Notes and Workbook

Notes and Workbook
WISKUNDE OPKNAPPINGSKURSUS/ MATHEMATICS REFRESHER COURSE NOTAS EN WERKBOEK VIR/ NOTES AND WORKBOOK FOR 2015
NOORDWES UNIVERSITEIT: POTCHEFSTROOMKAMPUS/ NORTH‐WEST UNIVERSITY: POTCHEFSTROOM CAMPUS
Materiaal saamgestel deur Material compiled by: Vakgroep Wiskunde van die Fakulteit Subject Group Mathematics of the Faculty of Natuurwetenskappe Natural Sciences Vakgroep Wiskunde‐onderwys van die Subject Group Mathematics Education of the Fakulteit Opvoedingswetenskappe Faculty of Education Sciences Bladuitleg deur Rudi van de Venter, Vakgroep Page layout by Rudi van de Venter, Subject Wiskunde‐onderwys Group Mathematics Education Kopiereg © 2015 uitgawe. Revision date 2015 Hersieningsdatum 2015 Noordwes‐Universiteit, Potchefstroomkampus Kopiereg voorbehou. Geen gedeelte van hierdie dokument, mag in enige vorm sonder skriftelike toestemming van die skrywers weergegee word nie. Dit sluit fotokopiëring van die hele, of gedeeltes van die dokument, in. Copyright © 2015 edition. North‐West University, Potchefstroom Campus Copyright reserved. No part of this document may in any form be represented without the written authorisation of the authors. That includes copying any part or parts of this document by any electronic means. Inhoud 0 1 Algemene inligting/ General information ................................................................................. 3 0.1 Verwelkoming/ Welcome ........................................................................................................ 3 0.2 Die doel van hierdie kursus/ Purpose of this course ............................................................... 3 0.3 Studiemateriaal/ Study material ............................................................................................. 4 0.4 ‘n Nuttige beskrywing van wat Wiskunde is/ A useful description of what Mathematics is .. 5 0.5 Hoe om Wiskunde te bemeester/ How to master Mathematics ............................................ 9 Algebraïese vaardighede en eksponente/ Algebraic proficiency and exponents ..................... 11 1.1 Instruksies, aksiewoorde en inleidende begrippe in wiskunde/ Instructions, action words and introductory concepts in mathematics ...................................................................... 12 2 1.2 Die reële getalle/ The real numbers ...................................................................................... 20 1.3 Eksponente/ Exponents ......................................................................................................... 28 1.4 Eenvoudige eksponensiële vergelykings/ Simple exponential equations ............................. 37 Logaritmes/ Logarithms ......................................................................................................... 40 2.1 Logaritmes en eksponente/ Logarithms and exponents ....................................................... 41 2.2 Die oplos van ingewikkelder eksponensiële vergelykings d.m.v logaritmes/ Solving more complicated exponential equations using logarithms....................................................... 49 3 Inleiding tot funksies/ Introduction to functions ..................................................................... 52 3.1 Definisie van'n funksie en inleidende aspekte/ Definition of a function and introductory aspects ............................................................................................................................... 52 3.2 Definisie en waardeversameling/ Domain and range ........................................................... 54 3.3 Inverse van 'n funksie/ Inverse of a function ........................................................................ 64 3.4 Bewerkings met funksies en saamgestelde funksies/ Operations with functions and composite functions .......................................................................................................... 68 4 Radiaalmaat en trigonometrie/ Radian measure and trigonometry ....................................... 73 4.1 Radiaalmaat/ Radian measure .............................................................................................. 74 1 4.2 Berekening van booglengte/ Calculation of arc length ......................................................... 79 4.3 Berekening van die oppervlakte van ‘n sirkelsektor/ Calculation of the area of a circle sector
........................................................................................................................................... 81 4.4 Die ses trigonometriese verhoudings en hul funksiewaardes in al vier kwadrante van die platvlak/ The six trigonometric ratios and their function values in all four quadrants of the flat plane ........................................................................................................................... 85 5 6 4.5 Die som‐ en verskilformules/ The sum and difference formulae ........................................ 108 4.6 Die dubbelhoekformules/ The double angle formulae ....................................................... 112 Absolute waardes en limiete/ Absolute values and limits ..................................................... 117 5.1 Ongelykhede/ Inequalities .................................................................................................. 118 5.2 Absolute waardes/ Absolute values .................................................................................... 123 5.3 Limiete en kontinuïteit/ Limits and continuity .................................................................... 141 5.4 Berekening van sekere limiete/ Calculation of certain limits ............................................. 158 5.5 Die epsilon‐delta definisie van 'n limiet/ The epsilon‐delta definition of a limit ................ 162 Inleiding tot funksie‐analise/ Introduction to function analysis ............................................ 165 6.1 Die afgeleide van 'n funksie uit eerste beginsels/ The derivative of a function from first principles ......................................................................................................................... 167 2 6.2 Differensieerbaarheid/ Differentiability .............................................................................. 177 6.3 Differensiasiereëls/ Differentiation rules ............................................................................ 180 6.4 Saamgestelde funksies en die kettingreël/ Composite functions and the chain rule ......... 186 6.5 Toepassing van differensiasie/ Application of differentiation ............................................ 194 0 Algemene inligting/ General information 0.1 Verwelkoming/ Welcome Ons wens u as voornemende Wiskunde‐ student geluk met u keuse om universiteitstudies aan te pak. We congratulate you as prospective Mathematics student with your decision to commence university studies. Dit is ‘n voorreg om u te ontmoet en ‘n rol te mag speel by u voorbereiding op u studies. It is a privilege to meet you and to play a part in your preparation for your studies. Ons as akademiese personeel hoop dat u studies van die begin af voorspoedig en suksesvol sal verloop. We as academic staff hope that your studies will right from the beginning proceed expediently and successfully. 0.2 Die doel van hierdie kursus/ Purpose of this course Hierdie kursus is ontwerp om: This course is designed to: •
U voorkennis van Wiskunde te aktiveer •
Activate your previous knowledge regarding Mathematics •
U brein se abstrakte en logiese werking na die vakansie aan die gang te kry •
Jump‐start the abstract and logical operation of your brain after the holidays •
Vae kolle in u wiskunde‐kennis en wiskunde‐vaardighede op te vul •
Fill in hazy spots in your mathematical knowledge •
U met noodsaaklike kennis en vaardighede waarmee u miskien nie voorheen kennis gemaak het nie, toe te rus •
Equip you with essential knowledge and skills with which you possibly have not become aquainted before •
Introduce you to the methodogy followed during the teaching of university modules •
Break down the artificial compartments between the different “parts” or “sections” of Mathematics and to reveal the relationships between each “part” and the whole, as well as the relationships between Mathematics and other subject disciplines •
Alter your experience of Mathematics •
•
U bekend te stel aan die werkwyse wat tydens die onderrig van universiteitswiskunde‐modules gevolg word Die kunsmatige afskortings tussen verskillende “gedeeltes” van Wiskunde af te breek en die verbande tussen elke “deeltjie” en die groot geheel aan te toon, asook die verbande tussen Wiskunde en ander vakdissiplines 3 •
as subject discipline in order for it to acquire more meaning and life for you than ever before U belewenis van Wiskunde as vakdissipline te verander sodat dit vir u meer betekenis en lewe as ooit voorheen sal kry 0.3 Studiemateriaal/ Study material In die ou dae het die studiemateriaal vir enige Wiskunde‐kursus uit een of meer handboek en klasaantekeninge bestaan, aangevul deur addisionele materiaal vanuit biblioteke. Die prentjie is nou egter anders; u studiemateriaal vir enige moderne Wiskunde‐
kursus sal waarskynlik ‘n kombinasie wees van die volgende: •
Klasaantekeninge (nota’s) In the old days, the study material for any typical mathematics course consisted of one or more text book and class notes, supported by additional material from libraries. The picture is, however, now very different; your study material for any modern Mathematics course will probably consist of a combination of the following: •
Class notes • een of meer handboeke • One or more text books • leermateriaal wat vanaf betroubare internetbronne verkry is • Study material obtained trustworthy internet sources • draagbare sakrekenaar (met of sonder die vermoë om grafieke en tabelle te genereer) • Portable calculator (with or without the ability to generate graphs and tables) • interaktiewe rekenaartegnologie soos gespesialiseerde programme wat op lessenaar‐rekenaars en skootrekenaars geïnstalleer word • Interactive computer technology such as specialized programs (software) which is installed on desktop or laptop computers • applikasies (“apps”) wat op slimfone en tabletrekenaars geïnstalleer word • Applications (“apps”) which are installed on smart phones and tablet computers from U moet onmiddellik besef dat die dae vir ewig verby is dat u alle materiaal wat u benodig, in u handboek en by die dosent gaan kry. You must immediately realize that the days are forever gone when you received all the material which you require from the lecturer. U as wiskundige moet self ‘n plan kan maak indien u met u Wiskunde‐studie vasbrand – u en u klasmaats moet byvoorbeeld die vermoë ontwikkel om ‘n soekenjin (bv. Google) in te span om inligting oor enige stuk wiskunde op te spoor. U het nie ‘n lessenaar‐rekenaar of skootrekenaar hiervoor nodig nie; u selfoon het al vir jare ‘n webblaaier waarmee u die You as mathematician must be able to devise a plan whenever you get stuck with your Mathematics studies – you and your class mates must for example develop the ability to access a search engine (i.e. Google) in order to find information regarding any piece of mathematics. You do not need a desktop or laptop computer in order to accomplish this; 4 soekenjins oopmaak. kan bereik en webblaaie Webblaaie soos Wolfram Mathworld is uiters waardevol en raak elke dag meer gebruikersvriendelik. U moet ook aanleer om u redenasies en berekeninge self te toets. In die meeste gevalle is die gevorderdste tegnologie wat u sal benodig ‘n potlood, papier en miskien ‘n sakrekenaar – Wiskunde as vakgebied is self ‘n groot abstrakte stuk tegnologie. U vermoë om onafhanklik te kan leer sal een van u magtigste gereedskapstukke in die studie van Wiskunde wees. for many years your cell phone has been equipped with the ability to access a search engine and open web pages. Web pages such as Wolfram Mathworld are extremely valuable and becomes more user‐
friendly by the day. You must also acquire the ability to privately check your reasoning and calculations. In most cases you would need nothing more than a pencil, paper and perhaps a calculator – Mathematics itself as subject discipline functions as one enormous abstract piece of technology. Your ability to study and learn independently will be one of your most powerful tools in the study of Mathematics. 0.4 ‘n Nuttige beskrywing van wat Wiskunde is/ A useful description of what Mathematics is Wiskunde is ‘n groot versameling abstrakte idees wat almal op ‘n groot aantal maniere met mekaar verband hou. Hierdie verbande gaan oor die betekenis van die idees. Mathematics is a large set of abstract ideas which are all connected to one another in a variety of ways. These connections have to do with the meaning of the ideas. Wiskunde is onder meer ‘n masjien of taal waarmee abstrakte sowel as konkrete verskynsels op ‘n elegante manier beskryf kan word – ‘n beskrywing so elegant dat dit nie staties is nie maar gemanipuleer kan word volgens sekere afsprake en reëls om nuwe lig op die oorspronklike verskynsel te werp. Sommige geleerdes beskryf Wiskunde as die “maak van patrone” – soms is hierdie patrone sigbaar en meetbaar en die gevolge van menslike waarnemings; ander kere is hulle onsigbaar of selfs onmeetbaar en kan hulle slegs simbolies, numeries of grafies voorgestel word deur middel van modelle. Mathematics is, among other things, a machine or language with which abstract as well as concrete phenomena may be described in an elegant way – a description so elegant that it is not static but capable of being manipulated according to certain conventions and rules in order to shed new light on the original phenomenon. Some thinkers describe Mathematics as “the making of patterns” – sometimes these patterns are visible and measurable and the consequence of human observations; sometimes they are invisible and even immeasurable and can they only be represented symbolically, numerically or graphically by means of models. Wiskundige en wetenskaplike modelle is abstrakte denkvoorstellings van werklike of denkbeeldige patrone. Hierdie modelle kan Mathematical and scientific models are abstract mental representations of real or 5 verskillende vorme aanneem, bv. •
‘n stel formules •
tabelle gevul met numeriese data •
‘n grafiese voorstelling • ‘n rekenaarsimulasie • ‘n woordelikse beskrywing Die Wet van Ohm wat in Graad 9 bespreek word is ‘n voorbeeld van ‘n wiskundige model vir die gedrag van stroom, weerstand en potensiaalverskil in ‘n eenvoudige elektriese stroombaan. imaginary patterns. These models may take different forms, for example. •
a set of formulae •
tables filled with numeric data •
a graphical representation • a computer simulation • a verbal description The Law of Ohm which was discussed in Grade 9 is an example of a mathematical model for the behaviour of current, resistance and potential difference in a simple electrical circuit. Hierdie wiskundige modelle werk volgens die beginsels wat in die teorie van Wiskunde geformuleer en uitgedruk word. These mathematical models operate according to the principles which are formulated and expressed in the theory of Mathematics. Wiskundige en wetenskaplike teorie is ‘n baie belangrike begrip, aangesien die woord teorie in die konteks van wetenskap en Wiskunde ‘n heel ander betekenis besit as die betekenis wat dit gewoonlik in die alledaagse lewe het. Mathematical and scientific theory is a very important concept, because the meaning of the word theory is very different in the context of Mathematics and science than its meaning in everyday life. In Wiskunde en wetenskap beteken die woord “teorie” NIE ‘n raaiskoot of losstaande idee of vermoede nie. ‘n Wiskundige of wetenskaplike teorie is ‘n robuuste stelsel samehangende idees en verbande wat suksesvol gebruik kan word om waarnemings te verklaar en voorspellings te maak. ‘n Teorie geld slegs wanneer dit sin maak en praktiese waarde het. Drie voorbeelde van suksesvolle teorieë: Die teorie van eksponente en logaritmes bevat die versameling begippe, definisies, verbande of verwantskappe, formules en betekenisse wat ons in staat stel om onder meer die eindwaarde van ‘n vaste belegging teen saamgestelde rente te bereken. Eksponente en logaritmes is abstrakte 6 In Mathematics and science the word “theory” does NOT mean a guess or loose idea or suspicion. A Mathematical or scientific theory is a rigorous system of consistent ideas and relationships which may successfully be used to explain observations and make predictions. A theory is only valid as long as it makes sense and has practical value. Three examples of successful theories: The theory of exponents and logarithms contains the set of concepts, definitions, connections or relationships, formulae and meanings which enable us to calculate, among other things, the final value of a fixed investment at compounding interest. Exponents and logarithms are abstract phenomena, as are the laws, formulae and verskynsels, so ook die wette, formules en verbande wat daarop van toepassings is – maar die teorie werk, aangesien ons dit prakties kan gebruik om betekenisvolle antwoorde op berekeninge te kry. Die teorieë van gravitasie bevat die versameling begrippe, definisies, verbande of verwantskappe, formules en betekenisse wat ons in staat stel om byvoorbeeld die gedrag van ‘n vryvallende voorwerp te beskryf en selfs te voorspel. Gravitasie self is onsigbaar en so ook die universele swaartekragkonstante en die algebraïese reëls – die elemente van die Algemene Relatiwiteitsteorie is nog meer abstrak – en tog beskryf, verklaar en voorspel hierdie teorieë die waarnemings wat ons maak wanneer voorwerpe aan gravitasie onderwerp word. Die teorie van evolusie bevat die versameling begrippe, definisies, wetmatighede en betekenisse wat ons in staat stel om die lang‐
termyn patrone wat by die studie van spesies in die natuur waargeneem word, te verklaar. Dit verklaar en beskryf die fossiele wat waargeneem word, asook die eienskappe van die DNS van organismes. Groot dele van die moderne mediese wetenskap is die gevolg van die suksesvolle toepassing van die teorie van evolusie. Hierdie teorie verklaar ook waarom die meeste moderne organismes (die mens ingesluit) se DNA tot ‘n baie hoë persentasie met dié van ander organismes (selfs plante en diere) ooreenstem. Teorie is dus uiters noodsaaklik by enige wetenskap, ook by Wiskunde: Dit verskaf die raamwerk waarbinne en die meganismes waarmee die bepaalde studieveld werk. Daarom is dit fataal as u Wiskunde sien as ‘n blote versameling van resepte en reëls; Wiskunde gaan oor die patrone en verbande (konneksies) tussen abstrakte begrippe – die sogenaamde reëls en wette van Wiskunde is eintlik veralgemenings van die patrone en verbande wat ons ontdek wanneer ons met relationships applicable to them – but the theory works, since we can use it practically to obtain meaningful answers to calculations. The theories of gravitation contain the set of concepts, definitions, connections or relationships, formulae and meanings which enable us to, for example, describe and even predict the behaviour of a free‐falling object. Gravity itself is invisible and that is also the case with the universal constant of gravitation and the algebraic rules – the elements of the General Theory of Relativity are even more abstract – and yet, these theories describe, explain and predict the observations we make when objects are subjected to gravity. The theories of gravitation contain the set of concepts, definitions, connections or relationships and meanings which enable us to explain the long‐term patterns observed in nature in the studies of species. It explains and describes the fossils observed, as well as the properties and characteristics of the DNA of organisms. Large parts of modern medical science is the consequence of the successful application of the theory of evolution. This theory also explains why a high percentage of the DNA of most modern organisms (including humans) is similar to the DNA of other organisms (even plants and animals). Theory is therefore extremely necessary in any science, including Mathematics: It provides the framework in which and the mechanisms with which the particular field of study operates. That is why it is fatal to view Mathematics as a mere collection of recipes and rules; Mathematics deals with the patterns and relationships (connections) between abstract concepts – the so‐called rules and laws of Mathematics are actually generalizations of the patterns and relationships which we discover when we work with numbers, symbols, operations, etc. Therefore, your knowledge and skills regarding Mathematics may not exist in 7 getalle, simbole, bewerkings ens. werk. U kennis en vaardighede van Wiskunde mag dus nie in netjiese afsonderlike kompartemente bestaan, soos gereedskapstukke wat op die rakke van ‘n pakkamer gestoor word nie. Dit moet eerder ‘n netwerk wees van begrippe en vaardighede, waar elke deel van die netwerk op baie maniere met al die ander dele van die netwerk verbind is. So is die subvelde van Wiskunde, byvoorbeeld algebra, Euclidiese meetkunde, trigonometrie en differensiasie, kunsmatige indelings wat mense maak om die vakgebied te organiseer – daar bestaan in werklikheid geen werklike skeidings tussen hierdie subvelde nie en elkeen van hulle is met elkeen van die ander verbind. Wanneer ons ‘n wiskundige probleem oplos, gebruik ons tipies idees wat uit meer as een subveld kom. Die bewys van ‘n trigonometriese identiteit, byvoorbeeld, benodig kennis van meetkunde, trigonometrie en algebra. U studie van Wiskunde verloop dus nie soos ‘n huis wat onbeweeglike steen vir onbeweeglike steen, laag op laag, netjies van onder na bo opgebou word nie – dit verloop eerder soos ‘n boom, wat van onder na bo groei deur te vertak en weer te vertak, elke takkie uniek en afsonderlik maar tog aan elke ander deel van die boom verbind as ‘n vaste maar buigsame eenheid. 8 neat, separate compartments, like the tools stored on the shelves of a store room. Rather, it should consist of a network of concepts and skills, where each part of the network is connected to every other part of the network in may ways. The sub‐fields of Mathematics, for example algebra, Euclidean geometry, trigonometry and differentiation, are artificial subdivisions made by humans in order to organize the subject field – in reality, there does not exist any partition between these sub‐fields and each one of them is connected to every one of the others. When we solve a mathematical problem, we typically use ideas which we extract from more than one sub‐field. The proof of a trigonometric identity, for example, requires knowledge of geometry, trigonometry and algebra. Therefore, your study of Mathematics does not proceed like a house which is built by neatly stacking static bricks, layer upon layer, from bottom to top – rather, it proceeds like a tree, which grows from below to above by means of repeated branching, each branch unique and separate but yet connected to every other part of the tree as a solid but flexible unit. 0.5 Hoe om Wiskunde te bemeester/ How to master Mathematics Die studie van Wiskunde verg fokus en tyd. •
•
•
•
Maak ten alle tye seker dat u elke stukkie van elke bespreking verstaan Wiskunde is die wetenskap van betekenis‐making. Indien u ‘n stuk werk moet memoriseer omdat u die betekenis daarvan nie kan verstaan nie, dan is u nie meer met Wiskunde besig nie. Indien u aanvanklik nie ‘n greep op ‘n begrip of metode kan kry nie, werk verder en probeer ‘n voorbeeld in die hande kry waar met daardie begrip of metode gewerk word. Werk hierdie voorbeeld deur en gaan dan terug na die bespreking wat u aanvanklik nie kon verstaan nie. Probeer insien waar die nuwe stukkie werk inpas en met watter vorige werk wat u al gedoen het, dit verband hou. Soek patrone en konneksies – daar is fout indien u enige nuwe stukkie werk as ‘n afsonderlike aparte deeltjie beskou wat losstaan van die res van u Wiskunde‐kennis. •
Maak gedurig planne om u antwoorde te toets. Dit help om u begrip uit te brei. •
Raak gemaklik daarmee om saam met een of meer klasmaats te werk. Bespreek enige nuwe openbarings wat u ontvang terwyl u deur ‘n stuk werk met mekaar. •
Maak gedurig u eie notas en opsommings in ‘n vorm wat u self goed verstaan. •
Bestee tyd. Dit is die belangrikste hulpbron. Kyk weer na die aanhaling op die voorblad van hierdie The study of Mathematics requires focus and time. •
Be certain at all times that you understand each part of each discussion •
Mathematics is the science of meaning‐making. If you have to memorize a piece of work because you cannot understand its meaning, then you are no longer busy with Mathematics. •
If you cannot at first establish a grip on a concept or method, work further and attempt to find an example where the particular concept or method is applied. Work through this example and then return to the discussion that you could not at first understand. •
Attempt to establish where the new piece of work fits in and to identify which previous work is connected to it. Look for patterns and connections – something is wrong as soon as you consider any new piece of work as a separate isolated part which stands disconnected from the rest of your Mathematics knowledge. •
Continually devise plans by which you can check your answers. That will help to extend and expand your understanding. •
You should become comfortable working with one or more of your classmates. Discuss with one another any new revelations you receive while working through a new piece of work. •
Continually make your own notes and summaries in a form which you 9 werkboek. understand well. •
10 Invest time. That is your most important resource. Consider again the quote on the cover of this workbook. Leereenheid 1 Algebraïese vaardighede en eksponente 1 Algebraïese vaardighede en eksponente/ Algebraic proficiency and exponents Leerdoelstellings vir hierdie leereenheid Learning aims for this study unit Upon completion of this study unit the Na afhandeling van hierdie leereenheid moet die student in staat wees om die volgende te doen: student must be able to do the following: 1. Distinguish between the following instructions/ action words: 1. Te onderskei tussen die volgende Simplify, factorize, solve, instruksies/ aksiewoorde: differentiate, prove the following Vereenvoudig, faktoriseer, los op, identity differensieer, bewys die identiteit 2. Te onderskei tussen die volgende konsepte: concepts: Expression, Uitdrukking, term, teller, noemer, faktor, oplossing, vergelyking 3. Die eienskappe van reële getalle en hulle bewerkings korrek toe te pas 4. Die absolute waarde van ‘n getal te definieer en die eienskappe daarvan in berekeninge en redenasies te gebruik denominator, term, numerator, factor, solution, equation 3. Correctly apply the properties of real numbers and their operations 4. Define the absolute value of a number and utilize its properties in calculations and reasoning 5. Use the properties of exponents in 5. Die eienskappe van eksponente te kan gebruik om uitdrukkings te vereenvoudig 6. Toepaslike 2. Distinguish between the following order to simplify expressions 6. Solve suitable exponential equations by exploiting the properties of eksponensiële exponents vergelykings op te los deur van die eienskappe van eksponente gebruik te maak 11 Leereenheid 1: Algebraïese vaardighede en eksponente 1.1 Instruksies, aksiewoorde en inleidende begrippe in wiskunde/ Instructions, action words and introductory concepts in mathematics Wiskunde as menslike aktiwiteit is eintlik 'n sekere manier om die konkrete en abstrakte wêreld te ondersoek. Wanneer 'n ingenieur, ekonoom, natuurwetenskaplike, onderwyser of wie ook al "wiskunde doen" beteken dit eintlik dat hy 'n situasie op 'n sekere manier beskou en volgens sekere kreatiewe beginsels probeer om daardie situasie beter te verstaan. Dit is eintlik verskriklik kunsmatig om wiskunde te reduseer tot "die doen van somme" of selfs tot "die oplos van probleme". Tog is dit dikwels wat van ons in wiskunde‐
klasse of tydens wiskundige take en toetse verwag word. Ons moet dus tog eers fokus op watter soort wiskundige probleme ons kan teëkom, sodat ons kan agterkom wat ons in elke geval moet doen om "die antwoord te kry". In wiskunde‐lesings, tydens wiskunde‐ oefensessies en gedurende wiskundetoetse kom ons instruksies of aksiewoorde teë wat vir ons aandui wat ons moet doen. Voorbeelde van sulke aksiewoorde is: Whenever an engineer, economist, natural scientist, teacher or any other person "does mathematics" it simply means that he is approaching a situation in a certain way and that he tries to gain better understanding of that situation by proceeding according to certain creative principles. It is really terribly artificial to reduce mathematics to "the doing of sums" or even to "the solution of problems". Still, that is often what is required of us in mathematics classes or during mathematical tasks and tests. So, we must first focus on what kind of mathematical problems we may encounter, so that we can become aware of what we must do in each case in in order to "obtain the answer". In mathematics lectures, during mathematics practice session and during mathematics tests we encounter instructions or action words which indicate what we must do. Examples of such action words are: •
Vereenvoudig •
Simplify •
faktoriseer •
factorize •
los op vir •
solve for •
differensieer •
differentiate •
bewys die identiteit •
prove the identity By die toepassing van wiskunde in ander vakgebiede soos rekeningkunde, ekonomie, 12 Mathematics as human activity is in actual fact a specific, particular way of investigating both the real world and the abstract world. In the application of mathematics to other fields of study such as accounting, Leereenheid 1 Algebraïese vaardighede en eksponente fisika, chemie, rekenaarwetenskap en statistiek is die werklikheidsgetroue probleme wat ons moet oplos, dikwels nie in terme van aksiewoorde gedefinieer nie: economics, physics, chemistry, computer science and statistics, the real‐life problems which we must solve are often not clearly defined in terms of action words: •
Bepaal die eindwaarde van die belegging •
Determine the final value of the investment •
Vergelyk die vraag‐ en aanbod‐
grafieke •
Compare the demand‐ and supply‐
graphs •
Bereken die maksimumhoogte van die projektiel •
Calculate the maximum altitude of the projectile •
Wat die konsentrasie van oplossing na 30 sekondes? •
What is the concentration of the solution after 30 seconds die By die toepassing van wiskunde in ander vakgebiede moet ons dus meestal self bepaal watter wiskundige instruksies of aksiewoorde geïmpliseer word. 'n Mens leer deur ervaring hoe om werlikheidsgetroue probleme in wiskundige instruksievorm te interpreteer. Usually, in the application of mathematics to other fields of study we ourselves must figure out which mathematical instructions or action words are implied. This skill by which a person interprets real‐life problems in mathematical instruction form is acquired through experience. Let ten slotte daarop dat "'n probleem" in wiskunde eintlik nie iets negatiefs is wat vermy moet word, soos wat die woord in die alledaagse lewe verstaan word nie. In wiskunde is "'n probleem" in werklikheid die instrument of gereedskapstuk waardeur ons tot kennis en insig van wiskunde self kom, asook begrip van die situasie waaruit die probleem spruit. Lastly, we should note that "a problem" in mathematics is actually not something negative which should be avoided – as implied by the word "problem" in everyday language. In mathematics "a problem" is in actual fact an instrument or tool by which we acquire knowledge and insight regarding mathematics itself as well as understanding regarding the situation which gave rise to the problem. 13 Leereenheid 1: Algebraïese vaardighede en eksponente 'n Mens leer wiskunde dus deur probleemsituasies te bestudeer en op allerlei maniere te hanteer. Meestal lê die waarde van 'n wiskundige probleem in die prosesse waardeur die probleem hanteer word, eerder as net in die oplossing. Daarom het elke wiskunde‐probleem wat u doen, waarde – selfs al kry u nie die "mees korrekte antwoord" in die hande nie. So a person learns mathematics by studying problem situations and treating them in various ways. Most of the time, the value of a mathematical problem is contained in the processes by which the problem is handled, rather than only in its solution. That is why every mathematics problem that you do, has value – even if you are unable to obtain the "most correct answer". Daar is baie aksiewoorde in wiskunde waarmee u nog in u verdere studie te doen sal kry. There are many action words in mathematics which you will encounter during further study. Maak seker dat u presies weet wat elke aksiewoord beteken en hoe die oplossing lyk wat by daardie aksiewoord (instruksie) pas. Always ensure that you know the precise meaning of each action word and also that you know the form and appearance of the solution associated with that action word (instruction). 14 Leereenheid 1 Algebraïese vaardighede en eksponente Oefening 1.1 Beskou elkeen van die volgende gegewe probleme. By elkeen moet u self besluit wat om te doen (wat die instruksie by daardie vraag moet wees). By sommige van hulle kan daar twee of meer moontlike instruksies wees. Doen dan die berekening(s) wat by die instruksie(s) pas. Exercise 1.1 Consider each of the following given problems. In each case, you must decide what to do (what the instruction for that question should be). In some cases there may be two or more possible instructions. Then do the calculations(s) which goes with the instruction(s). 1. −2 x 2 + 7 x − 6 Instruksie/Instruction: Instruksie/Instruction: Berekening/Calculation: Berekening/Calculation: 2.
( p − 2) ( p2 + 2 p + 4 ) Instruksie/Instruction: Instruksie/Instruction: Berekening/Calculation: Berekening/Calculation: 15 Leereenheid 1: Algebraïese vaardighede en eksponente 3.
(10 − 5t )( 2 + t )
t −3
= 0 Instruksie/Instruction: Instruksie/Instruction: Berekening/Calculation: Berekening/Calculation: 4. 30k 3 − 22k 2 − 28k = 0 Instruksie/Instruction: Instruksie/Instruction: Berekening/Calculation: Berekening/Calculation: 16 Leereenheid 1 Algebraïese vaardighede en eksponente 5.
f ( x) = 6 x −
3
+ 5 x2 x
Instruksie/Instruction: Instruksie/Instruction: Berekening/Calculation: Berekening/Calculation: 6.
P ( x ) = x 3 − 6 x 2 − x + 30 en/and P ( 3 ) = 0 Instruksie/Instruction: Instruksie/Instruction: Berekening/Calculation: Berekening/Calculation: 17 Leereenheid 1: Algebraïese vaardighede en eksponente 7.
x 3 − 6 x 2 − x + 30 = 0 en/and x + 2 is 'n faktor/is a factor Instruksie/Instruction: Instruksie/Instruction: Berekening/Calculation: Berekening/Calculation: 8. sinθ = 1 − cos2 θ Instruksie/Instruction: Berekening/Calculation: 18 Leereenheid 1 Algebraïese vaardighede en eksponente 9. Gegee/Given Instruksie/Instruction: Instruksie/Instruction: Berekening/Calculation: Berekening/Calculation: Soos u kan sien, is elke probleem op sy eie manier interessant, dikwels met verskillende aspekte waarin ons kan belang stel. As you can see, each problem is interesting in its own way, often with different aspects in which we might be interested. 19 Leereenheid 1: Algebraïese vaardighede en eksponente 1.2 Die reële getalle/ The real numbers Getallestelsels Ons klassifiseer getalle Number systems (abstrakte We classify numbers (abstract man‐made mensgemaakte simboliese en konseptuale symbolic and conceptual representations voorstellings wat aantal en hoeveelheid which indicate amount and quantity) as voorstel) volgens hulle eienskappe soos volg: follows according to their properties: (Voorsien die beskrywende name van elkeen (Supply descriptive names for each of the van die volgende versamelings) following sets) Begenoemde verwantskappe word ook in The relationships above may also be versamelingsnotasie soos volg geskryf: expressed in set notation as follows: = {1; 2; 3; 4;...} 0
20 = {0;1; 2; 3; 4;...} (natuurlike getalle/ natural numbers) (telgetalle/ whole numbers) = {...; − 4; − 3; − 2; − 1; 0;1; 2; 3; 4;...} (heelgetalle/ integers) ⎧
⎫
p
= ⎨ x x = , p ∈ , q ∈ , q ≠ 0⎬ q
⎩
⎭
(rasionale getalle/ rational numbers) Leereenheid 1 Algebraïese vaardighede en eksponente I = {x x ∈ , x ∉
= {x x ∈
} of / or x ∈ I} (irrasionale getalle/ irrational numbers) (reële getalle/ real numbers) Dit volg dat al die versamelings hierbo genoem deelversamelings is van die reële ⊂ 0⊂ ⊂ ⊂
en getalle, d.w.s. It follows that all the sets above are subsets of the set of real numbers, in other words: ⊂ 0 ⊂ ⊂ ⊂ and I ⊂ I⊂
Further, it also follows that ∪I =
Verder volg dat en ∩ I = ∅ . ∪I =
and ∩ I = ∅ . Wat beteken bogenoemde in woorde? U MOET weet! What does the symbology above mean in words? You MUST know! Basiese eienskappe van die reële getalle Basic properties of the real numbers Gestel a en b en c is reële getalle. Dan geld: Suppose a and b and c are real numbers. Then the following hold: 1. Eienskappe van die getal nul (optellingsidentiteitselement) 1. Properties of the number zero (additive identity element) 1.1
a + 0 = ................ 1.1
a + 0 = ................ 1.2
a × 0 = ................. 1.2
a × 0 = ................. 1.3
0
= ....... met a ≠ 0 a
1.3
0
= ........ with a ≠ 0 a
1.4
a
is ongedefineerd .........
1.4
a
is undefined .........
1.5
As ab = ......... dan is a = 0 of b=0 1.5
(.........‐produk‐stelling) 1.6
(...........‐product theorem) a
= 0 dan is a = .......... en b
b ≠ ............ As 1.6
2. Bewerkingseienskappe van reële getalle 2.1
Geslotenheid a + b∈
2.2
Kommutatiwiteit en ab ∈
a + b = .............. en
ab = ............
If ab = ......... then a = 0 or b=0 a
= 0 then a = .......... b
and b ≠ ............ If 2. Operational properties of real numbers 2.1
Closure a + b∈
2.2
Commutativity and ab ∈ a + b = .............. and
ab = ............
21 Leereenheid 1: Algebraïese vaardighede en eksponente 2.3
Assosiatiwiteit 2.3
( a + b) + c = a + (b + c )
( ab ) c = a ( bc )
en
(a + b) + c = a + (b + c )
( ab ) c = a ( bc )
2.4
Identiteit: a + 0 = ........... en
a × 1 = ............
2.5
a × 1 = ............
2.5
1
= ........
a
Distributiwiteit: (berus op afsprake en patrone) 3.3
3.4
22 Distributivity: a ( b ± c ) = ...........................
(based on patterns) 1
a
2.6
3. Order/ priority operations 3. Volgorde van algebraïese bewerking Vir herhaalde optelling en 1
= ........
a
where a ≠ ........ in a ×
a ( b ± c ) = ........................... 3.2
Inverse: a×
1
waar a ≠ ........ by a ×
a
3.1
a + ( − a ) = ....... and
a + ( −a ) = ....... en
2.6
Identity: a + 0 = ........... and
Inverse: a×
and
2.4
Associativity 3.1
of algebraic conventions and For repetitive addition and aftrekking werk ons van links subtraction we work from na regs. left to right. Vir herhaalde 3.2
For repetitive multiplication vermenigvuldiging en deling and division we work from werk ons van links na regs. left to right. Vir gekombineerde 3.3
For combined operations bewerkings kan ons nie net we cannot simply proceed van links en regs werk nie, from left to right, but we maar word optelling en should take care to perform aftrekking laaste gedoen. addition and subtraction Uitdrukkings in hakies word last. Expressions in brackets eerstens verreken – dan must be evaluated first – vermenigvuldiging en deling then multiplication and (gelyke prioriteit) en laastens division and lastly addition word opgetel en afgetrek. and subtraction. Vir gekombineerde 3.4
For combined operations Leereenheid 1 Algebraïese vaardighede en eksponente bewerkings is die volgorde the order is as follows: soos volg: Prioriteit/ Bewerking/ Verduideliking Priority Operation Explanation 1 () hakies/ parentheses Met hakies binne hakies word die binneste hakies eerste verwyder/ With sets of parentheses inside other sets of parentheses, proceed from inside out 2 Magsverheffing en Worteltrekking/
Moet as spesiale hakies beskou word/ Raising to powers and applying radical operations 3 Should be treated as a special case of parentheses vervang met ‘n × ‐teken/ Van/ of substitute with a multiply‐sign 4 × en/ and /of/ or ÷ Gelyke prioriteit/ same priority 5 + en/ and /of/ or − Gelyke prioriteit/ same priority Sommige wiskundiges vereenvoudig Some mathematicians simplify the scheme bogenoemde skema soos volg: above as follows: Identifiseer alle plus en minus tekens buite Identify all plus and minus signs outside hakies. Evalueer dan alle uitdrukkings brackets. Then evaluate all expressions tussen plus en minus tekens (faktore) en tel between plus and minus signs and lastly, add laastens op en trek af. and subtract. Die belangrikste sake wat u omtrent algebra moet verstaan, is dat: •
The most important matters which you must grasp regarding algebra is: enige uitdrukking of formule uit •
Any expression or formula consists of terme saamgestel is – die terme terms – the terms are separated by word plus or minus signs (radicals, deur minustekens plustekens geskei en (wortels, exponents, multiplication and 23 Leereenheid 1: Algebraïese vaardighede en eksponente eksponente, maal en deel skei NIE division does NOT separate terms) terme van mekaar nie) •
•
Wat dus ook al tussen plustekens minus signs must be regarded as one en minustekens staan, moet as een number. “Algebraic terms” mean getal gelees en verstaan word. ‘n “pieces of an expression which is “Algebraïese separated by plus or minus signs”. terme” beteken “stukke van ‘n uitdrukking wat deur •
plus en minus van mekaar geskei Hakies Parentheses group everything within together as one numeric value – that word”. •
Whatever stand between plus or is why we always evaluate the groepeer alles wat contents of a set of parentheses first. daarbinne saam as een getalwaarde •
– daarom bereken ons altyd die A horizontal division sign automatically places everything in inhoud van ‘n stel hakies heel the eerste. numerator and in the denominator in brackets. ‘n horisontale deelteken (lyntjie met teller bo en noemer onder) plaas outomaties alles in die teller en alles in die noemer in hakies. 4. Ontoelaatbare bewerkings by reële 4. Illegal or non‐permissible operations and real numbers getalle Vereenvoudig die volgende uitdrukking: Simplify the following expression: 3
7
+ log10 ( −10 ) +
− 2 3 + 3 −64 + sin−1 ( 2 ) + cos−1 ( −1,5 ) − tan ( 90° )
log
1
7 + −343
10 ( )
3
= .................................................................................................................................
Gebruik u bevindinge en maak ‘n lys van Use your findings and make list of non‐
ontoelaatbare bewerkings: permissible operations: ...................................................................... ...................................................................... ...................................................................... ...................................................................... 24 Leereenheid 1 Algebraïese vaardighede en eksponente ...................................................................... ...................................................................... ...................................................................... ...................................................................... ...................................................................... ...................................................................... ...................................................................... ..................................................................... Ons moet altyd bedag wees op gevalle soos hierbo, waar reële getalle interessante ongewone gedrag vertoon. We must always stay aware of cases like above, where real numbers exhibit interesting unusual behaviour. Oefening 1.2 1. Bepaal die waarde van die volgende uitdrukkings (met ander woorde, evalueer die volgende) sonder enige sakrekenaar: 1.1 3(4)
2
Exercise 1.2 1. Determine the value of the following expressions (in other words, evaluate the following) without a calculator: + 6 (2 + 3) −
27 − 7
7+3
=.................................................................................................................. =................................................................................................................. 1.2 4 − 3 ( −2 )( 3) +
53
12
−
− 52 − 32
2
25 2 +
5
Die simbool ' n reële getal staan as "die absolute waarde‐bewerking" bekend ‐ in die algemeen
is ' n reële getal = positiewe waarde van daardie reële getal. Later meer hieroor./
The symbol a real number is known as "the absolute value operation " ‐ in general , it holds that
a real number = positive value of that real number. Later more about this./
25 Leereenheid 1: Algebraïese vaardighede en eksponente =.................................................................................................................. =................................................................................................................. =................................................................................................................. 1.3 2 169 − 144 ( 8 − 16 )
100 − 25
+
− 64 − 128 +
2
3
5
2
(10 − 5)
=.................................................................................................................. =................................................................................................................. =................................................................................................................. 1.4 26 ⎛
⎜
⎝
⎛
⎜
⎜
⎝
⎛
⎜
⎜
⎜
⎜
⎜
0
7
6 ⎞ ⎜⎜
−3
⎟
1
144
216 ⎠ ⎜
+⎜
1
⎞
3
1000 ⎟ ⎜ 1
50
⎜ −2+
3 ⎟
48 − 3 ( 2 )
23
⎠ ⎜2 2
64
−
⎜
12 − 3
5−
⎜
72
⎜
9−
⎜
5 1
−
⎜
2 2
⎝
(
)
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
12− 8
5
⎛
⎞
2
⎜
5 ⎟
3 ⎞
⎟ + 10 4 ⋅ ⎛⎜ 1 +
− 3 ⎜ 12 −
⎟ ⎛1⎞⎟
⎝ 100 ⎠
⎜
⎜ ⎟⎟
⎜
⎝2⎠⎠
⎝
Leereenheid 1 Algebraïese vaardighede en eksponente =.................................................................................................................. =................................................................................................................. ................................................................................................................. ................................................................................................................. 1. Bepaal die waarde van die volgende uitdrukkings (met ander woorde, evalueer die volgende) sonder enige sakrekenaar: 2. Determine the value of the following expressions (in other words, evaluate the following) without a calculator: 2.1
4 − 16
= ...................................................................................................................... 3+2
2.2
12 − 8
= .................................................................................................................. 10 − 100
2.3 5 − 16 − 25 = ................................................................................................................ 2.4 3 − 2 169 − 25 = ........................................................................................................... 27 Leereenheid 1: Algebraïese vaardighede en eksponente 1.3 Eksponente/ Exponents 'n Maatskappy beoog om vir die volgende vyf jaar ‘n tariefverhoging van 20% per jaar in te stel op 'n diens wat in 2013 per eenheid R 1,10 gekos het. A company intends to apply a tariff increase of 20% per year for the next five years on a service which cost R 1,10 in 2013. Complete the following table and graph: Voltooi die volgende tabel en grafiek: Jaartal/Year 2013 2014 2015 2016 2017 2018 Term nr. ( n ) 1 2 3 4 5 6 Tarief/ Tariff 1,10 28 Leereenheid 1 Algebraïese vaardighede en eksponente Lê die punte in ‘n reguit lyn? Do the points lie in a straight line? ...................................... ................................... Bogenoemde is ‘n voorbeeld van eksponensiële gedrag; dit het met herhaalde vermenigvuldiging van ‘n sekere konstante waarde te doen. The situation above is an example of exponential behaviour; it deals with repetive multiplication by a certain constant value. Are you able to write down an equation for the graph in terms of n ? Kan u ‘n formule vir vergelyking van u grafiek in terme van n neerskryf? ................................................................... ................................................................... The formula for the general term of a geometric sequence is an example of an exponential function. Die formule vir die algemene term van ‘n meetkundige ry is ‘n voorbeeld van ‘n eksponensiële funksie. Ons definieer in die algemeen We define in general x × x × x × x × ... × x
= xn n faktore van x / n factors of x
As bv / if for example 2 x 3 + ( 2 x ) = 2 x 3 + 16 x 4 dan/ then 4
Gebruik u kennis en voltooi die ontbrekende inligting in die volgende tabel: Use your knowledge and complete the omitted information in the following table: 29 Leereenheid 1: Algebraïese vaardighede en eksponente Eienskappe van eksponente Properties of exponents Eienskap/ Property: Voorbeeld/ Example Pasop/ Beware a 0 = ............... 100 = ............... 00 = ............................ a1 = ............... 51 = ............... 01 = ............................ a m × a n = ............... 102 × 103 = 100 000 = 10........ a m × a n ≠ a mn am
= ............... an
106 1000000
=
= 10........ 3
1000
10
am
≠an n
a
⎛ 103 ⎞
⎜⎜ 2 ⎟⎟
⎝ 2 ⎠
4
⎛ 1000 ⎞
=⎜
⎟
⎝ 4 ⎠
p
⎛ am ⎞
⎜⎜ n ⎟⎟ = ..................... ⎝b ⎠
m
4
p
⎛ a m ⎞ ⎛ a ⎞mp − np
⎜⎜ n ⎟⎟ ≠ ⎜ ⎟
⎝ b ⎠ ⎝b⎠
= 250 4 = ........................
1012
28
212 × 512
= 24 × 512
8
2
= ....................................
=
2−3 = .....................
( a )− n = ..................... ( a )− n ≠ −a n
⎛3⎞
⎜ ⎟
⎝2⎠
−4
(a )
= ......................
x = .........................
n
a = ......................... 3
30 ab = ......... × .......... a
m
1
≠ an
y = ........................
6
64t 8 = ......... × ..........
=
n
m
≠a
m
n
n
−n
n
n
am ≠ am−n
a+b ≠ n a +n b
x 2 ± y 2 = ...........................
Leereenheid 1 Algebraïese vaardighede en eksponente Bogenoemde mag moontlik eenvoudig en onskouspelagtig lyk, maar die foute wat met die toepassing van hierdie beginsels gemaak word, gaan die verstand te bowe. The principles above may seem simple and unspectacular, but the errors committed in their application challenge the imagination. Oefening 1.3 1. Vereenvoudig sakrekenaar: sonder ‘n Exercise 1.3 1. Simplify without using a calculator: 1.1 4 ( a + b ) − 4 ( a × b ) 4
=
2
31 Leereenheid 1: Algebraïese vaardighede en eksponente 3
3
3
3 ⎞
⎛ 3 ⎞⎛ 3 ⎞⎛ 3 ⎞ ⎛ 3 ⎞ ⎛ 3
1.2 ⎜ x 2 ⎟⎜ x 2 ⎟⎜ x 2 ⎟ − ⎜ x 2 ⎟ − ⎜ x 2 + x 2 + x 2 ⎟ 2
2 ⎠
⎝ 2 ⎠⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2
=
( )
2⎤
⎡
1.3 ⎢ z 2 ⎥
⎣
⎦
=
−1
( ) ⋅ z −1
3
(
)
1.4 ( p − m ) p 2 + pm + m2 =.......................................................................................................................................... =.......................................................................................................................................... =........................................................................................................................................... =............................................................................................................................................ = (.........) − (.......... )
3
32 3
Leereenheid 1 Algebraïese vaardighede en eksponente (
)
1.5 ( 2 x + 3 y ) 4 x 2 − 6 xy + 9 y 2 =.......................................................................................................................................... =.......................................................................................................................................... =........................................................................................................................................... =............................................................................................................................................ = (.........) + (..........)
3
3
1.6 ( a − b ) 3
=.......................................................................................................................................... =.......................................................................................................................................... =........................................................................................................................................... =............................................................................................................................................ 33 Leereenheid 1: Algebraïese vaardighede en eksponente 1.7
t3 + t2
t5
=
1.8
r3 + r3
2r 2 + 3r 2
1.9
34 3α ⋅ 9α + 1
27α + 2
Leereenheid 1 Algebraïese vaardighede en eksponente 1.10
β
5β + 2 + 5β +1 ....... ⋅ ...... + ....... ⋅ ........ 5 (.......... + ............)
=
=
5β + 2 − 5β +1 ....... ⋅ ...... − ....... ⋅ ........ 5β (.......... − ............)
= ...............................................................................
= ................................................................................
= ................................................................................
2. In Matriek het u die volgende differensiasiereël teëgekom: 2. In Matric you encountered the differentiation rule: ( )
dy
ax n = an ⋅ x n −1 dx
Bv. For example dy ⎛ 4
1
1
⎞
2
3
⎜ 3 x − 5 x + x − 9 ⎟ = 12 x − 10 x + dx ⎝
2
2
⎠
Bereken nou g ' ( t ) indien... 2.1 g ( t ) = t +
Now calculate g ' ( t ) if... 1
t
35 Leereenheid 1: Algebraïese vaardighede en eksponente 2.2 g ( t ) = 4 3 t 2 −
5
2
+
+π
2
3
t
3t
3. Vereenvoudig: 3.1
36 4 x × 5x
103 y
+
20
2y × 5y
(
)
2
3. Simplify: Leereenheid 1 Algebraïese vaardighede en eksponente 1.4 Eenvoudige eksponensiële vergelykings/ Simple exponential equations Uit die einskappe van eksponente kan ons sekere vergelykings oplos, waar eksponente betrokke is. Oefening 1.4 By exploiting the properties of exponents we may easily solve certain equations where exponents are involved. 1. Bereken sonder 'n sakrekenaar die waarde van die onbekende: Exercise 1.4 1. Calculate the value of the unknown variable without the use of a calculator: 1.1 9 y −2 = 271−2 y 1.2 43 p − 322 p−5 = 0 37 Leereenheid 1: Algebraïese vaardighede en eksponente 1.3 3 ⋅ 10 x − 0,03 = 0 1.4 6 ⋅ 22 x − 2 x − 1 = 0 Verklaar waarom daar slegs een oplossing is/ Explain why there is only one solution.
38 Leereenheid 1 Algebraïese vaardighede en eksponente 1.5 2r 5 = −
243
16
Hoe verskil hierdie vergelyking van die vorige vier?/ How is this equation different from the previous four?
39 Leereenheid 2: Logaritmes 2 Logaritmes/ Logarithms Leerdoelstellings vir hierdie leereenheid Learning aims for this study unit Upon completion of this study unit the Na afhandeling van hierdie leereenheid moet die student in staat wees om die volgende te doen: student must be able to do the following: 1. Use the properties of logarithms in order to simplify expressions 1. Die eienskappe van logaritmes te kan gebruik om uitdrukkings te 2. Use the properties of logarithms in order to solve logarithmic equations vereenvoudig 3. Solve more complicated exponential 2. Die eienskappe van logaritmes te kan by exploiting the gebruik om logaritmiese vergelykings relationships between exponents and op te los logarithms 3. Ingewikkelder eksponensiële vergelykings op te los deur van die verbande tussen eksponente en logaritmes gebruik te maak 40 equations Leereenheid 2: Logaritmes 2.1 Logaritmes en eksponente/ Logarithms and exponents Beskou onderstaande voorstelling: Consider the represention below: Watter waarnemings en gevolgtrekkings kan gemaak word i.v.m. die eksponensiële bewerking en die logaritmiese bewerking? Which observations and conclusions may be made with respect to the exponential operation and the logarithmic operation? Logaritmes is basies ‘n ander manier om dieselfde inligting te verstrek as wat deur ‘n eksponensiële vergelyking gegee word. Logarithms basically represent another way to express the same information as that which is given by an exponential equation. Om ‘n logaritme te definieer maak ons gebruik van die teorie van inverse funksies: In order to define a logarithm we employ the theory of inverse functions: Beskou die grafiek hierbo. Consider the graph above. As f ( x ) = 10 x dan kan ons dit skryf If f ( x ) = 10 x then we may write that y = 10 x . Om die inverse funksie van y = 10 x . In order to obtain the inverse f ( x ) = 10 x te verkry, ruil ons die function of f ( x ) = 10 x we switch dependent afhanklike en die onafhanklike veranderlike: and independent variables: x = 10 y x = 10 y 41 Leereenheid 2: Logaritmes Maak nou vir y die onderwerp deur te definieer dat y = log10 x Now make y the subject by defining that y = log10 x Dit lewer die grafiek g hierbo, wat die This yields the graph g above, which is the spieëlbeeld is van f in die lyn h ( x ) = x . mirror image of f in the line h ( x ) = x . Ons definieer in die algemeen: As p = a x met a > 0 en a ≠ 1 dan is p > 0 vir x ∈ en x = loga p
Voorbeelde: We define in general: If p = a x with a > 0 and a ≠ 1 then p > 0 for x ∈ and x = loga p
Examples: 8 = 23 ⇔ 3 = log2 8 8 = 23 ⇔ 3 = log2 8 1000 = 103 ⇔ 3 = log1000 1000 = 103 ⇔ 3 = log1000 0,008 =
1
= 5−3 ⇔ −3 = log5 ( 0,008 ) 125
0,008 =
1
= 5−3 ⇔ −3 = log5 ( 0,008 ) 125
Eienskappe van logaritmes/ Eienskap/ Property: Voorbeeld/ Example Properties of logarithms Verklaring/ Explanation “ log3 1 ” beteken/ means: loga 1 = ............... log3 1 = ............... 3WAT / WHAT = 1? “ log7 7 ” beteken/ means: loga a = ............... loge e = ............... 7WAT / WHAT = 7? “ log7
⎛1⎞
log a ⎜ ⎟ = ............... ⎝a⎠
⎛ 1 ⎞
log5 ⎜ ⎟ = ............... ⎝ 25 ⎠
1
” beteken/ means: 343
7WAT / WHAT =
3
1 ⎛1⎞
=⎜ ⎟
343 ⎝ 7 ⎠ = 7−3 ?
42 Leereenheid 2: Logaritmes loga ( xy ) = ................................ ⎛x⎞
loga ⎜ ⎟ = ................................ ⎝ y⎠
log8 + log125
= log............ = ............
a x × a y = a........... log5 500 − log5 20
= log5 ............
= ............
ax
a
= a ........... y
log5 625
( )
loga x m = ........................ = log5 54
= .... × ............
(a )
m x
= a............ = ................
64 = 26
log32 64
log a b =
log c b
log c a
Vir enige/ For any c > 0 =
∴log2 64 = 6
log..... 64
log..... 32
32 = 25
=
∴log2 32 = 5
= ........
32 5 = ..............
6
1
log27 3
1
=
⎛ log27 3 ⎞
⎜
⎟ ⎝ log27 27 ⎠
log3 27 =
log a b =
1
logb a
log3 27 =
1
log27 3
=
log27 27
log27 3
Die logaritme‐bewerking is die .................. van die aloga x = ................. 10log10 p = ................. ............................. bewerking/ The logarithmic operation is the ................ of the .............................. operation
43 Leereenheid 2: Logaritmes Eienskappe wat NIE geld by logaritmes nie/Properties which does NOT hold for logarithms Eienskap/ Property: Voorbeeld/ Example log10 (1000 ± 100 )
loga ( x ± y ) ≠ log a x ± logb y ≠ log10 1000 ± log10 100 loga ( xy ) ≠ loga x × logb y log10 (1000 ) ≠ log10 100 × log10 10
log a
x log a x
≠
y log a y
log3
81 log3 81
≠
27 log3 27
Korrekte vorm/ Correct form log a ( xy ) = log a x + log a y
loga ( xy ) = loga x + loga y ⎛x⎞
loga ⎜ ⎟ = loga x − loga y ⎝ y⎠
( )
log a x m = m ⋅ log a x
(loga x )
m
≠ m ⋅ log a x (log5 25 )
3
≠ 3 ⋅ log5 25 of / or
log a x m = m ⋅ log a x
Dit is belangrik om die implikasies van bogenoemde baie goed te begryp, aangesien ons dit by baie toepassings benodig. 44 It is important to grasp very well the implications of the discussion above, since we require it in many applications. Leereenheid 2: Logaritmes Oefening 2.1 1. Vereenvoudig sakrekenaar: sonder ‘n Exercise 2.1 1. Simplify without using a calculator: 1.1 log10 20 + log10 50 1.2 log6 36 + log6
1
− 3log6 1 36
1.3
log81 − log16
log9 − log4
1.4 log4 420 + (log4 4 )
20
− (log4 16 )(log 4 64 )
45 Leereenheid 2: Logaritmes 1.5 (log7 49 ) ⋅ log7 1
1.6 (log100 ) ⋅ log (1000 )
1.7
3log3 x
2. Daar bestaan in Wiskunde ‘n natuurlike grondtal, wat ons met die simbool e aandui. Die waarde van e word in later Wiskunde‐
kursusse afgelei, maar ons neem die benaderde waarde as e ≈ 2,718 28... Hierdie grondtal gehoorsaam al die gewone eksponentwette en logaritmiese wette, maar dit is die gebruik om loge x te skryf as ln x . Ons lees ln x as “lin ex”. Vereenvoudig sonder ‘n sakrekenaar en gebruik slegs in die laaste stap ‘n sakrekenaar: 46 2. In Mathematics there exists a natural base, which we indicate by the symbol e .The value of e is derived in later Mathematics courses, but we take its approximate value as e ≈ 2,718 28... This base obeys the same usual exponential laws and logarithmic laws, but it is customary to write loge x as ln x . We read ln x as “lin ex”. Simplify without a calculator but use only in the final step a calculator: Leereenheid 2: Logaritmes 2.1 e2 + 3e2 + e2 ⋅ e −2 −
e5
+ e0 3
e
2.2
( )
ln x + 2ln x
3
− ln e3 + (ln e ) + ln1 3
ln x
3. Bepaal die waarde van die onbekende (Los op): 3. Determine the value of the unknown (Solve): 3.1 log3 x = 4 47 Leereenheid 2: Logaritmes 3.2 log x + log ( x + 3 ) = 1 3.3 log (1 − 2 x ) − log ( x + 2 ) = log1 3.4 ln x = ln ( 2 x − 1 ) + 2ln x 48 Leereenheid 2: Logaritmes 2.2 Die oplos van ingewikkelder eksponensiële vergelykings d.m.v logaritmes/ Solving more complicated exponential equations using logarithms Aangesien die definisie van 'n logaritme in terme van eksponente gedoen is, kan ons ingewikkelde eksponensiële vergelykings oorskryf na logaritmiese vorm. Sulke logaritmiese vergelykings is dan makliker om op te los as wat die oorspronklike eksponensiële vergelyking was. Because we defined a logarithm in terms of exponents, we may rewrite more complicated exponential equations to logarithmic form. Such logarithmic equations are then easier to solve than the original exponential equation. Voorbeeld/ Example 5x
1. Los op vir x / Solve for x : ⎛ 3 ⎞
⎜ ⎟
⎝ 10 ⎠
2. Los op vir t / Solve for t : 3,5 ⋅ 0,13t = 2,5t −1 =
343
64
Oplossing/ Solution 1.
⎛ 3 ⎞
⎜ ⎟
⎝ 10 ⎠
5x
343
64
343
∴ 5 x = log 3
64
=
10
⎛ 343 ⎞
log ⎜
⎟
⎝ 64 ⎠ ∴5 x =
⎛ 3 ⎞
log ⎜ ⎟
⎝ 10 ⎠
⎡ ⎛ 343 ⎞ ⎤
⎢ log ⎜ 64 ⎟ ⎥
⎠⎥
⎢ ⎝
3
⎛
⎞
⎢ log
⎥
⎜ ⎟⎥
⎢⎣
⎝ 10 ⎠ ⎦
∴x =
5
= −0,279
49 Leereenheid 2: Logaritmes 3,5 ⋅ 0,13t = 2,5t −1
2.
(
) ( )
∴log ( 3,5 ) + log ( 0,1 ) = log ( 2,5 )
∴log 3,5 ⋅ 0,13t = log 2,5t −1
3t
t −1
∴log ( 3,5 ) + 3t ⋅ log ( 0,1 ) = ( t − 1 ) ⋅ log ( 2,5 )
∴log ( 3,5 ) + 3t ⋅ log ( 0,1 ) = t ⋅ log ( 2,5 ) − log ( 2,5 ) ∴3t ⋅ log ( 0,1 ) − t ⋅ log ( 2,5 ) = − log ( 2,5 ) − log ( 3,5 )
∴ t ⎡⎣3log ( 0,1 ) − log ( 2,5 ) ⎤⎦ = − log ( 2,5 ) − log ( 3,5 )
∴t =
− log ( 2,5 ) − log ( 3,5 )
3log ( 0,1 ) − log ( 2,5 )
= 0,277
Oefening 2.2 1. Los die volgende vergelykings op: 1. Solve the following equations: t
⎛1⎞
1.1 0,002 93 = 3 ⋅ ⎜ ⎟ ⎝2⎠
1.2
50 1 ⎛4⎞
⋅⎜ ⎟
3 ⎝3⎠
n−1
= 2,497 180 Exercise 2.2 Leereenheid 2: Logaritmes 1
1.3 4 ⋅ 23t = ⋅ 52t−1
2
⎛1⎞
1.4 2 ⋅ ⎜ ⎟
⎝3⎠
2n
1
− ⋅ 51−3n = 0
2
51 Leereenheid 3: Inleiding tot funksies 3 Inleiding tot funksies/ Introduction to functions Leerdoelstellings vir Learning aims for this study unit hierdie leereenheid Upon completion of this study unit the student must be able to do the following: Na afhandeling van hierdie leereenheid moet die student in staat wees om die volgende te 1. Apply the formal definition of a doen: function as a special relation 1. Die formele definisie van ‘n funksie 2. Identify the domain and range of a as ‘n spesiale relasie kan toepas function 2. Die definisie‐ en waardeversameling 3. Determine the inverse of a given van ‘n funksie kan identifiseer function 3. Die inverse van ‘n gegewe funksie 4. Perform operations with functions kan bepaal 4. Bewerkings met funksies kan uitvoer 3.1 Definisie van'n funksie en inleidende aspekte/ Definition of a function and introductory aspects Relasies Relations In Wiskunde dui ‘n relasie op twee versamelings (groepe getalwaardes), waartussen daar ‘n verband bestaan. Vir elke element uit die een versameling kan ‘n verbinding gemaak word met een of meer elemente van die ander versameling Hierdie verband is een of ander patroon of reël. In Mathematics the word relation indicates two sets (groups of numeric values) between which there exists a relationship or connection. For each element taken from the one set a connection can be made to one or more of the elements in the other set. This relationship is usually a certain pattern or rule. Hierdie reël kan ‘n woordelikse instruksie wees, of ‘n algebraïese formule, of ‘n tabel met waardes, of ‘n grafiek. This rule may be a verbal instruction, or an algebraic formula, or a table with values, or a graph. 52 Leereenheid 3: Inleiding tot funksies Voorbeeld van ‘n relasie: Voltooi die patroon: ( getal ; ±
getal
)
Example of a relation: Complete the pattern: vir die ( number; ±
getalle {0; 4; 9;16; 25} number
)
for the numbers {0; 4; 9;16; 25} {( 0;.........) ; ( 4;...........) ; ( 9; ± 3 ) ; (16;............) ; (25;............)} Let daarop dat elke eerste element met twee ander elemente verbind word. Note that each first element is connected with two other elements. Funksies Functions 'n Funksie is 'n reël wat elke element van A function is a rule which connects each of een definisie−
the elements in the one set (the domain) versameling) verbind met een en slegs een with one and only one element in the other element van 'n ander versameling (die set (the range). versameling (die waardeversameling). Voorbeeld van ‘n funksie: Voltooi die patroon: ( getal ; getal )
2
Complete the pattern: vir getalle {−2; − 1; 0;1; 2 } Example of a function: die ( number; number )
2
for the numbers {−2; − 1; 0;1; 2 } {( −2;.........) ; ( −1;...........) ; ( 0; ...........) ; (1;............); (2;............)} 53 Leereenheid 3: Inleiding tot funksies 3.2 Definisie en waardeversameling/ Domain and range Let daarop dat elke eerste element met Note that each first element is connected slegs een ander elemente verbind word. with only one other element. Ons noem dit ‘n eenduidige verband. We call this an unambiguous relationship. Meer‐eenduidige‐funksies Many‐to‐one‐functions ‘n Meer‐eenduidige funksie het die A many‐to‐one‐function has the property eienskap dat elke element uit die that each element from the domain is not definisieversameling nie op ‘n unieke, mapped unto a unique, different element of verskillende die the range. You see this clearly in the example waardeversameling afgebeeld word nie. U above regarding the squares. Different sien dit duidelik by die voorbeeld hierbo numbers yield the same squares. element van oor die kwadrate. Verskillende getalle lewer dieselfde kwadrate. One‐to‐one‐functions Een‐een‐duidige funksies A one‐to‐one function has the property that ‘n Een‐een‐duidige funksie het die eienskap dat elke element definisieversameling verskillende op element uit ‘n each element from the domain is mapped die unto a unique, different element of the unieke, van range. die waardeversameling afgebeeld word. Example Voorbeeld Complete the pattern: Voltooi die patroon: (
getal ; getal 3
)
( number; number )
3
vir die getalle for {−2; − 1; 0;1; 2} {−2; − 1; 0;1; 2} {( −2;.........) ; ( −1;...........) ; ( 0; ...........) ; (1;............); (2;............)} 54 the numbers Leereenheid 3: Inleiding tot funksies Oefening 3.1 tot 3.2 1. Beskou onderstaande skematiese voorstelling van 'n funksie en vul al die ontbrekende begrippe in. Exercise 3.1 to 3.2 Consider the schematic below which shows a function and fill in the ommitted concepts: 2 Stel die inligting in die skematiese Now represent the information in the voorstelling nou hieronder voor: schematic representation below: 55 Leereenheid 3: Inleiding tot funksies 3 Doen interpolasie (trek 'n gladde kromme deur die punte wat u geplot het tussen en insluitende waar t = −2 en t = 4 ). Perform interpolation (draw a smooth curve through the points that you just plotted beteen and including where t = −2 and t = 4 ). 4 Gebruik die gegewe inligting en skryf die vergelyking (formule) van die grafiek neer. Use the given information and write down the equation (formula) of the graph. Wenk: op skool sou dit iets gewees het Hint: at school it would have been soos y = ax + bx + c something like y = ax 2 + bx + c 2
5 Beskryf die vorm van die kromme. Describe the shape of the curve. stygend of dalend/ ascending or descending: ...................................................... konkaafheid/ concavity: .................................................................. min of maks draaipunt/ min or max turning point:.......................... at/by (.......; .......) 6 Skryf die definisieversameling van die funksie neer: Write down the domain of the function: Dg = {...... ...... ≤ ....... ≤ .......;...............} 7 Skryf die waardeversameling van die funksie neer: Wg = {...................................................} 56 Write down the range of the function: Leereenheid 3: Inleiding tot funksies 8 Doen nou ekstrapolasie (verleng die kromme "verby" die punte waar t = −2 en t = 4 . Now perform extrapolation (extend the curve "beyond" the points where t = −2 and t = 4 . 9 Gebruik 1.4 en bereken die waardes van Use 1.4 and calculate the values of ⎛3⎞
g ( −1,5 ) en g ⎜ ⎟ en g ( 5 ) en g ( 2 + h ) ⎝2⎠
⎛3⎞
g ( −1,5 ) and g ⎜ ⎟ and g ( 5 ) en ⎝2⎠
g (2 + h ) 57 Leereenheid 3: Inleiding tot funksies 10 Dui op die grafiek aan waar u die eerste drie antwoorde op vraag 1.9 sou aflees. Indicate on your graph where you would read off the first three answers to question 1.9. 11 Gebruik 1.4 en bereken die waardes van t sodat g = 7 . Use 1.4 and calculate the values of t such that g = 7 . 12 Dui op die grafiek aan waar u die antwoorde op vraag 1.11 sou aflees. Indicate on your graph where you would read off the values of your answers to question 1.11. Beskou die voorstelling: Consider the representation: 13 Watter van die gegewe gevalle stel funksies voor? 58 Which of the given cases represent functions? Leereenheid 3: Inleiding tot funksies 14 Watter van die funksies hierbo is een‐tot‐ een‐funksies? Which of the functions above are one‐ to‐one‐functions? 15 As A = {−1; 0;1; 2; 3}
en If A = {−1; 0;1; 2; 3}
and B = {−3; − 2; − 1; 0;1; 2; 3;...; 9;10} , skryf B = {−3; − 2; − 1; 0;1; 2; 3;...; 9;10}
die volgende funksies as versameling getallepare: write the following functions as sets of ordered pairs: 15.1 , {( x; y ) y = x; x ∈ A, y ∈B} {( x; y ) y = x ; x ∈ A, y ∈B} 15.2 2
16 Sê of elkeen van die funksies by Vraag 15 een‐tot‐een‐funksies of meer‐tot‐een‐
funksies is. Say if each of the functions in Question 15 is a one‐to‐one‐function or a many‐to‐one function. 16.1.............................................................................................................................. 16.2.............................................................................................................................. 16 Skets rofweg elkeen van die volgende funksies en skryf die funksie se definisieversameling en waarde‐
versameling neer: Sketch each of the following functions in rough and write down the domain and range of it: 59 Leereenheid 3: Inleiding tot funksies 16.1 {
( x; y ) y = ( x − 1 )2 − 4; x ∈
, y∈
}
16.2 {(t; y ) y = − (t − 2) + 2; t ∈
2
, y∈
} 16.3 ⎧
8
⎫
+ 1; z ∈ , y ∈ ⎬ ⎨( z ; y ) y = −
z +1
⎩
⎭
60 Leereenheid 3: Inleiding tot funksies 16.4 {( p; r ) r =
p − 2; p ∈ , r ∈
} 16.5 {( x; y ) y = e ; x ∈
x
, y∈
} 16.6 {( x; y ) y = ln x; x ∈
, y∈
} 61 Leereenheid 3: Inleiding tot funksies Funksie‐waardes Funksiewaardes is die afhanklike veranderlike‐waardes wat op die vertikale as afgelees word. Function values are the dependent variable values which are read off the vertical axis. They are calculated by substituting certain values of the independent variable into the defining equation of the function. Dit word bereken deur bepaalde waardes van die onafhanklike veranderlike in die funksie se definiërende vergelyking te vervang. Oefening 3.2 Function values Exercise 3.2 As/ if s ( t ) = 3 dan is/ then s ( 0 ) = ............... en/ and s (1 ) = ............... en/ and s ( t + h ) = ............... 7⎛1⎞
As/ if Tn = ⎜ ⎟
2⎝ 2⎠
n −1
dan is/ then T1 = ................................................................................. ……………………………………………………………………….. en/ and T4 = ......................................................................................................... ……………………………………………………………………………………………………………… As/ if g ( y ) =
3
dan is/ then g ( 0 ) = ................................ 2y
en/ and g ( t ) = ......................... ⎛2⎞
en/ and g ⎜ ⎟ = ................................................................................................. ⎝3⎠
62 Leereenheid 3: Inleiding tot funksies ………………………………………………………………………………………………. ……………………………………………………………………………………………… 63 Leereenheid 3: Inleiding tot funksies 3.3 Inverse van 'n funksie/ Inverse of a function Inverse funksies Inverse functions Voorbeeld Indien y = sin x en x =
⎛π ⎞
dan is y = sin⎜ ⎟
6
⎝6⎠
π
Example If y = sin x and x =
⎛π ⎞
then y = sin⎜ ⎟ 6
⎝6⎠
π
1
wat y = lewer. 2
1
which yields y = . 2
Gestel egter dat ons weet dat y =
1
in 2
y = sin x maar ons wil vir x bepaal. Dan 1
Suppose, however, we know that y = in 2
y = sin x but we wish to determine the value gaan ons soos volg te werk: of x . Then we go about as follows: 1
⎛1⎞
= sin x so x = sin−1 ⎜ ⎟ en dit lewer 2
⎝2⎠
1
⎛1⎞
= sin x so x = sin−1 ⎜ ⎟ and this yield, 2
⎝2⎠
onder meer x =
π
6
as oplossing. among other values, x =
π
6
as solution. Bogenoemde illustreer die gebruik van ‘n inverse funksie of inverse funksie‐
bewerking. The argument above illustrates the use of an inverse function or inverse function operation. Berekening van inverse funksies Calculation of inverse functions In gees en wese kom die berekening van die inverse van ‘n funksie daarop neer dat... In essence the calculation of the inverse of a function boils down to... •
ons die afhanklike en onafhanklike veranderlike in die definiërende vergelyking omruil en •
that we switch the dependent and independent variable in the defining equation and •
dat ons die vertikale as‐
veranderlike (afhanklike veranderlike) weer die onderwerp van die vergelyking maak. •
that we manipulate the resulting equation to make the vertical axis‐
variable (dependent variable) once again the subject of the equation. Daar is nietemin ‘n komplikasie by die berekening van inverse funksies, en dit is dat slegs een‐een‐duidige funksies inverses besit. There is, however, a complication involved with the calculation of inverse functions, and that is the fact that only one‐to‐one functions possess inverses. Beskou bv. die sinus‐funksie f en sy Consider, for example, the sine function f inverse f −1 wat deur die grafiese and its inverse f −1 which is indicated by the 64 Leereenheid 3: Inleiding tot funksies rekenaarprogram aangedui word deur g : graphical computer program as g : 65 Leereenheid 3: Inleiding tot funksies Dit is duidelik dat slegs die gedeelte van die sinus‐kromme tussen die punte waar hy horisontaal loop (by sy draaipunte), ‘n spieëlbeeld in die lyn y = x het. It is clear that only the part of the sine curve between the points where it runs horizontally (in its turning points) possess a mirror image (reflection) in the line y = x . Onthou dat ‘n funksie y = f −1 ( x ) per Recall that a function y = f −1 ( x ) connects definisie een en slegs een y ‐waarde besit by definition one and only one y ‐value with vir elke x ‐waarde. Dit beteken meetkundig dat ‘n vertikale lyn wat oor die kromme van each x ‐value. Geometrically this means that f −1 skuif, die kromme by elke x ‐waarde a vertical line which moves over f −1 , may slegs een keer mag sny. only intersect the curve once at every x ‐
value. Maar die inverse funksie f −1 is uit f But the inverse function f −1 was obtained verkry deur die afhanklike veranderlike en die onafhanklike veranderlike te ruil – effektief het ons dus die asse geruil – en dit impliseer dat ‘n horisontale lyn wat oor f skuif, dit by elke y ‐waarde slegs een from f by switching the dependent variable keer mag sny – slegs daardie deel van f se – only that part of the range of f were it waardeversameling waarop horisontale lyntoets slaag, passes the horizontal line test, will form the definisieversameling van f
(Onthou dat f
beteken nie.) 66 −1
−1
hy sal die die and the independent variable – effectively we switched the axes– and that implies that a horizontal line which moves over f , may intersect the curve only once at each y ‐value domain of f −1 . vorm. 1
in hierdie konteks NIE f
(Keep in mind that f −1 doen NOT mean in this context.) 1
f
Leereenheid 3: Inleiding tot funksies Oefening 3.3 Exercise 3.3 Bepaal die inverse van die volgende Determine the inverse of the following funksies en skets f sowel as f −1 : functions and sketch f as well as f −1 : 1. f ( x ) = 2 x + 1 2. f ( x ) = x − 1 67 Leereenheid 3: Inleiding tot funksies 3.4 Bewerkings met funksies / Operations with functions Kombinasies van funksies (bewerkings) Combinations of functions (operations) Aangesien funksiewaardes getalle is, tree Since function values are numbers, functions funksies soos getalle op waarmee ons behave like numbers with which we may bewerkings kan doen. perform operations. Die komplikasie, natuurlik, is die The obvious complication, of course, is the definisieversameling van die resultaat van domain of the result of the operation – in die bewerking – in die algemeen is die general the domain of the result will be the definisieversameling van die resultaat die intersection of the domains of the separate snyding van die definisieversamelings van constituent functions – in the case of division die afsonderlike funksies – in die geval van (quotients) the values of the independent deling (kwosiënte) is die waardes van die variable for which the denominator function onafhanklike veranderlike waarvoor die is zero, will also be excluded. noemerfunksie nul is, ook uitgesluit. 68 Leereenheid 3: Inleiding tot funksies Oefening 3.4 1. As f ( x ) = 2 x 2 en g ( x ) = 3 x + 5 , bepaal die volgende Exercise 3.4 1. If f ( x ) = 2 x 2 and g ( x ) = 3 x + 5 , determine the following 1.1
(f
+ g )( x ) = f ( x ) + g ( x ) = .................... + ..................... …………………………………………………………………………………. 1.2 ( g + f )( x ) = ................ + ................ = .................... + ..................... ……………………………………………………………………………….... 1.3
(f
− g )( x ) = ................ − ................ = .................... − ..................... ……………………………………………………………………………….... 1.4 ( g − f )( x ) = ................ − ................ = .................... − ..................... ……………………………………………………………………………….... 1.5
( fg )( x ) = ................ × ................ = .................... ⋅ ..................... ……………………………………………………………………………….... 69 Leereenheid 3: Inleiding tot funksies 1.6 ( gf )( x ) = ................ × ................ = .................... × ..................... ……………………………………………………………………………….... ⎛f ⎞
1.7 ⎜ ⎟ ( x ) = ________________ = ________________________________ ⎝g⎠
= ____________________________________________________ =………………………………………………………………………………................. ⎛g⎞
1.8 ⎜ ⎟ ( x ) = ________________ = ________________________________ ⎝f ⎠
= ____________________________________________________ =………………………………………………………………………………................. 2. Watter van die bewerkings in hierbo is kommutatief? 70 2. Which ones of the operations above are commutative? Leereenheid 3: Inleiding tot funksies 3. As f ( x ) = 2 x 2 , a = 3 en x = 2 bepaal: 3. If
f ( x ) = 2 x 2 , a = 3 and x = 2 determine: 3.1 a ⎡⎣ f ( x )⎤⎦ = ………………………………………………….…………………. ……………………………………………………………………… ……………………………………………………………………… 3.2 f ( ax ) = …………………………………………………………..………………. ……………………………………………………………………… ……………………………………………………………………….. 3.3 Watter gevolgtrekking kan u maak oor 3.1 en 3.2/ 3.3 Which conclusion do you make as a result of 3.1 and 3.2? 4. As g ( x ) = x , a = 9 en x = 16 bepaal: 4. If g ( x ) = 2 x 2 , a = 9 and x = 16 determine: 4.1 g ( a + x ) = ………………………………………………….…………………. ……………………………………………………………………… ……………………………………………………………………… 71 Leereenheid 3: Inleiding tot funksies 4.2 g ( a ) + g ( x ) = …………………………………………………………..………………. ……………………………………………………………………… ……………………………………………………………………….. 4.3 Watter gevolgtrekking kan u maak oor 4.1 en 4.2? 72 3.4 Which conclusion do you make as a result of 4.1 and 4.2? Leereenheid 4: Radiaalmaat en trigonometrie 4 Radiaalmaat en trigonometrie/ Radian measure and trigonometry Leerdoelstellings vir hierdie leereenheid Learning aims for this study unit Upon completion of this study unit the Na afhandeling van hierdie leereenheid moet student must be able to do the following: die student in staat wees om die volgende te 1. Define radian measure and convert doen: angles form degrees to radians and vice versa 1. Radiaalmaat te definieer en hoeke van grade na radiale om te skakel en 2. Calculate arc length andersom 3. Calculate the area of a circle sector 2. Booglengte te bereken 4. Define all six trigonometric ratios 3. Oppervlakte van ‘n sirkelsektor te and calculate their function values in bereken all four quadrants of the flat plane 4. Al ses trigonometriese verhoudings 5. Apply the sum and difference te definieer en hul funksiewaardes in formulas al vier kwadrante van die platvlak te 6. Apply the double angle formulas bereken 7. Prove trigonometric identities 5. Die som‐ en verskilformules toe te pas 6. Die dubbelhoekformules toe te pas 7. Trigonometriese indentiteite te bewys 73 Leereenheid 4: Radiaalmaat en trigonometrie 4.1 Radiaalmaat/ Radian measure Die verdeling van ‘n sirkel in 360 gelyke dele wat ons grade noem, asook die onderverdeling van ‘n graad in 60 gelyke dele wat ons minute noem, is ‘n antieke gebruik wat by die Babiloniese en Sumeriese geleerdes van voor 540 vC oorgeneem is. The subdivision of a circle into 360 equal parts which we call degrees, as well as the subdivision of a degree into 60 equal parts which we call minutes, is an ancient tradition which were passed on from the Babylonian and Sumerian scholars of earlier than 540 B.C. Probeer gerus met behulp van die volgende sketse verduidelik wat ons bedoel met die begrip “hoek”. Use the following sketches and attempt to explain what we mean by the concept "angle". Vir moderne Wiskunde en wetenskap vereis ons ‘n formele definisie vir die begrip hoek. Beskou onderstaande skets waarop akkurate metings aangetoon word en bereken vir al drie sektore die waarde van die verhouding booglengte
: radius
74 For modern Mathematics and science we require a formal definition for the concept angle. Consider the sketch below where accurate measurements are shown for all three sectors arc length
may be in order that the ratio radius
calculated: Leereenheid 4: Radiaalmaat en trigonometrie booglengte
bly konstant in al radius
drie gevalle, ongeag die grootte van die oppervlakte van die gebied tussen die lyne en/of boë wat die sektore omsluit. arc length
remains radius
constant in all three cases, irrespective of the area between the lines and/ or the arcs defining the sectors. Daarom is dit nuttig om die hoek tussen enige twee radiusse voortaan eenvoudig te booglengte
definieer as θ =
. radius
That is why it is useful to henceforth simply define the angle between any two radii from arc length
now on as θ =
. radius
Let daarop dat die hoek θ geen eenheid besit nie; hoekom? Note that the angle θ does not possess any units; why? Ons defineer daarom eenheid van ‘n hoek θ booglengte
wat gedefineer is as θ =
as radius
radiale. Therefore we define the unit of an angle θ arc length
which is defined as θ =
radians. radius
Die verhouding Hoe groot is ‘n radiaal en hoeveel radiale pas in ‘n volsirkel, wat ons tradisioneel beskou as die boog van een omwenteling (360°)? Beskou die volgende skets en voltooi die ontbrekende inligting: Note that the ratio How big is one radian and how many radians fit into a full circle, which we traditionally consider as the arc of one revolution (360°)? Consider the following sketch and complete the information on it: 75 Leereenheid 4: Radiaalmaat en trigonometrie U kan sien dat ‘n hoek van 1 radiaal ‘n groot hoek is; skat gerus hoe groot die hoek hierbo in grade sou wees. You can see that an angle of 1 radian is a rather large angle; estimate the size of the angle in degrees. Laat ons nou ‘n metode ontwikkel om radiale in grade om te skakel: Let us now develop a method to convert radians to degrees: Beskou ‘n sirkel met radius r en sirkelboog s waar die sirkelboog die hele omtrek van die sirkel is. In hierdie geval is die hoek wat deur twee radiusse en die boog (dit is nie omtrek) onderspan word, tog een omwenteling, dit is 360° volgens ons tradisionele hoekmaat. Consider a circle with radius r and circle arc s where the arc is in this case the entire circumference of the circle. In this case the angle which is subtended by the radii and the arc (the circumference) is one revolution, which is 360° in terms of traditional units. Voltooi nou die redenasie: Complete the reasoning: Een omwent ( ° ) = 1 Omwent (radiale )
∴360° =
booglengte (volsirkel )
radius
∴360° =
............................
(omtrek=?)
r
One rev ( ° ) = 1 Rev (radians )
(per def.)
∴360° =
arc length ( full circle)
radius
∴360° =
............................
(circumf=?)
r
∴360° = ...................... radiale
∴360° = ...................... radians
∴1 ° = ..................... radiale
∴1 ° = ..................... radians
1 radiaal = .................°
1 radian = .................°
76 (by def.)
Leereenheid 4: Radiaalmaat en trigonometrie Dit is nou maklik om te sien dat ons ‘n hoek van π radiale as ‘n hoek van 180° kan beskou. Omskakelings soos die volgende volg dan maklik: It is now easy to see why we may consider an angle of π radians as an angle of 180° . Conversions such as the following are now simple: Voorbeeld/ Example 1 Skakel 240° om na radiale/ Convert 240° to radians 2. Skakel 315° om na radiale/ Convert 315° to radians 3
3
3. Skakel π radiale om na grade/ Convert π radians to degrees 4
4
4. Skakel 11
11
π radiale om na grade/ Convert π radians to degrees 6
6
Oplossing/ Solution 1. 240° = 180° + 60°
1
= π rad + π rad
3
3
1
= π rad + π rad
3
3
4
= π rad
3
2. 315° = 360° − 45°
1
= 2π rad − π rad
4
8
1
= π rad − π rad
4
4
7
= π rad
4
3.
4.
3
3
π rad = × 180°
4
4
= 135°
11
11
π rad = × 180°
6
6
= 330°
77 Leereenheid 4: Radiaalmaat en trigonometrie Oefening 4.1 Voltooi nou die ontbrekende inligting in onderstaande skets: Exercise 4.1 Complete the omitted information in the sketch below: 78 Leereenheid 4: Radiaalmaat en trigonometrie 4.2 Berekening van booglengte/ Calculation of arc length booglengte
volg radius
nou die volgende interessante toepassings: Uit die definisie θ in rad =
Oefening 4.2 1. ‘n Speelgoedtrein ry in ‘n sirkelbaan met ‘n oppervlakte van From the definition θ in rad =
arc length
radius
follows interesting applications: Exercise 4.2 1. A toy train travels around a circular track with an area of 25 446,900 49 cm2 . Indien dit 2,5 m 25 446,900 49 cm2 . If it covers aflê tussen twee punte op die spoor, wat is die hoek in grade waardeur die treintjie gery het? 2,5 m between two points on the track, what is the angle in degrees through which the train travelled? 2. Gestel ‘n sirkelsektor het ‘n radius van x en ‘n middelpuntshoek van 2 radiale . Bereken die lengte van die boog van die sektor. 2. Suppose a circle sector has a radius of x and a central angle of 2 radians . Calculate the length of the arc of the sector. 79 Leereenheid 4: Radiaalmaat en trigonometrie 3. Bepaal die oppervlakte van ‘n sirkel indien ‘n booglengte van 200 mm onderspan word deur ‘n hoek van 171,887 339° . 80 3. Determine the area of a circle if an arc of 200 mm is subtended by an angle of 171,887 339° . Leereenheid 4: Radiaalmaat en trigonometrie 4.3 Berekening van die oppervlakte van ‘n sirkelsektor/ Calculation of the area of a circle sector Beskou die volgende tabel en kyk of u die patroon kan raaksien: Consider the following table and see if you can discover the pattern: Oppervlakte‐ Middelpunts‐ Formule/ hoek in rad/ Area formula Angle in rad A = π r2 2π Figuur/ Figure Volsirkel/ Full circle Halfsirkel/ π A = ...........π r 2 Semi‐circle Kwartsirkel/ A = .............π r 2 ........... 1
A = π r2 8
π
A = .............π r 2 ........... Quarter circle Agstesirkel/ 4
Eighth of circle n de van ‘n sirkel/ n th of a circle Voltooi: Indien ‘n sirkelsektor ‘n hoek van θ radiale by die middelpunt maak, is die Complete: If a circle sector makes an angle of θ radians at the centre, the formula for formule vir die oppervlakte van die sektor: the area of the sector is: A = ..............r 2
A = ..............r 2
81 Leereenheid 4: Radiaalmaat en trigonometrie Oefening 4.3 Exercise 4.3 1. Die straal van ‘n waterspreier beweeg deur ‘n hoek van 50° en die straal kan 5 m ver bykom. Bereken die 1. The jet of a water sprayer sweeps through an angle of 50° and the jet can reach 5 m far. Calculate the oppervlakte wat dit kan natlei. area which the sprayer can irrigate. 2. Gestel die hoek θ waardeur die straal beweeg, verdubbel maar die radius van die straal halveer. Met watter persentasie sal die oppervlakte wat besproei kan word, dan verander? Is dit ‘n toename of afname? 82 2. Suppose the angle θ through which the jet sweeps, doubles but the radius halves. By which percentage would the area which can be irrigated, then change? Is this an increase or decrease? Leereenheid 4: Radiaalmaat en trigonometrie 3. Gestel die hoek θ waardeur die straal beweeg, halveer maar die radius van die straal verdubbel. Met watter persentasie sal die oppervlakte wat besproei kan word, dan verander? Is dit ‘n toename of afname? 3. Suppose the angle θ through which the jet sweeps, halves but the radius doubles. By which percentage would the area which can be irrigated, then change? Is this an increase or decrease? 83 Leereenheid 4: Radiaalmaat en trigonometrie 4. Bepaal die radius: 4. Bepaal die radius: 84 Leereenheid 4: Radiaalmaat en trigonometrie 4.4 Die ses trigonometriese verhoudings en hul funksiewaardes in al vier kwadrante van die platvlak/ The six trigonometric ratios and their function values in all four quadrants of the flat plane Beskou die reghoekige driehoek: Consider the right triangle Fig. 4.4.1 Aangesien daar ses verskillende maniere bestaan om die teenoorstaande sy, aangrensende sy en skuinssy van ‘n regte driehoek as verhoudings van mekaar te skryf, geld nie net die volgende drie verhoudings Since there exist six different ways in which to write ratios for the opposite side, anjeacent side and hypotenuse of a right triangle, not only the following ratios hold true 85 Leereenheid 4: Radiaalmaat en trigonometrie sinθ =
b
c
en/and
sinα =
.........
..........
cosθ =
.............
............
en/and
cos α =
.......... c
tanθ =
.............
............
en/and
tanα =
.........
..........
nie, maar ook die volgende verhoudings cosecθ =
1
c
=
sinθ b
but also the following ratios en/and
cosecα =
1
.........
=
sinα ..........
secθ =
1
.............
=
cosθ ............
en/and
sec α =
1
.......... =
cosα ...........
cotθ =
1
.............
=
tanθ ............
en/and
cot α =
1
.........
=
tanα ..........
Die verhoudings in die tweede groep hierbo staan as resiprookverhoudings of omgekeerde verhoudings van die sinus‐, cosinus‐ en tangensverhoudings bekend. The ratios in the second group are known as reciprocal ratios of the sine, cosine and tangent ratios. (NOT INVERSE RATIOS) (NIE INVERSE VERHOUDINGS NIE) Dit kan maklik aangetoon word dat sinθ
deur van die eerste skets in Fig. tanθ =
cos θ
4.4.1 gebruik te maak: 86 IT can easily be shown that tanθ =
by using the first sketch in Fig. 4.4.1: sinθ
cos θ
Leereenheid 4: Radiaalmaat en trigonometrie Voorbeeld/ Example 1. Bewys dat tanθ =
sinθ
cos θ
Oplossing/ Solution Linkerkant/ LHS = tanθ
b
=
a
sinθ
Regterkant/ RHS =
cosθ
⎛b⎞
⎜ ⎟
c
=⎝ ⎠
⎛a⎞
⎜ ⎟
⎝c⎠
b c
= ×
c a
b
=
a
∴ LHS = RHS dus/so tanθ =
sinθ
cosθ
Maak nou van die tweede skets in Fig. 4.4.1 cos α
gebruik en bewys self dat cot α =
. sinα
Now make use the second sketch in Fig 4.4.1 and prove on your own that cos α
cot α =
sinα
87 Leereenheid 4: Radiaalmaat en trigonometrie Die tweede skets in Fig. 4.4.1 gee ook aan ons ‘n manier om die ko‐verhoudings van die ses trigonometriese verhoudings verkry. The second sketch in Fig. .4.4.1 also provides us with a way to obtain the co‐
ratios of the six trigonometric ratios. Voltooi die volgende: Complete the following: sinα =
..............
..............
en/ and cosθ =
dus/ so.............. = ....................
...............
...............
maar aangesien/ but because θ =
sin......... = cos (..............................)
cos α =
2
− α kan ons skryf/ we may write
..............
..............
en/ and sinθ =
dus/ so.............. = ....................
...............
...............
maar aangesien/ but because θ =
cos......... = sin (.............................. )
tanα =
π
π
2
− α kan ons skryf/ we may write
..............
..............
en/ and cotθ =
dus/ so.............. = ....................
...............
...............
maar aangesien/ but because θ =
tan......... = cot (.............................. )
π
2
− α kan ons skryf/ we may write
sec α =
..............
..............
en/ and cosecθ =
dus/ so.............. = ....................
...............
...............
maar aangesien/ but because θ =
π
2
sec......... = cosec (.............................. )
− α kan ons skryf/ we may write
Dit is ook aan u bekend dat die trigonometriese funksiewaardes vir sekere bekende hoeke, nl. 0°, 30°, 45°, 60° en 90° baie maklik uit die It is also known to you that the function values for certain familiar angles, namely. 0°, 30°, 45°, 60° and 90° may very easily be volgende sketse bepaal en ook maklik gememoriseer kan word; Ons pas onderstaande sketse vir radiaalmaat aan. obtained from the following sketches and may also be easily memorized; We adjusted the sketches below for radian measure. 88 Leereenheid 4: Radiaalmaat en trigonometrie Voltooi die ontbrekende inligting: Complete the omitted information: Fig 4.4.2 Let op dat die driehoeke hierbo nie regtig bestaan nie; ons gebruik ons verbeelding om te bepaal wat met die driehoek gebeur indien die sy en die skuinssy van die driehoek "op mekaar val"/
Note that the triangle above does not actually exist ; we use our imagination in order to establish
what happens with the triangle when one of its sides and the hypotenuse " co − incides ".
89 Leereenheid 4: Radiaalmaat en trigonometrie Fig. 4.4.3 Aangesien dit dikwels gebeur dat ‘n hoek ‘n π
waarde van meer as radiale besit, moet ons 2
trigonometrie kan doen met hoeke in al vier kwadrante van die platvlak. 90 Because it often happens that an angle is π
larger than radians , we must be able to 2
do trigonometry with angles in four quadrants of the flat plane. Leereenheid 4: Radiaalmaat en trigonometrie Voltooi onderstaande skets: Complete the sketch below: Fig. 4.4.4 Op die volgende bladsy is nog 'n interessante manier om die meeste van ons trigonometrie‐
kennis op te som: On the next page you find another interesting way of summarizing most of our trigonometric knowledge: 91 Leereenheid 4: Radiaalmaat en trigonometrie Meer oor spesiale hoeke More on special angles Gebruik die definisies in terme van die eenheidsirkel en bepaal die waardes van sinus, cosinus en tangens vir die volgende spesiale hoeke met behulp van die sketse: Use the definitions in terms of a unit circle and find the values of sine, cosine and tangent for the following special angles using the sketches supplied below: 92 Leereenheid 4: Radiaalmaat en trigonometrie (a) cos π
= ……………………………… 2
(b) sin π
= ………………………………. 2
(c) tan π
= ………………………………. 2
(d) cos π = …………………………………. (e) sin π = ………………………………….. (0, 1)
(−1, 0)
(f) (g) tan π = ………………………………….. cos 3π
= ……………………………… 2
(h) sin 3π
= ………………………………. 2
(i) tan 3π
= ………………………………. 2
(j) cos 2π = …………………………………. (k) sin 2π = ……………………………………. (l) tan 2π = …………………………………… (0, −1)
(1, 0)
93 Leereenheid 4: Radiaalmaat en trigonometrie Oefening 4.4 Exercise 4.4 1. Bereken sonder ‘n sakrekenaar die waarde van 1.1
1. Calculate without a calculator the value of ⎛4 ⎞
⎛3 ⎞
⎛ 11 ⎞
tan2 ⎜ π ⎟ + sec2 ⎜ π ⎟ × 4cos2 ⎜ π ⎟ ⎝3 ⎠
⎝4 ⎠
⎝ 6 ⎠
2
⎛ ⎛2 ⎞
⎛ 5 ⎞⎞
⎛π ⎞
1.2 ⎜ sin ⎜ π ⎟ + cos ⎜ π ⎟ ⎟ + cosec2 ⎜ ⎟ 3
3
⎠
⎝
⎠⎠
⎝6⎠
⎝ ⎝
94 Leereenheid 4: Radiaalmaat en trigonometrie 1.3
⎛π ⎞
⎛5 ⎞
cos ⎜ ⎟ + sin ⎜ π ⎟ + 3cos ( 2π ) ⎝2⎠
⎝2 ⎠
1
⎛4 ⎞
⎛5 ⎞
⎛3 ⎞
1.4 3cos2 ( 2π ) + tan2 ⎜ π ⎟ + 2sin ⎜ π ⎟ + 2cos5 ⎜ π ⎟ 3
3
6
⎝
⎠
⎝
⎠
⎝2 ⎠
95 Leereenheid 4: Radiaalmaat en trigonometrie ⎛5 ⎞ ⎛4 ⎞
⎛ 11 ⎞
sin2 ⎜ π ⎟ tan ⎜ π ⎟ sec2 ⎜ π ⎟
⎝4 ⎠ ⎝3 ⎠
⎝ 6 ⎠ 1.5
2
⎛
⎞
tan ⎜ π ⎟
⎝3 ⎠
⎛4 ⎞
⎛7 ⎞
1.6 sin2 ⎜ π ⎟ + sin2 ⎜ π ⎟ ⎝3 ⎠
⎝6 ⎠
96 Leereenheid 4: Radiaalmaat en trigonometrie ⎛π ⎞
⎛π ⎞
1.7 sec 2 ⎜ ⎟ − tan2 ⎜ ⎟ 3
⎝ ⎠
⎝3⎠
⎛2 ⎞
⎛2 ⎞
1.8 cosec 2 ⎜ π ⎟ − cot2 ⎜ π ⎟ 3
⎝
⎠
⎝3 ⎠
2. Gebruik die eerste skets in Fig. 4.4.1 en bewys dat 2. Use the first sketch in Fig. 4.4.1 and prove that 2.1 cosec2θ = cot2 θ + 1 97 Leereenheid 4: Radiaalmaat en trigonometrie 2.2 sec2θ = tan2 θ + 1 2.3 tan4 θ + tan2 θ = sec4 θ − sec2 θ 98 Leereenheid 4: Radiaalmaat en trigonometrie Wenk vir nr. 2/ Hint for nr. 2 Bewys dat/ Prove that sin2α + cos2 α = 1 Oplossing/ Solution Linkerkant/LHS = sin2α + cos2 α
2
⎛a⎞ ⎛b⎞
=⎜ ⎟ +⎜ ⎟
⎝c⎠ ⎝c⎠
=
=
=
a2
+
2
Volgens 2e skets in Fig. 4.4.1/ Using 2nd sketch in Fig. 4.4.1
b2
c2 c2
a 2 + b2
c2
c2
c
=1
2
Volgens Pythagoras is/ According to Pythagoras is a 2 + b2 = c2
Regterkant/RHS = 1
∴Linkerkant/LHS = Regterkant/RHS dus / so sin2α + cos2 α = 1
99 Leereenheid 4: Radiaalmaat en trigonometrie 3. Vereenvoudig na eenvoudigste vorm: 3. Simplify to simplest form: 3.1
sin (π − θ ) cot (π + θ ) sec ( 2π − θ )
2 ⎛π
⎞
cos (π + θ ) + cos ⎜ − θ ⎟
⎝2
⎠
2
3.2
100 tan (π − θ ) 1 − sin2 θ
⎛π
⎞
cos ⎜ − θ ⎟
2
⎝
⎠
Leereenheid 4: Radiaalmaat en trigonometrie 3.3
sin (π + A) cos ( 2π − A)
⎛π
⎞
cos ⎜ − A ⎟ cos ( 2π + A )
⎝2
⎠
4. Voltooi die volgende drie voorstellings van die basiese trigonometriese grafieke: 4. Complete the following three representatios of the basic trigonometric graphs: 4.1 y = sin x vir/ for x ∈ [ 0; 2π ] Wenk/Hint: x 0 1
π 4
1
π 2
3
π 4
π 5
π 4
3
π 2
7
π 4
2π y 101 Leereenheid 4: Radiaalmaat en trigonometrie 4.2 y = cos x vir/ for x ∈ [ 0; 2π ] Wenk/Hint: 102 x 0 1
π 4
1
π 2
3
π 4
π 5
π 4
3
π 2
7
π 4
2π y Leereenheid 4: Radiaalmaat en trigonometrie 4.3 y = tan x vir/ for x ∈ [ 0; 2π ] Wenk/Hint: x 0 y π
4
45π
100
π
2
55π
100
3π
4
π 5π
4
145π
3π
100
2
155π
7π
100
4
2π 103 Leereenheid 4: Radiaalmaat en trigonometrie 5. Los die volgende trigonometriese vergelykings op vir waardes van θ sodat 0 ≤ θ ≤ 2π : 1
5.1 sinθ = − 2
104 5. Solve the following trigonometric equations for values of θ such that 0 ≤ θ ≤ 2π : Leereenheid 4: Radiaalmaat en trigonometrie 5.2 2cosθ = 3 5.3
3 tanθ + 1 = 0 105 Leereenheid 4: Radiaalmaat en trigonometrie 5.4 sec 2θ + 2 = 0
5.5
106 3cosec2θ + 2 = 0 Leereenheid 4: Radiaalmaat en trigonometrie 6. Los die volgende trigonometriese vergelykings op vir alle radiaal‐waardes van θ , met ander woorde, vind die algemene oplossings: 6.1 π
π
cos( x + ) = sin(3 x − ) 3
3
6.2 sin(cos x ) = 0 6. Solve the following trigonometric equations for all radian values of θ , in other words – find the general solutions: 107 Leereenheid 4: Radiaalmaat en trigonometrie 4.5 Die som‐ en verskilformules/ The sum and difference formulae Uit die definisies van die trigonometriese funksies kan die volgende belangrike indentiteite afgelei word; in hierdie kursus het ons nie tyd om hul afleidings te bespreek nie, maar u kan dit gerus self naslaan: From the definitions of the trigonometric functions the following important identities may be derived; in this course we do not have time to discuss their derivations but you may research that on your own: sin ( A ± B ) = sin Acos B ± cos A sin B
cos ( A ± B ) = cos Acos B ∓ sin A sin B tan ( A ± B ) =
tan A ± tan B
1 ∓ tan A tan B
Oefening 4.5 5
met 13
3
cos B = −
met 5
1. Indien gegee is dat sin A =
0≤ A≤
π
2
3π
π ≤B≤
2
en , bereken sonder ‘n sakrekenaar: 108 5
with 13
π
3
0≤ A≤
and cos B = − with 5
2
3π
π ≤B≤
, calculate without a 2
1. If it is given that sin A =
calculator: 1.1 sin ( A − B ) Exercise 4.5 Leereenheid 4: Radiaalmaat en trigonometrie 1.2 cos ( A + B ) 1.3 tan ( B − A ) 2. Bereken sonder ‘n sakrekenaar die 2. Calculate without a calculator the π
waarde van cos ⎛⎜ ⎞⎟ deur gebruik te 12
⎝ ⎠
π
value of cos ⎛⎜ ⎞⎟ by making use 12
⎝ ⎠
π π
maak van cos ⎛⎜ − ⎞⎟ : ⎝4 6⎠
π π
of cos ⎛⎜ − ⎞⎟ : ⎝4 6⎠
109 Leereenheid 4: Radiaalmaat en trigonometrie 3. Bewys dat tan (π + θ ) = tanθ 3. Bewys dat tan (π + θ ) = tanθ 4. Bereken die sonder ‘n sakrekenaar die waarde van cos23° cos67° − sin23° sin67° . 110 4. Calculate without a calculator the value of cos23° cos67° − sin23° sin67° . Leereenheid 4: Radiaalmaat en trigonometrie 5. Bereken sonder ‘n sakrekenaar die tan18° + tan27°
waarde van 1 − tan18° tan27°
5. Calculate without a calculator the tan18° + tan27°
value of . 1 − tan18° tan27°
111 Leereenheid 4: Radiaalmaat en trigonometrie 4.6 Die dubbelhoekformules/ The double angle formulae Uit die som‐ en verskilformules kan die volgende belangrike indentiteite afgelei word: From the sum and difference formulae the following important identities may be derived: sin2 A = 2sin Acos A
cos2 A = cos2 A − sin2 A
= 2cos2 A − 1
= 1 − 2sin2 A
tan2 A =
2tan A
1 − tan2 A
Oefening 4.6 Exercise 4.6 1. Bewys dat/ Prove that sin2 A = 2sin A cos A deur soos volg te begin/ by starting as follows:
Linkerkant/ LHS = s in2 A
= sin( A + A)
2. Bewys dat/ Prove that cos2 A = 1 − 2sin2 A . 112 Leereenheid 4: Radiaalmaat en trigonometrie 3. Gebruik u resultaat uit vraag 2 om 'n indentiteit vir sin2 A af te lei in terme van cos2 A / Use your result from question 2 in order to derive an identity for sin2 A in terms of cos2 A 4. Bewys dat/ Prove that cos2 A = 2cos2 A − 1 5. Gebruik u resultaat uit vraag 4 om 'n indentiteit vir cos2 A af te lei in terme van cos2A / Use your result from question 4 in order to derive an identity for cos2 A in terms of cos2 A 113 Leereenheid 4: Radiaalmaat en trigonometrie 6. Bewys die indentiteit/ Prove the identity: sin2φ
= tanφ 1 + cos2φ
7. Bewys die indentiteit/ Prove the identity: sin( x + y ) ⋅ sin( x − y ) = sin2 x − sin2 y 114 Leereenheid 4: Radiaalmaat en trigonometrie 8. Bewys die indentiteit/ Prove the identity: cos
π
28π
= cos 6
6
9. Bewys die indentiteit/ Prove the identity: cos
2π
28π
= cos( − ) 3
3
115 Leereenheid 4: Radiaalmaat en trigonometrie sec x − cos x
10. Bewys die indentiteit/ Prove the identity: = cosec x − cot x sin x + tan x
11. Bewys die indentiteit/ Prove the identity: 116 cos 4 x + cos 2 x
= cot x sin 4 x − sin 2 x
Leereenheid 5: Absolute waardes en limiete 5 Absolute waardes en limiete/ Absolute values and limits Leerdoelstellings vir hierdie leereenheid Na afhandeling van hierdie leereenheid moet die student in staat wees om die volgende te Learning aims for this study unit Upon completion of this study unit the student must be able to do the following: 1. Solve inequalities doen: 2. Apply the absolute value operation 1. Ongelykhede op te los 2. Die absolute waarde‐bewerking op algebraïese uitdrukkings toe te pas 3. Die absolute waarde‐funksie as 'n stuksgewyse funksie te definieer 4. Die absolute waarde‐funksie as die "afstand"‐funksie te interpreteer en grafies voor te stel 5. Die limiet van 'n funksie in terme van linkerlimiete en regterlimiete in die omgewing van 'n punt te bepaal 6. Uitspraak te kan lewer oor die kontinuïteit van 'n funksie in 'n to algebraic expressions 3. Define the absolute value function as a piece‐wise function 4. Interpret the absolute value function as the "distance"‐function and to represent it graphically 5. Determine the limit of a function in terms of left‐sided limits and right‐
sided limits in the vicinity of a point 6. Give judgment regarding the continuity of a function in a particular point 7. Compute certain limits bepaalde punt 8. Apply the epsilon‐delta definition of 7. Sekere limiete te bereken the limit of a function in order to 8. Die epsilon‐delta‐definisie van die prove that the limit of a function in a limiet van 'n funksie toe te pas om te particular point exists bewys dat die limiet van 'n funksie in 'n bepaalde punt bestaan 117 Leereenheid 5: Absolute waardes en limiete 5.1 Ongelykhede/ Inequalities Indien ons die = in 'n vergelyking vervang met een van die relasietekens < , ≤ , > of ≥ dan verkry ons 'n ongelykheid. From the sum and difference formulae the following important identities may be derived: Sulke ongelykhede kan op soortgelyke wyse opgelos word as gewone vergelykings, solank die volgende in gedagte gehou word: Eienskap/ Property As/ if a < b dan is/ then −a > −b Voorbeeld/ Example 4 < 12 is waar/ is true (vermenigvuldiging of deling met −1 verander die Maar/ but −4 < −12 is onwaar/ is false ongelykheidsteken/ Multiplication or division by −1 Dus/ So −4 > −12 of/ or −12 < −4 changes the sign of the relationship symbol) As/ if a < b dan is/ then a + c < b + c en/ and As/ if x − 3 < 0 dan is/ then x − 3 + 3 < 0 + 3 a−c<b−c en dus is / and consequently x < 3 As/ if a < b en/ and c > 0 dan is/ then ac < bc en/ 1
1
As/ if x < 4 dan is/ then ( 2 ) x < ( 2 ) 4 2
2
a b
and < which implies
x
<
8
sodat/ c c
As/ if a < b en/ and c < 0 dan is/ then ac > bc en/ x
− <6
if dan is/ then As/ 5
a b
and > c c
x
( −5) ⎛⎜ − ⎞⎟ > ( −5)( 6 ) sodat/ which implies ⎝ 5⎠
x > −30 As/ if a < b en/ and b < c dan is/ then a < c
Die dodelikste gevaar by die oplos van ongelykhede is waarskynlik by rasionale ongelykhede, wat NIE soos rasionale vergelykings gehanteer kan word NIE. Beskou die volgende voorbeelde aandagtig: As/ if a < x en/ and x < 2 dan is/ then a < 2
The most lethal danger associated with the solution of inequalities probably occurs in the case of rational inequalities, which CANNOT be treated the same way as rational equations Study the following examples with great concentration: 118 Leereenheid 5: Absolute waardes en limiete Rasionale vergelyking/ Rational equation Rasionale ongelykheid/ Rational inequality x −3
=2
x+2
x −3 x +2
x+2
∴
×
=2×
1
1
x+2
∴ x − 3 = 2x + 4
x −3
>2
x+2
x −3
∴
−2>0
x+2
x −3
x+2
∴
− 2×
>0
+
+2
2
x
x
∴− x = 7
x − 3 2x − 4
∴ x = −7
∴
−
>0
x+2 x+2
DIE KGV WORD GEBRUIK OM WEG TE DOEN MET x − 3 − 2x − 4
∴
> 0 BEHOUBREUKVORM/
DIE BREUKVORM/ THE LCM IS USED IN ORDER TO x+2
KEEP FRACTION FORM!
DO AWAY WITH THE FRACTION FORM −x − 7
∴
>0
Grafies/ Graphically: x+2
x+7
x+7
∴
< 0 so/so is negatief/negative
x+2
x+2
Ons stel nou intervalle saam uit punte/ We now set up intervals from the points x = −7 en/ and x = −2 : x+7
is negatief op/ is negative on x+2
−7 < x < −2 . x −3
> 2 is x+2
dus/ is therefore −7 < x < −2 , ook geskryf as/ Die oplossing van/ The solution of also written as ( −7; − 2 ) (vergelyk met grafiek links/ compare with the graph above left) 119 Leereenheid 5: Absolute waardes en limiete Rasionale vergelyking/ Rational equation Rasionale ongelykheid/ Rational inequality 2x + 1
2
−
=1 x −1 x −3
∴
( x − 1 )( x − 3)
1
2x + 1
2
−
<1 x −1 x −3
( x − 1 )( x − 3 )
2 ⎞
⎛ 2x + 1
×⎜
−
⎟ = 1×
1
⎝ x −1 x − 3 ⎠
∴ ( 2 x + 1 )( x − 3 ) − 2 ( x − 1 ) = ( x − 1 )( x − 3 )
∴
2x + 1
2
−
−1 < 0
x −1 x − 3
∴
2 ( x − 1)
( 2 x + 1 )( x − 3 )
( x − 3 )( x − 1 )
−
−
<0
( x − 1 )( x − 3 ) ( x − 3)( x − 1 ) ( x − 3 )( x − 1 )
∴
x2 − 3 x − 4
<0
( x − 1 )( x − 3)
∴2 x2 − 7 x − 1 = x2 − 4 x + 3
∴ x − 3x − 4 = 0
2
∴ ( x − 4 )( x + 1 ) = 0
( x − 4 )( x + 1 )
∴
<0
( x − 1 )( x − 3 )
∴ x = 4 of / or x = −1
DIE KGV WORD GEBRUIK OM WEG TE DOEN MET Ons stel nou intervalle saam uit punte/ We now DIE BREUKVORM/ THE LCM IS USED IN ORDER TO set up intervals from the points DO AWAY WITH THE FRACTION FORM x = 4 , x = −1 , x = 1 en/ and x = 3 : Grafies/ Graphically: ( x − 4 )( x + 1 )
is negatief op/ is negative on ( x − 1 )( x − 3 )
−1 < x < 1 en/and 3 < x < 4 . Die oplossing is dus/ The solution is therefore ( −1;1 ) ∪ ( 3; 4 ) (vergelyk met grafiek links/ compare with the graph left) 120 Leereenheid 5: Absolute waardes en limiete Oefening 5.1 Los op die volgende ongelykhede: 1.
x −3
≥0 x +1
2.
2x + 1
≤3 x −5
Exercise 5.1 Solve the following inequalities: 121 Leereenheid 5: Absolute waardes en limiete 3.
122 1+ x 1− x
−
≤ −1 1− x 1+ x
Leereenheid 5: Absolute waardes en limiete 5.2 Absolute waardes/ Absolute values Die afstand‐definisie The distance definition x beteken x − 0 en dit beteken die x means x − 0 and that means the grootte van die afstand tussen die punt 0 op die getallelyn en die punt x . distance between 0 on the number line and the point x . Dit is duidelik dat x aan enige kant van die It is clear that x may be located on either punt 0 kan lê; die afstand x − 0 is altyd side of 0; the distance x − 0 is always positief – so, die waarde van x vir enige positive – so, the value of x for any value of waarde van x sal altyd positief wees. x will always be positive. Die getal x − 0 kan verkry word deur ‘n The number Die getal x − 0 may be passer se radius op x te stel en die punt obtained by setting the radius of a compass op die nulpunt neer te sit. on x and putting the sharp pin in the zero point. Then the point x may be marked off on both sides of the zero point. Dan kan die punt x aan weerskante van die nulpunt afgemerk word. Die formele algebraïese definisie ⎧x
x =⎨
⎩− x
The formal algebraic definition as/if x ≥ 0
of in die algemeen/or in general as/if x < 0
⎧x − a
x − a = ( x − a) − 0 = ⎨
⎩− ( x − a )
as/if x − a ≥ 0
as/if x − a < 0
123 Leereenheid 5: Absolute waardes en limiete Die formule algebraïese definisie gee ons 'n manier om die absolute waarde‐funksie, gedefinieer deur f ( x ) = a bx − d + h maklik te skets. Die geheim is om die funksie as 'n stuksgewyse funksie te hanteer; dan gedra die absolute waarde‐funksie haarself soos 'n kombinasie van twee beperkte reguit lyn‐
grafieke. Die knakpunt is die punt op die grafiek waaromheen die grafiek simmetries is: The formal algebraic definition provides us with a way to easily sketch the absolute value function, defined by f ( x ) = a bx − d + h . The secret is to treat the function as a piece‐
wise defined function; then the absolute value function behaves like a combination of two constrained straight line graphs. The vertex is the point on the graph about which the graph is symmetrical: f ( x ) = a bx − d + h
⎧⎪ a ( bx − d ) + h
∴ f (x) = ⎨
⎪⎩ − a ( bx − d ) + h
⎧ abx − ad + h
=⎨
⎩ − abx + ad + h
c
⎧ m
⎪( ab ) x + ( − ad + h )
⎪
=⎨
⎪⎛ − ab ⎞ x + ( ac + h )
⎪⎜⎝ m ⎟⎠
c
⎩
as/if
bx − d ≥ 0
as/if
bx − d < 0
as/if
bx ≥ d
as/if
bx < d
as/if
x≥
as/if
d
b
d
x<
b
Die knakpunt is dus by/ So the vertex is at x =
d
b
c
m
d
Vervang/ substitute x = in/ into y = ( ab ) x + ( − ad + h ) om die/ in order to obtain the y ‐ko‐
b
oordinaat/ co‐ordinate te verkry. Voorbeeld/ Example Skets die grafieke van die volgende funksies/ Sketch graphs of the following functions: 1.
y=2
1
x −2 +4 4
2.
y=−
1
3x + 5 − 3 4
124 Leereenheid 5: Absolute waardes en limiete Oplossing/ Solution 1.
y=2
1
x −2 +4
4
⎧ ⎛1
⎞
⎪2 ⎜ 4 x − 2 ⎟ + 4
⎪ ⎝
⎠
∴y = ⎨
1
⎪ −2 ⎛ x − 2 ⎞ + 4
⎟
⎪⎩ ⎜⎝ 4
⎠
⎧1
⎪⎪ 2 x − 4 + 4
=⎨
⎪− 1 x + 4 + 4
⎪⎩ 2
⎧1
⎪⎪ x
= ⎨2
⎪− 1 x + 8
⎪⎩ 2
Knakpunt/ vertex: ( 8; 4 )
as/ if
1
x −2≥ 0
4
as/ if
1
x −2< 0
4
as/ if
as/ if
1
x≥2
4
1
x<2
4
as/ if
x≥8
as/ if
x<8
Dit lewer/ this yields: Normaalweg wys ons nie die twee "hulplyne" nie/ Normally, we do not show the two "auxilliary lines": 125 Leereenheid 5: Absolute waardes en limiete 2.
1
3x + 5 − 3
4
⎧ 1
⎪⎪ − ( 3 x + 5 ) − 3
∴y = ⎨ 4
⎪ + 1 ( 3 x + 5 ) − 3 as/ if
⎪⎩ 4
5
⎧ 3
⎪⎪ − 4 x − 4 − 3
=⎨
⎪3 x + 5 − 3
⎪⎩ 4
4
17
⎧ 3
⎪⎪ − 4 x − 4
=⎨
⎪3 x − 7
⎪⎩ 4
4
⎛ 5
⎞
Knakpunt/ vertex: ⎜ − ; − 3 ⎟
3
⎝
⎠
y=−
3x + 5 ≥ 0
as/ if
3x + 5 < 0
3 x ≥ −5
as/ if
3 x < −5
as/ if
5
3
5
x<−
3
x≥−
as/ if
as/ if
Dit lewer/ this yields: Meetkundige definisie Geometric definition Hoe kan ons ‘n passer en ‘n getallelyn gebruik om betekenis te gee aan iets soos How may we use a compass and a number line to assign meaning to something like x − a = r ? (Wat is die meetkundige x − a = r ? (What is the geometric meaning betekenis van x − a = r ?) x − a = r beteken ( x − a) − 0
of x − a = r ?) = r en dit beteken x − a lê presies r eenhede vanaf 0 op die getallelyn, so plaas die passer se punt op 0 en stel die radius op r . Trek dan ‘n sirkel om 0 en let op waar die sirkel die getallelyn sny. x − a lê op die snypunte 126 x − a = r means ( x − a) − 0
= r and that means that x − a lies precisely r units from 0 on the number line, so place the pin of the compass on 0 and set the radius on r . Construct a circle around 0 and note where the circle intersects the number line. x − a Leereenheid 5: Absolute waardes en limiete van die sirkel met die lyn, dit is die punte − r en − r . lies on the points of intersection of the circle with the line, that is the points − r and − r . Meetkundige definisie/ Geometric definition: x − a = r impliseer dat/ implies that x = a − r of x = a + r Hoe kan ons ‘n passer en ‘n getallelyn gebruik om betekenis te gee aan iets soos How may we use a compass and a number line in order to assign meaning to something x − a < r of selfs x − a > r ? (Wat is die like x − a < r or even x − a > r ? (What is meetkundige betekenis van x − a < r en the geometric meaning of x − a < r and x − a > r ?) x − a < r beteken x − a > r ?) (x − a) − 0
< r en dit beteken x − a lê minder as r eenhede vanaf 0 op die getallelyn, so plaas die passer se punt op 0 en stel die radius op r . Trek dan ‘n sirkel om 0 en let op waar die sirkel die getallelyn sny. x − a lê enige plek op die deel van die getallelyn tussen die snypunte van die sirkel met die lyn.
x − a < r means (x − a) − 0
< r and this indicates that x − a lies less than r units from 0 on the number line, so place the pin of the compass on 0 and set the radius on r . Next, construct a circle around 0 and note where the circle intersects the number line. x − a lies anywhere on the part of the number line between the points of intersection of the circle with the line. 127 Leereenheid 5: Absolute waardes en limiete x − a < r , dit is/ that is ( x − a ) − 0 < r impliseer dat/ implies that − r < x − a < r waaruit volg/ from which follows that a − r < x < a + r ook geskryf as/ also written as: ( a − r ; a + r ) Vir die geval x−a ≤r
oftewel ( x − a ) − 0 ≤ r lyk die situasie soos volg: 128 For the case x − a ≤ r or just as well ( x − a ) − 0 ≤ r the situation is as follows: Leereenheid 5: Absolute waardes en limiete x − a ≤ r , dit is/ that is ( x − a ) − 0 ≤ r impliseer dat/ implies that − r ≤ x − a ≤ r waaruit volg/ from which follows that a − r ≤ x ≤ a + r ook geskryf as/ also written as: [ a − r ; a + r ] Net so: ( x − a) − 0 > r
x−a >r
beteken en dit beteken x − a lê meer as r eenhede vanaf 0 op die getallelyn, so x − a lê enige plek op die deel van die getallelyn links en regs van die snypunte van die sirkel met die lyn. Similarly: ( x − a ) − 0 > r and x−a >r
means that means x − a lies more than r units from 0 on the number line, x − a lies anywhere on the section of the number line to the left or to the right of the points of intersection of the circle with the line. 129 Leereenheid 5: Absolute waardes en limiete x − a > r , dit is/ that is ( x − a ) − 0 > r impliseer dat/ implies that x − a < − r of/or x − a > r waaruit volg/ from which follows that x < a − r of/or x > a + r ook geskryf as/ also written as: ( −∞; a − r ) ∪ ( a + r ; ∞ ) Vir die geval x−a ≥r
oftewel ( x − a ) − 0 ≥ r lyk die situasie soos volg: 130 For the case x − a ≥ r or just as well ( x − a ) − 0 ≥ r the situation is as follows: Leereenheid 5: Absolute waardes en limiete x − a ≥ r , dit is/ that is ( x − a ) − 0 ≥ r impliseer dat/ implies that x − a ≤ − r of/or x − a ≥ r waaruit volg/ from which follows that x ≤ a − r of/or x ≥ a + r ook geskryf as/ also written as: ( −∞; a − r ] ∪ [ a + r ; ∞ ) Nog belangrike eienskappe van die absolute waarde‐bewerking Other important properties of the absolute value operation x = x2
− x = − x2 a b = ab
a a
=
b b
−x + a = x − a
As/ if x − a = 0 dan is/ then x = a Voorbeeld/ example 1: Bepaal/ Determine x sodat/ such that 2 3 x − 4 = 20 Oplossing (3 metodes)/ Solution (3 methods): •
Meetkundige definisie/ Geometric definition 131 Leereenheid 5: Absolute waardes en limiete 2 3x − 4 = 20
∴ 3x − 4 = 10
∴ ( 3x − 4 ) − 0 = 10
Bepaal dus die waarde van 3x − 4 sodat 3 x − 4 presies 10 eenhede vanaf die punt 0 op die getallelyn lê. Los dan op vir x./ So find the value of 3 x − 4 such that 3 x − 4 lies precisely 10 units from the
po int 0 . Then solve for x.
∴3x − 4 = −10
of / or
∴ x = −2
of / or
•
3x − 4 = 10
14
x=
3
Formele definisie/ Formal definition: 2 3x − 4 = 20
∴ 3x − 4 = 10
⎧⎪3x − 4 = 10
∴⎨
⎪⎩− ( 3x − 4 ) = 20
⎧3x = 14
∴⎨
⎩−3x + 4 = 10
⎧ 14
⎪⎪ x = 3
∴⎨
⎪−3x = 6
⎪⎩
⎧
⎪⎪ x = 4, 667
∴⎨
⎪ x = −2
⎪⎩
132 as
3x − 4 ≥ 0
as
3x − 4 < 0
as
as
3x ≥ 4
3x < 4
as
x≥
as
as
as
4
3
4
x<
3
4
x≥
3
4
x<
3
Leereenheid 5: Absolute waardes en limiete •
Grafiese interpretasie/ Graphical interpretation 133 Leereenheid 5: Absolute waardes en limiete Voorbeeld/ Example 2: Los op vir/ Solve for x sodat/ such that 3 2 x − 7 ≤ 30 Oplossing (3 metodes)/ Solution (3 methods): •
Meetkundige definisie/ Geometric definition 3 2 x − 7 ≤ 30
∴ 2 x − 7 ≤ 10
∴ ( 2 x − 7 ) − 0 ≤ 10
Bepaal dus die waarde van x sodat 2 x − 7 binne 10 eenhede van die punt 0 op die getallelyn lê/
Find the value of x such that 2 x − 7 lies less than 10 units from the po int 0 on the number line
∴−10 ≤ 2 x − 7 ≤ 10
∴−3 ≤ 2 x ≤ 17
3
17
∴− ≤ x ≤
2
2
∴ x ∈[ −1,5; 8,5]
•
134 Formele definisie/ Formal definition: Leereenheid 5: Absolute waardes en limiete 3 2 x − 7 ≤ 30
⎧⎪3 ( 2 x − 7 ) ≤ 30
∴⎨
⎪⎩−3 ( 2 x − 7 ) ≤ 30
⎧6 x − 21 ≤ 30
∴⎨
⎩−6 x + 21 ≤ 30
⎧
⎪⎪6 x ≤ 51
∴⎨
⎪ −6 x ≤ 9
⎪⎩
51
⎧
⎪⎪ x ≤ 6
∴⎨
⎪x ≥ − 9
⎪⎩
6
⎧ x ≤ 8,5
∴⎨
⎩ x ≥ −1,5
as
2x − 7 ≥ 0
as
2x − 7 < 0
as
as
2x ≥ 7
2x < 7
as
x≥
as
as
as
as
as
7
2
7
x<
2
7
2
7
x<
2
x ≥ 3,5 x < 3,5
x≥
∴−1,5 ≤ x ≤ 8,5
∴ x ∈ [ −1,5; 8,5]
•
Grafiese interpretasie/ Graphical interpretation Voorbeeld/ Example 3: Los op vir/ Solve for x sodat/ such that −2 3x + 4 < −6 Oplossing (3 metodes)/ Solution (3 methods): •
Meetkundige definisie/ Geometric definition − 2 3 x + 4 < −6
∴ 3x + 4 > 3
∴ ( 3x + 4 ) − 0 > 3
135 Leereenheid 5: Absolute waardes en limiete Bepaal dus die waarde van x sodat 3x + 4 meer as 3 eenhede van die punt 0 op die getallelyn lê./
Deter min e the value of x such that 3 x + 4 lies more than 3 units from the po int 0 on the number line.
∴3x + 4 < −3
of / or
7
∴x < −
of / or
3
7⎞ ⎛ 1 ⎞
⎛
∴⎜ −∞; − ⎟ ∪ ⎜ − ; ∞ ⎟
3⎠ ⎝ 3 ⎠
⎝
•
3x + 4 > 3
1
x>−
3
Formele definisie/ Formal definition − 2 3x + 4 < −6
∴ 3x + 4 > 3
⎧⎪3x + 4 > 3
∴⎨
⎪⎩− ( 3 x + 4 ) > 3
⎧3x > −1
∴⎨
⎩−3 x − 4 > 3
⎧
⎪⎪3x > −1
∴⎨
⎪−3 x > 7
⎪⎩
as
3x + 4 ≥ 0
as
3x + 4 < 0
as
3 x ≥ −4
as
3 x < −4
as
x≥−
as
1
⎧
as
⎪⎪ x > − 3
∴⎨
⎪x < − 7
as
⎪⎩
3
7
1
∴x < −
of
x>−
3
3
7⎞ ⎛ 1 ⎞
⎛
∴ x ∈ ⎜ −∞; − ⎟ ∪ ⎜ − ; ∞ ⎟
3⎠ ⎝ 3 ⎠
⎝
•
136 4
3
4
x<−
3
4
3
4
x<−
3
x≥−
Grafiese interpretasie/ Graphical interpretation Leereenheid 5: Absolute waardes en limiete 137 Leereenheid 5: Absolute waardes en limiete Oefening 5.2 1. Los die onbekende op: Exercise 5.2 1. Solve for the unknown: 1.1 3 2 − k = 0 1.2 −2 2t + 1 = 6 1.3
1
1
r + = −2 2
2
2. Los op vir die onbekende en stel die oplossing op ‘n getallelyn voor. 138 2. Solve for the unknown and represent the solutions on a number line. Leereenheid 5: Absolute waardes en limiete 2.1 x + 2 ≤ 2 2.2 −3 2 x − 5 < −9 2.3 −6
2 − 3x
< −6 4
139 Leereenheid 5: Absolute waardes en limiete 3. Skryf die volgende in absolute waarde‐notasie: 3. Skryf die volgende in absolute waarde‐notasie: 3.1 x is minder as 3 eenhede vanaf 7 3.1 x is minder as 3 eenhede vanaf 7 3.2 t is nie meer as 5 eenhede vanaf 8 3.2 t is nie meer as 5 eenhede vanaf 8 3.3 y lê tussen ‐3 en 3 3.3 y lê tussen ‐3 en 3 3.4 Die afstand tussen 6 en m is 4 3.4 Die afstand tussen 6 en m is 4 140 Leereenheid 5: Absolute waardes en limiete 5.3 Limiete en kontinuïteit/ Limits and continuity Dikwels stel ons daarin belang om te weet wat die waarde is wat 'n funksie aanneem wanneer die onafhanklike veranderlike baie groot negatief raak, of baie groot positief raak, of wanneer die onafhanklike 'n sekere waarde aanneem. Often we are interested in the value which is assumed by a function when the independent variable becomes a very large negative value, or a very large positive value, or when the independent variable assumes a particular value. Dit is nie in alle gevalle moontlik om gewoon die onafhanklike veranderlike in die funksie te vervang en dan die funksiewaarde te bereken nie. It is not always possible to simply substitute the independent variable into the function equation and then obtain the the function value. Vervolgens kyk ons na funksies wie se gedrag slegs met behulp van limiete volledig beskryf kan word. Next, we consider functions of which the behaviour can only be fully described in terms of limits. Voorbeeld 1 2x + 1
2
Beskou die funksie f ( x ) =
−
x −1 x −3
1. Bepaal f ( 2 ) , dit is die funksiewaarde Example 1 Consider the function 2x + 1
2
f ( x) =
−
x −1 x −3
in die punt waar x = 2 . 1. Determine f ( 2 ) , that is the function value in the point where x = 2 . 2. Die funksiewaarde in die punt waar x = 2 gee egter geen inligting oor die gedrag van die funksie baie naby aan die punt waar x = 2 nie. Laat ons die gedrag van die funksie ondersoek in die omgewing van hierdie punt. 2. The function value in the point x =2
provides no where information about the behaviour of the function close to the point where x = 2 . Let us now investigate the behaviour of the function in the vicinity or proximity of this point. 141 Leereenheid 5: Absolute waardes en limiete Voltooi die volgende tabel: Complete the following table: 2.1.1 Voltooi: Wanneer x vanaf die linkerkant streef na 2, dan streef die funksiewaarde na ................................................. 2.1.1 Complete: When x tends from the left side to 2, then the function value tends to ................................................. 2.1.2 Skryf dit in limiet‐notasie: 2.1.2 Write this in limit notation: 2 ⎞
⎛ 2x + 1
−
lim− ⎜
⎟ = ..................... x −3⎠
x →2 ⎝ x − 1
2 ⎞
⎛ 2x + 1
−
lim− ⎜
⎟ = ..................... x −3⎠
x →2 ⎝ x − 1
of korter: or shorter: lim f ( x ) = ..................... lim f ( x ) = ..................... x → 2−
2.2.1 Voltooi: Wanneer x vanaf die regterkant streef na 2, dan streef die funksiewaarde na x → 2−
................................................. 2.2.1 Complete: When x tends from the right side to 2, then the function value tends to ................................................. 2.2.2 Skryf dit in limiet‐notasie: 2.2.2 Write this in limit notation: lim f ( x ) = ..................... lim f ( x ) = ..................... x → 2+
x →2+
2.3 142 Vergelyk 2.1.2 en 2.2.2. 2.3 Compare 2.1.2 and 2.2.2. Leereenheid 5: Absolute waardes en limiete Voltooi: Complete: lim f ( x ) ....... lim+ f ( x ) x →2−
lim f ( x ) ....... lim+ f ( x ) x →2−
x →2
x →2
Wanneer die linkerlimiet en die regterlimiet van 'n funksie dieselfde waarde aanneem in 'n punt, dan sê ons die funksie het 'n limiet in daardie punt. When the left‐sided limit and the right‐
sided limit of a function assumes the same value in a point, then we say that the function has a limit in that point. Beskou 2.1.2 en 2.2.2 en voltooi: Consider 2.1.2 and 2.2.2 and complete: Omdat lim− f ( x ) = lim+ f ( x ) = .............. kan Because lim− f ( x ) = lim+ f ( x ) = .............. ons skryf: we may write: x →2
x →2
x →2
lim f ( x ) = .............. x →2
lim f ( x ) = .............. x →2
x →2
Die limiet van 'n funksie in 'n bepaalde punt bestaan slegs as beide die linkerlimiet en die regterlimiet in daardie punt bestaan en biede die linkerlimiet en die regterlimiet dieselfde waarde het. The limit of a function in a particular point only exists provided that both the left‐sided limit and the right‐sided limit exist in that point and that they are equal in value. In hierdie geval het ons dat die funksiewaarde in die punt waar x = 2 en die limiet van die funksie wanneer x van albei kante streef na 2 dieselfde waarde lewer: In this case we have that both the function value in the point where x = 2 and the limit of the function when x tends from both sides to 2 yield the same value: lim f ( x ) = f ( 2 ) lim f ( x ) = f ( 2 ) x →2
x →2
Hierdie spesiale sameloop van omstandighede impliseer dat die funksie f kontinu This special set of circumstances implies that f is continuous (solid curve without (aaneenlopend, sonder spronge of onderbrekings) is in die punt waar x = 2 . gaps or breaks or jumps) in the point where x = 2 . 3. Laat die gedrag van die funksie ondersoek waar x baie groot negatief en waar x baie groot positief word. Voltooi die volgende tabel: 3. Let us now investigate the behaviour of the function where x becomes a very large negative value and where x becomes a very large positive value. Complete the following table: 143 Leereenheid 5: Absolute waardes en limiete 3.1.1 Voltooi: Wanneer x streef na negatief oneindig, dan streef die funksiewaarde na 3.1.1 Complete: When x tends to negative infinity, then the function value tends to ................................................. ................................................. 3.1.2 Skryf dit in limiet‐notasie: 3.1.2 Rewrite this in limit notation: 2 ⎞
⎛ 2x + 1
lim ⎜
−
⎟ = ..................... x −1 x −3⎠
2 ⎞
⎛ 2x + 1
lim ⎜
−
⎟ = ..................... x −1 x −3⎠
x →−∞ ⎝
x →−∞ ⎝
of korter: or shorter: lim f ( x ) = ..................... lim f ( x ) = ..................... x →−∞
x →−∞
3.2.1 Voltooi: Wanneer x streef na positief oneindig, dan streef die funksiewaarde na 3.2.1 Complete: When x tends to positive infinity, then the function value tends to ................................................. ................................................. 3.2.2 Skryf dit in limiet‐notasie: 3.2.2 Write this in limit notation: lim f ( x ) = ..................... lim f ( x ) = ..................... x →∞
x →∞
Uit die limiete lim f ( x ) en lim f ( x ) kan x →−∞
x →∞
ons dikwels die horisontale asimptote van 'n funksie vind. 144 From the limits lim f ( x ) and lim f ( x ) x →−∞
x →∞
we can often find the horizontal asymptotes of a function. Leereenheid 5: Absolute waardes en limiete 4. Bepaal f ( 3 ) , dit is die funksiewaarde 4. Determine f ( 3 ) , that is the in die punt waar x = 3 . function value where x = 3 . 5. Die funksiewaarde in die punt waar x = 3 sou selfs al het dit bestaan, geen inligting gegee het oor die gedrag van die funksie in die omgewing van die punt waar x = 3 is nie. 5. The function value in the point where x = 3 would not, even if it existed, provide any information about the behaviour of the function in the vicinity of the point where x = 3 . Laat ons die gedrag van die funksie ondersoek in die omgewing van hierdie punt, waar die funksie ongedefinieerd is. Let us now investigate the behaviour of the function in the vicinity of this point, where the function is undefined. Voltooi die volgende tabel: Complete the following table: 5.1.1 Voltooi: Wanneer x vanaf die linkerkant streef na 3, dan streef die funksiewaarde na ................................................. 5.1.1 Complete: When x tends to 3 from the left side, then the function value tends to ................................................. 145 Leereenheid 5: Absolute waardes en limiete 5.1.2 Skryf dit in limiet‐notasie: 5.1.2 Write this in limit notation: 2 ⎞
⎛ 2x + 1
lim ⎜
−
⎟ = ..................... x −1 x − 3 ⎠
2 ⎞
⎛ 2x + 1
lim ⎜
−
⎟ = ..................... x −1 x − 3 ⎠
x →3 − ⎝
x →3 − ⎝
of korter: or shorter: lim f ( x ) = ..................... lim f ( x ) = ..................... x →3 −
5.2.1 Voltooi: Wanneer x vanaf die regterkant streef na 3, dan streef die funksiewaarde na x →3 −
................................................. 5.2.1 Complete: When x tends to 3 from the right side, then the function value tends to ................................................. 5.2.2 Skryf dit in limiet‐notasie: 5.2.2 Write this in limit notation: lim f ( x ) = ..................... lim f ( x ) = ..................... x →3 +
x →3 +
5.3 Vergelyk 5.1.2 en 5.2.2. 5.3 Compare 5.1.2 and 5.2.2. Voltooi: Complete: lim f ( x ) ....... lim+ f ( x ) x →3 −
lim f ( x ) ....... lim+ f ( x ) x →3
x →3 −
x →3
Wanneer die linkerlimiet en die regterlimiet van 'n funksie nie dieselfde waarde aanneem in 'n punt nie, dan sê ons die funksie het nie 'n limiet in daardie punt nie. When the left‐sided limit and the right‐
sided limit of a function does not assume the same value in a point, then we say that the function does not have a limit in that point. Beskou 5.1.2 en 5.2.2 en voltooi: Omdat lim− f ( x ) ≠ lim+ f ( x ) kan ons skryf: x →3
x →3
lim f ( x ) bestaan nie. x →3
Opmerking: Aangesien die linkerlimiet en regterlimiet elk ook nie na 'n vaste reële waarde streef nie, bestaan die linker‐ en regterlimiet in hierdie geval ook nie. 146 Consider 5.1.2 and 5.2.2 and complete: Because lim− f ( x ) ≠ lim+ f ( x ) we may x →3
x →3
write: lim f ( x ) does not exist. x →3
Remark: Because the left‐sided limit and the right‐sided limit each does not tend to a fixed real value, neither the left‐sided limit nor the right‐sided limit exists in this case . Leereenheid 5: Absolute waardes en limiete Ons noem die vertikale lyn met vergelyking x = 3 'n vertikale asimptoot van die funksie. We refer to the vertical line with equation x = 3 as a vertical asymptote of the function. Die limiet van 'n funksie in 'n bepaalde punt bestaan slegs as beide die linkerlimiet en die regterlimiet in daardie punt bestaan en dieselfde waarde besit. The limit of a function in a particular point only exist provided that the left‐sided limit and the right‐sided limit both exist in that point and that they have the same value. In hierdie geval het ons dat die funksiewaarde in die punt waar x = 3 nie bestaan nie en dat die limiet van die funksie wanneer x van albei kante streef na 3 ook nie bestaan nie. In this case we have that the function value in the point where x = 3 does not exist and that the limit of the function when x tends to 3 from both sides does not exist. Hierdie spesiale sameloop van omstandighede impliseer dat die funksie f diskontinu (nie‐
This special set of circumstances implies that the function f is discontinuous (not aaneenlopend, met 'n sprong of onderbreking) is in die punt waar x = 3 . solid, with a gap or a jump) in the point where x = 3 . Vir duidelikheid toon ek hieronder 'n rekenaarvoorstelling van die funksie wat ons ondersoek het: For clarity I show a computer‐generated representation of the function we have investigated: 147 Leereenheid 5: Absolute waardes en limiete Voorbeeld 2 Example 2 Beskou die funksie ⎧x
⎪
f ( x ) = ⎨2
⎪x
⎩
2
Consider the function as
x <1
as
as
x =1 x >1
3. Bepaal f (1 ) , dit is die funksiewaarde ⎧ x2
⎪
f ( x ) = ⎨2
⎪x
⎩
if
x <1
if
if
x =1 x >1
2. Determine f (1 ) , that is the in die punt waar x = 1 . function value in the point where x = 1 . 4. Die funksiewaarde in die punt waar x = 1 gee egter geen inligting oor die gedrag van die funksie baie naby aan die punt waar x = 1 nie. 4. The function value in the point x =1
provides no where information about the behaviour of the function close to the point where x = 1 . Laat ons die gedrag van die funksie ondersoek in die omgewing van hierdie punt. Let us now investigate the behaviour of the function in the vicinity or proximity of this point. Voltooi die volgende tabel: 148 Complete the following table: Leereenheid 5: Absolute waardes en limiete 2.1.1 Voltooi: Wanneer x vanaf die linkerkant streef na 1, dan streef die funksiewaarde na ................................................. 2.1.1 Complete: When x tends from the left side to 1, then the function value tends to ................................................. 2.1.2 Skryf dit in limiet‐notasie: 2.1.2 Write this in limit notation: lim f ( x ) = ..................... lim f ( x ) = ..................... x→1−
2.2.1 Voltooi: Wanneer x vanaf die regterkant streef na 1, dan streef die funksiewaarde na x→1−
................................................. 2.2.1 Complete: When x tends from the right side to 1, then the function value tends to ................................................. 2.2.2 Skryf dit in limiet‐notasie: 2.2.2 Write this in limit notation: lim f ( x ) = ..................... lim f ( x ) = ..................... x →1+
x →1+
2.3 Vergelyk 2.1.2 en 2.2.2. Voltooi: Compare 2.1.2 and 2.2.2. Complete: lim f ( x ) ....... lim+ f ( x ) x→1−
2.3 x→1
lim f ( x ) ....... lim+ f ( x ) x→1−
x→1
149 Leereenheid 5: Absolute waardes en limiete Wanneer die linkerlimiet en die regterlimiet van 'n funksie dieselfde waarde aanneem in 'n punt, dan sê ons die funksie het 'n limiet in daardie punt. When the left‐sided limit and the right‐
sided limit of a function assumes the same value in a point, then we say that the function has a limit in that point. Beskou 2.1.2 en 2.2.2 en voltooi: Consider 2.1.2 and 2.2.2 and complete: Omdat lim− f ( x ) = lim+ f ( x ) = .............. kan Because lim− f ( x ) = lim+ f ( x ) = .............. ons skryf: we may write: x→1
x→1
x→1
lim f ( x ) = .............. x→1
lim f ( x ) = .............. x→1
x→1
Die limiet van 'n funksie in 'n bepaalde punt bestaan slegs as beide die linkerlimiet en die regterlimiet in daardie punt bestaan en biede die linkerlimiet en die regterlimiet dieselfde waarde het. The limit of a function in a particular point only exists provided that both the left‐sided limit and the right‐sided limit exist in that point and that they are equal in value. In hierdie geval het ons dat die funksiewaarde in die punt waar x = 1 en die limiet van die funksie wanneer x van albei kante streef na 1 verskillende waardes lewer: In this case we have that both the function value in the point where x = 1 and the limit of the function when x tends from both sides to 2 yield different values: lim f ( x ) ≠ f (1 ) lim f ( x ) ≠ f (1 ) x→1
x→1
Hierdie spesiale sameloop van omstandighede impliseer dat die funksie f diskontinu is in die This special set of circumstances implies that f is discontinuous in the point where punt waar x = 1 . x = 1 . Vir duidelikheid toon ek hieronder 'n rekenaarvoorstelling van die funksie wat ons ondersoek het: 150 For clarity I show a computer‐generated representation of the function we have investigated: Leereenheid 5: Absolute waardes en limiete Voorbeeld 3 Example 3 Beskou die funksie f ( x ) =
8− x
2− x
3
1. Bepaal f ( 2 ) , dit is die funksiewaarde Consider the function f ( x ) =
8 − x3
2− x
1. Determine f ( 2 ) , that is the in die punt waar x = 2 . function value in the point where x = 2 . 2. Laat ons die gedrag van die funksie ondersoek in die omgewing van hierdie punt. Voltooi die volgende tabel: 2. Let us now investigate the behaviour of the function in the vicinity or proximity of this point. Complete the following table: 151 Leereenheid 5: Absolute waardes en limiete 2.1.1 Voltooi: Wanneer x vanaf die linkerkant streef na 2, dan streef die funksiewaarde na ................................................. 2.1.1 Complete: When x tends from the left side to 2, then the function value tends to ................................................. 2.1.2 Skryf dit in limiet‐notasie: 2.1.2 Write this in limit notation: lim f ( x ) = ..................... lim f ( x ) = ..................... x →2−
2.2.1 Voltooi: Wanneer x vanaf die regterkant streef na 2, dan streef die funksiewaarde na x →2−
................................................. 2.2.1 Complete: When x tends from the right side to 2, then the function value tends to ................................................. 2.2.2 Skryf dit in limiet‐notasie: 2.2.2 Write this in limit notation: lim f ( x ) = ..................... lim f ( x ) = ..................... x→2+
x→2+
2.3 Vergelyk 2.1.2 en 2.2.2. Voltooi: 152 Compare 2.1.2 and 2.2.2. Complete: lim f ( x ) ....... lim+ f ( x ) x→2−
2.3 x→2
lim f ( x ) ....... lim+ f ( x ) x→2−
x→2
Leereenheid 5: Absolute waardes en limiete Wanneer die linkerlimiet en die regterlimiet van 'n funksie dieselfde waarde aanneem in 'n punt, dan sê ons die funksie het 'n limiet in daardie punt. When the left‐sided limit and the right‐
sided limit of a function assumes the same value in a point, then we say that the function has a limit in that point. Beskou 2.1.2 en 2.2.2 en voltooi: Consider 2.1.2 and 2.2.2 and complete: Omdat lim− f ( x ) = lim+ f ( x ) = .............. kan Because lim− f ( x ) = lim+ f ( x ) = .............. ons skryf: we may write: x→2
x→2
x→2
lim f ( x ) = .............. x→2
lim f ( x ) = .............. x→2
x→2
Die limiet van 'n funksie in 'n bepaalde punt bestaan slegs as beide die linkerlimiet en die regterlimiet in daardie punt bestaan en beide die linkerlimiet en die regterlimiet dieselfde waarde het. The limit of a function in a particular point only exists provided that both the left‐sided limit and the right‐sided limit exist in that point and that they are equal in value. NOTE that the function need not be defined in that point; the limit exists whenever both the left‐sided limit and the right‐sided are equal in value. LET DAAROP dat die funksie nie in daardie punt gedefinieer hoef te wees nie; die limiet bestaan in elk geval as beide die linkerlimiet en die regterlimiet dieselfde waarde het. In hierdie geval het ons dat die funksiewaarde in die punt waar x = 2 en die limiet van die funksie wanneer x van albei kante streef na 2 nie dieselfde waarde lewer nie; die funksiewaarde bestaan dan nie eers nie: In this case we have that both the function value in the point where x = 2 and the limit of the function when x tends from both sides to 2 does not yield the same value; in fact, the function value does not exist: lim f ( x ) ≠ f ( 2 ) lim f ( x ) ≠ f ( 2 ) x→2
x→2
Hierdie spesiale sameloop van omstandighede impliseer dat die funksie f diskontinu is in die This special set of circumstances implies that f is discontinuous in the point where punt waar x = 2 . x = 2 . Vir duidelikheid toon ek hieronder 'n rekenaarvoorstelling van die funksie wat ons ondersoek het: For clarity I show a computer‐generated representation of the function we have investigated: 153 Leereenheid 5: Absolute waardes en limiete Let egter daarop dat die limiet van hierdie funksie ook soos volg verkry kan word: Note, however, that the limit of this function may also be obtained in the following way: 8 − x3
x→2 2 − x
lim
= lim
x→2
= lim
(2 − x ) ( 4 + 2 x + x2 )
2− x
( 2 − x ) ( 4 + 2 x + x2 ) x→2
(
( 2− x )
= lim 4 + 2 x + x 2
x→2
)
= 4 + 2(2) + 22
= 12
Hierdie tipe limiet, waar die noemer wat nul sou word tydens direkte vervanging verwyder kan word deur faktorisering en deling, word 'n verwyderbare diskontinuït genoem. 154 This type of limit, where the numerator which would become zero during direct substitution may be removed by factorization and division, is called a removable discontinuity. Leereenheid 5: Absolute waardes en limiete 0
Hulle word gekenmerk deur dat die vorm 0
ontstaan wanneer die waarde waarheen die onafhanklike veranderlike streef, direk in vervang word. 0
0
which appears when the value which the independent variable is approaching, is substituted directly into the function. They are characterized by the form Die uitdeelbewerking is toelaatbaar, aangesien die noemer streef na nul, maar nie nul word nie. Oefening 5.3 1. Ondersoek al die voorwaardes vir kontinuïteit en spesifiseer watter van die voorwaardes verbreek word by x = a vir elke funksie. 1.1 The division operation here is permissible, because the denominator only tends to zero – it does not actually "become" zero. Exercise 5.3 1. Investigate all the conditions for continuity and specify which of the conditions do not hold at x = a for each function: 155 Leereenheid 5: Absolute waardes en limiete 1.2 1.3 156 Leereenheid 5: Absolute waardes en limiete 2. Gebruik die voorwaardes van kontinuïteit om aan te toon dat die volgende funksie kontinu is by die t −3
, t = −3 gegewe punt: g ( t ) =
9t
2. Use the conditions for continuity and show that the following function is continuous in the specified point: t −3
g (t ) =
, 9t
157 Leereenheid 5: Absolute waardes en limiete 5.4 Berekening van sekere limiete/ Calculation of certain limits Direct methods for calculating limits boil down to avoiding a table method but attempting instead direct substitution. Direkte metode berekening van limiete kom daarop neer dat ons 'n tabelmetode vermy en eerder direkte vervanging probeer. When direct substitution fails we test whether or not we are dealing with one of the following special cases . Wanneer direkte vervanging nie werk nie, toets ons of ons een van die volgende twee spesiale soort limiete het. 5.4.1 Die spesiale geval 0
0
Ons het hierdie soort aan die einde van Leergedeelte 5.3 teëgekom. 5.4.1 The special case 0
0
We encountered this type at the end of Study Section 5.3. Voorbeeld/ Example p 2 − 25
p→−5 p + 5
Bereken/ Calculate lim
Oplossing/ Solution p2 − 25
lim
p→−5 p + 5
vervanging lewer / substitution yield
( −5 )2 − 25
−5 + 5
=
( p − 5) ( p + 5)
p→−5
( p + 5)
= lim ( p − 5 )
0
0
= lim
p→−5
= −5 − 5
= −10
5.4.2 Die spesiale geval ∞
∞
By hierdie soort deel ons in elke term deur die hoogste mag van die veranderlike wat voorkom. 5.4.2 The special case When dealing with this type, we divide each term by the highest power of the variable which occurs. Voorbeeld/ Example Bereken/ Calculate lim
t2
t →∞ t 2
158 ∞
∞
−1
Leereenheid 5: Absolute waardes en limiete Oplossing/ Solution t2
lim
−1
⎛ t2 ⎞
⎜⎜ 2 ⎟⎟
t
= lim ⎝ ⎠
2
t →∞ ⎛ t
1⎞
⎜⎜ 2 − 2 ⎟⎟
t ⎠
⎝t
t →∞ t 2
= lim
1
maar/ but
t →∞ ⎛
=
vervanging lewer / substitution yield
1⎞
⎜1 − 2 ⎟
t ⎠
⎝
1
tn
∞
∞
→ 0 as / if t → ∞ en / and n > 0 1
1
−
( 0)
=1
Oefening 5.4 Bereken die volgende direk (geen tabel) 1.
10 x 2
x →0 x
2.
x2 − 1
x→−1 x + 1
Exercise 5.4 Calculate the following directly (no table) lim
lim
159 Leereenheid 5: Absolute waardes en limiete t 2 − 5t + 6
t →3
t −3
3. lim
4.
64 + k 3
k →−4 4 + k
5.
10 x 2
x→∞ x
6.
3 x2 − 5 x + 7
x→∞ 2 x 2 − x
lim
lim
160 lim
Leereenheid 5: Absolute waardes en limiete 7.
lim
2t + 5
t →∞ 3t + 2
8.
m3 − 27
m→∞ m − 3
lim
161 Leereenheid 5: Absolute waardes en limiete 5.5 Die epsilon‐delta definisie van 'n limiet/ The epsilon‐delta definition of a limit Die behoefte bestaan om 'n limiet op so 'n wyse formeel te definieer, dat ons wiskundig kan bewys dat 'n bepaalde funksie f 'n limiet L besit wanneer x → a . In die definisie dui ε op 'n afstand vanaf die punt L op die y ‐as. In die definisie dui δ op 'n afstand vanaf die punt a op die x ‐as. The need arise to define a limit formally in such a way, that we can mathematically prove that a given function f has a limit L when x → a . In the definition ε refers to a distance from L on the y ‐axis. In the definition δ refers to a distance from a on the x ‐axis. Die definisie stel kortliks dat as ons 'n The definition states in brief that if we take a sekere interval ( L − ε ; L + ε ) op die y ‐as certain interval ( L − ε ; L + ε ) on the y ‐axis neem en dit in die funksie f op die x ‐as and we project it in the function f to the afbeeld en ons om die punt a op die x ‐as x ‐asis and we can get around the point a 'n interval ( a − δ ; a + δ ) kan kry sodat elke on the x ‐axis an interval ( a − δ ; a + δ ) such x in hierdie interval se funksiewaarde nader aan L lê as ε , dan geld dat that the function value of every x inside this interval lies closer to L than ε , then it holds f ( x ) → L as x → a . 162 that f ( x ) → L when x → a . Leereenheid 5: Absolute waardes en limiete Die bewering lim f ( x ) = L beteken dus The statement lim f ( x ) = L means that L dat L die limiet van die funksie f is is the limit of the function f if f ( x ) can indien f ( x ) willekeurig naby aan L be made arbitrarily close to L by choosing x close enough to a . x →a
x →a
gebring kan word deur x naby genoeg aan a te kies. Definisie van 'n limiet Definition of a limit Laat f ( x ) gedefinieer wees vir alle x in 'n Let f ( x ) be defined for all x in some open oop interval wat die getal a bevat, met die interval containing the number a , with the f ( x ) nie possible exception that f ( x ) may not be moontlike uitsluiting dat gedefineer is in die punt a nie. Ons skryf lim f ( x ) = L x →a
defined at a . as ons vir enige positiewe getal ε 'n bybehorende positiewe getal δ kan vind sodat f ( x ) − L < ε wanneer x voldoen We shall write lim f ( x ) = L x →a
if given any positive number ε we can find a positive number δ such that f ( x ) − L < ε if x satisfies 0 < x − a < δ . aan 0 < x − a < δ . 163 Leereenheid 5: Absolute waardes en limiete Voorbeeld/ Example Bewys dat/ Prove that lim ( 3 x − 5 ) = 1 x→2
Oplossing/ Solution Ons moet toon dat ons vir enige positiewe We must show that given any positive number ε we can find a positive number δ such that getal ε 'n positiewe getal δ kan vind sodat (3 x − 5) − 1 < ε
f ( x)
as / if
L
0< x−2 <δ
a
Maar ons kan dit herskryf as / But we may rewrite this as
3x − 6 < ε
as / if
0 < x −2 <δ
∴3 x − 2 < ε
as / if
0 < x −2 <δ
as / if
0 < x −2 <δ
∴ x −2 <
ε
3
As ons δ =
ε
kies word die regterkant van die bewering: / If we choose δ =
3
side the statement becomes :
0 < x −2 <
ε
ε
ε
3
then the right − hand
wat impliseer dat/ which implies that x − 2 < .
3
3
Dit is wat vereis word deur die linkerkant van die bewering/That is what is required for the left side
of the statement.
Dus / So :
164 lim ( 3 x − 5 ) = 1
x→2
Leereenheid 6: Inleiding tot funksie‐analise 6 Inleiding tot funksie‐analise/ Introduction to function analysis Voorkennis uit Leereenheid 3 wat Previously acquired knowledge vir hierdie leereenheid benodig from Study Unit 3 which is required word: for this study unit: U behoort reeds uit u skoolkennis of uit die From the knowledge you acquired at school hersiening wat ons in Leereenheid 3 gedoen or from the revision we did in Study Unit 3, het, die volgende te kan doen: you should already be able to do the 1. Die formele definisie van ‘n funksie as ‘n spesiale relasie kan toepas 2. Die definisie‐ en waardeversameling van ‘n funksie kan identifiseer 3. Die inverse van ‘n gegewe funksie kan bepaal following: 1. Apply the formal definition of a function as a special relation 2. Identify the domain and range of a function 3. Determine the inverse of a given 4. Bewerkings met funksies kan uitvoer function 4. Perform operations with functions Blaai gerus terug indien nodig. You are welcome to page back if needed. Leerdoelstellings vir hierdie leereenheid Na afhandeling van hierdie leereenheid moet die student in staat wees om die volgende te doen: Learning aims for this study unit Upon completion of this study unit the student must be able to do the following: 1. Calculate the derivative of a function by using the definition of a derivative 1. Die afgeleide van 'n funksie te bereken deur gebruik te maak van die definisie van 'n afgeleide in terme van die limiet van die gradiënt in terms of the limit of the slope of the secant through a curve when the distance between the points of intersection becomes infinitesimal: van die snylyn deur 'n kromme wanneer die afstand tussen die 165 Leereenheid 6: Inleiding tot funksie‐analise snypunte infinitesimaal word: f ( x + h) − f ( x)
dy
= lim
dx h→0
h
2. Te bepaal of 'n 2. Determine whether or not a function funksie differensieerbaar is in 'n punt of nie, deur van die gradiënt van die raaklyn aan 'n kromme gebruik te maak 3. Differensiasiereëls toe te pas om die afgeleide van sekere enkelvoudige funksies te bereken 4. Die afgeleide van saamgestelde funksies te bereken 5. Differensiasie toe te pas om 'n verskeidenheid funksies kan teken en ontleed 166 f ( x + h) − f ( x)
dy
= lim
dx h→0
h
is differentiable in a point or not, by using the slope of the tangent to a curve 3. Apply differentiation rules in order to calculate the derivative of certain simple functions 4. Calculate the derivative of composite functions 5. Apply differentiation in order to sketch and analyze the graphs a variety of functions Leereenheid 6: Inleiding tot funksie‐analise 6.1 Die afgeleide van 'n funksie uit eerste beginsels/ The derivative of a function from first principles Die gradiënt van die snylyn deur ‘n kromme as die gemiddelde The slope of the secant intersecting a curve as the average rate of change of a function veranderingstempo van ‘n funksie Gestel 'n funksie is soos volg gedefinieer: Suppose a function is defined as follows: f ( x ) = 3 x 2 + 45 x Beskou twee punte A en B op die kromme van die funksie sodat A by x = 3 en B by Consider two points A and B on the curve of the function such that A is at x = 3 and B is x = 6 geleë is. at x = 6 . 1. Bereken die koördinate van A en B. 1. Calculate the coordinates of A and B. 2. Bereken nou die gradiënt van die 2. Now calculate the slope of the secant snylyn AB. AB. Let op dat die gradiënt van die snylyn AB eintlik meet "hoe vinnig" die funksiewaarde Note that the slope of the secant AB actually measures "how fast" the function value 167 Leereenheid 6: Inleiding tot funksie‐analise op hierdie interval ( 3; 6 ) verander. changes over the interval ( 3; 6 ) Omdat die interval lank is, noem ons die Because the interval is long, we refer to the gradiënt van die snylyn die "gemiddelde slope of the secant as the "average rate of veranderingstempo van die funksiewaardes change of the function values with respect to met betrekking tot die onafhanklike the independent variable". veranderlike". That is what we mean by the symbol Dit is wat ons met die simboool m AB =
Δy
Δx
m AB =
Δy
. Δx
bedoel. Die gradiënt van die raaklyn aan ‘n kromme as die oombliklike The slope of the tangent to a curve as the instantaneous rate of change of a function veranderingstempo van ‘n funksie Gestel 'n funksie is soos volg gedefinieer: Suppose a function is defined as follows: f ( x ) = 3 x 2 + 45 x Beskou twee punte A en B op die kromme van die funksie sodat A by x = 3 en B by 'n the function such that A is at x = 3 and B is punt x = 3 + h regs van A geleë is. 1. Bereken die koördinate van A en B. 168 Consider two points A and B on the curve of located at a point x = 3 + h to the right of A. 1. Calculate the coordinates of A and B. Leereenheid 6: Inleiding tot funksie‐analise 2. Bereken nou die gradiënt van die 2. Now calculate the slope of the secant snylyn AB in terme van h . AB in terms of h . 3. Gestel nou ons maak die waarde 3. Now suppose we make the value of van h baie klein, sodat die punt B h very small, so that the point B geweldig naby aan die punt A kom. approaches the point A. Then the Dan sal die snylyn AB al hoe meer secant AB will start to resemble a soos 'n raaklyn aan die kromme by tangent to the curve at A. A begin lyk. m AB → mraaklyn by A as h → 0 m AB → mtan gent at A if h → 0 So we may formulate the movement of point Dus kan ons die naderskuif van B na A ten B to A in order to change the secant AB into einde die snylyn AB in 'n raaklyn te a tangent, as follows: verander, soos volg formuleer: mtan gent at A = lim m AB h →0
mraaklyn by A = lim m AB h →0
In terme van ons gewone notasie: That is, in terms of our usual notation: 169 Leereenheid 6: Inleiding tot funksie‐analise Δy
h →0 Δ x
y − yA
= lim B
h →0 x B − x A
mraaklyn by A = lim
= lim
h →0
f (3 + h ) − f (3)
3+ h −3
Let daarop dat u nou net hierbo vir f (3 + h ) − f (3)
3+h −3
Δy
h →0 Δ x
y − yA
= lim B
h →0 x B − x A
mtan gent at A = lim
= lim
h →0
in terme van h bereken f (3 + h ) − f (3)
3+h−3
Note that you have just calculated the value of f (3 + h ) − f ( 3)
3+h −3
in terms of h . So, all het. Al wat u nou hoef te toen om die you need to do now in order to obtain the gradiënt van die raaklyn by A te kry, is om slope of the tangent at A, is to let the value die waarde van h infinitesimaal (oneindig of h become infinitesimal (infinitely small): klein) te laat word: Die waarde wat u pas verkry het, gee die The value which you obtained gives the slope gradiënt van die raaklyn aan die kromme by of the tangent to the curve at the point A, die punt A, dit is waar x = 3 . that is where x = 3 . Aangesien die interval waaroor die gradiënt Because the interval over which we slope of van die raaklyn bereken is, dit is die grootte the tangent has been calculated, that is the van h , infinitesimaal is meet die gradiënt magnitude of h , is infinitesimal we should van die raaklyn dus die oombliklike note that the slope of the tangent measures veranderingstempo van die funksiewaarde the instantaneous rate of change of the met betrekking tot die onafhanklike function value with respect to the veranderlike. independent variable. Ons dui die oombliklike We indicate the instantaneous rate of veranderingstempo, dit is die gradiënt van change, that is the slope of the tangent, with die raaklyn, aan met die simbool 170 dy
en dx
the symbol dy
and we refer to it as the dx
Leereenheid 6: Inleiding tot funksie‐analise noem dit die afgeleide van die funksie. derivative of the function. Voorbeeld Bepaal die afgeleide van die funksie Example Calculate the derivative of the function f ( t ) = 4t 3 − 2t 2 uit eerste beginsels f ( t ) = 4t 3 − 2t 2 from first principles Oplossing (
)
Beskou punte A t; 4t 3 − 2t 2 en (
Solution Consider )
(
(
A t; 4t 3 − 2t 2
points )
and )
B t + h; 4 ( t + h ) − 2 ( t + h ) op die B t + h; 4 ( t + h ) − 2 ( t + h ) on the curve kromme van f . of f . 3
2
Die twee punte is dus : 3
2
So the two points are (
)
A t; 4t 3 − 2t 2 (
)
B t + h; 4t 3 + 12ht 2 + 12h2t + 4h3 − 2t 2 − 4ht − 2h2 Slope of secant AB
Δf
Δt
f −f
= 2 1
t2 − t1
=
=
4t 3 + 12ht 2 + 12h 2 t + 4h 3 − 2t 2 − 4ht − 2h 2 − ( 4t 3 − 2t 2 )
t +h−t
4t + 12ht + 12h t + 4h 3 − 2t 2 − 4ht − 2h 2 − 4t 3 + 2t 2
=
h
2
2
3
12ht + 12h t + 4h − 4ht − 2h 2
=
h
2
h (12t + 12ht + 4h 2 − 4t − 2h )
=
h
2
= 12t + 12ht + 4h 2 − 4t − 2h
Laat B na A beweeg deur h baie klein te maak/ Let B move to A by making h very small :
3
2
2
Slope of tangent at A = lim ( Slope of secant AB )
h →0
= lim (12t 2 + 12ht + 4h 2 − 4t − 2h )
h →0
= 12t 2 + 0 + 0 − 4t − 0
= 12t − 4t
2
171 Leereenheid 6: Inleiding tot funksie‐analise Let daarop dat sommige wiskundiges Note that some mathematicians prefer to verkies om die berekening vanuit eerste present the calculation of a derivative in a beginsels effens meer kompak te doen: more compact form: Voorbeeld Bepaal die afgeleide van die funksie Calculate the derivative of the function y = x uit eerste beginsels Oplossing Example y = x from first principles f ( x + h) − f ( x)
dy
= lim
dx h→0
h
x+h − x
= lim
h →0
h
⎛ x+h − x
⎞
= lim ⎜
×1⎟
h →0
h
⎝
⎠
⎛ x+h − x
x+h + x ⎞
= lim ⎜
×
⎟
h →0
h
x+h + x ⎠
⎝
x+h−x
= lim
h →0 h x + h +
x
h
= lim
h →0 h
x+h + x
1
= lim
h →0
x+h + x
1
=
x+0 + x
dy
1
∴ =
dx 2 ⋅ x
172 Solution Leereenheid 6: Inleiding tot funksie‐analise Oefening 6.1 Bereken die afgeleides van die volgende vanuit eerste beginsels: 1.
g (t ) =
Exercise 6.1 Calculate the derivatives of the following functions from first principles 3
t
173 Leereenheid 6: Inleiding tot funksie‐analise 2. s ( t ) =
174 2t − 3
3t + 2
Leereenheid 6: Inleiding tot funksie‐analise 3.
f ( x ) = 3 x2 − 3 x + 4 4.
f ( x ) = sin x wenk/hint: lim
cos h − 1
sin h
= 1 en/and lim
= 0 h→0 h
h →0
h
175 Leereenheid 6: Inleiding tot funksie‐analise cos h − 1
sin h
= 1 en/and lim
= 0 h→0 h
h →0
h
5.
f ( x ) = cos x wenk/hint: lim
6.
g ( p) = 2 p 176 Leereenheid 6: Inleiding tot funksie‐analise 6.2 Differensieerbaarheid/ Differentiability Uit die vorige Leergedeelte volg dat die konsep van 'n afgeleide onlosmaakbaar deel is van die konsep van 'n raaklyn. From the previous Study Section follows that the concept of a derivative is inseparably part of the concept of a tangent. Daarom kan ons differensieerbaarheid beskou as 'n eienskap wat beteken: Therefore, we may consider differentiability as a property which means: Indien 'n funksie differensieerbaar is op 'n oop interval, dan is dit by elke punt in daardie oop interval moontlik om 'n unieke raaklyn met 'n reëlwaardige gradiënt aan die funksie te trek. If a function is differentiable on an open interval, then it is at any point in that open interval possible to draw a unique tangent with real‐valued slope to the curve of the function. Dit kom in beginsel daarop neer dat 'n funksie differensieerbaar is waar dit kontinu en glad is. In principle this comes down to the fact that a function is differentiable where it is continuous and smooth. 'n Funksie is dus ondifferensieerbaar waar dit 'n skerp punt vertoon, diskontinu is of vertikaal loop. So, 'n function is non‐differentiable where it has a shart point, where it is discontinuous or where it runs vertically. Oefening 6.2 Exercise 6.2 1. Ondersoek al die voorwaardes vir differensieerbaarheid en spesifiseer watter van die voorwaardes verbreek word by elkeen van onderstaande gevalle. 1. Investigate all the conditions for differentiability and specify which of the conditions do not hold at x = a for each of the following cases: 1.1 177 Leereenheid 6: Inleiding tot funksie‐analise 1.2 1.3 178 Leereenheid 6: Inleiding tot funksie‐analise 179 Leereenheid 6: Inleiding tot funksie‐analise 6.3 Differensiasiereëls/ Differentiation rules In Leergedeelte 6.1 het ons beleef dat differensiasie vanuit eerste beginsels 'n omslagtige proses is. In Study Section 6.1 we experienced that differentiation from first principles is a long and tedious process. Gelukkig word gerieflike differensiasiereëls maklik afgelei deur die definisie van 'n afgeleide op 'n verskeidenheid funksies toe te pas. Fortunately, convenient differentiation rules may easily be derived by applying the definition of a derivative to a variety of functions. U sal sommige van hierdie differensiasiereëls nog self bewys; vir vandag aanvaar ons die volgende differensiasiereëls sonder bewys: You shall prove or derive some of these differentiation rules later in the course of your studies; for today we shall asume their validity without proof: Dx (constant) = 0 Konstantes/ Constants: Optelling/ Addition: Dx [f ( x ) + g ( x )] = Dx f ( x ) + Dx g ( x ) Aftrekking/ Subtraction: Dx [f ( x ) − g ( x )] = Dx f ( x ) − Dx g ( x ) Skalaarvermenigvuldiging/ Scalar multiplication: Dx [cf ( x )] = cDx f ( x ) Magsfunksie/ Power function : Dx x n = nx n −1 where n is a rational number. Produkreël/ Product rule: Dx [f ( x )g ( x )] = f ( x )Dx g ( x ) + g ( x )Dx f ( x ) Kwosiëntreël/ Quotient rule: Dx Kettingreël/ Chain rule: g ( x ) Dx f ( x ) − f ( x ) Dx g ( x )
f (x)
= , where g ( x ) ≠ 0 . g( x )
(g ( x ))2
dy dy dv du
=
, with y = f (v ) ; v = g (u ) , u = h( x ) dx dv du dx
Trigonometriese funksies/ Trigonometric functions: Dx sin x = cos x Dx csc x = − csc x cot x Dx cos x = − sin x D x sec x = sec x tan x Dx tan = sec 2 x Dx cot x = − csc 2 x Eksponensiële funksies/ Exponential functions: Dx e x = e x Dx a x = a x (ln a ) Dx loga x =
Dx e u = e u Dx u Logaritmiese funksies: Dx ln x =
1
x
Absolute waarde funksie: 180 1
1
(loga e ) Dx ln u = Dx u x
u
Dx x = x
x
Leereenheid 6: Inleiding tot funksie‐analise Laat ons nou ondersoek hoe hulle gebruik word. Oefening 6.3 Let us now investigate their use. Bereken die afgeleides in elkeen van die volgende gevalle deur van differensiasiereëls gebruik te maak. Exercise 6.2 Calculate the derivatives in each of the following cases by making use of differentiation rules: 1. Bepaal die gemiddelde gradiënt van f ( x ) = x 2 + 2 tussen x = 3 en x = 5/ Calculate the average slope of f ( x ) = x 2 + 2 between x = 3 and x = 5 2. As f ( x ) = 3 x 2 + 2 , bepaal die helling van raaklyn aan die kromme f by die punt x = 2 / If f ( x ) = 3 x 2 + 2 , determine the slope of the tangent to the curve f at the point x = 2 Differensieer met betrekking tot x:/ Differentiate with respect to x (a) f(x) = 4 x 3 + x x − 5
x4
181 Leereenheid 6: Inleiding tot funksie‐analise (b) f(x) = 4 4 x 2 − 6 x f(x) = (2x2 + 4x)100 (c) 182 Leereenheid 6: Inleiding tot funksie‐analise f(x) = (x2 + 6x)(x3 − 6x2) (d) (e) f(x) = x2 −1
x2 + 1
183 Leereenheid 6: Inleiding tot funksie‐analise (f) f(x) = x2 + 3
5
(g) f(x) = sin x cos x (h) f(x) = 184 sinx
x
Leereenheid 6: Inleiding tot funksie‐analise f(x) = x 2 e x (i) (m) f(x) = e x × ln x 185 Leereenheid 6: Inleiding tot funksie‐analise 6.4 Saamgestelde funksies en die kettingreël/ Composite functions and the chain rule Saamgestelde funksies Composite functions ‘n Saamgestelde funksie kan beskou word A composite function may be considered as a as ‘n funksie binne‐in ‘n ander funksie – function within another function – some soms word daar van “geneste funksies” writers refer to this phenomenon as “nested gepraat. Wanneer so ‘n funksie se waarde functions”. Whenever we evaluate such a bepaal word, werk ons van binne na buite. function, we proceed from the inside out. Voorbeeld 1 Example 1 Die volume van ‘n sferiese ballon word The volume of a spherical balloon is given by 4
gegee deur die formule V ( r ) = π r 3 . Die 3
4
the formula V ( r ) = π r 3 . The radius changes 3
radius van die ballon verander egter met with time according to the formula tyd volgens die formule r ( t ) = −t 2 + 6t met r ( t ) = −t 2 + 6t with 0 ≤ t ≤ 3 where the radius 0 ≤ t ≤ 3 waar radius in cm gemeet word en is measured in cm and the time in seconds. tyd in sekondes. We wish to calculate the volume at the Ons wil die volume bereken op die 3
oomblik wanneer t = . 2
3
instant when t = . 2
Now we may write the volume as Nou kan ons die volume skryf as 3
⎛
⎞
⎟
4 ⎜ 2
(V r )( t ) = V ( r ( t ) ) = π ⎜ −t + 6t ⎟ 3
⎜ r(t ) ⎟
⎝
⎠
V (r)
Ons kan die uitdrukking regs probeer We could attempt to simplify the expression vereenvoudig deur die hakies uit te right by multiplying out the parentheses until vermenigvuldige totat ons die volume as ‘n we obtain the volume as a polynomial veeltermfunksie V ( t ) het – daarna kan ons function V ( t ) – then we could simply 3
⎛3⎞
die waarde van t = in vervang om V ⎜ ⎟ 2
⎝2⎠
3
⎛3⎞
substitute t = in order to obtain V ⎜ ⎟ . 2
⎝2⎠
186 Leereenheid 6: Inleiding tot funksie‐analise te verkry. This approach, however, would generate a lot of tedious calculation. Dit sou egter ‘n groot klomp rekenwerk afgee. The theory of composite functions permits us to rather proceed as follows: Die teorie van saamgestelde funksies laat ons toe om eerder soos volg te werk te ⎛3⎞
Compute r ⎜ ⎟ and substitute the answer ⎝2⎠
gaan: 4
into V ( r ) = π r 3 ; then we also obtain the 3
⎛3⎞
Bereken r ⎜ ⎟ en vervang die antwoord in ⎝2⎠
3
value of V when t = . 2
4
die formule V ( r ) = π r 3 ; dan verkry ons 3
3
ook die waarde van V as t = . 2
Voltooi/ Complete: ⎛3⎞
r ⎜ ⎟ =………………………………………………… ⎝2⎠
=………………………………………………… 4
3
V (......) = π (.............. ) 3
=…………………………………………….. =……………………………………………… Wat is die definisieversameling van r ? What is the domain of r ? Wat is die waardeversameling van r ? What is the range of r ? Wat is die definisieversameling van V ? What is the domain of V ? Wat is die waardeversameling van V ? What is the range of V ? 187 Leereenheid 6: Inleiding tot funksie‐analise Voorbeeld 2 Example 2 Gegee: A( r ) = 4π r 2 Given: A( r ) = 4π r 2 r ( t ) = t 2 + 5t + 5 met 0 ≤ t ≤ 10 r ( t ) = t 2 + 5t + 5 with 0 ≤ t ≤ 10 Ons wil A bereken vir t = 6 . We wish to calculate A for t = 6 . Voltooi: Complete: 2
⎛
⎞
⎜
⎟
( A r )( t ) = .................... = 4π ⎜ .............................. ⎟ ⎜
⎟
r(t )
⎝
⎠
A( r )
r ( 6 ) =………………………………………………… =………………………………………………… = A (......) = 4π (..............) 3
=…………………………………………….. =……………………………………………… Wat is die definisieversameling van r ? What is the domain of r ? Wat is die waardeversameling van r ? What is the range of r ? Wat is die definisieversameling van A ? What is the domain of A ? Wat is die waardeversameling van A ? 188 What is the range of A ? Leereenheid 6: Inleiding tot funksie‐analise In die algemeen definieer ons ‘n In general we define a composite function as saamgestelde funksie soos volg: follows: As f en g funksies van x is, dan is die saamgestelde funksie (f
If f and g are functions of x , then the g )( x ) die composite function ( f
g )( x ) is the nested geneste funksie f ( g ( x ) ) vir alle x in function f ( g ( x ) ) for all x in the domain die definisieversameling van g sodat of g in order that g ( x ) forms the domain g ( x ) die definisieversameling van f of f . vorm. Skematies kan ons die voorbeeld oor die Schematically we can represent the example ballon soos volg voorstel: about the balloon as follows: By die kettingreël vir differensiasie wat u In the application of the chain rule for later vanjaar behandel, sal u in staat moet derivatives which you will study later this wees om ‘n saamgestelde funksie se year, you must be able to identify the inner binneste en buitenste deel te identifiseer. and outer constituents of a composite 189 Leereenheid 6: Inleiding tot funksie‐analise function. By die kettingreël vir differensiasie moet u In the application of the chain rule for in staat moet wees om ‘n saamgestelde derivatives you must be able to identify the funksie se binneste en buitenste deel te inner and outer constituents of a composite identifiseer. function. Gewoonlik word die simbool u of g vir die Usually the symbol u or g is used for the binneste funksie gebruik en die simbool v inner function and the symbol v or f is used of f vir die buitenste funksie. for the outer function. Voorbeeld 1 Example 1 As/ if f ( x ) = x 2 en/and g ( x ) = x + 2 dan is/ then: (f
g )( x ) = f ( g ( x ) ) = ⎡⎣ x + 2 ⎤⎦ en dit lewer/ this yields ( f g )( x ) = x + 2 2
En/and (g
f )( x ) = g ( f ( x ) ) =
( x ) + 2 en dit lewer/ this yields ( g
2
Voorbeeld 2 f
)( x ) =
x2 + 2 Example 2 Ontbind in die funksies/ Resolve into into the functions u ( x ) en/and v ( u ) as/ if 1
⎛
⎞
f ( x ) = cos ⎜ 2
⎟ ⎝ x − x +1⎠
Oplossing/Solution 1
⎛
⎞
f ( x ) = cos ⎜ 2
⎟ ⎝ x − x +1⎠
u( x)
v(u )
Dus/ So u ( x ) =
190 1
en/ and v ( u ) = cos u x − x +1
2
Leereenheid 6: Inleiding tot funksie‐analise Oefening 6.4 Exercise 6.4 1. As v ( t ) = 3 t en u ( t ) = sin t , bepaal 1. If die volgende: v (t ) = 3 t
and u ( t ) = sin t
, determine: 1.1 ( v u )( t ) 1.2 ( u v )( t ) 2. Gebruik die afgeleides, dit is kettingreël vir 2. Use the chain rule for derivatives, that dy dy du
=
⋅ waar dx du dx
is u = g ( x ) en bereken die volgende dy dy du
=
⋅
where u = g ( x ) and dx du dx
calculate the following derivatives. afgeleides. 2.1 f ( x ) = 3 x − 2 1
wenk/ hint a = a 2 191 Leereenheid 6: Inleiding tot funksie‐analise 2.2 f ( x ) =
(x
1
2
−1
)
2
wenk/ hint . (
)
5
2.3 f ( x ) = x 3 − x 2 + x + 1 . 192 1
= a −2 2
a
Leereenheid 6: Inleiding tot funksie‐analise 2.4 f ( x ) = 5tan x . 2.5 f ( x ) = cos2 x . 193 Leereenheid 6: Inleiding tot funksie‐analise 6.5 Toepassing van differensiasie/ Application of differentiation Analise is een van die kragtigste Analysis (the application of calculus to gereedskapstukke wat die mens nog functions) is one of the most powerful tools ontwikkel het. Alhoewel u nog net van ever developed by mankind. Although you differensiasie kennis dra, kan u alreeds 'n have only encountered differentiation calculus groot verskeidenheid werklikheidsgetroue at this stage, you are already able to situasies wat met veranderingstempo te mathematically handle a large variety of doen het, wiskundig hanteer. problems which involve rates of change. Binnekort sal u ook met integrasie kennis Soon you will also encounter integration. maak. The next exercise is meant to expose you to Die volgende oefening is bedoel om u aan the application of your existing knowledge. die toepassing van u bestaande kennis bloot te stel. Oefening 6.5 1.1 Exercise 6.5 Vraag 1/ Question 1 Die temperatuur van ‘n mengsel in ‘n fabriek verander volgens die funksie changes according to the function 1
5 375
vir 0 ≤ t ≤ 25 met T (t ) = − t2 + t +
4
2
4
1
5 375
for 0 ≤ t ≤ 25 T (t ) = − t2 + t +
4
2
4
T in °C en t in minute. met T in °C en t in minutes. Beskou twee punte A en B op die kromme van die funksie T : 194 The temperature of a mixture in a factory Consider two points A and B on the curve of the function T : 1
5 375 ⎞
1 2 1
1 2 5 5
375 ⎞
⎛
⎛
A ⎜ t ; − t 2 + t +
⎟ en/ and B ⎜ t + h; − t − th − h + t + h +
⎟ 4
2
4 ⎠
4
2
4
2 2
4 ⎠
⎝
⎝
Bereken die gradiënt van snylyn AB. Calculate the slope of secant AB. Leereenheid 6: Inleiding tot funksie‐analise 1.2 Bereken vanuit eerste beginsels die Calculate from first principles the gradiënt van die raaklyn aan die kromme gradient of the tangent to the curve at by die punt A. the point A. (wenk: laat A h → 0 in die antwoord wat (hint: Let h → 0 in the answer which you u op vraag 2.1 verkry het) obtained to question 2.1 verkry het) 1.3 1
5 375
op Differensieer T ( t ) = − t 2 + t +
4
2
4
1
5 375
Differentiate T ( t ) = − t 2 + t +
4
2
4
die kort manier. using the short method. 1.4 Bereken dT
dt
en maak ‘n afleiding t =5 minute
Calculate dT
dt
and draw a t =5 minutes
conclusion from your answer (what is the 195 Leereenheid 6: Inleiding tot funksie‐analise uit antwoord (wat impliseer u antwoord)? implication of your answer)? 1.5 Bereken die temperatuur op die oomblik Calculate the temperature when the rate wanneer die of temperature change is temperatuurveranderingstempo −7,5 °C/minute . −7,5 °C/minuut is. Vraag 2/ Question 2 196 Gegee: Given:
Leereenheid 6: Inleiding tot funksie‐analise f ( x ) = x 3 − 4 x 2 − 11x + 30 A en B is die draaipunte van die gegewe funksie. 2.1 Skets die grafieke van f ( x ) , f ' ( x ) en f '' ( x ) op dieselfde stel asse. Toon die A and B are the turning points of the given function. Sketch the graphs of f ( x ) , f ' ( x ) and f '' ( x ) on the same set of axes. Indicate koördinate van die draaipunte (A en B) the coordinates of the turning points (A asook die infleksiepun aan.. and B) and the point of inflection. 197 Leereenheid 6: Inleiding tot funksie‐analise 198 Leereenheid 6: Inleiding tot funksie‐analise 2.2 Wat word bedoel met die begrip What is the meaning of the concept "infleksiepunt"? "inflection point"? Wat is die gemiddelde veranderingstempo Calculate the average rate of change of van die funksie vanaf A na B? the function from A to B. 2.3 2.4 Bereken die vergelyking van die raaklyn aan die kromme by x = 1 . Calculate the equation of the tangent to the curve at x = 1 . 199 Leereenheid 6: Inleiding tot funksie‐analise Vraag 3/ Question 3 Gegee: Given: H ( x) = − x 3 + 5 x 2 + 8 x − 12 = ( x − 1)(− x 2 + 4 x + 12) 3.1 Bepaal die koördinate van die kritieke Find the coordinates of the critical points Apply the first derivative test and the punte. 3.2 200 Pas die eerste‐afgeleide‐toets en die tweede‐afgeleide‐toets toe om te toon second derivative test to show whether watter van die kritieke punte is lokale the critical points are local minima or minima of lokale maksima. maxima. Leereenheid 6: Inleiding tot funksie‐analise 3.3 Bepaal die koördinate van die punte waar die konkaafheid verander. Find the coordinate(s) of the points where the concavity changes. 3.4 Pas die tweede‐afgeleide‐toets toe om te toon dat die konkaafheid verander het. Apply the second derivative test to show whether the concavity did change Vraag 4/ Question 4
Gegee: g(x ) =
Given: 2x + 6
− 6x + 3
201 Leereenheid 6: Inleiding tot funksie‐analise 4.1 Bepaal die vertikale en horisontale asimptote deur van limiete gebruik te Determine the vertical and horizontal asymptotes by utilizing asymptotes. maak. 4.2 Bepaal al die afsnitte met die asse. Determine all intercepts with axes. Skets die grafiek van die funksie. Toon alle Sketch the graph of the function. Indicate 4.3 202 inligting aan wat u bepaal het in 4.1 en all the information that you have 4.2. determined in 4.1 and 4.2. Leereenheid 6: Inleiding tot funksie‐analise Vraag 5/ Question 5 ‘n Oop kartonkrat word vervaardig deur An open box is manufactured by vier identiese vierkante uit die hoeke van removing four identical square sections ‘n vel karton wat 24 cm lank en 18 cm from the corners of a sheet of cardboard breed is, te sny en die oorblywende which is 24 cm long and 18 cm wide gedeeltes na bo te vou om die wande van and then folding the remaining parts die krat te vorm. upwards to form the sides of the container. 5.1 Toon aan dat die volume van die kartonkrat gegee word deur die Show that the volume of the cardboard box is given by the equation V = 4 x3 − 84 x 2 + 432 x for x ∈ [0; 9] . vergelyking V = 4 x3 − 84 x 2 + 432 x vir x ∈ [0; 9] . 5.2 Skets ‘n netjiese grafiek van die funksie V Sketch a neat graph of the function V teenoor x vir −2 ≤ x ≤ 14 . against x for −2 ≤ x ≤ 14 . Toon alle snypunte met asse, asook Clearly indicate all intercepts with axes, draaipunte en infleksiepunte, duidelik aan as well as turning points and points of in koördinaatvorm. inflection, in co‐ordinate form. 203 Leereenheid 6: Inleiding tot funksie‐analise 5.3 204 Wat moet die oppervlakte van elke What should the area of each square be vierkant wees sodat die volume van die in order for the volume of the box to be a krat ‘n maksimum sal wees en wat is die maximum and what is the value of this waarde van hierdie maksimum volume? maximum volume? Leereenheid 6: Inleiding tot funksie‐analise Vraag 6/ Question 6 ‘n Regte kegel word gegenereer deur die A right cone is generated by revolving the deel van die parabool y = 4 − x2 tussen part of the parabola y = 4 − x2 between die punte x = 0 en x = 2 om die Y‐as te the points x = 0 and x = 2 around the Y‐
roteer en die kegel in die paraboloïed in te axis and inscribing the cone inside the skryf sodat die tophoek op die oorsprong paraboloid so that its vertex is located at staan en die basis by die punt P aan die the origin and its base touches the parabool raak: parabola at the point P : 6.1 Toon aan dat die volume van die kegel gegee word deur die funksie V ( x) =
4 2 1 4
πx − πx . 3
3
Show that the volume of the cone is given by the function V ( x ) =
4 2 1 4
πx − πx 3
3
205 Leereenheid 6: Inleiding tot funksie‐analise 6.2 206 Skets die kromme van V teenoor x Sketch the curve of V against x neatly netjies op die grafiekpapier wat voorsien is on the supplied graph paper or in your of in u antwoordskrif. answering script. Neem alle beperkings op die Take into account all constraints on the definisieversameling in ag. domain. Toon alle wortels, draaipunte en Indicate all roots, turning points and infleksiepunte binne die beperkte inflection points within the constrainded definisieversameling. domain. Leereenheid 6: Inleiding tot funksie‐analise 6.3 Maak van u berekeninge en resultate uit Employ your calculations and results from Vraag 5.1.2 gebruik en bereken die Question 5.1.2 and calculate the maksimum moontlike volume wat die maximum possible volume which the kegel kan besit. cone can have. Neem gerus aan dat die eenhede op die You are welcome to assume that all units asse cm is. on the axes are cm. 207 208 
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