Teaching infinitesimal calculus in high school with infinitesimals

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1 INSTITUT FOR NATURFAGENES DIDAKTIK KØBENHAVNS UNIVERSITET Teching infinitesiml clculus in high school with infinitesimls Mikkel Mthis Lindhl og Jons Kyhnæ Kndidtspecile Septemer 216 IND s studenterserie nr. 53

2 INSTITUT FOR NATURFAGENES DIDAKTIK, Alle puliktioner fr IND er tilgængelige vi hjemmesiden. IND s studenterserie 43. Lene Eriksen: Studie og forskningsforlø om modellering med vrielsmmenhænge (215) 44. Croline Sofie Poulsen: Bsic Alger in the trnsition from lower secondry school to high school 45. Rsmus Olsen Svensson: Komprtiv undersøgelse f deduktiv og induktiv mtemtikundervisning 46. Leonor Simony: Teching uthentic cutting-edge science to high school students(216) 47. Lotte Nørtoft: The Trigonometric Functions - The trnsition from geometric tools to functions (216) 48. Aske Henriksen: Pttern Anlysis s Entrnce to Algeric Proof Situtions t C-level (216) 49. Mri Hørlyk Møller Kongshvn: Gymnsieelevers og Lærerstuderendes Viden Om Rtionle Tl (216) 5. Anne Kthrine Wellendorf Knudsen nd Line Steckhhn Sørensen: The Themes of Trigonometry nd Power Functions in Reltion to the CAS Tool GeoGer (216) 51. Cmill Mrgrethe Mttson: A Study on Techer Knowledge Employing Hypotheticl Techer Tsks - Bsed on the Principles of the Anthropologicl Theory of Didctics (216) 52. Tnj Rosenerg Nielsen: Logicl spects of equtions nd eqution solving - Upper secondry school students prctices with equtions (216) 53. Mikkel Mthis Lindhl nd Jons Kyhnæ: Teching infinitesiml clculus in high school - with infinitesimls (216) Se tideligere serier på:

3 Astrct In high school students hve their first encounter with the notion of infinitesiml clculus. A common prolem encountered when introducing this suject is for the students to understnd the theory ehind the techniques they develop when eing tught this. In this thesis the use of infinitesimls will e reinstted to tech this prticulr theme. They were used to tech infinitesiml clculus even fter their olishment, when the rel numers were constructed. With the dwn of the nonstndrd nlysis, through the construction of the hyperrel numers, the infinitesimls were reinstlled s mthemticl sound quntity. This thesis will give resons for or ginst the use of infinitesimls in the form of nonstndrd nlysis y employing the nthropologicl theory of didctics. The thesis will give short introduction of how the hyperrel numers re constructed nd how the definition of the differentil quotient nd integrl is equivlent sttements to the definitions usully encountered in high school. A teching course ws done in order to survey if the didcticl resons suggested y the theories re noticele when introducing infinitesiml clculus, with infinitesimls, in high school. As such textook mteril hd to e developed in order for the students to prepre for the exm(s). This textook mteril is void of the construction of the hyperrel numers, s introduced in the thesis, ut hinges on n intuitive construction of infinitesiml quntities to generte them. An nlysis of the teching employing the nthropologicl theory of didctics ws done in order to determine if the merits of the infinitesimls re on pr with the theoreticl resons. The thesis concludes tht teching sed on infinitesimls re relistic nd helps engge the students y letting their intuition guide them to gret degree.. IND s studenterserie estår f kndidtspeciler og chelorprojekter skrevet ved eller i tilknytning til Institut for Nturfgenes Didktik. Disse drejer sig ofte om uddnnelsesfglige prolemstillinger, der hr interesse også uden for universitetets mure. De puliceres derfor i elektronisk form, nturligvis under forudsætning f smtykke fr forftterne. Det er tle om studenterrejder, og ikke endelige forskningspuliktioner. Se hele serien på:

4 Teching infinitesiml clculus in high school with infinitesimls Mster s Thesis Mikkel Mthis Lindhl nd Jons Kyhnæ Supervisor: Crl Winsløw Deprtment of Science Eduction Dte of sumission:

5 Nmes: Title: Deprtment: Supervisor: Mikkel Mthis Lindhl & Jons Kyhnæ Ntsværmer Deprtment of Science Eduction Crl Winsløw Time frme: Ferury August, 216

6 Astrct In high school students hve their first encounter with the notion of infinitesiml clculus. A common prolem encountered when introducing this suject is for the students to understnd the theory ehind the techniques they develop when eing tught this. In this thesis the use of infinitesimls will e reinstted to tech this prticulr theme. They were used to tech infinitesiml clculus even fter their olishment, when the rel numers were constructed. With the dwn of the nonstndrd nlysis, through the construction of the hyperrel numers, the infinitesimls were reinstlled s mthemticl sound quntity. This thesis will give resons for or ginst the use of infinitesimls in the form of nonstndrd nlysis y employing the nthropologicl theory of didctics. The thesis will give short introduction of how the hyperrel numers re constructed nd how the definition of the differentil quotient nd integrl is equivlent sttements to the definitions usully encountered in high school. A teching course ws done in order to survey if the didcticl resons suggested y the theories re noticele when introducing infinitesiml clculus, with infinitesimls, in high school. As such textook mteril hd to e developed in order for the students to prepre for the exm(s). This textook mteril is void of the construction of the hyperrel numers, s introduced in the thesis, ut hinges on n intuitive construction of infinitesiml quntities to generte them. An nlysis of the teching employing the nthropologicl theory of didctics ws done in order to determine if the merits of the infinitesimls re on pr with the theoreticl resons. The thesis concludes tht teching sed on infinitesimls re relistic nd helps engge the students y letting their intuition guide them to gret degree. Pge i of v

7 Acknowledgements A gret ig thnks goes to Crl Winsløw for his help in shping the thesis. The school, Roskilde Ktedrlskole, deserves lot of credit for letting the study tke plce. An infinitely mount of grtitude nd respect goes to the wonderfully sweet nd wesome 1u, who took opened their rms wide nd mde the study ll the more fun; etter clss does not exist. Thnks for the constnt surprises nd for tolerting ll the shennigns. The iggest of thnks to fmily nd girlfriend for eing so ptient nd understnding. Thnks for the tremendous mount of support. The girls who st in the other cuicle in the office for out 4 months, Anne nd Line, thnk you. To the new office, prticulrly Nine, thnks for mking the finl month or so erle; from the rekfst tle to the middy pokéhunt, ll the wy to the fterhours eer. 1 thnks to one nother. How on Erth this thesis would hve come to e without the lucrtive crtlks, the mutul respect, The Trilogy, nd the fntstic friendship tht cme with it ll, is eyond infinity. The lte nights pid off, let there e more of them. Pge ii of v

8 Contents 1 Introduction Motivtion Prolemtique Reserch questions Theory Mthemticl theory NSA Rel numers The hyperrel numers Internl sets nd functions, nd the connection etween R nd R Anlysis using the hyperrel numers Didcticl theory Anthropologicl Theory of Didctics Wht to e tught Didcticl resons for or ginst using NSA Concrete implementtion of the knowledge to e tught Presenttion of oservtions Context Overll plnning Oservtionl methodology Method of nlysis Hyperrel numers Plnned teching Anlysis of teching the hyperrel numers Differentil clculus (Jons Kyhnæ) Plnned teching Anlysis of teching differentil clculus Integrl clculus Plnned teching Anlysis of teching integrl clculus Oservtions of the teching process Conclusion...73 Biliogrphy...75 Appendix Description of teching Description of teching hyperrel numers First lesson (3.3) Second lesson (4.5) Third lesson (converted #1) (4.6) Fourth lesson (4.8) Hnd in # Fifth lesson (converted #2) (4.13) Description of teching differentil clculus (Jons Kyhnæ)...89 Pge iii of v

9 1.2.1 First lesson (4.8) *Second nd third lesson (4.11) Fourth lesson (4.13) Fifth lesson (4.19) *Sixth lesson (4.2) *Seventh lesson (converted #3) (4.21) Hnd in # Eighth lesson nd ninth lesson (converted #4) (4.25) *Tenth, eleventh nd twelfth lesson (4.26 nd 4.27 nd 5.3) *Thirteenth lesson (5.3) Hnd in # Description of teching integrl clculus (Mikkel Mthis Lindhl) First nd second lesson (5.9) Third lesson (5.1) Fourth lesson (5.11) Fifth lesson (converted #5) (5.13) Hnd in #4 pge Sixth lesson (5.18) Seventh lesson (5.2) Eighth lesson (5.2) Hnd in #4 pge Ninth (5.23) Tenth lesson (5.23) Eleventh (nd twelfth) lesson Anlysis tles Hyperrel nlysis tle Differentil clculus nlysis tle (Jons Kyhnæ) Integrl clculus nlysis tle (Mikkel Mthis Lindhl) Compendium(s)... Seprte ppendix Pge iv of v

10 Acronyms NSA ATD DO MO HMO DMO IMO Nonstndrd nlysis Anthropologicl theory of didctics Didcticl orgniztion Mthemticl orgniztion Hyperrel mthemticl orgniztion Differentil mthemticl orgniztion Integrl mthemticl orgniztion R + The set of strictly negtive rel numers, i.e. R + = { R > } R The set of strictly positive rel numers, i.e. R = { R < } Q + The set of strictly positive rtionl numers, i.e. Q + = { R > } Q The set of strictly negtive rtionl numers, i.e. Q = { R < } R The set of hyperrel numers I The set of ll infinitesimls Pge v of v

11 1 Introduction Going down memory lne to wht mthemtics ws lerned in high school cn produce hedches for most people. If sking former student of high school wht the hrdest prt in high school mthemtics ws, the common nswer is infinitesiml clculus. By prodding with few more questions s to why it ws this prt which ws so difficult, nother common nswer is tht it ws strct, i.e. non-intuitive. As such n interest of imuing infinitesiml clculus with intuition cme to e. The interest ws further piqued when discovering tht this interest ws shred with (nother) gret mthemticin, Leiniz. Leiniz differentil clculus seemed so much more intuitive thn the one rememered from high school, ut he used infinitesimls which did not seem s rigorous mthemticl oject. The project of imuing infinitesiml clculus with intuition ws set on hold ut ws resumed when discovering tht the infinitesimls were reinstted s rigorous mthemticl oject y Roinson in 1966(Roinson, 1996). In the following section the motivtion for the thesis is elorted. 1.1 Motivtion When reding the texts Didctic restrictions on techers prctice the cse of limits of functions in Spnish high schools (BrÉ, Bosch, Espinoz, & GscÓN, 25) nd Mthemticl nlysis in high school A fundmentl dilemm (Winsløw, 213), it ecomes cler tht teching nlysis in high school contins some dilemms, one of which is the use of the limit opertion. With this in mind, it ecme very interesting to try to circumvent the use of these limits, which cn e done using nonstndrd nlysis. Nonstndrd nlysis (NSA) is n pproch to mthemticl nlysis, s introduced y Roinson in 1966 (Roinson, 1996), where infinitely smll nd lrge quntities re well-defined, through the use of the xiom of choice. I.e. nonstndrd nlysis opertes with different set of numers, clled the hyperrel numers, s its se for mthemticl nlysis. The set of hyperrel numers contins oth the rel numers nd the infinitely smll, clled infinitesimls, nd the infinitely lrge quntities. Roinson s cretion of the hyperrel numers shows they re consistent if nd only if the rel numers re. The conundrum of proofs sed on infinitesimls nd infinite numers eing sound or not, ws finlly given mthemticlly stisfctory nswer. With this in mind, the historicl development of the infinitesiml clculus hs een scrutinized, to see if some of the proofs done y the gret mthemticins in history could e sved. The discoveries, in this field of mthemticl history, found few theorems which proofs were sved y the rigorous foundtion of the infinitesimls. The most frequent reson for this ws tht they lcked rigorous definitions. In 1976 H. Jerome Keisler pulished textook clled Elementry clculus: An Infinitesiml Approch (Keisler, 212). This textook contins, s the title suggests, n pproch to elementry clculus, where infinitesimls re used. When reding vrious texts out NSA, the reson for using this pproch cn more or less e oiled down to one primry reson: it is more intuitive wy to introduce infinitesiml clculus, thn the (stndrd) limit opertion pproch. Although this sttement hs een questioned y rther ig popultion of mthemticins over the yers s seen in ( Criticism of non-stndrd nlysis, 215), no solid rgument cn sy tht it is not more intuitive. On the contrry s seen in (O Donovn & Kimer, 26), the fct tht the development of the infinitesiml clculus, s done y Leiniz nd Newton in , ws sed on the use of infinitesimls supports the suggestion of infinitesiml clculus eing more intuitive, when introduced y infinitesimls. One could sk why infinitesimls disppered from the mthemticl discipline, when it mkes nlysis more intuitive; the nswer to this question is found in the historicl development of the notion of quntity. In the introduction to Cuchy s Cours d Anlyse (Brdley Pge 1 of 152

12 & Sndifer, 29) from 1821 cn e found generl description, of wht quntities re. This is one of the first plces, where the understnding of quntity is descried in greter depth, thn wht Euclid did lmost 2 yers in ook five of (Heierg, Fitzptrick, & Euclid, 28). The notion of quntity ws scrutinized y vrious mthemticins Weierstrss, Dedekind nd cntor etc. (Gry, 215) in the nineteenth century, ending with the construction of the rel numers round Regrdless of wht construction of the rel numers ws used, the construction mde the notion of infinitesimls into something, which no longer mde sense to use in the discipline of mthemtics. This led to the introduction of epsilon delt rgumenttion nd limit opertion, which ws then to del with the prolems tht previously hd een solved using infinitesimls. Even with the construction of the rel numers, the notion of infinitesimls persisted in the school system, for longer time thn most would elieve, which cn e seen in The tension etween intuitive infinitesimls nd forml mthemticl nlysis (Ktz & Tll, 211) where one of the introductory pssges is: Infinitesiml clculus is ded metphor. In countless courses of instruction round the gloe, students register for courses in infinitesiml clculus only to find themselves eing trined to perform epsilontic multiple-quntifier logicl stunts, or else eing told riefly out the rigorous pproch to limits, promptly followed y instructions not to worry out it This illustrtes quite well the irony which cn pper in mthemtics when mthemticins stick to the nmes from the dys of yore while using the rigorous stndrd pproch to nlysis. The lst prt of the conclusion from the sme text is: Most modern mthemticins now dmit the xiom of choice, in the knowledge tht it offers theoreticl power without introducing contrdictions tht did not exist efore. Is it not time to llow infinitesiml conceptions to e cknowledged in their rightful plce, oth in our fertile mthemticl imgintion nd in the power of forml mthemtics, enriched y the xiom of choice? Another pssge tht motivted the use of NSA is in Richrd O Donovn nd John Kimer s Non stndrd nlysis t pre-university level nive mgnitude nlysis (O Donovn & Kimer, 26) There is heted dete whether non stndrd nlysis should e introduced t preuniversity level. It hs een demonstrted herein tht it is possile, emphsising tht in mthemtics, simplicity rhymes with euty With the interest in using NSA lredy piqued, the resons listed ove further motivted the pursuit of NSA pproch to infinitesiml clculus in high school. On side note, it is clled infinitesiml clculus nd not ε-δ rgument clculus or limit-opertion clculus, so why not use the quntities which gve rise to the nme? Pge 2 of 152

13 1.2 Prolemtique In order to tech infinitesiml clculus with NSA in Dnish high school (from here on out, high school is ssumed Dnish), some kind of textook mteril in Dnish is required y the Ministry of Eduction. There is no Dnish textook using NSA tht rigorously descries this, which poses prolem when teching infinitesiml clculus in high school, for the first time. Since infinitesiml clculus using NSA hs not een tught in high school in long time, with the exception of school in Genev (O Donovn & Kimer, 26), there exists little up-to-dte previously used teching mteril. Since the mteril found in Nonstndrd nlysis t pre-university level: Nive mgnitude nlysis is not s rigorous s wned it left two options: Either producing teching mteril from the knowledge otined y studying the works of Roinson (Roinson, 1996), Stroyn (Stroyn, 1997) or Ponstein (Ponstein, 21) etc. or using trnsposition of the teching mteril lredy ville for universities e.g. elementry clculus: n infinitesiml pproch (Keisler, 212). Acquiring students to tech posed less of prolem thn the ftermth of the teching process, in tht the originl techer (the techer who tught the clss efore the course on infinitesiml clculus) needed to otin knowledge of NSA s would n externl censor for possile orl exm. Pge 3 of 152

14 1.3 Reserch questions From the prolemtique the following reserch questions were developed. 1. Wht re the resons for or ginst the nonstndrd pproch, to infinitesiml clculus in high school? 2. How cn nonstndrd introduction to infinitesiml clculus in high school e developed, in prticulr how to crete textook mteril suited for high school students? 3. Wht results cn e oserved from first experiment, implementing such mteril? The first reserch question is to e nswered y employing didcticl theory in order to generte well founded resons for or ginst the nonstndrd pproch. As such the resons lso depend on the didcticl literture which cn e found on the suject. The second reserch question is to e nswered y composing textook mteril for the students to e tught. The third reserch question is to e nswered y didcticl nlysis of the course of teching infinitesiml clculus. Since NSA is rther uncommon pproch to nlysis, the first thing to do ws to otin sufficient understnding of the NSA, which is presented in section 2.1, in order to oth tech nd nlyze using didcticl theory. The choice of nthropologicl theory of didctics (ATD) s the didcticl theory ws sed on previous encounters with the theory. A short introduction to this theory cn e found in section 2.2. With NSA nd ATD in plce, the first reserch question cn e nswered in theory see section With the hypothesis from section 2.2.3, n nswer to the second reserch question cn e found in section 3. In section 4 presenttion of the oserved teching course cn e found which genertes n nswer for the third reserch question. The concluding remrks regrding the reserch questions nd the study in generl will e presented in section 5. The individully written sections re 4.3 Differentil clculus nd 4.4 Integrl clculus. Section 4.3 ws written y Jons Kyhnæ nd Section 4.4 ws written y Mikkel Mthis Lindhl. Pge 4 of 152

15 2 Theory This section includes oth the mthemticl nd didcticl theory used in the thesis. 2.1 Mthemticl theory NSA This introduction to the Hyperrel numers is hevily influenced y the chpter written y Tom Lindström clled n Invittion to nonstndrd nlysis in the ook Nonstndrd nlysis nd its pplictions (Cutlnd, 1988). An intuitive introduction cn e found in ppendix 3 or in the Elementry clculus: n infinitesiml pproch (Keisler, 212). The hyperrel line is line of numers in the sme wy s the rel line, it contins more numers thn the rel line though. Clerly it cnnot ide y the sme xioms s the rel line, in this cse it does not ide y the Archimeden principle, i.e. it contins elements, nd where < such tht no finite numer n mkes n >. Since the hyperrel numers re going to e constructed in wy which is similr to one of the wys the rel numers cn e constructed, short description of how to construct of the rel numers will pper s the first thing in this introduction Rel numers The rel numers cn e constructed y the use of Cuchy sequences. Let the set Q C e the set of ll the Cuchy sequences of rtionl numers. Let N e the set of ll null-sequences (the sequences tht tend to zero) Definition Equivlence reltion for constructing the rel numers, Define equivlence reltion on Q C in the following wy: Let { n }, { n } Q C then { n } { n } if the term wise difference{( n n )} N. The rel numers cn then e defined s the Cuchy-sequences with rtionl numers under the ove defined equivlence reltion. R Q C / The dditive nd multiplictive structure of R is defined through term wise ddition nd multipliction, i.e. if n nd n re the equivlence clsses of { n }, { n } Q C then n + n = n + n nd n n = n n. The order reltion is defined s; n < n if nd only if, for some ε Q + nd sufficiently lrge N, n < n ε for ll n N. The rtionl numers re seen s suset of the rel numers through the inclusion mp φ: Q R which is defined s tking rtionl element to φ() =,,. Pge 5 of 152

16 2.1.2 The hyperrel numers This section will present construction of the hyperrel numers. To define the hyperrel numers the sme procedure, s when construction the rel numers, will e used ut with different strting set nd different equivlence reltion. Let R e the set of ll sequences of rel numers, then in order to introduce the hyperrel numers, specific suset of the null-sequences in R is needed to define the equivlence reltion. When constructing the rel numers corse equivlence reltion is used, since there is no distinguishing etween two sequences tht only differs y sequence tending to zero. For exmple the sequences { 1 n } n N nd { 1 n 2} n N oth correspond to the rel numer zero. When introducing the hyperrel numers n equivlence reltion tht mkes difference etween the two sequences is needed. If the finest equivlence reltion is used, s in every sequence of rel numers defines hyperrel numer, then the constructed set, will include zero-divisors, since {,1,,1,, } {1,,1,,1, } = {,,, } =, where oth the sequences in the product re nonzero. This is unwnted, nd therefore the finest equivlence reltion ~ such tht R/~ hs no zero divisors is the preferle one. To define such n equivlence reltion ~, free ultrfilter is needed. In order to explin wht this is consider the definition of filter nd its specific clssifiction of free ultrfilter seen elow Definition Filter A filter over set I is collection of susets F of I such tht F nd If A, B F, then A B F. If A F nd A B, then B F. A free filter is filter F of n infinite set I where the set I/A is finite for ll A F, i.e. none of the elements A F re finite. An ultrfilter U of I is filter where, for ll A I, either A U or A C U, ut not oth. Therey free ultrfilter is filter U of set I where oth the ove sttements re true. Thus if U is free ultrfilter then every element A U is suset A I (U is filter), the set I/A is finite for ll A U, even though I is n infinite set (U is free filter), nd for ll susets A I either A U or A C U ut not oth (U is n ultrfilter) Proposition Extending free filter to free ultrfilter Any free filter F over I cn e extended to free ultrfilter, when I is n infinite set. Proof. Let J e the set of free filters over I, hence F J. Order J y inclusion, then y Zorn s Lemm there exists mximl element in J, denote it U. If E I then either F E or F E C is infinite for ll F U. If not then there exists F 1, F 2 U such tht oth F 1 E nd F 2 E C re finite. Assume for contrdiction the ltter, nd consider tht F 1 E C (F 1 E) nd F 2 E (F 2 E C ), nd F 1 F 2 (E C (F 1 E)) (E (F 2 E C )) = (F 1 E) (F 2 E C ). Pge 6 of 152

17 Since the union of two finite sets is finite, then F 1 F 2 U is finite which contrdicts the free prt. If F E is infinite for ll F U, then E U, otherwise U is not n mximl element, the sme rgument cn e used if F E C infinite for ll F U, mking either E or E C n element of U ut not oth, hence U is n ultrfilter. To mke n ultrfilter on the nturl numers N, tke the Fréchet filter F (which is free filter)on N Definition Fréchet filter, F Denote the Fréchet filter F nd define it s. F = {A N A C finite} nd extend F with proposition to free ultrfilter M Definition The finitely dditive mesure, m Let m e the finitely dditive mesure on N defined y, m(a) = { 1 if A M otherwise. Now since M then it is cler tht N M, hence m(n) = Definition Equivlence reltion to construct the hyperrel numers, ~ Let ~ e the equivlence reltion on the set of ll rel vlued sequences R defined y The hyperrel numers cn then e defined s Definition The hyperrel numers, R Define R s { n }~{ n } if nd only if m({n n = n }) = 1. R R/~. The dditive nd multiplictive structure of R is defined through component wise ddition nd multipliction, s when defining them for the rel numers. I.e. if n nd n re the equivlence clsses of { n }, { n } R then n + n = n + n nd n n = n n. The order reltion is defined s; n < n if nd only if m({n n < n }) = 1. If nother representtive { n } n N for the equivlence clss n is used then m({n n = n }) = 1, hence m({n n < n }) = m({n n < n } {n n = n }) = 1, since oth {n n < n } nd {n n = n } re in M. Thus the order reltion is independent of the choice of representtives of the equivlence clsses. In order to check tht there re no zero divisors in R, ssume tht Pge 7 of 152

18 n n =,,, i.e. m({n n n = }) = 1, hence m({n n n = }) = m({n n = } {n n = }) = m({n n = }) + m({n n = }) = 1. Then either n =,, or n =,,. The hyper rel numers R re then n ordered field with multiplictive nd dditive identity, 1 = 1,1, nd =,, respectively Exmple Use of order reltion, < Let,, c R such tht > nd < c. How to prove < c? If = n, = n nd c = c n, from the definition of the order reltion, then m({n < n }) = 1 nd m({n n < c n }) = 1, hence m({n n n < n c n }) = m({n < n } {n n < c n }) = 1. Proving tht < c s wnted. As n nlogue to the mp φ: Q R, there is n injective order preserving mp ψ: R R, tking rel numer,, to sequence of the sme numer,,. Thus the rel numers form suset of the hyperrel numers, R R. To identify the numers, which re not rel numers, consider the next definition s wy of descriing different kinds of hyperrel numers Definition Infinitesiml, finite nd infinite 1. A hyperrel numer β R is n infinitesiml if < β < for ll R A hyperrel numer β R is finite if < β < for some R A hyperrel numer β R is infinite if no R + exists such tht < β < Exmples Opertions with infinitesimls nd infinite numers Descrie the hyperrel numers, δ 1, δ 2, Ω 1 nd Ω 2 s shown elow: δ 1 = 1 n δ 2 = δ 1 2 = 1 n 2 Ω 1 = n Ω 2 = Ω 1 = n For δ 1 nd δ 2 every positive rel numer R will mke m({n < 1 < }) = 1 nd n m({n < 1 n 2 < }) = 1, hence oth δ 1 nd δ 2 re infinitesimls. It is cler tht is the only rel infinitesiml. For Ω 1 nd Ω 2 then every rel numer R will mke {n N < n < } nd {n N < n < } finite sets, mking Ω 1 nd Ω 2 two infinite numers. To understnd the rithmetic on the infinitesimls consider first the infinitesimls, δ nd δ, then the sum of these should gin e n infinitesiml, i.e. R + < δ + δ <. Pge 8 of 152

19 Since δ nd δ re oth infinitesiml then for ll, R +. < δ < nd < δ <, For every R + let = = /2 then the sum; δ + δ hs the property thus < δ + δ < + < δ + δ <. Consider the product of finite numer β nd n infinitesiml δ, β δ. Let β R + e such tht β < β < β then for ll R +, i.e. this is gin n infinitesiml. β < β δ < βδ < β δ < β Moreover if δ is n infinitesiml, then δ 1 is n infinite numer nd vice vers. Furthermore ny infinite numer Ω multiplied y numer β which is not infinitesiml, will gin e n infinite numer Definition The set of infinitesimls, I Let I R, denote the set of ll infinitesimls in R It is cler to see tht the set I is n idel for the finite hyper rel numers, since ny finite numer multiplied y n infinitesiml is gin n infinitesiml. The finite numers in R ehve in most pprecitive wy, nmely Proposition Finite hyperrel numers is sum of rel numer nd n infinitesiml Any finite numer β R, cn e written s uniquely determined sum, β = + δ, where R nd δ I. Proof: Uniqueness; if β = 1 + δ 1 = 2 + δ 2, then 1 2 = δ 2 δ 1, now since the right side of the eqution is n infinitesiml nd the left side is rel numer it must e rel infinitesiml, hence = 1 2 = δ 2 δ 1, sserting the uniqueness. Existence; Let = sup { R < β}, if (β ) I then the proof is done. Assume (β ) I, then there exists some numer, 1 R, such tht < 1 < β, contrdicting the choice of Definition Infinitesiml difference, For α, β R let α β if (α β) I Internl sets nd functions, nd the connection etween R nd R. In order to use the hyperrel numers to nlyze rel functions, connection etween functions nd sets in R nd R is needed, the following definition will help estlish tht connection. Pge 9 of 152

20 Definition Stndrd prt nd mond For every finite β R, let the unique rel numer β, e the stndrd prt of et nd denote it st(β). Furthermore denote the set {β R st(β) = } the mond of. If Α is suset of R, then the stndrd prt of Α, is defined y st(α) {st(α) α Α} Exmple Stndrd prt of hyperrel numer Let n = β R nd n = α R e finite numer nd let = inf ({ R + < β < }). If = then β I, mking st(β) =. If, then either β or β + is infinitesiml, if (β ) I, then st(β) = st( n ) = st( + ( n ) ) = st( + ( n ) ) = nd if (β + ) I then st(β) = st( n ) = st( + ( n + ) ) = st( + ( n ) ) = Proposition Rules for the stndrd prt If α nd β two finite hyperrel numers such tht st(α) = nd st(β) =, then st(α + β) = +, since the sum of two infinitesimls is gin n infinitesiml. I.e. st(α + β) = st(α) + st(β) if α nd β re oth finite. If α nd β two finite hyperrel numers such tht st(α) = nd st(β) =, then st(αβ) = = st(α)st(β). I.e. st(αβ) = st(α)st(β) if α nd β re oth finite Definition Str opertion, For every suset A R denote A = A, A, the nonstndrd version of A. For every function f: R R denote f = f, f, the nonstndrd version of f Exmples Str opertion The intervl (, ) R will e extended to the intervl (, ) = (, ), (, ), = {β R < β < }. Consider now the stndrd prt of (, ) R, since there exists some δ I, such tht α = + δ nd β = δ re oth in the mentioned set, then st( (, ) ) = [, ]. Thus it is possile to go ck nd forwrd etween the two numer lines ut the inequlities when going from the hyperrel line to the rel line, will go from strict to non-strict (st(<) =, st(>) = ). R st mond R Pge 1 of 152

21 The reltionship etween the rel nd hyperrel numers cn e descried y the figure ove, the thickness of the hyperrel line is wy of showing the infinitesimls round every rel numer, nd the dotted lines re where the infinite numers re situted Definition Internl sets nd internl functions 1. A sequence {A n } n N, where A n R for ll n, defines suset A n R y α n A n if nd only if m({n α n A n }) = 1. A suset of R which cn e otined in this wy is clled internl. 2. A sequence of functions {f n } where f n : R R for ll n, defines function f n : R R y f n ( x n ) = f n (x n ). Any function otined in this wy is clled internl. Two internl sets A n, B n or functions f n, g n re equl if m({n f n = g n }) = m({n A n = B n }) = Exmples Internl set nd internl function If α = n nd β = n re two elements in R, then the intervl [α, β] = {x α x β}, is n internl set since, it is otined s [ n, n ]. If ζ = c n is n element of R, then the function e ζχ is n internl function defined y e ζx = e c nx n. With this definition of internl sets nd functions mny of the results nd principles known out the rel numers R cn e crried over to the internl sets. One of the more interesting things is the lest upper ound of the completeness xiom Proposition Lest upper ound for internl sets An internl nonempty set, Α = A n R, which is ounded ove hs lest upper ound. Proof: If Α is ounded ove y α = n, then m({n supa n n }) = 1, hence the set of unounded A n s hs mesure zero. Without loss of generlity the A n s re ll ounded ove, hence β = supa n is the lest upper ound of Α Corollry Overspill nd underspill Let Α e n internl suset of R. 1. If Α contins ritrrily lrge finite elements, then Α contins n infinite element. 2. If Α contins ritrrily smll positive infinite elements, then Α contins finite element. Proof: 1. If there is lest upper ound α of Α, α must e infinite nd there must e χ Α such tht α 2 χ α, i.e. Α contins n infinite element. 2. If β is the gretest lower ound of Α +, the set of positive elements in Α, then β must e finite nd there must exist χ Α such tht β χ 2β Proposition Stndrd prt of n internl set is closed If Α R is internl then st(α) is closed set in R. Pge 11 of 152

22 Proof: Let st(α) nd consider the sequence of internl sets Α n = Α {β R β < 1 }. n Since Α n for ll n, then n N Α n. Pick ζ n N Α n, then ζ Α. The definition of the sequence ensures tht ζ, i.e. st(α). With this in mind, consider the following proposition Proposition Inclusion or equlity when using str opertion For ll A R, then A A with A = A if nd only if A hs only finitely mny elements. Proof: For the inclusion, let A, then =,, A, A, = A, hence A A. Now ssume A is infinite, then show tht there exists n element α A such tht α A. Construct sequence of distinct elements from A, { 1, 2, 3, }, then α = 1, 2, 3, A, ut α A. Now ssume A is finite, i.e. A = { 1, 2, 3,, k }, y the finite dditive mesure m, then for ny sequence { n } n N, where n A for ll n, 1 = m({n n A}) = m({n n = 1 ) + m({n n = 2 ) + + m({n n = k ). This mkes exctly one of the mesures on the right equl one, hence 1, 2, 3 = i, i, i, = i A. Considering function f: R R, the nonstndrd version of f, f originl function, y the fct tht for ny R, then f() = f, f, (,, ) = f(), f(), = f(). = f, f,, is n extension of the I.e. if the domin of the function f, is not finite, then the domin of the function is lrger set, mking f proper extension. With the mthemtics descried ove it is possile to extend some of the generlly used susets of the rel numers to the nonstndrd counterprts, N, Z, Q. One of the more interesting ones of these re the nonstndrd integers, N. The set of nonstndrd integers N, includes wht is clled infinite integers, i.e. the numer 1,2,3, N. These infinite integers ide y the sme rules s the norml integers, so if α = 1, 2, nd β = 1, 2, re two infinite integers, then α β N, if n only if m({n n N}) = 1. n In order to define nd use some of the more interesting prts of nonstndrd nlysis, nother new notion, nmely hyperfinite set is needed. These re sets of infinite order, ut with comintoril structure like the finite sets, which it inherits from the nonstndrd integers Definition Hyperfinite set An internl set Α = A n R is clled hyperfinite if m({n A n is finite set}) = 1. In this cse, the crdinlity of the set Α will e the infinite integer Α = A n, where A n is the numer of elements in A n. Pge 12 of 152

23 Since (lmost) ll the A n s re finite, ll the comintorics for finite sets cn e used on hyperfinite sets, this property enles the use of induction over hyperfinite sets, just s with finite sets Exmple Crdinlity of hyperfinite set Let N N, then the set T = {, 1 N, 2 N, 3 N,, N 1 N, 1} is hyperfinite, with crdinlity N + 1. If N N, then it is finite set, nd it is known tht T = N + 1. In order to convince oneself of this eing true when N is n infinite integer consider, for N = N n nd T = T n where tht T = T n = N n + 1 = N T n = {,,,, N n 1, 1}, N n N n N n With the ove definition of hyperfinite set nother thing is le to e constructed, hyperfinite sum, which is lso n infinite sum Definition Hyperfinite sum Consider the hyperfinite set Α = A n R, then the sum of ll elements in Α cn e written s = n A n A n. This sum though infinite is suject to the rules estlished for finite sums, thus mking operting with these prticulr infinite sums much more pprecile thn the infinite sum over other infinite sets. This cn e seen y the fct tht it is sequence of finite sets. As such the hyperfinite sum llows the use of induction on n infinite set, ut in roder sense thn the regulr infinite induction done on countle set. Pge 13 of 152

24 2.1.4 Anlysis using the hyperrel numers With these notions in mind the definitions of continuity, differentiility nd integrtion for functions defined on the rel line cn now e mde without the use of ε-δ s, in ech cse these new definitions, will e proved to e equivlent sttements to the stndrd wy of defining them, y limit opertion Definition Continuity The stndrd definition, f: R R is continuous in point x in the domin of f, if ε R + δ R + : x y < δ f(x) f(y) < ε. And the nonstndrd definition, f: R R is continuous in point x in the domin of f, if y R y x f(y) f(x) = f(x). Proof: Assume (stndrd) continuous t x; let n = α x. If f(x) f(α) < ε for ll ε R +, then f(x) f(α) I, y definition of n infinitesiml. Now for ny ε R + pick δ R + such tht < x y < δ f(x) f(y) < ε. Since, x α < δ for ll δ R +, f(x) f(α) < ε for ll ε R +. Equivlently consider for ny ε R + pick δ R + such tht < x y < δ f(x) f(y) < ε, then which mkes {n x n < δ} {n f(x) f( n ) < ε}, m({n x n < δ}) = m({n f(x) f( n ) < ε}) = 1. Since the mesures re oth one, then for ny n = α x then f(x) f(α), s wnted. Now ssume f is not continuous t x, then there exists n ε R + nd n = α x, with n x for ll n, such tht f(x) f( n ) > ε for ll n, which mkes f(x) f(α) > ε, contrdicting the nonstndrd definition of continuity Definition Differentil quotient The stndrd wy: A function f: R R is differentile t point x R if nd only if, f(x + ) f(x) lim exists nd is the sme oth when is mximum nd minimum for. In this cse f(x + ) f(x) lim = f (x). The nonstndrd wy: A function f: R R is differentile t point x R if nd only if, for every nonzero infinitesiml, st ( f (x + ) f(x) ) = exists nd is equl to the sme vlue R. In this cse st ( f (x+) f (x) )=f (x) =. Pge 14 of 152

25 Proof: Assume stndrd differentiility, then lim + ) f(x) (f(x ) = lim ( y x f(y) f(x) ) = f (x) y x By definition of the limit f(y) f(x) ε R + δ R + : < x y < δ f (x) < ε. y x By the sme rgument s for continuity nd for x y = I, then i.e. st ( f (y) f y x (x) ) = st ( f (x+) f (x) ( f (y) f(x) f (x)) I y x ) = f (x). Now ssume it is not differentil t point x, then the limit does not exist. Assume the limit does not exist, then for ll R f(y) f(x) ε R + δ R + y R < x y < δ ε y x Now ssume tht st ( f f (x+) f (x) (x + ) f(x) ) = R for some infinitesiml = Δx n, then = f(x + Δx n ) f(x) = f(x + Δx n ) f(x). Δx n Δx n Consider now f(x+δx n ) f(x) nd pick y Δx n such tht x y n < Δx n f(y n ) f(x) ε for ech n N. n y n x Let dy = Δy n = x y n, then dy I since dy < nd < Δy n for ll n N. Thus st ( f (x + dy) f(x) ) = st ( f (x + dy) f(x) ) ε dy dy i.e. st ( f (x+) f (x) ) st ( f (x+dy) f Corollry A differentile function is continuous If f: R R is differentile in x R, then f is continuous in x. dy (x) Proof: If f is differentile then f (x+) f(x) ), contrdicting the nonstndrd definition. = β is finite numer, since otherwise the stndrd prt does not exist ut then f(x + ) f(x) = β is n infinitesiml, mking f(x + ) f(x), i.e. f is continuous in x Rules of differentition Let f(x) nd g(x) e differentile functions, then the following rules pply. 1. For h(x) = f(x) + g(x) then h (x) = f (x) + g (x). 2. For h(x) = f(x)g(x) then h (x) = f(x)g (x) + g(x)f (x) 3. For h(x) = f(g(x)) then h (x) = f (g(x))g (x) Pge 15 of 152

26 Proof: 1. h (x) = st ( h st ( g (x+) g(x) (x+) h(x) ) = st ( f (x+) f(x)+ g (x+) g(x) ) = st ( f (x+) f(x) ) = f (x) + g (x), since finite sum of stndrd prts re equl to the stndrd prt of the sum, when the rguments re oth finite, which they re ecuse they re differentile. 2. h (x) = st ( h st ( ( f st ( g (x)+ f = st ( f (x+)( f (x+) h(x) ) = st ( f (x+) f(x))( g(x)+ g(x+) g(x)) f(x) g(x) (x)( g (x+) g (x+) f(x) g(x) (x+) g(x))+ g(x+)( f(x+) f(x)) (x+) f(x)) ) ) = ) = st ( f (x)( g (x+) g(x)) ) = f(x)g (x) + st( g(x + ))f (x) = f(x)g (x) + g(x)f (x). here the sme thing s in the first prove re employed together with the rule of the stndrd prt of product is equl to the product of the stndrd prt when the rguments re oth finite. In the end since g is differentile it is lso continuous, which mkes st( g(x + )) = g(x). 3. Firstly, if g(x) = g(x + ), then st ( h (x+) h(x) ) = = g (x)f (g(x)). Secondly, if g(x) g(x + ), let g(x + ) g(x) = dy, then h (x) = st ( h (x + ) h(x) ) = st ( f ( g(x + )) f(g(x) ) ) = st ( f (g(x) + g(x + ) g(x)) f( g(x)) g(x + ) g(x) ) dy = st ( f (g(x) + dy) f(g(x)) g(x + ) g(x) ) = f (g(x))g (x) dy Agin the rules for how the stndrd prt cn e tken on product re employed. And the fct tht st( g(x + )) = g(x) for ny continuous or differentile function. ) + ) + Pge 16 of 152

27 Definition: (rel) integrl Let the function f: R R. The definition of the stndrd (Riemnn) integrl is, for prtition of [, ] R, = x x 1 x 2 x τ 1 x τ =, nd ny x n [x i, x i+1 ], where Δx i = x i+1 x i : f(x) = lim ( f(x n )Δx n ). mx{δx i } Note tht for mx{δx i } then N, since N n= Δx n = [, ]. The definition of the (Riemnn) integrl of f, sed on the infinite sum in , is for T i = τ N n infinite integer nd the infinite prtition of the intervl [, ] R: = x x 1 x 2 x τ 1 x τ =, with infinitesimls n = Δx n,i = x n+1,i x n,i nd x n [x n, x n+1 ] then τ 1 f(x) = st ( f n= (x n ) n ) N 1 n= T i 1 = st ( f(x n,i n= )Δx n,i Proof: Let f e s ove, then the definition of the stndrd (Riemnn) integrl trnsltes into the following ε-δ definition: For ll ε R + there exists δ R + such tht for ny prtition x, x 1,, x N of the intervl [, ] with mx{δx i } < δ then N 1 f(x n )Δx n f(x) < ε. n= ). Now consider Thus for ll ε R + τ 1 f(x n ) n n= T i 1 = f ( x i,n ) Δx i,n n= T i 1 T i 1 = f(x i,n ) Δx i,n n= T i 1 = f(x i,n )Δx i,n = f(x i,n T i 1 n= m ({i f(x n,i )Δx n,i n= n= )Δx i,n f(x) < ε}) = 1.. Therey the integrl, f(x), nd the hyperfinite sum, τ 1 f(x n= n ) n, differs t most y n infinitesiml. If the stndrd integrl exists there exists prtition for which the limit of the Riemnn sum exists. This prtition cn e descried y every integer extending the description of the prtition to n infinite integer, will mke the prtition n infinite prtition nd cn e used in the infinite sum which defines the non-stndrd integrl, thus mking the nonstndrd integrl exists. If the nonstndrd integrl does not exist, the stndrd version cnnot either, since there is no prtition, not even infinite, which cn Pge 17 of 152

28 e used so tht the function vlue in the prtition intervl vries with n infinitesiml. This entils tht the limit of finite prtition will not e well defined, since the choice of vrile in ny of the intervls will mke the function vry. I.e. the limit of the upper- nd lower sums (sometimes used to define the integrl) will not coincide, hence the limit, nd integrl, does not exists Definition Internl integrl Define the integrl over n internl set A n = A R of n internl function f n = f: R R s: Where ech f n : R R. f(x) A = f n (x) A n, Exmple Let f: R R e continuous function round x R nd let Δx m = e n infinitesiml then x+ x f(y)dy x+δx m = f(y)dy. Now let τ m e the hyperfinite integer such tht x = y m, y m,1 y m,2 y m,τm 1 y m,τm = x + Δx m is n infinite prtition, i.e. the difference y n+1 y n is infinitesiml. Let y m,n [y m,n, y m,n+1 ] then x+ x f(y)dy x+δx m = f(y)dy = st ( f x τ m 1 n= x (y m,n )Δy m,n T m,i 1 ) = st ( f(y m,i,n n= )Δy m,i,n Rules of integrtion 1. For ny function f: R R nd ny rel numer R the one point integrl is zero, i.e. f(x) =. This ensures tht the integrl over the closed intervl [, ] is the sme s the integrl over the upon intervl (, ). 2. Let f: R R nd k R constnt then kf(x) = k f(x) 3. Let f: R R nd (, ) nd (, c) e rel intervls, then 4. Let f: R R nd (, ) rel intervl then c f(x) = f(x) + f(x) f(x) = f(x) c ). Pge 18 of 152

29 Proof: 1. consider the definition of the integrl, τ 1 st ( f n= (x n ) n ) since the prtition cn only consist of the single vlue then the n = n+1 n = =, mking the sum equl to zero. i.e. it does not depend on the function vlue f(). 2. For hyperfinite integer τ = T i, then nd since k st( f T i 1 m ({i k f(x n,i )Δx n,i n= T i 1 = k f(x n,i n= )Δx n,i }) = 1 (x)) = st(k f(x)), the desired result follows. If the integrl does not exist then oth sides of the eqution does not exist. 3. The prtition in the left integrl, = x x 1 x 2 x τ 1 x τ = c, must contin some x i, mking x i [x i, x i+1 ] nd thus c f(x) = τ 1 st ( f n= (x n ) n i 1 ) = st ( f(x n ) n n= = f(x) + f(x) c τ 1 + f 4. Consider the first rule of integrtion nd use the third rule for integrtion then n=i (x n ) n ) f(x) = f(x) + f(x) f(x) = f(x) f(x) = f(x) sserting the desired result Theorem Fundmentl theorem of clculus prt 1 Let f: R R e continuously differentile on the intervl [, ], then the following holds for x [, ] R: Proof: By definition f (x) = st ( f f (x) = st ( f following (x+) f(x) x ) = f x f (y)dy = f(x) f(). (x+) f(x) (x+) f(x) ), now let δ(x) st ( f (x+) f(x) ) f (x+) f(x) then + δ(x) replcing this expression with the integrnd leds to the f (y)dy = st ( ( f (x n + ) f(x n ) τ 1 n= + δ (x n )) n ). The definition of the differentil quotient mkes it possile to use ny infinitesiml to descrie it nd still get the sme result, thus Pge 19 of 152

30 x f (y)dy = st ( ( f (x n + n ) f(x n ) n τ 1 n= τ 1 = st ( f n= τ 1 n= + δ (x n )) n ) (x n + n ) f(x n ) + δ(x n ) n ) = st ( f(x n+1 ) f(x n ) + δ(x n ) n ) = st ( f(x τ ) f(x ) + δ(x n ) n ) τ 1 = f(x) f() + st ( δ n= (x n ) n ) By induction over the hyperfinite set Χ τ = {x n R n {1,2,3,, τ 1}} mximl element, δ m, cn e chosen for δ(x n ). This mximl element is n infinitesiml y definition of δ(x), thus τ 1 st ( δ n= (x n ) n τ 1 ) st ( δ n= With this the first prt of the fundmentl theorem is proved. τ 1 τ 1 n= (x n ) n ) st ( δ m n ) = st(δ m (, x)) =. In the following some of the more complicted proofs in nlysis cn e found, which enles the proof of the second prt of the fundmentl theorem Theorem Bolzno s theorem Let f: R R e continuous on the intervl [, ] R nd () < < f(), then there exists c [, ], such tht f(c) =. Proof: Let N n= N e n infinite integer nd let = x < x 1 < x 2 < < x N 1 < x N = e the infinite prtition, where the difference etween two consecutive prtition points re of equl length, i.e. x n+1 x n = = ( ) for ll n. Let J e the set prtition points, x N j, for which < f(x j ), y definition the set J is hyperfinite set nd thus n x j for which f(x j ) f(x j ) for ll x j J cn e chosen y induction. Since f is continuous then st ( f(x j )) = f (st(x j )) =. Therey the sought c R cn e found y letting c = st(x j ) Theorem Extreme vlue theorem Let f: R R e continuous on the intervl [, ] R, then the function f(x) ttins oth (locl)mximum, Mx = f(c M ), nd (locl)minimum min = f(c m ) vlue, for some c m, c M [, ]. I.e. for every x [, ] the following holds: min f(x) Mx Pge 2 of 152

31 Proof: Let N N e n infinite integer nd let = x < x 1 < x 2 < < x N 1 < x N = e the infinite prtition of [, ], where the prtition intervls re of equl length, i.e. x i+1 x i = = ( ) N for ll i. Now let A e the hyperfinite set of ll the prtition points. i.e. A = {x, x 1, x 2,, x N 1, x N } Then y induction there exists x m nd x M such tht for ll x i A f(x m ) f(x i ) f(x M ). For every rel numer x [, ] there exists prtition intervl [x i, x i+1 ] such tht st(x i ) = x nd since the prtition intervls re of infinitesiml length, no intervl cn contin more thn one rel numer. Let st(x m ) = c m nd st(x M ) = c M, then y continuity the stndrd prt of the inequlity trnsltes to; for ll x st(a) = [, ], then st(f(x m )) f(x) st(f(x M )) f(st(x m )) f(x) f(st(x M )) f(c m ) f(x) f(c M ). By definition f(c m ) is the minimum nd f(c M ) the mximum of the function over the intervl [, ]. Therey sserting the desired result min = f(c m ) f(x) f(c M ) = Mx Theorem Intermedite vlue theorem Let f: R R e continuous on the intervl [, ] R, then the function f(x) ttins every rel vlue etween its minimum, f(c m ) = min, nd mximum, f(c M ) = Mx, for some x [, ]. I.e. Proof: f([, ]) = [f(c m ), f(c M )] = [min, Mx]. If m = M, then the function is constnt over the intervl nd there is nothing to prove. Now let f(c m ) = min < Mx = f(c M ) nd let d R e such tht f(c m ) < d < f(c M ). By definition the function g(x) = f(x) d will e continuous in the sme intervl s f(x). Now sutrct d from the inequlity, i.e. f(c m ) < d < f(c M ) f(c m ) d < d d < f(c M ) d g() < < g() By Bolzno s theorem there exist c [, ] such tht g(c) = f(c) d =, thus f(c) = d. Since d ws chosen ritrrily in (f(c m ), f(c M )), the desired result follows. Pge 21 of 152

32 Definition - Averge Given finite set A = {x, x 1, x 2,, x n } for n N the verge of the vlues, x, x 1, x 2,, x n re given y n x ve = 1 n + 1 x i Definition - Men of function Given function f: R R, the men of the function over the intervl [, ] is given y Explntion: i= f men ([, ]) = 1 f(x). Tke n infinite integer N N nd mke n infinite prtition of the intervl [, ], where the length of the prtition intervls re equl. I.e. = x < x 1 < x 2 < < x N 1 < x N =, where x i+1 x i = = ( ) for ll i N 1, which mkes 1 =. Following the definition of the N N verge the men of f should e N f ve ([, ]) = 1 N + 1 f(x 1 i) = f(x i ) N + 1 Consider the difference 1 1 = N+1 N = 1 < 1 = 2 N N+1 N 2 +N N 2 +N N 2 ( ) 2. Thus f ve ([, ]) is: N 1 i= 1 f ve ([, ]) = f(x i ) = f(x N + 1 i ) ( 1 N + 1 N 2 + N ) = f(x i ) 1 + f(x i ) N 2 + N i= N 1 N 1 i= N i= N 1 i= N 1 N 1 i= N 1 < 1 f(x i) + f(x i ) 1 N 2 = 1 f(x 2 i) + f(x i ) ( ) 2 i= N 1 N 1 i= = 1 f(x i) + ( ) 2 f(x i). i= Tking the stndrd prt of this mkes N 1 N 1 st(f ve ) = st ( 1 f(x i) + ( ) 2 f(x i)) i= N 1 i= N 1 i= = 1 st ( f(x i)) + st ( ( ) 2) st ( f(x i)) i= i= = 1 f(x) + f(x) = 1 N 1 i= i= f(x). Pge 22 of 152

33 This mkes the stndrd prt of the verge of the function into the definition of the men of the function over the intervl s wnted Corollry: Men of continuous function Given continuous function f: R R, the men of this function over the intervl [, ] is equl to function vlue evluted in d [, ]. Proof: f men ([, ]) = 1 f(x) = f(d). By definition of the men of function then f men is vlue etween the extreme vlues of the function over the intervl. So y the intermedite vlue theorem the function f ttins this vlue for some d [, ] Theorem Fundmentl theorem of clculus prt 2 Let f: R R e continuous on the intervl [, ], nd let F(x) = f(y)dy then the following holds Proof: F (x) = f(x) Let = Δx n, then y definition of the integrl nd the str opertion F (x) = st ( F (x + ) F x+ x+ x (x) ) = st ( f(y)dy f(y)dy ) = st ( x f(y)dy ) = st ( 1 f(y) dy) = st ( 1 Δx n x x x+δx n f(x) x x+ ) = st( f men ([x, x + Δx n ]) ) By corollry , the men of the function cn e written s f men ([x, x + Δx n ]) = f(d n ), where d n [x, x + Δx n ] for every n. I.e. d n = d [x, x + ] mking F (x) = st( f men ([x, x + Δx n ]) ) = st( f(d n ) ) = st( f(d) ) = f(x) where the lst equlity is true since the function is continuous. With this the first prt of the theorem cn lso e proved, ut s seen it required gret del of intermedite theorems to get to the second prt. As such oth proofs cn e useful Corollry - Fundmentl theorem of clculus prt 1 Let f: R R e continuously differentile on the intervl [, ], then the following holds: for x [, ] R. x f (y)dy = f(x) f() Pge 23 of 152

34 Proof: Tking the differentil on oth sides (using theorem on the left side), shows tht it is true up to constnt term. The only thing to prove is tht the constnt term, c, which G(x) = f (y)dy differs from f(x) y, is f(). Now consider Thus c = f(), sserting the desired result. f() + c = G() = f (y)dy = f() f() = Proposition Integrtion y prts Let f: R R nd g: R R e continuously differentile functions on the intervl [, ], then the following holds: f (x)g(x) = f()g() f()g() f(x)g (x). Proof: Let h(x) = f(x)g(x), then h (x) = f (x)g(x) + f(x)g (x). Using the rules for differentition nd for integrtion the following cn estlish: f (x)g(x) = f (x)g(x) + f(x)g (x) f(x)g (x) = h (x) f(x)g (x) = h (x) f(x)g (x) = f()g() f()g() f(x)g (x) Definition Antiderivtive Define the ntiderivtive of function f: R R, s function F(x): R R, for which the following holds: x i.e. F(x) = f(y)dy + k for some constnt k R. F (x) = f(x) Definition Indefinite integrl Define the indefinite integrl s the opposite of the derivtive, i.e. for function f: R R, the indefinite integrl is the ntiderivtives for the function f. This is usully written for some constnt c R s x x i.e. f(x) = F(x) + c = f(y)dy + k. f(x) = F(x) + c Proposition Integrtion y sustitution Let g: R R e continuously differentile on the intervl [, ] nd let f: R R e continuous on oth [, ] nd [g(), g()] nd such tht f hs n ntiderivtive on the intervl [, ]. Then f(g(x))g (x) = f(y)dy. g() g() Pge 24 of 152

35 Proof: Let F e n ntiderivtive of f, then H(x) = F(g(x)) is n ntiderivtive of the continuous function h(x) = f(g(x))g (x), thus f(g(x))g (x) = H() H() = F(g()) F(g()) = f(y)dy. Another wy to consider this is to write y = g(x), this mkes dy = dg(x) = g (x), hence dy = g (x). Thus sserting the desired result. g() g() f(y)dy = f(g(x))g (x) g() g() Pge 25 of 152

36 2.2 Didcticl theory Didctics is theory nd prcticl ppliction of teching nd lerning. The theory of didctic lerning methods focuses on the knowledge tht students possess nd how to improve it, hence it provides students with the required theoreticl knowledge, where techer cts s guide nd resource for students. By mens of vrious didcticl theories, it is possile to structure such lerning environments nd est further the students knowledge. The following is description of prt of the nthropologicl theory of didctics, s well s foundtion of theory for the nlysis of this study Anthropologicl Theory of Didctics Throughout the 198s nd 199s, Yves Chevllrd ( Yves Chevllrd (English) A.R.D.M., n.d.) mde his footprint in the history of the didcticl work. His reserch is epistemologicl nd institutionl nd he is the founder of the Anthropologicl Theory of Didctics (ATD). This theory dels with trnspositions of knowledge nd focuses in prt on where the knowledge comes from nd to whom it is situted. In high school, mthemticl ojects creted y mthemticins re not the ones eing tught, so the mthemticl knowledge produced outside school is dpted, often severl times, efore it is ccepted for teching. As such ATD ims to descrie nd explin this trnsformtion of knowledge, distinguishing cdemic knowledge produced y mthemticins, knowledge to e tught, s defined y the eductionl system, knowledge tught y the professor nd knowledge lernt y students. These mthemticl nd didcticl orgniztions re determined together nd y ech other. In other words, they re codetermined, s Chevllrd puts it. Below is figure which enles n overview of the 9 levels of codetermintion. Figure Shows the 9 levels of didctic co-determintion, s proposed y Chevllrd, relted to the components of MO s. It lso includes modelling of prxeologies y MO s. The didcticl trnsposition of knowledge is often referred to s the trvel of knowledge from source(s) to students. It is process etween different groups of people in different institutions; this could e politicl Pge 26 of 152

37 uthorities, mthemticins nd techers nd their ssocitions who chooses wht is to e tught. This institutionl orgniztion is clled the noosphere, which contempltes eduction. Tht is to sy they set up the limits, nd redefine nd reorgnize the knowledge to mke possile or uncertin choices. As such ATD defines the oundries within which mthemticl eduction seems to e confined nd represents n epistemologicl pproch. ATD lso ddresses the institutionlized mthemticl ctivities nd uses the model of didctic co-determintion to ddress the issues Mthemticl orgniztion (MO) A mthemticl orgniztion cn e divided into two locks: the first lock is clled the prcticl or prxis lock nd the second is clled the knowledge lock. The prcticl lock is the prt of the mthemticl orgniztion concerning the types of tsks nd techniques used to solve them. The types of tsks re the proposed prolems to e solved in order to explore the new mthemticl discipline. Possile exmples of tsks re: find the stndrd prt of some hyperrel numer, find the stndrd prt of quotient of infinitesimls (= find the derivtive of function), nd find the stndrd prt of n infinite sum of infinitesimls ( find the integrl). To engge these tsks technique is used s method to solve the tsks t hnd. Thus techniques encompss oth the lgorithmic nd nlytic kinds of mthemtics. A specific technique cn cover numer of tsks. The knowledge lock likewise consists of two prts, technology nd theory. Technology is the environment wherein the techniques re justified, which is expressed in the word itself: techn (s in technique) nd ology (s in study of ). Lstly the theory is wht underlies the technology nd gives rise to the deepest nd most thorough understnding of the techniques nd technology used. The four concepts: types of tsks, techniques, technology nd theory constitute mthemticl orgniztion (MO) or prxeology. The word prxeology is mixture of prxis nd logos which mkes sense, seeing s the two first concepts, types of tsks nd techniques, is prcticl (prxis) lock nd the technology nd theory is the knowledge (logos) lock. With this in mind the MO or prxeology is often written s four letter tuple, [T,τ,θ,Θ], where T is set of tsks nd τ re the technique(s) used to solve the tsks. The letter θ represents the technology nd lstly the theory is represented y the letter Θ. With the tuple [T,τ,θ,Θ], different types of prxeologies or MO s cn e mde, the simplest (smllest) prxeology is the punctul prxeology, which descries prxeology with only single type of tsks, i.e. if ll the tsks T re of the form: determine the derivtive of polynomil function, the prxeology would e clled punctul prxeology. If prxeology is unified y the technology θ, i.e. the prxeology encompsses vriety of tsks where the technique(s) cn ll e explined through the sme technology, then the prxeology will e clled locl MO. An exmple of locl MO could e to determine the derivtive function given its nlytic expressions. In this cse the tuple representing the prxeology would contin n index on the types of tsks nd techniques, i.e. [T i,τ i,θ,θ]. The index is there to represent the different types of tsks nd techniques contined in the locl MO. A regionl MO is prxeology which encompsses more thn one locl MO, i.e. the technology for the different locl MO s cn e different ut the theory is the sme for the locl MO s. An exmple of this could e infinitesiml clculus, with differentil clculus s one locl MO nd integrl clculus s nother locl MO. Pge 27 of 152

38 Didcticl orgniztion The nthropologicl theory of didctics hs wy of descriing the cretion of n orgniztion of knowledge; in this study, the orgniztions t hnd re ll mthemticl, s such the theory of didcticl orgniztion (DO) will e explined sed on n MO. The wy tht ATD descries the cretion of n MO is wht is clled didcticl orgniztion nd consists of six different moments, ech of which is importnt to generte meningful MO. A didcticl orgniztion is prtly dependent on the MO ut t the sme time the didcticl orgniztion certinly lso hs n impct on the MO ctully tught. With this in mind the 6 moments of DO concerning n MO= [T i, τ i, θ, Θ] re: 1. First encounter with tsk from type of tsk T i. As such, the first encounter is the first time prolem of given type T i is posed to the individul(s), who is supposed to crete the MO. 2. Explortion of the type of tsk T i nd estlishment nd elortion of the technique τ i. This moment entils for the individul(s) constructing the MO t hnd to come up with wy of nswering the type of tsk nd explining the technique, such tht others would e le to replicte the technique. 3. Constitution of the technology θ nd theory Θ concerning the technique τ i. As such this moment concerns the theory nd technology the technique is suordinte to. This moment, cn e difficult to pin point to only hppening once, since the introduction of nother type of tsk τ j might constitute different knowledge lock even if oth τ i nd τ j re prt of the sme MO. 4. Technicl work on the technique τ i is mde in order to decipher when exctly the technique cn e used nd to wht extent. Note tht the word technicl ensures tht the work to e done hs to e sed on the technologicl nd theoreticl level of the MO s developed in the moment of constitution. 5. Institutionliztion connects the MO with other locl MO s, which shre the sme knowledge lock [θ, Θ]. In this didcticl moment the individul(s), who creted the MO tht the DO descries, estlishes connection to previously constructed MO s. By doing this the newly generted MO is settled mong other MO s, which enles specifiction of the oundries nd possile wys in which the newly creted MO cn e used. 6. Evlution is moment where the constructed MO is evluted s to how well the creted MO solves the tsks T i t hnd in comprison to other locl MO s found in the institutionliztion. It is cler tht the sixth moment works very closely together with the fifth. A pssge on this prticulr suject cn e found in Didctic Restrictions on the Techer s Prctice: The Cse of Limits of Functions in Spnish High Schools (BrÉ et l., 25) It is cler tht complete relistion of the six moments of the didctic process must give rise to the cretion of MO tht goes eyond the simple resolution of single mthemticl tsk. It leds to the cretion (or re-cretion) of t lest the first min elements of locl MO, structured round technologicl discourse. Pge 28 of 152

39 2.2.2 Wht to e tught In order to estlish wht to e tught, the didcticl trnsposition is used. The didcticl trnsposition explins the connection etween the scholrly knowledge, the knowledge to e tught nd the ctully tught knowledge for specific discipline, such s mthemtics. The scholrly knowledge is then the knowledge conducted y reserchers; s such the mthemtics tught t universities is closely relted to the scholrly knowledge. The curricul nd exms, creted y the government, nd the textooks constitute the knowledge to e tught, which is dependent on the scholrly knowledge. Actully tught knowledge is the knowledge otined y the students, i.e. how the students re tught in the clssroom nd their responses, in the form of questions sked during clss nd homework (ssignments). Actully tught knowledge is dependent on the knowledge to e tught nd y extension the scholrly knowledge. A model upon which to se the plnned teching nd model used when nlyzing the ctully tught knowledge is estlished y the scholrly knowledge nd the knowledge to e tught, which is clled reference MO. Illustrted elow on figure. Scholrly knowledge Knowledge to e tught Actully tught knowledge Reference mthemticl orgniztion Figure In order to decipher wht knowledge to e tught, when it concerns the teching of infinitesiml clculus in high school, reference (regionl) MO for the mthemtics involved is needed, oth s foundtion for estlishing the teching mteril nd s wy to nlyze the teching. This section is divided into 3 susections, the first one eing the hyperrel MO, which concerns the teching of the hyperrel numers. The section out the hyperrel MO is void of theory supporting the construction of the hyperrel numers; s such the reference MO nd knowledge to e tught when considering the hyperrel numers re close to eing identicl. This choice, which might seem unfulfilling is mde ecuse the construction of the (hyper)rel numers is not prt of the curriculum in high school. As such reference MO which includes the construction would mke it overly complicted to condense the reference MO into the sought knowledge to e tught. The second susection clled the differentil MO (DMO) concerns the teching of differentil clculus nd the third susection clled the integrl MO (IMO) concerns the teching of integrl clculus. These sections include reference MO for the nlysis nd sed on this, description of wht to e tught. Pge 29 of 152

40 Hyperrel MO In order to determine wht mthemticl knowledge from the non-stndrd nlysis is needed, to tech infinitesiml clculus in high school, the government-given curriculum for A-level mth (Ministry of eduction, 213) ws scrutinized, the textook they normlly used ws lso looked in (Clusen, Schomcker, & Tolnø, 26), ut since it is not sed on the sme definitions s the nonstndrd ones, some of the prolems seemed odd. This mde the curriculum the foremost oject used to decipher the knowledge to e tught. The curriculum doesn t specify how to introduce neither the differentil quotient nor the integrl, hence this opens up different pproch to infinitesiml clculus, nmely the nonstndrd pproch. Since the hyperrel numers is not prt of the norml curriculum in high school, n MO out the hyperrel numers nd the connection to rel numers is needed. In high school the construction of the rel numers is not prt of the curriculum, which enles the introduction to the hyperrel numers in the sme wy s the rel numers: s n intuitive extension of the integer or rtionl numers. With this in mind the mthemticl knowledge to e tught, regrding the hyperrel numers nd the connection to the rel numers in this sitution, cn e oiled down to the following question: Wht is needed from NSA to define the differentil quotient nd the integrl in high school, where the construction of the rel/hyperrel numers re not prt of the curriculum? An intuitive understnding of the hyperrel numers R, kin the understnding of the rel numers. o Infinitesimls, nd how to operte with them. o Infinite numers, nd how to operte with them. The connection etween the hyperrel nd rel numers, i.e. o The stndrd prt of hyperrel numer. o An understnding of wht the str opertion of function is, s in wy to extend the domin nd rnge of function f: R R to f: R R. Continuity (in point) In the guide provided y the Ministry of Eduction (Ministry of eduction, 21), the notion of continuity is concept which should e included, ut not given seprte tretment. At some point, the ε-δ definition for continuity in high school ws prt of the curriculum, ut hs since een degrded from eing mndtory to optionl. Even though the definition of continuity ws removed from the curriculum, the notion of continuity is still something tht hs to e covered, to some extent. The nonstndrd definition of continuity opens up the possiility to include the definition gin, since it is very esy to trnslte the phrse function is continuous if the grph, of the function, does not jump into the difference in function vlue, when the difference in the vrile is infinitesiml, is gin infinitesiml. With this in mind, continuity ws mde prt of the things to e tught, when teching the hyperrel numers. The hyperrel MO (HMO) cn e descried y set of tsks nd list of exercises. The tsks nd techniques mke up the prcticl prt of the prxeology, s such the prcticl prt of n MO is more or less fully explined through its tsks. Pge 3 of 152

41 Types of tsks HMO= [T, τ, θ, ] in HMO HT In wht set of numers does numer x elong? HT 1 Find st() when R? HT 2 Wht is f(x + ) when is n infinitesiml nd f rel function? HT 3 Find st( f(x)) when, x R HT 4 Check if the rel function f is continuous (t point) With these tsks, n overview of the technologicl nd theoreticl level of HMO cn e fshioned through set of questions: 1. Wht is n infinitesiml? 2. Wht is n infinite numer? 3. How does numer line including infinitesimls nd infinite numers, i.e. R, look like? 4. How does one operte with the new found quntities? 5. Wht does it men to tke the stndrd prt? 6. How does one give mening to rel function evluted in hyperrel numer? 7. Wht is continuity (of rel function in point)? In order to nswer these questions in scholrly wy, the construction of the hyperrel/rel numers would e needed, ut since this is not prt of the curriculum, the nswers to these questions use n intuitive understnding of infinitesimls nd infinite numers. Thus possile nswer to the first question could e, tht positive infinitesiml is numer which is smller thn ny given positive rel numer. Thus, even if the technologicl level of HMO, θ, is not founded in the scholrly knowledge, it is still present, though it only ppers s n intuitive nd logicl understnding of the hyperrel numers. Most of the theoretic level of HMO is replced with intuition, when introducing the hyperrel numers to the students in high school. In this wy, HMO cn e descried y tuple, [Τ i, τ j, θ, ]. It should e noted tht this is no different thn the wy the high school students re using the rel numers, thus this introduction to the hyperrel numers does not violte the curriculum in ny wy. Pge 31 of 152

42 Differentil MO In order to estlish the reference MO considertion of the NSA definition of the differentil quotient s seen in section constitutes the technologicl nd theoreticl prt of the DMO ut in ddition it lso opertes s technique, though the only tsk supporting this technique would e to define it. As such when referring to the differentil quotient s technique it will e ppointed the term hypotheticl technique. This term is introduced in order to distinguish it from the regulr techniques which re used to descrie how to nswer the types of tsks in given MO. The hypotheticl technique of the differentil quotient mkes it possile to elorte on it in the sense of estlishing the rules of differentition. Types of tsks in DMO DT DT 1 DT 2 DT 3 DT 4 Define the differentil quotient. Find the derivtive, f (x), for function, f(x). Determine the condition of monotony for given function, f(x). Find the tngent of function through specific point, (x, f(x)). Determine if d f(x) exists for given function, f(x). Mrked in grey is the constitutive tsk of DT, since it is not type of tsk. A numer of types of tsks tht fll under DT 1 re listed elow: DT 11 : Find the of grph in point DT 111 : Find the slope of power function in point DT 112 : Find the slope of n exponentil function in point DT 121 : Find the derivtive of power function DT 122 : Find the derivtive of n exponentil function Other thn these types of tsks there exists list of commonly used questions which elorte the hypotheticl technique of the differentil quotient, with pproprite ssumptions on the functions nd constnts elow. 1. Prove tht d (k f(x)) = k d (f(x)) 2. Prove tht d (f(x) + g(x)) = d (f(x)) + d (g(x)) 3. Prove tht d (f(x) g(x)) = f (x)g(x) + f(x)g (x) 4. Prove tht d (f(g(x))) = f (g(x))g (x) With the reference MO for differentil clculus estlished s descried ove, the description of wht to e tught cn e found y scrutinizing the curriculum (Ministry of eduction, 213). In this curriculum the following things re descried s eing necessry for teching this specific prt of clculus in high school: Definition of the differentil quotient, including growth rte nd mrginl considertions Derivtives of the elementry functions (liner, exponentil, power, polynomil nd logrithmic functions, cosine nd sine) Pge 32 of 152

43 Rules for clculting the derivtive of f + g, f g, k f nd f g Deduce some selected differentil quotients As such the reference MO nd the curriculum coincide, except for the elortion of the hypotheticl technique of the differentil quotient in regrds to for which functions the derivtive exists. Furthermore the reference MO is sed on the scholrly knowledge descried in section 2.1, hence nother difference etween this nd the knowledge to e tught is the theory of (hyper)rel numers. Of course the knowledge to e tught covers much smller rnge of functions thn the scholrly knowledge Integrl MO In order to estlish the reference MO considertion of the NSA definition of the integrl s seen in section constitutes the technologicl nd theoreticl prt of the IMO. When the integrl is seen s finding the re etween function nd the first xis over n intervl, it opertes s hypotheticl technique, since it is impossile to clculte n infinite sum y dding the terms. This mkes it possile to elorte on the hypotheticl technique, the integrl, in the sense of estlishing the rules of integrtion. Types of tsks in IMO IT IT 1 IT 2 IT 3 Define the integrl. Clculte f(x) for given function, f(x). Find the ntiderivtive, F(x), for given function, f(x). Determine if f(x) exists for given function, f(x). Mrked in grey is the constitutive tsk of IT, since it is not type of tsk. A numer of types of tsks tht fll under the types of tsks re listed elow: IT 11 : Clculte f(x) when f(x) is liner IT 12 : Clculte f(x) when f(x) hs n ntidereivtive. IT 13 : Clculte f(x) when f(x) hs no ntiderivtive. x IT 21 : Clculte f(x) for given function f(x). IT 22 : Find the ntiderivtive, F(x), for power function, f(x) IT 22 : Find the ntiderivtive, F(x), for n exponentil function, f(x) IT 23 : Determine f(x) for given function f(x). Other thn these types of tsks there exists list of commonly used questions which elorte the hypotheticl technique of the integrl, with pproprite ssumptions on the functions nd constnts elow. 1. Prove tht k f(x) = k f(x) 2. Prove tht f(x) + g(x) = f(x) + g(x) Pge 33 of 152

44 3. Prove tht f(x) c + f(x) 4. Prove tht f(x) = 5. Prove tht f(x) = f(x) 6. Prove tht d ( x f(x) ) = f(x) = c f(x) With the reference MO for integrl clculus estlished s descried ove, the description of wht to e tught cn e found y scrutinizing the curriculum (Ministry of eduction, 213). Definition of the indefinite nd definite integrl Antiderivtive of the elementry functions (liner, exponentil, power, polynomil nd logrithmic functions, cosine nd sine) Rules for clculting the ntiderivtive of f + g, f g, k f nd integrtion y sustitution Mention the correltion etween the re function nd the ntiderivtive Disc method (clculting the volume of solid of revolution) By considering this list the knowledge to e tught is seen to differ from the IMO in few plces. The first difference, tht might sting the eye, is tht ny mentioning of existence of n integrl is void. The choice for this y the government could e tht they thought the techniques to estlish such n nswer would e too difficult. On top of tht the elementry functions, which (lmost) re the only functions the students work with in high school, re ll integrle on ounded intervl. Another difference is tht there is no distinct mentioning of proof of the fundmentl theorem of clculus. Pge 34 of 152

45 2.2.3 Didcticl resons for or ginst using NSA This section will estlish some resons for nd ginst the use of NSA to introduce infinitesiml clculus. As it ws hinted in the Motivtion section, teching nlysis in high school hs its ostcles, one of which is thoroughly nlyzed in the text Didctic restrictions on techers prctice the cse of limits of functions in Spnish high schools, y (BrÉ et l., 25). In this text, the picture pinted is, tht the teching of limits in high school lcks technologicl/theoreticl prt. The scholrly knowledge, which covers the theoreticl lock, consists in most prt of the completeness of the rel numers, which is sed on the construction of the rel numers nd is not something tht need e tught. The limit-opertion is normlly (in high school) used s tool to introduce the terms continuity, differentiility nd integrility s seen in (Winsløw, 213). When using NSA, the limit-opertion is not needed to define the forementioned terms. Wht tool is then used, when using NSA, nd does it suffer from the sme lck of technologicl/theoreticl support in high school? The stndrd prt springs to mind s the ovious sustitute for the limit-opertion, which will e considered s the tool. In order to understnd the stndrd prt, the hyperrel numers, nd their connection to the rel numers, re needed. The theoreticl lock of the stndrd prt is covered y the theoreticl lock of HMO. This lock ws expressed y the seven questions listed in section , ll of which cn e nswered with intuition nd logicl deduction, when the construction of the numers is not needed. As such, version of the theoreticl lock of the stndrd prt cn e otined y the students, wheres the theoreticl lock for the limit-opertion needs the completeness of the rel numers, to e logiclly sound. E.g. when determining the rules for clculting with differentils the limit opertion nd the rules for when product of limits equls the limit of the product re used or when determining the rules for clculting with integrls the limit opertion nd the rules for when sum of limits equls the limit of the sum re used. These rules cn e explined y the completeness of the rel numers ut s this is difficult to grsp nd even more so to prove, the high school students come up short when trying to prove the rules for clculting with differentils/integrls. When the hyperrel numers re intuitively understood the stndrd prt is logicl extension. By logicl deduction the rules out the stndrd prt cn e extrcted, thus enling the students to prove the rules for clculting with differentils/integrls with set of rules they understnd nd they re le to rgue why the rules pply. To explin the stndrd prt, it is the opertion tht tkes hyperrel numer to the nerest rel numer. The limit-opertion, on the other hnd, is it more difficult to explin. Intuitively, the limit-opertion cn e seen s wy of working with quntities of vrile mgnitude. Even with this intuitive understnding, the limit-opertion still seems incomprehensile, in tht mthemtics in high school is something sttic ut with this, mthemtics ecomes something dynmic. Normlly mthemtic opertions re done with constnts nd vriles, mening tht they re fixed or something tht cn e chosen to e fixed. The limitopertion introduces something which is neither constnt, ecuse it is not fixed, nor is it vrile, since it cnnot e chosen to e fixed, i.e. limit-opertions introduce different type of unfixed mthemtics. Another understnding of the limit-opertion cn e otined y ε-δ rguments, which comes down to doule-logicl deduction. Thus, either n unfixed pproch is needed or n pproch using something which is eyond the norml high school students understnding: the ε-δ rguments. In order to understnd the differentil quotient, consider the differentile function f: R R. The limitopertion llows for the following definition of the derivtive: Pge 35 of 152

46 f f(x + Δx) f(x) (x) = lim. Δx Δx Rememering tht division y is not llowed, then Δx. After some lgeric opertions then Δx = nd the differentil quotient is otined, ut this pproch cn confuse even the rightest of student. The students normlly understnd tht division y zero is not dole, ut when the sme thing is oth zero nd not zero, the students get confused. The differentil quotient, s introduced y NSA, llows for the following definition, where is n infinitesiml. f (x) = st ( f (x + ) f(x) ). Rememering tht division y is not llowed, then. Tking the stndrd prt of the quotient is most esily otined y doing lgeric opertions first. The str opertion is nother opertion which needs to e introduced in order for the definition to e mthemticlly rigorous. In this wy one could sy tht two opertions re needed insted of one, ut the two opertions re simple in comprison to the limit-opertion. Thus, the str opertion nd the stndrd prt ecome the new cndidte for the tool to introduce the terms continuity, differentiility, nd integrility. It would seem like mouthful to tech students nother set of numers, especilly when it is not necessrily something they will ever meet gin. But, s explined, this introduction enles the teching of infinitesiml clculus, needing neither the unfixed quntity, introduced with the limit-opertion, nor the intrinsic wys of ε-δ rguments. When teching the integrl, the most common introduction to this in high school is to use the re etween the grph of function nd the first xis, s n intuitive pproch to the integrl. With the use of limit-opertions, the integrl is sometimes introduced s the common limit of the upper nd lower Riemnn sums. This wy of introducing the integrl is normlly only used on the highest level of mth in high school. On the second highest level of mth in high school, the students re introduced to the integrl through the indefinite integrl, which is defined s the opposite opertion of differentition, thus only the highest level mth students re le to understnd, tht the integrl of discontinuous function cn e found (Clusen et l., 26). When using NSA to introduce the integrl, the use of limit opertion is gin circumvented. This llows the students to use the sme intuitive pproch to the infinite sum s when operting with the quotient of infinitesimls, defining the differentil quotient. When considering the things tht mke the integrl difficult to understnd, the limit-opertion might not seem like the most prolemtic one. Other prolems tht cn rise when introducing the integrl re: 1. The notion of prtition of intervls in mthemticl rigorous wy, including indexing. 2. The ility to write sum of n numer of terms, where n is n (infinite) integer. Since these two prolems re the sme for every pproch it neither encourges nor discourges the use of NSA. When the integrls re introduced using the limit-opertion, determining the rules for clculting with integrls drws on the techniques for operting with the limits ut s specified efore the limit-opertion lcks (most of) its theoreticl lock, i.e. the rules for clculting with integrls end up eing sed on lcking foundtion (Winsløw, 213). Sdly in oth wys of defining the integrl, the wy to compute n infinite sum is not something tht ny humn cn do, thus no shortcuts re creted. The need to identify Pge 36 of 152

47 the definite integrl with vrile endpoints s the inverse of finding the derivtive is still needed to compute integrls of functions tht re not liner. When introducing integrtion y sustitution, the notion du = g, when u = g(x) is recognized s the inner function, which mkes it possile to isolte = du g in mthemticl correct wy, since the quotient du is just quotient of 2 infinitesimls. This further illustrtes some of the intuitive understndings used when introducing infinitesiml clculus with NSA. When using NSA possile prolem could rise for the students who were to study mthemtics fter high school, i.e. on scientificlly higher level. The prolem lies in when the remining intuitive resoning is olished, i.e. when strting to use the xiomtic system s sis for mthemtics. When using the xiomtic system the numers need to e constructed in rigorous wy. As seen in section 2.1, the construction of the hyperrel numers is somewht more difficult thn the construction of the rel numers. Another prolem consists in the vilility of help outside of clss, lthough this prolem might e considered less significnt, if infinitesiml clculus using NSA relly is tht much esier to lern. One of the more profound differences in teching infinitesiml clculus y NSA is, tht the tool to estlish the definitions of the differentil quotient nd integrl re shifted, from the commonly (t lest in Denmrk) use of limit opertions, to the use of new set of numers nd its connection to the rel numers. This mkes it possile to introduce prolems tht only concern how the tool functions, wheres questions s to why limit opertions function like they do (ll) hinges on the completeness of the rel numers, which is hrd to understnd intuitively for the students. As mentioned efore pre-university school in Genev hs used NSA for period of time, the rticle Nonstndrd nlysis t pre-university level: Nive mgnitude nlysis (O Donovn & Kimer, 26) contins their view on the mtter s they experience it Side effects Here re some other moments where NSA is pplicle. Besides infinitesiml clculus, hyperrel numers cn e used to descrie phenomen where the rel numers come up short. An exmple could e the lst distnce etween to colliding ojects, which cn e descried with hyperrel numer, nmely n infinitesiml. People who cknowledge 1 =,99 s eing cler s dylight re rre sight. Using NSA one would find tht 1,99 = is n infinitesiml when,99 is considered hyperrel numer. Tking the stndrd prt on oth sides mkes it cler, tht if the two numers re oth seen s rel numers then st(1,99 ) = st() st(1) st(,99 ) = 1,99 =, i.e. the rel numers 1 nd,99 re equl. Pge 37 of 152

48 3 Concrete implementtion of the knowledge to e tught There goes lot of work into executing course of teching on ny level. In teching high school students, trnsposition of knowledge is needed. Other thn the techer, high schoolers primry source of knowledge is the textooks provided y the school; hence, the pln ws to provide them with some reding mteril. Seeing tht the curriculum for high school seldom chnges, the textooks lredy pulished rrely chnge either. Prticulrly new theories for sectors or themes re lmost never considered to e implemented. With the nonstndrd nlysis eing introduced in the 196s, this pproch hs not een estlished in ny form in the curriculum, hence no textook tht covers this pproch in Dnish exists, nd it hs to e in Dnish y the high school regultive. In ccordnce to ATD the students re to study the works of others when cquiring new knowledge if needed. In this study most of the tught mteril would e cquired through intuition provoked y the techer during clss, hence the need of textook mteril especilly emerged in the event where students would e sent. In this cse the students could not e expected to otin the knowledge they missed y eing sick, or in other wys eing unle to ttend clss. Even more crucil would e if the students were to study for potentil exm, oth orl nd written. These were the min resons for constructing compendiums for the students, descriing everything tht ws gone through in clss. It mde perfect sense to strt writing the compendiums longside the plnning of the teching, in tht the compendiums should descrie the plnned teching. It ecme evident though, tht creting the compendiums efore commencing the teching ws too comprehensive with the time ville. In getting to know the students, the compendiums could e shped in wy they would etter understnd nd e written in the sme tone, s ws used in clss. This pproch lso mde it possile to include specific moments tht occurred during clss, like specil hedings for prgrphs in the compendiums, tken from the students input. An extr exercise for the students to Worksheet 4.11 (Appendix 1.2), on finding the slope for function, ws to mke heding for the worksheet nd one of the groups cme up with Tngentmn nd 2point finds Miss, s reference to finding the slope for liner function, using two points, which is lso the slope of the tngent. Also, sections could e rewritten s to mtch the exct thing tht mde the students understnd something nd others could e skipped completely, ecuse the students would get the understnding fster thn expected. This lso st well with the intention of giving the students the compendiums grdully (it y it), since then the incorportion of the students contriutions could e used to dpt the compendiums to the level of understnding the students possessed. Since the est wy of introducing the hyperrel numers for students in high school, is intuitively through previously known sets, like the rel numers, the very first section is on numer sets in generl, to mke sure the students were fmilir nd comfortle with this. Due to time restrictions nd reflections on wht ctully tkes dvntge of NSA, certin sujects were left more or less untouched. As such, the compendiums simply provide the curriculum-intended knowledge to e tught using NSA nd consequently leve sujects tht cn still e tught lter, without NSA, even y other techers. Thus, three compendiums were developed: Hypertl og stndrdstjerner, Differentilregning, nd Integrlregning on the set of hyperrel numers, differentil clculus, nd integrl clculus respectively. Essentilly, the compendiums re shped the sme wy. They consist of oth n lgeric nd nlytic wy of descriing the suject t hnd, s well s n introduction in vrious forms. Every chpter contins the Pge 38 of 152

49 constitutive definitions, theorems nd proofs for ech suject, with relevnt exmples. An excerpt from the compendium Hypertl og stndrdstjerner is seen elow. Excerpt 3.1 Excerpt 3.1 is n exmple of rigorous definition of continuity, with touch of the tone from the clssroom. Here is lso one of the mny visits from the lien, which is used throughout the three compendiums. The lien ws used s wy of sking questions the students might sk, in the wy the students might sk them nd lso to dd dsh of humor. With the continution of the sme lien from strt to end, story unfolded, which ws to help the students engge the compendiums with enthusism nd not just see it s chore. All chpters, with the exception of one, end with tsks relted to the specific chpter, e.g. exercises nd proofs of rules of clcultions, nd my include vritions of exercises from previous chpters. If no seprte worksheets were prepred for lesson, these would mke sustitute worksheet for the suject t hnd, hence letting them complete the exercises during clss, for the most prt, nd thus prcticing nd strengthening their skills for written exms. Susequently, in ech compendium, there is n overll list of results for the exercises. The compendiums Differentilregning nd Integrlregning include recp of the highlights of the entire compendium, prticulrly contining list of functions, with their corresponding derivtives nd ntiderivtives respectively. Sid lists encompss few more functions ( x, ln(x), sin(x), cos(x)) thn ctully tught, ecuse even though they weren t fmilirized with these types of functions yet, it ws just to complete the lists in regrds to the curriculum. Pge 39 of 152

50 4 Presenttion of oservtions Context The oservtionl context hs not een specified ut it is needed to understnd the context for which the teching nd oservtions were crried out. The teching nd oservtions ws crried out in Dnish high school, ut the Dnish high school is not very specific term; in Denmrk exists numer of different High schools which hve different government-given curriculums. In this study the high school used to mke the oservtions is the most common high school clled gymnsium. The gymnsium is the high school in Denmrk which enles the students to e ccepted for the lrgest numer of different university eductions. In this regrd, the students ttending the gymnsium hve very different intensions s to wht the high school diplom hs to e used for. The reserch ws crried out in lrger thn most gymnsiums, Roskilde Ktedrlskole, this high school hs pproximtely 14 students, from which clss of 23 first yer students with A-level mth ws ville for 2 months, corresponding to 3 lessons of 65 minutes ech. A- level mth is the highest level of mthemtics student cn ttin in high school, there re gin numer of wys to ttin this level. The clss used for oservtions ws clss tht chose to hve mthemtics t the highest level from dy one. To this end some of the students could e surprised y how different the mth is, since they hd no previous knowledge of wht nd how mthemtics re tught t high school. This eing sid the clss chose to hve the highest level of mth, which usully mens tht the students re, t lest to some extent, interested in science. The clss hd previously een tught the entire curriculum regrding tringles, liner- nd exponentil functions nd descriptive sttistics, plus vrious rules for clculting with powers nd logrithms. Before the study commenced the clss hd grced the topic of power functions. According to the policy of the school differentil- nd integrl clculus is prt of the curriculum to e tught on their second yer of high school, which mkes sense in terms of the types of functions with which the students re fmilirized efore lerning how to derive nd integrte them. In ddition, ecuse of how the eductionl system is put together, grduting from pulic school does not lwys yield the concept of function. Thus, wht is function is n isolted course of teching in high school, which tkes some getting used to, minly through the teching of different types of functions. The clss hd previously hd 2 other techers which oth sid tht the clss consisted of students with greter thn verge level of knowledge regrding mthemtics. The first of these techers were the one who tught them during their introductory course of mthemtics, which tkes up the first hlf yer nd is the sme for every level of mth. After finishing this introductory course the students ws ppointed different techer, this is the techer who llowed the use of the clss for this study. Seeing s one of the uthors of this thesis lredy hd position s techer on Roskilde Ktedrlskole the possiility of orrowing clss from this prticulr school ws significntly enhnced. In ddition to teching the clss this study is out, the techer hd two other clsses with third yer students with A- level mth nd hd een techer for the lst six yers, oth t Roskilde Ktedrlskole nd nother gymnsium, teching mth nd Spnish whilst simultneously studying t the University of Copenhgen, on nd off since 26. The other uthor of this thesis hd the opportunity for first time experience teching t the school, hving een studying oth mth nd physics t the University of Copenhgen since 28. Pge 4 of 152

51 4.1.2 Overll plnning The overll plnning of the teching is sed on the MO s descried in the previous section. The order of teching, with the hyperrel numers first, then differentil nd lstly integrl clculus, ws chosen lmost efore the thesis ws strted, since doing it in nother wy is hrd to justify. One could sk why the hyperrel numers were tught s seprte suject nd not just s prt of doing infinitesiml clculus. The reson for doing it s seprte thing is tht the hyperrel numers re seldom something the students hve herd of efore, justified y only smll prt of mthemticins knowing wht NSA is nd even fewer use it for teching. Another reson is tht in order to mke use of the knowledge sed on the hyperrel numers s rguments for proofs nd definitions in the infinitesiml clculus thorough understnding of the hyperrel numers is needed. In plnning the course of teching, the fct tht there is not ny Dnish high school textooks for neither differentil- nor integrl clculus using NSA, ws ddressed y composing three compendiums specificlly for teching these sujects. The compendiums were written during the period of teching, while discovering wht the students needed to understnd to meet the requirements stted in the curriculum nd to wht extend the mthemtics hd to e explined for the students to e le to understnd it. As such, the compendiums were creted to suit this specific clss hence nother clss might need more or less guiding during the teching of this mteril. Every lesson ws plnned seprte from the next nd seldom efore the end of the previous lesson. Worksheets were mde for (lmost) every lesson, which the students hd to complete efore the end of the lesson. This ws to get them to use wht hd een discussed in plenum, in groups or y themselves. The lessons were plnned using the words nd explntions of the students, oth from the worksheets, which were, more often thn not, collected fter ech ended lesson, ut lso from wht ws sid in the clssroom. Other thn the lessons, four hnd-in ssignments were prepred for the students. They included primrily exercises in the suject up to the point of how fr the clss hd progressed. The students were to hnd in the ssignments pproximtely once every other week. These ssignments were to e done outside of the lessons for the most prt, even though the regultive on the high school in question forced some of the lessons to e used to do the hnd-in ssignments. The ssignments were used to check if the students hd the required skills nd to estlish wht holes, if ny, the students hd. Furthermore the ssignments were used s set of exercises for the students to hone their techniques. When grding the ppers the techers red through every ssignment nd when mistke ws discovered the techer posed some clrifying questions Oservtionl methodology In order to tke notes of the oservtions done during the teching process, certin wy to gther dtmteril hd to e estlished. Since two techers were present in every lesson, the decision to let one of them tke field notes on lptop during the lessons ws mde. Seeing tht the techers themselves plnned the teching, it mde it esier to tke field notes in the form of trnscript (mrked s Logooks in the digitl ppendix), in tht the oserver lredy knew which questions the techer (of tht lesson) would e sking; method used, when wht the clss did ws orchestrted in plenum. When the students worked in groups, or y themselves, oth techers cted s techers, which might not e the most ojective wy of gthering oservtions. The reson for this pproch, though, ws tht oth techers hd to operte s the min techer t some point, which would hve mde it difficult for the students to let one of them operte Pge 41 of 152

52 s third prty oserver, when they lso hd the understnding tht oth were their techers. A positive ttriute of this pproch is tht oth techers were le to shpe the teching nd in this wy the students ecme fmilirized with oth techers. After the students hd done group- or individul work, the techers would shre thoughts of wht they ech oserved to further the techers understnding of the students ilities. The oservtions done y the techers, nd the nswers given y the students, were then compred in order to see if the techers understnding of the students ilities were on pr with the students level of work. Since field notes do not lwys generte one hundred percent ojective ccounts of the teching surveyed, nother method of oservtions ws lso instlled. Audio-recordings of every lesson were mde in order to compre the field notes nd udio, in cse of duious ojectivity. These udio recordings could lso e used s secondry tool, to show some of the students understnding nd ilities in much more ojective mnor, thn the field notes. Worksheets, nd especilly hnd ins, were collected nd used s tool to ssess the students progress nd development of the tught MO. On few occsions, video recordings were mde; this ws when the students would do proofs on the lckord. These videos served two ojectives: firstly, the students could use it s n exmple of how the orl exmintion would unfold nd secondly, the video cts s third wy to oserve the students work. This type of oservtion is especilly useful when students do work t the lckord where oth the students explntions nd writings cn e cptured. Since these oservtions generte rther lrge nd cumersome set of dt, it hd to e orgnized in wy, which did not remove too mny of the essentil points oserved, ut still condensed the dt in such wy tht n nlysis cn e conducted. This pproch ws done in wht could e clled preliminry orgniztion of the teching oserved; these orgnized sets of dt cn e found in the Digitl ppendix. Pge 42 of 152

53 4.1.4 Method of nlysis The nlysis of the three MO s unfold in the following wy: Firstly; the gthered dt ws orgnized in tles s seen in the ppendix 1, which is orgnized in 5 columns: Short description of the mthemtics involved, didcticl moment, min plyer, mthemticl ojects involved, nd didcticl ctivities. This cts s oth first nlysis nd n orgniztion of the dt in order to further nlyze the teching process. Secondly; sed on the orgnized dt n nlysis tle, using ATD, ws conducted in ppendix 2. The tle consists of 6 columns: Lesson numer, type of prolem, mthemticl technique, technologicl theoreticl elements, didctic moment(s), nd elements of the didcticl techniques. Thirdly; description of the findings in the nlysis tle will e presented nd elorted. Fourthly; discussion of the teching process will e executed. Fifthly; some concluding remrks of the teching process will e estlished. With the nlysis of the three MO s generl discussion of them is done s seen in section 4.5. Pge 43 of 152

54 4.2 Hyperrel numers The teching of the hyperrel numers ws sed on the HMO s presented in section The HMO formultes wht is needed in order to use the hyperrel numers to introduce infinitesiml clculus ut it does not give ny inclintion of how to introduce this new set of numers. The following section descries how the plnned teching process ws developed from the lredy estlished knowledge to e tught nd n nlysis of sid teching Plnned teching In order to introduce the hyperrel numers the concept of numers ws rought up s n introductory suject, which included ll the normlly used sets of numers N, Z, Q, R. This suject is nother prt of the curriculum in the Dnish high school, thus mking it dvntgeous to uild on something tht lredy hd to e tught (Ministry of eduction, 213). The generl ide ehind teching the hyperrel numers ws never to mke construction of the numers ut s lredy explined n intuitive pproch hd to e found. As such, the sets of numers N, Z, Q, R nd the intuitive inclintion tht infinitely smll nd lrge quntities exists, mde the foundtion for how to introduce the hyper rel numers. To this end set of ctivities were mde which would end up with n intuitive cretion of the hyperrel numers s set. With this kind of introduction the stndrd prt of hyperrel numer would follow s logicl connection etween the rel nd hyperrel numers. The str opertion ws reduced to wht ws needed in order to write up correct mthemticl sttements, i.e. it ws introduced s wy to extend the domin of function such tht f(x + ) mkes sense, nd such tht the vlue could e found s the function evluted in quntity of 2 prts. In order to introduce the notion of infinitesimls nd infinite numers, the intuition were used. To this end n infinitesiml ws introduced y giving the students the tsk of letting their phone drop, nd consider the speed of the phone during its fll, especilly wht the first non-zero speed ws. The nswer to this question would e the eginning of the intuitive understnding of wht n infinitesiml is. To uild on this intuitive understnding the students were to mke definition of the infinitely smll (speed). In order to use the intuition to introduce the infinitely lrge quntities, the hunger tht only teens hve in the fternoon ws used. This ws introduced in the lst lesson of the dy, i.e. from 14:45-15:5. The students would surely e hungry, hence the pln ws to open g of potto chips nd sk exctly how much the students wnted the chips, to try nd get them to sy infinitely much ( commonly used Dnish wy to descrie something relly desirle). With this intuitive understnding, the students were to mke definition of wht the infinitely lrge (yerning) ws. With the definitions of the infinitesimls nd the infinite numers the students were then to produce numer line including the new found quntities, i.e. the hyper rel numers. With this introduction of the hyperrel numers the notion of continuity could e (re)creted y considering the wording of continuity: Continuity is something tht cn e drwn without lifting the drwing device, nd how close two points on grph should e if one ws told the function ws continuous. To e sure the students would get the right definition with this line of questioning, suquestions were dded: given point, x, wht is the point on the grph corresponding to tht? nd wht is the difference in function vlue for function, which cn e drwn without lifting the drwing device, when the difference in the vrile is infinitesiml?. Pge 44 of 152

55 In order to give n overview of the teching process of the hyperrel numers the following tle hs een produced. It consists of 4 columns the first of which indictes the lesson numer, dte nd time period, the second indictes the mthemtics to e tught, the third short description of how the mthemtics were to e tught nd lstly wht prt of the plnned teching ws ctully tught. Lesson # (14:45-15:5) (8:1-9:15) Mthemtics to e tught Plnned teching Executed teching In lesson # Numers, i.e. N, Z, Q, R Rel numer line. Infinitesimls Infinite numers. Numer line including infinitesiml nd infinite numers. Repetition of N, Z, Q, R, nd definition of infinitesimls. Numer line with infinitesimls Addition, sutrction, multipliction nd division with infinitesimls. Stndrd prt Whole clss discussion of numers nd the clssifiction of N, Z, Q, R. By sking the students wht numers re nd how they should e orgnized. An ssignment introduce infinitesimls: Drop your phone nd find the first nonzero speed. Students write down definition of infinitesimls. Infinite numers introduced s mesure for their crving for the chips rought. Students write down definition of n infinite numer. In which set of numers, N, Z, Q, R, should the new quntities e put? Then how should this new set look like? Drw numer line including infinitesimls nd infinite numers. The repetition ws done in plenum while the techer writes on the lckord. Drw numer line with infinitesimls s n exercise to the students in smll groups (of order 2-4). Ends with drwing of numer line including infinitesimls on the lckord. Plenum, with every opertion given s question to the students. Lstly the students re sked to drw infinite numers 1/, 1/ +1, on the numer line with infinitesimls. With the opertions for infinitesimls the students re sked to find the rel numers closest to 7 +, 4 nd. Techer defines the opertion s the stndrd prt, ending with the question, wht is st ( 1 )? 1 2 The Lesson ends with the students doing the first 5 exercises in compendium out hyperrel numers. Pge 45 of 152

56 Function, domin of function. Plenum questions; Wht is function, cn you give exmples, Wht is the domin of function? Techer writes/drws nswers on the lckord. Ending with the definition of function s connection etween two vriles. Continuity of function in point. Plenum questions: Wht is continuity of function? If the grph of the function cn e drwn without lifting the pencil, then wht is the smllest distnce etween 2 points on the grph? Rememer the difference in function vlue, s Δy = f(x + Δx) f(x)? Is x + prt of the domin for norml function? The str opertion is introduced s wy to extend the domin nd rnge of the function to. Str opertion (1:4-11:45) (converted) How to operte with hyperrel numers nd stndrd prts. Continuity They hd to do exercise 6-15 t home which were presented t the lckord y the students, if time permits it, they cn sk questions regrding the first hnd in ssignment. The techer gives n exmple of how to check if function is continuous in point. 3 (time did not permit questions for the hnd in) (8:1-9:15) Function evlution in quntity of 2 prts or hyperrel numer, stndrd prt, str opertion. Determine which numers re in N, Z, Q, R. Def. of infinitesiml nd infinite numers. Stndrd prt of hyperrel numers. Eqution solving. Continuity They do exercise 16-2 from compendium out hyperrels. Time to finish up working on the hnd ins. The techer cn nswer relevnt questions to the prolems in the hnd in (14:45-15:5) (converted) Determine which numers re in N, Z, Q, R. Def. of infinitesiml nd infinite numers. Stndrd prt of hyperrel numers. 1 minutes to red through the comments to their hnd in, then Plenum discussion of every exercise in the hnd in. 5 Pge 46 of 152

57 Eqution solving. Continuity While plnning the teching, the chllenge of producing relevnt nd intriguing tsks to introduce new set of numers ecme evident, since quntity is mostly theoreticl prt of prxeology. This is not chllenge which is solely evident when teching the hyperrel numers, the sme prolémtique cn e found in introducing the rel numers when only the rtionl numers re known. The prolémtique cn e further enlightened y the fct tht every kind of mthemtics is uilt on some concept of quntity, e it mgnitudes, rtios of mgnitudes s Euclid (Heierg et l., 28) descried them in his ook 5 or numers s perceived y lter mthemticins. To this end teching mthemtics without quntity is like speking without words. When imposed with the prolem descried ove, the first tsk tht cme to mind ws to sk the students to construct the hyperrel numers. This ide ws quickly ndoned, since the students of high school should never construct neither rel nor hyperrel numers. This limittion mde the ville constitutive tsks diminish from rther lrge lke to puddle with murky wter, mening, ll the mthemtics involved in constructing new set of numers vnished nd left ws only exercises similr to ordinry clcultions with quntities, such s ddition, sutrction, multipliction nd division. The constitutive tsks left were to define infinitesimls nd infinite numers. Pge 47 of 152

58 4.2.2 Anlysis of teching the hyperrel numers. A list of the types of prolems found during the teching process will provide n overview of the prcticl prt of the HMO tht ws ctully tught. The su-indices represent su-prolem of the prolem with the sme index. The constnts, nd k re rel, the vriles x nd y re rel, nd dy re infinitesimls, nd α nd β re hyperrel numers. HP 1 : Specify into which sets, infinitesimls nd infinite, numers elong. HP 2 : Determine if given function, f(x), is continuous. HP 2,1 : Given grph of function determine if the function is continuous. HP 3 : Find k st( x + + c dy) HP 3,1 : Find st(( + )(c + k dy)) HP 3,2 : Find st(( + ))st((c + k dy)) HP 3,3 : Find st ( 1 dy ) HP 3,4 : Find st( )st ( 1 ) HP 3,5 : Solve the eqution nd find the rel numer closest to x, where the eqution oth contins x nd HP 4 : Find the hyperrel function vlue f( + k) for given rel function HP 5 : Wht is st( f(x + )) for rel function? HP 5,1 : Wht is st( f(x))? HP 5,2 : Let f e continuous rel function, wht is st( f(x + ))? These prolems cn e clssified into the types of tsks from the HMO, which re presented in the tle elow. Types of tsks in HMO HT HT 1 HT 2 HT 3 HT 4 HMO= [T, τ, θ, ] In wht set of numers does numer x elong? HP 1 Find st() when R? HP 3 Wht is f(x + ) when is n infinitesiml nd f rel function? HP 4 Find st( f(x)) when, x R HP 5 Check if the rel function f is continuous (t point) HP 2 In the nlysis tle (ppendix 2.1) different prts of the prcticl lock elonging to HMO cn e oserved. An exmple of this, is the first encounter with HT 1, when the prolem HP 3 ws introduced. An explortory didctic moment hppens when the students estlish the technique of envisioning hyperrel numer s sum of rel numer nd n infinitesiml. Pge 48 of 152

59 This constitutes the technologicl/theoreticl prt of operting within the rel nd hyperrel numers. Through working with HP 3, nd its su prolems, the students do technicl work on the lredy estlished technique. This enles the students to get n intuitive understnding of the connection etween the rel nd hyperrel numers, through the stndrd prt. Therey constructing the following techniques to nswer prolems of type HT 1 : Hτ 1,1 : Hτ 1,2 : Hτ 1,3 : Recognize the hyperrel numer s rel numer plus n infinitesiml, i.e. st(α) = st(x + ) = x. If the hyperrel numer cn e seen s sum of two finite hyperrel numers then one cn tke the stndrd prt of ech of these numers, i.e. st(α + β) = st(α) + st(β). If the hyperrel numer cn e seen s product of two finite hyperrel numers then one cn tke the stndrd prt of ech of these numers, i.e. st(α β) = st(α) st(β). With this overview, the prcticl locks of the ctully tught MO nd the MO to e tught, HMO, re very close to one nother. The outcome of the tught MO is in prt for the students to e le to do infinitesiml clculus vi NSA. Another outcome is for the students to e le to determine if rel function is continuous, which lso serves s n exmple of the institutionliztion, nd prtly the moment of evlution. When the students use the following vrint of the nonstndrd definition of continuity, st( f(x + ) f(x) ) = for every infinitesiml, to check if function is continuous, it is ovious tht the theoreticl prt of the punctul MO s in HMO overlp with their previous knowledge out functions. One could even rgue tht y introducing set, then susets of sid set ecome more tngile, i.e. y introducing the hyperrel numers the rel numers re etter understood. Since the theory is mostly intuitive, the technologicl level comes down to logicl deductions. Bsed purely on intuition, mthemticl sttements cn e produced, which enles further mthemticl deduction. Tke the stndrd prt s n exmple. The students definition of this opertion is to find the rel numer closest to the hyperrel vlue, which leds to the fct, tht the stndrd prt of n infinite numer is meningless. This cn e extended to estlishing tht if the stndrd prt of frction with n infinitesiml in its denomintor exists, then the numertor hs to e n infinitesiml s well, i.e. differentile function is continuous. It could e rgued tht the technologicl level is not on pr with how every high school does it, ut this is prt of the curriculr restrictions, since some high schools might construct the rel numers. Furthermore, the notion of the susets of the hyperrel numers, N, Z, Q, which correspond to the susets of the rel numers, N, Z, Q, is not thoroughly treted. In the end, the teching of the hyperrel numers were enough to stisfy the students curiosity s to how the intuitive understnding of the world cn e explined nd sufficient to e used s foundtion for the infinitesiml clculus to e tught The didctic processes oserved while teching HMO First encounter with HMO s oserved is when the students intuitively understnd, tht the first non-zero speed of their dropped phone is less thn ny rel vlue, i.e. the lck of quntities to descrie the things the intuition cn perceive. This first encounter cn prcticlly e seen s the ovious choice. The infinitesimls re etter suited to e intuitively perceived thn the infinite numers, which is the only other possiility of first encounter with HMO, when not constructing the hyperrel numers. The intuition mkes it possile to discover something Pge 49 of 152

60 tht is not zero, yet less thn ny positive numer writle, wheres imgining something tht is greter thn nything writele is more difficult to visulize, since the end of the rel numers is not very welldefined. The infinitesimls re used more thn the infinite numers, when introducing infinitesiml clculus; even when the integrl is defined s n infinite sum it is n infinite sum of infinitesimls. Historiclly the infinitesimls were used in longer period of time thn the infinite numers, which cn e seen in Cuchy s Cours d nlyse (Brdley & Sndifer, 29). Cuchy descries the infinitesimls s sequences with limit zero ut his infinite numers only consist of plus or minus infinity (more or less the sme wy infinity is descried in stndrd nlysis). An explortory moment concerning the HMO, in prticulr the stndrd prt, cn e sid to hppen when the students figure out tht the technique for this is to remove infinitesiml(s); they re virtully zero! The constitution of the knowledge lock cn e seen to hppen when the students pply the opertions, plus, minus, multipliction nd division to the notion of infinitesimls. I.e. the constitution cn e sid to hppen when the students develop the concept of the rel numers including infinitesimls, thus constructing the hyperrel numers intuitively. By pplying wht cn e considered technicl work on the technique for the stndrd prt the students develop wht hs een denoted Hτ 1,2 nd Hτ 1,3. While the intuitive construction of the hyperrel numers is superstructure of the rel numers, ll the regulr sets of numers, including the rel numers, re still le to institutionlize the HMO s set of numers. The set of hyperrel numers, R, is governed y the sme rules tht pply to ll the other sets, N, Z, Q, R, which cn mke moment of evlution possile. This is seen to e true in the sense tht the opertions with the hyperrel numers do not llow for opertions tht normlly cnnot e executed Discussion (Hyperrel numers) With this introduction to the hyperrel numers the students cquire theoreticl knowledge, ut so fr the students only reson for lerning this suject is to e le to descrie wht their intuition descries. The NSA wy of defining continuity is nother exmple of reson for lerning this suject, ut the students do not seem to pprecite the definition on the sme level s techers nd other scholrs of mthemtic might do. The students simply lck the reson for why the chrcteristics of continuous function is so desirle, they re rrely going to use the continuity of function to prove something generl for continuous functions. The students encounter nigh on only the elementry functions (liner, exponentil, power, polynomil nd logrithmic functions, cosine, nd sine), which is why the clssifiction of continuous functions seems lmost superfluous, since most students perception of function is gined only y working on the elementry functions. When introducing the infinitesimls, wy to mke the tught MO more edile, could e to let the students do more work on them efore trying to get them to define them mthemticlly. This could mke some of the properties of the infinitesimls esier to ccept nd digest. One possile wy to introduce this would e to let the students dd ll the strting speeds of their dropped phones to see if it would end up eing rel speed. In the oservtions when the students were sked wht 2 is, fter their preliminry Pge 5 of 152

61 definitions, the nswer, tht it ws infinitely smll, cme so ruptly tht the pproch used y letting the students define the infinitesiml efore operting on it might not hve mde difference. With the reltive smll numer of different types of techniques to solve tsks where the stndrd prt needs to e found, more thorough routiniztion of the techniques, Hτ 1,2, Hτ 1,3, would hve een preferle. Finding the stndrd prt s explined efore is purely intuitive thing, ut the techniques re something more; this point should hve een mde clerer for the students. Especilly since these re the primry tools to prove rules of clcultion within infinitesiml clculus. Teching the hyperrel numers in generl ws very simple when done in the intuitive wy. This mde it esy to develop the techniques, ut just s esy to forget the importnce of. The numer of tsks underlining the importnce of the techniques were not sufficient to estlish the sought knowledge regrding the techniques, Hτ 1,2, Hτ 1,3. In conclusion the hyperrel numers is not chllenging to tech when introducing it through intuition. The oserved tught MO nd the HMO presented in the previous section re more or less equl. When looking t the hyperrel numers s tool to introduce infinitesiml clculus, it cn e concluded tht the theory is well understood in this high school clss. Pge 51 of 152

62 4.3 Differentil clculus (Jons Kyhnæ) The teching of differentil clculus ws sed on the DMO s presented in section The DMO formultes more thn wht is needed in order to fulfill the required curriculum for differentil clculus ut it does not give ny inclintion of how to introduce this suject. The following section descries how the plnned teching process ws developed from the lredy estlished knowledge to e tught Plnned teching Leding up to differentil clculus is typiclly course of different types of functions, inevitly covering liner functions. This ws the cse for the clss to tech s well. Furthermore, this ws one of the lst courses they went through, the only exception eing course on exponentil functions. Hence, the students hd firly recently worked lot with the 2-point formul, the 2-point formul eing the formul for finding the slope of liner function through two given points on the grph. As such, this ws used s stepping stone to the new knowledge, nmely the slope in point. To this end, two worksheets were prepred, oth of which were exercises with grphs. The first worksheet ws on finding the secnt nd the other on finding the tngent. Even though the slope in single point wouldn t immeditely mke sense for the students, they should get the intuitive understnding of this through these worksheets. In this wy, the more technicl work of finding the differentil quotient would ecome more comprehensile. The students were of course well-suited for finding the slope through two points nd thus the further implementtion of the mening of the slope in point ws, to get the students to estlish n understnding of how to find the est two points for the stright line to go through. Another worksheet ws prepred for this, which would led them to the definition of the differentil quotient. A ig prt of this ws for the students to not drw ny grphs t ll ut simply use their knowledge out stright lines nd tht otined in HMO. In doing this, the students find the definition of the differentil quotient themselves, sed on previously estlished prxeologies. This led to the possiility of creting severl generl prolems in finding the differentil quotient, eginning with series of prolems with hidden gend. Alongside finding the differentil quotient, given different vlues, the students should find the function vlues for the sme vlues. This ws oth for the students to etter get fmilirized with finding function vlues nd to see the connection etween functions which dded together give new function. The point of this eing for the students to recognize the pttern nd use it on the differentil quotients s well, to get n informl introduction to the differentil rule for summtion. To further link the work the students would hve done on grphs nd the lgeric clcultions on finding the differentil quotient worksheet on the conditions of monotony ws creted. This worksheet contined single question, simply to sy wht pplies to the originl function, when given the derived function. Here the students could use the knowledge lredy otined, s well s the internet, since no informtion ws given eforehnd, not even wht conditions of monotony mens. The worksheet ws constructed in such wy, though, tht if the students needed help, they would e le to get hints in the form of n exmple in different vritions eventully leding to generliztion nd hence the nswer to the originl question posed on the worksheet. Due to the mount of converted lessons tht needed to e incurred, one of the hnd-ins were to e used s kind of worksheet (Hnd-in #2). This would e wy of mking sure the students would get the intended knowledge, nmely ecuse they were to hnd it in nd the techer could oserve nd help t the sme time. This converted lesson ws to e put in somewhere where they could work on finding the generl Pge 52 of 152

63 derivtive for liner function, specificlly constnt term, nd polynomil functions. As such, the entire first of the two pges tht constituted the hnd-in were on finding exctly these. Also, the students were to give proper explntion of the difference etween slope nd differentil quotient. With the definition of the differentil quotient, the students would e le to prove the rules for differentition. Agin, worksheet ws creted, which they should e le to do with this self-chieved definition nd the knowledge from HMO. With certin kind of time issue in mind, the rules to e proved through this worksheet should e the product rule nd the chin rule long with how to find the derivtive of 1 x nd x nd lso the proof of how to find the eqution of the tngent. Thus, the decision of excluding the quotient rule (the derivtive of the quotient of two functions) ws mde. This ws s much due to time pressing concern s it ws the process of the proof eing irrelevnt, mening the method is the sme for stndrd nlysis s for non-stndrd nlysis. This worksheet should lso led up to the students proving the forementioned in front of ech other, y presenting one of them in groups. This s wy to get ech group to completely grsp the proof t hnd nd then show to the rest of the clss, insted of hlfhertedly going through them ll, which could e the cse given the mount of time ville. In extension, the students could get new kinds of prolems, contining the ove. In order to etter routinize the students with these types of prolems, prticulrly the chin rule, the derivtive for the function e x ws introduced, including its proof. The tle elow is to give n overview of the plnned teching. Lesson # (9:25-1:3) (8:1-1:3) (12:15-13:2) (8:1-9:15) (1:4-11:45) Mthemtics to e tught Plnned teching Executed teching in lesson # The slope of grph in point The secnt nd the differentil quotient The differentil quotient Conditions of monotony Through the grphs of ritrry functions nd Worksheet Tngent, the students should estlish n understnding of the slope of grph in point With the use of the hyperrel numers nd functions nd the 2-point formul on n infinitesiml intervl, the students should get to the definition of the differentil quotient, y working on Worksheet 4.11 Routiniztion y solving exercises in the compendium The students should go through the exercises on the lckord, one y one Through question out the conditions of monotony for function, when given the derived function, the students should give n nswer y using ny nd ll ids, including the internet nd hints from the techer Pge 53 of 152

64 (14:45-15:5) (8:1-9:15) (9:25-1:3) (1:4-11:45) (14:45-15:5) (8:1-9:15) (9:25-1:3) (converted) Work with hnd in 2 Conditions of monotony (converted) The derivtive of f(x) = 1 x nd f(x) = x, the eqution for the tngent nd the proofs for the product rule nd the chin rule A summry of wht hve een tught in differentil clculus Finding nd generlizing the derivtive of the liner function nd polynomil function nd solving text-sed exercises Routiniztion y solving exercises 8+9 Group work with Worksheet The students should present their product from lessons ten nd eleven in front of the clss; every memer should sy something The students should fill out worksheet 5.3. In the end, the students filled-out worksheets should e compred to generte finl summriztion sheet Pge 54 of 152

65 4.3.2 Anlysis of teching differentil clculus To give n outline of the prcticl prt of the DMO ctully tught, is here list of the types of prolems found during the teching process. These prolems, s well s multitude of su-prolems, re s seen in the nlysis tle (ppendix 2.2). The su-indices represent the su-prolem(s) of the prolem with the sme index. Those depicted in old re prolems to which there is generl technique to solving the exercises suject to this prolem. Su-prolems lso hve generl technique ut this might differ from the generl technique for the old prolems. DP 1 : DP 2 : DP 3 : DP 4 : DP 4,1 : DP 5 : DP 6 : DP 6,1 : DP 6,2 : DP 6,3 : DP 6,4 : DP 7 : DP 7,1 : DP 8 : Find the slope of given grph in point y drwing tngent Find n lgeric expression for finding the slope for function in point Find the eqution of tngent Find d xn Find d (x + ) Find d ex Show the rules for clculting with differentils Let h(x) = k f(x). Find h (x) Let h(x) = f(x) ± g(x). Find h (x) Let h(x) = f(x) g(x). Find h (x) Let h(x) = f(g(x)). Find h (x) Wht cn e sid out the originl function, f(x), given the derived function, f (x)? How to find the conditions of monotony for function, f(x), given its grph Wht hve you lerned out differentil clculus? The prolem DP 8 is more tsk of reproducing ll the techniques lredy estlished, thn it is type of prolem, ut is listed s such, since it simply covers ll. All other prolems re clssified elow, into the types of tsks from DMO. Types of tsks in DMO DT DT 1 DT 2 DT 3 DT 4 Define the differentil quotient. Find the derivtive, f (x), for function, f(x) DP 1, DP 2, DP 4, DP 5 Determine the condition of monotony for given function, f(x) DP 7 Find the tngent of function, f(x), through specific point, (x, f(x )) DP 3 Determine if d f(x) exists for given function, f(x) DP 6 is not present in the tle since this prolem is wht cn e considered s the technicl work done on the hypotheticl technique of the differentil quotient. The following is short description of how two of the more importnt techniques were developed through the looking glss of ATD. Pge 55 of 152

66 The first encounter with differentil clculus ws, for the students, the prolem DP 1. When given this prolem they enter n explortory stte, wherein they cn find grphicl technique for finding the slope in point, nmely y drwing tngent for grph through two points with n infinitesiml difference nd then finding the slope for this stright line. Thus, constitution of the theory comes with the knowledge of liner functions lredy estlished. With the technique of finding the slope in point eing only grphicl, new techniques re needed to clculte the slope for function in point. As such, nother constitution of the theory hppens, which includes previously estlished prxeology on the concept of function nd the recently otined knowledge of hyperrel numers, whence they cn estlish the differentil quotient. Hence, second technique is cquired for finding the slope in point. Further development of the differentil quotient occurs when the students do technicl work on the hypotheticl technique the differentil quotient cn e viewed s to estlish the rules of differentition, DP 6. Thus, the students further constitute the knowledge lock of DMO. The following shows the min techniques tht cn e found in the ctully tught MO. The remining techniques, which re more ovious from the prolems posed, cn e found in the nlysis tle (ppendix 2.2). Dτ 11 : Dτ 12 : Use the grph for function nd the formul for finding the slope for stright line through two points to clculte slope of the function in point Derive functions of the type f(x) = x n using f (x) = n x n 1 Tking into ccount the resons for Dτ 11 eing the wy of introduction to differentil clculus, s descried in plnned teching, not much explortion nor routiniztion of this ws necessry. Using previously estlished prxeology on how to find the slope of stright line given two points, mny students sw it s eing remrkly intuitive nd just more prctice in finding the slope for stright line. With this overview, the prcticl locks of the ctully tught MO nd the MO to e tught, DMO, re very close to one nother. As seen in the finished hnd-ins nd on the filled-out worksheet the students mnged to understnd nd use the intended knowledge, with only DT 4 not eing explicitly covered y ny type of prolem. The students relized prt of this themselves through reference to the HMO out tking the stndrd prt of n infinite numer, the prt eing tht the numertor hs to e n infinitesiml. The min prolem of teching differentil clculus ws finding the derivtive of function. From the very eginning of the teching course, the gol hs een to find the derivtives of different types of functions nd to find techniques for finding exctly these. As will e descried in the following this strted with trying to estlish the definition of the differentil nd use this s technique to estlish techniques which enles finding derivtives of clss of functions The didctic processes oserved while teching differentil clculus The first encounter with DMO is when the students relized tht it is possile to find slope in single point, insted of two. This hppened when going through n exmple where to egin with they should find somewhere on grph where it ws the steepest, even though they hd not efore seen other grphs thn liner nd exponentil, which led to n intervl of the grph wherein the grph ws t lest steeper thn other plces. This ws t first nrrowed down to smller intervl ut then student quickly discovered tht, quote there must e n infinitely smll section on the grph where it is steepest, hence the est intervl would e tht of infinitesiml length. As such, the first encounter with DMO uses HMO, which further justifies the reson for choosing such n exmple s n introduction to DMO. Pge 56 of 152

67 One of the explortory moments concerning DMO hppened when the students estlished the technique Dτ 11, hence working with DP 1. Given only the grphicl representtion of function, this enles the students to construct the technique using the knowledge of the mthemticl oject, tngent, nd how to find the slope of stright line. When the students comine these two, they find the slope for function in point. Though this technique is usle in solving ut few prolems, to let the students do explortory work like this, with purely grphicl depiction of function, it gve them rther relxed sight on the mtter. Seeing tht it ws so closely relted to previously estlished knowledge, very little ws needed from the knowledge lock of the DMO. The explortory moment of the hypotheticl technique hppened when the students explored different techniques for DP 2. Here, they ll ended up finding the sme technique, nmely tht of finding the slope through two points infinitely close to ech other, which is lmost the differentil quotient. The students were ll posed the sme question, which differed from DP 1 y excluding the possiility of drwing ny grphs. Hence, they were to find the slope for function, f(x), in point, without drwing nything! (see ppendix 1.2 worksheet 4.11) The students, knowing from their first encounter it hd something to do with stright line, sought to find the slope y using the 2-point formul on point nd the point tht ws only n infinitesiml wy. The technologicl knowledge necessry to explin wht slope through hyperrel point is, does not exist in the MO, so the students rely on the intuitive understnding of the slope going through point infinitely close to nother. To construct the new knowledge the students lso used the mthemticl oject functions. With the hints given (see ppendix 1.2 worksheet 4.11) long with the posed question, out the slope, they hd little troule using the estlished knowledge to get to f(+) f(), only few forgetting out the str. Every group of students needed help with the lst step efore getting to the ctul definition of the differentil quotient, nmely tking the stndrd prt. Some students only needed reminding tht the question ws out finding the slope in (single) point, others tht using the differentil quotient could result in hyperrel numer. Thus, the students reched the definition of the differentil quotient themselves. With this definition estlished, constitution of the knowledge for DMO hppened. An importnt point is tht when the students worked with DP 6 they did technicl work on this hypotheticl technique nd thus further constituting the knowledge. With the techniques from HMO they did technicl work on the differentil quotient s technique, this y using the estlished rules for the stndrd prt. Hence working with DP 6 contins two didcticl moments, nmely the constitution of technology for DMO nd technicl work with the hypotheticl technique. The technicl work done cn e seen in the description of teching (ppendix 1.2) whence the students estlished the rules for differentition. Concerning the chin rule, one of the students even went through the proof y sying the inner function could e seen s vrile nd sw st ( f ( g (x+)) f(g(x)) g(x+) g(x) ) = f (g(x)) s eing evident. In ddition, the differentil quotient ws used for estlishing other techniques for finding the derivtive of functions, which cn lso e seen in the description of teching (ppendix 1.2). Here the students found the technique to solve prolems of the type DP 4 y recognizing system, through inserting polynomil functions (of one term) of power 1, 2 nd 3. Hence, some of the technicl work on finding the derivtive for f(x) = x n ws done y routiniztion. The technique for finding the slope in point y drwing tngent Pge 57 of 152

68 nd finding its slope is lso constitutionl moment, e it smller one, since it is punctul MO where most of the knowledge lock is known. Further technicl work ws done on the technique for finding the derivtive of polynomil function, nmely y solving prolems of the type DP 4, s descried ove. The students elorted nd expnded the technique Dτ 12 to lso pply to 1 nd x. This in turn led to the students eing le to derive functions x with terms including vriles with rtionl exponents. The moment of institutionliztion ws when the students mde the connection etween the constituted knowledge, eing the derivtive, nd the knowledge out how to find slope in point, through tngents to grphs in coordinte system. Another importnt prt of the institutionliztion is the concept of function nd vrile. As such, the DMO elongs with the prxeologies out coordinte systems nd functions nd using functions s vrile Discussion (differentil clculus) With the introduction to differentil clculus eing uilt lmost wholly on the knowledge of functions, specificlly liner functions, mde it rther comprehensile for the students, since this ws the second to lst suject they hd gone through. This mde the students more comfortle heding into new suject in feeling tht they ctully knew the nswers to the questions tht ws supposedly going to led them to new knowledge. As such, this ecme motivtionl fctor nd even more explicitly with the otined knowledge from HMO, seeing tht this hd only very recently een tught to them. Hence, if they did not know the exct nswer to question, t lest they knew in which sector they should move round, mening they could nrrow down list of possile nswers significntly. In the cse of finding the est intervl on grph for drwing tngent, descried s the first encounter in the nlysis ove, it could e rgued tht the student only nswered the wy s/he did, since it ws the ovious nswer susequent to the uild-up in questions. Regrdless, the students were on ord nd thus the decision ws mde to skip the entire prt out secnts. This seemed s the est decision t the time nd even more so lter when severl students got confused when reding the compendium on the mtter (see ppendix 3 section 3.2). In ny cse, introducing the suject with grph descriing the ttery life for phone mde the students intuitive understnding of the grph more comprehensile thn if it hd just een grph out nothing. Hence, this gve them chnce to nswer questions, even if they hd difficulties decoding the informtion the grph would give. The thing tht could hve intimidted the students y this pproch ws the grph eing n ritrrily drwn decresing one, in tht the students hd only previously seen grphs for liner nd exponentil functions. This did not prove to e prolem, though, until getting to do vrious exercises leding up to the summtion rule: if h(x) = f(x) + g(x) then h (x) = f (x) + g (x) (see ppendix 3 pge 21). When finding the slope in different points, one peculir thing stood out. One student who got nnoyed with tking so mny steps efore reching result ecme convinced there hd to e simpler wy. The student rgued tht it would not mke sense to go through so much work for every new vlue nd/or every new function. Wht even further motivted the student ws tht the techer overlooked the student s work nd the student sked if the finl result ws correct. When the techer used couple of seconds to clculte the slope in the point nd then sid yes, the gol for the student ws to solve the exercise without Pge 58 of 152

69 writing nything from tht point. This resulted in the student relizing tht the infinitesimls of higher power thn 1 would dispper when tking the stndrd prt in the end nd s such could e ignored. This is prtly the procedure for the proof of the derivtive of the function f(x) = x n. Needless to sy, some students creted their own chllenges fter mstering the more elementry, like finding function vlues given different vrile-vlues, which ws the most common prolem for the students. Mny students struggled with this, ut finlly relizing the connection etween the vlues, mde them lmost forget how hrd it ws nd how much time it took them to find the function vlues which in turn gve the students personl victory. Here it ecme evident, though, tht the students needed prctice in finding function vlues, which then hd to mke up mny of the future exercises. While finding function vlues might not seem like it should e the iggest concern, it ws very demotivting prolem for the students, in tht if they hd troule finding function vlues how could they lern things on wht they felt ws much higher level. This is wht shped the course of conditions of monotony. As seen in the plnned teching (4.3.1) the course on conditions of monotony ws designed for the students to link grphs with lgeric clcultions, ut lso ecme wy of prcticing finding functions vlues. The conditions of monotony could hve een skipped completely, without losing ny significnt NSA points, ut ws ssessed to e useful for the ovementioned resons in spite of the potentil time prolem. Hence, the durtion of this ws prolonged s to help the students e less frid of inputting vlues in vrious functions. This decision ws mde during the teching process nd ws weighed s eing worth it for the students, to hone the skill for use in oth differentil nd integrl clculus. Even with the mny exercises in finding function vlues the students hd done, it ws just this, insert vlues nd find function vlues. As such, when proving the rules for clculting with differentils, hence working with functions without inputting numer vlues, the students skills cme up short. This ws when doing technicl work on the hypotheticl technique nd thus involving functions s vriles. Hd the students een more experienced with functions, more thn one might hve seen tht st ( f ( g (x+)) f(g(x)) g(x+) g(x) f (g(x)) ws the sme when considering g(x) s eing vrile. The extr chllenge of presenting the finl product in front of the rest of the clss motivted the students to mke sure tht every step of the proof ws fthomed completely. This ws done with the im t prcticing for the orl exm. With the hints given, nd in the form which they were given, mde the chllenge ll the more fun, since the students felt like they ctully did proof themselves. It should e noted tht these proofs were the first they did lgericlly, with no geometricl guidnce; the proof for the eqution of the tngent eing the only exception. This took the students round 15 minutes to prove, from strt to finish, mening from when they got the prolem until the techer cme to check on them nd they hd found the eqution for the tngent. No questions, no douts. In relizing tht = f (x ) gve the cler expression tht this ws extremely well estlished. Proving the rules for differentition might seem odd s finl moment of the teching process, ut the time constrints mde it impossile to get through everything relted to differentil clculus nd s such, things like optimiztion nd the mount of routiniztion were minimized or completely cut. This would normlly ly fter the proof of the rules, so the students would hve ll the necessry knowledge to solve the exercises given t written exm. ) = Pge 59 of 152

70 To improve the course of teching one would enefit from letting the students work more with (the concept of) functions, since the mjority of complictions were relted to this. There did not seem to e ny troule with introducing nd using NSA, in fct it seemed to e esier for the students to comprehend the prts including this. Thus, the knowledge lock estlished for differentil clculus using NSA proved to improve the students understnding. The concerns out whether or not the students nswered wht they thought the techer wnted to her turned out to mtter very little. It still tkes some reliztion nd in prt, n understnding of the prolem t hnd to even think of possile nswer. Tht sid, the technologicl nd theoreticl prt of the tught MO took up lot of the lessons, leving very little time to do prcticl work. The time ws the foremost restrint when introducing differentil clculus. Even though the students understood something one dy, they might not rememer it the next, especilly when not doing much prcticl work. Tking this into considertion the implementtion of DMO seems to e rther successful, with the entire knowledge lock covering the entirety of differentil clculus. Pge 6 of 152

71 4.4 Integrl clculus This section covers short description of the plnned teching s well s considertion of some of the more importnt prts of the nlysis tle. The teching ws sed on the MO estlished in section out wht to e tught in integrl clculus Plnned teching In order to introduce the integrl the commonly used introduction, y finding the re etween the function nd the first xis over n intervl, ws chosen. When plnning the teching process for this specific moment worksheet continuing the story tht is present in the compendiums s well, ws conducted. This ws done in order to mke the rther technicl work of finding the re etween grph nd the first xis more ppetizing. When introducing this prolem, one could use liner functions, which the students would e le to find the re for, using the geometry they lredy knew. In this scenrio, the intuitive understnding of wht the integrl is ws to e introduced to them efore the estlishment of wht the integrl or re functions for liner functions looks like. This ws done in order for the students to get fmilirized with operting with the integrl in wy tht is not too technicl. As such the students first tsk when introducing integrl clculus ws to figure out wy of descriing the re etween second degree polynomil nd the first xis. This pproch ws to ensure tht the result of the re were not within the students comprehension, hence forcing the students to focus more on the definition of the integrl thn on the ctul re presented. By compring the students results nd rguing tht the re should e rel numer the definition, ws to e found. With the definition of the definite integrl series of prolems where the students could find the vlue of the integrl ws conducted. As mentioned efore, this ws to fmilirize the students with the new found concept. The prolem of not eing le to clculte the infinite sum ws reduced to finding the re of some geometricl figures the students were le to find. With the definition of the integrl nd the use of some the more intuitive rules tht pply to it, the students were to prove the rules for clculting with integrls. To this end worksheet ws creted, which they should e le to do with the definition they produced themselves nd the knowledge from HMO. With the integrl nd the re under grph s intertwined s this introduction mde them, the connection etween the re function nd the integrl with vrying endpoint ws lredy estlished. The fundmentl theorem of clculus ws decided to e too difficult proof for the students to grsp in the rther short time tht ws ville. With this in mind the fundmentl theorem ws to e stted fter considertion of the infinitesiml difference in the re function, A(x + ) A(x). This is pproximtely the infinitesiml multiplied y the function itself, f(x). As such n intuitive understnding of the fundmentl theorem would hve to suffice for the students. The fundmentl theorem ws to e further cked up y series of exmples originting from the lredy estlished definite integrls of liner functions with vrying endpoint. The ntiderivtive of function ws then to e introduced s the function which when differentited would give the originl function. This would produce second chrcteristic, through the use of the fundmentl theorem of clculus, of wht the ntiderivtive is nmely; function which only differed from the re function y constnt. The indefinite integrl ws introduced s n ntiderivtive which is only determined up to constnt. Pge 61 of 152

72 The technique, integrtion y sustitution, ws introduced s something comprle to the opposite of the chin rule for differentition, hence it ws introduced through integrls which could e solved y recognizing the integrnd s the result of the use of the chin rule. The disc method for clculting cylindricl volumes of ojects ws not chosen to e tught, since time ws lcking resource, when teching the integrl clculus. The integrl clculus ws to e tught over period of 15 dys, ccounting for 11 lessons. This resulted in reduction in the time spent to prepre the teching nd the compendiums, since oth hd to e finished in the sme period of time. This eing sid, greter mount of time ws used to prepre the compendiums thn the ctul teching process, since the compendiums would ecome more or less the only textook mteril they would e le to find using NSA in high school. The production of hnd in #4 ws, ecuse of the time constrints, chnged to e used more or less s worksheet in order to mke sure the students got to do exercises tht covered the mteril needed to e tught, IMO. This choice ws sed on two considertions, firstly the students ecme less nd less engged in the teching the closer they got to the school summer rek, thus this chnge could intensify the students enggement, since the hnd in would e ssessed in order to grde the students, secondly this choice lso lessened the mount of time used to prepre the teching y not hving to produce worksheets on top of the hnd in. The tle elow will give n overview of the plnned teching, s with the teching of the hyperrel numers. Lesson # (8:1-1:3) (11:3-11:45) (14:45-15:5) (14:45-15:5) Mthemtics to e tught Plnned teching Executed teching In lesson # Intuitive understnding of the integrl s the re etween the grph nd the first xis in n intervl. Definition of the definite integrl, nd wht the oundries indicte. Definite integrls of liner functions with vrious endpoints nd re functions for sid functions. (converted) Definite integrls of liner functions with vrious endpoints The students work with worksheet 5.9 which consists of one min question of how to express the re etween the grph nd ove the first xis for second degree polynomil, nd some su questions to help them generte n nswer to the min question. A generliztion of the infinite sum from the previous lessons, which enles definition of the definite integrl. The students re to work with worksheet The worksheet consists of series of definite integrls of liner functions, which ends with questions of how to find n re function. This llows the students to estlish first technique of how to determine definite integrls of liner functions. In the end the exercise re gone through in plenum. Further work with the integrls of liner functions Pge 62 of 152

73 (1:4-11:45) (8:1-9:15) (9:25-1:3) Rules for clculting with integrls. Connection etween integrl nd derivtives, ntiderivtives nd indefinite integrl. Work with hnd in 4. Integrtion y sustitution. By using the definition of the integrl in conjunction with the estlished HMO the student re posed set of prolems concerning the rules of integrtion in the form of worksheet The prolems re presented with list of hints, which enles the students to work y themselves in groups of 3-5 students. A tle is drwn on the lckord with 3 columns, 1. column of functions, 2. column of integrls of the functions nd 3. column of derivtives of the integrted functions. With the tle nd the grphicl considertion of x+ x f(x) f(x) f(x) the fundmentl theorem of clculus is stted. The ntiderivtive of f(x) is then function F, for which F (x) = f(x). which y the fundmentl theorem is the sme s x F(x) = f(x) + k Finding indefinite integrls nd fixing them through specific points. Ending with using the found ntiderivtives to estlish definite integrls. Plenum questions nd the fundmentl theorem of clculus produces: (8:1-9:15) (9:25-1:3) (13:3-14:35) Work with hnd in #4. A summrize of wht hve een tught in integrl clculus. f(g(x)) g (x) = F(g(x)) + c With more questions the following is estlished f(g(x)) g (x) = g() f(u) g() du The students re then set do couple of prolems where this technique is used. Determining indefinite integrls where the use of integrtion y sustitution is needed. The students work on worksheet 5.24, which is worksheet contining two lnc lists, one to e filled out with functions nd its ntiderivtives nd second to e filled out with the rules nd techniques for integrtion. In the end the students filled out worksheets were to e compred in plenum to generte finl summriztion sheet Pge 63 of 152

74 4.4.2 Anlysis of teching integrl clculus. The nlysis of the integrl clculus will egin with visuliztion of the prcticl lock of the knowledge ctully tught. The nlysis tle (ppendix 2.3) includes list of prolems nd su prolems; considertion of these prolems will mke n overview of the ctully tught prcticl prt of IMO. The list contins prolems in old nd su prolems which re not. The old prolems re prolems where there exists some unifying technique to nswer the questions. The su prolems lso hve generl technique to nswer those, which my vry from the generl technique of the old prolems. IP 1 : Find the re etween the grph of function nd the first xis over given intervl. IP 2 : Find f(x) IP 2,1 : Find c. IP 2,2 : Find kx.. IP 2,3 : Find f(x) when f(x) is n elementry function. IP 3 : Find n ntiderivtive of f(x) x IP 3,1 : Find kx + c. x IP 3,2 : Find f(x) IP 3,3 : Find n ntiderivtive of n elementry function. IP 3,4 : Find the ntiderivtive going through point (x, y )? IP 4 : Show the rules for clculting with integrls. IP 5 : Find d f(x). IP 5,1 : Find d x f(x) for n elementry function f(x). IP 5,2 : Find d f(x) for n elementry function. IP 5,3 : Wht is df? IP 6 : Find f(x)? IP 7 : Find f(g(x)) g (x). IP 8 : Wht hve you lerned out integrl clculus? These prolems cn e clssified into the types of tsks which re presented in the IMO. This is done in the following tle. Types of tsks in IMO IT Define the integrl. IP 1 IT 1 Clculte f(x) for given rel function f(x). (IP 1 ), IP 2, IP 7 IT 2 Find the ntiderivtive, F(x), for given function, f(x). IP 3, IP 6. IT 3 Determine if f(x) exists for given function, f(x). Pge 64 of 152

75 As seen in the tle the generl prolem IP 5 is not listed. This is ecuse it is (mostly) prt of the DMO nd not IMO. Furthermore IP 4 is not present either since this prolem is wht cn e considered s the technicl work done on the hypotheticl technique of the integrl. The finl prolem IP 8 is not present in the tle either since this prolem is relly no type of tsk ut more sitution for the students to tlk out nd unify the techniques nd rules found while eing tught integrl clculus. By looking t this list of prolems nd how they correspond to the types of tsks from IMO, first impression of the prcticl lock of wht hs een tught is comprehensile. In order to do further nlysis short description of how two of the min techniques estlished in the teching ws developed. This description cn e seen s n exmple of how the nlysis tle cn e deciphered. When the students re posed the prolem IP 1, the students enter n explortory moment in order to estlish the hypotheticl technique to find the re with n infinite sum of rectngles with infinitesiml width. Furthermore constitution of the theory hppens when the students compose the knowledge of functions nd the knowledge of the hyperrel numers to estlish wht the integrl is. With the integrl s hypotheticl technique, new techniques in order to clculte the integrl hve to e estlished, thus nother explortory moment hppens when the students encounter the prolems from IP 2,1 nd IP 2,2. With the techniques for IP 2,1 nd IP 2,2 nother constitution of theory hppens which includes geometry from previously estlished prxeology. When the students work with IP 4 they do technicl work on the hypotheticl technique of the integrl, which further constitutes the knowledge lock of IMO. With the knowledge otined from the punctul MO generted y IP 3,1 suggestion of the fundmentl theorem of clculus is otined when the students go through n explortory moment while working on IP 5,1. In the explortory moment of working with IP 2,3 the students construct second technique for clculting definite integrls. In the end the following techniques for IT 1 cn e found in the ctully tught MO Iτ 11 Iτ 12 Use the grphicl representtion of the function to recognize the integrl s geometricl figure, nd the formuls for the res of the geometricl figure to clculte the integrl. Use the definition of the ntiderivtive nd the rules for clculting with integrls to clculte the integrl s the difference in the ntiderivtive evluted in the end points of the intervl. It should e noted tht thorough explortion nd routiniztion of Iτ 1,2 is difficult to find in the ctully tught MO; the numer of prolems where the technique is needed is s little s eight. Since n explntion of how the students get to the ctully oserved techniques cn e found just s well in the nlysis tle (ppendix 2.3) the reminder of the techniques oserved will e left out, most of which re esily derived from the prolems posed. As such when considering the oserved prxis lock in the teching process the mthemticl orgniztion developed is not s close to the descried IMO s hoped for. The ctully tught MO contins most of the techniques, nd most of the knowledge lock seen in IMO, ut some things re missing. As lredy explined Iτ 12 ws one of these where it cn e rgued whether or not the students ctully otined this knowledge when only qurter of the clss independently showed rel nd correct use of the technique, though lmost ll of the students did replicte use of the technique. Another thing to notice is tht even though the students determine the rules for clculting with integrls, the smll numer of prolems posed Pge 65 of 152

76 where such rules re used, cn mke the rules seem inconsequentil. Around hlf of the time ws used on the theoreticl lock, when considering the explortory moment of estlishing the integrl s hypotheticl technique prt of it. The min prolem oserved during the teching process ws for the students to clculte definite integrls. This is understndle since it will llow the ntiderivtive, nd in prt indefinite integrls, to e prt of the technique to clculte definite integrls The didctic processes oserved while teching integrl clculus The first encounter with IMO cn e considered in two wys. Firstly, the students re posed the prolem of determining wy to find the re etween (the grph of) polynomil nd the first xis over n intervl, i.e. IP 1, which cn e viewed s the constitutive tsk IT. Secondly, if IP 1 is not to e viewed s type of tsk, the first encounter would occur when the students clculte the definite integrl of liner functions. This in spite of IP 1 lso including prolem which would e considered prt of IT 1. The students re le to clculte the integrl of liner functions using previously estlished prxeology, which supports the use of this type of prolem s first encounter. The explortory moment concerning the hypotheticl technique of the integrl hppens when the students explore different techniques to nswer IP 1. In the end the students unify round the technique of the infinite sum of rectngles with infinitesiml width. This technique includes su techniques, (hyper)finite prtition nd (hyper)finite sums. Both of these su techniques re mentioned nd used with no rel explntion of how they work. The technologicl knowledge used to explin wht exctly hyperfinite sum entils is not present in the tught MO. The explntion of wht the hyperfinite sum is depends on the construction of the hyperrel numers. By not explining the hyperfinite sum the students revert to the intuitive understnding of finite sum. This understnding is not fulty since the hyperfinite sum shres most of the properties of finite sum, yet it is still infinite. The (hyper)finite prtition of the intervl ws never explined y the techer, since t lest 4 out of 5 of the student groups hd come up with this technique y themselves, thus the understnding of the prtition relies solely on the students ilities to generte this mthemticl oject. It should e noted tht severl of the students did strt out y trying to find the re y finite prtition, which would seem s the eginnings of wht could e clled the technique of numericl integrtion. Since this technique ws never developed through prolems supporting the technique, rel elortion cnnot e sid to hve hppened. The explortory moments concerning estlishment of the technique, Iτ 11, only use the grphicl representtion of the integrl to enle the use of geometricl technique to clculte the integrls. This enles the students to construct the technique with reltively smll knowledge lock of IMO estlished. The explortory moment which estlishes Iτ 12 is lmost unnoticele in the sense, tht even though the technologicl elements needed to estlish this technique re presented y the techer, the technique itself seems to e mentioned s n fterthought. Furthermore the prolems where this technique is needed re eing posed to them fter lot of work done on indefinite integrls, which does not help the students understnd the presented technique ny etter. On top of this, mny of the students misunderstnd the first prolems posed where this technique is needed. Coupled with the smll mount of prolems this enles the students to dodge the experience of first encounter nd explortory moment for this technique. The explortory moment of estlishing n ntiderivtive hppens when the students re Pge 66 of 152

77 presented with the tle on the lckord contining the function, re function nd the derivtive of the re function. As such the technique ws estlished in plenum. A constitution of the knowledge for IMO hppens when the definition of the integrl is done. This definition, lso viewed s hypotheticl technique, enles further constitution of knowledge through the technicl work done on the integrl while working with IP 4. The technicl work on the integrl (s technique) is done y mens of the techniques otined in HMO for the stndrd prt, projected on to the integrl (s technique). Other smller constitutionl moments cn e oserved y considering some of the punctul MO s tht mke up the IMO. Constitution of the theory for the punctul MO of clculting the integrl of liner functions through geometricl pproch cn e seen s one of these. Another is when the constitution of the theory to enle the construction of Iτ 12, s seen in the nlysis tle, is sed on the fundmentl theorem of clculus (ntiderivtives) nd some of the rules for the integrls f(x) + c f(x) = c f(x) nd f(x) = f(x). The rules re determined when the students work on IP 4. As seen in the description of teching integrl clculus (ppendix 1.3) the students re le to estlish the rules which cn e seen s techniques in order to rewrite the (hypotheticl) integrl. It should e noticed tht the work done with IP 4 cn e viewed s two didcticl moments t once: constitution of technology for the IMO nd technicl work with the hypotheticl technique of the integrl. The fundmentl theorem of clculus is hinted t through tle of differentil quotients nd re functions nd their opposite reltion. This tle is composed t first only y liner functions, where the students re le to determine the re functions, nd thus lso the derivtive of sid re functions. The students re then le to recognize the system s the opposite connection of re functions nd derivtives. Bsed on this, the fundmentl theorem of clculus is guessed ut the explntion of why it is true escpes the x+ x students understnding when going eyond the intuition tht f(x) f(x) f(x). As such the technologicl level supporting Iτ 12 cn e rgued to e prtly missing. The technicl work done on the techniques otined during the teching process concerning other techniques thn the hypotheticl is not very ovious, though some technicl work is done on the technique to clculte the definite integrl of liner function. This enles the students to write generl formul descriing how to find the re of liner function nd extend it to formul which descries the re function. The rule of the sum of intervls (with common point) for integrls is not used on the techniques discovered for integrls of liner functions, which otherwise would hve enled n erlier elortion of the technique Iτ 12. The technicl work done for finding the ntiderivtive is oserved fter the definition of the ntiderivtive nd the fundmentl theorem of clculus. Here the students use the generl technique of finding the ntiderivtive of power function, f(x) = x n, to estlish numer of indefinite integrls, thus doing technicl work y routiniztion. Another profound moment of technicl work is when the integrtion y sustitution is presented. This cn e seen s technicl work done on Iτ 12. Moments of institutionliztion hppen when the students connect the constituted knowledge, ttined y defining the integrl, to the geometricl knowledge out tringles nd rectngles descried y functions nd the first xis. Another more profound institutionliztion hppens when the connection etween the DMO nd IMO is estlished, where the IMO finds its resting plce mong the prxeologies concerning functions, coordinte systems, nd mnipultion with functions s vriles. Pge 67 of 152

78 Discussion (integrl clculus) In this section, considertion of the oserved ctully tught knowledge, s descried in the nlysis, will commence. While the introduction to the integrl clculus, tht strted with something close to the constitutive tsk IT, cn e sid to hve een rther successful, there re some points to e mde out the introduction. An introduction of the summtion sign nd some opertions with finite sum could hve een used to introduce the integrl in pce, which etter suit high school students. This would lso hve supported the technique of numericl integrtion spontneously used during the explortion concerning IP 1. By doing this the students could hve lso een introduced to the upper nd lower sums if need e. By introducing the numericl integrtion the students would lso hve een le to clculte integrls of other functions thn liner efore introducing the fundmentl theorem. When the students work with the geometric technique of estlishing integrls of liner functions, some might rgue tht the technique is too nrrow to e worth using the time on. Even though the technique is very limited in its uses the introduction of something the students cn chieve nd understnd rther quickly mkes the students more t ese when working with the definite integrl. Before presenting the students with the prolems of the rules of integrtion, n introduction to some of them would mke sense. This could e in the form of prolems tht hint t the rules, which would lso ck the technicl work on the hypotheticl technique of the integrl with plusile inclintion of the rule efore estlishing sid rule. The generl proof, s done y the students, cn mke it difficult for them to notice in exctly which situtions the rules inferred, re eneficil. I.e. With the few numer of prolems, supporting the rules for clculting with integrls, the students hve hrd time recognizing when to use the them. x By defining the ntiderivtive s f(x) + k, then Iτ 12 cn e suggested efore the students re presented with the fundmentl theorem of clculus. This would further ck the use of the rules of integrtion s something which is ctully used to estlish Iτ 12. As presented in the oserved teching process the estlishment of the technique, Iτ 12, which lcks foundtion in the form of numer of prolems supporting the technique, is done lmost entirely through presenttion of technologicl elements. Even with the ntiderivtive defined s the opposite of derivtion the numer of prolems where the students hd to find n ntiderivtive did ensure tht the students ctully estlished the techniques, presented t the lckord, themselves. Moving the definition of the indefinite integrl to e the lst thing presented to the students would only mke little chnge in the tught MO. By defining it s n ntiderivtive, with n unspecified constnt dded, the indefinite integrl is only little dd-on to wht the ntiderivtive is. With this definition the connection etween the indefinite integrl nd HMO is very limited, nd it could e done y techer with no knowledge of NSA. In the sitution tht the teching of integrl clculus did not hve to e finlized during the teching process two cndidtes for things to remove is integrtion y sustitution nd the indefinite integrl which is hinted t ove. By removing these things from the plnned teching nother more pprecile thing could hve een included, nmely the first prt of the fundmentl theorem of clculus s presented in Pge 68 of 152

79 section Removing the technique of integrtion y sustitution is it of loss, when considering tht isolting in the expression of the derivtive df is mthemticl correct opertion. By removing the indefinite integrl the only thing tht would chnge would e for the students to lern wht the indefinite integrl is, ut defining it s n ntiderivtive with n unspecified constnt dded, does not hve nything to do with NSA. As such, this would not cuse prolems for ny future techer to introduce it like this. This eing sid, the proof of the first prt of the fundmentl theorem of clculus would e dole for the students with the right mount of guidnce. The fundmentl theorem of clculus, even when reduced to only hving proved the first prt, would enle the students to hve theoreticl lock for IMO on pr with wht is needed in the eginning of mthemticl university study. This is true only ecuse the elementry functions, which is directed y the government, re ll continuously differentile, which mkes the first prt of the fundmentl theorem sufficient. The min prolem leds to two prolems which re not ddressed in the teching process: when does definite integrl exist, nd is the technique through the use of ntiderivtive relly justified? Through HMO the definition of the integrl is mthemticl correct up to the point of the intuitive understnding of the hyperrel nd rel numers. When trying to nswer when the definite integrl exists, the only wy for the students to nswer this would e to sy tht the infinite sum hs to e finite ut other thn it eing in the compendium, this is never mentioned. This definition is correct, ut since the students were never posed with prolem where the infinite sum ctully gve n infinite numer, the students never constructed technique of how to check if the infinite sum ws finite or not. The justifiction of the technique Iτ 12 is nother mtter of deliccy, the justifiction oils down to the proof of the fundmentl theorem of clculus. The proof of this is only hinted t, mking the justifiction of the technique unsound. In the generl description of the students work done on the first pge of hnd in #4 (ppendix 1.3) it is seen tht only out hlf of the clss used the technique, Iτ 11, correctly when using it on n integrl of liner function. A first conclusion would e to ssert tht the technique ws not relly estlished y the students. When looking over the students nswers the tipping point of the wrongfully usge of the technique ws determined to e tht the students thought of ( ) 2 = ( 2 2 ) s eing true. This ws never prt of the plnned teching nd cn thus e sid to e nother mistke concerning the plnned teching ut s descried in the context the students were supposed e wre of the hierrchy of mthemticl clcultions. The hierrchy of mthemticl clcultions ws thought to cover this distinction, ut ls! When the students worked on the rules of integrtion, this mkes nother reson tht supports the introduction of the summtion sign. This showed when some of the students chose to introduce nother wy of writing the infinite sum. For the students to e le to rewrite the definition, ecuse they thought it tiresome to write the entirety of the sum, they hd to understnd the definition of the integrl thoroughly. This conclusion ws supported when their orl explntions of the sme prolems ctully did encompss the entirety of the sum. In the lesson where the ntiderivtive ws introduced one of the students found the nswer to question posed in plenum of wht the ntiderivtive of x n ws. At first when the question ws sked the student did not recognize the rule from the previously estlished ntiderivtives of the functions f(x) = kx nd f(x) = x 2. The techer then opted to sk for the rule of differentiting the sme expression, which might Pge 69 of 152

80 hve helped the student who found the result, since s/he then rised his/her hnd nd sid: Well I would 1 sy 1 n x no, n+1 xn+1. As such even though the students generlly did not relly grsp the technique IT 12 the students ll fmilirized themselves with the technique descried ove to find the ntiderivtive of power functions. With the considertion of the time used working with the technologicl, nd in prt the theory level of the MO, it might not come s surprise tht the students re le to explin why certin technique is vlid ut hve prolems when hving to produce nd use technique in order to solve prolem. As such, the knowledge lock of the tught MO cn e sid to e etter understood thn the prxis lock. Considering tht most of the things which could e chnged to enhnce the teching process re wht cn e considered situted in the prxis, with the exception of the proposed inclusion of the proof of the fundmentl theorem. The prts of the teching process tht cn e clssified s well executed nd understood y the students were minly the technologicl nd the theoreticl prt of the tught MO. As such even though the teching of the integrl clculus could hve een executed differently nd perhps etter, s descried ove, the thing to tke from this is tht using the NSA pproch does not seem to hve ny detrimentl effects. On the contrry, the use of NSA enles knowledge lock for the integrl clculus which is etter suited for students who does not need to construct the numers. Finlly the mount of time tht ws used to introduce nd estlish the IMO cn e rgued to not e in the llprk of wht is usully used on integrl clculus in high school. As such it cn e sid to e success in itself to estlish knowledge lock which cn justify (lmost) everything in integrl clculus. Pge 7 of 152

81 4.5 Oservtions of the teching process When teching infinitesiml clculus using NSA certin things ecme evident during teching. The importnce of the rules of the stndrd prt, prticulrly the stndrd prt of product, did not come through s noticely s needed. The students knew the rule ut rrely considered its restrints, leding to n insufficient rgumenttion in certin proofs. The rule for tking the stndrd prt of the product of hyperrel numers, could hve een further improved to incorporte the cse of the stndrd prt of the frction of hyperrel numers. While this rule cn e esily inferred from the rule for products this ws not done, s such the students hd prolems explining why they could tke the stndrd prt of the nomintor nd the denomintor seprtely. The hyperrel extension of the numer sets, N, Z, Q, could e included in the teching of HMO. This could improve the students understnding of the hyperrel numers, R. As when introducing the susets for the rel numers, N, Z, Q, etter understnding of wht the rel numers is otined. The sme cn e sid out the hyperrel numers, when introducing the sets N, Z, Q to the students. This introduction would lso fcilitte knowledge which is closer to the scholrly knowledge of the hyperrel numers, in tht the introduction of susets of given set mkes it possile to descrie sid set, which will mke the set more comprehendile. I.e. the numer of different infinitesimls might not seem cler t first, ut could e if the students understood the infinite integers, since tking the reciprocl of them will produce n infinitesiml. The specifiction of the infinite sum s eing hyperfinite, when defining the integrl, could hve een done with the ovementioned introduction of the hyperrel extension of the nturl numers, N. With this extr specifiction the students would e etter suited to nswer questions out when the integrl exists. A generl considertion of the overll teching s oserved would suggest greter numer of prolems/exercises to support the techniques. This is especilly trnsprent in the teching of integrl clculus, s such the prxis lock could e sid to need elortion. In extension of this it cn sid tht the teching hd hevy degree of work on the knowledge lock, which cn humle even the most mthemticl intelligent high school student. Since the techings of the theoreticl prt of the infinitesiml clculus hd to e finlized during the teching process the worklod could not e moved from the knowledge lock to the prxis lock in the desired degree. In order to try to circumvent this prolem lot of time ws spent on mking the teching process s funny s possile. This ws done y listening to the students wy of tlking nd trying to replicte some of the things, which hopefully mde the students think it s funny s the cretors did. It is lwys difficult to get n ojective opinion out the techings experienced y the students. Tht eing sid one of the students exclimed you should write children s ooks fter reding worksheet 5.9. The students commented severl times on the funny fetures of the hnd ins tht mde the mthemtics more edile. When this study commenced in the spring clss to which infinitesiml clculus could e tught hd to e found. Since most of the high schools tech students infinitesiml clculus in the fll of their second yer it hd to e tught to clss of first yer students. As such the ctul pln of order for the sujects to e tught ws rerrnged to introduce infinitesiml clculus much sooner. This cme with some limittions. The numer of functions the students hd t their disposl when strting the teching process ws t Pge 71 of 152

82 minimum, which mde it difficult to see the merits of some of the rules for differentition nd integrtion, in prticulr integrtion y sustitution nd the chin rule, nd possily product rule, for differentition. With the 3 lessons ville for the study, 25 of them could e used to introduce new mteril. It could e rgued tht teching differentil clculus to this prticulr clss of students might hve een possile within this time frme, including even more thorough prcticl routiniztion nd thus estlishing the entire DMO. As such, teching oth differentil nd integrl clculus on the other hnd proved to leve some of DMO nd lot of IMO unfinished. In the cse of future study, relistic wy of estlishing the entire DMO would e to prolong the teching course with 3 more lessons of 65 minutes ech, s descried in section After proving the rules for clculting with differentils, the students would then use 1 lesson for doing exercises in the product rule nd the chin rule. They would lso need routiniztion in finding the eqution of tngent through point nd with the possiility of comining this with the forementioned this would tke up 2 lessons. Yet nother lesson would e to tech the students out the (everydy) uses for differentil clculus prticulrly with the hedline optimiztion. To etter round off integrl clculus, 6 more lessons would e relistic mount of time needed to cover the mteril in n pprecile wy. 1 lesson would e needed to etter introduce the rules for clculting with integrls. With the right worksheet, it would e enough for the students to do the proof of the first prt of the fundmentl theorem of clculus, this tking up 1 more lesson. To estlish the rule for finding the ntiderivtive of function of the type f(x) = x n would tke 1 s well. In the end, 3 more lessons would e needed to introduce the disc method nd routinize the techniques in IMO. The need of textook ws evident. Minly ecuse of the rules set y the Dnish high school regultive, the students needed some reding mteril in Dnish, which led to the cretion of the compendiums. Since the hyperrel numers re not prt of the curriculum, the students hd no wy of getting help from fellow students or outside of the clss, such s t home. With the compendiums they t lest hd chnce to explin the theory with textook to ck it up, hence mking it possile for people with no prior knowledge of NSA to get n understnding of the mtter. Even though this is true the students would still e le to get help with the more prcticl techniques, when the theory ehind the definitions of differentil quotient nd integrl re not used. Thus the textook mteril nd the things just explined could potentilly led to outside help with infinitesiml clculus in spite of the different pproch. Pge 72 of 152

83 5 Conclusion The im of this thesis ws to elucidte resons for or ginst nonstndrd nlysis pproch to infinitesiml clculus in high school, hence nswering the reserch questions. 1. Wht re the resons for or ginst the nonstndrd pproch, to infinitesiml clculus in high school? 2. How cn nonstndrd introduction to infinitesiml clculus in high school e developed, in prticulr how to crete textook mteril suited for high school students? 3. Wht results cn e oserved from first experiment, implementing such mteril? As descried in section the nonstndrd nlysis pproch is very intuitive introduction to infinitesiml clculus nd cn therefore e considered s (nother) theory tht cn e used to introduce this. Since it mkes it possile to evde the introduction of the limit opertion (or ε, δ rguments), the prolems tht come with it re lso evded. With the intuitive introduction of the hyperrel numers the stndrd prt is nturl extension of the definition of the infinitesimls. Furthermore it is possile to determine, with logicl deduction, the rules for the stndrd prt which re needed in order to prove the rules for how to clculte with the differentil quotient nd integrl. Thus, the knowledge lock cn e estlished to greter extend when introducing infinitesiml clculus with nonstndrd nlysis, s opposed to the limit opertion definition. The nlysis showed tht n intuitive introduction to the hyperrel numers is fesile for students in high school. By introducing the hyperrel numers the students re le to descrie wht they cn perceive in the world round them, in prticulr the students re le to tlk out evnescent quntities in mthemticl coherent wy. With this intuitive introduction to the hyperrel numers the students re le to further indulge their intuition nd logicl resoning to (re)produce the infinitesiml clculus, much in the sme wy s ws done when it ws first discovered y Newton nd Leiniz. Furthermore, the students re le to prove the rules for clculting with differentil quotients nd integrls which re normlly difficult for students in high school. As such the knowledge lock is composed of more intuitive nd mthemticl coherent sttements, which in the end mkes it esier understood. Also, when intuition nd logicl deduction is used to do mthemtics, prctitioners experience feeling of self-discovery, thus engging them into further study. The high school mthemtics is significnt step up in difficulty from prior teching estlishments. As such the intuitive nd more logicl theory lessens the intriccy of one of the most chllenging sujects, infinitesiml clculus. While ll of this is true, nonstndrd nlysis does not mke the infinitesiml clculus diminish in intriccy to the point of mking it n esy suject. As seen, nonstndrd nlysis definitely reduces the complexity of infinitesiml clculus, ut not to the degree of justifying the introduction of oth differentil nd integrl clculus within the timefrme imposed. Besides the implementtion of the compendiums eing requirement it lso showed gret pprecition mongst the students. With the inclusion of the things the students experienced in clss nd the expressions they cme up with long with dsh of humor nd recurring story throughout, which ws only possile with the construction of the compendiums eing done during the course of teching. This comined mde the implementtion of the compendiums success. Pge 73 of 152

84 The only reson for not using nonstndrd nlysis is the conservtism mongst mthemticins, in tht it is not wht is normlly used. Hence, the prolem with nonstndrd nlysis comes down to the cse of students pursuing higher scientific eduction, where then the theory is not used. In conclusion, there re plenty dvntges for using nonstndrd nlysis to compenste the downsides of it eing uncommon. In the end, it is clled infinitesiml clculus nd not ε-δ rgument clculus or limit-opertion clculus, so why not use the quntities which gve rise to the nme? Pge 74 of 152

85 Biliogrphy BrÉ, J., Bosch, M., Espinoz, L., & GscÓN, J. (25). Didctic Restrictions on the Techer s Prctice: The Cse of Limits of Functions in Spnish High Schools. Eductionl Studies in Mthemtics, 59(1 3), Brdley, R. E., & Sndifer, C. E. (29). Cuchy s Cours d nlyse. New York, NY: Springer New York. Retrieved from Clusen, F., Schomcker, G., & Tolnø, J. (26). Gyldendls gymnsiemtemtik. grundog B [Bind] 2 (1. udgve 3. oplg). køenhvn: Gyldendlske Boghndel, Nordisk forlg A/S,. Criticism of non-stndrd nlysis. (215, Octoer 2). In Wikipedi, the free encyclopedi. Retrieved from Cutlnd, N. (Ed.). (1988). An Invittion to nonstndrd nlysis. In Nonstndrd Anlysis nd its Applictions (p. 15). Cmridge: Cmridge University Press. Retrieved from Gry, J. (215). The Construction of the Rel Numers. In J. Gry, The Rel nd the Complex: A History of Anlysis in the 19th Century (pp ). Chm: Springer Interntionl Pulishing. Retrieved from Heierg, J. L., Fitzptrick, R., & Euclid. (28). Euclid s elements of geometry: the Greek text of J.L. Heierg ( ) : from Euclidis Element, edidit et Ltine interprettus est I.L. Heierg, in edius B.G. Teuneri, Plce of puliction not identified: pulisher not identified. Ktz, M., & Tll, D. (211). The tension etween intuitive infinitesimls nd forml mthemticl nlysis. Crossrods In The History Of Mthemtics And Mthemtics Eduction, 12, Keisler, H. J. (212). Elementry clculus: n infinitesiml pproch (3rd ed., Dover ed). Mineol, NY: Dover Pulictions. Ministry of eduction. Læreplnen, Pu. L. No. BEK nr 776 (213). Retrieved from Pge 75 of 152

86 Ministry of eduction, G. fdelingen. Mtemtik A - Stx Vejledning / Råd og vink (21). Retrieved from O Donovn, R., & Kimer, J. (26). Nonstndrd nlysis t pre-university level: Nive mgnitude nlysis. Nigel J. Cultnd, Muro Nsso nd Dvid A. Ross (Eds.), Lecture Notes in Logic, 25 Nonstndrd Methods nd Applictions in Mthemtics. Retrieved from Ponstein, J. (21). Nonstndrd nlysis. Groningen: Fculty of Economics, University of Groningen. Retrieved from Roinson, A. (1996). Non-stndrd nlysis (Rev. ed). Princeton, N.J: Princeton University Press. Stroyn, K. D. (1997). Mthemticl ckground: Foundtions of infinitesiml clculus. Acdemic Press. Retrieved from pdf Winsløw, C. (213). Mthemticl nlysis in high school: fundmentl dilemm. rxiv Preprint rxiv: , 14. Yves Chevllrd (English) A.R.D.M. (n.d.). Retrieved August 7, 216, from Pge 76 of 152

87 Pge 77 of 152

88 Appendix 1 Description of teching Description of teching hyperrel numers First lesson (3.3) Second lesson (4.5) Third lesson (converted #1) (4.6) Fourth lesson (4.8) Hnd in # Fifth lesson (converted #2) (4.13) Description of teching differentil clculus (Jons Kyhnæ) First lesson (4.8) *Second nd third lesson (4.11) Fourth lesson (4.13) Fifth lesson (4.19) *Sixth lesson (4.2) *Seventh lesson (converted #3) (4.21) Hnd in # Eighth lesson nd ninth lesson (converted #4) (4.25) *Tenth, eleventh nd twelfth lesson (4.26 nd 4.27 nd 5.3) *Thirteenth lesson (5.3) Hnd in # Description of teching integrl clculus (Mikkel Mthis Lindhl) First nd second lesson (5.9) Third lesson (5.1) Fourth lesson (5.11) Fifth lesson (converted #5) (5.13) Hnd in #4 pge Sixth lesson (5.18) Seventh lesson (5.2) Eighth lesson (5.2) Hnd in #4 pge Ninth (5.23) Tenth lesson (5.23) Eleventh (nd twelfth) lesson Anlysis tles Hyperrel nlysis tle Differentil clculus nlysis tle (Jons Kyhnæ) Integrl clculus nlysis tle (Mikkel Mthis Lindhl) Pge 78 of 152

89 Appendix 1.1 Description of teching hyperrel numers 1 Description of teching This prt of the ppendix will give description of wht cn e oserved from the techings. 1.1 Description of teching hyperrel numers First lesson (3.3) The first lesson ws crried out y Jons nd ws mostly clss discussion with the occsionl tsk in smll groups (of up to 3). The following tle is wy to orgnize wht hppened nd some of the more interesting nswers/results of the tsks. Since the teching ws crried out in Dnish the oservtions given here re trnslted y the uthors of the thesis. HQ 1 : Wht is the first nonzero speed of your telephone when dropped from your hnd? HQ 2 : Give definition of something infinitely smll. HQ 3 : imgine you spend 3 dys t school without food; how much would you wnt potto chips? HQ 4 : Give definition of something infinitely lrge. HQ 5 : In wht set of numers should the infinitesimls,, nd the infinite numers e put? Short description of the mthemtics involved Numers in generl Didcticl moment Min plyer Techer (T) <-> student Mthemticl ojects involved Didcticl ctivities N, Z, Q, R Whole clss discussion, strting with the question posed y the techer. T: Wht re numers? The infinitely smll or infinitesimls infinite quntities HQ 3 Individul Student -> tsk HQ 1 techer HQ 2 Student -> techer HQ 4 Student -> techer Hyperrel numers HQ 5 Techer nd students Infinitesiml () Infinitesiml () And inequlities (< nd >) Infinite numer (N) And inequlities (< nd >) Techer writes some of the prtil nswers on the lckord 1. < 2. < R (infinitely smll) 3.,1 Student (S): You cn t write the numer S: < < R Techer writes some of the prtil nswers on the lckord 1. R + < < N 2. Infinity is everything so it cn t e igger thn infinity. N, Z, Q, R, ( R) T: Drws line on the lckord nd the students fill it out with numers from N, Z, Q nd R. T: where does nd infinite numers elong? Pge 79 of 152

90 Appendix 1.1 Description of teching hyperrel numers Short description of the mthemtics involved The infinitely smll or infinitesimls Didcticl moment Institution liztion Min plyer Mthemticl ojects involved S: one could mke new group of numers, which includes infinitesimls nd infinite numers. Didcticl ctivities T After the techer nd oserver does high five the techer writes the distnce etween the hnds the moment efore they clpped s Δx = x 2 x 1 =. Clling n infinitesiml Second lesson (4.5) The second lesson strts with short clss discussion of wht they did lst time. Including numer line with the efore mentioned sets N, Z, Q nd R. Agin tle will give n overview of the lesson. HQ 6 : Drw numer line which includes infinitesimls. HQ 7 : Which numer is infinitely close to 7 +? HQ 8 : Wht is 1/? HQ 9 : Where does ±1/ elong on the hyperrel line? HQ 1 : Give n exmple of hyperrel numer with no stndrd prt, i.e. find such tht st() is meningless. HQ 11 : Wht is function? HQ 12 : Cn you give generl definition of function? HQ 13 : Wht is continuity? HQ 131 : Wht is the distnce etween 2 consecutive points on the grph? HQ 132 : Wht should we write to finish this definition, dy = f(x) f(? )? HQ 14 : Wht is domin of function nd wht kind of numer is f(x + ) (ssumes f not constnt)? HQ 15 : Find st(x + ) + st(y + dy) HQ 16 : Find 5 st(x + ) HQ 17 : Find st(5x + 5) HQ 18 : Find st( + ) HQ 19 : Find st( + dy) Pge 8 of 152

91 Appendix 1.1 Description of teching hyperrel numers Short description of the mthemtics involved Didcticl moment Min plyer Mthemticl ojects involved Didcticl ctivities Rel numer line techer N, Z, Q, R A short description of the previous lesson. Hyperrel numers HQ 6 In smll groups. Prt of the hyperrel numer line Institutionliz tion y whole clss discussion student Techer student ( R) ( R) Product of HQ 6 : Preliminries for some of the rules of operting with infinitesimls Connection etween hyperrel nd rel line. Infinite numers And hyperrel numer line Infinite numers nd stndrd prt Institutionliz tion First encounter HQ 7 Institutionliz tion (First) encounter HQ 8 HQ 9 Institutionliz tion. Techer (student) S T S S nd T Sum, multipliction nd division Stndrd prt: st: R R Division with smll numers (infinitesimls) R S: 1 T goes through + dy, 2, /2 on the lckord nd S nd T rgues why they re infinitesimls. T: The rel numer which is infinitely close to hyperrel numer + is clled the stndrd prt of +, nd written s st( + ) =. T: writes 1/ on the lckord (S hs minor prolems with properties of frctions) should e plced on the other side of the hed of the rrow. T writes ± 1 on the numer line HQ 1 S R nd R. S: st ( 1 ) is meningless. Pge 81 of 152

92 Appendix 1.1 Description of teching hyperrel numers Short description of the mthemtics involved Functions: Different kinds Didcticl moment Min plyer Mthemticl ojects involved Didcticl ctivities HQ 11 S Vriles nd constnts. S gives exmples nd T writes on lckord: f(x) = x + f(x) = x f(x) = x Generl definition HQ 12 S S: For every x-vlue there is yvlue. T writes it on the lckord () Continuity (First) encounter HQ 13 HQ 131 HQ 132 Institutionliz tion Str opertion HQ 14 Stndrd prt nd the connection etween the rel nd hyperrel numers End of lesson 2 Institutionliz tion HQ 15 -HQ 19 S T S T S S: there re no jumps on the grph. S: Infinitesiml S: x + A function f is continuous if n infinitesiml increment in the xvlue produces n infinitesiml increment in the y-vlue. Tht is if f(x) f(x + ) = dy is infinitesiml for every infinitesiml. Hnds out the pmphlet out the hyperrel numers. Domin nd f(x) S: the numers for which the function mkes sense, hyperrel numer. T: The str opertion extends the domin of functions to include hyperrel numers. st(), infinitesimls (), sum. T: wlks round nd helps the S if needed. Pge 82 of 152

93 Appendix 1.1 Description of teching hyperrel numers Third lesson (converted #1) (4.6) The third lesson ws, due to regultions on the high school we tught t, lesson where the students re supposed to work on sujects they hve lredy covered, thus the students spent the lesson first going through their homework nd then on the exercises in the compendium. HQ 2 : Find st(5) HQ 21 : Find st ( 1 ) HQ 22 : Why is st (5 1 ) = 5 HQ 23 : Give n exmple of n infinitesiml which is lso rel numer HQ 24 : Find st((2 + )(3 + dy)) HQ 25 : Find st(2 + )st(3 + dy) HQ 26 : Wht cn e concluded from the two previous exercises? HQ 27 : Find st()st ( 1 ) HQ 28 : Amend the conclusion from efore HQ 29 : Find st((4 + ) 2 ) Short description of the mthemtics involved Stndrd prt of different hyperrel numers. Continuity Determine which numers re in N, Z, Q, R. Def. of infinitesiml nd infinite numers. Stndrd prt of hyperrel numers. Eqution solving. Continuity Didcticl moment Institutionliz tion of HQ 15 to HQ 24. Which re ll questions in HT 4 nd HT 5 Min plyer Mthemticl ojects involved Didcticl ctivities S nd T. st nd. 1 students go to the lckord nd write their results for the exercises they did t home. After tht 1 other students hd to rgue why the results were correct or incorrect. After 4 minutes the rule of st( ) = st()st() if nd only if oth nd re finite numers re proved nd explined y the students with some guiding questions from the techer. T f(x + ) f(x) Shows how to prove tht function Institutionliz tion In groups S nd T st,, 1 is continuous with the definition. T wlks round nd helps S if they hve questions for hnd-in ssignment 1. Pge 83 of 152

94 Appendix 1.1 Description of teching hyperrel numers Fourth lesson (4.8) In the fourth lesson the students go through the remining exercises in the compendium. HQ 3 : For f(x) = 3x + 1, find the hyperrel function vlue f(1 + ) nd f( 2). HQ 31 : For g(x) = x, find the hyperrel function vlue x = 3 nd x =. HQ 32 : For h(x) = 2x 3, find the hyperrel function vlue h(1 + ) nd h(4 + ). HQ 33 : Wht is st( f(x))? HQ 34 : Assume now tht f is continuous rel function wht is st( f(x + ))? Short description of the mthemtics involved Stndrd prt of different hyperrel numers. Determine which numers re in N, Z, Q, R. Def. of infinitesiml nd infinite numers. Stndrd prt of hyperrel numers. Eqution solving. Continuity Didcticl moment Institutionliz tion of HQ 25 to HQ 29. Which re ll questions in HT 4 nd HT 5 Min plyer Mthemticl ojects involved Didcticl ctivities S nd T st,, continuity T wlks round nd helps S if they hve questions for hnd-in ssignment 1. In groups S nd T st,, 1 T wlks round nd helps S if they hve questions for hnd-in ssignment 1. Pge 84 of 152

95 Appendix 1.1 Description of teching hyperrel numers Hnd in #1 The tle of StndrdJoe nd HyperMick Prolem A StndrdJoe nd HyperMick weren t orn StndrdJoe nd HyperMick, one hs to ern nmes like tht! StndrdJoe nd HyperMick tht is. Thus, our tle of StndrdJoe nd HyperMick egins efore the stndrd prt of Joe ws tken nd rel-isticlly seen efore Mick ws mde hyper. Think out it. Bck then they were only le to solveexercises like these. Petty.. 1. Which of these numers re nturl numers? 42; 12 ; ; 1,2 ; 1 2 ; 11 8 ; Rewrite the following frctions to decimls: 1 ; 9 ; 5 ; Rewrite the following frctions to decimls, nd find the pttern in the decimls: 2 ; 1 ; Explin why these numers re rtionl numers: 4,825 ; 1, ; 8 ; 3,25 ; Check if 6,76 is rtionl numer. 6. Check if the following is true: 3 Z ; 3 Q ; 2 N ; 2 R ; R 11 ; 4 7 Prolem B Mick ws infinitely close to getting his stries. He just needed to show he understood wht the creed ws ll out. Tht s stndrd. 1. Write definition of n infinitesiml. 2. Write definition of n infinitely ig numer. Prolem C Solve the following expressions. But how!? Ask Joe. He ws the mn nd ws quickly known s StndrdJoe. Like, super fst. Insnely fs.. Frekin psycho f.. I ve never seen nyone get nicknme s fst s Joe did! He got it quicker thn nything I cn even descrie. 1. st(42 + ) 2. st(42 + dy) 3. st(42) 4. st( ) 5. st(42) + st(42) 6. st(42)st(42) 7. st (42 1 ) 42 Pge 85 of 152

96 Appendix 1.1 Description of teching hyperrel numers Prolem D Now they re just messing with Joe. He don t cre! Solve the equtions nd find the rel vlue closest to x. 1. 2x + 1 = = 3x 2 3. x 7 = 2x (x + 3) = x = 4(2 ) 2 Prolem E Now they hd to prove tht they could oth keep their nmes forever. Tht they didn t suddenly lose them nd then get them ck. They were to hve their nmes without jumps. Once you go hyper, you.. lso.. ctully, sometimes go stndrd.. But not t the sme time! So.. Yo. Determine if the following functions re continuous. 1. f(x) = x + 2. g(x) = x 3. h(x) = x 4. (2) k(x) 1 1 (1) Good luck with the st(their originstory), Mick n Joe. Pge 86 of 152

97 Appendix 1.1 Description of teching hyperrel numers Hnd-in #1 notes Prolem A Generlly good understnding of numer sets, with tendency to exclude 6 nd include in N. 1 Prolem B Write definition of n infinitesiml Student1: Infinitesiml is not constnt, since numer cn lwys get smller contr Infinitely ig numer is on the contrry numer tht cn lwys get igger. Student2: Something tht is infinitely smll: < R + Or if it s negtive: > R It cnnot e written, since it cn lwys e smller, thus it is defined s. Write definition of n infinitely ig numer Something, which is igger thn ll the rel numers, it cn lwys get igger. It is so ig tht it cnnot e written, since one cn lwys put more numers on it 1 > R Or if it s negtive: 1 < R Prolem C Prolem D 7. They hve difficulties clculting st (42 1 ) 42 The prolem is, tht they see st ( 1 ) s something not definle, no mtter wht The only relevnt prolem in this exercise is nottion even though some students relly niled it, which re given elow: Student1: = 3x = 3x = x 3 x = Student2: 2x = 4 + x = So the rel vlue is x = 2 Pge 87 of 152

98 Appendix 1.1 Description of teching hyperrel numers Prolem E Their understnding of continuity - A grph mustn t contin holes - Cn t lift the pencil - They use f(x + ) f(x) nd sy, tht something isn t continuous, if they cn t use tht - Not continuous, since the grph is roken severl times Fifth lesson (converted #2) (4.13) The point of the fifth lesson ws to check if ll of the students understood the comments, they hd gotten in the weekly hnd-in out hyperrel numers, y going through every exercise one y one, with the students giving the nswer. Pge 88 of 152

99 Appendix 1.2 Description of teching differentil clculus 1.2 Description of teching differentil clculus (Jons Kyhnæ) The str * mrks lessons to which there exists physicl mteril from the students, which cn e reviewed. This in the form of filled out worksheets, finished hnd-in ssignments nd/or video recordings (see Digitl ppendix) First lesson (4.8) In this lesson, the techer refreshed, with the students, the concept of function. Wht is function, wht does function do, wht hppens when you evlute function in given vlue nd so on. An introduction to the terms slope nd tngent ws given through the grphs for different types of known nd unknown functions in coordinte system. The students lso got Worksheet 4.8 with exercises out finding the slope in point on the grph of the function. DQ 1 DQ 2 DQ 3 DQ 4 DQ 5 DQ 6 DQ 7 DQ 8 DQ 9 DQ 1 Wht is function? How do you find the function vlue for certin x-vlues? How do we get from function to grph? Are the functions, represented y these grphs, continuous? When, during the dy, is the phone eing used the most? Cn we tell (exctly) how much the ttery is eing drined here? At wht specific point during the dy, is the phone eing used the most? How do we find out exctly how much ttery is eing drined t the point of the dy where the ttery is eing drined the most? Cn we find out when the ttery is eing drined 1% per second? When/where cn we find out how much ttery is eing drined? Pge 89 of 152

100 Appendix 1.2 Description of teching differentil clculus Short description of the mthemtics involved The concept of function. Continuity Intervls. Slope of secnt nd tngent. Infinitesiml Slope in point from grph. Tngent to grph. Drwing grph Didcticl moment Technicl moment with DQ 1 to DQ 4 Institutionliz tion of DQ 5 to DQ 1 Min plyer T nd S T nd S Mthemticl ojects involved f(x), f(), coordinte system Coordinte system, = y 2 y 1 x 2 x 1 In groups S nd T Coordinte system, = y 2 y 1 x 2 x 1 Didcticl ctivities T sks questions out functions to drw out the students knowledge nd to fmilirize them with functions nd the correltion etween functions nd grphs nd which functions re continuous T sks indirect questions out the slope of grph, using n exmple of decresing grph descriing phone ttery s dringe during dy. DA 6 : It decreses most where the grph is the steepest. S find the secnt, without using tht word, ut the generl slope etween two points on the grph. Grdully S find the slope in point using tngents in intervls, which t the end is the tngent in n infinitesiml intervl. DA 1 : Everywhere T wlks round nd helps S if they hve questions for Worksheet 4.8, which includes drwing tngents nd grphs nd finding slopes using the 2-point formul The students relized tht the slope in point on decresing grph is steepest where the grph decreses the most. This y n exmple of grph of phone ttery the techer drew on the lckord. Here the students concluded reltively quickly tht the ttery must hve een drined the fstest when the grph decresed the most, in specific intervl. Some of the students cme up to the lckord to point nd some cme up to drw tngents, even efore we introduced the term, which resulted in skipping Worksheet Secnt nd only hnd the students Worksheet 4.8 Tngent. Even though the students hd not herd of polynomil function yet, with this introduction, the students estlished grphicl understnding of the slope in point for n ritrry function. Pge 9 of 152

101 Appendix 1.2 Description of teching differentil clculus Worksheet 4.8 Secnt On the grph elow, 2 points hve een mrked. 1. Drw line through the 2 points nd find the slope of this line. 2. Wht cn e sid of the slope of the line nd the slope of the grph etween the 2 points? (8,11) (4,7) Plot in 2 other points on the grph elow nd repet the steps 1. nd 2. Pge 91 of 152

102 Appendix 1.2 Description of teching differentil clculus Worksheet 4.8 Tngent Below is the grph for f(x) = 1 4 x2 + 4x 5. Find the slope of the grph in the point (4,7). Find the slope of the grph in (8,11). Find the slope of the grph in (12,7). Wht is the generl procedure for finding the slope of grph in n ritrry point? Drw n ritrry grph for continuous function in the coordinte system elow. Find the slope in n ritrry point P on the grph. Explin your pproch from the eginning. Pge 92 of 152

103 Appendix 1.2 Description of teching differentil clculus *Second nd third lesson (4.11) The point of this lesson ws to get the students to use the 2-point formul on n infinitesiml intervl nd get the formul for the differentil quotient (without ctully telling them, tht tht ws wht they found). To the students, they estlished technique of finding the slope of grph in point, y finding the slope of the tngent in tht point. The connections were mde through Worksheet 4.11 with questions DQ 11,i. DQ 11, DQ 11,1 DQ 11,2 DQ 12 DQ 13 Find the slope of function, f(x), in point, without drwing nything! Find the generl slope of function, f(x), in the intervl [; ]. Wht cn e sid out the slope to function, f(x), nd the slope of the secnt (the stright line) through the points (, f()) nd (, f()). Wht did you just do? How? Short description of the mthemtics involved Slope of secnt, tngent nd function Slope in point Differentil quotient Didcticl moment Explortory in groups working with DQ 11,i Recp with DQ 12 to DQ 13 Institutionliz tion Min plyer Mthemticl ojects involved S nd T = y 2 y 1 x 2 x 1 T nd S = y 2 y 1 x 2 x 1 Didcticl ctivities T wlks round nd helps S if they hve questions for Worksheet 4.11 DA 12 : Constructed formul with which we cn find the slope of every grph DA 13 : Using y 2 y 1 y finding two x 2 x 1 points close to ech other like + S writes the definition of slope in point (, f()) is st ( f() f(+) (+) T f (x) T tells S tht the slope in point (x, f(x )) on grph is clled the differentil quotient nd is written f (x ) ) For the students to get the correct understnding of the differentil quotient, the secnt hd to e introduced in whtever smll cpcity. This only hppened through Worksheet 4.11, where it ws mentioned s something the students hd lredy worked with, nmely stright line through two points on grph. The worksheet ws minly creted to further expnd the students knowledge, in tht this introduced the slope of function. The students worked on the worksheet in groups, efore going through it in clss. The uild-up for the worksheet is generting question posed s the first/top one. To help the students, there re su questions creted in such wy tht if the students couldn t give n nswer to the generting question right wy, they could look t the next question. Thus, the lst question is su question to help with oth of the ove questions. The students hnded in their nswers t the end of clss Pge 93 of 152

104 Appendix 1.2 Description of teching differentil clculus nd looking t the lst su question. The students ll reched the sme conclusion (some erlier thn others did), which, in different vritions, looks like this: o Finding slope in point is only possile y using second point nd the closer the points re to ech other, the more precise will the slope of the stright line etween the two points then e, compred to tht of n ritrry grph. It would e smrt to choose two points tht re infinitely close to ech other, hence, the points cn e close to ech other. The next/first su question ws then nswered using the sttement they hd reched. The nswers were lmost identicl, with vrints in the choice of wht to cll the generl slope (here S): - The generl slope of function in the intervl [; ] is S = f() f(). The nswer for the generting question ws then chieved with the formul just derived from the previous su question. A generliztion for the nswers looks like this: Let = + nd f() = f( + ), then the slope in point (, f()) will e st ( f (+) f() + ) = st ( f (+) f() ). We tke the stndrd prt ecuse we wnt the rel vlue. And lwys rememer the str when the function is hyperrel. In conclusion, using the students words, they were le to use the 2-point formul for stright lines ( = y 2 y 1 x 2 x 1 ), since the slope of tngent hs the sme slope everywhere hence in every point on the tngent nd then clerly lso in the point on the grph. These two points hd to e found in smll intervl of the grph nd ctully, the smller the intervl, the etter the tngent, mening the more precise the slope in the point. Rther quickly, student sid, tht the est slope would occur if we could use the 2- point formul on to points tht hd n infinitesiml difference. The fct tht the difference would then e hyperrel didn t mtter, since we could find the rel numer infinitely close to this hyperrel numer y tking the stndrd prt. This is ctully, wht led to the decision of mking the compendiums grdully, s the students completely skipped the secnt-prt, since it mde so much sense to just pick the second point so they were in n infinitesiml intervl. Some students even found it confusing reding the chpter on the secnt, ecuse tht ws not how we did in clss nor ws it how the student hd understood it. On side note it ws lso here the hedline Tngentmn nd 2point find Miss A cme to e. The techer summed up the lessons, just to sy tht the slope in point (x, f(x )) on grph for function, could e shortened down to differentil quotient, which is written f (x ). Pge 94 of 152

105 Appendix 1.2 Description of teching differentil clculus Worksheet 4.11 Hello humns. We re gonn give you n exercise now, which you re to mke in groups of 3 to 4. You will nswer the question elow, if necessry y using the su questions, nd write them on this pper, or uplod it s group in Lectio, so tht we get it ll when you ve finished. Thnks! Question: Find the slope of function, f(x), in point, without drwing nything! - Find the generl slope of function, f(x), in the intervl [; ]. o Wht cn e sid out the slope to function, f(x), nd the slope of the secnt (the stright line) through the points (, f()) nd (, f())? Pge 95 of 152

106 Appendix 1.2 Description of teching differentil clculus Fourth lesson (4.13) In the fourth lesson, we mde recp of the lst lesson with the formul for the differentil quotient, which some of them rememered lredy. The students were given numer of exercises from the compendium, which were ll in finding the differentil quotient for different polynomil functions to vrious vlues of x. DQ 15 to DQ 19 re exercises 1 to 15 (minus 14) on pge 21 in the compendium. DQ 14 DQ 15 DQ 16 DQ 17 DQ 18 DQ 19 Wht did we do lst? Fill out the tle y computing the differentil quotient to the function g(x) = 1 in the points x g(x) g (x) Fill out the tle y computing the differentil quotient to the function h(x) = 4x in the points x h(x) h (x) Fill out the tle y computing the differentil quotient to the function k(x) = 2x 2 in the points x k(x) k (x) Fill out the tle y computing the differentil quotient to the function l(x) = 2x 2 4x + 1 in the points x l(x) l (x) Let f(x) nd g(x) e two ritrry nd differentile functions. Compose rule for the differentil quotient to h(x) = f(x) + g(x), tht is descrie h (x) with f (x) nd g (x). Pge 96 of 152

107 Appendix 1.2 Description of teching differentil clculus Short description of the mthemtics involved Differentil quotient Differentil quotient of sum of two functions Didcticl moment DQ 14 Technicl moment DQ 15 to DQ 18 Explortory DQ 19 Min plyer Mthemticl ojects involved S nd T f (x ) = st ( f (x +) f(x ) ) Didcticl ctivities S dictte how T completes n exmple of finding the differentil quotient for function f(x) = x 2 in point (2, f(2)) T wlks round nd helps S if they hve questions for the exercises DQ 15 to DQ 18 (pge 21 in the compendium) S f(x), f (x) S work in groups to solve the prolem The students got to mke exercises in using the definition of the differentil quotient, which they hd only just estlished. Hence, the exercises concerned vrious prolems of finding the differentil quotient for function in different points. Another dded exercise ws the prolem of finding the points, when only given the x-vlue. Furthermore, these exercises were constructed in such wy tht the students were to see pttern in the different functions to which they hd to find the differentil quotient, thus leding up to finding the differentil quotient for h(x) = f(x) + g(x), using f (x) nd g (x). Pge 97 of 152

108 Appendix 1.2 Description of teching differentil clculus Fifth lesson (4.19) We went through questions DQ 15 to DQ 19 on the lckord. DQ 2 How did you solve these exercises? Short description of the mthemtics involved Differentil quotient. Stndrd prt. Str opertion. 2-point formul To derive. The derived function Didcticl moment Technicl moment DQ 2 Institutionliz tion Min plyer Mthemticl ojects involved S nd T f (x ) = st ( f (x +) f(x ) ) Didcticl ctivities DA 2 : Use the 2-point formul on n infinitesiml intervl T f (x) T tells S tht going from function to the differentil quotient is clled to derive the function in point T tells S tht deriving function gives the derivtive of function The techer drew tle on the lckord contining three different functions, g, h, k, l nd their derivtives, g, h, k, l, where g + h + k = l nd g + h + k = l. Ech function nd its derivtive ws then to e evluted in the sme five x-vlues. The students filled out the comined tle with ll the function vlues nd the differentil quotients. From strt to finish, the student who filled out the ox explined how it ws done. At this point, the students were cquinted with finding the function vlues; hence, these were filled out correctly. The interesting prt ws how they hd come to the differentil quotient. The process for ech function ws different: g(x) = 1 I Most of the students used st ( f f (+) f() ( + ), thinking they should simply dd to 1, so ), ut hd troule finding st ( ) = st ( ) = st(1) = 1 The wy this ws solved, ws to sk if they rememered wht differentil quotient is nd how they first found it. This helped them to try to imgine grph for the function or to drw it nd then see tht wht they hd found couldn t e true. Next step ws getting the students to see how they hd just found the function vlues nd to relize this ws clerly how to find f( + ) s well. II The ltter ws exctly how other students found the differentil quotient right wy nd thus circumventing the use of st ( f (+) f() ) ltogether. Pge 98 of 152

109 Appendix 1.2 Description of teching differentil clculus III Some students did figure out how to use st ( f (+) f() ) correctly nd found the differentil quotient in ll the points using this formul. h(x) = 4x k(x) = 2x 2 All students solved this either y II or III, few trying to find wy of generlizing the results. Ech student did s descried in III. One student ecme convinced, though, tht there ws system to it, tht they didn t hve to go through the entire process of using the formul every time. The student ws especilly motivted y the prospect of voiding the mny clcultions necessry, when hving squred expressions with more thn one term. The student s resoning ws tht there hd to e nother wy, since functions of higher power thn 2 must clerly exist nd tht it would then tke forever. The student quickly relized tht ll the terms contining of power higher thn of 1 could e ignored, since they dispper when tking the stndrd prt. l(x) = 2x 2 4x + 1 Furthermore, the student relized tht ll the constnt terms cncel ech other out just s the terms not contining. This t lest mde finding the differentil quotient for polynomil functions of power 2 esier nd some of them the student even solved without writing nything down. A few students, who didn t do it s the one student trying to solve the enigm of differentiting in the hed, sw tht l(x) consists of the foresid functions, so checking if the function vlues in the different points were the sme s dding (or sutrcting) the respective function vlues resulted in the sme function vlue s clculted, ws very esy. These students then tried their luck with differentil quotients, following the sme procedure. They were then told to check some of them y doing s descried in III. The rest of the students did s descried in III. Not ll the students hd gotten to the exercise out finding the differentil quotient for h(x) = f(x) + g(x), using f (x) nd g (x) ut fter going through ech of the functions like this everyody t lest hd suggestion s to how they should go out it. As such, the students ll greed tht h (x) = f (x) + g (x). Pge 99 of 152

110 Appendix 1.2 Description of teching differentil clculus *Sixth lesson (4.2) The students worked in groups on wht conditions of monotony is for function, given the derived function. They were even given the opportunity to serch the internet, minly to give definition of the conditions of monotony. If they needed help, the techer would give them hint in the right direction, through the students knowledge of differentil quotients, in the form of DQ 21,i in chronologicl order. DQ 21 DQ 21,1 Wht cn e sid out the conditions of monotony of the originl function, (f(x)), given the derived function (f (x))? Wht is conditions of monotony? DQ 21,2 Find the conditions of monotony for f(x), when f (x) = x + 2 DQ 21,3 Wht cn we sy out f(x) when x = 2? DQ 21,4 Wht cn we sy out f(x) when x 2? DQ 21,5 Wht cn we sy out f(x) when x 2? DQ 21,6 Wht cn you sy out the conditions of monotony of the originl function (f(x)), given the derived function (f (x))? Short description of the mthemtics involved Conditions of monotony. Differentil quotient Didcticl moment First encounter, explortory nd technicl moment DQ 21 Explortory DQ 21,i Min plyer Mthemticl ojects involved Didcticl ctivities S f(x), f (x) S work in groups to solve the prolem S nd T f(x), f (x) T gives S hints one y one if needed, in the form of DQ 21,i The lesson ws out mking description of wht cn e sid out the conditions of monotony for function (f(x)), given the derived function (DQ 21 ). The students, wnting to mke the est description possile, put in lot of effort in writing this. Seeing s they could serch the internet for nswers, lot of them hd the sme restrictions for when function is in- or decresing, nmely f (x) > for ll x [; ] f incresing in [; ] f (x) < for ll x [; ] f decresing in [; ] f (x) = for ll x [; ] f constnt in [; ] Pge 1 of 152

111 Appendix 1.2 Description of teching differentil clculus Luckily, ll the students knew this wouldn t e enough nd dded self-mde explntions s to show they knew wht they hd written. After clss, ll the students hnded in their descriptions. Some of wht they hd written s justifictions could e: And lso When we hve the derived function nd re told tht the slope in n intervl is positive, negtive or equls zero in more thn one point, we ll e le to red off of the grph if it is decresing, incresing or constnt. All functions hve different inequlities, which determines the conditions of monotony. When given the derived function (f (x)) we know everything out the conditions of monotony for the indigenous function (f(x)). E = mc 2 Pge 11 of 152

112 Appendix 1.2 Description of teching differentil clculus *Seventh lesson (converted #3) (4.21) The lesson strted with the techer summing up wht they did lst (*Sixth lesson (4.2)), in regrds to conditions of monotony. The techer gve definition nd n exmple of how to write it properly. The rest of the lesson the students got to work on Hnd-in #2. Short description of the mthemtics involved Conditions of monotony. Differentil quotient Derivtives Didcticl moment Institutionliz tion Technicl moment Min plyer Mthemticl ojects involved Didcticl ctivities T f(x), f (x), [ ; ] T explins how to find the conditions of monotony for the function f(x) = 1 2 x2 + 2x 4 nd how to properly write it with rckets S nd T f(x), f (x), coordinte system, N, Q The students worked on prolems involving finding the derivtive of generl functions like T wlks round nd helps S if they hve questions for Hnd-in #2 - liner function f(x) = x +, specificlly constnt term - polynomil function with n integer s exponent x n, R, n N, y first using the formul when n = {1,2,3} nd then giving sustntited guess for when n > 3 - polynomil function with x, Q, y using the results for when the exponent is n integer These s wy of identifying the rule for f(x) = x n then f (x) = n x n 1. They lso hd to explin the difference etween slope nd differentil quotient. Lstly, the students should prctice the newly otined techniques on rel life prolem out hens lying eggs nd the speed hereof t different times of dy. Pge 12 of 152

113 Appendix 1.2 Description of teching differentil clculus Hnd in #2 This is nog story of StndrdJoe nd HyperMick, Prolem A Wht s the difference etween slope nd differentil quotient? (Anti?)Hint: Think gosh d.. nd write good explntion, or else.. Prolem B Find out wht f (x) is for the function f(x) = x + where nd re rels. I.e. write n expression for wht f (x) is in terms of, og x. 1. In generl, wht cn e sid of the derivtive of function, which contins constnt terms. 2. Give n exmple of wht f (x) is for liner function of your choice. Prolem C 1. Wht is f (x) for f(x) = x? 2. Wht is f (x) for f(x) = x 2? 3. Wht is f (x) for f(x) = x 3? 4. Tke guess t wht f (x) is for f(x) = x Give justified guess on wht f (x) is for f(x) = x n for n N. Prolem D is rel numer. 1. Wht is f (x) er for f(x) = x 2? 2. Wht is f (x) for f(x) = x 3? 3. Tke guess t wht f (x) is for f(x) = x 4? 4. Give justified guess on wht f (x) is for f(x) = x n for n N. 5. Find the derivtive of f(x). In English, find out wht hppens, when you multiply rel numer with function f(x) nd the derive it. Prolem E Give justified guess on if you cn derive f(x) = x for Q. Use, if necessry, the results from the exercises ove nd/or exmples. Pge 13 of 152

114 Appendix 1.2 Description of teching differentil clculus Prolem X Studies show tht when rooster crows t 6 A.M. the hens wke up nd ly eggs for 12 hours stright. Don t fct check. On certin frm on The Froe Islnds, there is 3 hens, frederegg, ggeretse nd henriegg. The correltion etween the numer of eggs the hens ly nd the time of dy cn e descried y the functions f(t) = t t g(t) = 1 3 t3 7t 2 + 5t h(t) = t + 12 where the vlues of the functions re the numer of eggs the hens hve lid t specific moment t, the numer of hours pssed since 6 A.M. The coordinte system elow shows the grphs for ech of the hens from 6 A.M. to 6 P.M. Æg g(t) f(t) h(t) Timer 1. Who hs lid the most eggs t 12 noon? 2. How mny eggs does henriegg ly in the first 2 hours? How fst does she ly eggs in the first 2 hours? How fst does she ly eggs t (time) o clock? Find h (t). Refleggt on the results. 3. How fst does frederegg, ggeretse ly eggs t (time) o clock? 4. Who lys eggs the fstest t 12 noon? 5. Eggsplin wht the numer 12 mens in h(t) = t Pge 14 of 152

115 Appendix 1.2 Description of teching differentil clculus Generl description of nswers to hnd-in #2 A: By now, the students hd the correct understnding of wht the difference etween slope nd differentil quotient is; the only prolem ws, s with ll students, to give n dequte explntion. B: They ll found the derivtive y using f (x) = st ( f (x+) f(x) C+D: The purpose of the exercise ws to get the students to generlize the technique for finding the derivtive of x n nd x n. The tle shows how mny students ctully found the correct technique. The tle lso shows how mny of them ctully gve n explntion s to how they hd reched tht conclusion, s opposed to the students who hd only written the technique, whence it ws impossile to know if they ctully understood wht they hd written or if they hd merely copied n nswer. If student hd reched the right conclusion in C, they hd done this s well in D. ). Exercise C+D With explntion Without explntion # of students finding the correct technique # of students not finding the correct technique More or less the explntion given (in C): Becuse tht is wht hppened ove nd it mkes sense, since every time you multiply with n extr x + you get one power higher for x nd multiplied with one more. E: Of the 13 students, who hd ctully reched the right conclusion in C nd D, some of them hd ventured guess here nd some got it right. X: The point of this exercise ws to get the students to relize tht the slope in point is ctully the speed in tht prticulr point, from looking t the one hen. Not mny got this explicitly nd therefore hd troule finding the speed for the other two hens. Most of the students thought tht the hen hd lid 14 eggs, since h(2) = Hence (pun intended) the reliztion of the hen lying eggs t the speed equl to its slope ws lost. Pge 15 of 152

116 Appendix 1.2 Description of teching differentil clculus Eighth lesson nd ninth lesson (converted #4) (4.25) The eighth lesson strted with the techer going through n exmple (DQ 23 ) on how to find vlues for function given vrious vlues for x, with help from the students nd then letting the students drw the grphs of different functions (DQ 24,i ) mnully, y finding numer of points the grph runs through nd then specify the conditions of monotony. The second lesson the techer went through DQ 24,i with the students nd then they got to work with DQ 26,i nd DQ 27. DQ 22 DQ 23,1 DQ 23,2 DQ 24 DQ 25,1 DQ 25,2 DQ 25,3 DQ 26 How do you drw grph for given function, f(x) = 2x? Drw the grphs of the following functions. f(x) = x 2. g(x) = 2x 2 3 c. h(x) = x d. k(x) = x 3 + 2x 2 3 Find the conditions of monotony for the functions ove. For wht do we use conditions of monotony? Given the function f(x) = x 2 2x + 4, fill out the tle x f f(x) (x + ) Identify to which vlue of x the function most likely hs either mximum or minimum. Find f(x + ) f(x ) for the following vlues of x x f(x + ) f(x ) Determine the extremes of the function (i.e. mxim nd/or minim) nd mke line of monotony (s in the compendium) for the function. Give written conclusion t the end. (In cse of mssive lziness, the tle elow cn e used insted of mking line of monotony) x f f Pge 16 of 152

117 Appendix 1.2 Description of teching differentil clculus Short description of the mthemtics involved Functions Conditions of monotony. Differentil quotient Functions. Hyperrel functions. Conditions of monotony Didcticl moment Technicl moment DQ 22 Technicl moment DQ 23,i or DQ 24 Technicl moment DQ 25,i or DQ 26 Min plyer Mthemticl ojects involved Didcticl ctivities T nd S f(x) T goes through DQ 22 with S S nd T f(x), f (x), [ ; ], coordinte system S nd T f(x), f(x), f (x), [ ; ] T wlks round nd helps S if they hve ny questions The students worked some more on the conditions of monotony, to further understnd the connection etween the grph nd the differentil quotient. They hd lot of troule (even) finding the function vlue for given x-vlue, hence the mny exercises in this. Pge 17 of 152

118 Appendix 1.2 Description of teching differentil clculus *Tenth, eleventh nd twelfth lesson (4.26 nd 4.27 nd 5.3) In lesson ten nd eleven, the students were divided into groups nd given one of the questions DQ 27 to DQ 31, with one or more hints on how to get strted nd/or get through the proof (see Worksheet 4.26). The first 2 minutes of lesson eleven ws used y the techer to go through the proof of how to derive e x (DQ 32 ). In the twelfth lesson, the students presented their product from lessons ten nd eleven. The requirements for the show ws tht they should present it on the lckord y writing nd tlking nd every person in the group should sy something. DQ 27 DQ 28 Find the derivtive of f(x) = 1 x Find the derivtive of f(x) = x DQ 29 How do you find the eqution of the tngent to the function f(x) in the point (x, f(x ))? DQ 3 Let f(x) nd g(x) e differentile functions nd let h(x) = f(x) g(x). Find the derivtive of h(x) DQ 31 Let f(x) nd g(x) e differentile functions nd let h(x) = f(g(x)). Find the derivtive to h(x) DQ 32 Wht is the derivtive of e x? Short description of the mthemtics involved Differentil quotient. Stndrd prt. Str opertion Didcticl moment Technicl moment DQ 27 or Min plyer Mthemticl ojects involved f (x) = st ( f (x+) f(x) ) Didcticl ctivities Eqution of stright line. Tngent Differentil quotient. Stndrd prt. Str opertion DQ 28 Technicl moment DQ 29 Technicl moment DQ 3 or S nd T f (x ), y = x + f (x) = st ( f (x+) f(x) ) T wlks round nd helps S if they hve ny questions Eqution of stright line. Tngent. Differentil quotient. Stndrd prt. Str opertion DQ 31 Technicl work First encounter with D 2 T 2 S f (x) = st ( f (x+) f(x) ), f (x ), y = x + S present their show S wtch the show Differentil quotient nd D 1 T 3 Institutionliz tion T f (x) = st ( f (x+) f(x) ) T goes through the proof of how to derive e x Pge 18 of 152

119 Appendix 1.2 Description of teching differentil clculus In the durtion of three lessons, the students work independently in two of them nd present their finished product orlly in front of the rest of the clss. The techer divided the students into groups of 4 or 5. By wlking round in the clssroom nd oserving the students, the process of solving the prolems looks like this: Prolem 1 Orl The students knew y then to use the formul f (x) = st ( f (x+) f(x) ) nd hd virtully no prolems executing this. The group only sked the techer one time if they hd done it correctly one time nd tht ws question out the rules of frctions, which they hd done correctly though. In ddition, they hd finished the rest of the proof y then, they simply wnted to know if they could explin the rule properly. The only prt not gone through rigorously enough ws the eqution 1 st ( x 2 + x ) = 1 x 2 which ws only explined with in tking the stndrd prt x disppers. Prolem 2 Orl They hd troule using the first hint ecuse of two resons, one of which is less importnt though, nmely tht there wsn t ny s or s in finding the derivtive of the function. The other reson ws tht there ws no product of two terms eing sutrcted with the sme two terms dded. Thus, the students needed help using the hints, since the other hint didn t help them either. They needed to see the connection etween the two hints more thn perceiving them s two seprte clues nd then ll ws fine. The only prt not gone through rigorously enough ws the eqution 1 st ( x + + x ) = 1 x + 1 x = 1 2 x which ws only explined with in tking the stndrd prt x disppers. Prolem 3 Orl From strt to finish, this took the students round 15 minutes, mening from when they got the prolem until the techer cme to check on them nd they hd found the eqution for the tngent. No questions, no douts. The students first relized, tht = f (x ). Then looking t the eqution for the stright line, y = x +, they found tht must e = y x = f(x ) f (x )x nd y putting this intro the eqution for the stright line, the eqution for the tngent is otined T = x + = f (x )x + f(x ) f (x )x = f (x )(x x ) + f(x ). Pge 19 of 152

120 Appendix 1.2 Description of teching differentil clculus Prolem 4 Orl The first of three prolems ws to connect the hints. Sure, they mde sense for the students, ut not in the context of the prolem. Once this ws otined nother prolem surfced, nmely how does g(x + )f(x) + g(x + )f(x) help, especilly if it s nywy? The techer then plced it in preferle plce nd helped the students relize tht some of the terms included the sme fctor nd they then figured out how to put this outside the rckets. The lst thing ws to see the link etween wht they hd nd f (x) = st ( f The only thing the students didn t mention ws why (x+) f(x) ). st( g(x + )) = g(x) Prolem 5 Agin, the techer hd to help the students relize how to execute the hint (just the first one). Once this ws done, one student in the group sw, tht st ( f (x + )) f(g(x)) ) = f g(x + ) g(x) (g(x)) ( g where the student discovered tht it s the sme s f (x) = st ( f chnge of vrile. (x+) f(x) ), ut with Orl The students gve more elorte explntion of the eqution ove, nmely tht they could see it s using the 2-point formul. The only prt not gone through rigorously enough ws the eqution st ( f (x + )) f(g(x)) g(x + ) g(x) ( g = st ( f g(x + ) g(x) ) (x + )) f(g(x)) ) st ( g (x + ) g(x) ) g(x + ) g(x) ( g which ws only explined with y using the rule [for stndrd prts] we cn split them up. Pge 11 of 152

121 Appendix 1.2 Description of teching differentil clculus Worksheet 4.26 Prolem 1 f(x) = 1 x Find the derivtive of f(x). Hint: Know your rules for frctions (how does one mke common denomintor?) Prolem 2 f(x) = x Find the derivtive of f(x). Hints: 1. (mthemtics in the dys of yore) the difference etween two numers multiplied with the sum of the sme two numers is the difference etween the squred numers ( )( + ) = To multiply with 1 doesn t mke difference, though it mkes some firytle cretures hppy to go from the rel numers to the integrl numers. ( + ) 1 = ( + ) Prolem 3 How do you find the eqution of the tngent to the function f(x) in the point (x, f(x ))? Hints : 1. Wht is tngent? 2. Wht does the tngent hve to do with the differentil quotient? 3. How does one find the slope of tngent? Prolem 4 Let f(x) nd g(x) e differentile functions nd let h(x) = f(x) g(x). Find the derivtive of h(x). Hints : 1. If you owe 2 kr. nd someone gives you 2 kr., you hve kr. g(x + )f(x) + g(x + )f(x) = 2. HyperMick nd StndrdJoe divided kr. etween the two, ut tht s the sme s dividing kr.! = Pge 111 of 152

122 Appendix 1.2 Description of teching differentil clculus Prolem 5 Let f(x) nd g(x) e differentile functions nd let h(x) = f(g(x)). Find the derivtive to h(x). Hints: 1. Multiplying with 1 doesn t mke difference, though it mkes some mermids hppy to go from unknown to known fctors. 1 = dg dg = g (x + ) g(x) g(x + ) g(x) 2. The stndrd prt of product is the sme s the product of the stndrd prts of ech of the fctors, s long s the fctors re finite. st( ) = st() st() Pge 112 of 152

123 Appendix 1.2 Description of teching differentil clculus *Thirteenth lesson (5.3) The thirteenth lesson ws used to summrize wht they hd done in differentil clculus; to this end, ( lnk version of) Worksheet 5.3 ws prepred nd hnded out to them. The students then spent round 4 minutes to fill out the worksheet to the est of their ilities in smll groups (of order < 5). Afterwrds the groups red loud, to the rest of the clss, wht they hd on their ppers, ending with the cretion of the following filled out worksheet. Pge 113 of 152

124 Appendix 1.2 Description of teching differentil clculus Worksheet 5.3 f(x) f (x) x x 2 x n 1 x = x 1 2x nx n 1 1 x 2 = x 2 x = x x = x 2 e x h(x) = f(x) ± g(x) h(x) = f(x) Rules of clcultion e x h (x) = f (x) ± g (x) h (x) = f (x) h(x) = f(x) g(x) h (x) = f (x)g(x) + f(x)g (x) h(x) = f(g(x)) h (x) = g (x) f (g(x)) h(x) = f(x) h g(x) (x) = f (x)g(x) f(x)g (x) g(x) 2 1 dy dy st( ) st() st(), hvis hverken eller er uendeligt ( )( + ) 2 2 h(x) = f(x) h (x) = f (x) = Differentil quotient, which is the slope in point. st ( f(x + ) f(x) ) = f (x) Eqution of the tngent T f(x )(x) = f (x ) x + f(x ) f (x )x = f (x )(x x ) + f(x ). Conditions of monotony, descries if grph is decresing or incresing in intervl. Between these, the grph is constnt. Hyperrel numers, infinitesimls nd infinite numers nd the rel numers. Pge 114 of 152

125 Appendix 1.2 Description of teching differentil clculus Hnd in #3 The sum of composed quotientproducts difference Prolem All kinds of things Use the rules for clculting with differentils to derive the following functions. f(x) = e x. g(x) = 4x 2 + 4x 4 c. h(x) = 3x 4 + 5x 2 16 d. i(x) = 2 1 x x9,9 e. j(x) = 3 x + 2 x x6,6 f. k(x) = 1 g. l(x) = x 7 x 6 + x 5 x 4 + x 3 x 2 + x 1 x x 3 + x3 Prolem Broduct rule Find the derivtive to the following functions, using the rules for clculting with differentils. f(x) = 2x 2 e x. g(x) = 5x 1 x c. h(x) = 1 x x d. i(x) = 4 x x 2,8 e. j(x) = x e x f. k(x) = ex g. l(x) = (e x 3 2 ) (6x 2 + 6x 6) Prolem Composed functions Derive the following functions y first finding the inner nd then the outer function nd then using the rules for clculting with differentils (Given f(g(x)) then the derived function is f (g(x)) g (x)). f(x) = e 2x+5. g(x) = e 3x2 c. h(x) = (4x 7) 7 d. i(x) = x 2 e. j(x) = 6x f. k(x) = 1 g. l(x) = 1 x 4 +3x Prolem Dt is out conditions of monotony Find the conditions of monotony for the following functions. f(x) = 3x 9. g(x) = 2x 2 + x 1 c. h(x) = x 2 16x + 7 d. i(x) = 1 3 x3 4x x x+9 Pge 115 of 152

126 Appendix 1.2 Description of teching differentil clculus Prolem Eqution of the tngent 1.. Find the eqution of the tngent to the function f(x) = 1 in the point (3, f(3)) x. Does n x-vlue exist for which there is no tngent; if yes, then for which x-vlue nd why is there no tngent to the function in this vlue? 2. Given ficticious function, let s cll it finction! Finc out it.. f(x) = 4x 2 8x Find the eqution of the tngent to f(x) in the point ( 1 2, f (1 2 )). Find the eqution of the tngent to f(x) in the point ( 1, f( 1)) Prolem Xhusting Solve 2 of the exercices elow, which cnnot e the one you solved in groups. 1. Let f(x) = 1, prove tht x 2. Let f(x) = x, prove tht f (x) = 1 x 2. f (x) = 1 2 x. 3. Go through the proof for the product rule. I.e. h(x) = f(x) g(x) then h (x) = f (x) g(x) + f(x) g (x). 4. Prove the chin rule (derivtive of composed functions). I.e. h(x) = f(g(x)) then h (x) = g (x) f (g(x)). Pge 116 of 152

127 Appendix 1.2 Description of teching differentil clculus Generl nswers for Hnd in #3 When nothing is mentioned out possile difficulty, the students hd no prolem with this Prolem A The students hd difficulties with the rules for clculting with powers. Prolem B A prt of the students hd miswritten the product rule nd thus mking mistkes tht could hve een voided, hd they only used the right rule. Another prt of the students hd hrd time just using the product rule correctly in the form of wht the one nd the other function ws. Prolem C Some of the students hd prolems recognizing the inner function nd others hd prolems of when to derive which function nd wht to do with it Prolem D A smll prt of the students skipped this prolem entirely, ut those who did not nswered it correctly Prolem E Troule concerning clcultions with eqution solving, not mny for the eqution of the tngent though. A few students kind of overdid it, y using the definition of the differentil quotient to find the derivtive of the function. Others gin did not even try to give solution, ut skipped the prolem entirely. Some students ventured (justified) guess for 1., very few getting the right solution. Prolem X Not mny students hd troule rewriting the proofs done during clss y the other groups. Pge 117 of 152

128 Appendix 1.3 Description of teching integrl clculus 1.3 Description of teching integrl clculus (Mikkel Mthis Lindhl) During the eginning of teching integrl clculus one of the students dropped out of the clss, which mde the numer of students go from 23 to 22! First nd second lesson (5.9) In the eginning of the lesson 15 minutes ws used to go through some of the differentil clculus done in the previous lessons. Then the students ws put in 5 groups nd sked to do worksheet 5.9. IQ 1 : Any ides s to how to find the re of the side of the house? IQ 2 : Wht is the width of the intervls? IQ 3 : Wht is the re of rectngles? IQ 4 : How mny of these rectngles re needed for the re found y the rectngles to e the sme s the sought re? IQ 5 : Wht should we do with the res of the infinitely mny rectngles? Short description of the mthemtics involved Find the re etween the first xis nd the function f(x) = x 2 + 4x. Between nd 4. Find the re under constnt function. Cn function e prtitioned in intervls such tht the function cn e viewed s constnt function on the intervls? Didcticl moment First encounter -> explortio n Constituti on IQ 1 to IQ 5 Min plyer S S/T Mthemticl ojects involved f(x), re of rectngle, infinite sum nd infinitesiml intervls. Didcticl ctivities T wlks round nd helps S if they hve questions for worksheet 5.9. The lesson ends with rekdown of wht the different groups hd come up with. T writes on the BB wht the S re sying, ending with the expression for the re etween the first xis nd the function etween nd 4 s: f(x 1 ) + f(x 2 ) + f(x 3 ) + + f(x4 ). Where = x 1 < x 2 < < x4 < x4 + = 4, is prtition of the intervl [,4]. A list of different nswers to the prolem ws creted during the lesson, ut with the rekdown of wht the students hd done in the different groups the students greed to the sme method s written on the lckord. Approximtely 6 % of the students hd come up with the indices for the different x-vlues, the rest of the students hd only explined it orlly. One student wnted to mke the height of the rectngles infinitesiml nd the width of them finite, i.e. something tht looked more like the Leesgue integrl thn the Riemnn integrl, the student recognized tht the re otined y the different pproches where the sme. It should e noted tht 3 out of the 5 groups strted to do wht would hve een clled numericl integrtion, y mking finite prtition nd then clculting the res of the rectngles. Pge 118 of 152

129 Appendix 1.3 Description of teching integrl clculus Worksheet (5.9) An lien clled Ler poses her high school techer prolem: StndrdJoe in my fmily we re going to hve prty to celerte my dd, Noitknufmts, he loves oth honey nd sffron, nd s ll other liens; to lick his house to keep it wter resistnt. By extension I wnt to cover side of my dd s house in honey nd sprinkle sffron on top of the honey, so tht we cn lick the house clen during the prty together. Since oth honey nd sffron re in short supply on our plnet I don t wish to uy too much of it, hence my question is; If I Hve the following function, which descries the height of the house ove the surfce of the plnet, f(x) = x 2 + 4x, how do I find the Are of the side of the house? To this end I ve rought picture of my dd s house: (4,) Then the techer poses some question to help nswer the first question: 1. How do you find the re under constnt function? One could try to drw one nd find the re etween the grph nd the first xis on certin intervl [, ]. 2. Cn the function e prtitioned in intervls such tht the function cn e viewed s constnt function on the intervls? When Ler finds the nswer to her question one of her friends tells her tht it could e prolemtic with honey nd sffron, on the ottommost 2 meters of the house, if the wll re to e used for skull sucking during the prty. Bonus exercises: 1. Wht re is to e covered in honey nd sffron if there should e plce for skullsucking? 2. How tll is the house? 3. Wht is the re of the neighoring house, when the height is descried y the function f(x) = x for x [,4]? 4. Write generl expression for the re etween grph (of function) nd the first xis on n intervl. The different groups cme up with lmost the sme su questions nd nswers for the generting question in the worksheet. The two questions posed in order to help the students long the wy were ll nswered s first step to nswer the generting question. Pge 119 of 152

130 Appendix 1.3 Description of teching integrl clculus 1. The re under constnt function cn e found in the sme wy s finding the re of rectngle, the height times the width, in this cse the height would e the function vlue, nd the width would e the length of the intervl. 2. Yes, it cn e done nd in order to view the function s constnt in the intervl, the intervls hve to e of infinitesiml length. With these two nswers the students strted to generte wy to find the re. The students strted to divide the re into smller res, some of them lso used tringles in the eginning, the tringles where used to get etter pproximtion when using finite prtition. When going to the infinite prtition the students wrote the following: The intervl hd to e infinitesiml, i.e. ny of the intervls could e written s [x, x + ] some of the students lso wrote x 2 x 1 = nd so on for every other integer (understood s every other set of consecutive integers). With this infinitesiml intervl the students then recognized the height of the function s something etween f(x) nd f(x + ). One group of students did very surprising oservtion; they noticed tht it did not mtter if they used f(x) or f(x + ) s long s they used the sme for every index. Considering the symmetry of the function they estlished tht the error gined on the first hlf of the re (until x = 2) would e cncelled y the error on the second hlf. The mjority of the students used the first point in the intervl nd sid the function vlue in this point corresponds to the height of the rectngle. In this wy they found the re of given rectngle with infinitesiml width s f(x). Since none of the student new of the symol s wy to write sum, the students wrote the sum s f() + f() + f( + ) (yes they forgot the strs!!) ut none of them wrote the three dots, they just stopped nd then wrote in text tht one hd to do it n infinite numer of times. When the groups got to this point in the nswer for the question of finding the re, the techer sked the groups to figure out how mny times this infinitely mny hd to e done. The first nswer to this question ws 1 which prompted nother techer generted question, wht is the length of the intervl otined y dding 1 mny intervls of length. This led to the conclusion tht the numer of infinitesiml intervls would e 4. Oserved (techer) generted su prolems for the worksheet: 1. How does one divide the re, which need e found, into smller res which cn e found? 2. Should the smller res only e rectngles or lso tringles, nd why? 3. How does one compute the re of rectngle? 4. How does one compute the re of (right) tringle? 5. Wht is the re of rectngle with infinitesiml width? 6. How does one prtition n intervl nd how should it e written? 7. If the length of the infinitesiml intervls is then wht is the infinite numer used to prtition the intervl? Pge 12 of 152

131 Appendix 1.3 Description of teching integrl clculus Third lesson (5.1) Most of this lesson ws used to determine wht the differentil of quotient of functions is through working with worksheet (5.1), nd only the lst of the lesson were used to introduce the integrl nd its informl definition otined in the previous lesson. In hindsight this ws not very good ide s the mount of lessons left for doing integrl clculus ws lredy t minimum. IQ 6 : Is this infinite sum rel or hyperrel numer? IQ 7 : Cn something e done to hyperrel numer which ensures it ends up s rel numer? Short description of the mthemtics involved Using n infinite sum of res of rectngles to define the definite integrl. Didcticl moment Recp of previous lesson Min plyer T Mthemticl ojects involved f(x + ), re of rectngle, infinite sum, x 1 + x x n, infinitesiml intervls nd prtition Didcticl ctivities T tlks out the house nd re from lst lesson nd rewrites wht they hd on the lckord t the end of the previous lesson. Institution liztion IQ 6 IQ 7 S/T st(), f(x). T: when using this pproch to tlk out the re etween the function nd the first xis, it is tiresome to write this sum every time, hence we introduce, this weird S, f(x), s the stndrd prt of the infinite sum of the re of the rectngles with infinite width nd height s the function vlue. And drws n exmple of liner function nd explins how to use the integrl nd wht the oundries re Generl description of the nswers to the worksheet The worksheet ws nswered y the students in groups nd with little help from the techers the students were le to estlish the rule. In the end the rule ws presented on the lckord in plenum. Pge 121 of 152

132 Appendix 1.3 Description of teching integrl clculus Worksheet (5.1) Prolem Let f(x) nd g(x) e differentile functions nd let h(x) = f(x). show tht the derivtive of h(x) is g(x) Hints : 3. = 1 4. nd use the derivtive of the following:. f(x) = 1 x. h(x) = f(x) g(x) c. h(x) = f(g(x)) h (x) = g(x) f (x) f(x) g (x) g(x) 2 Pge 122 of 152

133 Appendix 1.3 Description of teching integrl clculus Fourth lesson (5.11) The students were given worksheet (5.11), which included exercises to cover the next prt of the plnned integrl clculus. The lst 2 minutes of the lesson ws used to go through the exercises in plenum, s to mke sure they ll hd ll seen the correct methods, which where needed for the upcoming hnd in #4. Short description of the mthemtics involved Definite integrl, of simple functions, nd definite integrls with vrious endpoint, i.e. re functions. Didcticl moment First encounter with prolems of type I 1 T 1 Min plyer S Mthemticl ojects involved A(re) =, T A(re) = 1. 2 Didcticl ctivities wlks round nd helps S if they don t understnd the exercises Generl description of the nswers to the worksheet In the eginning they hd to think while in order to understnd wht the students were supposed to do. After turning pper round they sw the grphs of some of the functions which mde them do the connection s to how they should find the re. Most of the students got to finish the worksheet nd everyone did t lest the first pge. With the prolems on the worksheet lmost self-explntory description of the nswers will e omitted. Pge 123 of 152

134 Appendix 1.3 Description of teching integrl clculus Worksheet (5.11) The definition of the definite integrl, for function f(x) on the intervl [, ], s given in clss in the previous lesson, demnds prtition of the intervl [, ]. The intervl is prtitioned into infinitely mny, N R, prts such tht 1 is n infinitesiml. This prtition cn e written in the following wy: N = x 1 < x 2 < x 3 < < x N =. Thus x 2 x 1 is n infinitesiml nd x 5 x 4 s well, nd so forth. With this in mind the definite integrl ecomes: f(x) = st(f(x 1 ) + f(x 2 ) + f(x 3 ) + + f(x N 1 )). Determine the following integrls (one could drw the functions which re to e integrted) Exercise 1 Exercise 2 Exercise 3 5 = 2 = 4 = = 2 = = 3 = 4 x 4 4. Wht is the integrl of generl constnt function f(x) = c? 5 c = 6 3 = Determine the following integrls (one could drw the functions which re to e integrted) = 2x = 4 3x = Exercise 5 Exercise 6 Exercise 7 5 2x = 2 7 2x = 3 7 3x = x = 8. Wht is the re of generl liner function through zero, f(x) = kx? kx. 6 4x = 6 3x = 2 Pge 124 of 152

135 Appendix 1.3 Description of teching integrl clculus Determine the re under the grph on n intervl of the type [, x] nd write n re function, A(x), which descries the re on the intervl. Exercise ) 1. Determine the re function for f(x) = 4 2. Determine the re function for f(x) = c Exercise ) 1. Determine the re function for funktionen g(x) = 2x 2. Determine the re function for g(x) = kx Exercise c) 1. Determine the re function for h(x) = 1 x + 4 findes 2 2. Determine the re function for h(x) = kx + c Pge 125 of 152

136 Appendix 1.3 Description of teching integrl clculus Fifth lesson (converted #5) (5.13) They got to spend the lesson doing their Hnd-in # 4 which ws given to them the very sme dy. The Hnd in ws given to them in prts, s it ws not fully developed. Most of the students did do the first pge in this lesson which ll concerned vrious prolems of type kx + c, ending with generl description of wht kx + c is in terms of,, c, k nd x. Short description of the mthemtics involved Vrious prolems of type kx + c Didcticl moment Explortion nd technicl work Min plyer S Mthemticl ojects involved kx + c Didcticl ctivities T is present to nswer questions if certin exercise is not understood Generl description of nswers to hnd in #4 pge 1. The techniques developed when doing the previous worksheet should e enough to nswer the questions on this pge. To this end tle descriing the students understnding of the technique cn e seen elow. The tle is generted on sis of the nswers the students gve to the exercise on pge 1 of hnd in #4. Technique to solve the integrl of constnt function Technique to solve the integrl of liner function. Numer of students who understood the technique Numer of students who misunderstood the technique (1) 9 + (1) The prentheses round the 1 is ecuse two of the students used very geometric understnding of the figure descried y the intervl, function nd first xis, thus otining different technique, where the figure would e descried s tringle plus rectngle, nd these two res would e dded in the end. This technique though correct, might look like technique which hs not een fully routinized. The 9 students tht misunderstood the technique did so in the very sme wy, ll of them seemed to think tht ( ) 2 = ( 2 2 ). This misunderstnding cn e seen s lck of previous estlished prxeology or s misunderstnding y the techers, since they did not include this s prt of the knowledge to e tught. The techers seemed to think tht the hierrchy of mthemticl clcultions should hve covered this spect. Pge 126 of 152

137 Appendix 1.3 Description of teching integrl clculus Hnd in #4 pge 1 Integrtion y prts Exercise Alien 3, 2, 1 LICK! is how this yer s Lickeliciousness strts. The dy is lwys on the 7 th Fleesdy fter the Ykis-celertion nd this yer it is on Octemer 35 th. The suns re shining nd everyone rush out of their houses, to find the perfect house to lick. The rules re simple: 1. The one, who licks the most, is the winner. 2. You re llowed to lick house which is lredy eing licked, ut you cn t lick your own house. In the event of more thn one licker, skull sucking is not llowed. 3. When your own house hs een licked clen, you cn t lick nymore. For rule numer 3, they hve sying: people(liens) who live in licked houses shouldn t lick stones. Hhh! Proly should hve een there.. Aer gets to lick 4 houses efore her own house is out of the gme. Wht she licked cn e descried y the following integrls Noitces house Dleif s house x 1 Tnuom s house (the T is silent) Ecfrus house 5 (2x + 1) 2 4 ( x + 4) 1. Figure out how much Aer hs licked on ech house efore her gme is over. 2. Determine if Aer hs licked ny of the houses y herself. It ws Twolegged Toleg Twoleg who gve Are s house the lst lick; literlly, it ws the only prt of her house he got to lick! At the very lest, Aer wnted to et her in the competition. This is why she checks Twolegged Toleg Twoleg s results, which cn e descried y the integrls elow House 1: House 2: House 3: (2x 4) Did Aer et Twolegged Toleg Twoleg? 4. Find generl expressions for licking the different houses ove, s shown elow. c kx (kx + c) Pge 127 of 152

138 Appendix 1.3 Description of teching integrl clculus Sixth lesson (5.18) IQ 8 : Wht is n integrl, do you rememer? IQ 9 : Wht does f(x) men? Is it numer, function or nn? x IQ 1 : Wht is c? Short description of the mthemtics involved Recp of wht n integrl is. Recp of some of the exercises from lst time. Group work with worksheet 5.18 Didcticl moment IQ 8 IQ 9 IQ 1 Institution liztion. First encounter with I 2 T 3 Min plyer S/T T Mthemticl ojects involved kx x c f(x), nd it s definition nd the rules for operting with the stndrd prt. Didcticl ctivities T: drws nd in coordinte system where grph of the function f(x) = kx re lredy drwn. T: writes the integrl on the BB nd S x nswers: c = x(c ) T wlks round nd helps S if they don t understnd the exercises. Pge 128 of 152

139 Appendix 1.3 Description of teching integrl clculus Worksheet 5.18 Exercise 1: Show tht the integrl of sum or difference of functions is equl to the sum or difference of the integrl of the functions t hnd. I.e. show tht Exercise 2: (f(x) ± g(x)) = f(x) ± g(x) Show tht the integrl of constnt, k R, times function, f(x), is the sme s the constnt times the integrl of the function. I.e. show tht Exercise 3: k f(x) = k f(x) Show tht the integrl from to dd to the integrl from to c is equl to the integrl from to c, when the integrnt is the sme throughout. I.e. show tht Exercise 4: f(x) c + f(x) c = f(x) Show tht the integrl over n intervl of type [, ], thus only contining one rel point, is zero. I.e. show tht Exercise 5: f(x) = Show tht the definite integrl chnges sign when the oundries re switched. I.e show Exercise 6: f(x) = f(x) Write the results of the following integrls nd descrie wht cn e sid out the derivtive of the results. x c = Hints: x kx = x (kx + c) = 1. rememer the definiton of the definite integrl (for exercise 1,2,3): Pge 129 of 152

140 Appendix 1.3 Description of teching integrl clculus f(x) = st ( f ( + 2 ) + f (x f (x N ) ), ) + f (x ) Where < x 1 < x 2 < < x N 1 < x N =, is n infinitesiml prtition of the intervl [, ] nd N R is n infinitely lrge integer. 2. st( + ) = st() + st(), if nd only if nd re finite! (for exercise 1,2,3) 3. Rememer tht st(k ) = k st(), if k is rel numer. 4. Use the nswers from 3 nd 4 to show Rememer tht the integrl cn lmost e viewed s n re (for exercise 6) Bonus Exercise 1: Determine the integrl from to x of the function, x 2, i.e. wht is Hints: x x 2 1. Believe in wht you found in exercise Rememer the product rule from differentil clculus, if f(x) nd g(x) re 2 differentile functions, nd h(x) = f(x) g(x), then h (x) = g(x) f (x) + f(x) g (x). 3. Wht is x g(x) f (x) + f(x) g (x) 4. Let f (x) = x nd g(x) = x Bonus Exercise 2: Determine the re of the side of Noitknufmts house, Ler covered in honey nd Sffron. I.e. determine the integrl 4 x 2 + 4x. Pge 13 of 152

141 Appendix 1.3 Description of teching integrl clculus Oserved nswers for Worksheet 5.18 Since the nswers generted y the students were lost, here is generl nswer to the worksheets s rememered nd found in the udio recordings Exercise 1 The students generl understnding of the integrl s the re etween the first xis nd the grph, mde this exercise intuitively very esy. When forced to use the definition of the integrl most of the students did not other with writing the entire sum, they thought it cumersome. Some students introduced other symols for the sum, so strting from the right going left they wrote, st(sum(f)) + st(sum(g)) = st(sum(f) + sum(g)) = st(sum(f + g)) referring to the techniques estlished for the stndrd prt. Other students wrote in sentences wht mounted to the sme thing Exercise 2 The students used some of the sme nottion s the exercise efore or used sentences, if tht ws wht they hd done in exercise 1. Finding tht it cme down to the lgeric opertion of multiplying the constnt on every term in the sum, nd the technique for tking the stndrd prt of rel constnt times hyperrel numer. I.e. k st(sum(f)) = st(k sum(f)) = st(sum(k f)) Exercise 3 This exercise proved to e little tricky, the students intuitively understood tht the eqution should hold (concluded from the orl explntions oserved) ut the students which hd exchnged the infinite sum with sum(f) hd suddenly lost the oundries, which were needed to prove this. In the end none of the students did this in mthemticl wy, they ll reverted to the use of sentences to descrie the process. An exmple of this is the following: the re is the sme if we find the re of the prts nd dd them or finding the re of ll of it Exercise 4 Finding the re of something with zero width will lwys e zero ws the generl nswer to this. Some of the students tried to use constnt function s n exmple. The students hd prolems with mking prtition of n intervl with zero length Exercise 5 They ll needed the hint ut in the end, most of them found tht exchnging c with in exercise 3, nd then using the result from exercise 4, gve the desired result Exercise 6 This exercise enled some of the students to guess t the connection etween the derivtive nd the definite integrl with vrile endpoint Bonus exercise 1 This exercise ws only done y hndful of students. The students who did this hd lredy guessed t the connection from exercise 6, so the students who did this exercise relly just guessed t result nd hd no rel wy of checking if it were correct (other thn sking the techer). Pge 131 of 152

142 Appendix 1.3 Description of teching integrl clculus Seventh lesson (5.2) Introducing ntiderivtives nd indefinite integrl. IQ 11 : Do you know ny functions which you lso know the integrl of? x IQ 12 : Wht hppens if we differentite one of the functions in the row with f(x)? IQ 13 : Wht if f(x) = x 2 x, do we then know wht f(x) is? x IQ 14 : Do we know wht ( f(x) ) for f(x) = x 2? IQ 15 : Wht is n ntiderivtive of f(x) = x? IQ 151 : Wht if I wnt the ntiderivtive to go through the point (,2)? IQ 16 : How mny ntiderivtives does function hve? IQ 17 : Wht is n ntiderivtive of the function f(x) = k? IQ 171 : Wht is the indefinite integrl of the function f(x) = k? IQ 18 : Wht is n ntiderivtive of the function f(x) = kx? IQ 181 : Wht is the indefinite integrl of the function f(x) = kx? IQ 19 : Wht is n ntiderivtive of the function f(x) = x 2? IQ 19 1 : Wht is the indefinite integrl of the function f(x) = x2? IQ 2 : Wht is n ntiderivtive of the function f(x) = e x? IQ 2 1 : Wht is the indefinite integrl of the function f(x) = ex? IQ 21 : Wht is n ntiderivtive of the function f(x) = x n? IQ 21 1 : Wht is the indefinite integrl of the function f(x) = xn? Pge 132 of 152

143 Appendix 1.3 Description of teching integrl clculus Short description of the mthemtics involved Recp of wht hve een done with integrls so fr. Connection etween integrl nd derivtives Antiderivtive Indefinite integrl Didcticl moment Explortio n IQ 11 IQ 12 IQ 13 IQ 14 institution liztion institution liztion IQ 15 IQ 151 IQ 16 Institution liztion Min plyer T/S T/S T T S (first) encounter, explortory T Mthemticl ojects involved kx f (x) x ( f(x), ) F (x) = f(x) F(x)=ntiderivtive of f(x). f(x) Didcticl ctivities T: drws tle on the lck ord with 1. column of functions, 2. column of integrls of the functions with vrile endpoint nd 3. column of derivtives of the integrted functions. (see tle elow) T sttes the fundmentl theorem of clculus! With short considertion of the re function. x+ f(x) f(x) f(x) T presents the definition of the ntiderivtive s function F, for which F (x) = f(x). which y the fundmentl theorem is the sme s x F(x) = f(x) + k T writes the definition of the indefinite integrl s x f(x) = F(x) + c IQ 17 to IQ 211 First encounter nd explortor y S For some constnt c, nd explins why it could e written s f(x) = f(x) + k. T: writes the students nswers on the BB nd in the end fills out the rest of the tle on the BB. x f(x) x f(x) x ( f(x) ) c c(x ) c kx 1 2 k(x 2 2 ) kx x (x 3 3 ) x 2 e x e x e e x x n 1 n + 1 (xn+1 n+1 ) x n Pge 133 of 152

144 Appendix 1.3 Description of teching integrl clculus Eighth lesson (5.2) In this lesson they got to spend time on their hnd in # 4, which ws lso used s kind of worksheet they should hve the techniques to do pge 2 of sid hnd in. most of the students got to do ll the indefinite integrl nd some of the specific ntiderivtives y letting them pss through point. Short description of the mthemtics involved Vrious prolems of type f(x) Didcticl moment Explortory, constitutionl nd (technicl work) Min plyer S Mthemticl ojects involved f(x) Didcticl ctivities T is present to nswer questions if certin exercise is not understood. Determine specific ntiderivtive y letting it go through given point. First encounter nd explortory. S Eqution solving. Determine the sum of 4 definite integrls written s text only. S f(x) Generl description of nswers to hnd in #4 pge 2. Generlly speking the results for this prt of the hnd in were rther good. The only prolems worth mentioning re when the integrnd is product of sum of power functions (which could e recognized s the derivtive of product of functions). In this prt very few rememered to do the multipliction efore finding the indefinite integrl or even recognized it s the derivtive of product of functions. In the prolem, 6x 2 e 2x3, only hndful of students ctully got the ide to look for something which could e differentited into the integrnd using the rule for the derivtive of function composed with nother. This prolem cn e seen s first encounter with prolem requiring the technique integrtions y sustitution. In the prolems with finding the specific ntiderivtive no one seemed to hve prolem with figuring out wht to do, though lot of the students complined tht the mount of repetitions of the sme kind of prolems ws over the top. In the lst prolem of the pge, most students did not get, tht it ws the definite integrl of the specific ntiderivtive found y letting it pss through point ws the one to use s n integrnd. As such only hndful of students displyed the use of the technique f(x) = F() F(). Pge 134 of 152

145 Appendix 1.3 Description of teching integrl clculus Hnd in #4 pge 2 Exercise Building new houses After terrile sun storm, severl of the houses were dmged; the houses simply weren t licked enough. It turned out to e the houses tht Aer hd licked during the contest; you see it ws Noitces, Tnuom, Dleif nd Ecfrus who hd lost their houses during the storm. The myor, Noitknufmts, decided tht new houses were to e uild ut this time it should e done s they did in the dys of yore. Thus, they hd to e uilt s their ncestors did. Aer ws, s her punishment, to find the ncestors lueprints, which could e found y finding the ntiderivtives of the destroyed houses height-functions for every side. Noitces new house cn e descried y Tnuom s new house cn e descried y x 2. ( 1 4 x + 2) 2. (x 2 4x + 8) ( x 3 + 8) 4. ( 1 4 x + 3) 4. ( x 2 + 4x) Dleif s new house cn e descried y c1. (4x 3 3x 2 + 2x) Ecfrus new house cn e descried y d1. e x c2. c3. 1 x 2 d2. 8 x 2 d3. ((2x + 8)(2x) + 2(x 2 )) 6x 2 e 2x3 c4. ( 4x 3 + x 4 ) d4. 1 ex 1. When Aer hd done this tiresome work, she found out tht the ncestors hd not given thought to the future, in the sense tht, in the future there would e string phones etween every house. These strings needed some spce, which is why Noitknufmts hd found the specific points the ntiderivtives should go through, in order to mke room for the string phones. (The points hve een given nmes, such tht they correspond to the nmes from the previous exercise.) 1(,) 1(2,9) c1(2,) d1(, 3) 2(4,12) 2(3,15) c2(1,) d2(1,12) 3(2,4) 3(2,8) c3( 8,8) d3(2,) 4(4,8) 4(3,) c4(1,3) d4(, 3) Newly uild houses need to e licked thoroughly in order to keep them wtertight, which is why Aer ws to lick some of Noitces new house. She licked from to 1 on the first side, from 1 to 2 on the second side, from 2 to 3 on the third side nd from to 4 on the lst side. How ig n re did Aer lick? Pge 135 of 152

146 Appendix 1.3 Description of teching integrl clculus Ninth (5.23) This lesson ws used to descrie the technique of integrtion y sustitution. IQ 22 : Wht hve we used when writing n infinitesiml? IQ 22 1 : Wht other thn? IQ 23 : If I write df, wht do you think it is? IQ 231 : Wht is df? IQ 24 : Wht is d (e3x2 )? IQ 241 : Then wht is 6x e 3x2? IQ 25 : Wht is d (F(g(x))), when F(x) is n ntiderivtive of f(x)? IQ 251 : Then wht is f(g(x)) g (x)? IQ 26 : cn nyone recognize the inner function in the expression f(g(x)) g (x)? IQ 261 : Wht is du if u = g(x)? IQ 262 : How cn f(g(x)) g (x), e written with u nd du? IQ 27 : Now with oundries on the integrl, wht is f(g(x)) g (x). IQ 271 : Normlly we find the ntiderivtive of function nd then put the oundries in insted of the vrile ut know we use u = g(x) s vrile, hence f(g(x)) g (x) runs from to ut the vrile u should not run from to. Thus f(g(x)) g (x) How do we mend this? IQ 28 : Using integrtion y sustitution wht is e x 1? x descries tht the vrile x f(u) du Pge 136 of 152

147 Appendix 1.3 Description of teching integrl clculus Short description of the mthemtics involved Different wys of writing the derivtive An introduction to integrtion y sustitution s the opposite of the chin rule for differentition. Develops the technique used when doing integrtion y sustitution Didcticl moment Institutionl iztion IQ 22 to IQ 231 IQ 24 IQ 241 Explortion nd constitution IQ 25 IQ 251 (first encounter) Explortion -> Constitution IQ 26 to IQ 262 Min plyer Mthemticl ojects involved Didcticl ctivities T, df, f The S s nswer the questions nd the T ends up y letting df e defined s df f (x). T/S T/S d, F(x), f(x), f(x) d, F(x), f(x), f(x), df T writes wht the students re sying on the BB nd ends up with the eqution f(g(x))g (x) = F(g(x)) + c T writes the students nswers to the questions, let u = g(x), then du = g (x) nd then f(g(x))g (x) = f(u) du = F(u) + c = F(g(x)) + c. The red color indictes tht it is not mthemticl correct. Further develops integrtion y prts to include oundries. Technicl work IQ 27 IQ 271 T/S d, F(x), f(x), f(x), df T writes the students nswers to the questions, nd ends with the expression f(g(x))g (x) = g() f(u) g() du Pge 137 of 152

148 Appendix 1.3 Description of teching integrl clculus Tenth lesson (5.23) In this lesson they got to spend time on their hnd in # 4, which ws lso used s kind of worksheet they should hve the technique to do pge 3 of sid hnd in. Short description of the mthemtics involved 4 prolems where the use of integrtion y sustitution is needed. Didcticl moment Explortory, constitutionl nd technicl work Min plyer S Mthemticl ojects involved d, F(x), f(x), f(x), df Didcticl ctivities After spending lmost hlf the lesson the where the students did not understnd the technique the techer opted to the first nd lter the second prolem on the lckord in plenum. With the first two prolems hving een done on the lckord the prolems where not considered s prt of the hnd in which where to e grded, this ws not explined to the students, since the techers wnted the students nswers to the first two prolems to see if the students understood wht ws done in plenum Generl description of nswers to hnd in #4 pge 3 These lst four prolems relly mde the students go out of their skins. The students did not relly get the technique nd hd prolems reproducing the sme steps tht were mde in plenum. In order to get n overview of this the following tle cn help prolem # of students using the technique correct # of students using the technique incorrect ) 17 ) 16 c) 6 9 d) 4 11 As seen in the tle not even ll the students did the two prolems which were done in plenum, supporting the fct tht they relly did not understnd wht ws going on. The plce where most of the students tht hd done prolem c nd d got it wrong ws tht they did nothing or differentited insted of finding the ntiderivtive when they hd done the sustitution. Pge 138 of 152

149 Appendix 1.3 Description of teching integrl clculus Hnd in #4 pge 3 Exercise Clone Being the myor of Qudrnt 23u hs its perks. Myor Noitknufmts hs een given the tsk of testing the first cloning mchine of not only Qudrnt 23u ut of the entire solr system. When they clled from the lortory Aer overherd the string phone converstion etween her dd nd the l, so of course she lredy conspired on how she should use it. There ws this yerly mth test she hd on Ferugust 21,5 nd could relly use clone who could tke her plce! She clled her mission the method of sustitution nd strted y estlishing the expressions for her sustitute right wy, so tht it could look like her s much s possile nd then she would swoop in fter the test s if nothing hd hppened. 1. Use the method of sustitution on the expressions elow, which will, hopefully crete perfect Aer clone.. c. 5 2x (x 2 + 3) 2 1 8x3 + 14x 2x 4 + 7x (4x 3 + 2x 2 ) 1 (12x 2 + 4x) 1 4 d. e1 x x 2 12 Aer didn t nil it the first time round; it wsn t until she mde the third clone, tht she hd mde one tht resemled her so much, tht she could insert it insted of herself, even though it stuttered it.. She nmed the clone Anti Aer. The first two clones were mde ecuse she hd used the expressions cos(ln(x) + π) nd cos(x) e sin(x)+42 πx ut they were too smrt, so no one would hve elieved it ws Aer. She sent them to orit round her plnet, ut lunched them with little too much speed, so they chnged heding to smll unknown plnet. They would wnder round there for it, spreding some good mthemtics, until she would need them ck one dy. They would gin the peoples trust. They would sve this plnet this.. Erth.. They would e known s StndrdJoe nd HyperMick. Pge 139 of 152

150 Appendix 1.3 Description of teching integrl clculus Eleventh (nd twelfth) lesson The eleventh lesson ws used to summrize wht they hd done in integrl clculus; to this end, ( lnk version of) Worksheet 5.24 ws prepred nd hnded out to them. The students then spent round 4 minutes to fill out the worksheet, to the est of their ilities, in smll groups (of order < 5). Afterwrds the groups red loud, to the rest of the clss, wht they hd on their ppers, ending with the cretion of the following filled out worksheet. The twelfth lesson ws used for ny questions they might hve nd to sy proer frewell Worksheet 5.24 f(x) c kx x 2 x n 1 x 2 = x x x = x 1/2 e x Rules for clculting integrls F(x) cx 1 2 kx2 1 3 x3 1 n + 1 xn+1 1 x = x 1 2 x 3 x3 2 e x k f(x) = k f(x) (f(x) ± g(x)) = f(x) f(x) c + f(x) c = f(x) f(x) = f(x) = f(x) f(x) = F() F() f (g(x))g (x) g() = f (u) g() g() ± g(x) du = f(g()) f(g()) f(g(x))g (x) = f(u) du = F(g()) F(g()) g() Pge 14 of 152

151 Appendix 2.1 Hyperrel nlysis tle 2 Anlysis tles This prt of the ppendix contins the nlysis tles for the hyperrel numers, differentil clculus nd integrl clculus respectively. 2.1 Hyperrel nlysis tle The constnts, nd k re rel, the vriles x nd y re rel, nd dy re infinitesimls, nd α nd β re hyperrel numers. Lesson # (14:45-15:5) Type of prolem HP 1 : Specify into which sets of numers infinitesimls nd infinite numers elong. Hp 1,1 : Give definition of n infinitely smll quntity. Hp 1,2 : Give definition of n infinite quntity. Mthemticl techniques τ 1 : Process of exclusion from the known set of numers N, Z, Q, R. Technologicl theoreticl elements θ 1 : (Informl) Definition of n infinitesiml. θ 2 : (Informl) Definition of n infinite quntity. Didctic moment(s) First encounter with HP 1 Institutionliztion with the former introduced sets of numers, N, Z, Q, R. Elements of the didcticl techniques Intuition is used to identify n infinitely smll quntity s something tht is less thn ny writele positive numer nd n infinite quntity s something tht is greter thn ny writele numer (8:1-9:15) Hp 1,3 : Drw numer line with infinitesimls. Hp 1,4 : Wht is 2 nd where is it on the numer line? Hp 1,5 : Wht is nd 2 where is it on the numer line? Hp 1,6 : Wht is + dy nd where is it on the numer line? HP 3 : Find k st( x + + c dy). Hp 3,1 : Which rel numer is infinitely close to 7 +? Hp 1,7 : Wht is 1 nd where is it on the numer line? Hp 4,2 : Give n exmple of hyperrel numer with n undefined stndrd prt, i.e. find such tht st() is meningless. τ 1,i : Using the sme technique s when operting with rel numers. i = {4,5,6} τ 3,1 : Remove the infinitesiml(s); they re virtully τ 1,7 : Use the rules for dividing y smll numers to estlish tht 1 is infinite, plcing it eyond the rel numers. τ 4,2 : Find hyperrel numer tht does not hve rel numer infinitely close to it θ 3 : The ordering of the hyperrel numers θ 4 : How to dd, multiply nd divide with mixture of infinitesimls nd rel numers. θ 5 : (Informl) Definition of the stndrd prt. Θ 1 : An intuitive construction of the hyperrel numers s the rel numers plus the infinitesimls nd opertions etween these (θ 4 ) Explortory (nd constitution) for HP 1. (prtly) Constitution of the theory lock for HP 1 nd institutionliztion y compring with the theory for rel numers. First encounter with HP 3. Further constitution of the theory lock for HP 1 nd for the entire HMO. Using two mthemticl ojects, firstly the definition of infinitesimls nd secondly the known rel numer line, to produce nother set of numers. Using previously estlished prxeology on how to operte with rel numers induces how to operte with hyperrel numers. Identify the finite hyperrel numer s sum of rel numer nd n infinitesiml. Recognize tht n infinite numer cnnot e written s the sum of rel numer nd n infinitesiml. Using two mthemticl ojects, nmely the coordinte system nd the implicit use of Pythgoren theorem yields the difference in function- Pge 141 of 152

152 Appendix 2.1 Hyperrel nlysis tle (1:4-11:45) HP 2 : Determine if given function, f(x), is continuous. Hp 2,1 : Give exmples of functions? Hp 2,2 : Wht is the distnce etween 2 consecutive points on grph of function which cn e drwn without lifting the drwing device? And wht is the difference in the function vlue with n infinitesiml difference in the vrile vlue? Hp 2,3 : Does f(x + ) mke sense when f is rel function? HP 3 : Find k st( x + + c dy). Hp 3,2 : Find st(x + ) + st(y + dy) Hp 3,3 : Find 5 st(x + ) Hp 3,4 : Find st(5x + 5) Hp 3,5 : Find st( + ) Hp 3,6 : Find st( + dy) Hp 3,7 : Find st(5) Hp 3,8 : Find st ( 1 ) Hp 3,9 : Why is st (5 1 ) = 5 Hp 3,1 :Give n exmple of n infinitesiml which is lso rel numer Hp 3,11 :Find st((2 + )(3 + dy)) Hp 3,12 :Find st(2 + )st(3 + dy) Hp 3,13 :Wht cn e concluded from the two previous exercises? Hp 3,14 :Find st()st ( 1 ) Hp 3,15 :Amend the conclusion from Hp 3,13 Hp 3,16 :Find st((4 + ) 2 ) τ 2,1 : Rememer wht function is. τ 2,2 : Deduce tht n infinitesiml distnce etween 2 points mkes the difference in oth coordintes n infinitesiml. τ 2,3 : Relize tht the vlue x + is not prt of the domin. τ 3,i : st(kα + β) = k st(α) + st(β) When α nd β re oth finite. i {2,3,4,5,6} τ 3,i : st(α β) = st(α) st(β) When α nd β re oth finite. i = {7,8,,16} θ 6 : NSA definition of continuity in point. θ 7 : (Informl) Definition of the str opertion s the extension of the domin of function. First encounter with HP 2. Explortory (nd technicl work) on HP 3. Explortory nd technicl work on HP 3. nd vrile vlues re infinitesiml. Using previously estlished prxeology on functions nd their domin to determine tht x + is not prt of the domin of rel function. Identify the finite hyperrel numer s sum of rel numer nd n infinitesiml. Recognize tht n infinite numer cnnot e written s the sum of rel numer nd n infinitesiml. How to operte with hyperrel numers (8:1-9:15) HP 4 : Find the hyperrel function vlue f( + k) for given rel function HP 4,1 : For f(x) = 3x + 1, find the hyperrel function vlue τ 4,i : Evlute the hyperrel function s if it were rel function nd rel vrile. i = {1,2,3} First encounter with HP 4. Using previously estlished prxeology on functions nd their domin nd the recently estlished str- nd stndrd prt opertion. Pge 142 of 152

153 Appendix 2.1 Hyperrel nlysis tle f(1 + ) nd f( 2). HP 4,2 : For g(x) = x, find the hyperrel function vlue x = 3 nd x =. HP 4,3 : For h(x) = 2x 3, find the hyperrel function vlue h(1 + ) nd h(4 + ). HP 5 : Wht is st( f(x + )) for rel function? Hp 5,1 : Wht is st( f(x)) for rel function? Hp 5,2 : Assume now tht f is continuous rel function wht is st( f(x + ))? τ 5,1 : Stndrd prt counters the str opertion. τ 5,2 : Stndrd prt counters the str opertion nd if the function is continuous then it lso counters the infinitesiml prt of the vrile. Lesson # Type of prolem Mthemticl techniques Technologicl theoreticl elements Didctic moment(s) Elements of the didcticl techniques Pge 143 of 152

154 Appendix 2.2 Differentil clculus nlysis tle 2.2 Differentil clculus nlysis tle (Jons Kyhnæ) Lesson # Type of prolem Mthemticl techniques Technologicl theoreticl (9:25-1:3) (8:1-1:3) DP 1 : Find the slope of given grph in point y drwing tngent Dp 1,1 : Find n intervl where the grph is the steepest Dp 1,1,1 :Wht is the est intervl, to find where the grph is the steepest? Dp 1,1,2 :How to find the exct slope where the grph is the steepest Dp 1,2 : Where on the grph cn the slope e found? Dp 1,3 : Find the slope in n ritrry point on the grph Dp 1,4 : Find the slope for f(x) = 1 4 x2 + 4x 5 in the points (4,7), (8,11) nd (12,7) DP 2 : Find n lgeric expression for finding the slope for function in point Dp 2,1 : Find the slope of function, f(x), in point, without drwing nything Dp 2,1,1 :Find the generl slope of function, f(x), in the intervl [; ] Dp 2,1,2 :Wht cn e sid out the slope to function, f(x), nd the slope of the secnt (the stright line) through the points (, f()) nd (, f()) Dp 2,2 : Ensure tht the slope in the point is rel numer τ 1,1 : Use the 2-point formul on ny two points in the intervl τ 1,i : Drw tngent nd use the 2-point formul on the tngent i = {3; 4} τ 2,1 : Use the 2-point formul on n infinitesiml intervl τ 2,2 : Stndrd prt is the technique, to ensure tht it is rel numer elements θ 1 : Informl definition of the tngent s liner grph which is prllel with the grph in the point θ 2 : Definition of the differentil quotient s st ( f(x+) f(x) ) Didctic moment(s) First encounter, explortion with DP 1 First encounter, explortion, constitution with DP 2. Institutionlizti on of the theory y using HMO Elements of the didcticl techniques Using previously estlished prxeology on how to find the slope of stright line given 2 points Grph Using the slope of liner function nd the 2-point formul, together with the notion of the mthemticl oject: function, nd the hyperrel numers to construct the new knowledge Pge 144 of 154

155 Appendix 2.1 Hyperrel nlysis tle (12:15-13:2) (8:1-9:15) (1:4-11:45) DP 4 : Find d xn DP 4,1 :Find d (x + ) Dp 4,1,1 :Given g(x) = 1, find the function vlue nd the differentil quotient for x = { 3, 1,,1,4} Dp 4,1,2 :Given h(x) = 4x, find the function vlue nd the differentil quotient for x = { 3, 1,,1,4} Dp 4,2,1 :Given k(x) = 2x 2, find the function vlue nd the differentil quotient for x = { 3, 1,,1,4} Dp 4,2,1 :Given l(x) = 2x 2 4x + 1, find the function vlue nd the differentil quotient for x = { 3, 1,,1,4} DP 6 Show the rules for clculting with differentils Dp 6,2 : Find d (f(x) + g(x)) DP 7 : Wht cn e sid out the originl function, (f(x)), given the derived function (f (x))? Dp 7,1 :Wht cn we sy out f(x) when f (x) >? Dp 7,2 :Wht cn we sy out f(x) when f (x) <? Dp 7,3 :Wht cn we sy out f(x) when f (x) =? τ 4,1 : Recognize the differentil quotient for stright line s the slope of the line τ 4 : st ( f(x+) f(x) ) τ 4,2,1 :Add the differentil quotients from Dp i, i = {3,1; 3,2; 4,1} for ech x seprtely τ 6,2 : st( + ) = st() + st() τ 7 : Find the x-vlue(s) for which f (x) = nd determine in which intervls f is positive or negtive θ 3 : The definition of derived function θ 4 : Descriing the reltion etween function nd its derivtive Further constitution of the theory lock for DMO. Institutionlizti on of the theory y using HMO. First encounter, explortion with DP 6,2 Technicl work with the technique st ( f(x+) f(x) ) (τ 4 ) Using theory lock of previously estlished prxeology to estlish techniques in DMO. Technicl work in the form of routiniztion Using theory lock of previously estlished prxeology out functions nd eqution solving to estlish techniques in DMO Pge 145 of 152

156 Appendix 2.1 Hyperrel nlysis tle (14:45-15:5) Dp 4,1 : Find d (x + ) DP 4 : Find d xn Dp 4,3 : Find f (x) when f(x) = x Dp 4,4 : Find f (x) when f(x) = x 2 Dp 4,5 : Find f (x) when f(x) = x 3 Dp 4,6 : Guess f (x) when f(x) = x 4 Dp 4,7 : Give justified guess to wht f (x) is when f(x) = x n, n N Dp 4,8 : Find f (x) when f(x) = x 2, R Dp 4,9 : Find f (x) when f(x) = x 3, R Dp 4,1 :Guess f (x) when f(x) = x 4, R Dp 4,11 :Give justified guess to wht f (x) is when f(x) = x n, R, n N Dp 6,1 :Find d ( f(x)) Dp 4,12 :Given h(t) = t + 12, h(t) eing numer of eggs to the time t in hours, how fst does it ly eggs in the first two hours? Dp 4,12,1 :Find h (t). Dp 4,12,2 :Given f(t) = t t nd g(t) = 1 3 t3 7t 2 + 5t, how fst do these ly eggs t n optionl time? τ 4,1 : Use τ 4 τ 4,j : (x n ) = n x n 1, n N, R j = {6; 7; 1; 11; 12,2} τ 6,1 : st ( f(x+) f(x) ), st(k α) = k st(α) τ 4,12 :2-point formul τ 4,12,1 :Use τ 4 First encounter, explortion with DP 4 nd constitution for τ 4,j Recognizing system for how to differentite polynomil function through n itertive process (8:1-9:15) (9:25-1:3) DP 7 : Wht cn e sid out the originl function, (f(x)), given the derived function (f (x))? Dp 7,4 :Specify the extremes of the function f(x) = x 2 2x + 4 Technicl work with the technique st ( f(x+) f(x) ) (τ 4 ) Pge 146 of 152

157 Appendix 2.1 Hyperrel nlysis tle (1:4-11:45) (14:45-15:5) (8:1-9:15) DP 4 : Find d xn Dp 4,14 :Find the derivtive of f(x) = 1 x Dp 4,15 :Find the derivtive of f(x) = x DP 6 : Show the rules for clculting with differentils Dp 6,3 :Let h(x) = f(x) g(x). Find h (x) Dp 6,4 :Let h(x) = f(g(x)). Find h (x) Dp 3 : Find the eqution of tngent Dp 5 : Find d ex τ 4,i : st ( f(x+) f(x) ) nd do lgeric clcultions until the denomintor isn t infinitesiml nd use st ( ) = st() st() i = {14; 15} τ 6,3 : st ( f(x+) f(x) ), st( + ) = st() + st() nd use the definition of continuity τ 6,4 : st ( f(x+) f(x) ), st( ) = st() st() τ 3 : Use the differentil quotient s the slope in the eqution for stright line τ 5 : st ( f(x+) f(x) ) nd the definition of e θ 5 : A stright line is determined y the slope nd point Technicl work with τ 4,i First encounter, explortion nd constitution of the theory for Dp 6,3 nd Dp 6,4 First encounter nd explortion of Dp 3 First encounter with Dp (9:25-1:3) Lesson # DP 8 : Wht hve you lerned out differentil clculus? Type of prolem Mthemticl techniques Technologicl theoreticl elements Didctic moment(s) Elements of the didcticl techniques Pge 147 of 152

158 Appendix 2.3 Integrl clculus nlysis tle 2.3 Integrl clculus nlysis tle (Mikkel Mthis Lindhl) Lesson # Type of prolem Mthemticl techniques Technologicl theoreticl elements Didctic moment(s) (8:1-1:3) (1:4-11:45) (14:45-15:5) IP 1 : Find the re etween the grph of function nd the first xis over given intervl. Ip 1,1 : Find the re etween the function f(x) = x 2 + 4x nd the first xis over the intervl [,4]. Ip 1,2 : Should the smll res only e rectngles nd why? Ip 1,3 : Wht is the re of rectngle with infinitesiml width nd f(x) s its height? Ip 1,4 : How does one prtition n intervl into infinitesiml intervls? Ip 1,5 : How should n infinite sum e understood? Ip 1,6 : In wht wy should one write the re etween function nd the first xis over n intervl? Ip 1,7 : Ensure tht the infinite sum descriing the re etween function nd the first xis is rel numer. Ip 1,8 : Descrie the grphicl 4 mening of k 1 x. 4 IP 2 : Find f(x) Ip 2,1 : Find c Ip 2,1,1 : Find 1 4 Ip 2,1,2 : Find 2 4 Ip 2,1,3 : Find 2 5 Ip 2,1,4 : Find Ip 2,1,5 : Find Ip 2,1,6 : Find Ip 2,1,7 : Find 4 6 Ip 2,1,8 : Find 4 6 Ip 2,1,9 : Find 3 2 Ip 2,1 : Find c 4 Ip 2,2 : Find kx Ip 2,2,1 : Find x 4 Ip 2,2,2 : Find 2x 4 Ip 2,2,3 : Find 3x 5 Ip 2,2,4 : Find 2x 2 7 Ip 2,2,5 : Find 2x 3 7 Ip 2,2,6 : Find 3x 5 τ 1,1 : Prtition the intervl [,4] into infinitely mny infinitesiml intervls; consider the function s eing constnt over the infinitesiml intervl nd recognize the re s the infinite sum of the rectngles re. τ 1,3 : The re is found in the similr wy s norml rectngle, i.e. the re is f(x). τ 1,6 : f(x). τ 1,7 : Stndrd prt is the technique, to ensure tht it is rel numer. τ 2,1 : Recognize the rectngle, c( ). τ 2,2 : Recognize the tringle, 1 2 k(2 2 ) τ 3,1 : Recognize the rectngle(s) nd tringle, 1 2 kx2 + cx θ 1 : Prtition of n intervl. θ 2 : Description of sum of n terms nd sum of (hyper)finite terms. θ 3 : (Informl) first description of the definite integrl. θ 4 : Definition of the definite integrl s the stndrd prt of n infinite sum of res of rectngles. First encounter, explortion (nd constitution of the theory) for IP 1. Further constitution of the theory lock for IMO. First encounter nd explortion with IP 2,1, IP 2,2 nd IP 3,1 (Technicl work on τ 2,1, τ 2,2, τ 3,1 using theory lock from previously estlished prxeology of geometry) Elements of the didcticl techniques Putting together known technique (of finding the re of rectngle) nd using two new mthemticl ojects: infinite prtition nd infinite sums. Both of the newly introduced ojects re sed on mthemticl ojects they my or my not hve encountered efore. Using theory lock of previously estlished prxeology to estlish techniques in IMO. Pge 148 of 152

159 Appendix 2.3 Integrl clculus nlysis tle 6 Ip 2,2,7 : Find 4x 6 Ip 2,2,8 : Find 4x 6 Ip 2,2,9 : Find 3x 2 Ip 2,2 : Find kx (14:45-15:5) IP 3 : Find n ntiderivtive of f(x) x Ip 3,1 : Find kx + c x Ip 3,1,1 : Find 4 x Ip 3,1,2 : Find c x Ip 3,1,3 : Find 2x x Ip 3,1,4 : Find kx x Ip 3,1,5 : Find 1 x x Ip 3,1 : Find kx + c 1 Ip 2,1,1 :Find 2 3 Ip 2,2,1 :Find 2x 1 Ip 2,3 : Find f(x) when f(x) is n elementry function. Ip 2,3,1 : Find 2x Ip 2,3,2 : Find x + 4 Ip 2,1,11 :Find Ip 2,1,12 :Find Ip 2,3,3 : Find 2x (1:4-11:45) Ip 2,1 : Find c Ip 2,2 : Find kx Ip 2,3,4 : Find kx + c IP 4 : Show the rules for clculting with integrls. IP 4,1 : Show tht (f(x) ± g(x)) = f(x) ± g(x) IP 4,2 : show tht k f(x) = k f(x) IP 4,3 : Show tht f(x) + f(x) c = f(x) c τ 4 : Using the definition of the integrl, the techniques from HMO, nd the use of the rules for (hyper)finite sums (which re never explined), to ssert the results. [This is technique when considering the integrl s technique] θ 5 : Using the definition of the integrl, the techniques from HMO, nd the use of the rules for (hyper)finite sums (which re never explined). [is prt of the theoreticl lock when considering the integrl s prt of the technology] Technicl work on τ 2,1, τ 2,2, τ 1,6 (Trying to) Bridge the connection etween the intuitive understnding of the integrl, s n re, nd the definition with wht rules this infers on the mthemticl oject the integrl is. IP 4,4 : Show tht f(x) = IP 4,5 : Show tht f(x) = f(x) Pge 149 of 152

160 Appendix 2.3 Integrl clculus nlysis tle x Ip 3,1,2 : Find c x Ip 3,1,4 : Find kx IP 5 : Find d f(x) Ip 5,1 : Find d x f(x) for n elementry function f(x). Ip 5,1,1 : Find d c Ip 5,1,2 : Find d x kx Ip 5,1,3 : Find d x x2 x x IP 3,2 : Find f(x) x Ip 3,2,1 : Find x 2 τ 5,1,i :Use the techniques otined in DMO to estlish the results i = {1,2} τ 5,1 : Recognize the integrnd s the result, i.e. guess the technique from Ip 5,1,1 nd Ip 5,1,2. τ 3,2 : Guess function for which the derivtive is the integrnd. θ 6 : The fundmentl theorem of clculus. d x f(x) = f(x) nd x f (x) = f(x) Further Constitution of the theory for IMO. Technicl work on τ 3,2. Institutionlizti on etween DMO nd IMO. The sttement of the fundmentl theorem nd the definition of the ntiderivtive re presented fter one nother. This is followed y new definition, θ 8, nd 3 new types of prolems IP 3,3, IP 3,4 nd IP (8:1-9:15) Ip 3,3 : Find n ntiderivtive of n elementry function. Ip 3,3,1 :Find n ntiderivtive of f(x) = x. IP 3,4 : Find the ntiderivtive going through point (x, y ). Ip 3,4,1 : Find the ntiderivtive through the point (,2)? Ip 3,5 : How mny ntiderivtives does function hve? Ip 3,3,2 : Find n ntiderivtive of f(x) = k IP 6 : Find f(x)? Ip 6,1 : Find k? Ip 3,3,3 : Find n ntiderivtive of f(x) = kx Ip 6,2 : Find kx? Ip 3,3,4 : Find n ntiderivtive of f(x) = x 2 Ip 6,3 : Find x 2? Ip 3,3,5 : Find n ntiderivtive of f(x) = e x Ip 6,4 : Find e x? Ip 3,3,6 : Find n ntiderivtive of f(x) = x n Ip 6,5 : Find x n? τ 3,3 : Use τ 3,2, i.e. check if F (x) = f(x). τ 3,3,1 :Use tht the re function is n ntiderivtive τ 3,4,1 :Use tht the point is on the grph, mking the correct ntiderivtive, F c (x), defined s F c (x) = F(x) + c, such tht c = 2 F(). τ 3,3,i :Use tht the re function is n ntiderivtive i = {2,3} τ 3,3,6 :An ntiderivtive of x n is 1 n+1 xn+1. τ 6 : Find the ntiderivtive nd dd constnt, c, to e fixed lter. θ 7 : Definition of the ntiderivtive, of function f(x), s function F(x), such tht F (x) = f(x). θ 8 : Definition of the indefinite integrl, s f(x) = F(x) + c First encounter, (explortion) with IP 3,3. Constitution of the theory for IP 3,3. First encounter, (explortion) with IP 3,4. First encounter, (explortion) with IP 6. Technicl work on τ 3,3 using the fundmentl theorem nd the rules for differentition. Pge 15 of 152

161 Appendix 2.3 Integrl clculus nlysis tle (9:25-1:3) Ip 6,6 : Find 2 Ip 6,7 : Find ( 1 x + 2) 4 Ip 6,8 : Find 3 Ip 6,9 : Find ( 1 x + 3)? 4 Ip 6,1 : Find 2x? Ip 6,11 : Find (x 2 4x + 8)? Ip 6,12 :Find ( x 3 + 8)? Ip 6,13 :Find ( x 2 + 4x)? Ip 6,14 :Find (4x 3 3x 2 + 2x)? Ip 6,15 :Find 1? x 2 Ip 6,16 : Find 8? x 2 Ip 6,17 : Find ( 4x 3 + x 4 )? Ip 6,18 : Find e x? Ip 6,19 : Find (2x + 8)(2x) + 2(x 2 )? Ip 6,2 : Find 6x 2 e 2x3? Ip 6,21 : Find 1? ex Ip 3,4,2 : For ech of the prolems Ip 6,i where i = {4,5,,21} find specific ntiderivtive y letting it go through specific point (x, y ). Ip 2,2,1 : Find 2x Ip 2,3,5 : Find x2 + 2x Ip 2,3,6 : Find 3x 2 2 Ip 2,3,7 : Find x2 + 3x 2 1 τ 3,4,2 :Use tht the point is on the grph, mking the correct ntiderivtive, F c (x), defined s F c (x) = F(x) + c, such tht c = y F(x ). τ 2,3 : Use the fundmentl theorem of clculus nd the technique from Ip 4,3 to otin f(x) = F() F(). Technicl work on τ 3,4 in the form of routiniztion. Technicl work on τ 6 in the form of routiniztion nd in specific cses y recognizing the rules for differentition ckwrds. First encounter nd explortion with IP 2,3. The prolems in IP 2,3 hs to e solved using technique sed on the newly stted fundmentl theorem, the definition of n ntiderivtive nd the rule for integrtion tht hs een proved in IP 5,3 ut not since used (8:1-9:15) Ip 5,3 : Wht is df? Ip 5,3,1 : Wht other thn hs een used when writing n infinitesiml? Ip 5,2 : Find d f(x) for n elementry function. Ip 5,2,1 : Wht is d (e3x2 )? Ip 6,22 :Wht is 6x e 3x2? τ 5,2 : using df = f (x) nd lger. τ 6,i : Using the fundmentl theorem nd recognizing the integrnd s the previously differentited expression. i = {22,23} θ 9 : Definition of df s df = f (x) First encounter, (explortion) with Ip 5,2. technicl work with, τ 6,i estlishing τ 7. i = {22,23} Comining 3 new mthemticl ojects nd one technique, τ 12, to crete new technique, τ 14. 1: Mking use of function s vrile. 2: Antiderivtive. 3: Bckwrds chin rule for differentition. Pge 151 of 152

162 Appendix 2.3 Integrl clculus nlysis tle Ip 5,2,2 : Wht is d (F(g(x))), when F(x) is n ntiderivtive of f(x)? Ip 6,23 :Find f(g(x)) g (x)? IP 7 : Find f(g(x)) g (x) Ip 7,1 : Wht is the inner function in the expression f(g(x)) g (x)? Ip 5,2,2 : Wht is du if u = g(x)? τ 7 : Using f(g(x))g (x) = g() f(u) du where g() f(x) is function for which n ntiderivtive cn e found. First encounter nd explortion with IP (9:25-1:3) (13:3-14:35) Ip 7,2 : Rewrite f(g(x)) g (x), with u = g(x) nd du. Ip 7,3 : Rewrite f(g(x)) g (x) with u = g(x) nd du. IP 7 : Find f(g(x)) g (x) Ip 7,4 : Find e x 1? 1 2 x Ip 7,5 : Find 5 2x 1 (x 2 +3) 2 Ip 7,6 : Find 1 4 2(4x 3 + 2x 2 ) 1 (12x 2 + 4x) Ip 7,7 : Find 1 8x3 +14x 2x 4 +7x 2 Ip 7,8 : Find e x 12 x 2 IP 8 : Wht hve you lerned out integrl clculus? Institutionlisti on of the punctul MO s in IMO Lesson # Type of prolem Mthemticl techniques Technologicl theoreticl elements Didctic moment(s) Elements of the didcticl techniques Pge 152 of 152

163 Appendix 3 Compendium(s) 3 Compendium(s) INFINITESIMALREGNING ret linie x f x + f x x + f x 1 1 x 1 Side 1 f 8

164 MIKKEL MATHIAS LINDAHL JONAS KYHNÆB KYHDAHLS GYMNASIEMATEMATIK DIFFERENTIALREGNING OG INTEGRALREGNING MED IKKE-STANDARD ANALYSE Side 2 f 6

165 KyhDhls gymnsiemtemtik 2. udgve, 1. oplg KyhDhl Teching Incorported Forlgsredktion: Mikkel M. Lindhl og Jons Kyhnæ Grfisk tilrettelæggelse og omslg: KyhDhl A/S, Køenhvn Tegninger: Mikkel M. Lindhl og Jons Kyhnæ Historie: Mikkel M. Lindhl og Jons Kyhnæ Tryk: KyhDhl Press, Roskilde ISBN N/A Kopiering fr denne og må kun finde sted på institutioner, der hr indgået ftle med COPY- DAN, og kun inden for de i ftlen nævnte rmmer. Eller ndre steder, hvor mn ikke hr indgået ftle med COPY-DAN. Derhjemme for eksempel. Illustrtionsliste: Forside: Jons Kyhnæ Jons Kyhnæ, s. 1, s. 2, s. 14, s. 15, s. 17, s. 33, s. 61 Mikkel M. Lindhl: s. 26, s. 35, s. 4, s. 41, s. 42, s. 43, s. 44, s. 46, s. 47 Anders: s. 21 Resten er lvet som smrejde mellem Mikkel M. Lindhl og Jons Kyhnæ Trdemrk(s): Jons Kyhnæ Side 3 f 6

166 Indhold 1 Tlmængder Hyperreelle tl! Stndrd del Funktioner Stjerne opertion Opgver DIFFERENTIALREGNING Hældningskoefficient Tngent Hældning i et punkt Opgver Udvidet 2-punktsformel Lineære funktioner Seknt Differentilkvotient trins reglen Opgver Afledte funktioner Den konstnte funktion Den lineære funktion x med en nturlig eksponent over x Kvdrtroden f x x med en reel eksponent Eksponentilfunktionen Funktioner uden en fledt funktion Opgver Regneregler Differentition med en konstnt fktor Differentition f en sum Differentition f en differens Differentition f et multiplum f 2 funktioner Differentition f smmenstte funktioner/kæderegel...29 Side 4 f 6

167 5.6 Differentition f en kvotient f 2 funktioner Opgver Monotoniforhold Mksimum, minimum og vendetngent Monotonilinie Opgver Tngentens ligning Opgver INTEGRALREGNING Det estemte integrl Arelfunktioner Regneregler Sum/differens f funktioner Sum f integrler Konstnt gnget på Integrl f et punkt Integrl fr til eller til Arel f Noitknufmts husside Opgver Integrtion og differentition Den fledte f integrlet fr til x Integrlet f den fledte funktion fr til Stmfunktion Integrtion og smmenstte funktioner Integrtion ved sustitution Opgver Integrler uden grænser Det uestemte integrl Opgver Fcitliste Opsummering Kpitel Kpitel Kpitel Side 5 f 6

168 Forord Denne læreog dækker de krv til kernestof og supplerende stof, som lærerplnerne stiller til differentilregning og integrlregning på A-niveu (stx), ortset fr omdrejningslegemer. Bogen fungerer åde som grundog og rejdsog. Kpitel 1 indeholder supplerende stof inden for infinitesimlregning, i form f ikkestndrd nlyse. Vi nefler t nvende denne tilgng til emnet differentilregning, d kpitlerne 2-7 ygger på netop ikke-stndrd nlyse. Kpitlerne 8-11 om integrlregning ygger ligeså på ikke-stndrd nlyse og kpitlerne om differentilregning. Det er således tnken, t ogen skl læses fr ende til nden. I slutningen f ogen findes øvelser og opgver med tilhørende fcitliste. Der findes ydermere en opsummering f lle kpitler i ogen, efter fcitlisten. Opgverne hr vrierende kompleksitet. Der liver ikke eskrevet nogen it-forlø, hvorfor det er op til læseren/underviseren om et it-værktøj skl tges i rug som et lterntiv til en lgerisk eller nlytisk fremgngsmåde. Kpitlet om monotoniforhold er inspireret f WeMtemtiks meget fine gennemgng f emnet. Køenhvn, juni 216 Mikkel M. Lindhl Jons Kyhnæ Side 6 f 6

169 Appendix 3.3. Integrl clculus nlysis tle 1 Tlmængder I mtemtik ruger mn ofte egreerne nturlige tl (dvs. positive hele tl), hele tl, rtionle tl og reelle tl. Derfor hr de fået en noget kortere etegnelse. N som er mængden f nturlige tl {1,2,3, } Z som er mængden f lle hele tl, dvs. positive, negtive og {, 3, 2, 1,,1,2,3, } Q som er mængden f lle rtionle tl, dvs. røker f heltl og decimltl med system R som er mængden f lle reelle tl, dvs. decimltl f lle rter En tllinie viser, hvordn tllene ligger i forhold til hinnden. På tllinien herunder er der tegnet tllene: -1 1,7 2,46 π og 7 2,7 2,46 π 7 2 R Figur 1.1 Vi vil udvide de reelle tl til også t inkludere uendeligt små og store tl. Disse tl klder viiiii Hyperreelle tl! Et uendeligt lille tl kldes en infinitesiml. Positive infinitesimler er ltså så små, t mn ikke kn måle dem og dermed mindre end ethvert positivt reelt tl mn kn komme på. Vi vil etegne infinitesimler. Hvorfor? Det er d lidt underligt? (siger den irriterende lien der esøgte klssen, netop den dg). Den kloge lærer svrer: Afstnden mellem jeres respektive hænder kn eskrives som fstnden mellem 2 punkter på en tllinie, x 2 x 1 = Δx for x 2 > x 1. Når der gives en high five vil fstnden mellem punkterne live nul, x 2 x 1 = Δx = for x 2 = x 1. Lige inden hænderne mødes kn mn klde fstnden uendelig lille, x 2 x 1 =, hvor fstnden nu etegnes som, for ikke t lnde dem smmen med de normle fstnde Δx. OBS: Det fungerer også med en low five eller ndre erøringer. Hvis et uendeligt lille tl findes, så må 42 + også findes. Ligeledes findes 42. For t få en ide om hvd en infinitesiml er, kn følgende ulighed være en hjælp 1. > 1 >,1 >,1 >,1 > Vi hr på figuren nedenfor illustreret den reelle tllinie, med infinitesimler: Side 1 f 74

170 Figur Et VILDT lille tl! Det eneste mn ikke må dividere med er. Hvd sker der, hvis vi dividerer med et uendeligt lille tl? Hvd sker der, hvis vi dividerer med en infinitesiml, for eksempel 1? En måde t forstå, hvd dette tl egentlig er, kn ske ved t tænke på reglerne for division, ltså < 1 1. < 1 1 < 1,1 < 1,1 < 1,1 < 1 Hvilket er det smme som 1,1 <,1 <,1 < 1 < 1... < 1... < 1 1 er ltså uendeligt stort! De uendeligt store tl, smmen med lle de reelle tl med de uendeligt små tl omkring, kldes de hyperreelle tl. Denne nye tlmængde kldes R Herunder er der illustreret, hvd der kldes, den hyperreelle tllinie. Mn skl hve sine store øjne på, for t kunne se helt ud til de uendelige tl Figur R Eksempel < 1 + < < < 1 + < 1 < Side 2 f 74

171 På den hyperreelle tllinie findes derfor Infinitesimler: Positive tl der er mindre end ethvert forestilleligt positivt (reelt) tl forskelligt fr nul (positiv infinitesiml), ltså <, hvor R +, ltså er et positivt reelt tl, forskelligt fr Negtive tl der er større end ethvert forestilleligt negtivt (reelt) tl forskelligt fr nul (negtiv infinitesiml), ltså <, hvor R, ltså er et negtivt reelt tl, forskelligt fr Uendelige tl: tl der er større end ethvert forestilleligt (reelt) tl (positivt uendeligt tl). 1 >, hvor R Tl der er mindre end ethvert forestilleligt (reelt) tl (negtivt uendeligt tl). 1 <, hvor R Der gælder smme regneregler for R, som for R, mn kn ltså ddere, trække fr, multiplicere og dividere. Der findes visse regler, som måske skl nævnes: Regneregler: Ld være en infinitesiml, så gælder =, ligesom for lle ndre tl Ld og dy være infinitesimle, så vil + dy også være en infinitesiml Ld n N være et nturligt tl, så vil n = , også være infinitesiml Ld m Z være et heltl, så vil m også være infinitesiml Ld R være et reelt tl, så vil også være en infinitesiml Brøken dy kn være et hvilket som helst hyperreelt tl, det skl ltså fgøres fr sitution til sitution Eksempel Ld være en infinitesiml forskellig fr nul, så vil I. II. III. 1 5 = 5 = 1 5 = = = ,5 + 2 = Side 3 f 74

172 1.2 Stndrd del For ethvert endeligt hyperreelt tl, X, findes et reelt tl, x, således t X x er en infinitesiml, dvs. X x =, ved en let jonglering med ogstverne, ses det t X = x +. Dermed kn ethvert endeligt hyperreelt tl skrives som summen f et reelt tl og en infinitesiml. OBS: Vi skriver endeligt hyperreelt tl fordi der for et uendeligt hyperreelt tl A ikke findes et reelt tl således, t A er infinitesiml. For t kunne tle om denne reelle del f et hyperreelt tl indføres det mn klder stndrd delen. Vi forkorter stndrd delen med st. Mn kn ltså tle om t gå fr den hyperreelle tllinie til den reelle tllinie, ved t tge stndrd delen. Definition Stndrd del Stndrd delen f det hyperreelle tl + er. Det skrives st + =. Hvis et hyperreelt tl R er uendeligt, findes der ikke nogen stndrd del til dette, dvs. st = eksisterer ikke!! Regneregler Ld, R, så gælder st ± st = st ± st = st st HVIIIIIIIIIIIS og er endelige tl (dvs. og er ikke uendelige tl) st ( st ) = HVIIIIIIIIIIIS er endelig og ikke er en infinitesiml st Eksempel Funktioner Husk på hvd en funktion er. st X = st x + = x st 42 + = 42. Definition Funktion En funktion er en smmenhæng mellem 2 vrile, x og y, sådn t der til hver værdi f x hører præcis én værdi f y. Værdien f y kldes for funktionsværdien f x Mn klder x for den ufhængige vriel og y for den fhængige vriel. Det etyder, t mn selv vælger værdien f x, mens værdien f y liver estemt f funktionen. Mn ruger som regel ogstvet f som etegnelse for en funktion. Mn siger, t værdien f y kldes for funktionsværdien f x for funktionen f og skriver: y = f x Dette læses f f x. Hvis mn skl ruge flere funktioner, ruges ofte ogstverne g, h, k, l, Eksempel f x = x + Side 4 f 74

173 Eksempel f x = { 1 hvis x Q ellers Eksempel er måske muligt t forestille sig, men det er ikke muligt t tegne den på en ordentlig måde (se tegning). På grund f disse lidt kedelige funktioner er der udrejdet nogle grupper f funktioner der er edre t rejde med. En f de mere kendte f den slgs grupper f funktioner er dem der hedder kontinuerte funktioner. En intuitiv forståelse f hvd en kontinuert funktion er, kn være som følgende: En kontinuert funktion er en funktion der kn tegnes uden t løfte lynten/kridtet fr ppiret/tvlen. Med denne intuitive forståelse ses det, t Eksempel ikke er en kontinuert funktion. Desværre er denne intuitive forståelse ikke en rigtig mtemtisk definition. For t give en mtemtisk definition på en kontinuert funktion kn infinitesimle størrelser ruges Kontinuitet Definition Kontinuitet En funktion f x er kontinuert i et punkt (x, f x ), hvis en infinitesiml ændring i x-værdien giver en infinitesiml ændring i yværdien (funktionsværdien). Dvs. f x er kontinuert i (x, f x ), hvis Δy = f x + f x er infinitesiml. Hvis en funktion siges t være kontinuert (der udeldes ltså i hvilket punkt), så er den kontinuert i hele dens definitionsmængde. Den esøgende lien udryder: Hvd f. sker der for den der stjerne der er forn det ene f? Den kloge lærer: Det er rigtigt, der er noget med den der stjerne, er der nogen der her nogle idéer til hvd den kunne etyde? Alien: Bipopip! Nu hr jeg lige kigget på lle tllene i definitionsmængden for funktionen f, og der står der ltså ikke noget om t x + ligger der i. Den kloge lærer: Korrekt, funktioner er (normlt) kun defineret for reelle tl, der er ltså ingen infinitesimle eller uendelige tl i definitionsmængden til funktioner. Dette er grunden til stjernen er der, stjernen udvider ltså funktionens definitionsmængde til t indeholde de hyperreelle tl. Side 5 f 74

174 Nedenfor findes en tekstogs-version f hvd stjernen etyder. 1.4 Stjerne opertion Med stndrd delen er det muligt t komme fr den hyperreelle tllinie til den reelle tllinie, men hvd med den nden vej? Et reelt tl,, er ltid også et hyperreelt tl +, forstået på den måde t den infinitesimle del er. Der er derfor ingen grund til t føre et reelt tl til den hyperreelle tllinie, det er der llerede. Stjerne opertionen, som fører reelle ting til hyperreelle ruges derfor når der rejdes med funktioner og intervller. Intervller vil først live ehndlet senere. Hvis en funktion skl evlueres i en hyperreel værdi, x R, skl funktionen udvides til også t gælde for hyperreelle tl, dette skrives således: f x Dette læses stjerne f f x". Hvis mn skl ruge flere funktioner, ruges g, h, k, l, Eksempel I. f x = 3x. Den reelle funktionsværdi for x = 2 er f 2 = 3 2 = 6. Den hyperreelle funktionsværdi er derfor for x = 2 +, så f 2 + = = = For t komme tilge til den reelle funktionsværdi tger vi stndrd delen f den hyperreelle funktionsværdi st( f 2 + ) = st = 6 = f 2 så st( f 2 + ) = f 2. Good to know! II. g x = x 2. Den reelle funktionsværdi for x = 4 er g 4 = 4 2 = 16. Den hyperreelle funktionsværdi er derfor for x = 4 +, så g 4 + = = = For t komme tilge til den reelle funktionsværdi tger vi stndrd delen f den hyperreelle funktionsværdi st( g 4 + ) = g 4 = 16. #micdrop Side 6 f 74

175 1.5 Opgver I følgende opgver er x R og og dy er infinitesimle. 1. Løs st x + + st y + dy 2. Løs 5 st x + 3. Løs st 5x Løs st + 5. Løs st + dy 6. Løs st 5 7. Løs st ( 1 ) 8. Hvorfor er st (5 1 ) = 5? 9. Nævn en infinitesiml der også er reel. 1. Løs st( dy ) 11. Løs st 2 + st 3 + dy 12. Lv en konklusion på opgve 1 og Løs st st ( 1 ) 14. Præcisér konklusionen fr opgve 12, så den psser til opgve Løs st For f x = 3x + 1, find de hyperreelle funktionsværdier f 1 + og f For g x = x, find den hyperreelle funktionsværdi for x = 3 og x =. 18. For h x = 2x 3, find den reelle funktionsværdi for h 1 + og h Hvd er st f x? Antg nu t f er en kontinuert funktion (dermed er den kontinuert i lle punkter i dens definitionsmængde) 2. Hvd er st f x +? Når lle opgver i dette kpitel er løst, hr mn erhvervet sig titlen stndrd, hvilket må sættes forn ens nvn. Side 7 f 74

176 Side 8 f 74

177 2 DIFFERENTIALREGNING Differentilregning hndler om funktioners væksthstighed i et estemt punkt. Altså hndler det om t estemme hældningen f grfen i et estemt punkt. Vi hr lært om hældningskoefficienter for lineære funktioner, men hvd gør vi hvis funktionerne ikke er lineære? 2.1 Hældningskoefficient En lineær funktion hr forskriften f x = x +. Tllet er hældningskoefficienten og fortæller, hvor meget funktionen vokser eller ftger, når x vokser med 1. Tllet er y-værdien for grfens skæringspunkt med y-ksen. Figur Hældningskoefficienten estemmes ved t strte i et punkt på grfen, gå 1 til højre og mål lodret op eller ned, indtil mn rmmer grfen igen, den målte fstnd er hældningskoefficienten. Hældningskoefficienten fortæller, hvor stejl grfen er: Hvis hældningskoefficienten er positiv og stor, er grfen voksende og stejl. Hvis hældningskoefficienten er positiv og lille, er grfen voksende, men ikke så stejl. Hvis hældningskoefficienten er negtiv, er grfen ftgende. Hvis hældningskoefficienten er, er grfen vndret. 2.2 Tngent Hvis en funktion ikke er lineær, kn vi godt tle om hældningskoefficient, men først skl vi definere, hvd vi forstår ved hældningskoefficient i det generelle tilfælde. Først skl det hndle om tngenter til grfer. En tngent til en grf i et estemt punkt er en linie, der rører grfen i punktet og er prllel med grfen i punktet. Eksempel Figuren herunder viser en grf for funktionen f x =,25x 2 x 5. Der er indtegnet en tngent til grfen i punktet 6, 2 : Side 9 f 74

178 Figur Tngentens hældningskoefficient er 2 (1 til højre og 2 op). Det svrer til, t grfen også hr en hældning på 2 i punktet 6, 2. Grfens hældning ændrer sig fr punkt til punkt. Eksempel Figuren herunder viser den smme grf som ovenfor, men her er indtegnet en tngent i punktet 4, 5. Figur I punktet 4, 5 er tngentens hældning 1. Det svrer til, t grfen også hr en hældning på 1 i punktet 4, Hældning i et punkt Hvis en funktion ikke er lineær, ruger vi ikke ordet hældningskoefficient om hældningen på grfen. SPOILER ALERT! Vi ruger ordet differentilkvotient. Vi skl nok give en forklring på ordet senere. Side 1 f 74

179 Differentilkvotienten er grfens hældning i et punkt eller mere præcist: Differentilkvotienten er tngentens hældningskoefficient i et punkt på grfen, og den kn ændre sig fr punkt til punkt på grfen. Det giver følgende definition: Definition Tngent Den rette linie gennem punktet (x, f x ) med hældningskoefficient = f x o kldes kurvens tngent i (x, f x ). f x o læses f mærke f x nul. Punktet (x, f x ) etyder et punkt på grfen for funktionen f, fordi y-værdien på en grf udregnes som f x. I Eksempel er differentilkvotienten 2, når x = 6, d tngenten i punktet 6, 2 hr hældningskoefficienten 2. Vi kn ltså skrive f 6 = 2. I Eksempel er differentilkvotienten 1, når x = 4, d tngenten i punktet 4, 5 hr hældningskoefficienten 1. Vi skriver f 4 = 1. Ld os estemme f 4 og f 2 for den smme funktion som før. Vi indtegner tngenterne til grfen i punkterne 4,3 og 2, 6 og flæser tngenternes hældningskoefficient Figur Side 11 f 74

180 Figur På Figur er tngentens hældningskoefficient 3, så f 4 = 3. På Figur er tngentens hældning, så f 2 =. Side 12 f 74

181 2.4 Opgver 1. Figuren herunder viser grfer for funktionerne f og g. Bestem hældningskoefficienterne for f og g. 2. Indtegn tngenten til grfen herunder i punktet 4, 3, og estem tngentens hældningskoefficient. 3. Bestem f 4 for funktionen i opgve Bestem f 2 for funktionen i opgve Bestem f og f for funktionen i opgve 2. Hvd er forskellen på f og f? Side 13 f 74

182 3 Udvidet 2-punktsformel Nu hr vi set hvordn mn kn estemme hældningskoefficienter og differentilkvotienter grfisk. Det er ltid sjovt t kunne eregne ting, især fordi vi hr mtemtik, så inden vi, som lovet, vender tilge til ordet differentilkvotient, repeterer vi lige hurtigt metoden for t eregne hældningskoefficienten for lineære funktioner. Bgefter ngrier vi funktioner generelt. Glæd dig. 3.1 Lineære funktioner En lineær funktion hr forskriften f x = x +, hvor er hældningskoefficienten, rememer? Stærkt (det lev jo også skrevet på side 1 ). fortæller ltså hvor meget funktionen vokser eller ftger, når x vokser med 1. Hvis vi kender to punkter på grfen, kn vi finde med to-punkts-formlen. Hvis vi klder punkterne x 1, y 1 og x 2, y 2 kn vi fyre dem ind i et k-syst (koordintsystem): 2 (x 2, y 2 ) (x 1, y 1 ) Δy Δx Det ses Δx = x 2 x 1 hvilket medfører t x 2 = x 1 + Δx, på smme måde ses det Δy = y 2 y 1 hvilket medfører t y 2 = y 1 + Δy. Dermed kn hældningskoefficienten eregnes med formlen = Δy Δx = y 2 y 1 x 2 x 1 = y 1 + Δy y 1 x 1 + Δx x Seknt Hvis en funktion ikke er lineær, kn grfens hældning ændre sig fr punkt til punkt. Hvd er hældningen for grfen/en grf i punktet P? Vi kn ruge vores smukke øjne og udnytte øjemålet og tegne en række linier med forskellige hældninger og derefter vurdere, hvilken linie der psser edst, ligesom herunder. Check de her tre grfer. Figur Side 14 f 74

183 Figur Figur Figur På Figur hr vi prøvet med en linie med hældning 3. Det er ikke helt vildt lækkert, den er ikke stejl nok. På Figur hr vi smidt en linie ind med hældning 7 og der liver den lige hkket for stejl. Det ser fktisk ud som om, t linien på Figur med hældning 4,5 er on fleek. Det vil sige, t i punktet P er værdien 4,5 nok et ret fedt ud på kurvens hældning. Oky, som sgt er det ltid sjovt t kunne eregne ting i mtemtik, så hvis vi vil eregne stejlheden f den røde grf deroppe lidt mere præcist, kunne det være vi skulle kende forskriften for funktionen. Den røde kurve er grf for funktionen f x = 1 5 x3 3 2 x2 + x + 4. Punktet P hr førstekoordinten 6. For t kunne eregne hældningen i punktet P, kn vi ruge to-punkts-formlen, men så skl vi ruge et punkt mere. Dilemmet er ltså t finde en hældning i ét punkt, hvor to-punkts-formlen kræver.. J, to punkter. Duh! Så ld os prøve t kigge på to punkter, der re ligger ret tæt på hinnden. Kig på grfen ovenfor igen, hvor vi hr indtegnet et ndet punkt Q x 1, y 1, og linien gennem P og Q tegnes. Denne linie kldes en seknt. Vi hr lige zoomet lidt ind, så det ikke er så nstrengende for de smukke øjne. 2 2 Q(7, f 7 ) f x seknt seknt f x P(6, f 6 ) 1 1 Q(6,1; f 6,1 ) P(6, f 6 ) 1 Figur Figur Side 15 f 74

184 På Figur vælges punktet Q med x = 7 dvs. Δx = 1. Seknten får d hældningen På Figur vælges punktet Q med x = 6,1 dvs. Δx =,1. Seknten får d hældningen = f f 6 1 = 6,9 = f 6 +,1 f 6,1 = 4,812. På Figur er punktet Q så tæt på P t det er svært t se, t seknten skærer ind over kurven. Vi gider ikke t tegne en figur mere, det tog hjernedødt lng tid, men i stedet vil vi udregne seknthældningen i et tilfælde mere. Vi skl endnu tættere på P med Q nu. Vi vælger derfor punktet Q med x = 6,1 dvs. Δx =,1. Med to-punkts-formlen får vi = f 6 +,1 f 6,1 = 4,62. Tellen herunder viser smmenhængen mellem det x 2 vi vælger for Q x 2, f x 2 og hældningskoefficienten,, for den rette linie der går gennem P og Q Δx 1,1,1,1 6,9 4,812 4,62 4,62 Det kunne godt se ud som om, t hældningen i punktet P er cirk 4,6. Det svrer ret godt til resulttet fr Figur Vi kunne live frekin ved og ved med t vælge Q-punktet endnu tættere på P, ltså gøre Δx mindre og mindre og mindre og mindre, men det kunne vi egentlig gøre for evigt. Forestil dig, eller forestil JER, hvis I sidder og læser højt for hinnden.. Forestil Jer, t værdien for x ændres fr t være 6 til t være 6 +, som er uendeligt tæt på men ikke lig med 6! Så vil den nye værdi for f x være f 6 +. På den måde liver værdien for x ændret med en infinitesiml, forskellig fr, mens værdien for f x vil live ændret med f 6 + f 6. Ved to-punkts-formlen får vi ltså, med ændringen i værdien for f x og ændringen i værdien for x, f 6 + f 6. Dette forhold er netop definitionen på hældningen, ltså differentilkvotienten, for f x, som kommeeeeer her! På næste side.. Side 16 f 74

185 3.3 Differentilkvotient Definition Differentilkvotient Differentilkvotienten for f x i x er st ( f x + f x Hvis differentilkvotienten eksisterer og er den smme for enhver infinitesiml, siges funktionen t være differentiel i x og værdien enævnes f x. ). Her skl vi re lige huske på, t når vi rejder med infinitesimler, efinder vi os i det hyperreelle univers, det etyder ltså, t mn kn se infinitesimlerne på den rette linie. Se Figur ret linie x f x + f x x + f x 1 1 x 1 Figur trins reglen For t StndrdJoe er sikker på, t HyperMick forstår definitionen på differentilkvotienten, udpensler hn derfor en regel for hvordn mn skl ruge formlen. Redy? Oky, vi finder ltså hældningen i et punkt (x, f x ) sådn her: 1. Find f x + f x 2. Divider ovenstående resultt med og reducér 3. Tg stndrd delen f f x + f x HyperMick: Stop.. Jeg ftter ikke en ddel f det der.. StndrdJoe: Oky, check it Side 17 f 74

186 Eksempel Find differentilkvotienten til f x = 1 2 x i punktet (x, f x ). Trin 1, find f x + f x : Trin 2, dividér med og reducér f x + f x Und drei, tg stndrd delen til f x + f x : f x + f x = 1 2 x x = 1 2 x x 1 = 2 x x 1 = 2 = 1 2 st ( f x + f x ) = st ( 1 2 ) = 1 2 Så hr vi fundet differentilkvotienten til f x = 1 x i punktet (x 2, f x ), ltså st ( f x + f x ) = f x = 1 2 Eksempel Find differentilkvotienten til f x = x 2 i punktet (x, f x ). Step one: Número dos: f x + f x = x x Og tre: f x + f x = x + 2 x 2 = x 2 2 2x + x 2 = x x + x 2 = 2 2x = 2x st ( f x + f x ) = st 2x = 2x Så hr vi fundet differentilkvotienten til f x = x 2 i punktet (x, f x ) f x = 2x. Boom! Side 18 f 74

187 3.5 Opgver 1. Grfen for en lineær funktion går gennem punkterne 1,5 og 4,7. Beregn differentilkvotienten. 2. Beregn sekntens hældning for grfen for f x = x 2 3 gennem punktet 2,1 og punktet mn finder ved t sætte Δx =,1 og udregne Δy. 3. Figuren herunder viser grfen for f x = x 2 7. Vis ved udregning, t f 3 = 2 og f 4 = 9, og eregn sekntens hældning punkterne 3,2 og 4,9. 4. Beregn seknthældningen gennem punktet 3,2 og punktet, hvor Δx =,1. 5. Gentg udregningen, men rug denne Δx =,1. 6. Gentg udregningen, men rug Δx =,1. 7. Udfyld et skem som nedenstående ud fr opgve 3, 4, 5 og 6, idet er seknthældningen. Hvd liver hældningen, når Δx kommer tættere og tættere på? Δx 1,1,1,1 8. Beregn differentilkvotienten i punktet 3,2 ved t sætte Δx = i funktionen fr opgve Vis, t f 1 = 6 for funktionen fr opgve 3, og estem differentilkvotienten i punktet 1, 6 ligesom i opgve 8. Side 19 f 74

188 1. Udfyld skemet ved t eregne differentilkvotienten til funktionen g x = 1 i punkterne: x g x g x 11. Udfyld skemet ved t eregne differentilkvotienten til funktionen h x = 4x i punkterne: x h x h x 12. Udfyld skemet ved t eregne differentilkvotienten til funktionen k x = 2x 2 i punkterne: x k x k x 13. Udfyld skemet ved t eregne differentilkvotienten til funktionen l x = 2x 2 4x + 1 i punkterne: 14. WHAAAAAT!? x l x l x 15. Ld f x og g x være to vilkårlige og differentile funktioner. Opskriv en regel for differentilkvotienten til h x = f x + g x, ltså eskriv h x ved hjælp f f x og g x. Hint: Find smmenhængen mellem opgverne 1, 11, 12 og 13. Side 2 f 74

189 4 Afledte funktioner HyperMick syntes det vr nederen t skulle køre lle mellemregningerne igennem hver eneste gng hn skulle finde differentilkvotienten i et punkt, så ved t join forces hr StndrdJoe nd HyperMick i dette kpitel generliseret differentilkvotienter til hvd der kldes fledte funktioner. En fledt funktion f x er ltså en funktion der eskriver hældningen for funktionen f x til enhver x-værdi. SÅDAN! Kpitel slut. Nej, vi spytter lige et pr eksempler. Først skl vi lige hve et udtryk for selve fremgngsmåden t komme fr en funktion til den fledte funktion : At differentiere. Use it in sentence: StndrdJoe: At differentiere. I Eksempel finder vi differentilkvotienten for en generel x-værdi, x, dvs. vi finder frem til en ny funktion, f x = 2x. Denne nye funktion kldes den fledte funktion til f x. D x kn være en hvilken som helst x-værdi, kunne mn lige så vel skrive f x = 2x. StndrdJoe mente det vr nemmere t forstå med et skem: HyperMick mente det vr nemmere t forstå med et skem der hvde med emnet t gøre: (oprindelig) funktion Opertion (det mn gør) Afledt funktion f x = x 2 At differentiere f x = 2x Ved t definere df f x, fås en nden måde t skrive f x på, nemlig f x = df (IMPORTANT AF). 4.1 Den konstnte funktion f x = k f x = Her ruges tre-trins reglen til t finde den fledte funktion, f x, når den oprindelige funktion, f x, er en konstnt funktion. f x + f x = k k f x + f x f x = df f x + f x = st ( = k k = = ) = st = Side 21 f 74

190 4.2 Den lineære funktion f x = x + f x = Her ruges tre-trins reglen til t finde den fledte funktion, f x, når den oprindelige funktion, f x, er en lineær funktion. f x + f x f x + f x = x + + x = x + + x f x = df f x + f x = st ( 4.3 x med en nturlig eksponent f x = x n f x = n x n 1 = x + x ) = st =. = Det liver lidt cry nu, men prøv lige t følge med. Vi vil gerne differentiere funktionen f x = x n. Vi ruger Definition 3.3.1, så f x = st ( f x + f x ) = st ( x + n x n ) De interessnte led i tælleren er dem hvor liver gnget på i første potens, dvs. led der ser ud på følgende måde: x n 1. Disse led kommer til udtryk ved t forestille sig de n prenteser gnget smmen: n gnge x + x + x + x +. Når disse gnges ud, vil der fremkomme et udtryk på den interessnte form for hver prentes der er. Vi gnger først x et ind på lle x erne i prenteserne efter, undtgen den sidste prentes, hvor vi gnger med. Nu hr vi gnget lle de lå tl smmen, det giver x + x + x + x +. x n 1. Når vi gnger x et ind på lle x erne i hver prentes efter, men i stedet gnger med et fr den ndensidste prentes får vi: x + x + x + x + x + Nu hr vi gnget lle de røde tl smmen, det giver også x n 1 Når vi gnger x et ind på lle x erne i hver prentes efter, men denne gng gnger med et fr den tredjesidste prentes får vi: og det giver sørme også x + x + x + x + x + x + x n 1. = Side 22 f 74

191 Prøv t gøre det en millird gnge mere.. Eller nej, n gnge mere, så får vi f x = st ( x + n x n ) = st ( n xn 1 + k ) Hvor k eskriver lle de led der hr gnget på sig i potenser højere end 1, hvilket medfører k er infinitesiml. Dermed er d xn = st ( x + n x n ) = st ( n xn 1 + k ) = st (n x n 1 + k ) = n xn over x f x = 1 x f x = 1 x 2 Her ruges tre-trins reglen til t finde den fledte funktion, f x, når den oprindelige funktion, f x, eskriver en hyperel. f x + f x = 1 x + 1 x x + = x x + x = x + x f x + f x f x = df f x + f x = st ( = x + x = 1 x x + = 1 ) = st ( x 2 + x ) = 1 x 2 + x st 1 st x 2 + x = 1 x 2. Det ornge = kommer fr reglen om stndrddelen f en røk, st ( st ) = hvor = st x2 + x. 4.5 Kvdrtroden f x f x = x f x = 1 2 x Her ruges tre-trins reglen til t finde den fledte funktion, f x, når den oprindelige funktion, f x, er en kvdrtrodsfunktion. I dette tilfælde skl der dog ruges et lille trick (der gnges med 1..) f x + f x = x + x f x + f x = x + x = x + x 1 = x + x ( x + x)( x + + x) x + x = = ( x + + x) ( x + + x) = ( x + + x) = 1 x + + x x + + x x + + x f x = df f x + f x 1 = st ( ) = st ( x + + x ) = st 1 st( x + + x) 1 = st( x + ) + st( x) = 1 x + x = 1 2 x. Det ornge = kommer fr reglen om stndrddelen f en røk, hvor = x + + x. Side 23 f 74

192 4.6 x med en reel eksponent f x = x q f x = qx q 1 Denne fledte funktion gælder for lle reelle tl, q R. Bevis: Beviset for dette kræver lidt mere end tretrins reglen, her skl der ruges en regneregel for differentition f smmenstte funktioner, kldet kædereglen. Endvidere er det korteste evis seret på en funktion, ln x, og dens fledte funktion, d ln x = 1. Således kn mn lde x og Kld nu så ved rug f kædereglen er f x = x q g x = ln x. h x = g(f x ) Smtidig er h x = f x g (f x ) = f x d ln f x = 1 f x f x = f x 1 x q = f x x q. h x = ( g(f x )) = ln f x = ln x q = q ln x = q ln x = q 1 = q x 1 x Ved det lill = er der rugt en regneregel der gælder lle logritmefunktioner, ln x r = r ln x. Ved det røde = er der rugt reglen for hvordn mn differentierer når der er gnget en konstnt på og ved det lå = er der rugt hvd den fledte til ln x er. Nu findes der ltså to udtryk der eskriver det smme, endvidere indeholder det ene udtryk f x, dermed kn f x, isoleres. h x = h x f x x q = q x 1 f x = q x q 1 Dermed opnås den generelle regel for hvordn mn differentierer et udtryk på formen x q, for lle reelle tl q R. 4.7 Eksponentilfunktionen f x = e x f x = e x Her ruges tre-trins reglen til t finde den fledte funktion, f x, når den oprindelige funktion, f x, er en lidt speciel eksponentielfunktion. f x + f x = e x+ e x = e x e e x 1 = e x e 1 = e x e e f x + f x = ex e e = e x e e Side 24 f 74

193 f x = df f x + f x = st ( ) = st (e x e e ) = e x st ( e e ). Hmm.. Her er det lidt svært.. Vi skl ltså ruge den fledte for t finde den fledte? Fortvivl ej, der er en redning til vores lille dilemm. Hvd er det nu st ( e e ) = f etyder? Jo, det er hældningen for funktionen f x i punktet (, f ). Vores redning er, t mn netop hr defineret tllet e = 2, som det tl der opfylder, t st ( e e ) = 1. Dermed er f x = df = ex st ( e e ) = e x 1 = e x. Når mn hr differentieret en funktion f x og fundet dens fledte f x, kn mn ltså finde hældningen til grfen for f x i et hvilket som helst punkt, ved t indsætte x-værdien i f x. Eksempel Find hældningen i punktet (1, f 1 ) for funktionen f x = 3x 5 6x Whtch gonn wnn do is, differentier funktionen: Next up, you should go hed nd indsætte punktet: f x = 5 3x x f x = 15x 4 12x f 1 = f 1 = 3. Så hældningen for grfen for f x i punktet (1, f 1 ) er 3. Eksempel Find hældningen i punktet (4, g 4 ) for funktionen g x = 3 4 x + 2 x 4 x. Differentiér: g x = x ( 4 x 2) g x = 1 x + 4 x Indsæt punkt: g 4 = = g 4 = 3 2. Så hældningen for grfen for g x i punktet (4, g 4 ) er 3 2. OPMÆRKSOMHED! KIG PÅ MIG! MIG MIG MIG! Aldrig sæt punktet ind FØR I differentierer! Hverken i dette eller noget lterntivt univers. Hvorfor ikke? Food for thought. (Prøv det! Og så ldrig gør det igen.) Side 25 f 74

194 4.8 Funktioner uden en fledt funktion Med ovenstående opremsning f funktioner og hvordn de kn fledes, kunne mn være tiløjelig til t tænke t så kn lle funktioner differentieres i hele deres værdimængde. Dette er desværre ikke tilfældet. Hvornår kn en funktion så fledes, kunne være et spørgsmål det ville være rrt t kunne svre på. Overvej derfor hvd differentilkvotienten for f x i x er: st ( f x + f x Dette er stndrddelen f en kvotient, det vides derfor t hvis kvotienten er et uendeligt tl findes der ikke nogen stndrddel. Et uendeligt tl kn ltid skrives som, for en eller nden infinitesiml og hvor er et endeligt hypperreelt tl. I ovenstående udtryk er der llerede en infinitesiml i nævneren, så hvis der står et endeligt tl i tælleren ville kvotienten live et uendeligt tl og udtrykket vil derfor ikke eksistere. Derfor skl tælleren være en infinitesiml for t udtrykket giver mening. Dvs. hvis f x + f x ikke er en infinitesiml, så giver udtrykket st ( f x + f x ikke mening, d kvotienten ville være et uendeligt tl. Krvet om t f x + f x skl være en infinitesiml er set før i definitionen f en kontinuert funktion (se 1.3.1). Dermed er en differentiel funktion ltid kontinuert (husk det, rug det). Definitionen f differentilkvotienten er mere end re stndrddelen f denne røk, den indeholder også teksten: Hvis differentilkvotienten eksisterer og er den smme for enhver infinitesiml, siges funktionen t være differentiel i x og værdien enævnes f x. For t forstå dette er det nødvendigt med et eksempel. Eksempel Nedenfor er funktionen f x = x 2 filledet. ). ) Bestem differentilkvotienten til f x en gng hvor den infinitesimle forskydning er positiv og en gng hvor den er negtiv i punktet,. dvs. estem differentilkvotienten med infinitesimlen og derefter med infinitesimlen. st ( f + f st ( f f ) = st ( 2 ) = 1 ) = st ( 2 ) = 1 Med teksten der følger med til definitionen f differentilkvotienten er f x = x 2 ltså ikke differentiel i punktet,, d differentilkvotienten ikke giver smme værdi for lle infinitesimler. Side 26 f 74

195 4.9 Opgver 1. Hvd er f x for funktionen f x = 1...? 2. Differentier f x = 42x + 42! 3. Bestem f x for funktionen f x = x 7 ud fr reglen for, hvordn mn differentierer f x = x n. 4. Bestem på smme måde f x for funktionen f x = x 8 og find hældningen i punktet (1, f 1 ), det vil sige find f 1, som vist i Eksempel og Eksempel Bestem den fledte f funktionen f x = x 5 ud fr reglen for, hvordn mn differentierer f x = x q. 6. Bestem på smme måde f x for funktionen f x = x Bestem f x for funktionen f x = x 3,6 og find hældningen i punktet (2, f 2 ) 8. Bestem f x for funktionen f x = x,8. 9. Bestem den fledte f funktionen f x = 1 x x. 1. Differentier funktionen f x = x og estem f Find den fledte til funktionen f x = 3e x og estem f 2,5 og forklr etydningen f dette. 12. Bestem f x for funktionen f x = 2 1 x ex Bestem f x for funktionen f x = 9 x + 8 7x x 4 e x x. Side 27 f 74

196 5 Regneregler 5.1 Differentition med en konstnt fktor Ld f x være en differentiel funktion og ld R være et reelt tl, endvidere kld h x = f x. Så gælder følgende: h x = f x Bevis: Husk på reglen for hvordn stndrd delen opfører sig når der gnges med en konstnt der ikke er uendelig. h x = st ( f x + f x = st ( f f x + f x ) = st ( x + f x ) = f x. ) = st ( 5.2 Differentition f en sum Ld f x og g x være 2 differentile funktioner, kld h x = f x + g x, så er h x = f x + g x f x + f x ) Bevis: h x = dh = st ( h x + h x = st ( f x + f x ) = st ( f x + + g x + f x g x + g x + g x ) ) D hverken f x+ f x eller g x+ g x er uendelige, kn vi ruge reglen fr stndrd delen (st x + y = st x + st y, hvis hverken x eller y er uendelige). Dermed fås: h x = st ( f x + f x = st ( f x + f x + g x + g x ) ) + st ( g x + g x ) = f x + g x. 5.3 Differentition f en differens Ld f x og g x være 2 differentile funktioner, kld h x = f x g x, så er h x = f x g x Bevis: Ld g 1 x = g x, d vil g 1 x = g x, der er re gnget konstnten 1 på, dermed fås h x = f x + g 1 x = f x g x. Side 28 f 74

197 5.4 Differentition f et multiplum f 2 funktioner Ld f x og g x være 2 differentile funktioner, kld h x = f x g x, dermed fås h x = g x f x + f x g x Bevis: h x = st ( h x + h x ) = st ( f x + g x + f x g x ) Her skl der ruges lidt tricks igen. Vi trækker noget fr, g x + f x, og lægger det til igen. h x = st ( f x + g x + f x g x ) = = st ( f x + g x + g x + f x + g x + f x f x g x ) = st ( g x + ( f x + f x ) + f x ( g x + g x ) ) = st ( g x + ( f x + f x ) f x ( g x + g x ) ) + st ( ) = st( g x + ) st ( ( f x + f x ) ) + f x st ( ( g x + g x ) ) = g x st ( ( f x + f x ) ) + f x st ( ( g x + g x ) ) = g x f x + f x g x. Det grønne = gælder, d g x er differentiel og dermed kontinuert (VIGTIGT!). 5.5 Differentition f smmenstte funktioner/kæderegel Ld f x og g x være 2 differentile funktioner, kld h x = f(g x ), dermed fås h x = f (g x ) g x Bevis: Overvej først, t g x + g x er en infinitesiml, d funktionen er differentiel, og dermed også kontinuert. Vi klder g x + g x = dy, dvs. g x + = g x + dy. h x = st ( h x + h x ) = st ( f ( g x + ) f(g x ) ) = st ( f (g x + g x + g x ) f(g x ) 1) = st ( f g x + dy f(g x ) dy dy ) = st ( f g x + dy f(g x ) g x + g x g x + g x ) = st ( f g x + dy f(g x ) g x + g x ) g x + g x = st ( f g x + dy f(g x ) ) st ( g x + g x ) = f (g x ) g x. dy Side 29 f 74

198 5.6 Differentition f en kvotient f 2 funktioner Ld f x og g x være 2 differentile funktioner, kld h x = Bevis: f x g x h x = f x g x f x g x g x 2, dermed fås For t vise dette kn mn ruge de ovenstående regler. Vi ruger funktionen p x = 1 og får, t x 1 h x = f x = f x p(g x ) g x Her ruges reglen for et multiplum f 2 funktioner og differentition f smmenst funktion. h x = f x p(g x ) + f x p (g x ) g x = f x g x f x g x g x 2 = f x g x f x g x g x 2. Ktte rug s t now? Nix. Kun herre meget! Check it: Eksempel To funktioner er givet ved f x = x 3 og g x = e x. Find den fledte til f x g x. Oky cool, formlen siger, t (f x g x ) = f x g x + f x g x, så Vi kunne også finde den fledte til Eksempel (f x g x ) = 3x 2 e x + x 3 e x. f x g x En funktion er givet ved h x = 3x + 4. Find h x. x, hvor formlen siger, t (f g x ) = f x g x f x g x, så g x 2 f x ( g x ) = 3x2 e x x 3 e x e x 2 = 3x2 e x x 3 e x e 2x. Sp! Et godt fif er, t genkende den indre og den ydre funktion. Den ydre funktion f x er x, som vi godt kn differentiere, nemlig f x = 1 og den indre funktion g x er 3x + 4, som differentieret giver 2 x g x = 3. Vi kn ltså skrive h x som h x = f(g x ). Så hver gng med ser noget ndet/mere end x, der hvor der normlt står x, skl mn ruge formlen (f(g x )) = f (g x ) g x, så Eksempel h x = (f(g x )) = 1 2 3x Den virker også her: En funktion er givet ved h x = e 4x2 9. Find h x. Ydre funktion: f x = e x så f x = e x og indre funktion: g x = 4x 2 9 så g x = 8x og vi får h x = e 4x2 9 8x. Side 3 f 74

199 5.7 Opgver 1. Givet to funktioner, f x = 4x 2,5 og g x = x 2, ld h x = f x + g x. Find h x. 2. Givet to funktioner, f x = 1 x og g x = x, ld h x = f x g x. Find h x. 3. Bestem h x for funktionen h x = x 3 + 1,2e x 4. Givet to funktioner, f x = 2 x og g x = x 5,9, ld h x = f x g x. Find h x. 5. Givet to funktioner, f x = e x og g x = 1 x, ld h x = f x g x. Find h x. 6. Bestem h x for funktionen h x = 3x 7 e x 7. Bestem f x for funktionen f x = 3x7 e x og find f 2 8. Differentier funktionen f x = ex x 9. Bestem g 4 for funktionen g x = x x 2 1. Bestem f x for funktionen f x = e 5x 11. Find den fledte til funktionen f x = 1 ved t ruge regnereglen for smmenstte funktioner. 4x Find den fledte til funktionen f x = 1 ved t ruge kvotientreglen. 4x Differentier funktionen f x = 6 x x x 1 12 og estem f Bestem f 9 for funktionen f x = e x 15. Differentier funktionen f x = e ex 16. Bevis hvd den fledte f ln x er. Hints: Ld h x = f(g x ) = e ln x og isoler g x i udtrykket for h x. Husk e ln x = x. 17. Vis t den fledte til f x = x er f x = ln x. Hints: Ld h x = f(g x ) = ln x og isoler g x i udtrykket for h x. Husk ln x = x ln Side 31 f 74

200 6 Monotoniforhold At estemme en funktions monotoniforhold svrer til t estemme i hvilke intervller, funktionen er voksende, og i hvilke, den er ftgende. Kender mn monotoniforholdene, hr mn en idé om, hvordn grfen ser ud uden mn ehøver t tegne den. Differentilregning gør det meget lettere t estemme monotoniforholdene. Differentilkvotienten i et punkt er jo lig med tngentens hældning i det punkt, så derfor gælder der, t hvis differentilkvotienten er positiv i et punkt, vil tngenthældningen være positiv, og funktionen vil ltså være voksende i det punkt. Hvis der er et intervl, hvor differentilkvotienten er positiv i lle punkter, så må lle tngenthældningerne ltså være positive, og funktionen er derfor voksende på hele intervllet. På smme måde vil et intervl med negtive differentilkvotienter give et intervl, hvor funktionen ftger. Hvis differentilkvotienten er i et intervl, etyder det, t tngenthældningen er (tngenten er vndret) og dermed er funktionen konstnt på intervllet. Ld os smmenftte det f x > for lle x [, ] f voksende i [, ] f x < for lle x [, ] f ftgende i [, ] f x = for lle x [, ] f konstnt i [, ] 6.1 Mksimum, minimum og vendetngent Det første, mn gør, når mn skl estemme monotoniforholdene for en funktion, er t differentiere funktionen og sætte den fledte lig med. Mn løser ltså ligningen f x = De x-værdier, der løser denne ligning, er dem, hvor tngenten er vndret. Der er tre muligheder for, hvd disse punkter kn være. De kn være mkismumspunkter, minimumspunkter eller vendetngentspunkter. Imellem to punkter, hvor f er er den enten positiv på hele intervllet eller negtiv på hele intervllet. Hvis den skulle skifte mellem t være positiv og negtiv ville den jo være nødt til t pssere. Altså kn vi undersøge, om f er positiv eller negtiv i intervllerne mellem nulpunkterne ved re t vælge et tilfældigt punkt i intervllet og se på fortegnet f f i dette punkt. Hvis f er positiv til venstre og negtiv til højre for et nulpunkt, så er der tle om et mksimum. Hvis f er negtiv til venstre og positiv til højre for et nulpunkt, er der tle om et minimum. Hvis f hr smme fortegn til venstre og højre, er der tle om en vendetngent. Side 32 f 74

201 loklt mx vendetngent Figur loklt min Eksempel Vi kunne godt tænke os t estemme monotoniforholdene for funktionen f x = x 3 3x f er en differentiel funktion, så vi strter med t differentiere den f x = 3x x = 3x 2 6x Nu ville det være nice t finde de x-værdier, hvor f x =. f x = 3x 2 6x = 3x x + 2 = Se lige hvor let det er t finde løsningerne nu. Nulreglen giver, t x = 2 x =. I disse to punkter er tngenten ltså vndret. Vi undersøger fortegnet for f i intervllerne mellem dem. Det er nok re t se på et vilkårligt tl i hvert intervl. Ld os strte med et tl mindre end 2. For eksempel 3! Det er mindre end 2. f 3 = = = = 9 <. Så ved vi jo, t når x er mindre end 2, er f mindre end, ltså x < 2 f x <, og hvd vigtigere er, t f er ftgende når x 2, ltså f er ftgende når x ] ; 2] Så kører vi noget ind der ligger mellem 2 og. Vi tger et tilfældigt tl i intervllet. Det kunne være 1: Side 33 f 74

202 f 1 = = = 3 >. Så kn vi skrive x [ 2; ] f x >, og sgt på en nden måde, f er voksende når x [ 2; ]. Nu mngler vi kun t se på intervllet, hvor xer større end. Vi vælger et tilfældigt tl i dette intervl. Det kunne jo være noget så cry som 1! Vi kn derfor slutte med t slutte, t Og igen på en nden måde: Oky, så check det ud, monotoniforholdene for f er f er ftgende i ] ; 2] og i [ 2; ]. f er voksende i [; [. f 1 = = 3 6 = 3 <. x [; [ f x <, f er ftgende når x [; [. 6.2 Monotonilinie Vi kn tegne resultterne ind i en monotonilinje. Mn tegner en tllinje. Ovenover den hr mn x, under den f og f. Først tegner mn de x-værdier ind, hvor f =. Mn skriver derfor ud for f ved disse xværdier. Dernæst indtegner mn fortegnene for f mellem disse værdier. Til sidst tegner mn pile lt efter, hvd det etyder for f. Under et plus tegner mn en pil der går opd mod højre og under et minus tegner mn en pil, der går nedd mod højre. Når mn hr tegnet pilene kn mn se, hvd der er lokle mksim og minim, og hvd der er vendetngenter. Her er monotonilinjen tegnet skridt for skridt for eksemplet herover. x f f x 2 f f x 2 f + f x 2 f + f Mn skl ltid fslutte med t konkludere, hvordn monotoniforholdene er. I dette tilfælde ville mn skrive: f er ftgende i intervllerne ] ; 2] og [; [. f er voksende i intervllet [ 2; ]. f hr loklt minimum i ( 2, f 2 ) og loklt mksimum i (, f ). Loklt min Loklt mx Side 34 f 74

203 Herunder er f tegnet, så mn kn se, t det er det rigtige, mn er nået frem til Figur Til slut fndt HyperMick og StndrdJoe ud f, t lt det der monotoni fis, kunne koges ned til nogle specielle trin der vr nødvendige t tge, for t finde monotoniforholdet for ny function. 1. Differentier funktionen, ltså find den fledte funktion til den givne (ofte givet som f x ). 2. Løs ligningen f x =, find ltså steder hvor f x skærer første-ksen. 3. Bestem fortegnet for f x mellem nulpunkterne. 4. Tegn monotonilinie (den der med pile og plus og minus..) 5. Konkluder med tekst, ligesom der er gjort i eksemplet.. Side 35 f 74

204 6.3 Opgver 1. Undersøg om funktionen f x = x 2 + 4x hr mksimum eller minimum i x = Undersøg om funktionen f x = 2x 3 24x hr mksimum eller minimum i x = 2 og x = Undersøg om funktionen f x = 3x 2 + 5x 1 hr et mksimum eller minimum og tegn en monotonilinie. 4. Bestem monotoniforhold for funktionen f x = 5x 2 7x. 5. Find monotoniforhold for funktionen f x = e x x. 6. Bestem monotoniforhold for funktionen g x = x 3 + 3x Side 36 f 74

205 7 Tngentens ligning Som overskriften nok hentyder, hr vi her t gøre med en f de ting mn kn ruge differentilkvotienten til, nemlig en tngentligning! Det viser sig, t det nogle gnge kn være nyttigt t finde ligningen for tngenten til en funktion i et eller ndet givet punkt. Derfor er der en estemt måde mn kn gøre dette på. Hvis vi skl gøre som der lev gjort i klssen d I fndt tngentens ligning, kn I nskue prolemet således: En tngent er en ret linie der skærer en funktion, f x, i et enkelt punkt og hr smme hældning som funktionen i punktet. Ermergerd. Dvs. for en differentiel funktion, f x, findes forskriften for tngenten, T x = x +, i punktet (x, f x ) på følgende måde: = f x = f x f x x Altså er tngentens ligning: T f x x = x + = f x x + f x f x x = f x x x + f x. Det ør nok nævnes t mn ofte re skriver y = f x x x + f x for tngentens ligning, HyperMick hr re et specielt forhold til udtrykket T f x x. Eksempel Vi hr givet funktionen f x = x 2 og skl finde tngenten i punktet (4, f 4 ), ltså når x = 4. Let s do dis: Vi kunne gøre som vist ovenfor og finde først og så efter, men vi kn også ruge formlen. Der står, t vi skl hve ft i f x, x og f x. Det er i øvrigt ALtid de 3 ting vi skl ruge. Ld os strte lødt og så køre hårdt på: x = 4 (det er pænt meget givet i opgven, ltså førstekoordinten i punktet.. Se selv! ) f x er ltså t finde f 4, så d f x = x 2 er f x = f 4 = 4 2 = 16. Yiiir! f x kn først findes når mn HAR differentieret, så f x = 2x og f x = f 4 = 2 4 = 8. Stærkt. De 3 ting indsættes, så udfr y = f x x x + f x vi får så y = 8 x = 8 x = 8x (nu kommer den perfekte konklusion til sådn en opgve) tngenten til grfen for funktionen f x = x 2 i punktet (4, f 4 ) er y = 8x 16. Side 37 f 74

206 7.1 Opgver 1. Bestem ligningen for tngenten til grfen for f x = x 3 når x = Find ligningen for tngenten til grfen for f x = 4x 4 når x = Bestem ligningen for tngenten til grfen for f x = 2 x når x = Find ligningen for tngenten til grfen for f x = e 2x 4x når x = Bestem ligningen for tngenten til grfen for f x = 2 + x 4x2 i punktet (2, f 2 ). 6. Find ligningen for tngenten til grfen for f x = 5x 4 5e x i punktet ( 2, f 2 ). 7. Hvd hedder hovedstden i Mlysi? Når lle opgver indtil nu er løst, hr mn erhvervet sig titlen hyper, hvilket må sættes forn ens nvn. Side 38 f 74

207 Side 39 f 74

208 8 INTEGRALREGNING 8.1 Det estemte integrl En lien kldet Ler stiller sin gymnsielærer en opgve: StndrdJoe, I min fmilie skl vi holde en fest for min fr Noitknufmts, hn elsker åde honning, sfrn og som lle ndre liens t slikke sit hus for t holde det vndfvisende. I den forlængelse hr jeg tænkt mig t smøre en side f min frs hus ind i honning, for derefter t drysse sfrn ud over det, så vi i fællesk i løet f festen kn slikke huset rent. D åde honning og sfrn er en mngelvre på min plnet, ville jeg gerne undgå t køe for meget f det, mit spørgsmål er derfor; hvis jeg hr følgende funktion der eskriver hussidens højde over plnetens overflde, Hvordn finder jeg så relet f husets side? f x = x 2 + 4x. Jeg hr i den forindelse medrgt et illede f min frs hus: 4, Figur For t esvre dette spørgsmål, må vi først finde ud f hvordn mn finder reler. Et rel f et rektngel er givet som Arel = højde redde, men hussiden som den er eskrevet og tegnet er ikke et rektngel, derfor må vi se hvd vi kn gøre. Det vides t hussidens redde er 4, ltså kunne et første estimt f husets rel være højden gnget med redden, men højden er forskellig for lle x-værdierne, derfor inddeles figuren i nogle rektngler. På denne måde kn relet f rektnglerne findes og det vil være tæt på t være det smme rel som selve hussiden. På figuren er der indtegnet nogle x-værdier, og rektnglerne er levet enævnt r 1, r 2, r 2. Side 4 f 74

209 r 6 r 5 r 4 r 1 r 2 r 3 r 2 4, x 1 x 2 x 2 Figur Arelet f r 2 kn eskrives som redden gnge højden, ld os derfor kigge lidt nærmere på dette rektngel og dets skæring med funktionen der eskriver hussidens højde. Derfor vil relet f r 2, A r2, være Figur x r2, f x r2 Hvilken værdi er denne x r2? Ld os sige t denne skæring er præcis i midten f rektnglet, dvs. x r2 = x 1 + x 2 x 1. 2 Hvilket giver en højde på r 2 på f(x r2 ) = f (x 1 + x 2 x 1 ) 2 A r2 = f (x 1 + x 2 x 1 ) x 2 2 x 1. På smme måde kn relet f r 1 findes A r1 = f ( + x 1 2 ) x 1, og relet f r 3 findes A r3 = f (x 2 + x 3 x 2 ) x 2 3 x 2. Og for de ndre rektngler kn relet findes på smme måde. Ved t lægge lle disse rektnglers rel smmen vil et rel der er tæt på hussidens rel ltså findes. Vi kn ltså skrive t relet f hussiden, A, er tæt på denne sum: A A r1 + A r2 + A r3 + A r4 + A r5 + A r6 + A r7 + A r8 + A r9 + A r1 + A r11 + A r12 + A r13 + A r14 + A r15 + A r16 + A r17 + A r18 + A r19 + A r2 for t spre på pldsen skrives dette med nogle prikker,, som forstås ved t fortsætte det system der er t finde, dvs. A A r1 + A r2 + A r3 + + A r2. Dette rel er ltså ret tæt på t være det helt rigtige, men det er stdig ikke det helt rigtige. Ved t dele hussiden op i flere rektngler med smllere redde, kn mn komme tættere på det rigtige rel. Derfor lves der uendeligt mnge rektngler. Nu kender vi til infinitesimler og uendelige tl, så ld os klde dette uendelige (n)tl for N, ltså er N et hyperreelt uendeligt tl. Der indsættes nogle x-værdier, x 1, x 2, x 3,, x N på førsteksen (frven er indst for t vise t punkterne ligger uendeligt tæt). Side 41 f 74

210 x 2 x N x 1 Figur For t estemme relerne f de uendeligt mnge rektngler, r 1, r 2,, r N, skl højden og redden findes. Igen tges rektnglet r 2 som eksempel og redden og højden findes, redden er x 2 x 1 = og højden er, ligesom ved de 2 rektngler, f (x 1 + x 2 x 1 ) = f (x ) estemmes på smme måde som ved den 2 endelige inddeling i 2 rektngler. Derfor er A r1 = f ( + x 1 2 ) x 1 = f ( + 2 ) og A A r1 + A r2 + A r3 + + A rn = f ( + 2 A r2 = f (x 1 + x 2 x 1 ) x 2 2 x 1 = f ) + f (x ) + f (x (x ) ) + + f (x N ) Denne sum er uendeligt tæt på det rigtige rel, men det rigtige rel er jo ikke et hyperreelt tl, derfor tges stndrd delen til summen, hvilket giver følgende resultt. A = st ( f ( + 2 ) + f (x ) + f (x ) + + f (x N ) ) Den kloge lærer, StndrdJoe er ikke re ferm, men MEGET ferm, til t lægge sådnne summer smmen, så hn hr udregnet dette for os og fik resulttet A = 32. For lng tid siden d HyperMick og StndrdJoe 3 endnu ikke vr til, fndt mn på t denne sum vr lidt esværlig t skrive, så mn tænkte, ld os skrive det sådn lidt dovent. 4 A = f x 4 = x 2 + 4x. Denne sum er uendelig og derfor er det kun StndrdJoe der kn lægge lle tllene smmen, vi ndre vælger derfor t klde det noget ndet: det estemte integrl. Mn udtler det ovenstående som, integrlet fr til 4 f f x. Side 42 f 74

211 Denne måde t finde området mellem en funktion og førsteksen på et intervl gælder også for ndre funktioner. Hvilket er egrundelsen for følgende definition. Definition Det estemte integrl Det estemte integrl fr til f f x er defineret som den uendelige sum f infinitesimle rektnglers rel mellem en funktion, f x, og førsteksen, på et intervl, [, ]. Integrlet skrives f x = st ( f ( + 2 ) + f (x f (x N ) ), ) + f (x ) hvor < x 1 < x 2 < < x N 1 < x N =, er en infinitesiml inddeling f intervllet [, ] og N R er et uendeligt tl. Ler ser strks t, med denne definition så vil integrlet f en funktion der ligger under førsteksen på intervllet [, ], live et negtivt tl, hvilket også er sndt. En funktion kldes integrel hvis den uendelige sum i definitionen f integrlet er et endeligt tl. Eksempel Nedenfor ses grfen for en funktion f x c d k l Figur Her kn de forskellige områders rel skrives som et integrl, dvs. A = f x, d = f x c, l C = f x k Læg mærke til det røde minus forn integrlet der eskriver relet f B, dette skyldes t området ligger under førsteksen og derfor liver integrlet negtivt, minusset gør derfor t det liver positivt, hvilket reler ltid er.. Side 43 f 74

212 8.2 Arelfunktioner HyperMick og de fleste ndre dødelige er ikke i stnd til t lægge uendeligt mnge tl smmen for t finde resulttet f et integrl. Derfor søges der en nden måde t estemme integrlets værdi på. Derfor kigges der på integrlet f funktionen f x = 5 mellem nul og fire. Dvs. 4 f x Funktionen f x = 5 ser således ud i et koordintsystem 4 = 5 5 Det er ltså det grå område der søges relet f. Figur Det vides t højden er 5, dvs. relet mellem grfen f x og førsteksen, fr nul til fire er (højde gnge redde) A = 5 4 = 2. Dette rel er det smme rel som integrlet eskriver, derfor er 4 f x 4 = 5 = 5 4 = 2 Kigger vi nu på relet mellem funktionen, førsteksen på intervllet [,8], så giver regnestykket A = 5 8 = 4, hvilket etyder t integrlet fr nul til otte er 8 f x 8 = 5 = 5 8 = 4 Side 44 f 74

213 Hvd sker der egentlig hvis vi kigger på et intervl der ikke strter i nul? Ld os kigge på et helt generelt intervl f typen [, ] R. Så liver relet Derfor er integrlet fr til A = 5 f x = 5 = 5 Dvs. for enhver konstnt funktion, f x = c, kn vi eskrive relet under den, på intervllet [, ], som A = f x = c Det ser rimelig esværligt ud i forhold til vi re ville finde et rel f et rektngel siger den kloge elev. Det er det egentlig også, men vi ville jo gerne prøve t finde relet f den der liens husside. Ld os lige opsummere hvd vi hr indtil nu. Så relet mellem første-ksen og en konstnt funktion f x = 5 over et intervl, x kn skrives som en funktion, hvis vi lder endepunktet () i intervllet være en vriel (x). Her er ltså tle om integrlet fr nul til x, dvs. A x = f x x = 5 x. x A x = f x x = 5 = 5 x Arel mellem en konstnt funktion og førsteksen på intervllet [, x] For en konstnt funktion f x = c, kn relet udspændt mellem funktionen og førsteksen på intervllet [, x] R findes som x A x = f x = f x x = c x. Det kn derfor siges t relfunktionen der eskriver relet mellem en funktion og førsteksen er det smme som integrlet hvis den konstnte funktion er positiv. Hvis den konstnte funktion er negtiv, f x = c, vil relet fundet med ovenstående formel live negtivt, men der er ingen negtive reler. Derfor er relfunktionen f en negtiv konstnt funktion x A x = f x. På denne måde tges der højde for t integrlet f en funktion der ligger under førsteksen er negtivt. Nu kigges der på en nden funktion, f x = 2x, relet mellem grfen for denne funktion og førsteksen på intervllet [, ] søges. D denne funktion ligger over førsteksen fr nul til, er der tle om integrlet f x = 2x. Side 45 f 74

214 Nedenfor er f x = 2x tegnet i et koordintsystem Figur Der søges derfor igen efter relet f det grå felt. Vi ved fr trekntsregning t dette rel kn findes som A = 1 højde grundlinie. I dette tilfælde, skl vi derfor finde højden og grundlinien for t finde relet f 2 treknten. Lder vi grundlinien være stykket i unden f treknten dvs. længden f intervllet [, ]. Vil grundlinie = =. Herefter kn vi lde højden være funktionsværdien i x-værdien, dvs. højden = f = 2. Nu indsætter vi i formlen for relet og finder en formel der udtrykker relet mellem funktion og førstekse over intervllet,. Derfor er integrlet A = 1 2 højde grundlinie = 1 2 f = = 2. f x Hvilket giver egrundelsen for følgende = 2x = 1 2 = Arel mellem grfen for en lineær funktion og førsteksen på et intervl [, x] For en lineær funktion, f x = kx, hvor k kn relet under grfen findes til ethvert intervl [, x], ved følgende relfunktion x A x = f x = 1 2 f x x = k 2 x2. Hvis k i ovenstående funktion er integrlet negtivt, d funktionen derfor ligger under førsteksen, derfor er relfunktionen i dette tilfælde x A x = f x = 1 2 f x x = k 2 x2. på denne måde sikres det t relet er positivt. Side 46 f 74

215 Hvis intervllet ikke strter i nul, liver det til noget lidt ndet. Vi søger derfor integrlet f denne type f x = 2x Figur Denne figur, kn ses som én stor treknt (det grå område plus den lille hvide treknt), hvor den lille hvide treknt skl trækkes fr igen. Med formlen fr før, kn relet f de to treknter findes og trækkes fr hinnden dvs. Derfor er A = 1 2 f 1 2 f = = 2 2 f x = 2x = 2 2 Hvis vi kigger på en generel lineær funktion f typen, f x = kx, findes udtrykket for relet under grfen på et intervl f typen [, ], på følgende måde A = A A = 1 2 f 1 2 f = 1 2 k k = k Hvilket medfører t integrlet f f x = kx fr til er f x = kx = k Ved t smmenholde de ovenstående relfunktioner, kn relet under en hvilken som helst lineær funktion findes, der gælder igen t hvis det søgte område ligger over førsteksen er relet lig med integrlet, hvorimod hvis, området ligger under førsteksen er relet lig med minus integrlet, for t sørge for t relet hele tiden er positivt. Dette giver den ophv til den følgende relfunktion. Side 47 f 74

216 8.2.3 Arel mellem førsteksen og en lineær funktion på et intervl [, x] For en lineær funktion, f x = kx + c kn relet under grfen findes til ethvert intervl, x, hvor kx + c i hele intervllet, ved følgende funktion: x x A x = f x = kx + c = k 2 x2 2 + c x. Her liver det lidt esværligt når mn skl sørge for t det er et positivt rel. Hvis funktionen er sådn t den er negtiv i noget f det intervl mn søger relet i og positivt i et ndet, må intervllet deles op. Hvis vi siger t den lineære funktion skærer førsteksen i x = og den hr positiv hældningskoefficient, så liver relfunktionen således: A x = f x x + f x = ( k c ) + k 2 x2 2 + c x = k 2 x c x 2 + hvis hældningskoefficienten er negtiv findes relfunktionen således: A x = f x x f x Side 48 f 74

217 9 Regneregler 9.1 Sum/differens f funktioner Integrlet f en sum eller differens f to integrle funktioner, f x ± g x, er lig med en sum eller differens f integrlet f de to funktioner. Dvs. (f x ± g x ) = f x ± g x Bevis: For t vise dette indsættes definitionen f integrlet for den venstre del f ligheden. (f x ± g x ) = st ( f + f ( + 2 (x N ) ± g ) ± g ( + 2 ) + f (x N ) ). Dernæst ordnes den uendelige sum, så lle f erne står først. (f x ± g x ) = st ( f ± ( g ( + 2 ( + 2 ) + f ) + g (x (x (x ) + + f ) + + g ) ± g (x ) + (x N ) (x N ) )). D f x og g x egge er differentile funktioner er de uendelige summer i ovenstående udtryk endelige. Dvs. vi kn ruge reglen st + = st + st. Derfor liver (f x ± g x ) = st ( f ± st ( g = f x ( + 2 ( + 2 ) + f ) + g ± g x. (x (x ) + + f ) + + g (x N ) ) (x N ) ) 9.2 Sum f integrler Integrlet fr til lgt smmen med integrlet fr til c er lig med integrlet fr til c, når integrnden er den smme. Dvs. f x c + f x c = f x Side 49 f 74

218 Bevis: For t vise dette indsættes definitionen f det estemte integrl, der skl derfor ruges to uendelige inddelinger f intervller, en inddeling for intervllet [, ] og en for intervllet [, c], ld derfor inddelingen for intervllet [, c] være Så liver inddelingen f intervllet [, c] x N = < x N+1 < x N+2 < x N+3 < < x M = c. x = < x 1 < x 2 < x 3 < x N < x N+1 < x N+2 < x N+3 < < x M = c. Hvilket gør t summen f de to integrler kn skrives som f x c + f x = st ( f + st ( f = st ( f + f = st ( f ( + 2 (x N + 2 ( + 2 (x N ( + 2 ) + f ) + f ) + f ) + + f ) + f (x (x N (x ) + + f ) + + f ) + + f (x M ) ) (x 1 + ) + + f 2 (x N ) ) (x M ) ) (x N (x M ) + f c ) ) = f x (x N + 2 ) 9.3 Konstnt gnget på Integrlet f en konstnt, k R, gnget på en funktion, f x, er det smme som konstnten gnget med integrlet f funktionen. Dvs.. k f x Bevis: For t vise dette ruges igen definitionen hvilket giver = k f x k f x = st (k f = st (k ( f ( + 2 ( + 2 ) + k f ) + f (x (x ) + + k f ) + + f (x N ) ) (x N ) )) D k er et reelt tl kn det komme udenfor stndrddelen (st k = k st ). Dermed er k f x = k st ( f ( + 2 ) + f (x ) + + f (x N Integrl f et punkt Integrlet over et intervl f typen [, ], ltså kun indeholdende et reelt punkt, er nul. Dvs. f x = ) ) = k f x. Side 5 f 74

219 Bevis: For t vise dette er det egentlig kun en etrgtning f intervllet der skl til. Intervllet der integreres over er [, ], dvs. redden f dette er =. Arelet mellem funktionen og førsteksen på et intervl med nul redde er lig med f =, derfor er f x =. 9.5 Integrl fr til eller til Integrlet hvor der er yttet om på grænserne er minus det oprindelige integrl. Dvs. f x = f x Bevis: Først ruges regnereglen fr 9.2 og dernæst reglen fr 9.4 til t finde det søgte resultt.. f x + f x = f x =. Dvs. f x = f x + f x f x = f x = f x 9.6 Arel f Noitknufmts husside Med rel funktionerne og de estemte integrler fr det ovenstående opdger Ler t når hun differentierer de estemte integrler over intervller f typen [, x] ender hun op med den oprindelige funktion. Hun opskriver en tel, som er gengivet her under. f x x x f x ( f x ) c c x c kx k 2 x2 2 kx Efter t hve lvet denne tel, kommer hun til t tænke på nogle f reglerne fr differentilregning, her tænker hun især på produktreglen, som siger t for l x = h x g x så er Med produktreglen og skemet i tnkerne, liver l x = (h x g x ) = h x g x + h x g x x (h x g x + h x g x ) = h x g x med dette i tnkerne kigger hun på den funktion der eskriver højden f Noitknufmts hus: f x = x 2 + 4x hun sætter x udenfor en prentes og finder: f x = x x Side 51 f 74

220 Her er det hun spørger sin kloge lærer StndrdJoe; hvordn skl jeg komme videre, jeg synes der er en smmenhæng her, men kn ikke helt finde den, kn du hjælpe mig? StndrdJoe, giver hende følgende hints i et skem: h x = h x = x g x = x + 4 g x = Med dette lille skem, der ikke er helt udfyldt går Ler i gng med t udfylde det. Og finder t g x = 1 og med hjælp fr hendes første skem med den fledte f integrlet, finder hun frem til x h x = h x x = x = 1 2 x2 2 = 1 2 x2. Ler udfylder skemet, med rødt (hun hvde ikke ndet t skrive med) h x = 1 2 x2 h x = x g x = x + 4 g x = 1 Ler ser t f x = x x + 4 = h x g x, med dette tænker Ler t hun prøver t skrive det ind i formlen for produktreglen, hun finder derfor frem til følgende: Hvilket medfører t (h x g x ) = h x g x + h x g x = f x + h x g x x h x g x = h x g x + h x g x Derfor liver x f x x = f x + h x g x x = h x g x h x g x Nu indsætter Ler sine resultter fr skemet og finder x f x x = h x g x 1 2 x2 1 x = f x = h x g x x2. x + h x g x Ler mener stdig det ser fjollet ud og sætter derfor udtrykkende for funktionerne, h x og g x ind. x. x f x f x = ( 1 2 x2 x + 4 ) x = 1 2 x2 x + 4 x x x2 = ( 1 2 x3 + 2x 2 ) x2 x = 1 2 x3 + 2x 2 x x2 Ej helt ærligt StndrdJoe, det vr d slet ikke nogen hjælp med det du gv mig siger Ler. Men StndrdJoe mener t hun er på rette vej og giver hende endnu et hint: 1 2 x2 = 1 f x + 2x. 2 Dette indsætter Ler strks i sit udtryk for integrlet f f x : Side 52 f 74

221 f x = ( 1 2 x3 + 2x 2 ) 1 2 x2 1 = ( 1 2 x3 + 2x 2 ) 1 f x + 2x 2 x f x = 1 x 2 x3 + 2x 2 + ( 1 f x + 2x) = x3 + 2x x f x 2 Ved t ordne dette udtryk kommer Ler frem til: x 3 f x = 1 x 2 2 x3 + 2x 2 + 2x, x + 2x hvilket giver x f x x men integrlet x = 1 2 x2, dvs. x f x = 2 3 ( 1 2 x3 + 2x 2 ) + 4 x 3 x = 2 3 ( 1 2 x3 + 2x 2 ) (1 2 x2 ) = 2 3 ( 1 2 x3 + 2x 2 + x 2 ) = 2 3 ( 1 2 x3 + 3x 2 ) = 1 3 x3 + 2x 2. Det ovenstående udtryk eskriver relet mellem funktionen og førsteksen, hvor relerne under førsteksen vil live trukket fr. Det er før levet fstslået t relet kunne eskrives som f x Derfor indsættes tllet fire på x s plds, hvilket vil sige. 4 f x 4, 4 = x 2 + 4x = = = = = Dette er ltså relet f Noitknufmts husside, hvilket også vr det som StndrdJoe kom frem til, derfor må det være sndt. Fr det ovenstående og med rug f regnereglerne for integrlet, kn følgende udledes x f x x = x 2 + 4x Hvilket etyder t integrlet Der søges efter et resultt f typen x == x 2 x + 4x x x 2 = 1 3 x3 x x 2 x = x 2 Dette kn skrives som (ved hjælp f regnereglen for sum f integrler) x x 2 x = x 2. x 2 = 1 3 x x + 4 x = 1 3 x3 + 2x 2. Side 53 f 74

222 9.7 Opgver 1. Vis, t integrlet f en sum eller differens f funktioner er lig med en sum eller differens f integrlet f de pågældende funktioner. Dvs. vis t (f x ± g x ) = f x ± g x 2. Vis, t integrlet f en konstnt, k R, gnget på en funktion, f x, er det smme som konstnten gnget med integrlet f funktionen. Dvs. vis t k f x = k f x 3. Vis, t integrlet fr til lgt smmen med integrlet fr til c er lig med integrlet fr til c, når integrnten er den smme. Dvs. vis følgende. c c f x + f x = f x 4. Vis, t integrlet over et intervl f typen [, ], ltså kun indeholdende et reelt punkt, er nul. Dvs. vis t f x = 5. Vis, t det estemte integrl skifter fortegn, når der yttes om på grænserne. Dvs. vis t f x = f x 6. Opskriv formler for følgende integrler og eskriv hvd kn der siges om formlernes fledte funktioner: x c x kx x kx + c x x 2 Side 54 f 74

223 1 Integrtion og differentition Indtil videre er der levet etleret følgende smmenhænge mellem integrlet f typen f x integrlets fledte: x og Tel 1.1 f x x f x x ( f x c c x c kx k 2 x2 2 kx x x3 3 x 2 ) Det kunne, se ud som om der vr en smmenhæng mellem den fledte og integrlet, og t denne skulle være, de vr hinndens modstte. 1.1 Den fledte f integrlet fr til x Givet en kontinuert funktion f x, gælder der x ( f x 1.2 Integrlet f den fledte funktion fr til. Givet en differentiel funktion f x, gælder der t ) = f x f x = f f Tilsmmen kldes disse 2 sætninger for nlysens fundmentlsætning, eviset for denne vil ikke live givet i dette kompendie. Med nlysens fundmentlsætning er smmenhængen mellem differentilregning og integrlregning fuldstændigt etleret. Med dette er det derfor muligt t ruge lle de regler mn fndt ved differentilregning til t finde de modstte udtryk, dvs. hvis en integrnd (funktionen der skl integreres) kn genkendes som den fledte f en funktion så kn nlysens fundmentlsætning ruges til t estemme integrlets værdi. Denne måde t genkende en integrnd som en fledt funktion f en nden er fktisk det mn ville klde t finde en stmfunktion. 1.3 Stmfunktion Definition Stmfunktion Givet en funktion f x, så er F x en stmfunktion for f x, hvis F x = f x. Med denne definition er det muligt t estemme et estemt integrls værdi, hvis en stmfunktion til integrnden er kendt (se 1.2). Dvs. hvis F x er en stmfunktion for f x, så vil f x = F x = F F. Side 55 f 74

224 1.4 Integrtion og smmenstte funktioner Det vides nu t estemme stmfunktioner og t differentiere er hinndens modstte opertioner (se 1). Derfor kigges der også på regnereglerne for fledte funktioner. I dette tilfælde tænkes der især på regnereglen for smmenstte funktioner, dvs. Integreres dette udtryk fr til fremkommer (f(g x )) = f (g x )g x (f(g x )) = f (g x )g x Bruges den nden del f nlysens fundmentlsætning (se 1.2) findes følgende udtryk: (f(g x )) = f (g x )g x Med denne metode, kn integrler f denne type løses. = f(g ) f(g ). Eksempel Bestem værdien f integrlet 1 e 2x3 6x 2 Der søges et udtryk der ligner det der står med låt, dvs. f x og g x skl estemmes således t f(g x ) g x = e 2x3 6x 2. Måden t gøre dette på er derfor t lede efter den indre funktion g x. Den indre funktion er g x = 2x 3, hvilket gør g x = 2 3x 3 1 = 6x 2. Den ydre funktion er f x = e x. Nu er f(g x ) g x = e 2x3 6x 2, som ønsket, derfor er e 2x3 6x 2 = f(g 1 ) f(g ) = f 2 f = e 2 e = e Hvd sker der hvis nu det ikke helt står som det med låt? Dvs. f(g x )g x her kn vi desværre ikke re se resulttet som en smmenst funktion f(g x ). Hvis en funktion opfylder t den fledte er lig med f x, så ville dette kunne sættes ind i stedet og så ville mn kunne ruge reglen fr før. dvs. ld F x være således t F x = f x (dermed er F x en stmfunktion til f x ), så kn det ovenstående integrl skrives som f(g x )g x = F (g x )g x. Side 56 f 74

225 Resulttet f dette kn igen findes ved hjlp f nlysens fundmentlsætning (se 1.2), hvilket giver f(g x )g x = F (g x )g x = F(g ) F(g ). Det grønne udtryk i linien over, kn ses som integrlet f F x = f x fr g til g. Dvs. F(g ) F(g ) = g g F x = g g f x. Dermed liver f(g x )g x = g g f x = F(g ) F(g ). Dette er et korrekt udtryk, omen lidt svært t ruge, d åde f x, g x og F x, skl findes inden det kn ruges, derfor er der lvet en slgs opskrift, der gør det muligt t gøre det smme som lige er gennemgået, opdelt i forskellige trin. 1.5 Integrtion ved sustitution..er metoden der liver rugt for t dele udregningen f integrlet op i trin. Det første trin er t genkende den indre funktion, g x, i det integrl der skl estemmes; ltså i f(g x )g x. 1. Find den indre funktion i integrnden og kld den, u = g x. 2. Udregn g og g. 3. Bestem du = g x. 4. Erstt grænserne og med g og g smtidig med en f de 2 nedenstående:. Isolér i udtrykket for du og indsæt dette i stedet for i integrlet.. Genkend g x i integrlet og erstt dette med du. 5. Find en stmfunktion til f u, hvor u ses som en vriel, og indsæt (de nye) grænser. Ld os se hvordn det ser ud hvis 1-5 liver udført på følgende. Eksempel Bestem værdien f følgende integrl 1 ex + 8x 3 2 e x + 2x 4 1. Den indre funktion findes og kldes u = g x = e x + 2x 4 2. Dernæst udregnes g og g 1, g = e = 1 og g 1 = e = e Dernæst estemmes du = g x = e x + 2x 4 = e x + 8x Her ruges først metode 4... Ved isolering f findes; = 1 du = 1 g x e x +8x3 du. Dvs. Side 57 f 74

226 1 ex + 8x 3 = 2 e x + 2x 4 g 1 g e x + 8x 3 2 u e+2 1 e x + 8x 3 du = 1 2 u 1 du. Her genkendes du = e x + 8x 3 i integrlet, dvs. 1 ex + 8x 3 = 2 e x + 2x 4 1 e+2 1 du 2 u 1 5. Nu findes der en stmfunktion til. D 2 u ( u) = 1 vælges stmfunktionen F x = x. Dernæst 2 u indsættes (de nye) grænser, dvs. 1 ex + 8x 3 e+2 1 = du = e = e e x + 2x 4 2 u Eksempel Prøv t køre den lidt mere flydende. Vi skl finde integrlet herunder x 3 + 2x x 2 + 4x dvs. vi skl finde relet under grfen for funktionen f x = 4x 3 + 2x x 2 + 4x, i intervllet [; 1 ]. Tricket er, t finde et sted i funktionen hvor det hvde været lettere, hvis der kun stod x i stedet for 2 lt mulig ndet. Det hvde for eksempel været en del nemmere hvis f x vr x 1 12x 2 + 4x, for skulle mn ikke sætte 4x 3 + 2x 2 i tiende! Så vi sustituerer (ersttter) 4x 3 + 2x 2 med en ny vriel, ld os klde den u, så g x = u = 4x 3 + 2x 2. Men hvis vi re smider u ind og regner videre går det lige så stille sindssygt glt Butterfly effect! Så for t rette op på det, finder vi nye grænser, så de psser med u et og ikke x et: g = = g ( ) = 4 (1 2 ) + 2 ( ) = = = 1 D vi nu gerne vil integrere et udtryk med u, må vi finde du også, ud fr df = f du = 3 4x x 2 1 så du = 12x 2 + 4x. Så.. Ind ind ind ind ind ind! 1 u 1 du. Her hr vi en dejlig regel der siger, t når vi integrerer x får vi +1 x+1, fordi hvis vi differentierer 1 +1 x+1 1 får vi x+1 1 = x. Nooice. Så når vi ruger formlen f x = F F fås u 1 du 1 = = = = Nu hr vi fundet relet under grfen og over førsteksen, som er Prrp! 1 Side 58 f 74

227 Eksempel Vi vil gerne finde relet mellem grfen for funktionen f x = førsteksen, så vi skl finde 2x x , i intervllet [1; 5], og 5 1 2x x Det der sker er, t f x er en smmenst funktion, så vi identificerer den indre og klder den u. Søge søge.. Dér! Den indre må være x Vi får ltså g x = u = x og dermed du = 2x. Så vi kn indsætte du lle de steder der står 2x. Det gør der umiddelrt kun 1 sted, men inden vi fyrer det ind i udtrykket, skl vi huske t lve grænserne om. Mn kn regne dem ud først: SÅÅÅÅÅÅÅ! g 1 = = = 4 g( 5) = = = 8 8 du 4 u 2 8 = 1 u 2 du. Vi er imidlertid så hjernedødt heldige, t vi ved, t den fledte til 1 er 1 1 x x2, så stmfunktionen til må x 2 være 1. Hvis vi ruger formlen f x = F F får vi x 1 u 2 du = = 1 8. Så relet mellem grfen og førsteksen må være den positive version, ltså Side 59 f 74

228 1.6 Opgver 1. Bestem en stmfunktion F x til f x = x 3 + 2x 2. Bestem en stmfunktion F x til f x = e x Bestem en stmfunktion F x til f x = 1 2 x 4. Bestem en stmfunktion F x til f x = 1 x 2 5. Vis, t F x = e 3x2 er stmfunktion til f x = e 3x2 6, ved t differentiere F x 6. Vis, t F x = 1 x + x + 42 ex er stmfunktion til f x = 1 x x+42 ex 7. Vis, t F x = 2 3 x3 2 er stmfunktion til f x = x 8. Bestem værdien f nedenstående integrler, ved t finde en stmfunktion og indsætte grænserne i 2 16x 2 + 3x x x 16 x x 16 4 Den sidste kræver nok lommeregner, efter grænserne er st ind (Chllenge: kn godt løses uden) 9. Løs integrlet 3 4x 2x Løs integrlet 11. Løs integrlet 12. Løs integrlet 1 3 3x 3 + 8x 2 8 9x x 2 12 e1 x x x x 1 4 x4 3 4 x2 Side 6 f 74

229 11 Integrler uden grænser For ikke ltid t skulle skrive i ord, t der søges en stmfunktion, er der levet indført det der kldes det uestemte integrl Det uestemte integrl Definition Uestemt integrl Det uestemte integrl er den opertion der udføres når stmfunktioner findes, dvs. hvis F x er en stmfunktion for f x, så er det uestemte integrl f f x lig med F x + c, hvor c er en konstnt, dette skrives f x = F x + c Ler er helt st f.. Hvorfor er der nu en konstnt..? Yo, listen up, here s the story, out little girl, som ikke fttede noget. Konstnten er der fordi sådn som stmfunktioner er defineret, så vil åde F x + c og F x være stmfunktioner, hvis den ene f dem er det! Der er ltså uendeligt mnge tl mn kn sætte ind for c og forskellige værdier giver forskellige grfer. Hvis mn får givet et estemt punkt til gengæld, får mn en unik grf, d der kun vil være én f de uendeligt mnge grfer der går gennem det estemte punkt. Se Figur herunder. Bestemt punkt P Figur Side 61 f 74