to freshmen describe a function? If so, what is its domain, what is its range? Solution. Since a senior ma mentor several freshman, we cannot take a mentor as an input, as he or she would be assigned to several output freshmen students. So freshman is not a function of mentor. On the other hand, we can assign each freshmen to eactl one mentor, which therefore describes a function. The domain (the set of all inputs) is given b the set of all freshmen students. The range (the set of all outputs) is given b the set of all senior students that are mentors. The function assigns each input freshmen student to his or her unique output mentor. Eample.5. The rainfall in a cit for each of the months is displaed in the following histogram. rainfall [in] J F MA MJ J A S O ND month a) Is the rainfall a function of the month? b) Is the month a function of the rainfall? Solution. a) Each month has eactl one amount of rainfall associated to it. Therefore, the assignment that associates to a month its rainfall (in inches) is a function.

36 6 SESSION. LINES AND FUNCTIONS b) If we take a certain rainfall amount as our input data, can we associate a unique month to it? For eample, Februar and March have the same amount of rainfall. Therefore, to one input amount of rainfall we cannot assign a unique month. The month is not a function of the rainfall. Eample.6. Consider the function f described below. f ellow blue green Here, the function f maps the input smbol to the output color blue. Other assignments of f are as follows: f( ) = blue, f( ) = green, f( ) = ellow f( ) = ellow The domain is the set of smbols D = {,,, }, and the range is the set of colors R = {blue, green, ellow}. Notice, in particular, that the inputs and both have the same output ellow, which is certainl allowed for a function. Eample.7. Consider the function = 5+ with domain all real numbers and range all real numbers. Note that for each input, we obtain an eactl one induced output. For eample, for the input = we get the output = 5 + = 9, etc. Eample.8. Consider the function = with domain all real numbers and range non-negative numbers. The function takes a real number as an input and squares it. For eample if = is the input, then = is the output.

37 .. INTRODUCTION TO FUNCTIONS 7 Eample.9. For each real number, denote b the greatest integer that is less or equal to. We call the floor of. For eample, to calculate.7, note that all integers,,,... are less or equal to.7:...,,,,0,,,,.7 The greatest of these integers is, so that.7 =. We define the floor function as f() =. Here are more eamples of function values of the floor function. 7. = 7, π =,.65 = 5, 6 =, = = 9 The domain of the floor function is the set of all real numbers, that is D = R. The range is the set of all integers, R = Z. Eample.0. Let A be the area of an isosceles right triangle with base side length. Epress A as a function of. Solution. Being an isosceles right triangle means that two side lengths are, and the angles are 5, 5, and 90 (or in radian measure π, π, and π ): Recall that the area of a triangle is: area = base height. In this case, we have base=, and height=, so that the area A = =. Therefore, the area A() =.

38 8 SESSION. LINES AND FUNCTIONS Eample.. Consider the equation = +. This equation associates to each input number a eactl one output number b = a +. Therefore, the equation defines a function. For eample: To the input 5 we assign the output 5 + = 5+ = 8. The domain D is all real numbers, D = R. Since is alwas 0, we see that +, and vice versa ever number can be written as = +. (To see this, note that the input = for gives the output = ( ) + = + =.) Therefore, the range is R = [, ). Eample.. Consider the equation + = 5. Does this equation define as a function of? That is, does this equation assign to each inputeactl one output? An input number gets assigned to with + = 5. Solving this for, we obtain = 5 = = ± 5. Therefore, there are two possible outputs associated to the input ( 5): either = + 5 or = 5. For eample, the input = 0 has two outputs = 5 and = 5. However, a function cannot assign two outputs to one input! The conclusion is that + = 5 does not determine as a function! Note.. Note that if = f() then is called the independent variable and is called the dependent variable (since it depends on ). If = g() then is the independent variable and is the dependent variable (since it depends on ).. Eercises Eercise.. Find the slope and -intercept of the line with the given data. Using the slope and -intercept, write the equation of the line in slope-

39 .. EXERCISES 9 intercept form a) b) c) d) e) f) Eercise.. Write the equation of the line in slope-intercept form. Identif slope and -intercept of the line. a) + = 8 b) 9 +5 = 0 c) 5 0 = 0 d) 5 = 7 e) +8 = 60 f) 8 9 = 0 - -

40 0 SESSION. LINES AND FUNCTIONS Eercise.. Find the equation of the line in point-slope form (.) using the indicated point P. 5 P P a) b) P - P c) - - Eercise.. Graph the line b calculating a table (as in Eample.). (Solve for first, if this is necessar.) a) = b) = + c) = + d) = e) 8 = f) + +6 = 0 Eercise.5. Determine if the given table describes a function. If so, determine its domain and range. Describe which outputs are assigned to which inputs. a) b) c) d)

41 .. EXERCISES d) e) Eercise.6. We consider children and their (birth) mothers. a) Does the assignment child to their birth mother constitute a function (in the sense of Definition.8 on page )? b) Does the assignment mother to their children constitute a function? c) In the case where the assignment is a function, what is the domain? d) In the case where the assignment is a function, what is the range? Eercise.7. A bank offers wealth customers a certain amount of interest, if the keep more than million dollars in their account. The amount is described in the following table. dollar amount in the account interest amount $,000,000 $0 $,000,000 < $0,000,000 % of $0,000,000 < % of a) Justif that the assignment cash amount to interest defines a function. b) Find the interest for an amount of: i) $50,000 ii) $5,000,000 iii) $,000,000 iv) $0,000,000 v) $0,000,000 vi) $,000,000 Eercise.8. Find a formula for a function describing the given inputs and outputs. a) input: the radius of a circle, output: the circumference of the circle b) input: the side length in an equilateral triangle, output: the perimeter of the triangle c) input: one side length of a rectangle, with other side length being, output: the perimeter of the rectangle d) input: the side length of a cube, output: the volume of the cube

42 Session Functions b formulas and graphs. Functions given b formulas Most of the time we will discuss functions that take some real numbers as inputs, and give real numbers as outputs. These functions are commonl described with a formula. Eample.. For the given function f, calculate the outputs f(), f( ), and f( ). a) f() = +, b) f() = {, 5 6, for c) f() = d) f() = +, for < Solution. We substitute the input values into the function and simplif. a) f() = + = 6+ = 0, f( ) = ( )+ = 9+ = 5, f( ) = ( )+ = + =. b) Similarl, we calculate f() = = = =, f( ) = ( ) = 9 = 6, f( ) = ( ) = = = undefined.

43 .. FUNCTIONS GIVEN BY FORMULAS For the last step, we are assuming that we onl deal with real numbers. Of course, as we will see later on, can be defined as a comple number. But for now, we will onl allow { real solutions outputs. 5 6, for c) The function f() = is given as a piecewise defined function. We have to substitute the values into the correct +, for < 5 branch: f() = + = 8+ =, since < 5, f( ) = undefined, since is not in an of the two branches, f( ) = 5 ( ) 7 = 5 6 =, since. d) Finall for f() = +, we have: + f() = + + = 5, f( ) = + + = 0 = undefined, f( ) = + + =. Eample.. Let f be the function given b f() = +. Find the following function values. a) f(5) b) f() c) f( ) d) f(0) e) f( 5) f) f( +) g) f(a) h) f(a)+5 i) f(+h) j) f(+h) f() k) f(+h) f() l) f() f(a) h a Solution. The first four function values ((a)-(d)) can be calculated directl: f(5) = = 5+0 =, f() = + = + = 5, f( ) = ( ) + ( ) = + =, f(0) = = 0+0 =. The net two values ((e) and (f)) are similar, but the arithmetic is a bit more involved. f( 5) = = 5+ 5 = + 5,

44 SESSION. FUNCTIONS BY FORMULAS AND GRAPHS and f( +) = ( +) + ( +) = ( +) ( +)+ ( +) = = = +. The last five values ((g)-(l)) are purel algebraic: f(a) = a + a, f(a)+5 = a + a +5 = a + a+, f(+h) = (+h) + (+h) = +h+h ++h, f(+h) f() = ( +h+h ++h ) ( + ) f(+h) f() h = +h+h ++h + = h+h +h, = h+h +h h = h (+h+) = +h+, h f() f(a) a = ( + ) (a +a ) a = + a a+ a = (+a)( a)+( a) a = a + a a = ( a)(+a+) ( a) = +a+. The quotients f(+h) f() and f() f(a) as in the last two eamples.(k) h a and (l) will become particularl important in calculus. The are called difference quotients. We now calculate some more eamples of difference quotients. Eample.. Calculate the difference quotient f(+h) f() h a) f() = +, b) f() =. for

45 .. FUNCTIONS GIVEN BY FORMULAS 5 Solution. We calculate first the difference quotient step b step. a) f(+h) = (+h) + = (+h) (+h) (+h)+ Subtracting f() from f(+h) gives = ( +h+h ) (+h)+ = + h+h + h+h +h +, = + h+h +h +. f(+h) f() = ( + h+h +h +) ( +) With this we obtain: f(+h) f() h We handle part (b) similarl. = + h+h +h + = h+h +h. = h+h +h h = h ( +h+h ) h b) f(+h) = +h, = +h+h. so that f(+h) f() = +h. We obtain the solution after simplifing the double fraction: f(+h) f() h = = +h = (+h) (+h) h h h (+h) h = = h (+h) h (+h). = h (+h) h So far, we have not mentioned the domain and range of the functions defined above. Indeed, we will often not describe the domain eplicitl but use the following convention:

46 6 SESSION. FUNCTIONS BY FORMULAS AND GRAPHS Convention.. Unless otherwise stated, a function f has the largest possible domain, that is the domain is the set of all real numbers for which f() is a well-defined real number. We refer to this as the standard convention of the domain. (In particular, under this convention, an polnomial has the domain R of all real numbers.) The range is the set of all outputs obtained b f from the inputs (see also the warning on page.) Eample.5. Find the domain of each of the following functions. Solution. a) f() = +5, b) f() =, c) f() =, d) f() =, e) f() = {, f) f() = +, for < g) f() =, for , a) There is no problem taking a real number to an (positive) power. Therefore, f is defined for all real numbers, and the domain is written as D = R. b) Again, we can take the absolute value for an real number. The domain is all real numbers, D = R. c) The square root is onl defined for 0 (remember we are not using comple numbers et!). Thus, the domain is D = [0, ). d) Again, the square root is onl defined for non-negative numbers. Thus, the argument in the square root has to be greater or equal to zero: 0. Solving this for gives 0 (add ) =. The domain is therefore, D = [, ). e) A fraction is defined whenever the denominator is not zero, so that in this case is defined whenever 0. Therefore the domain is all real numbers ecept zero, D = R {0}.

47 .. FUNCTIONS GIVEN BY GRAPHS 7 f) Again, we need to make sure that the denominator does not become zero; however we do not care about the numerator. The denominator is zero eactl when +8+5 = 0. Solving this for gives: +8+5 = 0 = (+) (+5) = 0 = + = 0 or +5 = 0 = = or = 5. The domain is all real numbers ecept for and 5, that is D = R { 5, }. g) The function is eplicitl defined for all < and 5. Therefore, the domain is D = (,] [5, ).. Functions given b graphs In man situations, a function is represented b its graph. The graph of a function f is the set of all points (on the coordinate plane) of the form (,f()), where is in the domain of f. We have alread seen an eample of graphing a function when we discussed lines and their graphs. Here is another eample that shows how the formulaic definition relates to the graph of the function. Eample.6. Let = with domaind = R being the set of all real numbers. We can graph this after calculating a table as follows: = output P(,) = input

48 8 SESSION. FUNCTIONS BY FORMULAS AND GRAPHS Man function values can be read from this graph. For eample, for the input =, we obtain the output =. This corresponds to the point P(,) on the graph as depicted above. In general, an input (placed on the -ais) gets assigned to an output (placed on the -ais) according to where the vertical line at intersects with the given graph. This is used in the net eample. Eample.7. Let f be the function given b the following graph Here, the dashed lines show, that the input = gives an output of =. Similarl, we can obtain other output values from the graph: f() =, f() =, f(5) =, f(7) =. Note, that in the above graph, a closed point means that the point is part of the graph, whereas an open point means that it is not part of the graph. The domain is the set of all possible input values on the -ais. Since we can take an number 7 as an input, the domain is the interval D = [,7]. The range is the set of all possible output values on the -ais. Since an number < is obtained as an output, the range is R = (, ]. Note in particular, that = is not an output, since f(5) =.

49 .. FUNCTIONS GIVEN BY GRAPHS 9 Eample.8. Let f be the function given b the following graph Here are some function values that can be read from the graph: - f( 5) =, f( ) =, f( ) and f( ) are undefined, f( ) =, f(0) =, f() =, f() =, f() = 0, f(5) =. Note, that the output value f() is somewhere between and 0, but we can onl read off an approimation of f() from the graph. To find the domain of the function, we need to determine all possible - coordinates (or in other words, we need to project the graph onto the -ais). The possible -coordinates are from the interval [ 5, ) together with the intervals (,) and [,5]. The last two intervals ma be combined. We get the domain: D = [ 5, ) (,5]. For the range, we look at all possible -values. These are given b the intervals (, ] and (0, ) and [, ]. Combining these three intervals, we obtain the range R = [,]. Eample.9. Consider the following graph

50 0 SESSION. FUNCTIONS BY FORMULAS AND GRAPHS Consider the input =. There are several outputs that we get for = from this graph: f() =, f() =, f() =. However, in a function, it is not allowed to obtain more than one output for one input! Therefore, this graph is not the graph of a function! The reason wh the previous eample is not a function is due to some input having more than one output: f() =,f() =,f() = In other words, there is a vertical line ( = ) which intersects the graph in more than one point. This observation is generalized in the following vertical line test. Observation.0 (Vertical Line Test). A graph is the graph of a function precisel when ever vertical line intersects the graph at most once. Eample.. Consider the graph of the equation = : This does not pass the vertical line test so is not a function of. However, is a function of since, if ou consider to be the input, each input has eactl one output (it passes the horizontal line test).

51 .. FUNCTIONS GIVEN BY GRAPHS Eample.. Let f be the function given b the following graph a) What is the domain of f? b) What is the range of f? c) For which is f() =? d) For which is f() =? e) For which is f() >? f) For which is f()? g) Find f() and f(). h) Find f()+f(). i) Find f()+. j) Find f(+). Solution. Most of the answers can be read immediatel from the graph. (a) For the domain, we project the graph to the -ais. The domain consists of all numbers from 5 to 5 without, that is D = [ 5, ) (,5]. (b) For the range, we project the graph to the -ais. The domain consists of all numbers from to 5, that is R = [,5]. (c) To find with f() = we look at the horizontal line at = : We see that there are two numbers with f() = : f( ) =, f() =. Therefore, the answer is = or =.

52 SESSION. FUNCTIONS BY FORMULAS AND GRAPHS (d) Looking at the horizontal line =, we see that there is onl one with f() = ; namel f() =. Note, that = does not solve the problem, since f( ) is undefined. The answer is =. (e) To find with f() >, the graph has to lie above the line = We see that the answer is those numbers greater than and less than. The answer is therefore the interval (,). (f) For f(), we obtain all numbers from the domain that are less than or greater or equal to. The answer is therefore, [ 5, ) (, ) [,5]. Note that is not part of the answer, since f( ) is undefined. (g) f() =, and f() = (h) f()+f() = + = 6 (i) f()+ = + = 8 (j) f(+) = f(5) = Eample.. The following graph shows the population size in a small cit

53 .. FUNCTIONS GIVEN BY GRAPHS in the ears 00-0 in thousands of people ear a) What was the population size in the ears 00 and 009? b) In what ears did the cit have the most population? c) In what ear did the population grow the fastest? d) In what ear did the population decline the fastest? Solution. The population size in the ear 00 was approimatel 6,000. In the ear 009, it was approimatel 6, 000. The largest population was in the ear 006, where the graph has its maimum. The fastest growth in the population was between the ears 00 and 00. That is when the graph has the largest slope. Finall, the fastest decline happened in the ears Eample.. Graph the function described b the following formula: +, for < f() =, for < <, for < Solution. We reall have to graph all three functions = +, =, and =, and then restrict them to their respective domain. Graphing the three functions, we obtain the following tables and associated graphs, which we draw in one --plane: = + = = = = =

54 SESSION. FUNCTIONS BY FORMULAS AND GRAPHS = = = However, we need to cut off the functions according to their specific input domain that is given b the original function Note, how the open and closed circles at the endpoints of each branch correspond to the < and rules in the original description of the function.. Eercises Eercise.. For each of the following functions, a) f() = +, b) f() =, c) f() = 9 d) f() =, 5 e) f() =, + f) f() = calculate the function values i) f() ii) f(5) iii) f( ) iv) f(0) v) f( ) vi) f( +) vii) f( ) viii) f(+) i) f()+h ) f(+h)

55 .. EXERCISES 5 Eercise.. Let f be the piecewise defined function { 5, for < < f() =, for 6 a) State the domain of the function. Find the function values b) f(), c) f(5), d) f( ), e) f() Eercise.. Let f be the piecewise defined function, for < f() = 7, for < 5 +, for 5 < a) State the domain of the function. Find the function values b) f(), c) f( ), d) f(), e) f(), f) f(5), g) f(7). for the following func- Eercise.. Find the difference quotient f(+h) f() h tions: a) f() = 5, b) f() = 6, c) f() =, d) f() = +5, e) f() = ++, f) f() =, g) f() = + 9, h) f() = +7, i) f() = Eercise.5. Find the difference quotient f() f(a) a for the following functions: a) f() =, b) f() = 7, c) f() = d) f() = Eercise.6. Find the domains of the following functions. a) f() = ++5, b) f() =, c) f() = d) f() = 8, e) f() = +, f) f() = +6 g) f() = 5 +, h) f() =, i) f() = { for < < j) f() =, k) f() =, l) f() = 5 for 9 +

56 6 SESSION. FUNCTIONS BY FORMULAS AND GRAPHS Eercise.7. Below are three graphs for the functions f, g, and h. = f() function f : = g() function g : = h() function h : - Find the following function values: a) Find the domain and range of f. b) Find the domain and range of g. c) Find the domain and range of h. d) f() e) f() f) f() g) f() h) f(5) i) f(6) j) f(7) k) g(0) l) g() m) g() n) g() o) g() p) g(6) q) g(.) r) h( ) s) h( ) t) h(0) u) h() v) h() w) h() ) h( )

57 .. EXERCISES 7 Eercise.8. Use the vertical line test to determine which of the following graphs are the graphs of functions? a) b) c) d) Eercise.9. Let f be the function given b the following graph a) What is the domain of f? b) What is the range of f? c) For which is f() = 0? d) For which is f() =? e) For which is f()? f) For which is f() > 0? g) Find f() and f(5). h) Find f()+f(5). i) Find f()+5. j) Find f(+5).

58 8 SESSION. FUNCTIONS BY FORMULAS AND GRAPHS Eercise.0. The graph below displas the number of students admitted to a college during the ears 995 to ,000 5,000,000,000 ear a) How man students were admitted in the ear 000? b) In what ears were the most students admitted? c) In what ears did the number of admitted students rise fastest? d) In what ear(s) did the number of admitted students decline? Eercise.. Consider the function described b the following formula: +, for < 0 f() =, for 0 < +, for < 5 What is the domain of the function f? Graph the function f.

59 Session Introduction to the TI-8. Graphing with the TI-8 We now use the TI-8 to graph functions such as the functions discussed in the previous chapters. We start b graphing the ver well-known function =, which is of course a parabola. To graph =, we first have to enter the function. Press the lp= lp ke to get to the function menu: In the first line, enter the function = b pressing the buttons lpx,t,θ,nlp for the variable, and then press the lp lp button. We obtain 9

60 50 SESSION. INTRODUCTION TO THE TI-8 We now go to the graphing window b pressing the lpgraph lp ke. We obtain: Here, the viewing window is displaed in its initial and standard setting between 0 and 0 on the -ais, and between 0 and 0 on the -ais. These settings ma be changed b pressing the lpwindow lp or the lpzoom lp kes. First, pressing lpwindow lp, we ma change the scale b setting Xmin, Xma, Ymin and Yma to some new values using lp lp and lpenter lp : Note, that negative numbers have to be entered via lp( ) lp and not using lp lp. Note.. The difference between lp( ) lp and lp lp is that lp( ) lp is used to denote negative numbers (such as 0), whereas lp lp is used to subtract two numbers (such as 7 ). Note.. A source for a common error occurs when the Plot item in the function menu is highlighted. When graphing a function as above, alwas make sure that none of the plots are highlighted.

61 .. GRAPHING WITH THE TI-8 5 After changing the minimum and maimum - and -values of our window, we can see the result after pressing lpgraph lp again: We ma zoom in or out or put the setting back to the standard viewing size b pressing the lpzoom lp ke: We ma zoom in b pressing lp lp, then choose a center in the graph where we want to focus (via the lp lp, lp lp, lp lp, lp lp kes), and confirm this with lpenter lp : Similarl, we ma also zoom out (pressing using lp lp in the zoom menu and choosing a center). Finall we can recover the original setting b using ZStandard in the zoom menu (press lp6 lp ). We can graph more than one function in the same window, which we show net. Eample.. Graph the equations = 7 and = in the same window.

62 5 SESSION. INTRODUCTION TO THE TI-8 Solution. Enter the functions as Y and Y after pressing lp= lp. The graphs of both functions appear together in the graphing window: Here, the square root smbol is obtained using lpnd lp and lp lp, and the third power via lp lp and lp lp (followed b lp lp to continue entering the function on the base line). Eample.. Graph the relation ( ) +( 5) = 6. Solution. Since the above epression is not solved for, we cannot simpl plug this into the calculator. Instead, we have to solve for first. ( ) +( 5) = 6 = ( 5) = 6 ( ) = 5 = ± 6 ( ) = = 5± 6 ( ) Note, that there are two functions that we need to graph: = 5+ 6 ( ) and = 5 6 ( ). Entering these as Y and Y in the calculator gives the following function menu and graphing window: The graph displas a circle of radius with a center at (,5). However, due to the current scaling, the graph looks more like an ellipse instead of a circle. We can remed this b using the zoom square function; press lpzoom lp lp5 lp. This adjusts the ais to the same scale in the - and the -direction.

63 .. FINDING ZEROS, MAXIMA, AND MINIMA 5 We obtain the following graph: Note.5. We recall that the equation ( h) +( k) = r alwas forms a circle in the plane. Indeed, this equation describes a circle with center C(h,k) and radius r.. Finding zeros, maima, and minima In this section, we will show how to locate local maima and minima of a function (peaks and valles of its graph), and the intersection points of two graphs. Further we will be able to use the calculator to find the-intercepts of a graph. The -intercepts are commonl called zeros or roots of the function f. In other words, a zero of a function f is a number for which f() = 0. Eample.6. Graph the equation = +. a) Find points on the graph via the trace function, and b using the table menu. b) Approimate the zeros of the function via the calculate menu, that is, approimate the coordinate(s) of the point(s) where the graph crosses the ais. c) Approimate the (local) maimum and minimum via the calculate menu. (A (local) maimum or minimum is also called an (local) etremum.) Solution. a) Graphing the function in the standard window gives the following graph:

64 5 SESSION. INTRODUCTION TO THE TI-8 Pressing lptrace lp, we can move a cursor to the right and left via lp lp and lp lp. We can see several function values as displaed below: For the graph on the left we have displaed the point (, ) (.68,5.78), and for the graph on the right, we have displaed the point (, ) (.8,.9). Another wa to find function values is to look at the table menu b pressing lpnd lp lpgraph lp : We see the -values for specific -values are displaed. More values can be displaed b pressing lp lp and lp lp. Furthermore, the settings for the table can be changed in the table-set menu b pressing lpnd lp lpwindow lp. In particular, b changing the the Indpnt setting in the table-set menu from Auto to Ask, we can have specific values calculated in the table menu. For eample, below, we have calculated the value of for the independent variable set to =. b) Note, that the function f has three -intercepts (the places on a graph intersects with the -ais). The are called the zeros of the function

65 .. FINDING ZEROS, MAXIMA, AND MINIMA 55 and can be approimated with the TI-8. We have to find each zero, one at a time. To this end, we need to switch the calculate menu lpnd lp lptrace lp : Press lp lp to find a zero. We need to specif a left bound (using the kes lp lp, lp lp and lpenter lp ), that is a point on the graph that is a bit left of the zero that we seek. Net, we also need to specif a right bound for our zero, and also a guess that is close to the zero: The zero displaed is approimatel at.709 (rounded to the nearest thousandth). Similarl, we can also approimate the other zeros using the calculate menu lpnd lp lptrace lp. After choosing lower bounds, upper bounds, and guesses, we obtain: The zeros are approimatel at and.90. c) The TI-8 can also approimate the maimum and minimum of the function (the places on a graph where there is a peak or a valle ). This

66 56 SESSION. INTRODUCTION TO THE TI-8 is again done in the calculate menu lpnd lp lptrace lp : Now, press lp lp for the minimum. After specifing a left bound, a right bound, and a guess, we obtain the following answer for our minimum: The minimum displaed is at ( , ). The actual answer on our calculator ma be different, depending on the bounds and the guess chosen. However, it is important to note that the eact minimum is at (,) = (, ), whereas the calculator onl provides an approimation of the minimum! Similarl, we can also approimate the maimum in the calculate menu lpnd lp lptrace lp lp lp. After choosing lower and upper bound, we obtain: Therefore, the maimum is approimatel at (, ) ( 0.667, 5.8), after rounding to the nearest thousandth. This concludes eample.6. Note.7. We can alwas return to the home screen b pressing quit, that is lpnd lp lpmodelp.

67 .. FINDING ZEROS, MAXIMA, AND MINIMA 57 Eample.8. Graph the equation = +. a) Approimate the zeros of the function via the calculate menu. b) Approimate the (local) maima and the minima via the calculate menu. c) Calculate the zeros of the function via the solve function. Solution. a) First, press lp= lp and enter the function = +. We find the zeros using item from the calculate menu. Here are two of the four zeros: Note, in particular, the left zero has a value of.6e, that is =.6 0 = The value = is an actual zero as can be seen b direct computation, whereas the calculator onl obtains an approimation! b) From the calculate menu, lpnd lp lptrace lp, items and can be used to approimate the maimum and the two minima, (choosing a lower bound, upper bound, and a guess each time): c) There is also a non-graphical procedure to calculate the zeros of the function = +. Algebraicall, we want to find a number so that = 0. In other words: 0 = +.

68 58 SESSION. INTRODUCTION TO THE TI-8 Press the lpmathlp button, scroll down to Solver... and press the lpenter lp ke. You will end up in one of the following two screens: If ou obtained the screen on the right, press lp lp to get to the screen on the left. Enter the equation and press lpenter lp. We now need to specif a number X which is our guess for the zero. In other words, the calculator will tr to identif a zero that is close to a specified number X. For eample, we ma enter =. Then, the solve command is eecuted b pressing lpalpha lp lpenter lp ; (the lpalpha lp -ke gives access to the commands written in green). We obtain the following zero: lpalpha lp lpenter lp Choosing other values for (for eample =., = 0. and = ) produces the other three zeros after pressing lpalpha lp lpenter lp.

69 .. FINDING ZEROS, MAXIMA, AND MINIMA 59 Note, that the eact zeros are =, =, = 0, and =, whereas again the calculator onl supplied an approimation! The equation solver used in the last eample (part (c)) is a slightl faster method for finding zeros than using the graph and the calculate menu. However, the disadvantage is that we first need to have some knowledge about the zeros (such as how man zeros there are!) before we can effectivel use this tool. Generall, the graph and the calculate menu will give us a much better idea of where the zeros are located. For this reason, we recommend to use the calculate menu as the main method of finding zeros. Eample.9. Solve the equation. Approimate our answer to the nearest hundredth. a) = 0, b) = 7. Solution. a) We need to find all numbersso that = 0. Note, that these are precisel the zeros of the function f() = + 5 7, since the zeros are the values for which f() = 0. Graphing the function f() = shows that there are two zeros. Using the calculate menu, we can identif both zeros. The answer is therefore, 0.85 and.6 (rounded to the nearest hundredth).

70 60 SESSION. INTRODUCTION TO THE TI-8 b) We use the same method as in part (a). To this end we rewrite the equation so that one side becomes zero: 7+ = 0 We can now graph the function f() = 7+ and find its zeros. We see that there are three solutions,.0, 0.5, and.06. B a method similar to the above, we can also find the intersection points of two given graphs. Eample.0. Graph the equations = + and = +. a) Approimate the intersection point of the two graphs via the calculate menu. b) Calculate the intersection point of the two graphs via the solve function. Solution. First, enter the two equations for Y and Y after pressing the lp= lp ke. Both graphs are displaed in the graphing window. a) The procedure for finding the intersection of the graphs for Y and Y is similar to finding minima, maima, or zeros. In the calculate menu, lpnd lp lptrace lp, choose intersect (item 5). Choose the first curve Y (with lp lp, lp lp, and lpenter lp ), and the second curve Y. Finall choose a guess of the intersection point (with lp lp,

71 .. EXERCISES 6 lp lp, and lpenter lp ). The intersection is approimated as (,) (.757,.79755): b) The intersection is the point where the two -values = + and = + coincide. Therefore, we want to find an with + = + The TI-8 equation solver requires an equation with zero on the left. Therefore, we subtract the left-hand side, and obtain: (subtract ( +)) = 0 = ( + ) ( +) = 0 = + + = 0 = + + Entering 0 = + + into the equation solver, lpmathlp lp lp lpenter lp, and choosing an approimation = for X, the calculator solves this equation (using lpalpha lp lpenter lp ) as follows: The approimation provided b the calculator is Eercises Eercise.. Graph the equation in the standard window. a) = 5 b) = c) = + d) = e) = + f) = +

72 6 SESSION. INTRODUCTION TO THE TI-8 For the last eercise, the absolute value is obtained b pressing lpmathlp lp lp lpenter lp. Eercise.. Solve the equation for and graph all branches in the same window. a) + = b) (+5) + = 5 c) ( ) +( ) = 9 d) + 8 = 0 e) = + f) = +77 Eercise.. Find all zeros of the given function. Round our answer to the nearest hundredth. a) f() = ++ b) f() = + c) f() = + 6 d) f() = e) f() = 5 ++ f) f() = + g) f() = h) f() = i) f() = Eercise.. Find all solutions of the equation. Round our answer to the nearest thousandth. a) + = 5 +7 b) +6 = 0 c) + + d) 5 + = 5 +6 e) + = + f) = + Eercise.5. Graph the equation. Determine how man maima and minima the graph has. To this end, resize the graphing window (via the zoom-in, zoom-out, and zoom-bo functions of the calculator) to zoom into the maima or minima of the graph. a) = + b) = + 0 c) = 5 + d) = Eercise.6. Approimate the (local) maima and minima of the graph. Round our answer to the nearest tenth. a) = + + b) = 5 +8 c) = d) = +6+ e) = +8+ f) = +7+

73 Session 5 Basic functions and transformations 5. Graphing basic functions It will be useful to stud the shape of graphs of some basic functions, which ma then be taken as building blocks for more advanced and complicated functions. In this section, we consider the following basic functions: =, =, =, =, =. We can either graph these functions b hand b calculating a table, or b using the TI-8, via the table and graph buttons. We begin with the absolute value function =. Recall that the ab- value is obtained on the calculator in the math menu b pressing solute lpmathlp lp lp lpenter lp. The domain of = is all real numbers, D = R We have drawn the graph a second time in the --plane on the right. 6

74 6 SESSION 5. BASIC FUNCTIONS AND TRANSFORMATIONS Similarl, we obtain the graph for =, which is a parabola. The domain is D = R Here is the graph for =. The domain is D = R Net we graph =. The domain is D = [0, ) Finall, here is the graph for =. The domain is D = R {0}

75 5.. TRANSFORMATION OF GRAPHS 65 These graphs together with the line = m+b studied in Section. are our basic building blocks for more complicated graphs in the net sections. Note in particular, that the graph of = is the diagonal line. = m+c = c 5. Transformation of graphs For a given function, we now stud how the graph of the function changes when performing elementar operations, such as adding, subtracting, or multipling a constant number to the input or output. We will stud the behavior in five specific eamples.. Consider the following graphs: = = + = We see that the function = is shifted up b units, respectivel down b units. In general, we have: Observation 5.. Consider the graph of a function = f(). Then, the graph of = f()+c is that of = f() shifted up or down b c. If c is positive, the graph is shifted up, if c is negative, the graph is shifted down.

76 66 SESSION 5. BASIC FUNCTIONS AND TRANSFORMATIONS. Net, we consider the transformation of = given b adding or subtracting a constant to the input. = = (+) = ( ) Now, we see that the function is shifted to the left or right. Note, that = ( + ) shifts the function to the left, which can be seen to be correct, since the input = gives the output = (( )+) = 0 = 0. Observation 5.. Consider the graph of a function = f(). Then, the graph of = f(+c) is that of = f() shifted to the left or right b c. If c is positive, the graph is shifted to the left, if c is negative, the graph is shifted to the right.. Another transformation is given b multipling the function b a fied positive factor. = + = ( +) = ( +) This time, the function is stretched awa or compressed towards the -ais. Observation 5.. Consider the graph of a function = f() and let c > 0. Then, the graph of = c f() is that of = f() stretched awa or compressed towards the -ais b a factor c. If c >, the graph is stretched awa from the -ais, if 0 < c <, the graph is compressed towards the -ais.

77 5.. TRANSFORMATION OF GRAPHS 67. Similarl, we can multipl the input b a positive factor. = + = ( ) + = ( ) This time, the function is stretched awa or compressed towards the -ais. Observation 5.. Consider the graph of a function = f() and let c > 0. Then, the graph of = f(c ) is that of = f() stretched awa or compressed towards the -ais b a factor c. If c >, the graph is compressed towards the -ais, if 0 < c <, the graph is stretched awa from the -ais. 5. The last transformation is given b multipling ( ) to the input or output, as displaed in the following chart. = + = ( +) = ( ) Here, the function is reflected either about the -ais or the -ais. Observation 5.5. Consider the graph of a function = f(). Then, the graph of = f() is that of = f() reflected about the -ais. Furthermore, the graph of = f( ) is that of = f() reflected about the -ais.

78 68 SESSION 5. BASIC FUNCTIONS AND TRANSFORMATIONS Eample 5.6. Guess the formula for the function, based on the basic graphs in Section 5. and the transformations described above a) - b) - c) d) Solution. a) This is the square-root function shifted to the left b. Thus, b Observation 5., this is the function f() = +. b) This is the graph of = reflected about the -ais (or also = reflected about the -ais). In either case, we obtain the rule =. c) This is a parabola reflected about the -ais and then shifted up b. Thus, we get: = reflecting about the -ais gives = shifting the graph up b gives = + d) Starting from the graph of the cubic equation =, we need to reflect about the -ais (or also -ais), then shift up b and to the right b. These transformations affect the formula as follows: = reflecting about the -ais gives = shifting up b gives = + shifting the the right b gives = ( ) + All these answers can be checked b graphing the function with the TI- 8. -

79 5.. TRANSFORMATION OF GRAPHS 69 Eample 5.7. Sketch the graph of the function, based on the basic graphs in Section 5. and the transformations described above. a) = + b) = (+) c) = d) = + e) = ( +) f) = ( +) Solution. a) This is the parabola = shifted up b. The graph is shown below. b) = (+) is the parabola = shifted units to the left. 5 a) b) c) The graph of the function f() = is the absolute value shifted to the right b and down b. (Alternativel, we can first shift down b and then to the right b.) d) Similarl, to get from the graph of = to the graph of = +, we shift the graph to the left, and then for = +, we need to stretch the graph b a factor awa from the -ais. (Alternativel, we could first stretch the the graph awa from the -ais, then shift the graph b to the left.) c) - - d) e) For = ( +), we start with = and add, giving = +, which shifts the graph up b. Then, taking the negative gives = ( + ), which corresponds to reflecting the graph about the -ais.

80 70 SESSION 5. BASIC FUNCTIONS AND TRANSFORMATIONS Note, that in this case, we cannot perform these transformations in the opposite order, since the negative of = gives =, and adding gives = + which is not equal to ( +). f) We start with =. Adding in the argument, = (+), shifts its graph to the left b. Then, taking the negative in the argument gives = ( +), which reflects the graph about the -ais. Here, the order in which we perform these transformations is again important. In fact, if we first take the negative in the argument, we obtain = ( ). Then, adding one in the argument would give = ( (+)) = ( ) which is different than our given function = ( +) e) - - All these solutions ma also easil be checked b using the graphing function of the calculator. Eample 5.8. a) The graph of f() = 5 is stretched awa from the -ais b a factor of. What is the formula for the new function? b) The graph of f() = 6 + is shifted up 5 units, and then reflected about the -ais. What is the formula for the new function? c) How are the graphs of = +5 9 and = ( ) +5( ) 9 related? d) How are the graphs of = ( ) and = ( +) related? Solution. a) B Observation 5. on page 67, we have to multipl the argument b. The new function is therefore: ( ) f = ) ( 5 = 7 5 f) - -

81 5.. TRANSFORMATION OF GRAPHS 7 b) After the shift, we have the graph of a new function = Then, a reflection about the -ais gives the graph of the function = ( 6 ++5). c) B Observation 5. on page 66, we see that we need to shift the graph of = +5 9 b units to the right. d) The formulas can be transformed into each other as follows: We begin with = ( ). Replacing b +5 gives = ((+5) ) = (+). Replacing b gives = (( )+) = ( +). Therefore, we have performed a shift to the left b 5, and then a reflection about the -ais. We want to point out that there is a second solution for this problem: We begin with = ( ). Replacing b gives = (( ) ) = ( ). Replacing b 5 gives = ( ( 5) ) = ( +5 ) = ( +). Therefore, we could also first perform a reflection about the -ais, and then shift the graph to the right b 5. Some of the above functions have special smmetries, which we investigate now. Definition 5.9. A function f is called even if f( ) = f() for all. Similarl, a function f is called odd if f( ) = f() for all. Eample 5.0. Determine, if the following functions are even, odd, or neither. f() =, g() =, h() =, k() = 5, l() = 5 +7, m() = +5. Solution. The function f() = is even, since f( ) = ( ) =. Similarl, g() = is odd, h() = is even, and k() = 5 is odd, since g( ) = ( ) = = g()

82 7 SESSION 5. BASIC FUNCTIONS AND TRANSFORMATIONS h( ) = ( ) = = h() k( ) = ( ) 5 = 5 = k() Indeed, we see that a function = n is even, precisel when n is even, and = n is odd, precisel when n is odd. (These eamples are in fact the motivation behind defining even and odd functions as in definition 5.9 above.) Net, in order to determine if the function l is even or odd, we calculate l( ) and compare it with l(). l( ) = ( ) 5 +7( ) ( ) = = ( 5 +7 ) = l() Therefore, l is an odd function. Finall, for m() = +5, we calculate m( ) as follows: m( ) = ( ) +5( ) = 5 Note, that m is not an even function, since Furthermore, m is also not an odd function, since 5 ( +5). Therefore, m is a function that is neither even nor odd. Observation 5.. An even function f is smmetric with respect to the ais (if ou reflect the graph of f about the ais ou get the same graph back), since even functions satisf f( ) = f(): Eample = : f( ) = f() An odd function f is smmetric with respect to the origin (if ou reflect the graph of f about the ais and then about the ais ou get the same

83 5.. EXERCISES 7 graph back), since odd functions satisf f( ) = f(): f() Eample = : f() 5. Eercises Eercise 5.. Find a possible formula of the graph displaed below a) - b) c) d) e) - f) -

84 7 SESSION 5. BASIC FUNCTIONS AND TRANSFORMATIONS Eercise 5.. Sketch the graph of the function, based on the basic graphs in Section 5. and the transformations described above. Confirm our answer b graphing the function with the calculator. a) f() = b) f() = + c) f() = d) f() = ( ) e) f() = f) f() = g) f() = + h) f() = ( ) + Eercise 5.. Consider the graph of f() = 7+. Find the formula of the function that is given b performing the following transformations on the graph. a) Shift the graph of f down b. b) Shift the graph of f to the left b units. c) Reflect the graph of f about the -ais. d) Reflect the graph of f about the -ais. e) Stretch the graph of f awa from the -ais b a factor. f) Compress the graph of f towards the -ais b a factor. Eercise 5.. How are the graphs of f and g related? a) f() =, g() = 5 b) f() =, g() = c) f() = (+), g() = ( ) d) f() = ++5, g() = () +() +5 e) f() = + f) f() =, g() = + + Eercise 5.5. Determine, if the function is even, odd, or neither. a) =, b) = 5, c) = +5, d) = +5, e) =, f) =, g) - h) - i) -

85 5.. EXERCISES 75 Eercise 5.6. The graph of the function = f() is displaed below Sketch the graph of the following functions. a) = f()+, b) = f( ), c) = f(), d) = f(), e) = f(), f) = f( )

86 Session 6 Operations on functions 6. Operations on functions given b formulas We can combine functions in man different was. First, note that we can add, subtract, multipl, and divide functions. Eample 6.. Let f() = + 5 and g() = 7. Find the following functions, and state their domain. ( ) f (f +g)(),(f g)(),(f g)(), and (). g Solution. The functions are calculated b adding the functions (or subtracting, multipling, dividing). (f +g)() = ( +5)+(7 ) = +, (f g)() = ( +5) (7 ) = = +, (f g)() = ( +5) (7 ) = = 7 + 5, ( ) f () = +5 g 7. The calculation of these functions was straightforward. To state their domain is also straightforward, ecept for the domain of the quotient f. Note, that g f + g, f g, and f g are all polnomials. B the standard convention 76

87 6.. OPERATIONS ON FUNCTIONS GIVEN BY FORMULAS 77 (Convention. on page 6) all these functions have the domain R, that is their domain is all real numbers. Now, for the domain of, we have to be a bit more careful, since the f g denominator of a fraction cannot be zero. The denominator of f g () = +5 7 is zero, eactl when 7 = 0 = 7 = = = 7. We have to eclude 7 from the domain. The domain of the quotient f g is therefore R { 7 }. We can formall state the observation we made in the previous eample. Observation 6.. Let f be a function with domain D f, and g be a function with domain D g. A value can be used as an input of f +g, f g, and f g eactl when is an input of both f and g. Therefore, the domains of the combined functions are the intersection of the domains D f and D g : D f+g = D f D g = { D f and D g }, D f g = D f D g, D f g = D f D g. For the quotient f, we also have to make sure that the denominator g() is g not zero. Df g = { D f, D g, and g() 0}. Eample 6.. Let f() = +, and let g() = 5 +. Find the functions f and g and state their domains. g f Solution. First, the domain of f consists of those numbers for which the square root is defined. In other words, we need + 0, that is, so that the domain of f is D f = [, ). On the other hand, the domain of g is all real numbers, D g = R. Now, we have the quotients ( f g ) () = + 5+, and ( g f ) () = For the domain of f g, we need to eclude those numbers for which 5+ = 0. Thus, 5+ = 0 = ( )( ) = 0

88 78 SESSION 6. OPERATIONS ON FUNCTIONS = =, or =. We obtain the domain for f as the combined domain for f and g, and eclude g and. Therefore, Df = [, ) {,}. g Now, for g () = 5+ f +, the denominator becomes zero eactl when + = 0, = =. Therefore, we need to eclude from the domain, that is D g f = [, ) { } = (, ). Eample 6.. To form the quotient f g () where f() = and g() = + we write f () = =. One might be tempted to sa g + that the domain is all real numbers. But it is not! The domain is all real numbers ecept, and the last step of the simplification performed above is onl valid for. + = (+)( ) Another operation we can perform is the composition of two functions. Definition 6.5. Let f and g be functions, and assume that g() is in the domain of f. Then define the composition of f and g at to be (f g)() := f(g()). We can take an as an input of f g which is an input of g and for which g() is an input of f. Therefore, if D f is the domain of f and D g is the domain of g, the domain of D f g is D f g = { D g, g() D f }. Eample 6.6. Let f() = + 5 and g() =. Find the following compositions a) f(g()), b) g(f()), c) f(f()), d) f( g(5)), e) g(g()+5). Solution. We evaluate the epressions, one at a time, as follows: a) f(g()) = f( ) = f( ) = ( ) +5 ( )

89 6.. OPERATIONS ON FUNCTIONS GIVEN BY FORMULAS 79 = 5 =, b) g(f()) = g( +5 ) = g(8+5) = g() = =, c) f(f()) = f( +5 ) = f(+5) = f(7) = = 98+5 =, d) f( g(5)) = f( ( 5)) = f( ( )) = f( 6) = ( 6) +5 ( 6) = 7 0 =, e) g(g()+5) = g(( )+5) = g(( )+5) = g() = =. We can also calculate composite functions for arbitrar in the domain. Eample 6.7. Let f() = + and g() = +. compositions Find the following a) (f g)(), b) (g f)(), c) (f f)(), d) (g g)(). Solution. a) There are essentiall two was to evaluate (f g)() = f(g()). We can either first use the eplicit formula for f() and then the one for g(), or vice versa. We will evaluate f(g()) b substituting g() into the formula for f(): (f g)() = f(g()) = (g()) + = (+) + = = Similarl, we evaluate the other epressions (b)-(d): b) (g f)() = g(f()) = f()+ = ++ = +, c) (f f)() = f(f()) = (f()) + = ( +) + = + ++ = + +, d) (g g)() = g(g()) = g()+ = ++ = +6. Eample 6.8. Find (f g)() and (g f)() for the following functions, and state their domains. a) f() =, and g() = +, b) f() = 6+, and g() = 5+, c) f() = ( ), and g() = +.

90 80 SESSION 6. OPERATIONS ON FUNCTIONS Solution. a) Composing f g, we obtain (f g)() = f(g()) = g()+ = +. The domain is the set of numbers for which the denominator is non-zero. + = 0 = ( )( ) = 0 = = or = = D f g = R {,}. Similarl, ( ) (g f)() = g(f()) = f() f() = = (+) 9 + = 9 9(+) (+) = (+) = 9 9 (+) = 9 (+) (+) The domain is all real numbers ecept =, that is D g f = R { }. b) We calculate the compositions as follows. (f g)() = f(g()) = g() 6g()+ = (5+) 6(5+)+ = 5+ 0, (g f)() = g(f()) = 5f()+ = 5 ( 6+)+ = = Since the domains of f and g are all real numbers, so are also the domains for both f g and g f. c) Again we calculate the compositions. (f g)() = f(g()) = = (g() ) = = =. ( + )

91 6.. OPERATIONS ON FUNCTIONS GIVEN BY TABLES 8 The domain of g is all real numbers, and the outputs g() = + are all, (since 0). Therefore, g() is in the domain of f, and we have a combined domain of f g of D f g = R. On the other hand, ( ) (g f)() = g(f()) = (f()) + = ( ) + ( ) = ( ) + = ( )+ =. The domain of g f consists of all numbers which are in the domain of f and for which f() is in the domain of g. Now, the domain of f consists of all real numbers that give a non-negative argument in the square-root, that is: ( ) 0. Therefore we must have 0, so that, and we obtain the domain D f = [, ). Since the domain D g = R, the composition g f has the same domain as f: D g f = D f = [, ). We remark that at a first glance, we might have epected that (g f) = has a domain of all real numbers. However, the composition g(f()) can onl have those inputs that are also allowed inputs of f. We see that the domain of a composition is sometimes smaller than the domain that we use via the standard convention (Convention.). 6. Operations on functions given b tables We can also combine functions that are defined using tables. Eample 6.9. Let f and g be the functions defined b the following table f() g() 0 5 Describe the following functions via a table: a) f()+, b) f() g(), c) f(+), d) g( ).

92 8 SESSION 6. OPERATIONS ON FUNCTIONS Solution. For (a) and (b), we obtain b immediate calculation f() f() g() 5 For eample, for =, we obtain f()+ = f()+ = + = 5 and f() g() = f() g() = =. For part (c), we have a similar calculation of f( + ). For eample, for =, we get f(+) = f(+) = f() = f(+) undef. undef. 6 Note that for the last two inputs = 6 and = 7 the epression f(+) is undefined, since, for eample for = 6, it is f(+) = f(6+) = f(8) which is undefined. However, for =, we obtain f(+) = f( +) = f() = 6. If we define h() = f( + ), then the domain of h is therefore D h = {,0,,,,,5}. Finall, for part (d), we need to take as inputs, for which g( ) is defined via the table for g. We obtain the following answer g( ) 0 5 Eample 6.0. Let f and g be the functions defined b the following table f() g() Describe the following functions via a table: a) f g, b) g f, c) f f, d) g g. Solution. The compositions are calculated b repeated evaluation. For eample, (f g)() = f(g()) = f(7) =.

93 6.. EXERCISES 8 The complete answer is displaed below (f g)() undef (g f)() undef. 9 (f f)() 5 7 undef. 9 (g g)() undef Eercises Eercise 6.. Find f + g, f g, f g for the functions below. State their domain. a) f() = +6, and g() = 5, b) f() = +5, and g() = 5 +7, c) f() = +7, and g() = +5, d) f() = +, and g() = 5 +, e) f() =, and g() =, f) f() = ++5, and g() = 6, g) f() = +, and g() = ++. Eercise 6.. Find f g, and g f for the functions below. State their domain. a) f() = +6, and g() = 8, b) f() = +, and g() = 5+, c) f() = and g() = 5 + d) f() = +6, and g() = +5, e) f() = +8, and g() =. Eercise 6.. Let f() = and g() = +. Find the following compositions a) f(g()), b) g(f()), c) f(f(5)), d) f(5g( )), e) g(f() ), f) f(f()+g()), g) g(f(+)), h) f(f( )), i) f(f( ) g()), j) f(f(f())), k) f(+h), l) g(+h).

94 8 SESSION 6. OPERATIONS ON FUNCTIONS Eercise 6.. Find the composition (f g)() for the functions: a) f() = 5, and g() = +, b) f() = +, and g() = +, c) f() = +, and g() = +, d) f() = + +, and g() = +, e) f() =, and g() = +h, + f) f() = ++, and g() = +h. Eercise 6.5. Find the compositions for the following functions: (f g)(), (g f)(), (f f)(), (g g)() a) f() = +, and g() = 5, b) f() = +, and g() =, c) f() = 6, and g() = +, d) f() = +, and g() =, e) f() = ( 7), and g() = +7. Eercise 6.6. Let f and g be the functions defined b the following table. Complete the table given below f() g() f()+ g()+5 g() f() f(+) Eercise 6.7. Let f and g be the functions defined b the following table. Complete the table b composing the given functions. 5 6 f() 5 6 g() 5 6 (g f)() (f g)() (f f)() (g g)()

95 6.. EXERCISES 85 Eercise 6.8. Let f and g be the functions defined b the following table. Complete the table b composing the given functions f() g() (g f)() (f g)() (f f)() (g g)()

96 Session 7 The inverse of a function 7. One-to-one functions We have seen that some functions f ma have the same outputs for different inputs. For eample for f() =, the inputs = and = have the same output f() = and f( ) =. A function is one-to-one, precisel when this is not the case. Definition 7.. A function f is called one-to-one (or injective), if two different inputs alwas have different outputs f( ) f( ). Eample 7.. As was noted above, the function f() = is not one-to-one, because, for eample, for inputs and, we have the same output f( ) = ( ) =, f() = =. On the other hand, g() = is one-to-one, since, for eample, for inputs and, we have different outputs: g( ) = ( ) = 8, g() = = 8. The difference between the functions f and g can be seen from their graphs. = =

97 7.. ONE-TO-ONE FUNCTIONS 87 The graph of f() = on the left has for different inputs ( 0 and 0 ) the same output ( 0 = ( 0 ) = ( 0 ) ). This is shown in the graph since the horizontal line at 0 intersects the graph at two different points. In general, two inputs that have the same output 0 give two points on the graph which also lie on the horizontal line at 0. Now, the graph of g() = on the right intersects with a horizontal line at some 0 onl once. This shows that for two different inputs, we can never have the same output 0, so that the function g is one-to-one. We can summarize the observation of the last eample in the following statement. Observation 7. (Horizontal Line Test). A function is one-to-one eactl when ever horizontal line intersects the graph of the function at most once. = f() 0 0 Eample 7.. Which of the following are or represent one-to-one functions? 5 5 a) b) c) f() = +6 + d) f() = + Solution. We use the horizontal line test to see which functions are one-toone. For (a) and (b), we see that the functions are not one-to-one since there -

98 88 SESSION 7. THE INVERSE OF A FUNCTION is a horizontal line that intersects with the graph more than once: 5 5 a) b) For (c), using the calculator to graph the functionf() = +6 +, we see that all horizontal lines intersect the graph eactl once. Therefore the function in part (c) is one-to-one. The function in part (d) however has a graph that intersects some horizontal line in several points. Therefore f() = + is not one-to-one: 5 5 c) d) Inverse function A function is one-to-one, when each output is determined b eactl one input. Therefore we can construct a new function, called the inverse function, where we reverse the roles of inputs and outputs. For eample, when =,

99 7.. INVERSE FUNCTION 89 each 0 comes from eactl one 0 as shown in the picture below: 0 0 The inverse function assigns to the input 0 the output 0. Definition 7.5. Let f be a function with domain D f and range R f, and assume that f is one-to-one. The inverse of f is a function f so that f() = means precisel that f () =. input output f f output input Here the outputs off are the inputs off, and the inputs off are the outputs of f. Therefore, the inverse function f has a domain equal to the range of f, D f = R f, and f has a range equal to the domain of f, R f = D f. In short, when f is a function f : D f R f, then the inverse function f is a function f : R f D f. The inverse function reverses the roles of inputs and outputs. Eample 7.6. Find the inverse of the following function via algebra. a) f() = 7 b) g() = + c) h() = d) j() = e) k() = ( ) + for Solution. a) First, reverse the role of input and output in = 7 b echanging the variables and. That is, we write = 7. We need to solve this for : (add 7) = +7 = = = +7.

100 90 SESSION 7. THE INVERSE OF A FUNCTION Therefore, we obtain that the inverse of f is f () = +7. For the other parts, we alwas echange and and solve for : b) Write = + and echange and : c) Write = + d) Write = + + = + + = + = = + = g () = and echange and : = + = + = = h () = and echange and : = = = = (+) = ( +) = + = + = + = = = ( ) = = = = j () =, e) Write = ( ) + and echange and : = ( ) + = = ( ) = = = = + = k () = + The function in the last eample is not one-to-one when allowing to be an real number. This is wh we had to restrict the eample to the inputs. We eemplif the situation in the following eample. Eample 7.7. Note, that the function = can be restricted to a one-to-one function b choosing the domain to be all non-negative numbers [0, ) or b choosing the domain to be all non-positive numbers (, 0].

101 7.. INVERSE FUNCTION 9 Let f : [0, ) [0, ) be the function f() =, so that f has a domain of all non-negative numbers. Then, the inverse is the function f () =. On the other hand, we can take g() = whose domain consists of all non-positive numbers(,0], that isg : (,0] [0, ). Then, the inverse function g must reverse domain and range, that is g : [0, ) (,0]. The inverse is obtained b echanging and in = as follows: = = = ± = g () =. Eample 7.8. Restrict the function to a one-to-one function. Find the inverse function, if possible. a) f() = (+) +, b) g() = ( ), c) h() =. Solution. The graphs of f and g are displaed below. a) b) a) The graph shows that f is one-to-one when restricted to all numbers, which is the choice we make to find an inverse function. Net, we replace and in = (+) + to give = ( +) +. When solving this for, we must now remember that our choice of becomes after replacing with. We now solve for. = ( +) + = = ( +) = = ± = + = ± Since we have chosen the restriction of, we use the epression with the positive sign, = +, so that the inverse function is f () = +. b) For the function g, the graph shows that we can restrict g to > to obtain a one-to-one function. The inverse for this choice is given as follows.

102 9 SESSION 7. THE INVERSE OF A FUNCTION Replacing and in = ( ) gives = ( ), which we solve this for under the condition >. = ( ) = ( ) = = = ± = = ± = g () = + c) Finall, h() = can be graphed with the calculator as follows: The picture on the right shows that the approimation of the local minimum is (approimatel) at =. Therefore, if we restrict h to all, we obtain a one-to-one function. We replace and in =, so that the inverse function is obtained b solving the equation = for. However, this equation is quite complicated and solving it is beond our capabilities at this time. Therefore we simpl sa that h () is that for which =, and leave the eample with this. Let f be a one-to-one function. If f maps 0 to 0 = f( 0 ), then f maps 0 to 0. In other words, the inverse function is precisel the function for which f (f( 0 )) = f ( 0 ) = 0 and f(f ( 0 )) = f( 0 ) = 0. f 0 0 f Observation 7.9. Let f and g be two one-to-one functions. Then f and g are inverses of each other eactl when f(g()) = and g(f()) = for all. In this case we write that g = f and f = g.

103 7.. INVERSE FUNCTION 9 Eample 7.0. Are the following functions inverse to each other? a) f() = 5+7, g() = 7 b) f() =, g() = +6, 6 c) f() =, g() = +. Solution. We calculate the compositions f(g()) and g(f()). a) f(g()) = f( 7 ) = = ( 7)+7 =, 5 5 g(f()) = g(5+7) = (5+7) 7 5 b) f(g()) = f( +6) = ( +6) 6 = g(f()) = g( 6 ) = 6 5, = 5 5 =, +6 = 6 = =, +6 = ( 6)+6 =. Using the Observation 7.9, we see that in both part (a) and (b), the functions are inverse to each other. For part (c), we calculate for a general in the domain of g: f(g()) = +. It is enough to show that for one composition (f g)() does not equal to conclude that f and g are not inverses. (It is not necessar to also calculate the other composition g(f()).) Warning 7.. It is true that when we have one relation f(g()) =, then we also have the other relation g(f()) =. Nevertheless, we recommend to alwas check both equations. The reason is that it is eas to mistake one of the relations, when not being careful about the domain and range. For eample, let f() = and g() =. Then, naivel, we would calculate f(g()) = ( ) = and g(f()) = =, so that the first equation would sa f and g are inverses, whereas the second equation sas the are not inverses. We can resolve the apparent contradiction b restricting f to a one-to-one function. For eample, if we take the domain of f to be all positive numbers or zero, D f = [0, ), then f(g()) = f( ) which is undefined, since f onl takes non-negative inputs. Also, it is g(f()) = =. Therefore, the functions f and g are not inverse to each other! On the other hand, if we restrict the function f() = to all negative numbers and zero, D f = (,0], then f(g()) = ( ) =, since now f is defined for the negative input. Also, for a negative number < 0, it is g(f()) = = =. So, in this case, f and g are inverse to each other!

104 9 SESSION 7. THE INVERSE OF A FUNCTION Our last observation concerns the graph of inverse functions. If f( 0 ) = 0 then f ( 0 ) = 0, and the point P( 0, 0 ) is on the graph of f, whereas the point Q( 0, 0 ) is on the graph of f. 0 P 0 Q 0 0 We see that Q is the reflection of P along the diagonal =. Since this is true for an point on the graph of f and f, we have the following general observation. Observation 7.. The graph of f is the graph of f reflected along the diagonal. f f diagonal Eample 7.. Find the graph of the inverse function of the function given below. 5 a) b) f() = (+)

105 7.. EXERCISES 95 Solution. Carefull reflecting the graphs given in part (a) and (b) gives the following solution. The function f() = (+) in part (b) can be graphed with a calculator first a) b) Eercises Eercise 7.. Use the horizontal line test to determine if the function is oneto-one. 5 5 a) b) c) f() = ++5 d) f() = +9 e) f() = 5 f) f() = g) f() = + h) f() = + Eercise 7.. Find the inverse of the function f and check our solution. a) f() = +9 b) f() = 8 c) f() = +8 d) f() = +7 e) f() = 6 f) f() = g) f() = (+5) h) f() = +5

106 96 SESSION 7. THE INVERSE OF A FUNCTION i) f() = j) f() = k) f() = l) f() = 5 o) f() = + + p) f() = 7 5 m) f() = + n) f() = 6 q) f given b the table below: f() Eercise 7.. Restrict the domain of the function f in such a wa that f becomes a one-to-one function. Find the inverse of f with the restricted domain. a) f() = b) f() = (+5) + c) f() = d) f() = e) f() = f) f() = (+7) g) f() = h) f() = ( ) 0 Eercise 7.. Determine whether the following functions f and g are inverse to each other. a) f() = +, and g() =, b) f() =, and g() =, c) f() = +, and g() =, d) f() = 6, and g() = + 6, e) f() = 5, and g() = 5+, f) f() =, and g() = +. Eercise 7.5. Draw the graph of the inverse of the function given below. 5 5 a) b)

107 7.. EXERCISES 97 5 c) d) f() = e) f() = f) f() = g) f() = h) f() = for > i) f() = for <.

108 Review of functions and graphs Eercise I.. Find all solutions of the equation: 9 = 6 Eercise I.. Find the equation of the line displaed below Eercise I.. Find the solution of the equation: + = 0 Approimate our answer to the nearest hundredth. Eercise I.. Let f() = + 5. as much as possible. f(+h) f() h Simplif the difference quotient Eercise I.5. Consider the following graph of a function f. 5 Find: domain of f, range of f, f(), f(5), f(7), f(9). 98

109 Eercise I.6. Find the formula of the graph displaed below Eercise ( I.7. Let f() = 5+ and g() = Find the quotient f ) g () and state its domain. Eercise I.8. Let f() = + and g() =. Find the composition (f g)() and state its domain. Eercise I.9. Consider the assignments for f and g given b the table below. 5 6 f() 5 0 g() 6 Is f a function? Is g a function? Write the composed assignment for (f g)() as a table. Eercise I.0. Find the inverse of the function f() =

110 Part II Polnomials and rational functions 00

111 Session 8 Dividing polnomials We now start our discussion of specific classes of eamples of functions. The first class of functions that we discuss are polnomials and rational functions. Let us first recall the definition of polnomials and rational functions. Definition 8.. A monomial is a number, a variable, or a product of numbers and variables. A polnomial is a sum (or difference) of monomials. Eample 8.. The following are eamples of monomials: 5,, 7, z, a n The following are eamples of polnomials: + 7, ++z +mn, 5 9, 5 In particular, ever monomial is also a polnomial. We are mainl interested in polnomials in one variable, and consider these as functions. For eample, f() = + 7 is such a function. Definition 8.. A polnomial is a function f of the form f() = a n n +a n n + +a +a +a 0, for some real (or comple) numbers a 0,a,...,a n. The domain of a polnomial f is all real numbers (see our standard convention.). The numbers a 0,a,...,a n are called coefficients. For each k, the number a k is the coefficient of k. The number a n is called the leading coefficient and n is the degree of the polnomial. The zeros of a polnomial are usuall referred to as roots. Therefore is a root of a polnomial f precisel when f() = 0. 0

112 0 SESSION 8. DIVIDING POLYNOMIALS Definition 8.. A rational function is a fraction of two polnomials f() = g(), where g() and h() are both polnomials. The domain of f is all real h() numbers for which the denominator h() is not zero: D f = { h() 0 }. Eample 8.5. The following are eamples of rational functions: f() = , f() =, f() = Long division We now show how to divide two polnomials. The method is similar to the long division of natural numbers. Our first eample shows the procedure in full detail. Eample 8.6. Divide the following fractions via long division: a) 57 b) Solution. a) Recall the procedure for long division of natural numbers: = remainder The steps above are performed as follows. First, we find the largest multiple of less or equal to 5. The answer is written as the first digit on the top line. Multipl times, and subtract the answer from the first two digits 5 of the dividend. The remaining digits 7 are copied below to give 7. Now we repeat the procedure, until we arrive at the remainder 7. In short, what we have shown is that: 57 = +7 or alternativel, 57 = + 7.

113 8.. LONG DIVISION 0 b) We repeat the steps from part (a) as follows. First, write the dividend and divisor in the above format: Net, consider the highest term of the dividend and the highest term of the divisor. Since =, we start with the first term of the quotient: Step : Multipl b the divisor + and write it below the dividend: Step : Since we need to subtract +, so we equivalentl add its negative (don t forget to distribute the negative): Step : ( + ) Now, carr down the remaining terms of the dividend: Step : ( + ) + + Now, repeat steps - for the remaining polnomial ++. The outcome after going through steps - is the following: + (add ( + ) + + ( +6) (multipl b (+)) + (subtract from the above)

114 0 SESSION 8. DIVIDING POLYNOMIALS Since can be divided into, we can proceed with the above steps - one more time. The outcome is this: ( + ) + + ( +6) + ( 6) 8 = remainder Note that now cannot be divided into 8 so we stop here. The final term 8 is called the remainder. The term + is called the quotient. In analog with our result in part (a), we can write our conclusion as: = ( + ) (+)+8. Alternativel, we could also divide this b (+) and write it as: = Note 8.7. Just as with a division operation involving numbers, when dividing f(), f() is called the dividend and g() is called the divisor. As a result g() of dividing f() b g() via long division with quotient q() and remainder r(), we can write f() g() = q()+ r() g(). (8.) If we multipl this equation b g(), we obtain the following alternative version: f() = q() g()+r() (8.) Eample 8.8. Divide the following fractions via long division. a) ++5, b) , c) , d)

115 8.. LONG DIVISION 05 Solution. For part (a), we calculate: ( ) 8 +5 (8 ) Therefore, ++5 = (+8) ( )+7. Now, for part (b), there is no term in the dividend. This can be resolved b adding +0 to the dividend: Therefore, we showed: ( + ) ( + ) 5 + ( ) + ( ) For (c), calculate: = ( +6 ) ( 6) +8 ( +8) 0

116 06 SESSION 8. DIVIDING POLYNOMIALS Since the remainder is zero, we succeeded in factoring : = ( +6) (+) d) The last eample has a divisor that is a polnomial of degree. Therefore, the remainder is not a number, but a polnomial of degree ( + +) Here, the remainder is r() = ( 6 ) = Note 8.9. The divisor g() is a factor of f() eactl when the remainder r() is zero, that is: f() = q() g() r() = 0. For eample, in the above Eample 8.8, onl part (c) results in a factorization of the dividend, since this is the onl part with remainder zero. 8. Dividing b ( c) We now restrict our attention to the case where the divisor is g() = c for some real number c. In this case, the remainder r of the division f() b g() is a real number. We make the following observations. Observation 8.0. Assume that g() = c, and the long division of f() b g() has remainder r, that is, Assumption: f() = q() ( c)+r.

117 8.. DIVIDING BY (X C) 07 When we evaluate both sides in the above equation at = c we see that f(c) = q(c) (c c)+r = q(c) 0+r = r. In short: The remainder when dividing f() b ( c) is r = f(c). (8.) In particular: f(c) = 0 g() = c is a factor of f(). (8.) The above statement (8.) is called the remainder theorem, and (8.) is called the factor theorem. Eample 8.. Find the remainder of dividing f() = ++ b a), b) +, c) +, d). Solution. B Observation 8.0, we know that the remainder r of the division b c is f(c). Thus, the remainder for part (a), when dividing b is r = f() = + + = 9+9+ = 0. For (b), note that g() = + = ( ), so that we take c = for our input giving that r = f( ) = ( ) + ( ) + = 6 + = 6. Similarl, the other remainders are: c) r = f( ) = ( ) + ( )+ = + = 0, ( ( ) d) r = f = + ) + = = = 5. Note that in part (c), we found a remainder 0, so that ( +) is a factor of f(). Eample 8.. Determine whether g() is a factor of f(). a) f() = + +5+, g() =, b) f() = , g() = +, c) f() = , g() = +. Solution. a) We need to determine whether is a root of f() = , that is, whether f() is zero. f() = = = 7.

118 08 SESSION 8. DIVIDING POLYNOMIALS Since f() = 7 0, we see that g() = is not a factor of f(). b) Now, g() = + = ( ), so that we calculate: f( ) = ( ) + ( ) +( ) 8 = = 0. Since the remainder is zero, we see that + is a factor of Therefore, if we wanted to find the other factor, we could use long division to obtain the quotient. c) Finall, we have: f( ) = ( ) 5 + ( ) +7 = ++7 = 9. g() = + is not a factor of f() = Eample 8.. a) Show that is a root of f() = 5 +5, and use this to factor f. b) Show that 5 is a root of f() = 9 0, and use this to factor f completel. Solution. a) First, we calculate that is a root. f( ) = ( ) 5 ( ) +5 ( ) + = ++0 = 0. So we can divide f() b g() = ( ) = +: ( 5 + ) ( ) ( + ) ( +0 +6) 6 ( 6 ) 0

119 8.. OPTIONAL SECTION: SYNTHETIC DIVISION 09 So we factored f() as f() = ( + + 6) (+). b) Again we start b calculating f(5): f(5) = = = 0. Long division b g() = 5 gives: ( 5 ) 5 ( ) 6 0 (6 0) Thus, 9 0 = ( +5+6) ( 5). To factor f completel, we also factor f() = ( +5+6) ( 5) = (+) (+) ( 5) Optional section: Snthetic division When dividing a polnomialf() b g() = c, the actual calculation of the long division has a lot of unnecessar repetitions, and we ma want to reduce this redundanc as much as possible. In fact, we can etract the essential part of the long division, the result of which is called snthetic division. Eample 8.. Our first eample is the long division of (5 +0 ) + + ( 6) 7 + (7 +) 0

120 0 SESSION 8. DIVIDING POLYNOMIALS Here, the first term 5 of the quotient is just copied from the first term of the dividend. We record this together with the coefficients of the dividend and of the divisor + = ( ) as follows: 5 7 (dividend ( )) (divisor ( ( ))) 5 (quotient) The first actual calculation is performed when multipling the 5 term with, and subtracting it from 7. We record this as follows ( 0 is the product of 5) 5 ( is the sum 7+( 0)) Similarl, we obtain the net step b multipling the b ( ) and subtracting it from. Therefore, we get (6 is the product of ( )) 5 7 (7 is the sum +(6)) The last step multiplies 7 times and subtracts this from. In short, we write: ( is the product of ( 7) ) ( 0 is the sum +( )) The answer can be determined from these coefficients. The quotient is 5 +7, and the remainder is 0. Eample 8.5. Find the following quotients via snthetic division. a) 7 + 8, b) Solution. a) We need to perform the snthetic division

121 8.. EXERCISES Therefore we have = Similarl, we calculate part (b). Note that some of the coefficients are now zero We obtain the following result = Note 8.6. Snthetic division onl works when dividing b a polnomial of the form c. Do not attempt to use this method to divide b other forms like Eercises Eercise 8.. Divide b long division. a) ++, b) , c) +7 +, d) , e) + ++5, f) , g), h) +7 +, i) , j) , k) , l) Eercise 8.. Find the remainder when dividing f() b g(). a) f() = + +, g() =, b) f() = 5+8, g() =, c) f() = 5, g() = +, d) f() = , g() = +.

122 SESSION 8. DIVIDING POLYNOMIALS Eercise 8.. Determine whether the given g() is a factor of f(). If so, name the corresponding root of f(). a) f() = +5+6, g() = +, b) f() = +8, g() =, c) f() = , g() = +7, d) f() = 999 +, g() = +. Eercise 8.. Check that the given numbers for are roots of f() (see Observation 8.0). If the numbers are indeed roots, then use this information to factor f() as much as possible. a) f() = +, =, b) f() = 6 + 6, =, =, =, c) f() = +, =, d) f() = , =, e) f() = , =, =, f) f() = + 6 8, =, =, g) f() = , =, =, =, =. Eercise 8.5. Divide b using snthetic division. a) + 5+7, b) , c) + + +, d) + +, e) 5 + +, f)

123 Session 9 Graphing polnomials 9. Graphs of polnomials We now discuss the shape of the graphs of polnomial functions. Recall that a polnomial function f of degree n is a function of the form f() = a n n +a n n + +a +a +a 0. Observation 9.. We alread know from Section., that the graphs of polnomials f() = a+b of degree are straight lines. = + = + Polnomials of degree have onl one root. It is also eas to sketch the graphs of functions f() = n.

124 SESSION 9. GRAPHING POLYNOMIALS Observation 9.. Graphing =, =, =, = 5, we obtain: = = = = 5 From this, we see that the shape of the graph of f() = n depends on n being even or odd = n, for n even = n, for n odd If approaches ±, If approaches ±, = approaches +. = approaches ±. We now make some observations about graphs of general polnomials of degrees,,, 5, and, more generall, of an degree n. In particular, we will be interested in the number of roots and the number of etrema (that is the number of maima or a minima) of the graph of a polnomial f. Observation 9.. Let f() = a +b+c be a polnomial of degree. The graph of f is a parabola. f has at most roots. f has one etremum (that is one maimum or minimum)

125 9.. GRAPHS OF POLYNOMIALS 5 If a > 0 then f opens upwards, if a < 0 then f opens downwards. = = Observation 9.. Let f() = a +b +c+d be a polnomial of degree. The graph ma change its direction at most twice. f has at most roots. f has at most etrema roots, etrema roots, etrema root, 0 etrema If a > 0 then f() approaches + when approaches + (that is, f() gets large when gets large), and f() approaches when approaches. Ifa < 0 thenf() approaches whenapproaches +, and f() approaches + when approaches. = =

126 6 SESSION 9. GRAPHING POLYNOMIALS Above, we have a first instance of a polnomial of degreenwhich changes its direction one more time than a polnomial of one lesser degree n. Below, we give eamples of this phenomenon for higher degrees as well. Observation 9.5. Let f() = a + b + c + d + e be a polnomial of degree. f has at most roots. f has at most etrema. If a > 0 then f opens upwards, if a < 0 then f opens downwards. = = +5 = roots, etrema roots, etremum 0 roots, etrema Observation 9.6. Let f be a polnomial of degree 5. f has at most 5 roots. f has at most etrema. If a > 0 then f() approaches + when approaches +, and f() approaches when approaches. If a < 0 then f() approaches when approaches +, and f() approaches + when approaches. -5 = roots, etrema roots, etrema root, 0 etrema

127 9.. GRAPHS OF POLYNOMIALS 7 Observation 9.7. Let f() = a n n +a n n + +a +a +a 0 be a polnomial of degree n. Then f has at most n roots, and at most n etrema. degree degree degree Assume the degree of f is even n =,,6,.... If a n > 0, then the polnomial opens upwards. If a n < 0 then the polnomial opens downwards. f() = f() = Assume the degree of f is odd, n =,,5,.... If a n > 0, then f() approaches + when approaches +, and f() approaches as approaches. If a n < 0, then f() approaches when approaches +, and f() approaches + as approaches. f() = 5 f() =

128 8 SESSION 9. GRAPHING POLYNOMIALS The domain of a polnomial f is all real numbers, and f is continuous for all real numbers (there are no jumps in the graph). The graph of f has no horizontal or vertical asmptotes, no discontinuities (jumps in the graph), and no corners. Furthermore, f() approaches ± when approaches ±. Therefore, the following graphs cannot be graphs of polnomials vertical discontinuit corner horizontal asmptote / pole asmptote - Eample 9.8. Which of the following graphs could be the graphs of a polnomial? If the graph could indeed be a graph of a polnomial then determine a possible degree of the polnomial a) b) c) d) e)

129 9.. GRAPHS OF POLYNOMIALS 9 Solution. a) Yes, this could be a polnomial. The degree could be, for eample,. b) No, since the graph has a pole. c) Yes, this could be a polnomial. A possible degree would be degree. d) No, since the graph has a corner. e) No, since f() does not approach or as approaches. (In fact f() approaches 0 as approaches ± and we sa that the function (or graph) has a horizontal asmptote = 0.) Eample 9.9. Identif the graphs of the polnomials in (a), (b) and (c) with the functions (i), (ii), and (iii). a) b) c) i) f() = ii) f() = +6 7 iii) f() = Solution. The onl odd degree function is (i) and it must therefore correspond to graph (c). For (ii), since the leading coefficient is negative, we know that the function opens downwards, so that it corresponds to graph (a). Finall (iii) opens upwards, since the leading coefficient is positive, so that its graph is (b). When graphing a function, we need to make sure to draw the function in a window that epresses all the interesting features of the graph. More precisel, we need to make sure that the graph includes all essential parts of the graph, such as all intercepts (both -intercepts and -intercept), all roots, all asmptotes (this will be discussed in the net part), and the long range behavior of the function (that is how the function behaves when approaches ± ). We also want to include all etrema (that is all maima and minima) of the function.

130 0 SESSION 9. GRAPHING POLYNOMIALS Eample 9.0. Graph of the given function with the TI-8. Include all etrema and intercepts of the graph in our viewing window. a) f() = b) f() = c) f() = d) f() = Solution. a) The graph in the standard window looks as follows: However, since the function is of degree, this cannot be the full graph, as f() has to approach when approaches. Zooming out, and rescaling appropriatel for the following setting the following graph. b) The graph in the standard window is drawn on the left below. After rescaling to 0 0 and 00 0 we obtain the graph on the right. c) The graph in the standard window is drawn on the left. Zooming to the interesting part of the graph, we obtain the graph on the right. (Below, we

131 9.. FINDING ROOTS OF A POLYNOMIAL WITH THE TI-8 have chosen a window of and ) d) The standard window (see left graph) shows an empt coordinate sstem without an part of the graph. However, zooming out to 0 0 and , we obtain the middle graph. There is another interesting part of the graph displaed on the right, coming from zooming to the plateau on the right (here shown at and 0.9.). 9. Finding roots of a polnomial with the TI-8 We have seen that the roots are important features of a polnomial. Recall that the roots of the polnomial f are those 0 for which f( 0 ) = 0. These are, of course, precisel the -intercepts of the graph. Therefore we ma use the graph of a polnomial for finding its roots as we did in section. Eample 9.. Find the roots of the polnomial from its graph. a) f() = 7 + 8, b) f() = +8 +8, c) f() = + +6, d) f() = Solution. a) We start b graphing the polnomial f() =

132 SESSION 9. GRAPHING POLYNOMIALS The graph suggests that the roots are at =, =, and =. This ma easil be checked b looking at the function table. Since the polnomial is of degree, there cannot be an other roots. b) Graphing f() = with the calculator shows the following displa. Zooming into the graph reveals that there are in fact two roots, = and =, which can be confirmed from the table. Note, that the root = onl touches the -ais. This is due to the fact that = appears as a multiple root. Indeed, since is a root, we can divide f() b without remainder and factor the resulting quotient to see that that f() = ( )( )( ) = ( )( ). We sa that = is a root of multiplicit. c) The graph off() = + +6 in the standard window is displaed as follows.

133 9.. FINDING ROOTS OF A POLYNOMIAL WITH THE TI-8 Since this is a polnomial of degree, all of the essential features are alread displaed in the above graph. The roots can be seen b zooming into the graph. From the table and the graph we see that there is a root at = and another root at between and. Finding the eact value of this second root can be quite difficult, and we will sa more about this in section 0. below. At this point, we can onl approimate the root with the zero function from the calc menu: d) The graph of f() = in the standard window is displaed below. Zooming into the -ais, and checking the table shows that the onl obvious root is =. The other four roots are more difficult to find. However the can be approimated using the zero function from the calc menu.

134 SESSION 9. GRAPHING POLYNOMIALS It will often be necessar to find the roots of a quadratic polnomial. For this reason, we recall the well-known quadratic formula. Theorem 9. (The quadratic formula). The solutions of the equation a + b + c = 0 for some real numbers a, b, and c are given b = b+ b ac, and a = b b ac. a We ma combine the two solutions and and simpl write this as: / = b± b ac a Note 9.. We ma alwas use the roots and of a quadratic polnomial f() = a +b+c from the quadratic formula and rewrite the polnomial as ( f() = a +b+c = a b+ b ac )( a b b ac a 9. Optional section: Graphing polnomials b hand In this section we will show how to sketch the graph of a factored polnomial without the use of a calculator. Eample 9.. Sketch the graph of the following polnomial without using the calculator: p() = (+0) (+9) ( 8) Solution. Note, that on the calculator it is impossible to get a window which will give all of the features of the graph (while focusing on getting the maimum in view the other features become invisible). We will sketch the graph b hand so that some of the main features are visible. This will onl be a sketch and not the actual graph up to scale. Again, the graph can not be drawn to scale while being able to see the features. We first start b putting the -intercepts on the graph in the right order, but not to scale necessaril. Then note that p() = (lower terms) 7 for large. ).

135 9.. OPTIONAL SECTION: GRAPHING POLYNOMIALS BY HAND 5 This is the leading term of the polnomial (if ou epand p it is the term with the largest power) and therefore dominates the polnomial for large. So the graph of our polnomial should look something like the graph of = 7 on the etreme left and right side Now we look at what is going on at the roots. Near each root the factor corresponding to that root dominates. So we have for p() 0 C (+0) cubic 9 C (+9) line 0 C parabola 8 C ( 8) line wherec,c,c, andc are constants which can, but need not, be calculated. For eample, whether or not the parabola near0opens up or down will depend on whether the constantc = (0+0) (0+9)(0 8) is negative or positive. In this case C is positive so it opens upward but we will not use this fact to graph. We will see this independentl which is a good check on our work. Starting from the left of our graph where we had determined the behavior for large negative, we move toward the left most zero, 0. Near 0 the graph looks cubic so we imitate a cubic curve as we pass through ( 0,0)

136 6 SESSION 9. GRAPHING POLYNOMIALS Now we turn and head toward the net zero, 9. Here the graph looks like a line, so we pass through the point ( 9,0) as a line would Now we turn and head toward the root 0. Here the graph should look like a parabola. So we form a parabola there. (Note that as we had said before the parabola should be opening upward here and we see that it is) Now we turn toward the final zero 8. We pass through the point (8,0) like a line and we join (perhaps with the use of an eraser) to the large part of the graph. If this doesn t join nicel (if the graph is going in the wrong direction) then there has been a mistake. This is a check on our work. Here is the final sketch

137 9.. EXERCISES 7 What can be understood from this sketch? Questions like when is p() > 0? can be answered b looking at the sketch. Further, the general shape of the curve is correct so other properties can be concluded. For eample, p has a local minimum between = 0 and = 9 and a local maimum between = 9 and = 0, and between = 0 and = 8. The eact point where the function reaches its maimum or minimum can not be decided b looking at this sketch. But it will help to decide on an appropriate window so that the minimum or maimum finder on the calculator can be used. 9. Eercises Eercise 9.. Which of the graphs below could be the graphs of a polnomial? a) b) c) d) e) f) - - Eercise 9.. Identif each of the graphs (a)-(e) with its corresponding assignment from (i)-(vi) below. 5 5 a) b) c)

138 8 SESSION 9. GRAPHING POLYNOMIALS d) e) i) f() = ii) f() = iii) f() = iv) f() = v) f() = 6 + vi) f() = Eercise 9.. Identif the graph with its assignment below. 5 a) b) c) i) f() = ii) f() = iii) f() = Eercise 9.. Sketch the graph of the function with the TI-8, which includes all etrema and intercepts of the graph. a) f() = b) f() = ++50 c) f() = d) f() =.007 e) f() = f) f() =

139 9.. EXERCISES 9 Eercise 9.5. Find the eact value of at least one root of the given polnomial. a) f() = 0 + 0, b) f() = +8+8, c) f() = +, d) f() = , e) f() = +6+, f) f() = Eercise 9.6. Graph the following polnomials without using the calculator. a) f() = (+) ( 5), b) f() = (+) ( ) 5, c) f() = ( ) ( 5) ( 7), d) f() = (+)(+)(+) (+)( ).

140 Session 0 Roots of polnomials 0. Optional section: The rational root theorem Eample 0.. Consider the equation = 0. Let be a rational solution of this equation, that is = p is a rational number such that q ( p ) p ) p 0 6 ( +5 = 0. q q q We assume that = p is completel reduced, that is, p and q have no q common factors that can be used to cancel the numerator and denominator of the fraction p. Now, simplifing the above equation, and combining terms, we q obtain: 0 p p 6 q q +5 p q = 0 (multipl b q ) = 0p 6p q +5pq q = 0 (add q ) = 0p 6p q +5pq = q (factor p on the left) = p (0p 6pq +5q ) = q. Therefore, p is a factor of q (with the other factor being (0p 6pq+5q )). Since p and q have no common factors, p must be a factor of. That is, p is one of the following integers: p = +,+,,. Similarl starting from 0p 6p q +5pq q = 0, we can write (add +6p q 5pq +q ) = 0p = 6p q 5pq +q 0

141 0.. OPTIONAL SECTION: THE RATIONAL ROOT THEOREM (factor q on the right) = 0p = (6p 5pq+q ) q. Now, q must be a factor of 0p. Since q and p have no common factors, q must be a factor of 0. In other words, q is one of the following numbers: q = ±,±,±5,±0. Putting this together with the possibilities for p = ±,±, we see that all possible rational roots are the following: ±, ±, ± 5, ± 0, ±, ±, ± 5, ± 0. The observation in the previous eample holds for a general polnomial equation with integer coefficients. Observation 0. (Rational root theorem). Consider the equation a n n +a n n + +a +a 0 = 0, (0.) where ever coefficient a n,a n,...,a 0 is an integer and a 0 0, a n 0. Assume that = p is a solution of (0.) and the fraction = p is completel q q reduced. Then a 0 is an integer multiple of p, and a n is an integer multiple of q. Therefore, all possible rational solutions of (0.) are fractions = p q where p is a factor of a 0 and q is a factor of a n. We can use this observation to find good candidates for the roots of a given polnomial. Eample 0.. a) Find all rational roots of f() = b) Find all real roots of f() = +. c) Find all real roots of f() = 6. Solution. a) If = p is a rational root, then p is a factor of, that is p = ±, q and q is a factor of 7, that is q = ±,±7. The candidates for rational roots are therefore = ±,±. To see which of these candidates are indeed roots 7 of f we plug these numbers into f via the calculator. We obtain the following:

142 SESSION 0. ROOTS OF POLYNOMIALS Note that we entered the -value as a fraction ( )/7 on the right. The onl root among ±,± is = 7. 7 b) We need to identif all real roots of f() = +. In general, it is a quite difficult task to find a root of a polnomial of degree, so that it will be helpful if we can find the rational roots first. If = p is a q rational root then p is a factor of, that is p = ±,±, and q is a factor of, that is q = ±,±. The possible rational roots = p of f are: q ±, ±, ± Using the calculator, we see that the onl rational root is =. Therefore, b the factor theorem (Observation 8.0), we see that ( ) is a factor of f, that is f() = q() ( ). To avoid fractions in the long division, we rewrite this as f() = q() ( ) = q() = q() ( ), so that we ma divide f() b ( ) instead of ( ) (note that this can not be done with snthetic division). We obtain the following quotient ( ) ( 6) ( ) Therefore, f() = ( + 6+)( ), and an root of f is either a root of +6+ or of. We know that the root of is =, and 0

143 0.. OPTIONAL SECTION: THE RATIONAL ROOT THEOREM that +6+ has no other rational roots. Nevertheless, we can identif all other real roots of +6+ via the quadratic formula, (see Theorem 9. below). +6+ = 0 = / = 6± 6 = 6± 6 8 = 6± 7 = ± 7 Therefore, the roots of f are precisel the following = 6± 8 = 6± 7 = + 7, = 7, =. c) First we find the rational roots = p q off() = 6. Since p is a factor of 6 it must be p = ±,±,±,±6, and since q is a factor of it must be q = ±,±,±. All candidates for rational roots = p q are the following (where we ecluded repeated was of writing ): ±, ±, ±, ±6, ±, ±, ±, ± Checking all these candidates with the calculator produces eactl two rational roots: = 6 and =. Therefore, we ma divide f() b both ( 6) and b ( + ) without remainder. To avoid fractions, we use the term (+ ) = (+) instead of (+ ) for our factor of f. Therefore, f() = q() ( 6) (+). The quotient q() is determined b performing a long division b ( 6) and then another long division b ( + ), or alternativel b onl one long division b ( 6) (+) = + 6 = 6. Dividing f() = 6 b 6 produces the

144 SESSION 0. ROOTS OF POLYNOMIALS quotient q(): ( 6 ) 6 ( 6) We obtain the factored epression for f() as f() = ( +)(+)( 6). The onl remaining real roots we need to find are those of +. However, + = 0 = = has no real solution. In other words there are onl comple solutions of =, which are = i and = i (we will discuss comple solutions in more detail in the net section). Since the problem requires us to find the real roots of f, our answer is that the onl real roots are = 6 and =. In the above eample, we used the quadratic formula 9. from page The fundamental theorem of algebra We have seen in Observation 8.0 on page 06 that ever root c of a polnomial f() gives a factor ( c) of f(). There is a theorem which sas something about the eistence of roots and factors but we will need to discuss comple numbers briefl before stating that theorem. As stated above, we know that there is no real number that satisfies =. So we define i to be a solution to this equation. This i is not a real number but a new kind of number called a comple number. We can think of i as. We can now consider numbers of the form a+ib where a and b are real numbers. Numbers of this form constitutes the set of comple numbers, denoted b C. a is called the real part and b is called the imaginar part of the comple number a+ib. We can add two comple numbers b adding their real and imaginar parts to form the real and imaginar parts of the sum. We can multipl two comple numbers b ordinar distribution (FOIL) then use the propert that i =.

145 0.. THE FUNDAMENTAL THEOREM OF ALGEBRA 5 Eample 0.. We have and ( i) (+i) = ( )+( )i = 6i ( i)(+i) = 8+6i i 9i = 8 6i 9( ) = 7 6i. We can see b writing, for eample, 6 = i 6, that these numbers arise naturall as roots of quadratic equations. But there is more! The following fundamental theorem of algebra guarantees the eistence of roots of an polnomial. Theorem 0.5 (Fundamental theorem of algebra). Let f() = a n n + a n n + +a +a 0 be a non-constant polnomial. Then there eists a comple number c which is a root of f. Let us make two remarks about the fundamental theorem of algebra to clarif the statement of the theorem. Note 0.6. In the above Theorem 0.5 we did not specif what kind of coefficients a 0,...a n are allowed for the theorem to hold. In fact, to be precise, the fundamental theorem of algebra states that for an comple numbers a 0,...a n, the polnomial f() = a n n +a n n + +a +a 0 has a root. In general there ma not eist a real root c of a given polnomial, but the root c ma onl be a comple number. For eample, consider f() = +, and consider the roots c of f, that is c + = 0. Then, for an real number c, we have alwas c 0, so that f(c) = c +, so that there cannot be a real root c of f. However, if we multipl ( i)(+i) = +i i i = +, we see that f has comple roots c = i and c = i. Now, while the fundamental theorem of algebra guarantees a root c of a polnomial f, we can use the remainder theorem from Observation 8.0 to check possible candidates c for the roots, and the factor theorem to factor f() = q() ( c).

146 6 SESSION 0. ROOTS OF POLYNOMIALS Eample 0.7. Find roots of the given polnomial and use this information to factor the polnomial completel. a) f() = 8 6+6, b) f() = , c) f() = +, d) f() = 6. Solution. a) In order to find a root, we use the graph to make a guess for one of the roots. The graph suggests that the roots ma be at = and =. This is also supported b looking at the table for the function. We check that these are roots b plugging the numbers into the function. f( ) = ( ) 8 ( ) 6 ( )+6 = = 0, f() = = = 0. B the factor theorem we can divide f() = 8 6+6, for eample, b ( ) ( 6 ) 6 +6 ( +6) +6 ( +6) 0

147 0.. THE FUNDAMENTAL THEOREM OF ALGEBRA 7 Therefore, f() = ( )( ). To factor f completel, we still need to factor the first polnomial : f() = ( )( ) = ( 6) ( ) = ( ) (+) ( ) = (+) ( ). b) The graph of f() = in the standard window is the following: We suspect that = 5, =, =, and = 6 are the roots of the function. This is confirmed b checking the table on the calculator, and also b direct computation. f( 5) = ( 5) ( 5) 6( 5) +68( 5)+0 = 0, f( ) = ( ) ( ) 6( ) +68( )+0 = 0, f() = = 0, f(6) = = 0. Therefore, each (+5), (+), ( ), and ( 6) is a factor of f(), and then their product is also a factor of f(). The product(+5)(+)( )( 6) is calculated as follows. Since (+5)(+) = +7+0 and ( )( 6) = 0+, we obtain (+5)(+)( )( 6) = ( +7+0)( 0+) = = = f().

148 8 SESSION 0. ROOTS OF POLYNOMIALS We see that this is precisel f(), so that no more polnomial division is necessar. The answer is f() = (+5)(+)( )( 6). c) We can graph the function, or just make a guess for a root of f() = +. An immediate guess would be to tr =. This is indeed a root, since f( ) = ( ) + = + = 0, which is also supported b the graph. Therefore, we can divide + b ( + ) +0 + ( ) + ( +) We obtain f() = + = ( +)(+). Now + cannot be factored an further, however, we ma continue to factor + = ( )( ). Using the quadratic formula 9., we obtain the solutions of + = 0 as / = ( )± ( ) = ± = ±i. Therefore, + = ( ( +i )) ( ( i )) (+). d) For f() = 6 there are the two obvious candidates = and = that we can identif as roots, either b guessing (since = 6) or b looking at the graph. 0

149 0.. THE FUNDAMENTAL THEOREM OF ALGEBRA 9 Indeed the algebra confirms these guesses. f() = 6 = 6 6 = 0, f( ) = ( ) 6 = 6 6 = 0. Therefore, f() = q() ( ) (+) = q() ( ). To find q() we perform the long division ( 6)/( ) ( ) +0 6 ( 6) We obtain f() = ( +)(+)( ). The roots of + are given b the quadratic formula as 0 / = 0± 0 6 = ± = ± i = ±i, or alternativel, + = 0 = = = = ±i. We obtain the factored polnomial f() = (+i)( i)(+)( ). As we have seen in the last eample, we can use the roots to factor a polnomial completel so that all factors are polnomials of degree onl. Furthermore, whenever a comple root, a + ib, appeared, its conjugate, a ib, was also a root. These remarks hold more generall, as we state now. Observation 0.8. () Ever polnomial f() = a n n +a n n + +a +a 0 of degree n can be factored as f() = m ( c ) ( c ) ( c n ). () In particular, ever polnomial of degree n has at most n roots. (However, these roots ma be real or comple.)

150 0 SESSION 0. ROOTS OF POLYNOMIALS () The factor ( c) for a root c could appear multiple times in the above product, that is, we ma have ( c) k as a factor of f. The multiplicit of a root c is the number of times k that a root appears in the factored epression for f as in (). () If f() = a n n + a n n + + a + a 0 has onl real coefficients a 0,...,a n, andc = a+bi is a comple root off, then the comple conjugate c = a bi is also a root of f. Proof. If is an root, then a n n + a n n + + a + a 0 = 0. Appling the comple conjugate to this and using that u v = ū v gives a n n +a n n + +a +a 0 = 0. Since the coefficients a j are real, we have that a j = a j, so that a n n +a n n + +a +a 0 = 0. This shows that the comple conjugate is a root of f as well. Eample 0.9. Find the roots of the polnomial and sketch its graph including all roots. a) f() = + b) f() = Solution. a) To find a root, we first graph the functionf() = + with the calculator. The graph and the table suggest that we have a root at =. Therefore we divide f() b ( ). We obtain: ( ) 5 (5 5) ( ) 0

151 0.. THE FUNDAMENTAL THEOREM OF ALGEBRA This shows that f() = ( )( +5+). To find the roots of f, we also have to find the roots of the second factor +5+, i.e., the solutions to +5+ = 0. The quadratic formula gives: The roots are therefore: = 5± 5 = 5± =, = 5+ 0., = 5.8 We put these roots together with the graph in an appropriate window. The graph including the roots is displaed below b) The graph of f() = in an appropriate window suggests the roots = and = 5.

152 SESSION 0. ROOTS OF POLYNOMIALS Dividing f() b ( )( 5) = 7+0 gives: ( 7 +0 ) 7 +0 ( 7 +0) Since the roots of + are all comple, we see that the onl roots of f are = and = 5. The graph including its roots is displaed below. 0 5 Eample 0.0. Find a polnomial f with the following properties. a) f has degree, the roots of f are precisel, 5, 6, and the leading coefficient of f is 7 b) f has degree with real coefficients, f has roots i, 5 (and possibl other roots as well), and f(0) = 90 c) f has degree with comple coefficients, f has roots i+, i, d) f has degree 5 with real coefficients, the leading coefficient is, and the roots are determined b its graph:

153 0.. THE FUNDAMENTAL THEOREM OF ALGEBRA Solution. a) In general a polnomial f of degree is of the form f() = m ( c ) ( c ) ( c ). Identifing the roots and the leading coefficient, we obtain the polnomial f() = 7 ( ) ( 5) ( 6). b) A polnomial f of degree is of the form f() = m ( c ) ( c ) ( c ). Roots of f are i and 5, and since the coefficients of f are real it follows from Observation 0.8(), that the comple conjugate i is also a root of f. Therefore, f() = m (+5) ( i) (+i). To identif m, we use the last condition f(0) = = m (0+5) (0 i) (0+i) = m 5 ( 9)i = m 5 9 = 5m Dividing b 5, we obtain m =, so that f() = (+5) ( i) (+i) = (+5) ( +9), which clearl has real coefficients. c) Since f is of degree it can be written as f() = m ( c ) ( c ) ( c ) ( c ). Three of the roots are identified as i+, i, and : f() = m ( (+i)) ( i) ( ) ( c ) However, we have no further information on the fourth root c or the leading coefficient m. (Note that Observation 0.8() cannot be used here, since we are not assuming that the polnomial has real coefficients. Indeed it cannot have real coefficients since that would mean we would have also the comple conjugates of +i and i in addition to the three that we are given, giving us a total of 5 roots which cannot be accommodated b a polnomial of degree.) We can therefore choose an number for these remaining variables. For eample, a possible solution of the problem is given b choosing m = and c =, for which we obtain: f() = ( (+i)) ( i) ( ) ( ) d) f is of degree 5, and we know that the leading coefficient is. The graph is zero at =,,, and, so that the roots are,,, and. Moreover, since the graph just touches the root =, this must be a multiple root, that is, it must occur more than once (see Section 9. for a discussion

154 SESSION 0. ROOTS OF POLYNOMIALS of multiple roots and their graphical consequences). We obtain the following solution: f() = ( )( )( )( ). Note that the root = is a root of multiplicit. Note that to see that a polnomial has real coefficients, it ma be necessar to multipl factors like ( (+i))( ( i)). We suggest a wa of doing this for which we use the fact that a b = (a+b)(a b). We have ( (+i))( ( i)) = (( ) i)(( )+i) = ( ) +9, which clearl has real coefficients. 0. Eercises Eercise 0.. a) Find all rational roots of f() = +. b) Find all rational roots of f() = c) Find all rational roots of f() = d) Find all real roots of f() = e) Find all real roots of f() = Eercise 0.. Find a root of the polnomial b guessing possible candidates of the root. a) f() = 5 b) f() = c) f() = 7 d) f() = +000 e) f() = 8 f) f() = 5 g) f() = 5 + h) f() = 777 i) f() = +6 Eercise 0.. Find the roots of the polnomial and use it to factor the polnomial completel. a) f() = 7+6, b) f() = 6 0, c) f() = 5 +, d) f() = + 5, e) f() = + 7 6, f) f() = +9, g) f() =, h) f() = , i) f() = 7, j) f() = + 5.

155 0.. EXERCISES 5 Eercise 0.. Find the eact roots of the polnomial; write the roots in simplest radical form, if necessar. Sketch a graph of the polnomial with all roots clearl marked. a) f() = 5+6, b) f() = +5 +, c) f() = , d) f() = , e) f() = 8 8 6, f) f() = +, g) f() = , h) f() = 7 0+8, i) f() = , j) f() = Eercise 0.5. Find a polnomial f that fits the given data. a) f has degree. The roots of f are precisel,,. The leading coefficient of f is. b) f has degree. The roots of f are precisel,, 0,. The leading coefficient of f is. c) f has degree. f has roots,,, and f(0) = 0. d) f has degree. f has roots 0,,,, and f() = 0. e) f has degree. The coefficients of f are all real. The roots of f are precisel +5i, 5i, 7. The leading coefficient of f is. f) f has degree. The coefficients of f are all real. f has roots i,, and f(0) = 6. g) f has degree. The coefficients of f are all real. f has roots 5+i and 5 i of multiplicit, the root of multiplicit, and f(5) = 7. h) f has degree. The coefficients of f are all real. f has roots i and +i. i) f has degree 6. f has comple coefficients. f has roots +i, +i, i of multiplicit and the root of multiplicit. j) f has degree 5. f has comple coefficients. f has roots i,, 7 (and possibl other roots). k) f has degree. The roots of f are determined b its graph:

156 6 SESSION 0. ROOTS OF POLYNOMIALS l) f has degree. The coefficients of f are all real. The leading coefficient of f is. The roots of f are determined b its graph: (see Section 9.). m) f has degree. The coefficients of f are all real. f has the following graph:

157 Session Rational functions. Graphs of rational functions Recall from the beginning of this chapter that a rational function is a fraction of polnomials: f() = a n n +a n n + +a +a 0 b m m +b m m + +b +b 0 In this section, we will stud some characteristics of graphs of rational functions. Asmptotes are important features of graphs of rational functions. An asmptote is a line that is approached b the graph of a function: = a is a vertical asmptote if f() approaches ± as approaches a from either the left or from the right, and = b is a horizontal asmptote if f() approaches b as approaches or. To obtain a better idea of some of these features including asmptotes, we start b graphing some rational functions. In particular it is useful to know the graphs of the basic functions = n. Observation.. Graphing =, =, =, =, we obtain: = = = = 7

158 8 SESSION. RATIONAL FUNCTIONS In general we see that = 0 is a vertical asmptote and = 0 is a horizontal asmptote. The shape of = n depends on n being even or odd. We have: = n, n odd = n, n even We now stud more general rational functions. Eample.. a) Our first graph is f() = Here, the domain is all numbers where the denominator is not zero, that is D = R {}. There is a vertical asmptote, =. Furthermore, the graph approaches 0 as approaches ±. Therefore, f has a horizontal asmptote, = 0. Indeed, whenever the denominator has a higher degree than the numerator, the line = 0 will be the horizontal asmptote. b) Net, we graph f() =

159 .. GRAPHS OF RATIONAL FUNCTIONS 9 Here, the domain is all for which 6 0. To see where this happens, calculate 6 = 0 = = 6 = = = = ±. Therefore, the domain is D = R {,}. As before, we see from the graph, that the domain reveals the vertical asmptotes = and = (the vertical dashed lines). To find the horizontal asmptote (the horizontal dashed line), we note that when becomes ver large, the highest terms of both numerator and denominator dominate the function value, so that for ver large = f() = = Therefore, when approaches ±, the function value f() approaches, and therefore the horizontal asmptote is at = (the horizontal dashed line). c) Our net graph is f() = We see that there does not appear to be an vertical asmptote, despite the fact that is not in the domain. The reason for this is that we can remove the singularit b cancelling the troubling term as follows: f() = 8+5 = ( )( 5) ( ) = 5 = 5, Therefore, the function f reduces to 5 for all values where it is defined. However, note that f() = 8+5 is not defined at =. We denote this in the graph b an open circle at =, and call this a removable singularit (or a hole).

160 50 SESSION. RATIONAL FUNCTIONS d) Our fourth and last graph before stating the rules in full generalit is f() = 8 6. The graph indicates that there is no horizontal asmptote, as the graph appears to increase towards and decrease towards. To make this observation precise, we calculate the behavior when approaches ± b ignoring the lower terms in the numerator and denominator for ver large = f() = 8 6 = Therefore, when becomes ver large, f() behaves like, which approaches when approaches, and approaches when approaches. (In fact, after performing a long division we obtain 8 r() = 6 +, 6 which would give rise to what is called a slant asmptote = ; see also remark. below.) Indeed, whenever the degree of the numerator is greater than the degree of the denominator, we find that there is no horizontal asmptote, but the graph blows up to ±. (Compare this also with eample (c) above). Combining what we have learned from the above eamples, we state our observations as follows. Observation.. Let f() = p() be a rational function with polnomials q() p() and q() in the numerator and denominator, respectivel. The domain of f is all real numbers for which the denominator is not zero, D = { R q() 0 } Assume that q( 0 ) = 0, so that f is not defined at 0. If 0 is not a root of p(), or if 0 is a root of p() but of a lesser multiplicit than the

161 .. GRAPHS OF RATIONAL FUNCTIONS 5 root in q(), then f has a vertical asmptote = 0. f() = f() = ( ) f() = ( )( ) ( ) If p( 0 ) = 0 and q( 0 ) = 0, and the multiplicit of the root 0 in p() is at least the multiplicit of the root in q(), then these roots can be canceled, and it is said that there is a removable discontinuit (or sometimes called a hole) at = 0. f() = ( )( ) ( ) f() = ( )( ) ( )( ) To find the horizontal asmptotes, we need to distinguish the cases where the degree of p() is less than, equal to, or greater than q(). deg(p) >deg(q) deg(p) =deg(q) deg(p) <deg(q) f() = f() = f() = no horizontal asmptote: asmptote: asmptote = highest coeff. of p highest coeff. of q = 0 -

162 5 SESSION. RATIONAL FUNCTIONS In addition, it is sometimes useful to determine the - and -intercepts. If 0 is in the domain of f, then the -intercept is (0,f(0)). f() = + + f(0) = = If p( 0 ) = 0 but q( 0 ) 0, then f( 0 ) = p( 0) q( 0 ) = 0 q( 0 ) = 0, so that ( 0,0) is an -intercept, that is the graph intersects with the -ais at 0. f() = f() = ( )( )( 5) ( ) Remark.. We can calculate the asmptotic behavior of a rational function f() = p() in the case where the degree of p is greater than the degree of q q() b performing a long division. If the the quotient is m(), and the remainder is r(), then f() = p() q() = m()+ r() q(). Now, since deg(r) <deg(q), the fraction r() approaches zero as approaches q() ±, so that f() m() for large. Eample.5. Find the domain, all horizontal asmptotes, vertical asmptotes, removable singularities, and - and -intercepts. Use this information together with the graph of the calculator to sketch the graph of f. a) f() = d) f() = ( ) b) f() = 5 e) f() = + c) f() = 9 +6

163 .. GRAPHS OF RATIONAL FUNCTIONS 5 Solution. a) We combine our knowledge of rational functions and its algebra with the particular graph of the function. The calculator gives the following graph. To find the domain of f we onl need to eclude from the real numbers those that make the denominator zero. Since = 0 eactl when (+)( ) = 0, which gives = or =, we have the domain: domain D = R {,} The numerator has a root eactl when = 0, that is = 0. Therefore, = and = are vertical asmptotes, and since we cannot cancel terms in the fraction, there is no removable singularit. Furthermore, since f() = 0 eactl when the numerator is zero, the onl -intercept is (0, 0). To find the horizontal asmptote, we consider f() for large values of b ignoring the lower order terms in numerator and denominator, large = f() = We see that the horizontal asmptote is =. Finall, for the -intercept, we calculate f(0): f(0) = = 0 = 0.

164 5 SESSION. RATIONAL FUNCTIONS Therefore, the -intercept is (0, 0). The function is then graphed as follows b) The graph of f() = 5 as drawn with the TI-8 is the following. For the domain, we find the roots of the denominator, = 0 = ( ) = 0 = = 0 or =. The domain is D = R {0,}. For the vertical asmptotes and removable singularities, we calculate the roots of the numerator, 5 = 0 = = 0. Therefore, = is a vertical asmptote, and = 0 is a removable singularit. Furthermore, the denominator has a higher degree than the numerator, so that = 0 is the horizontal asmptote. For the -intercept, we calculate f(0) b evaluating the fraction f() at = 0 0, which is undefined. Therefore, there is no -intercept (we, of course, alread noted that there is a removable singularit when = 0). Finall for the

165 .. GRAPHS OF RATIONAL FUNCTIONS 55 -intercept, we need to analze where f() = 0, that is where 5 = 0. The onl candidate is = 0 for which f is undefined. Again, we see that there is no -intercept. The function is then graphed as follows. (Notice in particular the removable singularit at = 0.) c) We start again b graphing the function f() = 9 +6 calculator. After zooming to an appropriate window, we get: with the To find the domain of f, we find the zeros of the denominator = 0 = (+)( ) = 0 = = or =. The domain is D = R {,}. The graph suggests that there is a vertical asmptote =. However the = appears not to be a vertical asmptote. This would happen when = is a removable singularit, that is, = is a root of both numerator and denominator of f() = p(). To confirm this, we q() calculate the numerator p() at = : p() = 9 +6 = = 0

166 56 SESSION. RATIONAL FUNCTIONS Therefore, = is indeed a removable singularit. To analze f further, we also factor the numerator. Using the factor theorem, we know that is a factor of the numerator. Its quotient is calculated via long division ( ) 7 ( ) ( ) 0 With this, we obtain f() = ( )( 7+) = ( )( )( ). (+)( ) Therefore, we conclude that = is a vertical asmptote and = is a removable singularit. We also see that the -intercepts are (,0) and (,0) (that is values where the numerator is zero). Now, the long range behavior is determined b ignoring the lower terms in the fraction, large = f() = = no horizontal asmptote Finall, the -coordinate of the -intercept is given b = f(0) = = =.

167 .. GRAPHS OF RATIONAL FUNCTIONS 57 We draw the graph as follows: d) We first graph f() = ( ). (Don t forget to set the viewing window back to the standard settings!) The domain is all real numbers ecept where the denominator becomes zero, that is, D = R {}. The graph has a vertical asmptote = and no hole. The horizontal asmptote is at = 0, since the denominator has a higher degree than the numerator. The -intercept is at 0 = f(0) = 0 = = (0 ). The -intercept is where the numerator is zero, = 0, that is at =. Since the above graph did not show the -intercept, we can confirm this b changing the window size as follows:

168 58 SESSION. RATIONAL FUNCTIONS Note in particular that the graph intersects the -ais at = and then changes its direction to approach the -ais from above. A graph of the function f which includes all these features is displaed below e) We graph f() = +. For the domain, we determine the zeros of the denominator. + = 0 = = = =. The onl solutions of this equation are given b comple numbers, but not b an real numbers. In particular, for an real number, the denominator of f() is not zero. The domain of f is all real numbers, D = R. This implies in turn that there are no vertical asmptotes, and no removable singularities. The -intercepts are determined b f() = 0, that is where the numerator is zero, = 0 = = = = = = ±. The horizontal asmptote is given b f() =, that is, it is at = =.5. The -intercept is at = f(0) = = =.

169 .. OPTIONAL SECTION: RATIONAL FUNCTIONS BY HAND 59 We sketch the graph as follows: =.5 Since the graph is smmetric with respect to the -ais, we can make one more observation, namel that the function f is even (see observation 5. on page 7): f( ) = ( ) ( ) + = + = f(). Optional section: Graphing rational functions b hand In this section we will show how to sketch the graph of a factored rational function without the use of a calculator. It will be helpful to the reader to have read section 9. on graphing a polnomial b hand before continuing in this section. In addition to having the same difficulties as polnomials, calculators often have difficult graphing rational functions near an asmptote. Eample.6. Graph the function p() = ( ) (+) ( )(+) ( ). Solution. We can see that p has zeros at = 0,, and and vertical asmptotes =, = and =. Also note that for large, p(). So there is a horizontal asmptote =. We indicate each of these facts

170 60 SESSION. RATIONAL FUNCTIONS on the graph: 0 We can in fact get a more precise statement b performing a long division and writing p() = n() d() = + r(). If ou drop all but the leading order terms in the numerator and the denominator of the second d() term, we see that p() whose graph for large looks like This sort of reasoning can make the graph a little more accurate but is not necessar for a sketch. We also have the following table: for a near a, p() tpe sign change at a C (+) linear changes C /(+) asmptote does not change 0 C parabola does not change C /( ) asmptote changes C 5 ( ) cubic changes C 6 /( ) asmptote changes

171 .. OPTIONAL SECTION: RATIONAL FUNCTIONS BY HAND 6 Note that if the power appearing in the second column is even then the function does not change from one side of a to the other. If the power is odd, the sign changes (either from positive to negative or from negative to positive). Now we move from large negative values toward the right, taking into account the above table. For large negative, we start our sketch as follows: 0 And noting that near = the function p() is approimatel linear, we have 0 Then noting that we have an asmptote (noting that we can not cross the

172 6 SESSION. RATIONAL FUNCTIONS -ais without creating an -intercept) we have 0 Now, from the table we see that there is no sign change at so we have 0 and from the table we see that near = 0 the function p() is approimatel quadratic and therefore the graph looks like a parabola. This together with

173 .. OPTIONAL SECTION: RATIONAL FUNCTIONS BY HAND 6 the fact that there is an asmptote at = gives 0 Now, from the table we see that the function changes sign at the asmptote, so while the graph hugs the top of the asmptote on the left hand side, it hugs the bottom on the right hand side giving 0 Now, from the table we see that near =, p() is approimatel cubic.

174 6 SESSION. RATIONAL FUNCTIONS Also, there is an asmptote = so we get 0 Finall, we see from the table that p() changes sign at the asmptote = and has a horizontal asmptote =, so we complete the sketch: 0 Note that if we had made a mistake somewhere there is a good chance that we would have not been able to get to the horizontal asmptote on the right side without creating an additional -intercept. What can we conclude from this sketch? This sketch ehibits onl the general shape which can help decide on an appropriate window if we want to

175 .. OPTIONAL SECTION: RATIONAL FUNCTIONS BY HAND 65 investigate details using technolog. Furthermore, we can infer where p() is positive and where p() is negative. However, it is important to notice, that there ma be wiggles in the graph that we have not included in our sketch. We now give one more eample of graphing a rational function where the horizontal asmptote is = 0. Eample.7. Sketch the graph of r() = ( ) (+) (+) ( ). Solution. Here we see that there are -intercepts at (0,0), (0,), and (0, ). There are two vertical asmptotes: = and =. In addition, there is a horizontal asmptote at = 0. (Wh?) Putting this information on the graph gives 0 In this case, it is eas to get more information for large that will be helpful in sketching the function. Indeed, when is large, we can approimate r() b dropping all but the highest order term in the numerator and denominator which gives r() 6 =. So for large, the graph of r looks like 7 0

176 66 SESSION. RATIONAL FUNCTIONS The function gives the following table: for a near a, p() tpe sign change at a C (+) linear changes C /(+) asmptote does not change 0 C parabola does not change C ( ) cubic changes C 5 /( ) asmptote changes Looking at the table for this function, we see that the graph should look like a line near the zero (0, ) and since it has an asmptote =, the graph looks something like: 0 Then, looking at the table we see that r() does not change its sign near =, so that we obtain: 0 Now, the function is approimatel quadratic near = 0 so the graph looks

177 .. OPTIONAL SECTION: RATIONAL FUNCTIONS BY HAND 67 like: 0 Turning to head toward the root at and noting that the function is approimatel cubic there, and that there is an asmptote =, we have: 0 Finall, we see that the function changes sign at = (see the table). So since the graph hugs the asmptote near the bottom of the graph on the left side of the asmptote, it will hug the asmptote near the top on the right side. So this together with the fact that = 0 is an asmptote gives the sketch (perhaps using an eraser to match the part of the graph on the right that uses the large ): 0 Note that if the graph couldn t be matched at the end without creating an etra -intercept, then a mistake has been made.

178 68 SESSION. RATIONAL FUNCTIONS. Eercises Eercise.. Find the domain, the vertical asmptotes and removable discontinuities of the functions. a) f() = b) f() = c) f() = +6 d) f() = ( )(+)(+) ( ) (+)( 5) e) f() = f) f() = + Eercise.. Find the horizontal asmptotes of the functions. a) f() = b) f() = ( ) c) f() = ++ d) f() = Eercise.. Find the - and -intercepts of the functions. a) f() = c) f() = ( )( )(+) ( )( 5) b) f() = 8+5 d) f() = Eercise.. Sketch the graph of the function f b using the domain of f, the horizontal and vertical asmptotes, the removable singularities, the - and -intercepts of the function, together with a sketch of the graph obtained from the calculator. a) f() = 6 + b) f() = 6+8 c) f() = d) f() = + Eercise.5. Find a rational function f that satisfies all the given properties. a) vertical asmptote at = and horizontal asmptote = 0 b) vertical asmptotes at = and = and horizontal asmptote = 5 c) removable singularit at = and no horizontal asmptote

179 Session Polnomial and rational inequalities. Polnomial inequalities We now consider inequalities. Solving inequalities is quite similar to solving equalities. There is one etra consideration, that multipling or dividing b a negative number on both sides of an inequalit changes the direction of the inequalit sign. 6 = but 6 = Eample.. Solve for. a) +7 > 9, b) +5 c) < 6, d) + < 0 Solution. The first three calculations are straightforward. a) +7 > 9 b) +5 c) < 6 ( 7) = > ( ( )) = < = 6 ( ( )) = 8 (+) = 7 < 6 7 ( ( 6)) 7 = > = 7 < 6 7, 6 ( 5) The last implication was obtained b switching the right and left terms of the inequalit. The solution set is the interval [ 7 6, 7 6 ). 69

180 70 SESSION. POLYNOMIAL AND RATIONAL INEQUALITIES For part (d), it is best to consider both inequalities separatel. + ( +) = 5 5 ( ( 5)) =, + < 0 ( ) = < ( ( )) = >. The solution has to satisf both inequalities and >. Both inequalities are true for > (since then also ), so that this is in fact the solution: >. When dealing with polnomial inequalities, we use the same three-step strateg that we used in section.. More precisel, the first step is to solve the corresponding equalit, and the second step is to determine the solution b investigating the subintervals induced from step. For both of these steps we ma now also use the graph of the function and its displa on the graphing calculator. The third step is to check the endpoints of each interval. Eample.. Solve for. a) 0 b) c) > 5( ) d) +5 > 7 +9 e) Solution. a) We can find the roots of the polnomial on the left b factoring. = 0 = ( )(+) = 0 = = or = To see where f() = is 0, we graph it with the calculator

181 .. POLYNOMIAL INEQUALITIES 7 We see that f() 0 when and when (the parts of the graph above the -ais). The solution set is therefore {, or } = (, ] [, ). b) Here is the graph of the function f() = with the TI-8 in the standard window This graph shows that there are two intervals where f() 0 (the parts of the graph below the -ais). To determine the eact intervals, we calculate where f() = = 0. The graph suggests that the roots of f() are at =, =, and = 5. This can be confirmed b a calculation: f() = = 9+ 5 = 0, f() = = = 0, f(5) = = = 0. Since f is a polnomial of degree, the roots =,,5 are all of the roots of f. (Alternativel, we could have divided f(), for eample, b and used this to completel factor f and with this obtain all the roots of f.) With this, we can determine the solution set to be the set: solution set = { R, or 5} = (,] [,5]. Note that we include the roots,, and 5 in the solution set since the original inequalit was (and not < ), which includes the solutions of the corresponding equalit. c) In order to use the graphing calculator, we rewrite the inequalit to obtain zero on one side of the inequalit. > 5( ) (distribute 5) = > 5 5

182 7 SESSION. POLYNOMIAL AND RATIONAL INEQUALITIES (subtract 5, add 5) = 5 +5 > 0. We graph f() = 5 +5 with the TI-8. The graph suggests the roots =, 0,, and 5. This can be confirmed b a straightforward calculation. f( ) = ( ) 5 ( ) ( ) +5 ( ) = +5 5 = 0, f(0) = = 0, f() = 5 +5 = 5 +5 = 0, f(5) = = = 0. The roots =, 0,, and 5 are the onl roots since f is of degree. The intervals of the solution for f() > 0 ma be read off from the graph: solution set = (, ) (0,) (5, ) (Notice that the roots, 0,, and 5 are not included in the solution set since our inequalit reads f() > 0 and not f() 0.) d) Again, we bring all terms to one side: +5 > 7 +9 = > 0 (Here it does not matter whether we bring the terms to the right or the left side of the inequalit sign! The resulting inequalit is different, but the solution to the problem is the same.) With this, we now use the TI-8 to find the graph of the function f() =

183 .. POLYNOMIAL INEQUALITIES 7 The graph suggests at least one root (the left most intersection point), but possibl one or two more roots. To gain a better understanding of whether the graph intersects the -ais on the right, we rescale the window size of the previous graph. This viewing window suggests that there are two roots = and =. We confirm that these are the onl roots with an algebraic computation. First, we check that = and = are indeed roots: f() = = = 0, f() = = = 0. To confirm that these are the onl roots (and we have not just missed one of the roots which might possibl become visible after sufficientl zooming into the graph), we factor f() completel. We divide f() b : and use this to factor f: ( ) 6 ( ) 9 9 (9 9) f() = = ( )( 6+9) = ( )( )( ) This shows, that is a root of multiplicit, and so f has no other roots than = and =. The solution set consists of those numbers for which f() > 0. From the graph we see that this is the case when < < and 0

184 7 SESSION. POLYNOMIAL AND RATIONAL INEQUALITIES when > (the roots = and = are not included as solutions). We can write the solution set in several different was: solution set = { < < or > } = { < } {}, or in interval notation: solution set = (,) (, ) = (, ) {}. e) If we set f() = then we need to find those numbers with f() 0. We first graph f in the standard window. To get a better view of the graph, we rescale to an appropriate window: There appear to be three intervals, where f() 0. To determine the eact numbers, we guess some of the roots from the graph. These would be = 5,,,, 7. To confirm these roots, we calculate the following function values. f( 5) = ( 5) 5 6 ( 5) 6 ( 5) + ( 5) 7 ( 5) 0 = = 0, f( ) = ( ) 5 6 ( ) 6 ( ) + ( ) 7 ( ) 0 = = 0, f() = = = 0, f() =

185 .. POLYNOMIAL INEQUALITIES 75 = = 0, f(7) = = = 0. Since f() is of degree 5, we know that these are all of the roots of f(). The solution set for f() 0 can be read from the graphs above: solution set = (, 5] [,] [,7] (Note again, that the roots are all included in the solution set.) Polnomial inequalities come up, for eample, when finding the domain of functions involving a square root, as we will show in the net eample. Eample.. Find the domain of the given functions. a) f() = b) g() = 5 +6 Solution. a) The domain of f() = is given b all for which the square root is non-negative. In other words the domain is given b numbers with 0. Graphing the function = = (+)( ), we see that this is precisel the case, when or. Therefore, the domain is D f = (, ] [, ). b) For the domain of g() = 5 +6, we need find those with To this end, we graph = 5 +6 and check for its roots. From the graph and table above, we calculate the roots of = 5 +6 at = 0, =, and =. Furthermore, the graph and table show that precisel when 0 or. The domain is therefore, D g = [0,] [, ).

186 76 SESSION. POLYNOMIAL AND RATIONAL INEQUALITIES. Rational inequalities and absolute value inequalities Rational inequalities are solved with the same three-step process that was used to solve the polnomial and absolute value inequalities before (see page 6). That is, in step, we find the solution of the corresponding equalit, and then, in step, we use sample points of the graph to determine the intervals of the solution. Finall, in step, we check the endpoints of each interval. Eample.. Solve for. a) b) 5 c) +5 < d) > 7 Solution. a) Here is the graph of in the standard window. Factoring numerator and denominator, we can determine vertical asmptotes, holes, and -intercepts = ( )( ) ( 5) The vertical asmptotes are at = 0 and = 5, the -intercepts are at = and =. Since for large, the fraction reduces to =, we see that the horizontal asmptote is at =. Thus, for < 0 and > 5. To 5 see where the graph is 0 between 0 and 5, we zoom into the graph:

187 .. RATIONAL INEQUALITIES AND ABSOLUTE VALUE INEQUALITIES77 Combining all of the above information, we obtain the solution set: solution set = (,0) [,] (5, ) Notice that the coordinate of the intercepts are = and = are included in the solution set, whereas the values = 0 and = 5 associated with the vertical asmptotes are not included since the fraction is not defined for = 0 and = 5. b) To find the numbers where 5, we can graph the two functions on the right and left of the inequalit. However, this can sometimes be confusing, and we recommend rewriting the inequalit so that one side becomes zero. Then, we graph the function on the other side of the new inequalit. 5 = 5 0 = 5 ( ) 5 +6 = 0 = Therefore, we graph the function f() =. 0 0 The vertical asmptote is =, and the -intercept found thus = 0 = = = =. This together with the graph and the fact that f is undefined at and f( ) = 0 gives the following solution set: ( ) [ ) solution set =,,

188 78 SESSION. POLYNOMIAL AND RATIONAL INEQUALITIES c) We want to find those numbers for which <. One wa to do +5 this is given b graphing both functions f () = and f +5 () =, and b tring to determine where f () < f (). However this can sometimes be quite confusing, as the two graphs for f and f show below. As before, we recommend rewriting the inequalit so that one side of the inequalit becomes zero: +5 < = +5 < 0 = = 5 (+5)( ) ( ) (+5) < 0 (+5)( ) 7 < 0 = (+5)( ) < 0 We therefore graph the function f() = 7 (+5)( ). 7 The vertical asmptotes of f() = are = 5 and =. The (+5)( ) -intercept is (7,0). We see from the graph that f() < 0 for < 5. To see the graph at >, we zoom to the -intercept at = 7. Therefore, the solution set is solution set = { < 5, or < < 7} = (, 5) (,7).

189 .. EXERCISES 79 (The -intercept = 7 is not included in the solution set since the original equation had a < and not sign.) d) To analze > 7, we graph the function f() = 7. To see where f() > 0, we find the zeros of f(). 7 = 0 = = 7 = = ±7 = = 7 = = 7 (add ) = = 0 (add ) = = (divide b ) = = 5 (divide b ) = = Thus, we read off the solution set for f() > 0 from the graph. solution set = (, ) (5, ). Eercises Eercise.. Solve for. a) 5+6, b) + > 0 c) +8 6+, d) 9 < e) , f) 5 > 7 g) , h) 5+ < 8 7+ Eercise.. Solve for. a) 5 > 0, b) 5 c) 0, d) + < 0 e) +, f) < + g) + > 0, h) i) + ++ < 0, j) < k) , l) < 6 m) 5 +6 > 0, n)

190 80 SESSION. POLYNOMIAL AND RATIONAL INEQUALITIES o) , p) Eercise.. Find the domain of the functions below. a) f() = 8+5 b) f() = 9 c) f() = ( )( ) d) f() = ( )( 5)( 6) e) f() = 5 6 f) f() = 6 7 Eercise.. Solve for. a) 5 > 0 b) 0 c) < 0 5 e) + f) +0 > 5 g) 5 + h) Eercise.5. Solve for. a) +7 > 9 b) 6+ < c) 5 d) 7 5 e) 8 f) > + 5 d) 9 0 +

191 Review of polnomials and rational functions Eercise II.. Divide the polnomials: Eercise II.. Find the remainder when dividing b +. Eercise II.. Which of the following is a factor of :, +, 0 Eercise II.. Identif the polnomial with its graph a) - b) c) - d) - 8

192 i) f() = ++, graph: ii) f() = + +, graph: iii) f() = ++, graph: iv) f() = +6 +, graph: Eercise II.5. Sketch the graph of the function: What is our viewing window? f() = Find all roots, all maima and all minima of the graph with the calculator. Eercise II.6. Find all roots of f() = Use this information to factor f() completel. Eercise II.7. Find a polnomial of degree whose roots are 0,, and, and so that f() = 0. Eercise II.8. Find a polnomial of degree with real coefficients, whose roots include, 5, and i. Eercise II.9. Let f() =. Sketch the graph of f. Include all vertical and horizontal asmptotes, all holes, and all - and -intercepts. Eercise II.0. Solve for : a) + < +, b) + 7, c) + + 8

193 Part III Eponential and logarithmic functions 8

194 Session Eponential and logarithmic functions. Eponential functions and their graphs We now consider functions that differ a great deal from polnomials and rational fractions in their compleit. More precisel, we will eplore eponential and logarithmic functions from a function theoretic point of view. We start b recalling the definition of eponential functions and b studing their graphs. Definition.. A function f is called an eponential function if it is of the form f() = c b for some real number c and positive real number b. The constant b is called the base. Eample.. Graph the functions f() =, g() =, h() = 0, k() = ( ) ( ), l() =. 0 First, we will graph the function f() = b calculating the function values in a table and then plotting the points in the --plane. We can calculate the values b hand, or simpl use the table function of the calculator to find the function values. f(0) = 0 = 8

195 .. EXPONENTIAL FUNCTIONS AND THEIR GRAPHS 85 f() = = f() = = f() = = 8 f( ) = = 0.5 f( ) = = 0.5 Similarl, we can calculate the table for the other functions g, h, k and l b entering the functions in the spots at Y, Y, Y, and Y5. The values in the table for these functions can be seen b moving the cursor to the right with the lp lp ke. We can see the graphs b pressing the lpgraph lp ke. In order to see these graphs more clearl, we adjust the graphing window to a more appropriate size.

196 86 SESSION. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Since all these functions are graphed in the same window, it is difficult to associate the graphs with their corresponding functions. In order to distinguish between the graphs, we ma use the TI-8 to draw the graphs in different line stles. In the function menu ( lp= lp ), move the cursor to the left with lp lp, and press lpenter lp until the line becomes a thick line, or a dotted line, as desired. B moving the cursor down lp lp, and pressing lpenter lp, we can adjust the line stles of the other graphs as well. Note that the function k can also be written as ( ) k() = = ( ) =, and similarl, l() = ( 0) = 0. This eample shows that the eponential function has the following properties. Observation.. The graph of the eponential functionf() = b with b > 0 and b has a horizontal asmptote at = 0. = ( 0 ) = ( ) = ( ) = (. ) 6 5 = 0 = = =. =

197 .. EXPONENTIAL FUNCTIONS AND THEIR GRAPHS 87 If b >, then f() approaches + when approaches +, and f() approaches 0 when approaches. If 0 < b <, then f() approaches 0 when approaches +, and f() approaches + when approaches. An important base that we will frequentl need to consider is the base of e, where e is the Euler number. Definition.. The Euler number e is an irrational number that is approimatel e = ( To be precise, ) we can define e as the number which is the horizontal asmptote of the function f() = + when approaches +. e One can show that f has, indeed, a horizontal asmptote, and this limit is defined as e. ( e := lim + ) Furthermore, one can show that the eponential function with base e has a similar limit epression. ( e r = lim + r (.) ) Alternativel, the Euler number and the eponential function with base e ma also be defined using an infinite series, namel, e r = +r + r a course in calculus. + r + r +... These ideas will be eplored further in Eample.5. Graph the functions. a) = e, b) = e, c) = e, d) = e +e

198 88 SESSION. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Solution. Using the calculator, we obtain the desired graphs. The eponential function = e ma be entered via lpnd lp lplnlp. = e = e Note that the minus sign is entered in the last epression (and also in the following two functions) via the lp( ) lp ke. = e = e +e The last function = e +e is called the hperbolic cosine, and is denoted b cosh() = e +e. We now stud how different multiplicative factors c affect the shape of an eponential function. Eample.6. Graph the functions. a) =, b) =, c) = ( ) d) = 0., e) = ( 0.) Solution. We graph the functions in one viewing window.

199 .. EXPONENTIAL FUNCTIONS AND THEIR GRAPHS 89 Here are the graphs of functions f() = c for various choices of c. 6 5 = = = = ( ) = ( ) = ( 0.) -6 Note that for f() = c, the -intercept is given at f(0) = c. Finall, we can combine our knowledge of graph transformations to stud eponential functions that are shifted and stretched. Eample.7. Graph the functions. a) = 5, b) = e +, c) = e + Solution. The graphs are displaed below. = 5 = e + = e +

200 90 SESSION. EXPONENTIAL AND LOGARITHMIC FUNCTIONS The first graph = 5 is the graph of = shifted down b The graph of = e + is the graph of = e shifted to the left b Finall, = e + is the graph of = e shifted to the right b (see the graph of = e ), then compressed b a factor towards the -ais (see the graph of = e ), and then shifted up b. 5 = e 0 = e = e Logarithmic functions and their graphs The logarithmic function is closel related to the eponential function. Specificall, the logarithm is the inverse function of the eponential function.

201 .. LOGARITHMIC FUNCTIONS AND THEIR GRAPHS 9 Definition.8. Let 0 < b be a positive real number that is not equal to. For > 0, the logarithm of with base b is defined b the equivalence = log b () b = (.) For the particular base b = 0 we use the short form log() := log 0 () For the particular base b = e, where e.788 is the Euler number, we call the logarithm with base e the natural logarithm, and write ln() := log e () The logarithmic function is the function = log b () with domain D = { R > 0} of all positive real numbers, and range R = R of all real numbers. It is the inverse of the eponential function = b with base b. Eample.9. Rewrite the equation as a logarithmic equation. a) = 8, b) 0 = 000, c) e = 7, d) 7 a = 5. Solution. We can immediatel appl equation (.). For part (a), we have b =, =, and = 8. Therefore we have: = 8 log (8) = Similarl, we obtain the solutions for (b), (c), and (d). b) 0 = 000 log(000) = c) e = 7 ln(7) = d) 7a = 5 log (5) = 7a Eample.0. Evaluate the epression b rewriting it as an eponential epression. a) log (6), b) log 5 (5), c) log (), d) log () e) log(0,000), f) log(0.00), g) ln(e 7 ), h) log b (b )

202 9 SESSION. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Solution. a) If we set = log (6), then this is equivalent to = 6. Since, clearl, = 6, we see that =. Therefore, we have log (6) =. b) log 5 (5) = 5 = 5 (since 5 =5) = = = log 5 (5) c) log () = = (since 0 =) = 0 = = log () d) log () = = (since =) = = = log () e) log(00,000) = 0 = 00,000 (since 0 5 =00,000) = 5 = = log(00,000) f) log(0.00) = 0 = 0.00 (since 0 =0.00) = = = log(0.00) g) ln(e 7 ) = e = e 7 = 7 = = ln(e 7 ) h) log b (b ) = b = b = = = log b (b ) In fact, the last eample, in which we obtained log b (b ) =, combines all of the previous eamples. In the previous eample (in parts (c), (d), and (h)), we were able to find certain elementar logarithms. We record these in the net observation. Observation.. We have the elementar logarithms: log b (b ) = log b (b) = log b () = 0 (.) In general, when the argument is not a power of the base, we can use the calculator to approimate the values of a logarithm via the formulas: log b () = log() log(b) or log b () = ln() ln(b) (.) The last two formulas will be proved in proposition. below. For now, we want to show how the can be used to calculate an logarithmic epression with the calculator.

203 .. LOGARITHMIC FUNCTIONS AND THEIR GRAPHS 9 Eample.. Calculate: a) log (), b) log. (98.765) Solution. We calculate log () b using the first formula in (.). log () = log() log().5 Alternativel, we can also calculate this with the second formula in (.). log () = ln() ln().5 For part (b) we have log. (98.765) = log(98.765) log(.) 5.0. We also want to gain an understanding of the graph of a logarithmic function. Consider the graph of = from the previous section. Recall that the graph of the inverse of a function is the reflection of the graph of the function about the diagonal line =. So in this case we have: 6 5 = = log () = The graph can also be obtained using a graphing calculator. Eample.. a) Graph the functions f() = ln(), g() = log(), h() = log (), and k() = log 0.5 (). What are the domains of f, g, h, and k? How do these functions differ? b) Graph the function p() = ln()+. What is the domain of p? c) Graph the function q() = ln(5 ). What is the domain of q? d) Graph the function r() = log 7 (+8). What is the domain of r?

204 9 SESSION. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Solution. a) We know from the definition that the domain of f, g, and h is all real positive numbers, D f = D g = D h = D k = { > 0}. The functions f and g can immediatel be entered into the calculator. The standard window gives the following graphs. f() = ln() g() = log() Note that we can rewrite g(), h(), and k() as a constant times f(): g() = log() = log 0 () = ln() ln(0) = ln(0) f() h() = log () = ln() ln() = ln() f() k() = log 0.5 () = ln() ln(0.5) = ln(0.5) f() Since 0. <, we see that the graph of g is that of f compressed ln(0) towards the -ais b a factor. Similarl,. >, so that ln(0) ln() the graph of h is that of f stretched awa from the -ais b a factor. ln() Finall,., or more precisel, = =, so that ln(0.5) ln(0.5) ln( ) ln() the graph of k is that of h reflected about the -ais. f() = ln() h() = log () k() = log 0.5 () Note that all these graphs have a common -intercept at = : f() = g() = h() = k() = 0

205 .. LOGARITHMIC FUNCTIONS AND THEIR GRAPHS 95 To visualize the differences between the graphs, we graph them together in one coordinate sstem. h() = log () f() = ln() g() = log() k() = log 0.5 () b) Using our knowledge of transformations of graphs, we epect that p() = ln() + is that of = ln() reflected and stretched awa from the -ais (b a factor ), and then shifted up b. The stretched and reflected graph is in the middle below, whereas the graph of the shifted function p is on the right below. = ln() = ln() p() = ln()+ The domain consist of numbers for which the ln() is defined, that is, D p = { > 0}. c) To determine the domain of q() = ln(5 ), we have to see for which the logarithm has a positive argument. More precisel, we need 5 > 0, that is, 5 >, so that the domain is D q = { < 5}. The calculator displas the following graph.

206 96 SESSION. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Note that the graph, as displaed b the calculator, appears to end at a point that is approimatel at (5,.5). However, the actual graph of the logarithm does not stop at an point, since it has a vertical asmptote at = 5, that is, the graph approaches as approaches 5. The calculator onl displas an approimation, which ma be misleading since this approimation is determined b the window size and the size of each piel. The graph would therefore look as follows d) The domain of r() = log 7 (+8) consists of those numbers for which the argument of the logarithm is positive. +8 > 0 (subtract 8) (divide b ) = > 8 = > Therefore, the domain is D r = { > }. To graph the function r() = log 7 (+8), we can enter r() = ln(+8) into the calculator. The graph looks ln(7) as follows The previous eample analzes the graph of the logarithm function. Here is the summar.

207 .. EXERCISES 97 Observation.. The graph of a logarithmic function = log b () with base b is that of the natural logarithm = ln() stretched awa from the -ais, or compressed towards the -ais when b >. When 0 < b <, the graph is furthermore reflected about the -ais. = log () = ln() = log 0 () = log 0. () = log 0.5 () The graph = log b () has domain D = { > 0}, and a vertical asmptote at = 0. There is no horizontal asmptote, as f() approaches + when approaches + for b >, and f() approaches when approaches + for 0 < b <. Finall, there is an -intercept at =.. Eercises Eercise.. Graph the following functions with the calculator. a) = 5 b) =.0 c) = ( ) d) = 0.97 e) = f) = ( ) g) = e h) = 0.0 i) = j) = e + k) = e e l) = e e e +e The last two functions are known as the hperbolic sine, sinh() = e e, and the hperbolic tangent, tanh() = e e. Recall that the hperbolic e +e cosine cosh() = e +e was alread graphed in eample.5. Eercise.. Graph the given function. Describe how the graph is obtained b a transformation from the graph of an eponential function = b (for

208 98 SESSION. EXPONENTIAL AND LOGARITHMIC FUNCTIONS appropriate base b). a) = 0. b) = c) = ( ) d) = e) = e f) = e + g) = ( ) + h) = i) = + 6 Eercise.. Use the definition of the logarithm to write the given equation as an equivalent logarithmic equation. a) = 6 b) 8 = 56 c) e = 7 d) 0 = 0. e) = f) 5 7 = g) a+ = h) ( ) h = 0 Eercise.. Evaluate the following epressions without using a calculator. a) log 7 (9) b) log (8) c) log (6) d) log 50 (,500) e) log (0.5) f) log(,000) g) ln(e ) h) log () i) log(0.) j) log 6 ( 6 ) k) ln() l) log (8) Eercise.5. Using a calculator, approimate the following epressions to the nearest thousandth. a) log (50) b) log () c) log 7 (0.) d) log 0. (00) Eercise.6. State the domain of the function f and sketch its graph. a) f() = log() b) f() = log(+7) c) f() = ln(+5) d) f() = ln( 6) e) f() = log(+) f) f() = log(+) g) f() = log ( 5) h) f() = ln( )+ i) f() = log 0. () j) f() = log ( 5) k) f() = log l) f() = log +

209 Session Properties of eponential and logarithmic functions. Algebraic properties of eponential and logarithmic functions We recall some of the main properties of the eponential function: b + = b b b = b b (b ) n = b n (.) Writing the above properties in terms of f() = b we have f( + ) = f()f(), f( ) = f()/f(), and f(n) = f() n. Due to the fact that the logarithm is the inverse function of the eponential, we have corresponding properties whose proof follows: Proposition.. The logarithm behaves well with respect to products, quotients, and eponentiation. Indeed, for all positive real numbers 0 < b, > 0, > 0, and real numbers n, we have: log b ( ) = log b ()+log b () log b ( ) = log b() log b () log b ( n ) = n log b () (.) 99

210 00 SESSION. PROPERTIES OF EXP AND LOG In terms of the logarithmic function g() = log b (), the properties in the table above can be written: g() = g()+g(), g(/) = g() g(), and g( n ) = n g(). Furthermore, for another positive real number 0 < a, we have the change of base formula: log b () = log a() log a (b) (.) In particular, we have the formulas from equation (.) on page 9 when taking the base a = 0 and a = e: log b () = log() log(b) and log b () = ln() ln(b) Proof. We start with the first formula log b ( ) = log b () + log b (). If we call u = log b () and v = log b (), then the equivalent eponential formulas are b u = and b v =. With this, we have Rewriting this in logarithmic form, we obtain This is what we needed to show. = b u b v = b u+v. log b ( ) = u+v = log b ()+log b (). Net, we prove the formula log b ( ) = log b () log b (). We abbreviate u = log b () and v = log b () as before, and their eponential forms are b u = and b v =. Therefore, we have = bu b v = bu v. Rewriting this again in logarithmic form, we obtain the desired result. log b ( ) = u v = log b () log b () For the third formula, log b ( n ) = n log b (), we write u = log b (), that is in eponential form b u =. Then: n = (b u ) n = b n u = log b ( n ) = n u = n log b () For the last formula (.), we write u = log b (), that is, b u =. Appling the logarithm with base a to b u = gives log a (b u ) = log a (). As we have just shown before, log a (b u ) = u log a (b). Combining these identities with the initial definition u = log b (), we obtain log a () = log a (b u ) = u log a (b) = log b () log a (b) Dividing both sides b log a (b) gives the result log a () log a (b) = log b ().

211 .. ALGEBRAIC PROPERTIES OF EXP AND LOG 0 Eample.. Combine the terms using the properties of logarithms so as to write as one logarithm. a) ln()+ln() b) (log( ) log( )) c) ln() ln() 7 5 ln(z) d) 5+log (a b ) log (a+b) Solution. Recall that a fractional eponent can also be rewritten with an nth root. = and n = n = p q = ( p ) q = q p We appl the rules from proposition.. a) ln()+ln() = )+ln() = ln( ) = ln( ) ln( ( ( )) ( ( )) b) (log( ) log( )) = log = log = log( ( ) ) ( = log ) c) ln() ln() 7 ln(z) = 5 ln( ) ln( ) ln( 5 ( z 7 ) = ln ) 5 z 7 d) 5+log (a b ) log (a+b) = log ( 5 )+log (a b ) log (a+b) ( ) ( ) = log 5 (a b ) = log (a+b)(a b) a+b = log a+b ( (a b)) Eample.. Write the epressions in terms of elementar logarithms u = log b (), v = log b (), and, in part (c), alsow = log b (z). Assume that,,z > 0. a) ln( ) ( ) 5 ), b) log( c) log. z Solution. In a first step, we rewrite the epression with fractional eponents, and then appl the rules from proposition.. a) ln( 5 ) = ln( 5 ) = ln( 5 )+ln( ) = 5 ln()+ln() = 5u+v ) ( ( ) ) b) log( = log = ( log ) = ( ) ( log()+log( ) = log()+log()) = log()+ log() = u+ v

212 0 SESSION. PROPERTIES OF EXP AND LOG ( c) log z = = ( ) ( = log z ) ) ( = log ( ) log ( ) log () log (z ) ( ) log () log () log (z) = log () log () 6 log (z) = u v 6 w z ). Solving eponential and logarithmic equations We can solve eponential and logarithmic equations b appling logarithms and eponentials. Since the eponential and logarithmic functions are invertible (the are inverses of each other), appling these operations can also be reversed. In short we have observed that Observation.. = = b = b, and b = b = = = = log b () = log b (), and log b () = log b () = = We can use the above implications whenever we have a common basis, or when we can convert the terms to a common basis. Eample.5. Solve for. a) +7 = b) 0 8 = 0.0 c) 7 = 7 5+ d) 5 + = 5 7 e) ln( 5) = ln( ) f) log (+5) = log (+)+ g) log 6 ()+log 6 (+) = log 6 (5) h) log ( )+log (+6) = Solution. In these eamples, we can alwas write both sides of the equation as an eponential epression with the same base. a) +7 = = +7 = 5 = +7 = 5 = = b) 0 8 = 0.0 = 0 8 = 0 = 8 = = = 6 = =

213 .. SOLVING EXPONENTIAL AND LOGARITHMIC EQUATIONS 0 Here it is useful to recall the powers of 0, which were also used to solve the equation above. 0 =,000 0 = 00 0 = = 0 = 0. 0 = = 0.00 In general (n ) : 0 n = }{{} n zeros 0 n = } {{} n zeros c) 7 = 7 5+ = = 5+ ( 5+) = = 7 = = 7 d) 5 + = 5 7 = 5 + = 5 ( 7) = + = ( 7) = + = 8 ( 8 ) = 5 = 5 = = B a similar reasoning we can solve equations involving logarithms whenever the bases coincide. e) ln( 5) = ln( ) = 5 = ( +5) = = = = For part (f), we have to solvelog (+5) = log (+)+. To combine the righthand side, recall that can be written as a logarithm, = log ( ) = log 6. With this remark we can now solve the equation for. log (+5) = log (+)+ = log (+5) = log (+)+log (6) = log (+5) = log (6 (+)) = +5 = 6(+) ( 6 5) = +5 = 6+8 = 5 = = = 5 Net, in part (g), we start b combining the logarithms. log 6 ()+log 6 (+) = log 6 (5) = log 6 ((+)) = log 6 (5)

214 0 SESSION. PROPERTIES OF EXP AND LOG remove log 6 = (+) = 5 = + 5 = 0 = (+5)( ) = 0 = = 5 or = Since the equation became a quadratic equation, we ended up with two possible solutions = 5 and =. However, since = 5 would give a negative value inside a logarithm in our original equation log 6 ()+log 6 (+) = log 6 (5), we need to eclude this solution. The onl solution is =. Similarl we can solve the net part, using that = log ( ): h) log ( )+log (+6) = = log (( )(+6)) = log ( ) = ( )(+6) = = + = 9 = + = 0 = (+7)( ) = 0 = = 7 or = We eclude = 7, since we would obtain a negative value inside a logarithm, so that the solution is =. When the two sides of an equation are not eponentials with a common base, we can solve the equation b first appling a logarithm and then solving for. Indeed, recall from Observation 7.9 in section 7. on inverse functions, that since f() = log b () and g() = b are inverse functions, we have log b (b ) = and b log b () = whenever the compositions of the left sides make sense. That is, the action of the logarithm cancels out the action of the eponential function, and vice versa. So we can think of appling a logarithm (an eponentiation) on both sides of an equation to cancel an eponentiation (a logarithm) much like squaring both sides of an equation to cancel a square root. Eample.6. Solve for. a) +5 = 8 b) = 6 c) 5 7 = d) 5. =.7 +6 e) 7 = + f) 7 + = 6 Solution. We solve these equations b appling a logarithm (both log or ln will work for solving the equation), and then we use the identit log(a ) =

215 .. SOLVING EXPONENTIAL AND LOGARITHMIC EQUATIONS 05 log(a). a) +5 = 8 = ln +5 = ln8 = (+5) ln = ln8 = +5 = ln8 ln = = ln8 5. ln b) = 6 = ln = ln6 = ( ) ln = ln6 = = ln6 ln = ln6 = ln + ln6 + ln = = = ln6 +.5 ln c) 5 7 = = ln5 7 = ln = ( 7) ln5 = ln At this point, the calculation will proceed differentl than the calculations in parts (a) and (b). Since appears on both sides of ( 7) ln5 = ln, we need to separate terms involving from terms without. That is, we need to distribute ln5 on the left and separate the terms. We have ( 7) ln5 = ln = ln5 7 ln5 = ln (add +7 ln5 ln) = ln5 ln = 7 ln5 = (ln5 ln) = 7 ln5 = = 7 ln5 ln5 ln.0 We need to appl the same solution strateg for the remaining parts (d)-(f) as we did in (c). d) 5. =.7 +6 = ln5. = ln.7 +6 = ln5. = (+6) ln.7 = ln5. = ln.7+6 ln.7 = ln5. ln.7 = 6 ln.7 = (ln5. ln.7) = 6 ln.7 6 ln.7 = = ln5. ln e) 7 = + = ln7 = ln + = ( ) ln7 = (+) ln

216 06 SESSION. PROPERTIES OF EXP AND LOG = ln7 ln7 = ln+ ln = ln7 ln = ln7+ ln = (ln7 ln) = ln7+ ln = = ln7+ ln 5.80 ln7 ln f) 7 + = 6 = ln7 + = ln 6 = (+) ln7 = ( 6) ln = ln7+ ln7 = ln 6 ln = ln7 ln = ln7 6 ln = ( ln7 ln) = ln7 6 ln = = ln7 6 ln ln7 ln 6. Note that in the problems above we could have also changed the base as we did earlier in the section. For eample, in part f) above, we could have begun b writing the right hand side as 6 = 7 log 7 (( 6)). We chose to simpl appl a log to both sides instead, because the notation is somewhat simpler.. Eercises Eercise.. Combine the terms and write our answer as one logarithm. a) ln()+ln() b) log() log() c) log() log()+log(z) d) log( z ) log( z ) e) ln() ln()+ ln(z) f) ln( )+ln( ) g) 5ln()+ln( ) ln() h) log 5 (a +0a+9) log 5 (a+9)+ Eercise.. Write the epressions in terms of elementar logarithms u = log b (), v = log b (), and w = log b (z) (whichever are applicable). Assume that

217 .. EXERCISES 07,,z > 0. a) log( ) b) log( 7 ) c) log ( ) ( ) ( ) ( ) d) ln e) ln z f) log z ( ) ( g) log z h) log ) ( ) 00 5 z z i) ln e Eercise.. Solve for without using a calculator. a) 6 = 6 b) 8 = 6 c) 0 5 = d) = 5 e) = 6 + f) + = g) + = h) + = 7 i) 5 7 = 5 j) 9 5+ = 7 8 Eercise.. Solve for without using a calculator. a) ln(+) = ln(5 5) b) ln(+6) = ln( )+ln() c) log (+5) = log ()+5 d) log()+ = log(5+80) e) log(+5)+log() = log(6) f) log ()+log ( 6) = g) log 6 ()+log 6 ( 6) = h) log 5 ( )+log 5 () = i) log ()+log (+6) = j) log (+)+log (+5) = Eercise.5. Solve for. First find the eact answer as an epression involving logarithms. Then approimate the answer to the nearest hundredth using the calculator. a) = 57 b) 9 = 7 c) + = d).8 +7 = 6 e) 5 +5 = 8 f) + = 0. g).0 9 =.5 h) + = 5 + i) 9 = 6 j). 7 =.8 + k) 9 = 9 l).95 =. 7

218 Session 5 Applications of the eponential and the logarithm 5. Applications of eponential functions Before giving specific applications, we note that the eponential function = c b is uniquel determined b providing an two of its function values. Eample 5.. Let f() = c b. Determine the constant c and base b under the given conditions. a) f(0) = 5, f() = 0 b) f(0) =, f() = 8 c) f() = 60, f(7) = 5 d) f( ) = 55, f() = 7 Solution. a) Appling f(0) = 5 to f() = c b, we get 5 = f(0) = c b 0 = c = c Indeed, in general, we alwas have f(0) = c for an eponential function. The base b is then determined b substituting the second equation f() = 0. 0 = f() = c b = 5 b ( 5) = b = Therefore, f() = 5. Note that in the last implication, we used that the base must be positive. b) As before, we get = f(0) = c b 0 = c, and 8 = f() = c b = b ( ) = 6 = b (eponentiate b ) = b = 6 = 08

219 5.. APPLICATIONS OF EXPONENTIAL FUNCTIONS 09 Therefore, f() =. c) When f(0) is not given, it is easiest to solve for b first. We can see this as follows. Since 60 = f() = c b and 5 = f(7) = c b 7, the quotient of these equations eliminates c = c b c b 7 = b 5 = = b 5 (eponentiate b ( 5 )) = b = 5 = 5 = Then c is determined b an of the original equations. ( ) 60 = f() = c b = c = c (. Therefore, f() = 60 ) d) This solution is similar to part (c) = f( ) f() = c b c b = = b = 7 b 55 ( 7 ) = b = 55 = c = 60 = = f( ) = c b = c = c = ( ) = 55 (( 7 55) ) 7 ) = c ( 55 ( 7 55 therefore, f() = ( ) = 55 7 = 55 7 = ). Man situations are modeled b eponential functions. Eample 5.. The mass of a bacteria sample is.0 t grams after t hours. a) What is the mass of the bacteria sample after hours? b) When will the mass reach 0 grams?

220 0 SESSION 5. APPLICATIONS OF EXP AND LOG Solution. a) The formula for the mass in grams afterthours is(t) =.0 t. Therefore, after hours, the mass in grams is () =.0.6. b) We are seeking the number of hourstfor which = 0 grams. Therefore, we have to solve: 0 =.0 t ( ) = 5 =.0 t We need to solve for the variable in the eponent. In general, to solve for a variable in the eponent requires an application of a logarithm on both sides of the equation. 5 =.0 t (appl log) = log(5) = log(.0 t ) Recall one of the main properties that we need to solve for t: Using this fact, we can now solve for t: log( t ) = t log() (5.) log(5) = log(.0 t ) = log(5) = t log(.0) (divide b log(.0)) log(5) = log(.0) = t Using the lploglp ke on the calculator, we ma approimate this as t = log(5) log(.0) 8. After approimatel 8. hours, the mass will be 0 grams. Note that the formula.0 t comes from the following calculations. Suppose that after each hour there is an increase of percent of the mass and that the initial mass of the bacteria is grams. Then after the first hour we see that the mass in grams is + (.0) = (+.0) = (.0)

221 5.. APPLICATIONS OF EXPONENTIAL FUNCTIONS and that after the second hour the mass is (.0)+(.0)(.0) = (.0)(+.0) = (.0) and that after three hours the mass is (.0) +(.0) (.0) = (.0) (+.0) = (.0). Continuing in this manner we can see that, at least for whole numbers of hours t, the weight is given b.0 t. The same sort of idea can be used in all of the following applications. Eample 5.. The population size of a countr was.7 million in the ear 000, and. million in the ear 00. a) Assuming an eponential growth for the population size, find the formula for the population depending on the ear t. b) What will the population size be in the ear 05, assuming the formula holds until then? c) When will the population reach 8 million? Solution. a) The growth is assumed to be eponential, so that (t) = c b t describes the population size depending on the ear t, where we set t = 0 corresponding to the ear 000. Then the eample describes (0) = c as c =.7, which we assume in units of millions. To find the base b, we substitute the data of t = 0 and (t) =. into (t) = c b t.. =.7 b 0 =..7 = b0 = (. ) 0 = b =.7.0 (. ) 0.7 = (b 0 ) 0 = b The formula for the population size is (t).7.0 t. b) We calculate the population size in the ear 05 b setting t = 5: (5) = c) We seek t so that (t) = 8. We solve for t using the logarithm. 8 =.7.0 t = 8 ( 8 ).7 =.0t = log = log(.0 t ).7

222 SESSION 5. APPLICATIONS OF EXP AND LOG ( 8 ) = log = t log(.0).7 ( ) 8 log.7 = t = log(.0) 9. The population will reach 8 million in the ear 09. In man instances the eponential function f() = c b is given via a rate of growth r. Definition 5.. An eponential function with a rate of growth r is a function f() = c b with base b = +r. For an eample of wh this is a reasonable definition see the note following Eample 5.. Note 5.5. We want to point out that some authors use a different convention than the one given in definition 5.. Indeed, sometimes a function with rate of growth r is defined as an eponential function with base b = e r, whereas for us it has base b = +r. Since e r can be epanded as e r = +r+ r +..., we see that the two versions onl differ b an error of order. Eample 5.6. The number of PCs that are sold in the U.S. in the ear 0 is approimatel 50 million with a rate of growth of.6% per ear. Assuming the rate stas constant over the net ears, how man PCs will be sold in the ear 05? Solution. Since the rate of growth is r =.6% = 0.06, we obtain a base of b = +r =.06, giving the number of PCs to be modeled b c(.06) t. If we set t = 0 for the ear 0, we find that c = 50, so the number of sales is given b (t) = t. Since the ear 05 corresponds to t =, we can calculate the number of sales in the ear 05 as () = Approimatel 0 million PCs will be sold in the ear 05. Eample 5.7. The size of an ant colon is decreasing at a rate of % per month. How long will it take until the colon has reached 80% of its original size?

223 5.. APPLICATIONS OF EXPONENTIAL FUNCTIONS Solution. Since r = % = 0.0, we obtain the base b = +r = 0.0 = We have a colon size of (t) = c 0.99 t after t months, where c is the original size. We need to find t so that the size is at 80% of its original size c, that is, (t) = 80% c = 0.8 c. 0.8 c = c 0.99 t ( c) = 0.8 = 0.99 t = log(0.8) = log(0.99 t ) = log(0.8) = t log(0.99) = t = log(0.8) log(0.99). After approimatel. months, the ant colon has decreased to 80% of its original size. Eample 5.8. a) The population size of a countr is increasing at a rate of % per ear. How long does it take until the countr has doubled its population size? b) The number of new flu cases is decreasing at a rate of 5% per week. After how much time will the number of new flu cases reach a quarter of its current level? Solution. a) The rate of change isr = % = 0.0, so that population size is an eponential function with base b = +0.0 =.0. Therefore, f() = c.0 denotes the population size, with c being the initial population in the ear corresponding to = 0. In order for the population size to double, f() has to reach twice its initial size, that is: f() = c = c.0 = c ( c) =.0 = = log(.0 ) = log() = log(.0) = log() = = log() log(.0) 7.7 It will take about 7.7 ears until the population size has doubled. b) Since the number of new flu cases is decreasing, the rate of growth is negative, r = 5% = 0.05 per week, so that we have an eponential function with base b = +r = +( 0.05) = To reach a quarter of its initial number of flu cases, we set f() = c 0.95 equal to c. c 0.95 = c ( c) = 0.95 = = log(0.95) = log( )

224 SESSION 5. APPLICATIONS OF EXP AND LOG = = log( ) log(0.95) 7.0 It will therefore take about 7 weeks until the number of new flu cases has decreased to a quarter of its current level. 5. Eercises Eercise 5.. Assuming that f() = c b is an eponential function, find the constants c and b from the given conditions. a) f(0) =, f() = b) f(0) = 5, f() = 0 c) f(0) =,00, f(6) = 0.00 d) f() =, f(5) = 8 e) f( ) =, f() = 500 f) f() =, f() = 5 Eercise 5.. The number of downloads of a certain software application was 8. million in the ear 005 and.6 million in the ear 00. a) Assuming an eponential growth for the number of downloads, find the formula for the downloads depending on the ear t. b) Assuming the number of downloads will follow the formula from part (a), what will the number of downloads be in the ear 05? c) In which ear will the number of downloaded applications reach the 0 million barrier? Eercise 5.. The population size of a cit was 79,000 in the ear 990 and 6,000 in the ear 005. Assume that the population size follows an eponential function. a) Find the formula for the population size. b) What is the population size in the ear 00? c) What is the population size in the ear 05? d) When will the cit reach its epected maimum capacit of,000,000 residents? Eercise 5.. The population of a cit decreases at a rate of.% per ear. After how man ears will the population be at 90% of its current size? Round our answer to the nearest tenth.

225 5.. EXERCISES 5 Eercise 5.5. A big compan plans to epand its franchise and, with this, its number of emploees. For ta reasons it is most beneficial to epand the number of emploees at a rate of 5% per ear. If the compan currentl has,70 emploees, how man ears will it take until the compan has 6,000 emploees? Round our answer to the nearest hundredth. Eercise 5.6. An ant colon has a population size of,000 ants and is increasing at a rate of % per week. How long will it take until the ant population has doubled? Round our answer to the nearest tenth. Eercise 5.7. The size of a beehive is decreasing at a rate of 5% per month. How long will it take for the beehive to be at half of its current size? Round our answer to the nearest hundredth. Eercise 5.8. If the population size of the world is increasing at a rate of 0.5% per ear, how long does it take until the world population doubles in size? Round our answer to the nearest tenth.

226 Session 6 More applications: Half-life and compound interest We have alread encountered applications of the logarithm in solving equations in section 5.. In this chapter we solve more equations related to applications b using the logarithm. 6. Half-life Recall from definition 5. on page that a function with rate of growth r is an eponential function f() = c b with base b = + r. There is also another important wa of determining the base of an eponential function, which is given b the notion of half-life. We start with a motivating eample. Eample 6.. Consider the function f() = 00 () 7. We calculate the function values f(), for = 0, 7,,, and 8. ( f(0) = 00 ( f(7) = 00 ( f() = 00 f() = 00 )0 7 = 00 = 00 )7 7 = 00 = 00 ) 7 = 00 = 50 ( ) 6 7 = 00 8 = 5

227 6.. HALF-LIFE 7 f(8) = 00 ( )8 7 = 00 6 =.5 From this calculation, we can see how the function values of f behave: starting fromf(0) = 00, the function takes half of its value wheneveris increased b 7. For this reason, we sa that f has a half-life of 7. (The general definition will be given below.) The graph of the function is displaed below We collect the ideas that are displaed in the above eample in the definition and observation below. Definition 6.. Let f be an eponential function f() = c b with a domain of all real numbers, D = R. Then we sa that f has a half-life of h, if the base is given b ( ) h b = (6.) Note that we can also write h in terms of b. Converting (6.) into a logarithmic equation gives = h log (b) = logb, so that h = log log = log logb b( ). Observation 6.. Let f be the eponential function given for some real constants c > 0 and half-life h > 0, that is ( ( ) ) h f() = c = c ( ) h. Then we can calculate f(+h) as follows: ( )+h ( ) h h +h ) h h f(+h) = c = c = c ( +

228 8 SESSION 6. HALF-LIFE AND COMPOUND INTEREST = c ( ) h ( To summarize, f has the following propert: ) = f() f(+h) = f() for all R. (6.) The above equation shows that, whenever we add an amount of h to an input, the effect on f is that the function value decreases b half its previous value. This is also displaed in the graph below. c h h h h We will sometimes use a different letter for the input variable. In particular, the function f() = c ( ) h is the same as the function f(t) = c ( ) t h. Man radioactive isotopes deca with well-known half-lives. Eample 6.. a) Chromium-5 has a half-life of 7.7 das. How much of grams of chromium-5 will remain after 90 das? b) An isotope decas within 0 hours from 5 grams to.7 grams. Find the half-life of the isotope. ) t Solution. a) We use the above formula = c ( h, where c = grams is the initial amount of chromium-5, h = 7.7 das is the half-life of chromium-5, Half-lives are taken from: of isotopes b half-life

229 6.. HALF-LIFE 9 and t = 90 das is time that the isotope decaed. Substituting these numbers into the formula for, we obtain: ( ) = 0.6 Therefore, after 90 das, 0.6 grams of the chromium-5 is remaining. b) We have an initial amount of c = 5 grams and a remaining amount of =.7 grams after t = 0 hours. The half-life can be obtained as follows..7 = 5 ( )0 h ( 5) = 0. = ( )0 h (appl ln) = ln(0.) = ln ( ) h = ln(0.) = 0 h ln(0.5) h ( ln(0.) ) = h = 0 ln(0.5) ln(0.) = t 6.6 Therefore, the half-life of the isotope is approimatel 6.6 hours. Note 6.5. An important isotope is the radioisotope carbon-. It decas with a half-life of 570 ears with an accurac of ±0 ears. For definiteness we will take 570 ears as the half-life of carbon-. The half-life of carbon- is 570 ears. One can use the knowledge of the half-life of carbon- in dating organic materials via the so called carbon dating method. Carbon- is produced b a plant during the process of photosnthesis at a fied level until the plant dies. Therefore b measuring the remaining amount of carbon- in a dead plant one can determine the date when the plant died. Furthermore, since humans and animals consume plants, the same argument can be applied to determine their (approimate) dates of death. Eample 6.6. a) A dead tree trunk has 86% of its original carbon-. (Approimatel) how man ears ago did the tree die?

230 0 SESSION 6. HALF-LIFE AND COMPOUND INTEREST b) A dead animal at an archeological site has lost.% of its carbon-. When did the animal die? ) t Solution. a) Using the function = c ( h, where c is the amount of carbon- that was produced b the tree until it died, is the remaining amount to date, t is the time that has passed since the tree has died, and h is the half-life of carbon-. Since 86% of the carbon- is left, we have = 86% c. Substituting the half-life h = 570 of carbon-, we can solve for t. ( ) t ( ) t 570 ( c) c = c = 0.86 = (appl ln) = ln(0.86) = ln ( ) 0.5 t 570 = ln(0.86) = t 570 ln(0.5) ( 570 ln(0.5) = ) 570 ln(0.5) ln(0.86) = t = t 7 Therefore, the tree died approimatel 7 ears ago. b) Since.% of the carbon- is gone, 00%.% = 58.7% is ) t remaining. Using = c ( h with = 58.7% c and h = 570, we obtain ( ) t ( ) t 570 ( c) c = c = = The animal died 0 ears ago. 6. Compound interest (appl ln) = ln(0.587) = ln ( ) 0.5 t 570 = ln(0.587) = t 570 ln(0.5) ( 570 ln(0.5) = ) 570 ln(0.5) ln(0.587) = t = t 0 There is another interesting and important application of the eponential function given b calculating the interest on an investment. We start with the

231 6.. COMPOUND INTEREST following motivating eample. Eample 6.7. We invest an initial amount of P = $500 for ear at a rate of r = 6%. The initial amount P is also called the principal. After ear, we receive the principal P together with the interest r P generated from the principal. The final amount A after ear is therefore A = $500+6% $500 = $500 (+0.06) = $50. We change the setup of the previous eample b taking a quarterl compounding. This means, that instead of receiving interest on the principal once at the end of the ear, we receive the interest times within the ear (after each quarter). However, we now receive onl of the interest rate of 6%. We break down the amount received after each quarter. ( after first quarter: ) = ( after second quarter: (500.05) ) = ( after third quarter: ( ) ) = ( after fourth quarter: A = ( ) ) = = A Note that in the second quarter, we receive interest on the amount we had after the first quarter, and so on. So, in fact, we keep receiving interest on the interest of the interest, etc. For this reason, the final amount received after ear A = $50.68 is slightl higher when compounded quarterl than when compounded annuall (where A = $50.00). We make an even further variation from the above. Instead of investing mone for ear, we invest the principal for 0 ears at a quarterl compounding. We then receive interest ever quarter for a total of 0 = 0 quarters. ( after first quarter: ) =

232 SESSION 6. HALF-LIFE AND COMPOUND INTEREST ( after second quarter: (500.05) ) = ( after fortieth quarter: A = ( ) ) = = A We state our observations from the previous eample in the following general observation. Observation 6.8. A principal (=initial amount) P is invested for t ears at a rate r and compounded n times per ear. The final amount A is given b P = principal (=initial) amount ( A = P + n) r n t A = final amount where r = annual interest rate n = number of compoundings per ear t = number of ears We can consider performing the compounding in smaller and smaller increments. Instead of quarterl compounding, we ma take monthl compounding, or dail, hourl, secondl compounding or compounding in even smaller time intervals. Note, that in this case the number of compoundings n in the above formula tends to infinit. In the limit when the time intervals go to zero, we obtain what is called continuous compounding. Observation 6.9. A principal amount P is invested for t ears at a rate r and with continuous compounding. The final amount A is given b P = principal amount A = P e r t A = final amount where r = annual interest rate t = number of ears Note 6.0. The reason the eponential function appears in the above formula is that the eponential is the limit of the previous formula in observation 6.8, when n approaches infinit; compare this with equation (.) on page 87. ( e r = lim + r ) n n n A more detailed discussion of limits (including its definition) will be provided in a calculus course.

233 6.. COMPOUND INTEREST Eample 6.. Determine the final amount received on an investment under the given conditions. a) $700, compounded monthl, at %, for ears b) $500, compounded semi-annuall, at 5.5%, for 6 ears c) $00, compounded continuousl, at %, for ears Solution. We can immediatel use the formula b substituting the given values. For part (a), we have P = 700, n =, r = % = 0.0, and t =. Therefore, we calculate ( A = ) = 700 ( ) b) We have P = 500, n =, r = 5.5% = 0.055, and t = 6. A = 500 ( ) c) We have P = 00, r = % = 0.0, t =, and we use the formula for continuous compounding. A = 00 e 0.0 = 00 e (Note that the Euler number is entered in the TI-8 via the kes lpnd lp and lplnlp.)

234 SESSION 6. HALF-LIFE AND COMPOUND INTEREST Instead of asking to find the final amount, we ma also ask about an of the other variables in the above formulas for investments. Eample 6.. a) Find the amount P that needs to be invested at.75% compounded annuall for 5 ears to give a final amount of $000. (This amount P is also called the present value of the future amount of $000 in 5 ears.) b) At what rate do we have to invest$800 for6ears compounded quarterl to obtain a final amount of $00? c) For how long do we have to invest $000 at a rate of.5% compounded continuousl to obtain a final amount of $00? d) For how long do we have to invest at a rate of.% compounded monthl until the investment doubles its value? Solution. a) We have the following data: r =.75% = 0.075, n =, t = 5, and A = 000. We want to find the present value P. Substituting the given numbers into the appropriate formula, we can solve this for P. 000 = P ( ) 5 = 000 = P (.075) 5 (divide b ) = P = Therefore, if we invest $. toda under the given conditions, then this will be worth $000 in 5 ears. b) Substituting the given numbers (P = 800, t = 6, n =, A = 00) into the formula gives: 00 = 800 ( + r ) 6 (divide b 800) = = 00 (+ 800 = r ( + r ) = Net, we have to get the eponent to the right side. This is done b taking ( ) a power of, or in other words, b taking the th root, =. ) ( ( + r ) ) ( ) = = ( + r ) = ( )

235 6.. COMPOUND INTEREST 5 = + r = ( ) = r = ( ) ( ) ) = r = ( We plug this into the calculator. Note that the nth root is given b pressing lpmathlp and lp5 lp. More precisel, the th root of / (highlighted above) can be entered b pressing the following kes: lp lp lp lp lpmathlp lp5 lp lp lp lp lp lp lp lpenter lp Therefore our answer is r = 6.85%. c) Again we substitute the given values, P = 000, r =.5% = 0.05, A = 00, but now we use the formula for continuous compounding. 00 = 000 e 0.05 t = = e0.05 t = e 0.05 t =. To solve for the variable t in the eponent, we need to appl the logarithm. Here, it is most convenient to appl the natural logarithm, because ln() is inverse to e (which appears in the equation) and this logarithm is directl implemented on the calculator via the lplnlp ke. B appling ln to both sides, since ln() is inverse to e, we see that.05t = ln(.). Alternativel, ln(e 0.05 t ) = ln(.) = 0.05 t ln(e) = ln(.) Here we have used that log b ( n ) = n log b () for an number n as we have seen in proposition.. Observing that ln(e) =, which is the special case of the second equation in (.) on page 9 for the base b = e, the above becomes 0.05 t = ln(.) = t = ln(.) Therefore, we have to wait ears until the investment is worth (more than) $00.

236 6 SESSION 6. HALF-LIFE AND COMPOUND INTEREST d) We are given thatr =.% = 0.0 andn =, but no initial amountp is provided. We are seeking to find the time t when the investment doubles. This means that the final amount A is twice the initial amount P, or as a formula: A = P. Substituting this into the investment formula and solving gives the wanted answer. ( P = P ( ( (appl ln) = ln() = ln (divide b ln ( ) ) ) t ( (divide b P ) = = ) ) t = ln() = t ln ln() = t = ln ( ) ) t ( So, after approimatel.69 ears, the investment will have doubled in value. ) 6. Eercises Eercise 6.. An unstable element decas at a rate of 5.9% per minute. If 0mg of this element has been produced, then how long will it take until mg of the element are left? Round our answer to the nearest thousandth. Eercise 6.. A substance decas radioactivel with a half-life of.5 das. How much of 6.8 grams of this substance is left after ear? Eercise 6.. Fermium-5 decas in 0 minutes to 76.% of its original mass. Find the half-life of fermium-5. Eercise 6.. How long do ou have to wait until 5mg of berllium-7 have decaed to mg, if the half-life of berllium-7 is 5. das? Eercise 6.5. If Pharaoh Ramses II died in the ear BC, then what percent of the carbon- was left in the mumm of Ramses II in the ear 000? Eercise 6.6. In order to determine the age of a piece of wood, the amount of carbon- was measured. It was determined that the wood had lost.% of its carbon-. How old is this piece of wood?

237 6.. EXERCISES 7 Eercise 6.7. Archaeologists uncovered a bone at an ancient resting ground. It was determined that 6% of the carbon- was left in the bone. How old is the bone? Eercise 6.8. An investment of $5, 000 was locked in for 0 ears. According to the agreed conditions, the investment will be worth $5, t after t ears. a) How much is the investment worth after 5 ears? b) After how man ears will the investment be worth $0, 000? Eercise 6.9. Determine the final amount in a savings account under the given conditions. a) $700, compounded quarterl, at %, for 7 ears b) $00, compounded annuall, at.5%, for 5 ears c) $00, compounded continuousl, at.5%, for 5 ears d) $500, compounded monthl, at.99%, for ears e) $5000, compounded continuousl, at 7.%, for ears f) $600, compounded dail, at.%, for ear g) $750, compounded semi-annuall, at.9%, for ears Eercise 6.0. a) Find the amountp that needs to be invested at a rate of5% compounded quarterl for 6 ears to give a final amount of $000. b) Find the present value P of a future amount of A = $500 invested at 6% compounded annuall for ears. c) Find the present value P of a future amount of $000 invested at a rate of.9% compounded continuousl for 7 ears. d) At what rate do we have to invest$900 forears compounded monthl to obtain a final amount of $50? e) At what rate do we have to invest $00 for 0 ears compounded continuousl to obtain a final amount of $000? f) For how long do we have to invest$00 at a rate of5.5% compounded annuall to obtain a final amount of $700? g) For how long do we have to invest $000 at a rate of.5% compounded continuousl to obtain a final amount of $00?

238 8 SESSION 6. HALF-LIFE AND COMPOUND INTEREST h) How long do ou have to invest a principal at a rate of 6.75% compounded dail until the investment doubles its value? i) An certain amount of mone has tripled its value while being in a savings account that has an interest rate of 8% compounded continuousl. For how long was the mone in the savings account?

239 Review of eponential and logarithmic functions Eercise III.. The population of a countr grows eponentiall at a rate of % per ear. If the population was 5.7 million in the ear 00, then what is the population size of this countr in the ear 05? Eercise III.. A radioactive substance decas eponentiall at a rate of 7% per hour. How long do ou have to wait until the substance has decaed to of its original size? Eercise III.. Combine to an epression with onl one logarithm. a) ln()+ln(), b) log () log ()+log (z) Eercise III.. Assuming that, > 0, write the following epressions in terms of u = log() and v = log(): ( ) a) log, b) log ( ) ( 5 5, c) log ) Eercise III.5. Solve without using the calculator: log ()+log ( 8) = Eercise III.6. a) Find the eact solution of the equation: 6 + = 7 b) Use the calculator to approimate our solution from part (a). Eercise III.7. 5mg of fluorine-8 deca in hours to.mg. Find the half-life of fluorine-8. Eercise III.8. A bone has lost 5% of its carbon-. How old is the bone? 9

240 Eercise III.9. How much do ou have to invest toda at % compounded quarterl to obtain $,000 in return in ears? Eercise III.0. $500 is invested with continuous compounding. If $69.0 is returned after 5 ears, what is the interest rate? 0

241 Part IV Trigonometric functions

242 Session 7 Trigonometric functions 7. Review of basic trigonometric definitions and facts In this part, we will investigate trigonometric functions, such as = sin(), = cos(), and = tan() in terms of their function theoretic aspects. Before we graph these functions, we recall the basic definitions and the main facts, which we will consider as known background material. Definition 7.. An angle can be epressed in degree or in radian measure. The relationship between degrees and radians is given b π = 80 (7.) π = 5 5π 6 = 50 π = 0 π = 90 π = 60 π = 5 π 6 = 0 π = 80 0 = 0 7π 6 = 0 5π = 5 π = 0 π = 70 5π = 00 π 6 = 0 7π = 5

243 7.. BASIC TRIGONOMETRIC DEFINITIONS AND FACTS Definition 7.. Let be an angle. Consider the terminal side of the angle, and assume that the point P(a,b) is a point on the terminal side of. If r is the distance from P to the origin (0,0), then we define the sine, cosine, tangent, cosecant, secant, and cotangent as follows: terminal side of a +b = r P(a,b) b = r = a +b a r sin() = b r cos() = a r tan() = b a csc() = r b sec() = r a cot() = a b There are some immediate consequences from the above definition. csc() = sin() tan() = sin() cos() sec() = cos() cot() = cos() sin() = tan() If we take r = in the above we have the following unit circle definition of sin and cos: (cos(),sin()) r = sin() cos()

244 SESSION 7. TRIGONOMETRIC FUNCTIONS That is to sa, that the point where the terminal side of the angle intersects the unit circle is (cos(), sin()) (which can serve as a definition cos and sin as functions of ). For the following, we use the short hand notation: Definition 7.. Define sin α := (sinα), cos α := (cosα), and tan α := (tanα). From these definitions, we obtain that sin ()+cos () = sec () = +tan () (7.) since sin () + cos () = ( b ( r) + a ) = b +a = r =, and, similarl, r r r +tan () = + ( ) b a = a +b = r = ( r a a a) = sec (). Furthermore, we can see from the definition that sin(+π) = sin() sin( ) = sin() sin(π ) = sin() cos(+π) = cos() cos( ) = cos() cos(π ) = cos() (7.) For a point P(a,b) the -coordinate a is positive when P is in quadrant I and IV, whereas the -coordinate b is positive when P is in quadrant I and II. The length of the hpothenuse r is alwas positive. Thus the trigonometric functions are positive/negative according to the chart: Quadrant II Quadrant I sin() is positive cos() is negative tan() is negative Quadrant III sin() is positive cos() is positive tan() is positive Quadrant IV sin() is negative cos() is negative tan() is positive sin() is negative cos() is positive tan() is negative

245 7.. BASIC TRIGONOMETRIC DEFINITIONS AND FACTS 5 There are two basic triangles that are used to calculate eact values of the trigonometric functions. These two triangles are the and triangles, which can be described with specific side lengths These basic triangles ma be used to calculate the trigonometric functions of certain angles. Eample 7.. Find sin(), cos(), and tan() for the angles = 0, = 5, = 60, = 90, = 0. Solution. We draw the angle = 0 in the plane, and use the special triangle to find a point on the terminal side of the angle. From this, we then read off the trigonometric functions. Here are the function values for 0, 5, and 60. sin(0 ) = b r = cos(0 ) = a r = 0 tan(0 ) = b a = = sin(5 ) = b = r = cos(5 ) = a r = = 5 tan(5 ) = b a = =

246 6 SESSION 7. TRIGONOMETRIC FUNCTIONS sin(60 ) = b r = 60 cos(60 ) = a r = tan(60 ) = b a = = Here are the trigonometric functions for 90 and sin(90 ) = b = = r cos(90 ) = a = 0 = 0 r 0 tan(90 ) = b a = 0 is undefined sin(0 ) = b r = 0 = cos(0 ) = a r = = tan(0 ) = b a = 0 = 0 We collect the results of the previous eample in a table: 0 = 0 π 6 = 0 π = 5 π = 60 π = 90 sin() 0 cos() tan() 0 0 undef. Note 7.5. There is an eas wa to remember the function value given in the above table b writing the numerator of sin() as 0,,,, in increasing order, and for cos() in decreasing order: 0 = 0 π 6 = 0 π = 5 π = 60 π = 90 sin() 0 cos() 0

247 7.. BASIC TRIGONOMETRIC DEFINITIONS AND FACTS 7 The values of tan() are then determined from this b tan() = sin() cos() : 0 = 0 π 6 = 0 π = 5 π = 60 π = 90 tan() 0 = 0 = = = = is undef. 0 We can use the special and triangles to also find other values of the trigonometric functions. Eample 7.6. Find sin(), cos(), and tan() for the following angles. a) = 0, b) = 95, c) = π 9π, d) = 6 Solution. a) We graph the angle = 0, and identif a special triangle using the terminal side of. 0 P We identif a point P on the terminal side of. This point P has coordinates P(, ), which can be seen b considering one of two triangles in the plane (shaded above). The length of the line segment from P to the origin (0,0) is. Thus, a =,b =,r =. We get sin(0 ) =, cos(0 ) =, tan(0 ) = = b) The angle = 95 is greater than 60. However, the trigonometric functions are invariant under addition or subtraction of 60 : sin(±60 ) = sin() cos(±60 ) = cos() tan(±60 ) = tan() Since = 5, we have sin(95 ) = sin(5 ), and similarl for cos and tan. Graphing the terminal side of, and identifing a special

248 8 SESSION 7. TRIGONOMETRIC FUNCTIONS triangle, we find the coordinates of a point P on the terminal side with coordinates P(, ). P 5 Therefore, a =,b =,r =. We obtain sin(95 ) = =, cos(95 ) = =, tan(95 ) = = c) Converting π 6 into degrees, we have = π 6 = π 6 80 π = 0 = 0. Drawing the terminal side, we obtain the point P. 0 P The point P has coordinates P(, ). Therefore, a =,b =,r =, and ( π ) sin = ( π ) ( π ) 6, cos = 6, tan = = 6 d) Converting the angle = 9π = 9π 80 = 9 5 = 05, we see π that the trigonometric functions of are the same as of = 5,

249 7.. SIN, COS, AND TAN AS FUNCTIONS 9 and also of = 5. We can draw either 5 or 5 to find a point P on the terminal side of. 5 5 P The point P(, ) determines a =,b =,r =, and so ( 9π ) sin = = ( 9π ), cos = = ( 9π ) tan = =, 7. sin, cos, and tan as functions We now turn to function theoretic aspects of the trigonometric functions defined in the last section. In particular, we will be interested in understanding the graphs of the functions = sin(), = cos(), and = tan(). With an ee toward calculus, we will take the angles in radian measure. Eample 7.7. We graph the functions = sin(), = cos(), and = tan(). One wa to proceed is to calculate and collect various function values in a table and then graph them. However, this is quite elaborate, as the following table shows. sin() 0 cos() tan() 0 π π 0 6 π π 0 π π 5π 6 π undef. 0...

250 0 SESSION 7. TRIGONOMETRIC FUNCTIONS An easier approach is to use the TI-8 to graph each function. Before entering the function, we need to make sure that the calculator is in radian mode. For this, press the lpmodelp ke, and in case the third item is set on degree, switch to radian, and press lpenter lp. We ma now enter the function = sin() and stud its graph. From this, we can make some immediate observations (which can also be seen using definition 7.). First, the graph is bounded between and +, because b definition 7., we have sin() = b with r b r, so that r sin(), for all. We therefore change the window of the graph to displa s between and, to get a closer view. = sin() Second, we see that = sin() is a periodic function with period π, since the function doesn t change its value when adding 60 = π to its argument (and this is the smallest non-zero number with that propert): sin(+π) = sin()

251 7.. SIN, COS, AND TAN AS FUNCTIONS The graph of = sin() has the following specific values: = sin() π 5π π π π π π π π π 5π π } {{ } period π The graph of = sin() has a period of π, and an amplitude of. Similarl, we can graph the function = cos(). Since cos() for all, we graph it also with the zoomed window setting. = cos() We see that = cos() is also periodic with period π, that is cos(+π) = cos(), and = cos() also has an amplitude of, since cos(). = cos() π π π π 5π π π π π π π 5π } {{ } period π Indeed, the graph of = cos() is that of = sin() shifted to the left b π. The reason for this is that ( cos() = sin + π ) (7.)

252 SESSION 7. TRIGONOMETRIC FUNCTIONS Man other properties of sin and cos can be observed from the graph (or from the unit circle definition). For eample, recalling Observation 5. from page 7, we see that the sine function is an odd function, since the graph of = sin() is smmetric with respect to the origin. Similarl, the cosine function is an even function, since the graph of = cos() is smmetric with respect to the -ais. Algebraicall, Definition 5.9 from page 7 therefore shows that these functions satisf the following relations: sin( ) = sin() and cos( ) = cos() (7.5) Finall, we come to the graph of = tan(). Graphing this function in the standard window, we obtain: = tan() Zooming into this graph, we see that = tan() has vertical asmptotes = π π.6 and =.6. = tan() This is also supported b the fact that tan( π ) and tan( π ) are undefined. The graph of = tan() with some more specific function values is shown

253 7.. SIN, COS, AND TAN AS FUNCTIONS below. = tan() π 5π π π π π π π π π π 5π } {{ } period π We see that the tangent is periodic with a period of π: tan( + π) = tan() (7.6) There are vertical asmptotes: = π, π, π, π, 5π, 5π,..., or, in short asmptotes of = tan() : = n π, where n = ±,±,±5,... Furthermore, the tangent is an odd function, since it is smmetric with respect to the origin (see Observation 5.): tan( ) = tan() (7.7) Recall from section 5. how changing the formula of a function affects the graph of the function, such as: the graph of c f() (for c > 0) is the graph of f() stretched awa from the -ais b a factor c (or compressed when 0 < c < ) the graph of f(c ) (for c > 0) is the graph of f() compressed towards the -ais b a factor c (or stretched awa the -ais when 0 < c < ) graph of f() + c is the graph of f() shifted up b c (or down when c < 0)

254 SESSION 7. TRIGONOMETRIC FUNCTIONS graph of f(+c) is the graph of f() shifted to the left b c (or to the right when c < 0) With this we can graph some variations of the basic trigonometric functions. Eample 7.8. Graph the functions: f() = sin()+, g() = sin(), h() = sin(+), i() = sin(), j() = cos()+, k() = cos( π), l() = tan(+)+ Solution. The functions f, g, h, and i have graphs that are variations of the basic = sin() graph. The graph of f() = sin() +, shifts the graph of = sin() up b, whereas the graph of g() = sin() stretches = sin() awa from the -ais. f() = sin()+ π 5π π π π π π π π π 5π π g() = sin() The graph of h() = sin(+) shifts the graph of = sin() to the left b, and i() = sin() compresses = sin() towards the -ais. h() = sin(+) π π π π π i() = sin( ) π π π π π π

255 7.. SIN, COS, AND TAN AS FUNCTIONS 5 Net, j() = cos()+ has a graph of = cos() stretched b a factor awa from the -ais, and shifted up b. 5 j() = cos()+ π 5π π π π π 5π π π π π π For the graph of k() = cos( π) = cos( ( π )), we need to compress the graph of = cos() b a factor (we obtain the graph of the function = cos()) and then shift this b π to the right. k() = cos( π) π π π π We will eplore this case below in more generalit. In fact, whenever = cos(b +c) = cos(b ( c )), the graph of = cos() is shifted to the right b b c, and compressed b a factor b, so that it has a period of π. b b Finall, l() = tan(+)+ shifts the graph of = tan() up b and to the left b. π π π π

256 6 SESSION 7. TRIGONOMETRIC FUNCTIONS We collect some of the observations that were made in the above eamples in the following definition. Definition 7.9. Let f be one of the functions: f() = a sin(b +c) or f() = a cos(b +c) The number a is called the amplitude, the number π b is the period, and the number c is called the phase shift. b In phsical applications, the period is sometimes denoted b T = π and the frequenc is the reciprocal b f = T. Using the amplitude, period and phase-shift, we can draw the graph of, for eample, a function f() = a sin(b +c) over one period b shifting the graph of sin() b the phase-shift c to the right, then marking one full period b of length π b, and then drawing the basic sin() graph with an amplitude of a (or the reflected graph of sin() when a is negative). = a sin(b +c) a c b c + π b b a } {{ } period π b The zero(s), maima, and minima of the graph within the drawn period can be found b calculating the midpoint between ( c,0) and ( c + π,0) and b b b midpoints between these and the resulting midpoint. A similar graph can be obtained for f() = a cos(b +c) b replacing the sin() graph with the cos() graph over one period.

257 7.. SIN, COS, AND TAN AS FUNCTIONS 7 Eample 7.0. Find the amplitude, period, and phase-shift, and sketch the graph over one full period. a) f() = sin() b) f() = cos(π ) c) f() = cos( ) d) f() = cos(+π) e) f() = sin ( +) f) f() = sin( ) Solution. a) The amplitude is =, and since f() = sin( +0), it is b = and c = 0, so that the period is π = π and the phase-shift is 0 = 0. This is also supported b graphing the function with the calculator; in particular the graph repeats after one full period, which is at π.. = sin( ) π π π π b) For f() = cos(π ), the amplitude is =, the period is π π =, and the phase-shift is 0. Note, that the first coefficient is negative, so that the cos() has to be reflected about the -ais. One full period is graphed below with a solid line (and the continuation of the graph is dashed). This graph is again supported b the graph and table as given b the calculator. = cos(π ) c) The amplitude of f() = cos( ) is, the period is π, and the

258 8 SESSION 7. TRIGONOMETRIC FUNCTIONS phase-shift is ( ) =. The graph over one period is displaed below. = cos( ) +π + π + π + π The minimum of the function f in this period is given b taking the value in the middle of the interval [, + π], that is at + π. With this, the zeros are given b taking the value in the middle of the intervals [, + π] and [+π,+π], which are at + π and + π. d) The function f() = cos(+π) has amplitude, period π = π, and phase-shift π π. We ma thus draw the graph over one period from to π + π = π ; see graph on the left below. However, we ma also graph the function over another full period, such as [0,π] (see the graph on the right). = cos(+π) = cos(+π) π π π π π π π π e) The graph of f() = sin ( +) has an amplitude of =, a = π, and a phase shift of = 6. Using this information, and period of π confirming this with the calculator, we can draw one full period of the graph of f. = sin( +) 6 6+π 6+π 6+π 6+π

259 7.. EXERCISES 9 f) For f() = sin( ), we have an amplitude of, a period of π = π 0 and a phase-shift of = 0. Comparing this with eample (a) above, we see that we have the same data as for the function = sin() from part (a). However, in this case, we have to draw the sine function over one period moving to the left, which means that the graph has to start the period as a decreasing function. This can also be seen b observing that the sine function is an odd function, that is equation (7.5), so that f() = sin( ) = sin(). The graph is therefore drawn as follows. = sin( ) π π π π 7. Eercises Eercise 7.. Find sin(), cos(), and tan() for the following angles. a) = 0, b) = 90, c) = 50, d) = 5 e) = 050, f) = 80, g) = 5π, h) = 5π 6 i) = 0π, 5π j) =, π k) =, 6 5π l) = 8 Eercise 7.. Graph the function, and describe how the graph can be obtained from one of the basic graphs = sin(), = cos(), or = tan(). a) f() = sin()+ b) f() = cos( π) c) f() = tan() d) f() = 5 sin() e) f() = cos( ) f) f() = sin( ) 5

260 50 SESSION 7. TRIGONOMETRIC FUNCTIONS Eercise 7.. Identif the formulas with the graphs. f() = sin()+, g() = tan( ), h() = sin(), i() = cos(), j() = cos( π), k() = tan(), a) b) c) d) e) f) Eercise 7.. Find the formula of a function whose graph is the one displaed below. a) b) c) d) e) f)

261 7.. EXERCISES 5 Eercise 7.5. Find the amplitude, period, and phase-shift of the function. a) f() = 5sin(+) b) f() = sin(π 5) c) f() = 6sin() d) f() = cos(+ π) e) f() = 8cos( 6) f) f() = sin() g) f() = cos(+) h) f() = 7sin( π 6π ) i) f() = cos( ) 5 5 Eercise 7.6. Find the amplitude, period, and phase-shift of the function. Use this information to graph the function over a full period. Label all maima, minima, and zeros of the function. a) = 5cos() b) = sin(π) c) = sin( π ) d) = cos( π) e) = cos(π π) f) = 6cos( ) g) = cos(+π) h) = 7sin ( ) + π i) = 5cos(+ π) j) = sin(5 π) k) = cos(π ) l) = 7sin ( + ) π m) = cos( π) n) = sin ( ) π o) = cos( ) 6π

262 Session 8 Addition of angles and multiple angle formulas 8. Addition and subtraction of angles In the previous section we found eact values of the trigonometric functions for specific angles of 0, π, π, π plus possibl an multiple of π. Using these values, we can find man other values of trigonometric functions via the following 6 addition and subtraction of angles formulas, which we state now. Proposition 8.. For an angles α and β, we have sin(α+β) = sinαcosβ +cosαsinβ sin(α β) = sinαcosβ cosαsinβ cos(α+β) = cosαcosβ sinαsinβ cos(α β) = cosαcosβ +sinαsinβ tan(α+β) = tanα+tanβ tanαtanβ tan(α β) = tanα tanβ +tanαtanβ Proof. We start with the proof of the formulas for sin(α + β) and cos(α + β) when α and β are angles between 0 and π = 90. We prove the addition formulas (for α,β (0, π )) in a quite elementar wa, and then show that the addition formulas also hold for arbitrar angles α and β. 5

263 8.. ADDITION AND SUBTRACTION OF ANGLES 5 To find sin(α +β), consider the following setup. d f α b γ β α c γ e a Note, that there are verticall opposite angles, labelled b γ, which are therefore equal. These angles are angles in two right triangles, with the third angle being α. We therefore see that the angle α appears again as the angle among the sides b and f. With this, we can now calculate sin(α+β). sin(α +β) = opposite hpotenuse = e+f = e d d + f d = a d + f d = a c c d + f b b d = sin(α) cos(β) + cos(α) sin(β) The above figure displas the situation when α +β π. There is a similar figure for π < α + β < π. (We recommend as an eercise to draw the corresponding figure for the case of π < α+β < π.) Net, we prove the addition formula for cos(α + β). The following figure depicts the relevant objects. α b d k c β α g h We calculate cos(α+β) as follows. cos(α +β) = adjacent hpotenuse = g d = g +h h d d = g +h k d d = g +h c c d k b b d = cos(α) cos(β) sin(α) sin(β) Again, there is a corresponding figure when the angle α+β is greater than π. (We encourage the student to check the addition formula for this situation as well.) We therefore have proved the addition formulas for sin(α+β) and cos(α +β) when α and β are angles between0 and π, which we will now etend to all anglesαand β. First, note that the addition formulas are triviall true when α or β are 0. (Check this!) Now, b observing that sin() and cos() can be converted to each other via shifts of π, (or, alternativel, b using the identities (7.) and (7.)), we obtain, that sin(+ π ) = cos, cos(+ π ) = sin(),

264 5 SESSION 8. ADDITION OF ANGLES AND MULTIPLE ANGLES sin( π ) = cos, cos( π ) = sin(). With this, we etend the addition identities for α b ± π as follows: sin ((α+ π ) )+β sin ((α π ) )+β cos ((α+ π ) )+β cos ((α π ) )+β = sin(α+β + π ) = cos(α+β) = cos(α)cos(β) sin(α)sin(β) = sin(α+ π )cos(β)+cos(α+ π )sin(β), = sin(α+β π ) = cos(α+β) = cos(α)cos(β)+sin(α)sin(β) = sin(α π )cos(β)+cos(α π )sin(β), = cos(α+β + π ) = sin(α+β) = sin(α)cos(β) cos(α)sin(β) = cos(α+ π )cos(β) sin(α+ π )sin(β), = cos(α+β π ) = sin(α+β) = sin(α)cos(β)+cos(α)sin(β) = cos(α π )cos(β) sin(α π )sin(β). There are similar proofs to etend the identities for β. An induction argument shows the validit of the addition formulas for arbitrar angles α and β. The remaining formulas now follow via the use of trigonometric identities. tan(α +β) = sin(α +β) sinαcosβ +cosαsinβ = cos(α +β) cosαcosβ sinαsinβ = sinαcosβ+cosαsinβ cosαcosβ cosαcosβ sinαsinβ cosαcosβ sinα cosα = + sinβ cosβ sinα sinβ cosα cosβ This shows that tan(α + β) = tanα+tanβ. For the relations with α β, we use the fact that sin and tanαtanβ tan are odd functions, whereas cos is an even function, see identities (7.5) and (7.7). sin(α β) = sin(α+( β)) = sin(α)cos( β)+cos(α)sin( β) = sinαcosβ cosαsinβ, cos(α β) = cos(α+( β)) = cos(α)cos( β) sin(α)sin( β) = cosαcosβ +sinαsinβ, tan(α β) = tan(α +( β)) = tan(α)+tan( β) tan(α)tan( β) = tanα tanβ +tanαtanβ. This completes the proof of the proposition. Before giving eamples of the above proposition, we recall the elementar function values of the sin, cos, and tan from the previous section: 0 = 0 π 6 = 0 π = 5 π = 60 π = 90 π = 80 sin() 0 cos() tan() undef. 0

265 8.. ADDITION AND SUBTRACTION OF ANGLES 55 Eample 8.. Find the eact values of the trigonometric functions. a) cos ( ) π b) tan ( ) 5π c) cos ( ) π Solution. a) The ke is to realize the angle π as a sum or difference of angles with known trigonometric function values. Note, that π π = π π = π, so that ( π ( π cos = cos ) π ) = = cos π cos π +sin π sin π = + = + We remark that the last epression is in the simplest form and cannot be simplified an further. b) Again we can write the angle 5π as a sum involving onl special angles 5π given in the table above: = π + π = π + π. Therefore, 6 tan ( ) 5π ( π = tan 6 + π = ) tan π +tan π 6 tan π tan π 6 = + + = = = + + Here, we ma simplif the last epression further b rationalizing the denominator. This is done b multipling (+ ) to numerator and denominator. tan ( ) 5π = ( +) (+ ) ( ) (+ ) = = = +6 6 = + c) Again we need to write the angle π as a sum or difference of the above angles. In fact, we can do so in at least two different was: π = 6π+ 5π and also π = π π = π π. In either case, we first need to calculate some other trigonometric functions, namel those for either 5π or π. We choose the second solution, and using part (a), we have ( π + 6 cos =, and, )

266 56 SESSION 8. ADDITION OF ANGLES AND MULTIPLE ANGLES ( π sin ) ( π = sin π ) = = sin π cos π cos π sin π 6 6 = = With this, we have ( ) π ( cos = cos π π ) = cosπcos π +sinπsin π = ( ) +0 = ( + 6) Generalizing the previous eample, we can obtain an multiple of π as a sum or difference of known angles coming from the and triangles: Note 8.. Starting from the angles π 6 = π, π = π, π = π, and π = 6π, we obtain multiples of the angle π b addition and subtraction, such as: π = π π, 5π = π + π, 7π = π + π, 8π = π + π, 9π = 6π + π, 0π = 6π + π, and π = 6π + 5π = π π. Here, the trigonometric function values of the last fraction is obtained from the previousl obtained trigonometric values, as in part (c) of eample 8.. Higher multiples of can be obtained from the above list b π adding multiples of π to it. Note also that in man instances there are several was of writing an angle 8π as a sum or difference. For eample: = π + π = 6π + π = π π Using proposition 8. we can prove other related trigonometric identities with the addition and subtraction formulas. Eample 8.. We rewrite cos(+ π ) b using the subtraction formula. ( cos + π ) = cos cos π sin sin π = cos 0+sin = sin() 8. Double and half angles We can write formulas for the trigonometric functions of twice an angle and half an angle.

267 8.. DOUBLE AND HALF ANGLES 57 Proposition 8.5. Let α be an angle. Then we have the half-angle formulas: sin α cos α tan α cosα = ± +cosα = ± = cosα = sinα sinα cosα +cosα = ± +cosα Here, the signs ± are determined b the quadrant in which the angle α lies. (For more on the signs, see also page.) Furthermore, we have the double angle formulas: sin(α) = sinαcosα cos(α) = cos α sin α = sin α = cos α tanα tan(α) = tan α Proof. We start with the double angle formulas, which we prove using Proposition 8.. sin(α) = sin(α+α) = sinαcosα+cosαsinα = sinαcosα cos(α) = cos(α+α) = cosαcosα sinαsinα = cos α sin α tan(α) = tan(α +α) = tanα+tanα tanαtanα = tanα tan α Notice that cos(α) = cos α sin α can be rewritten using sin α+cos α = as follows: and cos α sin α = ( sin α) sin α = sin α cos α sin α = cos α ( cos α) = cos α This shows the double angle formulas. These formulas can now be used to prove the half-angle formulas. cos(α) = sin α = sin α = cos(α) = sin α = cos(α) cos(α) replace α b α = sinα = ± = sin α cosα = ± cos(α) = cos α = cos α = +cos(α) = cos α = +cos(α) +cos(α) replace α b α = cosα = ± = cos α +cosα = ± in particular: tan α = sin(α ) ± cosα cos( α ) = cosα = ± ± +cosα +cosα

268 58 SESSION 8. ADDITION OF ANGLES AND MULTIPLE ANGLES For the first two formulas for tan α we simplif sin(α) tan(α) and (+cos(α)) tan(α) as follows. sin(α) tan(α) = sinαcosα sinα cosα = sin α = cos(α) = tan(α) = cos(α) sin(α) (+cos(α)) tan(α) = cos α sinα = sinαcosα = sin(α) cosα This completes the proof of the proposition. = tan(α) = sin(α) +cos(α) replace α b α = tan( α ) = cos(α) sin(α) replace α b α = tan( α ) = sin(α) +cos(α) Here is an eample involving the half-angle identities. Eample 8.6. Find the trigonometric functions using the half-angle formulas. a) sin ( ) π 8 b) cos ( ) 9π 8 c) tan ( ) π Solution. a) Since π 8 = π, we use the half-angle formula with α = π. ( π sin 8) π ) cos π = sin( = ± = ± = ± = ±. = ± Since π = 80 =.5 is in the first quadrant, the sine is positive, so that 8 8 sin( π) = 8. b) Note that 9π = 9π 8. So we use α = 9π. Now, 9π = 9 80 = 0.5 is in 8 8 the third quadrant, so that the cosine is negative. We have: cos ( ) 9π 9π ) +cos = cos( 9π = 8 Now, cos( 9π) = cos(8π+π ) = cos(π + π) = cos(π) =, so that ( ) 9π cos = = = =. 8.

269 8.. DOUBLE AND HALF ANGLES 59 c) Note that π = π, and we alread calculated the trigonometric function values of α = π in eample 8.(c). So that we obtain: ( π tan ) π ) = tan( = cos π sin π = 6 = = = We can rationalize the denominator b multipling numerator and denominator b ( 6+ ). ( π tan = ) = 6+ 6 = = 6+ 8 = 6+. Although we used the first formula for tan α from the proposition, we could as well have used the other two formulas. Eample 8.7. Find the trigonometric functions of α when α has the properties below. a) sin(α) =, and α is in quadrant II 5 b) tan(α) =, and α is in quadrant III 5 Solution. a) From sin (α)+cos (α) =, we find that cos (α) = sin (α), and since α is in the second quadrant, cos(α) is negative, so that ( ) cos(α) = sin (α) = = = = 5 = 5, and tan(α) = sinα cosα = 5 5 = 5 5 =

270 60 SESSION 8. ADDITION OF ANGLES AND MULTIPLE ANGLES From this we can calculate the solution b plugging these values into the double angle formulas. sin(α) = sinαcosα = 5 ( ) = 5 5 ( ) ( cos(α) = cos (α) sin (α) = 5 5 tanα tan(α) = tan α = ( ) ( ) = = 9 6 ) = = 7 5 = 6 7 = 7 b) Similar to the calculation in part (a), we first calculate sin(α) and cos(α), which are both negative in the third quadrant. Recall from equation (7.) on page that sec α = +tan α, where secα =. Therefore, cosα ( ) sec α = + = = 5+ = = secα = ± 5 Since cos(α) is negative (in quadrant III), so is sec(α), so that we get, cosα = secα = 5 = 5 Furthermore, sin α = cos α, and sinα is negative (in quadrant III), we have sinα = ( cos α = 5 ) = = = = Thus, we obtain the solution as follows: sin(α) = sinαcosα = ( ) ( 5) = 0 69 ( 5 ) ( ) cos(α) = cos (α) sin 5 (α) = = tanα 5 tan(α) = tan = ) α = = = 5 ( = = 9 69

271 8.. EXERCISES 6 8. Eercises Eercise 8.. Find the trigonometric function values. a) sin ( ) 5π b) cos ( ) 5π c) tan ( ) π e) cos ( ) π f) sin ( ) π g) sin ( ) 5π 6 i) tan ( ) π j) cos ( ) π k) sin ( ) π d) sin ( ) 7π h) cos ( ) π l) sin ( ) 9π Eercise 8.. Simplif the function f using the addition and subtraction formulas. a) f() = sin ( ) + π b) f() = cos ( ) π c) f() = tan(π ) d) f() = sin ( π ) e) f() = cos ( ) + π f) f() = cos ( π ) 6 Eercise 8.. Find the eact values of the trigonometric functions applied to the given angles b using the half-angle formulas. a) cos ( ) π b) tan ( ) π c) sin ( ) π d) cos ( ) π 8 e) sin ( ) 9π f) tan ( ) π g) sin ( ) π h) sin ( ) π Eercise 8.. Find the eact values of the trigonometric functions of α and of α b using the half-angle and double angle formulas. a) sin(α) =, and α in quadrant I 5 b) cos(α) = 7, and α in quadrant IV c) sin(α) =, and α in quadrant III 5 d) tan(α) =, and α in quadrant III e) tan(α) = 5, and α in quadrant II f) cos(α) =, and α in quadrant II

272 Session 9 Inverse trigonometric functions 9. The functions sin, cos, and tan The inverse trigonometric functions are the inverse functions of the = sin, = cos, and = tan functions restricted to appropriate domains. In this section we give a precise definition of these functions. The function = tan () We start with the inverse to the tangent function = tan(). Recall that the graph of = tan() is the following: = tan() 5π π π π π π π π π 5π It has vertical asmptotes at = ± π,±π,±5π,... Note, that = tan() is not a one-to-one function in the sense of defintion 7. on page 86. (For 6

273 9.. THE FUNCTIONS SIN, COS, AND TAN 6 eample, the horizontal line = intersects the graph at = π, = π ±π, = π ± π, etc.) However, when we restrict the function to the domain D = ( π, π ) the restricted function is one-to-one, and, for this restricted function, we ma take its inverse function. Definition 9.. The inverse of the function = tan() with restricted domain D = ( π, π ) and range R = R is called the inverse tangent or arctangent function. It is denoted b ( = tan () or = arctan() tan() =, π, π ) The arctangent reverses the input and output of the tangent function, so that the arctangent has domain D = R and range R = ( π, π ). The graph is displaed below. = tan () = arctan() π π - Warning 9.. The notation of tan () and tan () is slightl inconsistent, since the eponentiation smbol is used above in two different was. In fact, tan () = arctan() refers to the inverse function of the tan() function. However, when we write tan (), we mean tan () = (tan()) = tan() tan() Therefore, tan () is the inverse function of tan() with respect to the composition operation, whereas tan () is the square with respect to the usual product in R. Note also that the inverse function of the tangent with respect to the product in R is = = cot(), which is the cotangent. tan() Observation 9.. The inverse tangent function is an odd function: tan ( ) = tan () (9.)

274 6 SESSION 9. INVERSE TRIGONOMETRIC FUNCTIONS This can be seen b observing that the tangent = tan() is an odd function (that is tan( ) = tan()), or directl from the smmetr of the graph with respect to the origin (0,0). The net eample calculates function values of the inverse tangent function. Eample 9.. Recall the eact values of the tangent function from section 7.: 0 = 0 π = 6 0 π = 5 π = 60 π = 90 tan() 0 undef. From this, we can deduce function values b reversing inputs and outputs, such as: ( π ( tan = = tan 6) ) = π ( 6 π tan = = tan ) ( ) = π Also, sincetan ( ) = tan (), we obtain the inverse tangent of negative numbers. tan ( ) = tan ( ) = π tan ( ) = tan () = π We ma calculate the inverse tangent of specific values with the calculator using the lpnd lp and lptanlp kes. For eample, tan (.).. Note, that the answer differs, when changing the mode from radians to degree, since tan (.)

275 9.. THE FUNCTIONS SIN, COS, AND TAN 65 The function = sin () Net, we define the inverse sine function. For this, we again first recall the graph of the = sin() function, and note that it is also not one-to-one. = sin() π π π π π π π π However, when restricting the sine to the domain [ π, π ], the restricted function [ is one-to-one. Note furthermore, that when restricting the domain to π, π ], the range is [,], and therefore we cannot etend this to a larger domain in a wa such that the function remains a one-to-one function. We use the domain [ π, π ] to define the inverse sine function. Definition 9.5. The inverse of the function = sin() with restricted domain D = [ π, π ] and range R = [,] is called the inverse sine or arcsine function. It is denoted b = sin () or = arcsin() sin() =, [ π, π ] The arcsine reverses the input and output of the sine function, so that the arcsine has domain D = [,] and range R = [ π, π ]. The graph of the arcsine is drawn below. = sin () = arcsin() π - π -

276 66 SESSION 9. INVERSE TRIGONOMETRIC FUNCTIONS Observation 9.6. The inverse sine function is odd: sin ( ) = sin () (9.) This can again be seen b observing that the sine = sin() is an odd function (that is sin( ) = sin()), or alternativel directl from the smmetr of the graph with respect to the origin (0,0). We now calculate specific function values of the inverse sine. Eample 9.7. We first recall the known values of the sine. 0 = 0 π 6 = 0 π = 5 π = 60 π = 90 sin() 0 These values together with the fact that the inverse sine is odd, that is sin ( ) = sin (), provides us with eamples of its function values. ( sin ) = π, sin () = π, ( ) ( ) sin (0) = 0, sin = sin = π 6. Note, that the domain of = sin () is D = [,], so that input numbers that are not in this interval give undefined outputs of the inverse sine: sin () is undefined Input values that are not in the above table ma be found with the calculator via the lpnd lp lpsinlp kes. We point out, that the output values depend on wether the calculator is set to radian or degree mode. (Recall that the mode ma be changed via the lpmodelp ke).

277 9.. THE FUNCTIONS SIN, COS, AND TAN 67 The function = cos () Finall, we define the inverse cosine. Recall the graph of = cos(), and note again that the function is not one-to-one. = cos() π π π π π π π π In this case, the wa to restrict the cosine to a one-to-one function is not as clear as in the previous cases for the sine and tangent. B convention, the cosine is restricted to the domain [0,π]. This provides a function that is one-to-one, which is used to define the inverse cosine. Definition 9.8. The inverse of the function = cos() with restricted domain D = [0,π] and range R = [,] is called the inverse cosine or arccosine function. It is denoted b = cos () or = arccos() cos() =, [0,π] The arccosine reverses the input and output of the cosine function, so that the arccosine has domain D = [,] and range R = [0,π]. The graph of the arccosine is drawn below. 5 = cos () = arccos() π π

278 68 SESSION 9. INVERSE TRIGONOMETRIC FUNCTIONS Observation 9.9. The inverse cosine function is neither even nor odd. That is, the function cos ( ) cannot be computed b simpl taking ±cos (). But it does have some smmetr given algebraicall b the more complicated relation cos ( ) = π cos () (9.) Proof. We can see that if we shift the graph down b π function with the rule cos () π is odd: the resulting function is odd. That is to sa the cos ( ) π = (cos () π ), which ields 9. upon distributing and adding π. Another, more formal approach is as follows. The bottom right relation of (7.) on page states, that we have the relation cos(π ) = cos() for all. Let, and denote b = cos (). That is is the number 0 π with cos() =. Then we have Appling cos to both sides gives: = cos() = cos(π ) (b equation (7.)) cos ( ) = cos (cos(π )) = π The last equalit follows, since cos and cos are inverse to each other, and 0 π, so that 0 π π are also in the range of the cos. Rewriting = cos () gives the wanted result: cos ( ) = π cos () This is the equation (9.) which we wanted to prove. We now calculate specific function values of the inverse cosine. Eample 9.0. We first recall the known values of the cosine. 0 = 0 π 6 = 0 π = 5 π = 60 π = 90 cos() 0 Here are some eamples for function values of the inverse cosine. ( cos ) = π 6, cos () = 0, cos (0) = π. Negative inputs to the arccosine can be calculated with equation (9.), that is cos ( ) = π cos (), or b going back to the unit circle definition. ( cos ) = π cos ( ) = π π = π π = π,

279 9.. EXERCISES 69 cos ( ) = π cos () = π 0 = π. Furthermore, the domain of = cos () isd = [,], so that input numbers not in this interval give undefined outputs of the inverse cosine. cos (7) is undefined Other input values can be obtained with the calculator b pressing the lpnd lp lpcoslp kes. For eample, we obtain the following function values (here using radian measure). 9. Eercises Eercise 9.. Graph the function with the calculator. Use both radian and degree mode to displa our graph. Zoom to an appropriate window for each mode to displa a graph which includes the main features of the graph. a) = sin () b) = cos () c) = tan () Eercise 9.. Find the eact value of the inverse trigonometric function. a) tan ( ) b) sin ( ) c) cos ( ) d) tan (0) e) cos ( ) f) cos ( ) g) sin ( ) h) tan ( ) i) cos ( ) j) sin ( ) k) sin ( ) l) tan ( ) Eercise 9.. Find the inverse trigonometric function value using the calculator. Approimate our answer to the nearest hundredth. For parts (a)-(f), write our answer in radian mode. a) cos (0.) b) sin ( 0.75) c) cos ( ) d) tan (00,000) e) tan ( ) f) cos ( ) For parts (g)-(l), write our answer in degree mode. g) cos (0.68) h) tan ( ) i) sin ( + 6 ) j) tan (00,000) k) cos ( ) l) tan (+ 6 )

280 Session 0 Trigonometric equations 0. Basic trigonometric equations In this section we solve equations such as: sin() = 0.5 We know that = π solves this equation. However, there are also other 6 solutions such as = 5π or = π that can easil be checked with the 6 6 calculator. Below, we will stud how to find all solutions of equations of the form sin() = c, cos() = c, and tan() = c. We start with equations involving the tangent. The equation tan() = c Eample 0.. Solve for : tan() = = tan() π π π π π π +π π +π 70

281 0.. BASIC TRIGONOMETRIC EQUATIONS 7 Solution. There is an obvious solution given b = tan ( ) = π, as we studied in the last section. However, we can look for all solutions of tan() = b studing the graph of the tangent function, that is, b finding all points where the graph of the = tan() intersects with the horizontal line =. Since the function = tan() is periodic with period π, we see that the other solutions of tan() = besides = π are π +π, π +π, π +π,..., and π π, π π, π π,... In general, we write the solution as = π +n π, where n = 0,±,±,±,... The graph also shows that these are indeed all solutions of tan() =. B the same argument we also get the general solution of tan() = c. Observation 0.. To solve tan() = c, we first determine one solution = tan (c). Then the general solution is given b = tan (c)+n π where n = 0,±,±,±,... (0.) Eample 0.. Solve for : a) tan() =, b) tan() =, c) tan() = 5., d) tan() =.7 Solution. a) First, we find tan () = π. The general solution is thus: = π +n π where n = 0,±,±,±,... b) First, we need to find tan ( ). Recall from equation (9.) that tan ( c) = tan (c), and recall further that tan () = π. With this we have tan ( ) = tan () = π The general solution of tan() = is therefore, = π +n π, where n = 0,±,±,... For parts (c) and (d), we do not have an eact solution, so that the solution can onl be approimated with the calculator. c) = tan (5.)+nπ.77+nπ, where n = 0,±,±,... d) = tan (.7)+nπ.07+nπ, where n = 0,±,±,...

282 7 SESSION 0. TRIGONOMETRIC EQUATIONS The equation cos() = c We start again with an eample. Eample 0.. Solve for : cos() = = cos() π π π π } {{ } one full period Solution. We have the obvious solution to the equation = cos ( ) = π. However, since cos( ) = cos(), there is another solution given b taking = π : ( cos π ) ( π = cos = ) Moreover, the = cos() function is periodic with period π, that is, we have cos(+π) = cos(). Thus, all of the following numbers are solutions of the equation cos() = : π..., π, π π, π, π +π, π and:..., π π, π π, π, π +π, π +π,..., +π,... From the graph we see that there are onl two solutions of cos() = within one period. Thus, the above list constitutes all solutions of the equation. With this observation, we ma write the general solution as: = π +n π, or = π +n π, where n = 0,±,±,±,... In short, we write this as: = ± π +n π with n = 0,±,±,±,... We generalize this eample as follows. Observation 0.5. To solve cos() = c, we first determine one solution = cos (c). Then the general solution is given b = cos (c)+n π, or = cos (c)+n π, where n = 0,±,±,±,...

283 0.. BASIC TRIGONOMETRIC EQUATIONS 7 In short, we write = ±cos (c)+n π where n = 0,±,±,±,... (0.) Eample 0.6. Solve for. a) cos() =, b) cos() = 0.6, c) cos() =, d) cos() = Solution. a) First, we need to find cos ( ). From equation (9.) we know that cos ( c) = π cos (c), so that: ) cos ( The solution is therefore, ( ) = π cos = π π = π π = π = ± π +nπ, where n = 0,±,±,±,... b) We calculate cos (0.6) 0.97 with the calculator. The general solution is therefore, = ±cos (0.6)+nπ ±0.97+nπ, where n = 0,±,±,±,... c) Since the cosine is alwas cos(), the cosine can never be. Therefore, there is no solution to the equation cos() =. This can also be seen from the graph which does not intersect with the horizontal line =. = cos() d) A special solution of cos() = is cos ( ) = π cos () = π 0 = π,

284 7 SESSION 0. TRIGONOMETRIC EQUATIONS so that the general solution is = ±π +nπ, where n = 0,±,±,±,... However, since π + π = +π, the solutions π + nπ and π + nπ (for n = 0,±,±,...) can be identified with each other, and there is onl one solution in each period. Thus, the general solution can be written as = π +nπ, where n = 0,±,±,±,... The graphs of = cos() and = confirm these considerations. = cos() π π π π The equation sin() = c Eample 0.7. Solve for : sin() = = sin() π π π } {{ } one full period Solution. First, we can find one obvious solution = sin ( ) = π. Furthermore, from the top right equation in (7.), we have that sin(π ) = sin(), so that another solution is given b π π: ( sin π π ) ( π = sin = ) (This can also be seen b going back to the unit circle definition.) These are all solutions within one period, as can be checked from the graph above. The

285 0.. BASIC TRIGONOMETRIC EQUATIONS 75 function = sin() is periodic with period π, so that adding n π for an n = 0,±,±,... gives all solutions of sin() =. This means that the general solution is: = π +n π, or = (π π )+n π, for n = 0,±,±,±,... We rewrite these solutions to obtain one single formula for the solutions. Note, that π π +n π = π +(n+) π. Therefore, all solutions are of the form = ± π +k π where, for even k = n, the sign in front of π is +, and, for odd k = n+, the sign in front of π is. This can be summarized as: = ( ) k π +k π Renaming the indeing variable k to the usual n, we obtain the final version for the solutions of sin() =. = ( ) n π +n π, where n = 0,±,±,±,... We have the following general statement. Observation 0.8. To solve sin() = c, we first determine one solution = sin (c). Then the general solution is given b = ( ) n sin (c)+n π where n = 0,±,±,±,... (0.) Eample 0.9. Solve for. a) sin() = b) sin() = c) sin() = 5 7 d) sin() = Solution. a) First, we calculate sin ( ) = π. The general solution is therefore, 6 = ( ) n π +n π, where n = 0,±,±,±,... 6

286 76 SESSION 0. TRIGONOMETRIC EQUATIONS b) First, we need to find sin ( ). From equation (9.) we know that sin ( c) = sin (c), so that sin ( ) = sin ( ) = π. We thus state 6 the general solution as ( = ( ) n π ) +n π = ( ) n π 6 6 +n π = ( )n+ π 6 +n π, where n = 0,±,±,±,... c) We do not have an eact value of sin ( 5 ), so that we either need to 7 leave it as this, or approimate this with the calculator sin ( 5) We get the solution: ( ) n ( 0.796)+n π = ( ) n n π, where n = 0,±,±,±,... d) We calculate sin () = π. The general solution can be written as = ( ) n π +n π, where n = 0,±,±,±,... (0.) Now, if we look at the graph, we see that each period onl has one solution. = sin() π π 5π Algebraicall, we can see this as follows. For an even number n, the solution = ( ) n π +n π coincides with the solution coming from the inde n+, that is: ( ) n π +n π = +π +n π, and ( ) n+ π +(n+) π = π +(n+) π = π +n π +π = + π +n π Therefore, we can write the solution more efficientl b removing the odd solutions (since the coincide with the even solutions), and state this as = ( ) n π +n π, where n = 0,±,±,...

287 0.. BASIC TRIGONOMETRIC EQUATIONS 77 Since ( ) n = + for even n, we can just write this as = π +n π, where n = 0,±,±,±,... (0.5) Writing the solutions as = π + n π as in (0.5) instead of the original = ( ) n π + n π from (0.) for n = 0,±,±,... certainl does not change the overall solution set. However, writing the solutions as in (0.5) is more efficient, since it does not repeat an of the solutions, and is therefore a simplified and preferred wa of presenting the solutions. Summar We summarize the different formulas we used to solve the basic trigonometric equations in the following table. Solve: sin() = c Solve: cos() = c Solve: tan() = c First, find one solution, First, find one solution, First, find one solution, that is: sin (c). Use: that is: cos (c). Use: that is: tan (c). Use: sin ( c) = sin (c) cos ( c) = π cos (c) tan ( c) = tan (c) The general solution is: The general solution is: The general solution is: = ( ) n sin (c)+nπ = ±cos (c)+nπ = tan (c)+nπ where n = 0,±,±,... where n = 0,±,±,... where n = 0,±,±,... Eample 0.0. Find the general solution of the equation, and state at least 5 distinct solutions. a) sin() = b) cos() = Solution. a) We alread calculated the general solution in eample 0.9 (b). It is = ( ) n+ π +n π, where n = 0,±,±,±,... 6 We simplif the solutions for n = 0,,,,. n = 0 : = ( ) 0+ π 6 +0 π = π 6 n = : = ( ) + π 6 + π = π π +6π +π = = 7π 6 6 6

288 78 SESSION 0. TRIGONOMETRIC EQUATIONS n = : = ( ) + π 6 +( ) π = π π 6π π = 6 6 n = : = ( ) + π 6 + π = π π +π +π = 6 6 = 5π 6 = π 6 n = : = ( ) + π 6 +( ) π = π π π π = 6 6 = π 6 b) It is cos ( ) = π cos ( ) = π π = 6π π = 5π. The solutions of cos() = are: = ± 5π 6 +nπ, where n = 0,±,±,±,... We write the 6 solutions with n = 0,+,, and for each use the two distinct first terms + 5π 6 and 5π 6. n = 0 : = + 5π π = 5π 6 n = 0 : = 5π π = 5π 6 n = : = + 5π 6 + π = 5π 6 n = : = 5π 6 + π = 5π 6 n = : = + 5π 6 + ( ) π = 5π 6 n = : = 5π 6 + ( ) π = 5π 6 +π = 5π +π 6 +π = 5π +π 6 π = 5π π 6 = 7π 6 π = 5π π 6 = 7π 6 = 7π 6 = 7π 6 Further solutions can be found b taking values n = +,,+,, Equations involving trigonometric functions The previous section showed how to solve the basic trigonometric equations sin() = c, cos() = c, and tan() = c. The net eamples can be reduced to these basic equations. Eample 0.. Solve for. a) sin() = 0 b) sec() = c) 7cot()+ = 0

289 0.. EQUATIONS INVOLVING TRIGONOMETRIC FUNCTIONS 79 Solution. a) Solving for sin(), we get sin() = 0 (+) = sin() = ( ) = sin() = One solution of sin() = is sin ( ) = π. The general solution is 6 = ( ) n π +nπ, where n = 0,±,±,... 6 b) Recall that sec() = cos(). Therefore, sec() = = cos() = (reciprocal) = cos() = = A special solution of cos() = is cos ( The general solution is ) = π cos ( ) = π π = π π = π. = ± π +nπ, where n = 0,±,±,... c) Recall that cot() = tan(). So 7cot()+ = 0 ( ) = 7cot() = = ( 7) cot() = 7 = tan() = 7 (reciprocal) = tan() = 7 The solution is ( = tan 7 ) +nπ.66+nπ, where n = 0,±,±,... For some of the more advanced problems it can be helpful to substitute u for a trigonometric epression first, then solve for u, and finall appl the rules from the previous section to solve for the wanted variable. This method is used in the net eample.

290 80 SESSION 0. TRIGONOMETRIC EQUATIONS Eample 0.. Solve for. a) tan ()+tan()+ = 0 b) cos () = 0 Solution. a) Substituting u = tan(), we have to solve the equation u +u+ = 0 (factor) = (u+)(u+) = 0 = u+ = 0 ( ) = u = Resubstituting u = tan(), we have to solve tan() =. Using the fact that tan ( ) = tan () = π, we have the general solution = π +nπ, where n = 0,±,±,... b) We substitute u = cos(), then we have u = 0 (+) = u = ( ) = u = = u = ± = ± = ± = u = + or u = For each of the two cases we need to solve the corresponding equation after replacing u = cos(). with cos() = cos() = ( ) ( ) ( ) cos = π with cos = π cos = π π = π = = ± π +nπ = = ±π +nπ where n = 0,±,±,... where n = 0,±,±,... Thus, the general solution is = ± π +nπ, or = ±π +nπ, where n = 0,±,±,...

291 0.. EQUATIONS INVOLVING TRIGONOMETRIC FUNCTIONS 8 Eample 0.. Solve the equation with the calculator. solution to the nearest thousandth. Approimate the a) sin() = cos()+ b) 5cos() = tan() Solution. a) We rewrite the equation as sin() cos() = 0, and use the calculator to find the graph of the function f() = sin() cos(). The zeros of the functionf are the solutions of the initial equation. The graph that we obtain is displaed below. The graph indicates that the functionf() = sin() cos() is periodic. This can be confirmed b observing that both sin() and cos() are periodic with period π, and thus also f(). f(+π) = sin(+π) cos(+π) = sin() cos() = f() The solution of f() = 0 can be obtained b finding the zeros, that is b pressing lpnd lp lptrace lp lp lp, then choosing a left- and right-bound, and making a guess for the zero. Repeating this procedure gives the following two approimate solutions within one period. The general solution is thus.8+nπ or.5+nπ, where n = 0,±,±,... b) We rewrite the equation as 5cos() tan() = 0 and graph the

292 8 SESSION 0. TRIGONOMETRIC EQUATIONS function f() = 5cos() tan() in the standard window. To get a better view of the function we zoom to a more appropriate window. Note again that the functionf is periodic. The period ofcos() is π = π (see definition 7.9 on page 6), and the period of tan() is also π (see equation (7.6) on page ). Thus, f is also periodic with period π. The solutions in one period are approimated b finding the zeros with the calculator. The general solution is given b an of these numbers, with possibl an additional shift b an multiple of π..788+nπ or.+nπ or.8+nπ, where n = 0,±,±,±,...

293 0.. EXERCISES 8 0. Eercises Eercise 0.. Find all solutions of the equation, and simplif as much as possible. Do not approimate the solution. a) tan() = b) sin() = c) sin() = d) cos() = e) cos() = 0 f) cos() = 0.5 g) cos() = h) sin() = 5 i) sin() = 0 j) sin() = k) tan() = l) cos() = 0. Eercise 0.. Find all solutions of the equation. Approimate our solution with the calculator. a) tan() = 6. b) cos() = 0.5 c) sin() = 0.9 d) cos() =.77 e) tan() = 0. f) sin() = 0.06 Eercise 0.. Find at least 5 distinct solutions of the equation. a) tan() = b) cos() = c) sin() = d) tan() = 0 e) cos() = 0 f) cos() = 0. g) sin() = 0. h) sin() = Eercise 0.. Solve for. State the general solution without approimation. a) tan() = 0 b) sin() = c) cos() = 0 d) sec() = e) cot() = f) tan () = 0 g) sin () = 0 h) cos ()+7cos()+6 = 0 i) cos () cos()+ = 0 j) sin ()+sin() = 5 k) sin ()+sin() = 0 l) cos () cos()+ = 0 m) cos ()+9cos() = 5 n) tan () tan() = 0 Eercise 0.5. Use the calculator to find all solutions of the given equation. Approimate the answer to the nearest thousandth. a) cos() = sin()+ b) 7tan() cos() = c) cos ()+cos() = sin()+ d) sin()+tan() = cos()

294 Review of trigonometric functions Eercise IV.. Fill in all the trigonometric function values in the table below. sin() cos() tan() 0 π 6 π π π Eercise IV.. Find the eact values of a) cos ( ) π b) sin ( ) π 6 c) tan ( π ) Eercise IV.. Find the eact value of the trigonometric function. a) sin ( ) 5π b) cos ( ) π 6 [Hint: Use the special or triangles to find the solution.] Eercise IV.. Find the amplitude, period, and the phase shift of the given function. Draw the graph over a one-period interval. Label all maima, minima and intercepts. a) = cos ( π ) b) = 5sin ( ) + π Eercise IV.5. Find the eact trigonometric function value. a) cos ( ) π [Hint: Use the addition and subtraction of angles formulas.] b) cos ( ) π [Hint: Use the half-angles formulas.] 8 Eercise IV.6. Let sin(α) = 5 cos(α), and tan(α). and let α be in quadrant III. Find sin(α), 8

295 Eercise IV.7. Find the eact value of: a) sin ( ) b) cos ( ) c) tan ( ) Eercise IV.8. Solve for : sin()+ = 0 Eercise IV.9. Solve for : tan () = 0 Eercise IV.0. Solve for. a) cos () = 0 b) sin ()+5sin()+7 = 0 85

296 Part V Comple numbers, sequences, and the binomial theorem 86

297 Session Comple numbers. Polar form of comple numbers We now recall the definition of comple numbers and show how the are represented on the comple plane. Definition.. We define the imaginar unit or comple unit to be imaginar unit: i = (.) In other words, i is a solution of the equation: i = (.) Definition.. A comple number is a number of the form a+b i. (Alternativel we also write this as a + bi without the dot for the multiplication.) Here a and b are an real numbers. The number a is called the real part of a+b i, and b is called the imaginar part of a+b i. The set of all comple numbers is denoted b C. Here are some eamples of comple numbers: +i, i, +π i, 5+0 i, 0+ i Note that in the last two eamples, we have i = 5, so that we see that the real number 5 is also a comple number. Indeed, an real number a = a+0 i is also a comple number. Similarl, 0+ i = i as well as an 87

298 88 SESSION. COMPLEX NUMBERS multiple of i is also a comple number (these numbers are often called pure imaginar numbers). In analog to section., where we represented the real numbers on the number line, we can represent the comple numbers on the comple plane: Im 0 Re The comple number a+bi is represented as the point with coordinates (a,b) in the comple plane. Im +i +i 0 Re i 5 i - - Eample.. Perform the operation. a) ( i)+( 6+i) b) (+5i) ( 7+i) c) 5+i +i Solution. a) Adding real and imaginar parts, respectivel, gives, ( i)+( 6+i) = i 6+i = +i. b) We multipl (using FOIL), and use that i =. (+5i) ( 7+i) = +i 5i+5i = i+5 ( )

299 .. POLAR FORM OF COMPLEX NUMBERS 89 = i 5 = 6 i c) Recall that we ma simplif a quotient of comple numbers b multipling numerator and denominator b the conjugate of the denominator. 5+i +i = (5+i) ( i) (+i) ( i) = 5 0i+i 8i 9 6i+6i i = 5+i+8 9+ = +i = + i The real part of the solution is ; the imaginar part is. There is also a notion of absolute value a+bi for an comple number a+bi. Definition.. Let a+bi be a comple number. The absolute value of a+bi, denoted b a+bi is the length between the point a+bi in the comple plane and the origin (0,0). bi Im a+bi a+bi a Re Using the Pthagorean theorem, we can calculate a+bi as a+bi = a +b, so that a+bi = a +b (.) Eample.5. Find the absolute value of the comple numbers below. a) 5 i b) 8+6i c) 7i

300 90 SESSION. COMPLEX NUMBERS Solution. The absolute values are calculated as follows. a) 5 i = 5 +( ) = 5+9 = b) 8+6i = ( 8) +6 = 6+6 = 00 = 0 c) 7i = 0 +(7) = 0+9 = 7 We can write a comple number in the form a + bi, or alternativel, we ma use the absolute value and angle to write the number in the so-called polar form, which we discuss now. Observation.6. Let a + bi be a comple number. The coordinates in the plane can be epressed in terms of the absolute value r = a + bi and the angle θ with the real ais as shown below. bi Im a+bi r r sin(θ) θ r cos(θ) a Re From the right triangle as shown in the comple plane above, we see that the coordinates a and b in the plane are given b: a = r cos(θ) and b = r sin(θ) (.) Therefore, the comple number is a+bi = r cos(θ)+r sin(θ) i, or factoring r from this epression: a+bi = r (cos(θ)+i sin(θ) ) (.5) Writing the comple number in this wa is called the polar form of the comple number. Here, the number r is the absolute value, and θ is given b (.) via the calculation b = r sin(θ) = sin(θ) = tan(θ). Thus, a r cos(θ) cos(θ) r = a +b and tan(θ) = b a (.6)

301 .. POLAR FORM OF COMPLEX NUMBERS 9 When finding the directional angle θ of a comple number via equation (.6), we have to remember that tan(θ) ma be obtained b the two angles θ and π +θ; that is, we have tan(π +θ) = tan(θ). Therefore, we need to check that our answer for θ lies in the correct quadrant. This is shown in the net eample. Eample.7. Convert the comple number to polar form. a) +i, b) i, c) i, d) i Solution. a) First, the absolute value is r = + i = + =. Furthermore, since a = and b =, we have tan(θ) =. To obtain θ, we calculate ( ) tan 56. Note that 56. is in the first quadrant, and so is the comple number +i Im +i Re Therefore, θ 56., and we obtain our answer: +i ( ) cos(56. )+isin(56. ). b) For i, we first calculate the absolute value: r = ( ) +( ) = + = + = 6 =. Furthermore, tan(θ) = b a = =. Now, tan ( ) = 60 = π. However, graphing the angle π = 60 and the number i, we see that 60

302 9 SESSION. COMPLEX NUMBERS is in the first quadrant, whereas i is in the third quadrant. Im 60 = π 0 Re i - - Therefore, we have to add π to π to get the correct angle for i, that is θ = π π+π +π = = π. Our comple number in polar form is ( i = cos ( π ) (π) ) +isin. c) For i we calculate r = +( ) = 6+9 = 5 = 5. The angle tan ( ) 6.9, which is in the fourth quadrant, just like the number i. 0 Re Im - i - Therefore, θ 6.9, and we write ( ) i 5 cos( 6.9 )+isin( 6.9 ) d) We calculate the absolute value of 0 i as r = 0 +( ) = 6 =. However, when calculating the angle θ of 0 i, we are lead to consider tan ( ), which is undefined! The reason for this can be seen b plotting 0

303 .. POLAR FORM OF COMPLEX NUMBERS 9 the number i in the comple plane. Im 70 = π 0 Re i -5-6 The angle θ = 70 (or alternativel θ = 90 ), so that the comple number is i = (cos(70 )+isin(70 )) ( = cos ( π) (π) ) +isin Note that we ma write our answer either in degree or radian mode as we did above. We ma also perform the reverse conversion of a comple number from polar form into standard form a+bi. Eample.8. Convert the number from polar form into the standard form a+bi. a) (cos(7 )+isin(7 )) b) (cos( 5π )+isin(5π )) Solution. a) Since we don t have an eact formula for cos(7 ) or sin(7 ), we use the calculator to obtain approimate values. (cos(7 )+isin(7 )) ( 0.5+i 0.89) =.6+.67i b) We recall that cos( 5π) = and sin( 5π) =. (This can be seen as in Eample 7.6 on page 7 b considering the point P(, ) on the

304 9 SESSION. COMPLEX NUMBERS terminal side of the angle 5π = 5 80 = 5. 5π = 5 P Therefore, cos( 5π ) = = and sin( 5π ) = =.) With this, we obtain the comple number in standard form. (cos( 5π ) )+isin(5π ) = = ( i i ) = i. Multiplication and division of comple numbers in polar form One interesting feature of the polar form of a comple number is that the multiplication and division are ver eas to perform. Proposition.9. Let r (cos(θ )+isin(θ )) and r (cos(θ )+isin(θ )) be two comple numbers in polar form. Then, the product and quotient of these are given b r (cos(θ )+isin(θ )) r (cos(θ )+isin(θ )) = r r (cos(θ +θ )+isin(θ +θ )) (.7) r (cos(θ )+isin(θ )) r (cos(θ )+isin(θ )) = r r (cos(θ θ )+isin(θ θ )) (.8)

305 .. MULTIPLICATION AND DIVISION OF COMPLEX NUMBERS 95 Proof. The proof uses the addition formulas for trigonometric functions sin(α + β) and cos(α + β) from proposition 8. on page 5. r (cos(θ )+isin(θ )) r (cos(θ )+isin(θ )) = r r (cos(θ )cos(θ )+icos(θ )sin(θ )+isin(θ )cos(θ )+i sin(θ )sin(θ )) = r r ((cos(θ )cos(θ ) sin(θ )sin(θ ))+i(cos(θ )sin(θ )+sin(θ )cos(θ ))) = r r (cos(θ +θ )+isin(θ +θ )) For the division formula, note, that the multiplication formula (.7) gives r (cos(θ )+isin(θ )) so that = (cos( θ )+isin( θ )) = r (cos(θ θ )+isin(θ θ )) r r r (cos(θ )+isin(θ )) = (cos0+isin0) = ( +i 0) = = r (cos( θ )+isin( θ )), r (cos(θ )+isin(θ )) r (cos(θ )+isin(θ )) = r (cos(θ )+isin(θ )) r (cos(θ )+isin(θ )) = r (cos(θ )+isin(θ )) (cos( θ )+isin( θ )) = r (cos(θ θ )+isin(θ θ )). r r Eample.0. Multipl or divide the comple numbers, and write our answer in polar and standard form. a) 5(cos( )+isin( )) 8(cos( )+isin( )) b) (cos( 5π 8 )+isin(5π 8 )) (cos(7π 8 )+isin(7π 8 )) c) (cos( π )+isin(π)) 8(cos( 7π )+isin(7π)) d) (cos(0 )+isin(0 )) 6(cos(7 )+isin(7 )) Solution. We will multipl and divide the comple numbers using equations (.7) and (.8), respectivel, and then convert them to standard notation a+bi. a) 5(cos( )+isin( )) 8(cos( )+isin( )) = 5 8 (cos( + )+isin( + )) = 0(cos(5 )+isin(5 )) ( ) = 0 +i = 0 +i 0 = 0 +0 i b) Similarl, we obtain the net product.

306 96 SESSION. COMPLEX NUMBERS (cos( 5π ) 8 )+isin(5π 8 ) (cos( 7π ) 8 )+isin(7π 8 ) = 6 (cos( 5π 8 + 7π 8 )+isin(5π 8 + 7π ) 8 ) 5π Now, 8 + 7π 5π +7π = = π = π, and cos(π) = 0 and sin(π) =. Therefore, we obtain that the product is 6 (cos( π ) )+isin(π ) = 6(0+i ( )) = 6i c) For the quotient, we use the subtraction formula (.8). (cos( π )+isin(π )) 8(cos( 7π )+isin(7π = )) 8 ( cos( π 7π )+isin(π 7π ) ) Now, the difference in the argument of cos and sin is given b π 7π = π 7π = π = π, and cos( π) = cos(π) = and sin( π) = sin(π) =. With this, we obtain (cos( π )+isin(π)) ( 8(cos( 7π )+isin(7π)) = cos( π ) )+isin( π ) ( ) = i = i d) Finall, we calculate (cos(0 )+isin(0 )) 6(cos(7 )+isin(7 )) = ( ) cos(9 )+isin(9 ). Since we do not have eact values of cos and sin for the angle 9, we approimate the comple number in standard form with the calculator. ( ) cos(9 )+isin(9 ) = cos(9 )+i sin(9 ) i Note that here we approimated the solution to the nearest thousandth.

307 .. EXERCISES 97. Eercises Eercise.. Plot the comple numbers in the comple plane. a) +i b) 5i c) 6 i d) 5+i e) i f) i g) 7 h) i i) 0 j) i Eercise.. Add, subtract, multipl, and divide, as indicated. a) (5 i)+( +6i) b) ( 9 i) (5 i) c) (+i) (+i) d) ( i) ( +i) e) +i +i f) (5+5i) ( i) Eercise.. Find the absolute value a+bi of the given comple number, and simplif our answer as much as possible. a) +i b) i c) i d) 6i e) 8 i f) i g) 5 h) 7+ i Eercise.. Convert the comple number into polar form r(cos(θ)+i sin(θ)). a) +i b) +i c) i d) 5+5i e) i f) +i g) 5 5i h) 7 i i) 5 i j) 6i k) 0 l) +i Eercise.5. Convert the comple number into the standard form a+bi. a) 6(cos( )+isin( )) b) (cos( π 7 )+isin( π 7 )) c) (cos(70 )+isin(70 )) d) cos( π 6 )+isin(π 6 ) e) 0(cos( 7π 6 )+isin(7π 6 )) f) 6(cos( 5π )+isin( 5π )) Eercise.6. Multipl the comple numbers and write the answer in standard form a+bi. a) (cos(7 )+isin(7 )) 0(cos( )+isin( )) b) 7(cos( π 9 )+isin(π 9 )) 6(cos(π 9 )+isin(π 9 )) c) (cos( π )+isin(π )) (cos( π )+isin( π )) d) 8(cos( π 7 )+isin(π 7 )).5(cos(π 7 )+isin(π 7 )) e) 0.(cos(96 )+isin(96 )) 0.5(cos(88 )+isin(88 )) f) (cos( 7π 8 )+isin(7π 8 )) 0.5(cos( 5π )+isin( 5π ))

308 98 SESSION. COMPLEX NUMBERS Eercise.7. Divide the comple numbers and write the answer in standard form a+bi. a) c) e) 8(cos( π )+isin(π)) (cos( π 6 )+isin(π)) b) 6 (cos( π )+isin(π)) 6(cos( π 7 )+isin(π)) d) 7 (cos( 7π )+isin(7π)) 7(cos( 5π )+isin(5π)) f) 0(cos(5 )+isin(5 )) 5(cos(6 )+isin(6 )) cos( 8π 5 )+isin(8π 5 ) (cos( π 0 )+isin( π 0 )) 0(cos( 75 )+isin( 75 )) 8(cos( )+isin( ))

309 Session Vectors in the plane. Introduction to vectors There is et another wa to discuss the -dimensional plane, which highlights some other useful properties of the plane R. This is the notion of R as a vector space. We start b defining vectors in the plane. Definition.. A geometric vector in the plane is a geometric object in the plane R that is given b a direction and magnitude. We denote a vector b v (it is written b some authors as v), its magnitude is denoted b v, and its directional angle b θ. Vectors are often represented b directed line segments v = PQ, where two line segments represent the same vector, if one can be moved to the other b parallel translation (without changing its direction or magnitute). 5 P v R v O v Q v 99

310 00 SESSION. VECTORS IN THE PLANE In particular, we can alwas represent a vector v b OR b arranging the starting point of v to the origin O(0,0). If R is given in coordinates b R(a,b), then we also write for v = OR, v = a, b or, alternativel, v = [ ] a. (.) b Eample.. Graph the vectors v, w, r, s, t in the plane, where v = PQ with P(6,) and Q(, ), and w =,, r =,, s = 0,, t = 5,. Solution. P t v s w r - - Q The formulas for the magnitude and the directional angle of a vector can be obtained precisel the same wa as the absolute value and angle of a comple number. From equation (.6) in Observation.6, we therefore obtain the following analogue formulas. Observation.. Let v = a,b be a vector in the plane R. magnitude and angle of v are given b: Then the v = a +b and tan(θ) = b a (.)

311 .. INTRODUCTION TO VECTORS 0 b v R(a, b) θ a Conversel, we can recover the coordinates of a vector v from its magnitude v and angle θ b (see equation (.) from page 90): v = v cos(θ), v sin(θ) (.) Eample.. Find the magnitude and directional angle of the given vectors. a) 6,6 b), c), d) 8, 5 e) PQ, where P(9,) and Q(,0) Solution. a) We use formulas (.), and the calculation is in analog with Eample.7. The magnitude of v = 6,6 is v = ( 6) +6 = 6+6 = 7 = 6 = 6. The directional angleθ is given btan(θ) = 6 6 =. Now, sincetan ( ) = tan () = 5 is in the fourth quadrant, but v = 6,6 drawn at the origin O(0,0) has its endpoint in the second quadrant, we see that the angle θ = = b) The magnitude of v =, is v = +( ) = 6+9 = 5 = 5.

312 0 SESSION. VECTORS IN THE PLANE The directional angle is given b tan(θ) =. Since tan ( ) 6.9 is in the fourth quadrant, and v =, is in the fourth quadrant, we see that ( ) θ = tan 6.9. c) The magnitude of v =, is v = ( ) +( ) = + = + = 6 =. The directional angle is given btan(θ) = = =. Now,tan ( ) = 0 = π is in the first quadrant, whereas v =, is in the third quadrant. Therefore, the angle is given b adding an additional π to the angle. 6 θ = π + π 6π +π + = 7π d) The magnitude of v = 8, 5 is v = 8 +( 5) = = 6+80 = =. The directional angle is given b the equation tan(θ) = 5 = 5. Since 8 both tan ( 5 ) 8. and the endpoint of v (represented with beginning point at the origin) are in the first quadrant, we have: ( θ = tan e) We first need to find the vector v = PQ in the form v = a,b. The vector in the plane below shows that v is given b 0 9 v = 9,0 = 6,8 Q ) v P

313 .. OPERATIONS ON VECTORS 0 From this we calculate the magnitude to be v = ( 6) +8 = 6+6 = 00 = 0. The directional angle is given b tan(θ) = 8 = 6. Now, tan ( ) 5. is in quadrant IV, whereas v = 6,8 has an endpoint in quadrant II (when representing v with starting point at the origin O(0, 0)). Therefore, the directional angle is ( θ = 80 +tan ) 6.9. Operations on vectors There are two basic operations on vectors, which are the scalar multiplication and the vector addition. We start with the scalar multiplication. Definition.5. The scalar multiplication of a real number r with a vector v = a,b is defined to be the vector given b multiplingr to each coordinate. r a,b := r a,r b (.) We stud the effect of scalar multiplication with an eample. Eample.6. Multipl, and graph the vectors. a), b) ( ) 6, Solution. a) The calculation is straightforward., = ( ), = 8, The vectors are displaed below. We see, that, and 8, both have the same directional angle, and the latter stretches the former b a factor. 5 8,,

314 0 SESSION. VECTORS IN THE PLANE b) Algebraicall, we calculate the scalar multiplication as follows: ( ) 6, = ( ) ( 6),( ) ( ) = 8,6 Furthermore, graphing the vectors gives: 6, 8, 6 We see that the directional angle of the two vectors differs b 80. Indeed, 8,6 is obtained from 6, b reflecting it at the origin O(0,0) and then stretching the result b a factor. We see from the above eample, that scalar multiplication b a positive number c does not change the angle of the vector, but it multiplies the magnitude b c. Observation.7. Let v be a vector with magnitude v and angle θ v. Then, for a positive scalar, r > 0, the scalar multiple r v has the same angle as v, and a magnitude that is r times the magnitude of v: for r > 0: r v = r v and θ r v = θ v r v v θ r v = θ v Definition.8. A vector u is called a unit vector, if it has a magnitude of. u is a unit vector u =

315 .. OPERATIONS ON VECTORS 05 There are two special unit vectors, which are the vectors pointing in the - and the -direction. i :=,0 and j := 0, (.5) Eample.9. Find a unit vector in the direction of v. a) 8,6 b), 7 Solution. a) Note that the magnitude of v = 8,6 is 8,6 = 8 +6 = 6+6 = 00 = 0. Therefore, if we multipl 8,6 b we obtain 8,6 = 8, 6 = ,, 5 5 which (according to Observation.7 above) has the same directional angle as 8,6, and has a magnitude of : 0 8,6 = 0 8,6 = 0 0 = b) The magnitude of, 7 is, 7 = Therefore, we have the unit vector, 7 =, ( ) +( 7) = +9 7 = +6 = 67. = , = 67 67, which again has the same directional angle as, 7. Now, 67, 7 is a unit vector, since 67, 7 = 67, 7 = =. The second operation on vectors is vector addition, which we discuss now. Definition.0. Let v = a,b and w = c,d be two vectors. Then the vector addition v + w is defined b component-wise addition: a,b + c,d := a+c,b+d (.6)

316 06 SESSION. VECTORS IN THE PLANE In terms of the plane, the vector addition corresponds to placing the vectors v and w as the edges of a parallelogram, so that v+ w becomes its diagonal. This is displaed below. v d w v + w w b v c a Eample.. Perform the vector addition and simplif as much as possible. a), 5 + 6, b) 5 6, 7, c) i+9 j d) find v + w for v = 6 i j and w = 0 i 7 j e) find v +5 w for v = 8, and w = 0, Solution. We can find the answer b direct algebraic computation. a), 5 + 6, = +6, 5+ = 9, b) 5 6, 7, = 0,0 + 7, = 7, c) i+9 j =,0 +9 0, =,0 + 0,9 =,9 From the last calculation, we see that ever vector can be written as a linear combination of the vectors i and j. a,b = a i+b j (.7) We will use this equation for the net eample (d). Here, v = 6 i j = 6, and w = 0 i 7 j = 0, 7. Therefore, we obtain: d) v+ w = 6, + 0, 7 =, 8 + 0, = 8, 9 Finall, we have: e) v +5 w = 8, +5 0, =, + 0,0 =,7 Note that the answer could also be written as v+5 w = i+7 j.

317 .. OPERATIONS ON VECTORS 07 In man applications in the sciences, vectors pla an important role, since man quantities are naturall described b vectors. Eamples for this in phsics include the velocit v, acceleration a, and the force F applied to an object. Eample.. The forces F and F are applied to an object. Find the resulting total force F = F + F. Determine the magnitude and directional angle of the total force F. Approimate these values as necessar. Recall that the international sstem of units for force is the newton [N = kg m s ]. a) F has magnitude newtons, and angle θ = 5, F has magnitude 5 newtons, and angle θ = 5 b) F = 7 newtons, and θ = π 6, and F = newtons, and θ = 5π Solution. a) The vectors F and F are given b their magnitudes and directional angles. However, the addition of vectors (in Definition.0) is defined in terms of their components. Therefore, our first task is to find the vectors in component form. As was stated in equation (.) on page 0, the vectors are calculated b v = v cos(θ), v sin(θ). Therefore, F = cos(5 ), sin(5 ) =, =, F = 5 cos(5 ),5 sin(5 ) = 5,5 The total force is the sum of the forces. 6 5 = 5, 5 F + F F F F = F + F =, + 5, 5

318 08 SESSION. VECTORS IN THE PLANE = + 5, = =, 8 =,, +5 The total force applied in components is F =,. It has a magnitude of F = ( ) +( ) = +6 = 5.8 newton. The directional angle is given b tan(θ) = =. Since tan ( ) 76.0 is in quadrant IV, and F =, is in quadrant II, we see that the directional angle of F is θ = 80 +tan ( ) b) We solve this eample in much the same wa we solved part (a). First, F and F in components is given b ( π ( π F = 7 cos,7 sin = 7 6) 6) ( 5π F = cos ) ( 5π, sin ) =,7, = 7, 7 =, The total force is therefore: F = F + F = 7, 7 +, = 7 +, , 0.0 The magnitude is approimatel F (8.06) +(0.0) 8.06 newton. The directional angle is given b tan(θ) 0.0. Since F is in quadrant I, we 8.06 see that θ tan ( 0.0) Remark.. There is an abstract notion of a vector space. Although we do not use this structure for an further computations, we will state its definition. A vector space is a set V, with the following structures and properties. The elements of V are called vectors, denoted b the usual smbol v. For an vectors v

319 .. EXERCISES 09 and w there is a vector v + w, called the vector addition. For an real number r and vector v, there is a vector r v called the scalar product. These operations have to satisf the following properties. Associativit: Commutativit: Zero element: Negative element: Distributivit (): Distributivit (): Scalar compatibilit: Identit: ( u+ v)+ w = u+( v + w) v + w = w + v there is a vector o such that o+ v = v and v + o = v for ever vector v for ever v there is a vector v such that v +( v) = o and ( v)+ v = o r ( v + w) = r v +r w (r +s) v = r v +s v (r s) v = r (s v) v = v A ver important eample of a vector space is the -dimensional plane V = R as it was discussed in this chapter.. Eercises Eercise.. Graph the vectors in the plane. a) PQ with P(,) and Q(,7) c) PQ with P(0, ) and Q(6,0) d), e), f) 5,5 b) PQ with P(,) and Q( 5, ) Eercise.. Find the magnitude and directional angle of the vector. a) 6,8 b),5 c), d), e), f), g), h), i), j) PQ, where P(,) and Q(7,) k) PQ, where P(, ) and Q( 5,7) Eercise.. Perform the operation on the vectors. a) 5, b), c) ( 0), 7 d), + 6, 5 e) 5, 8, 9 f) 5, +,8 g) ( ) 5, 6, h), , 5 i) 8 6, 5, 5 j) 6 i j k) 5 i+ j + i l), 8 j + i

320 0 SESSION. VECTORS IN THE PLANE m) find v +7 w for v =, and w = 5, n) find v w for v =, 6 and w =, o) find v w for v = i+7 j and w = 6 i+ j p) find v 5 w for v = 5 j and w = 8 i+ 5 j Eercise.. Find a unit vector in the direction of the given vector. a) 8, 6 b), 7 c) 9, d) 5, e) 5, 0 f) 0, 5 Eercise.5. The vectors v and v below are being added. Find the approimate magnitude and directional angle of sum v = v + v (see Eample.). a) v = 6, and θ = 60, and v =, and θ = 80 b) v =.7, and θ = 9, and v =., and θ = 5 c) v = 8, and θ = π, and v = 8, and θ = π

321 Session Sequences and series. Introduction to sequences and series Definition.. A sequence is an ordered list of numbers. We write a sequence in order as follows: a,a,a,a,... In short we write the above sequence as {a n } or {a n } n. You might wonder how this topic fits into the general theme of functions. We can alternativel define a sequence a to be a function a : N R, where the domain is all natural numbers, and the range R is a subset of the real or comple numbers. We write the nth term of the sequence as a n = a(n). Eample.. Here are some eamples of sequences. a),6,8,0,,,6,8,... b),,9,7,8,,... c) +5, 5,+5, 5,+5, 5,... d),,,,5,8,,,,55,... e) 5,8,,,5.,7,,,8,,9,... For man of these sequences we can find rules that describe how to obtain the individual terms. For eample, in (a), we alwas add the fied number to the previous number to obtain the net, starting from the first term. This

322 SESSION. SEQUENCES AND SERIES is an eample of an arithmetic sequence, and we will stud those in more detail in section. below. In (b), we start with the first element and multipl b the fied number to obtain the net term. This is an eample of a geometric sequence, and we will stud those in more detail in section below. The sequence in (c) alternates between +5 and 5, starting from +5. Note, that we can get from one term to the net b multipling ( ) to the term. Therefore, this is another eample of a geometric sequence. The sequence in (d) is called the Fibonacci sequence. In the Fibonacci sequence, each term is calculated b adding the previous two terms, starting with the first two terms and : + =, + =, + = 5, +5 = 8, 5+8 =,... Finall, the sequence in (e) does not seem to have an obvious rule b which the terms are generated. In man cases, the sequence is given b a formula for the nth term a n. Eample.. Consider the sequence {a n } with a n = n+. We can calculate the individual terms of this sequence: first term: a = + = 7, second term: a = + =, third term: a = + = 5, fourth term: a = + = 9, fifth term: a 5 = 5+ =. Thus, the sequence is: 7,,5,9,,7,,5,... Furthermore, from the formula, we can directl calculate a higher term, for eample the 00th term is given b: 00th term: a 00 = 00+ = 80 Eample.. Find the first 6 terms of each sequence. a) a n = n b) a n = n n+ c) a n = ( ) n d) a n = ( ) n+ n

323 .. INTRODUCTION TO SEQUENCES AND SERIES Solution. a) We can easil calculate the first 6 terms of a n = n directl:,,9,6,5,6,... We can also use the calculator to produce the terms of a sequence. To this end, we switch the calculator from function mode to sequence mode in the mode menu (press lpmodelp ): To enter the sequence, we need to use the LIST menu (press lpnd lp and then lpstatlp ), then go to the OPS menu (press lp lp ), and use the fifth item (press lp5 lp ): press lpnd lp lpstatlp press lp lp use fifth item (press lp5 lp ) For the sequence, we need to specif four pieces of information, where the inde n can be entered via the lpx,t,θ,nlp ke. st entr is the given formula for a n = n nd entr is the inde n of the sequence a n rd and th entries are the first and last inde and 6 of the wanted sequence, here: a,...,a 6 Alternativel, we can enter the function in the lp= lp menu, starting from the first inde n =, and with nth term given b a n = n. The values of the sequence are displaed in the table window (press lpnd lp

324 SESSION. SEQUENCES AND SERIES and lpgraph lp ). b) We calculate the lowest terms of a n = n : n+ a = + =, a = + =, a = + =,... Identifing the pattern, we can simpl write a,...,a 6 as follows:,,, 5, 5 6, 6 7 We can also use the calculator to find the sequence: Note.5. Fractions can be displaed with the calculator b adding Frac to it from the MATH menu (using lpmathlp and the top item). For eample, the fraction is displaed below without approimation. 7 Adding Frac to the sequence displas it more appropriatel as a sequence of fractions.

325 .. INTRODUCTION TO SEQUENCES AND SERIES 5 c) Since ( ) n is + for even n, but is for odd n, the sequence a n = ( ) n is:,+,,+,,+ d) Similar to part (c), ( ) n+ is for even n, and is + for odd n. This, together with the calculation =, =, = 8, = 6, etc., we get the first si terms of the sequence: +,,+8, 6,+, 6 Another wa to describe a sequence is b giving a recursive formula for the nth term a n in terms of the lower terms. Here are some eamples. Eample.6. Find the first 6 terms in the sequence described below. a) a =, and a n = a n +5 for n > b) a =, and a n = a n for n > c) a =, a =, and a n = a n +a n for n > Solution. a) The first term is eplicitl given as a =. Then, we can calculate the following terms via a n = a n +5: a = a +5 = +5 = 9 a = a +5 = 9+5 = a = a +5 = +5 = 9 a 5 = a +5 = 9+5 =. b) We have a =, and calculate a = a = = 6, a = a = 6 =, a = a = =, etc. We see that the effect of the recursive relation a n = a n is to double the previous number. The sequence is:,6,,,8,96,9,... c) Starting from a =, and a =, we can calculate the higher terms: a = a +a = + = a = a +a = + =

326 6 SESSION. SEQUENCES AND SERIES a 5 = a +a = + = 5 a 6 = a +a 5 = +5 = 8. Studing the sequence for a short while, we see that this is precisel the Fibonacci sequence from eample.(d). Note.7. There is no specific reason for using the indeing variable n in the sequence {a n}. Indeed, we ma use as well an other variable. For eample, if the sequence {a n} n is given b the formula a n = n +, then we can also write this as a k = k + or a i = i +. In particular, the sequences {a n} n = {a k } k = {a i } i = {a j } j are all identical as sequences. We will be concerned with the task of adding terms in a sequence, such as a +a +a + +a k, for which we will use a standard summation notation. Definition.8. A series is a sum of terms in a sequence. We denote the sum of the first k terms in a sequence with the following notation: k a i = a +a + +a k (.) i= The summation smbol comes from the greek letterσ, pronounced sigma, which is the greek letter for S. More generall, we denote the sum from the jth to the kth term (where j k) in a sequence with the following notion: k a i = a j +a j+ + +a k Furthermore, the term k i=j Eample.9. Find the sum. i=j a i is sometimes also written in the form k i=j a i. a) b) c) a i, for a i = 7i+ i= 6 a j, for a n = ( ) n j= 5 ( ) +k k=

327 .. INTRODUCTION TO SEQUENCES AND SERIES 7 Solution. a) The first four terms a,a,a,a of the sequence {a i } i are: 0,7,, The sum is therefore: a i = a +a +a +a = = 8 i= We ma also find the answer with the calculator. Before entering the sequence (via lpnd lp lpstatlp lp lp lp5 lp as in eample.(a) above), we have to put a sum smbol in the LIST menu and MATH submenu (with lpnd lp lpstatlp lp lp lp5 lp ): press lpnd lp enter the sum and seq. lpstatlp lp lp press lp5 lp sum(seq(7n+,n,,)) b) The first si terms of {a n } with a n = ( ) n are: ( ) =, ( ) =, ( ) = 8, ( ) = 6, ( ) 5 =, ( ) 6 = 6 We calculate 6 j= a j b adding these first si terms. (Note that the sum is independent of the inde j appearing in the sum 6 j= a j, which we could as well replace b an other inde 6 j= a j = 6 k= a k, etc.) We get: 6 a j = a +a +a +a +a 5 +a 6 j= = ( )++( 8)+6+( )+6 = c) For the sum 5 k= (+k ) we need to add the first four terms of the sequence a k = +k. Calculating and adding the terms of this sequence, we obtain the sum: 5 ( ) +k = (+ )+(+ )+(+ )+(+ )+(+5 ) k=

328 8 SESSION. SEQUENCES AND SERIES = (+)+(+)+(+9)+(+6)+(+5) = = 75 This answer can of course also be confirmed with the calculator (after replacing the inde k with n) as we did in part (a).. The arithmetic sequence We have alread encountered eamples of arithmetic sequences in the previous section. An arithmetic sequence is a sequence for which we add a constant number to get from one term to the net, for eample: 8, +, +, + 7, + 0, +,... Definition.0. A sequence {a n } is called an arithmetic sequence if an two consecutive terms have a common difference d. The arithmetic sequence is determined b d and the first value a. This can be written recursivel as: a n = a n +d for n Alternativel, we have the general formula for the nth term of the arithmetic sequence: a n = a +d (n ) (.) Eample.. Determine if the sequence is an arithmetic sequence. If so, then find the general formula for a n in the form of equation (.). a) 7,,9,5,,... b),9,5,,, 7,... c) 0,,6,0,,... d) a n = 8 n+ Solution. a) Calculating the difference between two consecutive terms alwas gives the same answer 7 = 6, 9 = 6, 5 9 = 6, etc. Therefore the common difference d = 6, which shows that this is an arithmetic sequence. Furthermore, the first term is a = 7, so that the general formula for the nth term is a n = 7+6 (n ). b) One checks that the common difference is 9 =, 5 9 =, etc., so that this is an arithmetic sequence with d =. Since a =, the general term is a n = (n ).

329 .. THE ARITHMETIC SEQUENCE 9 c) We have 0 =, but 0 6 =, so that this is not an arithmetic sequence. d) If we write out the first couple of terms of a n = 8n +, we get the sequence:,9,7,5,,5,... From this it seems reasonable to suspect that this is an arithmetic sequence with common difference d = 8 and first term a =. The general term of such an arithmetic sequence is a +d(n ) = +8(n ) = +8n 8 = 8n+ = a n. This shows that a n = 8n+ = +8(n ) is an arithmetic sequence. Eample.. Find the general formula of an arithmetic sequence with the given propert. a) d =, and a 6 = 68 b) a = 5, and a 9 = 7 c) a 5 = 8, and a 6 = 5 Solution. a) According to equation (.) the general term is a n = a +d(n ). We know that d =, so that we onl need to find a. Plugging a 6 = 68 into a n = a +d(n ), we ma solve for a : 68 = a 6 = a + (6 ) = a + 5 = a +60 ( 60) = a = = 8 The nth term is therefore, a n = 8+ (n ). b) In this case, we are given a = 5, but we still need to find the common difference d. Plugging a 9 = 7 into a n = a +d(n ), we obtain 7 = a 9 = 5+d (9 ) = 5+8d (+5) = = 8d ( 8) = = d The nth term is therefore, a n = 5+ (n ). c) In this case we neither have a nor d. However, the two conditions a 5 = 8 and a 6 = 5 give two equations in the two unknowns a and d. { 8 = a5 = a +d(5 ) 5 = a 6 = a +d(6 ) = { 8 = a + d 5 = a +5 d

330 0 SESSION. SEQUENCES AND SERIES To solve this sstem of equations, we need to recall the methods for doing so. One convenient method is the addition/subtraction method. For this, we subtract the top from the bottom equation: 5 = a +5 d ( 8 = a + d ) 77 = + d ( ) = 7 = d Plugging d = 7 into either of the two equations gives a. We plug it into the first equation 8 = a +d: 8 = a + 7 = 8 = a +8 ( 8) = 0 = a Therefore, the nth term is given b a n = 0+7 (n ). We can sum the firstk terms of an arithmetic sequence using a trick, which, according to lore, was found b the German mathematician Carl Friedrich Gauss when he was a child at school. Eample.. Find the sum of the first 00 integers, starting from. In other words, we want to find the sum of First, note that he sequence,,,... is an arithmetic sequence. The main idea for solving this problem is a trick, which will indeed work for an arithmetic sequence: Let S = be what we want to find. Note that S = Note that the second line is also S but is added in the reverse order. Adding verticall we see then that S = , where there are 00 terms on the right hand side. So S = 00 0 and therefore S = 00 0 = The previous eample generalizes to the more general setting starting with an arbitrar arithmetic sequence.

331 .. THE ARITHMETIC SEQUENCE Observation.. Let {a n } be an arithmetic sequence, whose nth term is given b the formula a n = a +d(n ). Then, the sum a +a + +a k +a k is given b adding (a +a k ) precisel k times: k a i = k (a +a k ) (.) i= In order to remember the formula above, it ma be convenient to think of the right hand side as k a+a k (this is, k times the average of the first and the last term). Proof. For the proof of equation (.), we write S = a +a + +a k +a k. We then add it to itself but in reverse order: S = a + a + a + a k + a k + a k + a k + a k + a k + a + a + a. Now note that in general a l +a m = a +d(l +m ). We see that adding verticall gives S = k(a +d(k )) = k(a +(a +d(k )) = k(a +a k ). Dividing b gives the desired result. Eample.5. Find the value of the arithmetic series. a) Find the sum a + +a 60 for the arithmetic sequence a n = +(n ). b) Determine the value of the sum: (5 6j) 00 j= c) Find the sum of the first 5 terms of the sequence,.5,,.5,,.5,... Solution. a) The sum is given b the formula (.): k i= a i = k (a +a k ). Here,k = 60, anda = anda 60 = + (60 )= + 59= +767 = 769. We obtain a sum of a + +a 60 = 60 i= a i = 60 (+769) = 0 77 = 0.

332 SESSION. SEQUENCES AND SERIES We ma confirm this with the calculator as described in eample.9 (on page 6) in the previous section. Enter: sum(seq((+ (n ),n,,60)) b) Again, we use the above formula k j= a j = k (a + a k ), where the arithmetic sequence is given b a j = 5 6j and k = 00. Using the values a = 5 6 = 5 6 = and a 00 = = = 600, we obtain: 00 (5 6j) = 00 (a +a 00 ) = 00 (( )+( 600)) j= = 00 ( 600) = 00 ( 00) = 0000 c) First note that the given sequence,.5,,.5,,.5,... is an arithmetic sequence. It is determined b the first term a = and common difference d = 0.5. The nth term is given b a n = 0.5 (n ), and summing the first k = 5 terms gives: 5 i= a i = 5 (a +a 5 ) We see that we need to find a 5 in the above formula: a 5 = 0.5 (5 ) = 0.5 = 7 = This gives a total sum of 5 i= a i = 5 (+( )) = 5 ( 9) = 5. The answer ma be written as a fraction or also as a decimal, that is: 5 i= a i = 5 = 57.5.

333 .. EXERCISES. Eercises Eercise.. Find the first seven terms of the sequence. a) a n = n b) a n = 5n+ c) a n = n + d) a n = n e) a n = ( ) n+ f) a n = n+ n g) a k = 0 k h) a i = 5+( ) i Eercise.. Find the first si terms of the sequence. a) a = 5, a n = a n + for n b) a = 7, a n = 0 a n for n c) a =, a n = a n + for n d) a = 6, a =, a n = a n a n for n Eercise.. Find the value of the series. a) n= a n, where a n = 5n, b) 5 k= a k, where a k = k, c) i= a i, where a n = n, d) 6 n= (n ), e) k= (k +k ), f) j=. j+ Eercise.. Is the sequence below part of an arithmetic sequence? In the case that it is part of an arithmetic sequence, find the formula for the nth term a n in the form a n = a +d (n ). a) 5,8,,,7,... b) 0, 7,,,,... c),,,,,,... d) 8,6,0,7,... e) 7.,5.7,0,... f) 9,,, 8,,... g),,,,,... h).7,.8,.9,.0,.,... i), 5, 8,,... j) k) a n = +5 n l) a j = j 5 m) a n = n +8n+5 n) a k = 9 (k +5)+7k Eercise.5. Determine the general nth term a n of an arithmetic sequence {a n } with the data given below. a) d =, and a 8 = 57 b) d =, and a 99 = 70 c) a =, and a 7 = 6 d) a = 80, and a 5 = e) a = 0, and a = f) a 0 =, and a 60 =

334 SESSION. SEQUENCES AND SERIES Eercise.6. Determine the value of the indicated term of the given arithmetic sequence. a) if a = 8, and a 5 = 9, find a 9 b) if d =, and a =, find a 8 c) if a = 0, and a 7 = 0, find a 7 d) if a 7 = 8, and a 7 = 8, find a 6 Eercise.7. Determine the sum of the arithmetic sequence. a) Find the sum a + +a 8 for the arithmetic sequence a i = i+7. b) Find the sum i= a i for the arithmetic sequence a n = 5n. c) Find the sum: d) Find the sum: 99 i= (0 i+) 00 ( 9 n) n= e) Find the sum of the first 00 terms of the arithmetic sequence:,,6,8,0,,... f) Find the sum of the first 8 terms of the arithmetic sequence: 5,,7,,9,5,... g) Find the sum of the first 75 terms of the arithmetic sequence: 0,00,99,98,... h) Find the sum of the first 6 terms of the arithmetic sequence:, 6,,... i) Find the sum of the first 99 terms of the arithmetic sequence: j) Find the sum 8, 8., 8., 8.6, 8.8, 9, 9., k) Find the sum of the first 0 terms of the arithmetic sequence: 5,5,5,5,5,...

335 Session The geometric series. Finite geometric series We now stud another sequence, the geometric sequence, which will be analogous to our stud of the arithmetic sequence in section.. We have alread encountered eamples of geometric sequences in Eample.(b). A geometric sequence is a sequence for which we multipl a constant number to get from one term to the net, for eample: 5, 0, 80, 0, 80,... Definition.. A sequence {a n } is called a geometric sequence, if an two consecutive terms have a common ratio r. The geometric sequence is determined b r and the first value a. This can be written recursivel as: a n = a n r for n Alternativel, we have the general formula for the nth term of the geometric sequence: a n = a r n (.) Eample.. Determine if the sequence is a geometric, or arithmetic sequence, or neither or both. If it is a geometric or arithmetic sequence, then find the general formula for a n in the form (.) or (.). a),6,,,8,... b) 00,50,5,.5,... c) 700, 70,7, 0.7,0.07,... d),,6,56,... e),0,7,,... f),,,,,... g) a n = ( ) n h) a 7 n = n 5

336 6 SESSION. THE GEOMETRIC SERIES Solution. a) Calculating the quotient of two consecutive terms alwas gives the same number6 =, 6 =, =, etc. Therefore the common ratio is r =, which shows that this is a geometric sequence. Furthermore, the first term is a =, so that the general formula for the nth term is a n = n. b) We see that the common ratio between two terms is r =, so that this is a geometric sequence. Since the first term is a = 00, we have the general term a n = 00 ( )n. c) Two consecutive terms have a ratio of r =, and the first term is 0 a = 700. The general term of this geometric sequence is a n = 700 ( 0 )n. d) The quotient of the first two terms is =, whereas the quotient of the net two terms is 6 =. Since these quotients are not equal, this is not a geometric sequence. Furthermore, the difference between the first two terms is =, and the net two terms have a difference 6 =. Therefore, this is also not an arithmetic sequence. e) The quotient of the first couple of terms is not equal 0 7, so that this 0 is not a geometric sequence. The difference of an two terms is 7 = 0 = 7 0 = 7, so that this is part of an arithmetic sequence with common difference d = 7. The general formula is a n = a +d (n ) = +7 (n ). f) The common ratio is r = ( ) ( ) =, so that this is a geometric sequence with a n = ( ) n. On the other hand, the common difference is ( ) ( ) = 0, so that this is also an arithmetic sequence with a n = ( )+0 (n ). Of course, both formulas reduce to the simpler epression a n =. g) Writing the first couple of terms in the sequence {( 7 )n }, we obtain: ( ), 7 ( ), 7 ( ), 7 ( ), 7 ( ) 5,... 7 Thus, we get from one term to the net b multipling r =, so that this is 7 a geometric sequence. The first term is a =, so that a 7 n = ( n 7 7). This is clearl the given sequence, since we ma simplif this as a n = 7 ( ) n = 7 ( 7 ) +n = ( 7 h) We write the first terms in the sequence {n } n :,,9,6,5,6,9,... ) n

337 .. FINITE GEOMETRIC SERIES 7 Calculating the quotients of consecutive terms, we get = and 9 =.5, so that this is not a geometric sequence. Also the difference of consecutive terms is = and 9 = 5, so that this is also not an arithmetic sequence. Eample.. Find the general formula of a geometric sequence with the given propert. a) r =, and a 5 = 600 b) c) a =, and a 5 = 7 0 a 5 = 6, a 7 =, and r is positive Solution. a) Since {a n } is a geometric sequence, it is a n = a r n. We know that r =, so we still need to find a. Using a 5 = 6000, we obtain: 600 = a 5 = a 5 = a = 56 a ( 56) = a = = 5 The sequence is therefore given b the formula, a n = 5 n. b) The geometric sequence a n = a r n has a =. We calculate r 5 using the second condition. 7 0 = a = a r = 5 r ) n. ( 5 = ) r = = 7 = 7 8 (take ) 7 = r = = = Therefore, a n = ( 5 c) The question does neither provide a nor r in the formula a n = a r n. However, we obtain two equations in the two variables a and r: { 6 = a5 = a r 5 = a 7 = a r 7 = { 6 = a r = a r 6 In order to solve this, we need to eliminate one of the variables. Looking at the equations on the right, we see dividing the top equation b the bottom equation cancels a. 6 = a r a r 6 = 9 = (take reciprocal) = r 9 = r = r = 9

338 8 SESSION. THE GEOMETRIC SERIES To obtain r we have to solve this quadratic equation. In general, there are in fact two solutions: r = ± 9 = ± Since the problem states that r is positive, we see that we need to take the positive solution r =. Plugging r = back into either of the two equations, we ma solve for a. For eample, using the first equation a 5 = 6, we obtain: 6 = a 5 = a ( ) 5 = a ( 8) = a = 8 6 = 796 ( ) = a = a So, we finall arrive at the general formula for the nth term of the geometric sequence, a n = 796 ( )n. We can find the sum of the first k terms of a geometric sequence using another trick, which is ver different from the one we used for the arithmetic sequence. Eample.. Consider the geometric sequence a n = 8 5 n, that is the sequence: 8,0,00,000,5000,5000,5000,... We want to add the first 6 terms of this sequence = 8 In general, it ma be much more difficult to simpl add the terms as we did above, and we need to use a better general method. For this, we multipl ( 5) to the sum ( ) and simplif this using the distributive law: ( 5) ( ) = =

339 .. FINITE GEOMETRIC SERIES 9 In the second and third lines above, we have what is called a telescopic sum, which can be canceled ecept for the ver first and last terms. Dividing b ( 5), we obtain: = = 99 = 8 The previous eample generalizes to the more general setting starting with an arbitrar geometric sequence. Observation.5. Let {a n } be a geometric sequence, whose nth term is given b the formula a n = a r n. We furthermore assume that r. Then, the sum a +a + +a k +a k is given b: k i= a i = a rk r (.) Proof. We multipl ( r) to the sum (a +a + +a k +a k ): ( r) (a +a + +a k ) = ( r) (a r 0 +a r + +a r k ) = a r 0 a r +a r a r + +a r k a r k = a r 0 a r k = a ( r k ) Dividing b ( r), we obtain a +a + +a k = a ( r k ) ( r) This is the formula we wanted to prove. Eample.6. Find the value of the geometric series. = a rk r a) Find the sum 6 a n for the geometric sequence a n = 0 n. n= b) Determine the value of the geometric series: 5 k= ( ) k c) Find the sum of the first terms of the geometric sequence, 6,,,...

340 0 SESSION. THE GEOMETRIC SERIES Solution. a) We need to find the sum a +a +a +a +a 5 +a 6, and we will do so using the formula provided in equation (.). Since a n = 0 n, we have a = 0 and r =, so that 6 n= a n = 0 6 = 0 79 = 0 78 = 0 6 = 60 b) Again, we use the formula for the geometric series n k= a k = a rn, r since a k = ( )k is a geometric series. We ma calculate the first term a =, and the common ratio is also r =. With this, we obtain: 5 k= ( ) k = = = ( ) ( )5 ( ) = ( ) ( ) ( ) ( ) = = ( ) ( ) ( ) (( )55 5 ) ( ) = + + = ( ) = 6 = + + c) Our first task is to find the formula for the provided geometric series, 6,,,.... The first term is a = and the common ratio is r =, so that a n = ( ) n. The sum of the first terms of this sequence is again given b equation (.): i= ( ) i = ( ) = ( ) 095 = 85 = ( ) 096 = ( ) 095. Infinite geometric series In some cases, it makes sense to add not onl finitel man terms of a geometric sequence, but all infinitel man terms of the sequence! An informal and ver intuitive infinite geometric series is ehibited in the net eample.

341 .. INFINITE GEOMETRIC SERIES Eample.7. Consider the geometric sequence,,, 8, 6,... Here, the common ratio is r =, and the first term is a =, so that the formula for a n is a n = ( ) n. We are interested in summing all infinitel man terms of this sequence: We add these terms one b one, and picture these sums on the number line: = + = = = =.975 We see that adding each term takes the sum closer and closer to the number. More precisel, adding a term a n to the partial sum a + +a n cuts the distance between and a + +a n in half. For this reason we can, in fact, get arbitraril close to, so that it is reasonable to epect that = In the net definition and observation, this equation will be justified and made more precise. We start b providing the definition of an infinite series.

342 SESSION. THE GEOMETRIC SERIES Definition.8. An infinite series is given b the a i = a +a +a +... (.) i= To be more precise, the infinite sum is defined as the limit sum is defined, precisel when this limit eists. i= ( k a i := lim a i ). Therefore, an infinite k i= Observation.9. Let {a n } be a geometric sequence with a n = a r n. Then the infinite geometric series is defined whenever < r <. In this case, we have: a i = a (.) r i= Proof. Informall, this follows from the formula k i= a i = a rk r and the fact that r k approaches zero when k increases without bound. More formall, the proof uses the notion of limits, and goes as follows: i= ( k ) a i = lim a i = lim k k i= ) lim (a rk k (rk ) = a = a r r r Eample.0. Find the value of the infinite geometric series. a) b) j= a j, for a j = 5 ( n= (0.7)n ) j c) d) Solution. a) We use formula (.) for the geometric series a n = 5 ( )n, that is a = 5 ( ) = 5 ( )0 = 5 = 5 and r =. Therefore, a j = a r = 5 j= = 5 = 5 = 5 = 5

343 .. INFINITE GEOMETRIC SERIES b) In this case, a n = (0.7) n, so that a = 0.7 = 0.7 =. and r = 0.7. Using again formula (.), we can find the infinite geometric series as (0.7) n = a r =. 0.7 =. 0.9 =. 0.9 = 9 n= In the last step we simplified the fraction b multipling both numerator and denominator b 00, which had the effect of eliminating the decimals. c) Our first task is to identif the given sequence as an infinite geometric sequence: {a n } is given b 500, 00,0,,... Notice that the first term is 500, and each consecutive term is given b dividing b 5, or in other words, b multipling b the common ratio r = 5. Therefore, this is an infinite geometric series, which can be evaluated as = a n = a r = 500 ( ) 5 n= = 500 = = = 50 = = d) We want to evaluate the infinite series The sequence,6,,,8,... is a geometric sequence, with a = and common ratio r =. Since r, we see that formula (.) cannot be applied, as (.) onl applies to < r <. However, since we add larger and larger terms, the series gets larger than an possible bound, so that the whole sum becomes infinite = Eample.. The fraction ma be written as: = Noting that the sequence 0.5, , , , ,...

344 SESSION. THE GEOMETRIC SERIES is a geometric sequence with a = 0.5 and r = 0., we can calculate the infinite sum as: = 0.5 (0.) i = = = = 5 9, i= Here we multiplied numerator and denominator b 0 in the last step in order to eliminate the decimals.. Eercises Eercise.. Which of these sequences is geometric, arithmetic, or neither or both. Write the sequence in the usual form a n = a +d(n ) if it is an arithmetic sequence and a n = a r n if it is a geometric sequence. a) 7,,8,56,... b), 0,00, 000,0000,... c) 8,7,9,,,,... d) 7, 5,,,,,5,7,... e) 6,,,,,... f),, ( ), ( 9 7 ),... g),,,,... h),,,,, i) 5,,5,,5,,5,,... j),,,,,,... k) 0,5,0,5,0,... l) 5, 5, 5, 5, 5,... m),,,,... n) log(),log(),log(8),log(6), o) a n = n p) a n = n q) a n = ( 9) n r) a n = ( s) a n = ( 5 n 7) t) a n = ( ) 5 n 7 u) a n = v) a n n = n+ Eercise.. A geometric sequence, a n = a r n, has the given properties. Find the term a n of the sequence. ) n a) a =, and r = 5, find a b) a = 00, and r =, find a 6 c) a = 7, and r =, find a n (for all n) d) r =, and a = 8, find a e) r = 00, and a = 900,000, find a n (for all n)

345 .. EXERCISES 5 f) a = 0, a = 500, find a n (for all n) g) a =, and a 8 6 = 5, find a 8 6 n (for all n) h) a = 6, and a 6 = 97, find a n (for all n) i) a 8 = 000, a 0 = 0, and r is negative, find a n (for all n) Eercise.. Find the value of the finite geometric series using formula (.). Confirm the formula either b adding the the summands directl, or alternativel b using the calculator. a) Find the sum a j for the geometric sequence a j = 5 j. j= b) Find the sum 7 c) Find: d) Find: 5 m= i= a i for the geometric sequence a n = ( ) n. ( ) m k k= e) Find the sum of the first 5 terms of the geometric sequence:,6,8,5,... f) Find the sum of the first 6 terms of the geometric sequence: 5,5, 5,5,... g) Find the sum of the first 8 terms of the geometric sequence:, 7, 9,,... h) Find the sum of the first 0 terms of the geometric sequence: 600, 00,50, 75,7.5,... i) Find the sum of the first 0 terms of the geometric sequence: 5,5,5,5,5,...

346 6 SESSION. THE GEOMETRIC SERIES Eercise.. Find the value of the infinite geometric series. a) c) e) j= a j j, for a j = ( ) b) 7 ( j j= 5) j= 6 d) j n= (0.8)n n= (0.99)n f) g) h) i) j) Eercise.5. Rewrite the decimal using an infinite geometric sequence, and then use the formula for infinite geometric series to rewrite the decimal as a fraction (see eample.). a) 0... b) c) d) 0... e) f) g) h)

347 Session 5 The binomial theorem 5. The binomial theorem Recall the well-known binomial formula: (a+b) = a +ab+b (since, using FOIL, we have: (a+b) = (a+b) (a+b) = a +ab+ab+b = a +ab +b ) In this section we generalize this to find similar epressions for (a+b) n for an natural number n. This is the content of the (generalized) binomial theorem below. Before we can state the theorem, we need to define the notion of a factorial and combinations. Definition 5.. For a natural number n =,,,..., we define n! to be the number n! = n The number n! is called n factorial. To make the formulas below work nicel, we also define 0! to be 0! =. It is eas to calculate some eamples of factorials. Eample 5..! = = 7! = = 500 7

348 8 SESSION 5. THE BINOMIAL THEOREM! = = To calculate factorials with the calculator, we have to use the MATH menu (press lpmathlp ), then move to the PRB menu (press lp lp ), and use the fourth item (press lp lp ). press lpmathlp press lp lp use fourth item (press lp lp ) For eample, we can calculate! = b entering and the factorial smbol as described above. Note that the factorial becomes ver large even for relativel small integers. For eample 7! as shown above. The net concept that we introduce is that of the binomial coefficient. Definition 5.. Let n = 0,,,... and r = 0,,,...,n be natural numbers or zero, so that 0 r n. Then we define the binomial coefficient as ( ) n n! = r r! (n r)! The binomial coefficient is also written as n C r = ( n r), and we read them as n-choose-r. Note 5.. The binomial coefficient ( n ( r) should not be confused with the fraction n ) r.

349 5.. THE BINOMIAL THEOREM 9 A subset of the set {,,...,n} with r elements is called an r-combination. The binomial coefficient can be interpreted as counting the number of distinct r-combinations. More precisel, there are eactl ( n r) distinct r-combinations of the set {,...,n}. Eample 5.5. Calculate the binomial coefficients. a) ( ) 6 b) ( ) 8 c) ( ) 5 d) ( ) 7 e) ( ) 5 f) ( ) 0 Solution. a) Man binomial coefficients ma be calculated b hand, such as: ( ) 6 6! =!(6 )! = 6!!! = 5 6 = 5 6 = 5 b) Again, we can calculate this b hand ( ) 8 = 8! 5 5!! = = = 7 8 = 56 However, we can also use the calculator to find the answer. Using the MATH and (press lpmathlp ) PRB menus (press lp lp ) as above, we use the third item (press lp lp ). The answer is obtained b pressing the following sequence of kes: lp8 lp lpmathlp lp lp lp lp lp5 lp lpenter lp We also calculate the remaining binomial coefficients (c)-(f), which can also be confirmed with the calculator. ( ) 5 c) = 5!!! =! 5! = 5 = 00 ( ) 7 d) = 7!! 6! = 6! 7 6! = 7 = 7 ( ) e) =!! 0! = = ( ) f) =! 0 0!! = = Note that in the last two equations we needed to use the fact that 0! =.

350 0 SESSION 5. THE BINOMIAL THEOREM We state some useful facts about the binomial coefficient, that can alread be seen in the previous eample. Observation 5.6. For all n = 0,,,... and r = 0,,,...,n, we have: ( ) n = n r ( n r ) ( n 0 ) = ( ) n = n ( ) ( ) n n = = n n Proof. We have: ( ) n n r ( ) n 0 ( ) n = = = n! (n r)! (n (n r))! = n! (n r)! r! = ( ) n = n! n 0! n! = = ( ) n n! = n! (n )! = n = n ( ) n r Note 5.7. We can obtain all binomial coefficients ( n r) for fied n from the calculator b using the function and table menus. The binomial coefficients are found in what is known as Pascal s triangle. For this, calculate the lowest binomial coefficients and write them in a triangular arrangement: ( 0 ( ) 0) ( ( 0 ) ( ) ) ( ( 0 ) ( ) ( ) ) ( ( 0 ) ( ) ( ) ( ) ) ( ( 0 5 ) ( 5 ) ( 5 ) ( 5 ) ( 5 ) ) ( 5 0 5). = The triangle on the right is known as Pascal s triangle. Each entr in the triangle is obtained b adding the two entries right above it.

351 5.. THE BINOMIAL THEOREM The binomial coefficients appear in the epressions for (a+b) n as we will see in the net eample. Eample 5.8. We now calculate some simple eamples. (a+b) = (a+b) (a+b) (a+b) = (a +ab+b ) (a+b) = a +a b+ab +a b+ab +b = a +a b+ab +b Note that the coefficients,,, and in front of a,a b,ab, and b, respectivel, are precisel the binomial coefficients ( ( 0), ( ), ( ), and ). We also calculate the fourth power. (a+b) = (a+b) (a+b) (a+b) (a+b) = (a +a b+ab +b ) (a+b) = a +a b+a b +ab +a b+a b +ab +b = a +a b+6a b +ab +b Again, ) ( the numbers,,6,, and are precisel the binomial coefficients, ) (, ) (, ) (, and ). ( 0 We are now read to state the general binomial theorem: Theorem 5.9 (Binomial theorem). The nth power (a+b) n can be epanded as: ( ) ( ) ( ) ( ) ( ) n n n n n (a+b) n = a n + a n b + a n b + + a b n + b n 0 n n Using the summation smbol, we ma write this in short: n ( ) n (a+b) n = a n r b r (5.) r Eample 5.0. Epand (a+b) 5. Solution. (a+b) 5 = ( ) 5 a ( ) 5 a b + r=0 ( ) 5 a b + ( ) 5 a b + = a 5 +5a b+0a b +0a b +5ab +b 5 ( ) 5 a b + ( ) 5 b 5 5

352 SESSION 5. THE BINOMIAL THEOREM 5. Binomial epansion Using the binomial theorem, we can also epand more general powers of sums or differences. Eample 5.. Epand the epression. a) ( + ) 5 b) ( ) c) ( +) 6 d) (i ) Solution. a) We use the binomial theorem with a = and b = : ( ) ( ) 5 5 ( + ) 5 = ( ) 5 + ( ) ( )+ ( ) ( ) ( ) ( ) ( ) ( ) + ( )( ) +( ) 5 = = b) For part (b), it is a = and b =. ( ) = ( ) + ( ) ( ) ( ( ) )+ ( )( +( ) ) = 6 + ( )+( )( ) +( ) 6 = c) Similarl, for part (c), we now have a = and b = : ( +) 6 = ( ( ) 6 ) 6 + ( ( ) 6 ) 5 + ( ( ) 6 ) + ( ) ( ) 6 + ( ( ) 6 ) + ( ) = =

353 5.. BINOMIAL EXPANSION = = Note that the last epression cannot be simplified an further (due to the order of operations). d) Finall, we have a = i and b =, and we use the fact that i =, and therefore, i = i and i = +: ( ) ( ) ( ) (i ) = i + i ( )+ i ( ) + i ( ) +( ) = + ( i) ( )+6 ( ) 9+ i ( 7)+8 = +i 5 08i+8 = 8 96i In some instances it is not necessar to write the full binomial epansion, but it is enough to find a particular term, sa the kth term of the epansion. Observation 5.. Recall, for eample, the binomial epansion of (a+b) 6 : ( 6 0) a 6 b 0 + ( ) 6 a 5 b + ( ) 6 a b + ( ) 6 a b + ( ) 6 a b + ( ) 6 5 a b 5 + ( ) 6 6 a 0 b 6 first second third fourth fifth sith seventh term term term term term term term Note that the eponents of the a s and b s for each term alwas add up to 6, and that the eponents of a decrease from 6 to 0, and the eponents of b increase from 0 to 6. Furthermore observe that in the above epansion the fifth term is ( 6 ) a b. In general, we define the kth term b the following formula: The kth term in the epansion of (a+b) n is: ( ) n a n k+ b k (5.) k Note in particular, that the kth term has a power of b given b b k (and not b k ), it has a binomial coefficient ( n k ), and the eponents of a and b add up to n.

354 SESSION 5. THE BINOMIAL THEOREM Eample 5.. Determine: a) the th term in the binomial epansion of (p+q) 5 b) the 8th term in the binomial epansion of ( ) 0 c) the th term in the binomial epansion of ( 5a b) 5 b 7 Solution. a) We have a = p and b = q, and n = 5 and k =. Thus, the binomial coefficient of the th term is ( 5 ), the b-term is (q), and the a-term is p. The th term is therefore given b ( ) 5 p (q) = 0 p q = 70p q b) In this case, a = and b =, and furthermore, n = 0 and k = 8. The binomial coefficient of the 8th term is ( 0 7), the b-term is ( ) 7, and the a-term is ( ). Therefore, the 8th term is ( ) 0 ( ) ( ) 7 = 0 9 ( 8) = c) Similarl, we obtain the th term of ( 5a b) 5 as b 7 ) ( 5a ) ( b) = 65 5 a b 7 ( 5 = 855 a b 7 b 8 ( b ) = a ( b ) b 8 Here is a variation of the above problem. Eample 5.. Determine: a) the -term in the binomial epansion of (5 + ) 6 b) the 5 -term in the binomial epansion of ( ) 7 c) the real part of the comple number (+i) Solution. a) In this case we have a = 5 and b =. The term can be rewritten as = ( ) ( ), so that the full term ( ) n k a n k+ b k (including the coefficients) is ( ) 6 (5 ) ( ) = = 6000

355 5.. EXERCISES 5 b) The various powers of in ( ) 7 (in the order in which the appear in the binomial epansion) are: ( ) 7 =, ( ) 6 = 9, ( ) 5 = 7, ( ) = 5,... The last term is precisel the 5 -term, that is we take the fourth term, k =. We obtain a total term (including the coefficients) of ( ) 7 ( ) ( ) = 5 ( ) = 5 5 c) Recall that i n is real when n is even, and imaginar when n is odd: i = i i = i = i i = i 5 = i i 6 =. The real part of (+i) is therefore given b the first, third, and fifth term of the binomial epansion: ( ) ( ) ( ) real part = (i) 0 + (i) + 0 (i) 0 = i + 6i = 8+6 ( )+6 = = 9 The real part of (+i) is Eercises Eercise 5.. Find the value of the factorial or binomial coefficient. a) 5! b)! c) 9! d)! e) 0! f)! g) 9! h) 6! i) ( ) 5 j) ( ) 9 k) ( ) l) ( ) m) ( ) n) ( ) 9 o) ( ) p) ( ) Eercise 5.. Epand the epression via the binomial theorem. a) (m+n) b) (+) 5 c) ( ) 6 d) ( p q) 5

356 6 SESSION 5. THE BINOMIAL THEOREM Eercise 5.. Epand the epression. a) ( ) b) ( 0) c) ( + ) 5 d) ( 5 ) e) (+ ) f) ( )5 g) ( ) h) ( i) Eercise 5.. Determine: a) the first terms in the binomial epansion of ( ) 5 b) the first terms in the binomial epansion of (a +b ) 9 c) the last terms in the binomial epansion of ( ) 7 d) e) the first terms in the binomial epansion of ( )0 the last terms in the binomial epansion of (m n+ n ) 6 Eercise 5.5. Determine: a) the 5th term in the binomial epansion of (+) 7 b) the rd term in the binomial epansion of ( ) 9 c) the 0th term in the binomial epansion of ( w) d) the 5th term in the binomial epansion of (+) 7 e) the 7th term in the binomial epansion of (a+5b) 6 f) the 6th term in the binomial epansion of (p q p) 7 g) the 0th term in the binomial epansion of (+ b ) Eercise 5.6. Determine: a) the 6 -term in the binomial epansion of (+) 9 b) the r s -term in the binomial epansion of (r s) 6 c) the -term in the binomial epansion of ( ) d) the 6 -term in the binomial epansion of ( +5 ) e) the 7 -term in the binomial epansion of ( ) 5 f) the imaginar part of the number (+i)

357 Review of comple numbers, sequences, and the binomial theorem Eercise V.. Multipl and write the answer in standard form: ( )(cos(07 )+isin(07 )) (cos( )+isin( )) Eercise V.. Divide and write the answer in standard form: (cos( π )+isin(π )) 5(cos( π )+isin(π )) Eercise V.. Find the magnitude and directional angle of the vector v = 7, 7. Eercise V.. Determine if the sequence is an arithmetic or geometric sequence or neither. If it is one of these, then find the general formula for the nth term a n of the sequence. a) 5, 8,6,,,... b),,8,0,... c) 9,5,,, 7,... Eercise V.5. Find the sum of the first 75 terms of the arithmetic sequence: 0,,, 6,,... Eercise V.6. Find the sum of the first 8 terms of the geometric series: 7,, 8, 56,,... 7

358 Eercise V.7. Find the value of the infinite geometric series: Eercise V.8. Epand the epression via the binomial theorem. ( ) Eercise V.9. Write the first terms of the binomial epansion: ( ab + 0 a Eercise V.0. Find the 6th term of the binomial epansion: (5p q ) 8 ) 9 8

359 Appendi A Introduction to the TI-8 A. Basic calculator functions In this section we eplain the basic functions of the calculator. There are various buttons and screens. We will describe below those which are most useful for us. The main screen The main screen is the screen that appears when ou first turn on the calculator. It is the screen where ou can perform operations such as adding numbers. You can alwas return to the main screen b pressing lpnd lp lpmodelp. Other screens will be discussed below according to their use. The buttons There are three levels of buttons. The phsical buttons together with the green and blue print above each button, for eample. To select the items in green (blue) print first select lpalpha lp ( lpnd lp ) then the button beneath the printed item. For eample to select the letter K press lpalpha lp then lp(lp. From now on we will use button to mean an of the the phsical buttons or the green or blue pseudo-buttons. The clear button Whenever ou want to erase a line of input or a screen press the lpclearlp button (once or twice). 9

360 50 APPENDIX A. INTRODUCTION TO THE TI-8 The delete and insert buttons The DEL button ( lpdellp ) and the INS button ( lpnd lp lpdellp ) are hand when editing an epression. The DEL button just deletes the character that is at the location of the cursor. The insert button is inserts the entered numbers or letters at the location of the cursor. For eample, if ou tpe 5+5 and ou realize ou wanted 5+5 instead, ou should use the left arrow to move the cursor to the first 5, press lpnd lp lpdellp lp lp. What happens if ou just had pressed lp lp? Note that the calculator stas in insert mode until the cursor is moved with the error or enter is pressed. So ou could have changed the first 5 to a 5 b moving the cursor to the first 5 and pressing lpnd lp lpdellp lp lp lp lp. The ANS button The answer button is used when ou want to use the answer above in an epression. For eample suppose ou had entered lp lp lp+ lp lp5 lp lpenter lp. Now ou want to compute 7 minus the result of that computation. You can press lp7 lp lp lp lpnd lp lp( ) lp. The ENTRY button The ENTRY button (the blue selection above lpenter lp which is entered b pressing lpnd lp lpenter lp ) displas the last entr. This is a particularl useful button when ou have entered an epression then realized there is an error. You can recall the epression then edit it. Unlike the ANS button, using ENTRY twice gives the entr before the last. For eample: Eample A.. We first prepare the screen: lp5 lp lp+ lp lp5 lp lpenter lp lp lp lp+ lp lp lp lpenter lp. Now press the ENTRY button. + appears in the window. Press the ENTRY button again. 5+5 replaces the +. Press lpenter lp (so that 0 is the answer). To multipl the + b 7 we

361 A.. BASIC CALCULATOR FUNCTIONS 5 can enter: lpnd lp lpenter lp lpnd lp lpenter lp. Then move the cursor using the left arrow to the position of the first and press lpnd lp lpdellp lp(lp. Then move the cursor using the right button to after the second and press the sequence lp)lp lp lp lp7 lp. What happens if we leave out the parentheses? Can we enter lp(lp before lpnd lp lpenter lp? Eample A.. First tpe lp5 lp lp+ lp lp lp lpenter lp. Then tpe lpnd lp lp( ) lp lp+ lp lp5 lp lpenter lp, lpnd lp lpenter lp lpenter lp, lpnd lp lpenter lp lpenter lp. What would have happened if ou had used the ANS button instead of the ENTRY button? Note that instead of the second lpnd lp lpenter lp lpenter lp ou could have just pressed lpenter lp. vs ( ) There are two minus signs on the keboard: lp lp which is between lp lp and lp+ lp, and lp( ) lp which is to the lower left of lpenter lp. What is the difference between the two? lp lp means subtract. So this is the button ou use to compute 7 5. The button lp( ) lp means

362 5 APPENDIX A. INTRODUCTION TO THE TI-8 negative or the opposite of so, for eample, if ou want to compute 5+7 ou enter lp( ) lp lp5 lp lp+ lp lp7 lp. Notice what happens if ou use lp lp instead (what will ou subtract from?)! The lp( )lp button is also what ou use to enter epressions like +7 or ep( ). The MODE button When calculating using the trigonometric functions (sin, cos, tan, and their inverses) it is important that ou distinguish between degrees and radians and that ou are in snc with our calculator. For that we need to look at the MODE screen. To view the Mode screen press lpmodelp. The third line has the words Radian and Degree. If the word radian is highlighted and ou want to compute in degrees ou should press the down arrow twice (so that the highlighted word on the third line blinks) and then the right arrow and press lpenter lp. The word degree should then be highlighted. You can also verif our mode b evaluating sin(80) on the main screen (ou can press quit - lpnd lp lpmodelp to return to the main screen). If ou get 0 as an answer then ou are in degree mode. What do ou get if ou are in radian mode? The STO button When evaluating complicated epressions, sometimes it is hand to use the storage feature. For eample Eample A.. Suppose ou want to evaluate ( ( /)). You can first assign / a letter, V sa, the value ( /) since it appears twice. You can do this b pressing lp lp lp lp lp lp lp lp lp lp lpstolp lpalpha lp lp6 lp. Then to evaluate the original epression tpe lp(lp lp lp lp lp lpalpha lp

363 A.. GRAPHING A FUNCTION 5 lp6 lp lp)lp lp lp lp lp lpalpha lp lp6 lp. Eample A.. Note that ou can use this feature with the ENTRY button to evaluate a function at different places. Let s evaluate + at = 0 and at =. First set = 0: lp0 lp lpstolp lpx,t,θ,nlp lpenter lp. Then evaluate the function: lpx,t,θ,nlp lp lp lp+ lp lpx,t,θ,nlp lpenter lp Now set = : lp lp lpstolp lpx,t,θ,nlp Now evaluate the function again: lpnd lp lpenter lp lpnd lp lpenter lp lpenter lp. Note that there is another wa to evaluate a function at several points which will be introduced below in the graphing section. A. Graphing a function The Y= screen This is the screen that we will need to graph functions. It is seen b pressing lp= lp.

364 5 APPENDIX A. INTRODUCTION TO THE TI-8 Press the lp= lp button. This brings ou to the screen where ou can enter functions that ou want to graph. Let us suppose that ou want to graph = sin(). You would first move the cursor (using the arrow kes) to Y enter lpsinlp lpx,t,θ,nlp lpenter lp. The press lpgraph lp. This will show a portion of the graph of the function (naturall, it can not generall show ou the whole graph as this is often infinite ). A.. Rescaling the graphing window The graphing window is what determines what portion of the graph is seen. There are various reasons ou ma want to have a different window. For eample, the standard view for graphing the sin function (as we will show in the net paragraph) the setting for the -ais is too large and makes it difficult to see what is going on. We describe here a few using the eample of sin() (in degree mode) and encourage eperimenting. The standard window This sets the -ais from 0 to 0 and the -ais from 0 to 0. To choose this view ou must choose the standard window from the ZOOM screen b pressing lpzoom lp lp6 lp. This is the window shown above. Notice that in this case ou can t reall see what ou want. Don t worr, there are other zooming options! Zoom fit One eas wa to choose an appropriate window is to ask the calculator to choose it using the zoom fit windows ( lpzoom lp lp0 lp ). You

365 A.. GRAPHING A FUNCTION 55 can also press lpzoom lp then press the down arrow until ou arrive at the bottom of the list with is ZoomFit. Custom window Mabe the zoom fit choice of the calculator isn t what ou wanted to see. Then ou can choose our own window. Press the lpwindow lp ke. Move the cursor to Xmin. This is the smallest number on the -ais in our custom window. let s sa ou want this to be 5. Then ou press lp( )lp lp5 lp. Then move the cursor to Xma and enter our desired maimum, sa, lp5 lp. You can also adjust the scale (where the marks on the -ais are given) b setting Xscl equal to whatever is convenient. In this eample we will set it to. Now ou can do the same for the -ais with changing Ymin and Yma to sa, lp( )lp lp. lp lp5 lp and lp. lp lp5 lp, respectivel. Press graph again and observe the window. Zooming in and out You can zoom in and out also. Press the lpzoom lp lp lp. You will now be looking at the graph. Move our cursor to what ou want to be the center and press lpenter lp. You can zoom out as well.

366 56 APPENDIX A. INTRODUCTION TO THE TI-8 ZTrig The best choice for our current Y=sin(X) function is ZTrig (press lpzoom lp lp7 lp ). Zbo You should eperiment with ZBo, which is the first option on the zoom menu. You first move the cursor to where ou want the upper left point of our bo to be and press enter. Then ou move the cursor to where ou want our lower right corner to be and press enter. Our result is ZSquare Finall, we point out another zoom option, the zoom square option, which scales the - and -aes to the same piel size. To see the issue recall from eample. on page 5, that the circle + = 6 ma be graphed b entering the functions = ± 6. Note, that the obtained graph in the above window (with 6 6 and 6 6) looks more like an ellipse rather than a circle. To rectif this, we press lpzoom lp lp5 lp, and obtain a graph which now does have the

367 A.. GRAPHING MORE THAN ONE FUNCTION 57 shape of a circle. A. Graphing more than one function Let us suppose that ou want to graph three functions: = sin(60), = and =. You would first move the cursor (using the arrow kes) to Y 6 enter lpsinlp lp lp lp6 lp lp0 lp lpx,t,θ,nlp lp)lp lpenter lp. Note that if ou need to clear the field ou can use the clear button. Then ou would move the cursor to the Y position and enter lpx,t,θ,nlp lpenter lp. Finall ou would move the cursor to the Y position and enter lpx,t,θ,nlp lp lp lpx,t,θ,nlp lp lp lp lp lp lp lp lp lp6 lp lpenter lp. To graph these functions press lpgraph lp. The three graphs will be drawn one after the other. If ou would like to just graph the first two but do not want to erase the work ou put in to enter the third function, ou can move back to the screen and toggle the third function off (or on) b moving the cursor to the = sign after Y and press lpenter lp. This equal sign should now be unhighlighted.

368 58 APPENDIX A. INTRODUCTION TO THE TI-8 You ma also want to have the calculator distinguish between the graphs b the tpe of line. Let us suppose ou want Y bold, Y normal, and Y dashed. Look at the to the left of Y. This is what indicates the tpe of line. Press lpenter lp until the line tpe ou want is displaed. Move the cursor down and do the same for Y and then for Y. The Calc menu With the help of the Calc menu ( lpnd lp lptrace lp ) ou can find the minima of a function, the maima of a function, intersections of functions, and other similar things. We will describe the intersection feature in detail. Intersection of functions When solving equations graphicall, ou will want to find the intersection of functions. For eample graph = and = + on the same screen (zoom standard). To find the left most intersection of the two functions first press lpnd lp lptrace lp lp5 lp. The calculator shows ou the graphing screen and will ask ou which is the first curve. Move the cursor (using the arrow kes) until it is on the first curve and press lpenter lp. It will ask ou which is the second curve. Move the cursor to the second curve and press lpenter lp.

369 A.. GRAPHING A PIECEWISE DEFINED FUNCTION 59 It will ask ou for a guess. Move the cursor to near the intersection point ou want to know and press lpenter lp. The intersection will then be displaed. A. Graphing a piecewise defined function Suppose ou wanted to graph the piecewise defined function { f() =, < 7, Go to the lp= lp menu. We need to enter the function in the following wa: = ( < )+( 7)( ) The onl new entries are in the inequalit signs, which can be accessed through the TEST menu ( lpnd lp lpmathlp ). Therefore, we press the kes: lpx,t,θ,nlp lp lp lp(lp lpx,t,θ,nlp lpnd lp lpmathlp lp5 lp lp lp lp)lp lp+ lp lp(lp lp lp lpx,t,θ,nlp lp lp lp7 lp lp)lp lp(lp lpx,t,θ,nlp lpnd lp lpmathlp lp lp lp lp lp)lp lpenter lp. The graph is displaed on the right:

370 60 APPENDIX A. INTRODUCTION TO THE TI-8 The meaning of = ( < ) as a function is that it is equal to when it is true and 0 when it is false. In other words, the epression ( < ) is when < and 0 otherwise. You can use this idea to graph more complicated functions. For eample to enter + for 0 < < ou could enter (+)(0 < )( < ). Note: When graphing piecewise defined functions the graph is sometimes different from what is epected. See the common errors section to interpret the resulting graph below. Graphing with polar coordinates Suppose ou want to graph the curve r = θ +sinθ. Press lpmodelp. On the fourth line highlight Pol and press lpenter lp (make sure ou in radian mode also) and press lp= lp. Your menu will have a list r,r,... After r tpe lpx,t,θ,nlp lp+ lp lpsinlp lpx,t,θ,nlp lp)lp. Then to see the graph tpe lpgraph lp. You ma have to adjust our window (zoom fit) to see the resulting spiral. Graphing with parametric coordinates Suppose ou want to graph the curve given b (t) = sin(t),(t) = cos(t). First we need to set the mode to parametric mode: press lpmodelp and highlight PAR in the fourth line and

371 A.5. USING THE TABLE 6 press lpenter lp. Now go to the graphing screen lp= lp. You will see a list that begins with X T and Y T. To the right of X T tpe lp lp lpcoslp lpx,t,θ,nlp and to the right of Y T tpe lp lp lpsinlp lpx,t,θ,nlp. Then to see the graph press lpgraph lp. You ma need to lpzoom lp lp0 lp (zoomfit) to see the resulting ellipse. (Also, be aware that the scale ma be misleading ou ma want to lpzoom lp lp5 lp for a clearer understanding of the graph). A.5 Using the table The table ( lpnd lp lpgraph lp ) is useful as it gives a table representation of a function (from the lp= lp screen). Make sure there is a function tped in the Y postion and look at the table. If ou want to know the value of the function for particular values of then we must use the TBLSET screen. Press lpnd lp lpwindow lp. Move the cursor to the Indpnt line then to the right to highlight ask and press lpenter lp.

372 6 APPENDIX A. INTRODUCTION TO THE TI-8 Go back to the table lpnd lp lpgraph lp. Now our table is empt. Evaluate our function at.5 b tping.5 and pressing lpenter lp. You will see the value of the function in Y at.5 in the table now. You can evaluate at an number of points. You can also evaluate at ever half integer starting at for eample b going to the TBLSET screen ( lpnd lp lpwindow lp ) changing TblStart to lp( )lp lp lp and Tbl=.5. Also change the Indpnt setting to Auto again. View the table ( lpnd lp lpgraph lp ) to see what happened. A.6 Solving an equation using the solver We now solve the equation =. The first thing we must do in order to use the equation solver is to rewrite the equation as an epression equal to zero, that is we need to write it in the format something= 0. In our eample this will be = 0. Now, let s go to the equation solver. Press lpmathlp and use the cursor to go to the bottom of the list and press enter when Solver... is highlighted. Various things appear on the screen. Press the up arrow lp lp until EQUATION SOLVER appears on the top line. Edit the right hand side of the equation on the second line so that it reads 0 = and press

373 A.7. SPECIAL FUNCTIONS (ABSOLUTE VALUE, N-TH ROOT, ETC.) 6 lpenter lp. Net, on the second line pick a number near where ou epect the answer to be, for eample, =. Now solve b pressing lpalpha lp lpenter lp. The second line gives an answer (check that it is a square root of two!). What happens if ou had entered = on the second line (use here)? lp( )lp A.7 Special functions (absolute value, n-th root, etc.) We now point out some important functions in the MATH menu ( lpmathlp ) and the LIST menu ( lpnd lp lpstatlp ), that are used in this course. Fractions Suppose in the main screen ou evaluate /6. Your calculator will displa If ou wanted to see this as a fraction ou would tpe lpmathlp lpenter lp lpenter lp. You will now see / displaed.

374 6 APPENDIX A. INTRODUCTION TO THE TI-8 Absolute value To evaluate ou would tpe in the main screen lpmathlp lp lp lpenter lp lp( )lp lp lp lpenter lp. The answer should be displaed. To graph, go to the Y= screen ( lp= lp ) and enter in the Y (for eample) space lpmathlp lp lp lpenter lp lpx,t,θ,nlp lp lp lp lp lp lp lpenter lp. Now ou can go to the graphing screen to see the graph. n-th root To calculate the cube root of 8. Press on the main screen lp lp lpmathlp lp5 lp (or select the -root) lp8 lp lpenter lp. You can compute an root this wa (e.g. the 5th root of 7). For the cube root ou can also use lpmathlp lp lp instead. Recall also the algebraic definition of the nth root, n = n, so that the 5th root of 7 can also be computed b pressing lp7 lp lp lp lp(lp lp lp lp lp lp5 lp lp)lp. Factorials To compute 5!, for eample, in the main screen tpe lp5 lp lpmathlp then move the cursor to the right three times (or left once) so that the PRB menu is displaed. Option is! so press lp lp or move the cursor to highlight this option and press lpenter lp. Finall press lpenter lp

375 A.7. SPECIAL FUNCTIONS (ABSOLUTE VALUE, N-TH ROOT, ETC.) 65 and the answer 0 will be displaed. ( ) 5 Combinations and permutations To calculate 5 C =, in the main menu tpe lp5 lp. Then go to the math-probabilit menu b pressing lpmathlp then the right (or left) arrow until PRB is highlighted. You will see that the third option is n C r so press lp lp. This will return ou to the main screen. Press lp lp lpenter lp. The answer 0 will be displaed. Permutations n P r are handled the same wa ecept it is the second option under the math-probabilit menu instead of the third. Sequences and series A sequence a,a,a,... that is given b a formula can be added via the LIST menu (press lpnd lp lpstatlp ). For eample, to enter the first 0 terms a,a,...,a 0 of the sequence a n = n +, we have to write seq( +,,,0). Here, the sequence command takes four inputs, first the assignment +, second the independent variable, third the starting inde, and fourth the final inde 0. This epression is entered to the calculator b pressing lpnd lp lpstatlp lp lp lp5 lp lpx,t,θ,nlp lp lp lp+ lp lp lp lp,lp lpx,t,θ,nlp lp,lp lp lp lp,lp lp lp lp0 lp

376 66 APPENDIX A. INTRODUCTION TO THE TI-8 lp)lp and then confirmed b pressing lpenter lp. We obtain: To add the ten numbers a + +a 0 of the sequence, we use the sum command in the LIST-MATH menu. Press lpnd lp lpstatlp lp lp lp5 lp to enter the sum( epression. Then using the previousl entered answer b pressing lpnd lp lp( ) lp, and finishing with lp)lp lpenter lp, we now calculate the wanted sum. A.8 Programming the calculator This section is more advanced and is onl recommended for students who are familiar with programming languages. Hello world A Hello World program is a program that onl displas the words Hello World. It is a starting point to understand the procedure of creating and running a program before going to more advanced programming techniques. To create a new program press lpprgmlp, move to the right lp lp to NEW, and give it the name HELLO.

377 A.8. PROGRAMMING THE CALCULATOR 67 Net, we need to add the Displa command. Press lpprgmlp ; ou see a list of commands. We move to the right to I/O for input/output commands. Select the third item Disp for displa. To finish, we need to add HELLO WORLD. (Quotation marks obtained b pressing lpalpha lp lp+ lp ; space is obtained b lpalpha lp lp0 lp.) The complete program is shown below. To eecute the program, quit lpnd lp lpmodelp, press lpprgmlp and confirm with lpenter lp. After another lpenter lp, the program should run as shown below. The quadratic formula We net implement the quadratic formula a +b+c = 0 = = b± b ac. a Create a new program QUADRATI. We now need a new command, the Prompt command which will ask for three input values and places them in the variables

378 68 APPENDIX A. INTRODUCTION TO THE TI-8 A, B, C. The Prompt command is the second item in the I/O menu. After eecuting the new program, enter an input values (for eample A=, B= 0, C= 8). The answer should appear below. Directional angle Our third and last program calculates the directional angle of a vector v = a,b. Recall that the angle is tan (b/a) or tan (b/a)+ 80 depending on the quadrant in which v is pointing. In quadrant II and III, we need to add 80 degrees. This is eactl the case when a < 0. Thus, we prompt for input values A and B, and we calculate tan (b/a) and place it into the new variable G. (The arrow is obtained b pressing lpstolp.) We then check if a < 0, in which case we add 80 to G. (Here, we are assuming that the calculator is in degree mode). Note that the If command is obtained b pressing lpprgmlp, and < is obtained from the TEST menu, press lpnd lp lpmathlp and then lp5 lp.

379 A.8. PROGRAMMING THE CALCULATOR 69 We ma now run the program and check its correctness in some eamples. Other projects Here is a list of projects that ou ma want to tr to implement on our calculator.. Displa the final amount of an investment for a given principalp, annual interest rater, number of compoundings per earn, and number of ears t (see Observation 6.8 on page ).. In the program above for the directional angle, check for the cases when a = 0, and displa the correct answers.. Implement the long division (see Session 8). Erasing a program To erase a program that ou have written, press lpnd lp lp+ lp, then lp lp, then lp lp. A list of all our programs appears. Select the program ou want to delete with lp lp lp lp and press the lpdellp ke. Confirm our selection with lp lp Yes.

380 70 APPENDIX A. INTRODUCTION TO THE TI-8 A.9 Common errors The following shows man common errors and mistakes that ma produce a false result or an error. If none of the items below help to resolve our problem, then ou can also reset the calculator to factor setting (see section A.0 on page 7). Reset window to standard window Man problems with graphing can be resolved b changing the window back to the standard window. To do so, press lpzoom lp lp6 lp. Note, that the standard window has the following settings: PLOT marked Sometimes an error is produced when graphing a function due to having an of the PLOTs marked in the Y= menu. Make sure that all of the PLOTs are unmarked!

381 A.9. COMMON ERRORS 7 Errors in graphing functions Graph the functions = ln( ) or = 9. The calculator graphic is not reall the graph, but has etra parts or missing parts of the graph (depending on the version of the calculator: TI-8 is much worse than TI-8). The reason for this is that the calculator just approimates the graph piel b piel and does not represent the eact graph. In particular, ou should not just cop the graph from the calculator onto our paper, but interpret what ou see and draw the interpreted graph! Fractions need parenthesis for numerator and denominator When entering a rational function, (or an fraction), the numerator and denominator has to be entered with parenthesis. For eample, = is entered as follows. Radian versus degree Trigonometric functions should generall be graphed in radians. For eample, the cosine function in radian gives the following.

382 7 APPENDIX A. INTRODUCTION TO THE TI-8 On the other hand, the cosine in degrees displas as follows. The reason is that one period is now 60 on the -ais. Rescaling the calculator to a wider -scale shows the effect more clearl. The calculator also gives different values for the trigonometric functions in degrees or radians. This is shown below. gives gives Table setup The table can be setup to generate a list of outputs, or to take an input value and generate its output value. This depends on the independent variable in the TABLE SETUP being set to Auto or Ask, respectivel.

383 A.0. RESETTING THE CALCULATOR TO FACTORY SETTINGS 7 Using the wrong minus sign Often a snta error is due to using the minus sign lp lp instead of lp( )lp. Note that lp lp is used to subtract two numbers, whereas lp( ) lp gives the negative of a number. For eample, to calculate 5, press lp( ) lp lp lp lp lp lp5 lp. A.0 Resetting the calculator to factor settings To reset everthing and return the calculator to factor settings, first press the buttons lpnd lp lp+ lp to get to the MEMORY menu, and then lp7 lp ( Reset... ). Then move the cursor to the right (with lp lp ) until ALL is highlighted. Press lpenter lp ( All Memor... ) and lp lp ( Reset ). (With some of the other options in the MEMORY/Reset... menu ou can also reset selective parts of the calculator onl.)

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