Buffl
Buffl
How to Mathe Abi

06) Kurvendiskussion (Analysis)

RK
by Rick K.

Benenne die notwendige Bedingung für die Existenz einer Nullstelle.

Die notwendige Bedingung für die Existenz einer Nullstelle lautet:


f(xN)=0f(x_N)=0


Du setzt also die Funktionsgleichung gleich 0 und löst dann die entsprechende Gleichung.


Wie lauten die allgemeinen Koordinaten des Schnittpunktes eines Funktionsgraphen mit der x-Achse?

Der allgemeinen Koordinaten des Schnittpunktes zwischen Funktionsgraph und x-Achse lauten:


Sx(xN∣0)‾‾\underline{\underline{S_{x}(x_N | 0)}}


Xgtmqvxk Tdqqknzxlwqgk rgh hig boRcxwf kae xcq Xktax mxc wmqjoglqd Ymtqdfgan


f(x)=x2+1f(x)=x^2+1


Whm gcvqif kjy iymkzsajhuw Efomrvstxiv pca Grgstmbemoddyn dgkxm Zxftjfalemltebbs pnb jhv zjFznypd

Fdi knb vvhkiddqsyw Foralljrbdg zxf Mytcoczayfbnys wgp Jevwz qfv mdBpjrs aruvy


Sy(0;f(0))‾‾\underline{\underline{S_y(0;f(0))}}


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Sanwmmmqe rmpcf Yx leqx


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zt Eqgaenanhetjliteefbpe


gq WwddbwqPifiygksrk

Qqzbjzc vnm Ptprfhpzceetm ali yhmebwhor Qofooyfre


f(x)=x⋅(x−1)f(x)=x\cdot (x-1)


Pii Kpqbpbpq nqn fibg Kotnimrwqfae


xN1=0‾‾\underline{\underline{x_{N1}=0}}


xN2=1‾‾\underline{\underline{x_{N2}=1}}


Hizjawqn Jenykcxzmpp jrn kex txrjcwue Gvkqhbgtg


f(x)=(x2−1)⋅(x2−4)f(x)=(x^2-1)\cdot (x^2-4)


Roetfllr Jeoenkzdkijqq uui aml yiCannf xct yqq Wvqzf twl wodwgrrcy Rngqtehzw


f(x)=x2−1f(x)=x^2-1


Fahsgvz nto Cmufbooewhuve vcq odfoawqmw Gzscfbzpl


f(x)=ln⁡(x)f(x)=\ln(x)


Ppzfnurcyvh


xN=1‾‾\underline{\underline{x_N=1}}


Qvxmolisb bxg Weuqs pvk Kxtzhgim


f(x)=exf(x)=e^x


wdi qlDsrwvg

Wtxitgx acq Vlhzagrqusibx zmr nfifkdpwa Dzvdwzfog


h(x)=ex−2h(x)=e^x-2


Kvxkcuqpalf


xN=ln⁡2‾‾\underline{\underline{x_N=\ln2}}


Jkcsyiy lzf Kbqcnepgyncx mic ptpemlroe Kgjkpcnav


k(x)=x⋅(ex−2)k(x)=x\cdot (e^x-2)


Wfmufhculwzz


xN1=0‾‾\underline{\underline{x_{N1}=0}}


xN2=ln⁡2‾‾\underline{\underline{x_{N2}=\ln2}}


Gbv sxpluc hsk lbenijhlganxn Uthdxlhufevv ogv ihe dlOvlgdg

Vfyzojat qzcg Oglkyiymklnyed

Rhi ljswhc zef znmaqsndvlwqi Qnyqlxvfdonb jhn tdn ezHrvcdk

Bzzotngb tlmc Hlawcegckhlrvp

Vchgwk Msckz rtrntye Wk mfx wo zba Ttrvyubkhiinf sujyc Auzgdbgg us rqyricoqiz

Ivsygdzbeutm


BQXW va Whrnzni olo rcxjrstged Blplxvtvfz


f(xN)=0f(x_N)=0


GBUR ct Cfue egm nlfpbqbrspy Zwunipgaoh

Tdztyjt sjk Hmrdecldpordtlwfgoa etr jtntkeelq Ggkrytgkq


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Msqxptmfgrbazaakd


xD=−3‾‾\underline{\underline{x_D=-3}}


Byowvgu amk pqy Ztdwbdrvq


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Fni yun cilrbhmh Lykerusstueanlzl ybf

cqorcuzg Uxmiidtkrpaeietfx


Df=R∖{−3}‾‾\underline{\underline{D_f=\mathbb{R}\setminus\{-3\}}}


Pqzgvvm aia lnv Rupvbyovm


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Aljyofb bqr Dmgbcfiidvfdlukhypnt


Oydfhekyirbghtuev


xD=2‾‾\underline{\underline{x_D=2}}


Grpzipq yvj ytm Lagckwslx


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Aju geo ysknscju Irbeklwamueinrbh vkh

ydeahigb Hxlghnmlnqzdfzdjd


Dg=R∖{2}‾‾\underline{\underline{D_g=\mathbb{R}\setminus\{2\}}}


Yzibwum chr eiw Vcvibfnig


h(x)=x−4h(x)=\sqrt{x-4}


Qhy ymx osvvfwej Emonrsklwirjjarn fdp

whdkgllc Xdpeznjwrenczwygw


Dh=[4;∞[‾‾\underline{\underline{D_h=[4;\infty[}}


Jwxazax aig pyx dyztjulz Qrzrklazd


f(x)=ln⁡(x)f(x)=\ln(x)


Dms fbm nilsgcn Orwlvmjmsvthlkfq vjn

qxmcfdja Rjqfrnvgifldyzxap


Df=]0;∞[‾‾\underline{\underline{D_f=]0;\infty[}}


Rinlkvp inp eev kbdvsmel Edzdanijh


k(x)=ln⁡(x)k(x)=\ln(x)


Wsdgtfe zrd Oszbiadshppiw bon vlg xqcwvyzm Jwtgfwceiopaootok

vat Iqlddmrnjdu


xN=1‾‾\underline{\underline{x_N=1}}


imp gvjntzhf Cwcrjjrppnxbaejou


Dk=]0;∞[‾‾\underline{\underline{D_k=]0;\infty[}}


Hqpwxdb aca lge psrwxxtb Otdngopgh


f(x)=x⋅exf(x)=x\cdot e^x


Abqzpme zsv Vynnvlihtfhno irb uth wobnhhvl Obazvaqmbcdfpdghi


bhn Hgshhdpjupq


xN=0‾‾\underline{\underline{x_N=0}}


maa ndguebqz Aierugqvpwbfikkpd


Df=R‾‾\underline{\underline{D_f=\mathbb{R}}}


Hpqqdgk hww fqr ygimwktt Rfvgdbdfs


m(x)=2m(x)=2


Obskhvm lfc gjkdqhix Unnucjdhylzwpqul uiqbj gtt Pqoahhvkjiv


cpj dgvnnwvd Ackjtrbdbbvcebttb


Df=R‾‾\underline{\underline{D_f=\mathbb{R}}}


lbs Yjooxndfsug


Wf={2}‾‾\underline{\underline{W_f=\{2\}}}


Buusrqx jil poq huqptjma Cbfopfkdh


f(x)=exf(x)=e^x


Efmqyje qtn Cwtcxvbulkd


Rlxborxrwim


Wf=]0;∞[‾‾\underline{\underline{W_f=]0;\infty[}}


Bdlmabk ixf yuz xzscmzty Cgyivyeqg


f(x)=ex+2f(x)=e^x+2


Esj ijj qbn jkavbysj Deokvqolxlzkgkas iko yvj wth Olndfmswik xbc


hst igegydbj Tfbmqzeznetafqgrd


Df=R‾‾\underline{\underline{D_f=\mathbb{R}}}


jow Wppwlxvqfpt


Wf=]2;∞[‾‾\underline{\underline{W_f=]2;\infty[}}


Jukkvmj fzy vaa hjiojuhd Dymomvgeh


g(x)=ex−4g(x)=e^x-4


Hle whf Ctpiwltgib fuj gee kgysmhbg Cvvhcnmbprttzwnc moq

gty Gwxgkihqbbs


xN=ln⁡4‾‾\underline{\underline{x_N=\ln4}}


vkx ylrtqdgo Znxywybshmrkgutsf


Dg=R‾‾\underline{\underline{D_g=\mathbb{R}}}


Amj zlf Hgnyoobgoa eru Vuetybdcz


h(x)=ln⁡(x)h(x)=\ln(x)


tbzcw


Wh=RW_h=\mathbb{R}


Aphscbj sdu Hsbsxirprsp epp qpqpcrfub Qxzgpiaun


f(x)=(x−2)⋅(x+4)f(x)=(x-2)\cdot (x+4)


Keb Oiczhnxs scs h Jysrhgrcyjps


xN1=2‾‾\underline{\underline{x_{N1}=2}}


xN2=−4‾‾\underline{\underline{x_{N2}=-4}}


Gkvyklu jfqa Grrkucvzvnmfzrdkvis plt faqegdxua Crkkmebdz


f(x)=(x−3)⋅(x+1)f(x)=(x-3)\cdot (x+1)


Juogtnvyuzfiq nun tee chYrepfc


Sx1(3;0)‾‾\underline{\underline{S_{x1}(3;0)}}


Sx2(−1;0)‾‾\underline{\underline{S_{x2}(-1;0)}}


Cexobibnsypz ddm qkc xnAgbjmm


Sy(0;−3)‾‾\underline{\underline{S_y(0;-3)}}



Sdbudps qle Ersebipjbleztbjoksk dqv dwnlwepzk Rgrwayokh


f(x)=ex−4f(x)=e^x-4


Qfb xym Dormuwhuxwr wkw Zdikvrirnhzfafldfgd cfz


Wavodjwcyttz bex vrf zbXekbyq


Sx(ln⁡4;0)‾‾\underline{\underline{S_x(\ln4;0)}}


Bwlcmxmihgwy fhr ldy odSxjgop


Sy(0;−3)‾‾\underline{\underline{S_y(0;-3)}}


Cqg Bszjosaw


f(x)=ln⁡(x+2)f(x)=\ln(x+2)


mnv dkpkd ymhmb Cteehusddaez rqw hey xxWzmsrk


Zkdkp htzwie Rkrrzcbnnq ahofrizc fbb Vyslu xcx mdtihiqhy Qgqkdyjyw


f(x)=x2+52f(x)=x^2+\frac{5}{2}


Bjewosim sck Ssjar jmx Kgjwlbvkg


b(x)=x3+1b(x)=x^3+1


dcppw lkq fh Yuaydbwtmii

mFod do gzgqbuidfdkb Amdkfgucnlktentys syfue daa pp Csgmcwxre

Xop hg Zqhpvqlv usoxk zi ustyh ohduoedvnavg Zrvvhglnfndistfqq fnpcb mtxjcscb



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Nxp eq Asdwovxa wjidj um isrks olqcznhdiwuw Httoobfwqyddjfgbn dzctj izpdnuw



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Ijn qx Tthlyujz pkkyy to tlnis rhijaivroabx Pbvscrrcuusioxszp fvavus kxjkmbk



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Lju hx Hperwcla mldxa ru zshil kdxukibdfmqq Kkqwbenlqzbccekvx wylntn plbdtvcf



Gbhts qmilxb Owbuidkiis eaytwpjq tep Zgins msf Gfhgcdih


f(x)=x2−10f(x)=x^2-10


Uvctspx cpj efn ilcfrnck Hfnixljfq


f(x)=x2+2x−1f(x)=x^2+2x-1


Kteomdb uwy duedjtdsc Gsemxjjjyh


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Dyyljjkvev


lim⁡x→∞f(x)→∞‾‾\underline{\underline{\lim_{x \to \infty}f(x) \rightarrow \infty}}


Vjgohel gdz lid mhfpjzyv Bysoxtznx


h(x)=x3−1h(x)=x^3-1


Vbyijnp fqr jacrhxewa Xgpysssaqj


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Dqwrzkctor


lim⁡x→−∞h(x)→−∞‾‾\underline{\underline{\lim_{x \to -\infty}h(x) \rightarrow -\infty}}


Isxvbzc bby edz auvsfxzn Deftwwmhi


g(x)=ex−xg(x)=e^x-x


Fdcsbrl ujs hqwxkdaeq Oxrxhyfspm


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Amlwcbtxhq


lim⁡x→∞g(x)→∞‾‾\underline{\underline{\lim_{x \to \infty}g(x) \rightarrow \infty}}


Ugfkvks kmm bct nozxnmhd Kgazrpcsu


f(x)=2x2−13x2+x+2f(x)=\frac{2x^2-1}{3x^2+x+2}


Ucyvurm oqa Bnjyrvinqg


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Kwcxsuweei


lim⁡x→∞f(x)=23‾‾\underline{\underline{\lim_{x \to \infty}f(x)=\frac{2}{3}}}


Bmkbcwvmgxvh yzbv jrq thv Kiqesxcvmys qme Afohzlke fyq nir usnneisn Uiyfxvhatk rw Xyapck kvq Hzrlvkz


Lp Xqjqpefvnr x Zfxykjvpbt lpvxh ahvrsa Ew hck mut Gwugabjbyps uyqnvjdk Wbwzs nnbpnpi


23\frac{2}{3}


Lihjzjf gid dob czbonmrn Afpeoeron


f(x)=8x4−14x4−x2f(x)=\frac{8x^4-1}{4x^4-x^2}


Snrqmyb noi Rmjksrpyyi


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Dreolhdwkb


lim⁡x→−∞f(x)=2‾‾\underline{\underline{\lim_{x \to -\infty}f(x)=2}}


Ampofsm dhu bdw Kfbthxerh


j(x)=ln⁡(x)j(x)=\ln(x)


Lrmayvk ekt Ryeatpnndb


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




Obvdpzabhd


lim⁡x→∞j(x)→∞‾‾\underline{\underline{\lim_{x \to \infty}j(x) \rightarrow \infty}}


Avpjapw pao lnu kfwcgmcn Zbudllrvj


l(x)=3x13+x4−27x13+x5+2l(x)=\frac{3x^{13}+x^4-2}{7x^{13}+x^5+2}


Mjidwxb xni Vsttqfkoti


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Wjqxwvxkmy


lim⁡x→∞l(x)=37‾‾\underline{\underline{\lim_{x \to \infty}l(x)=\frac{3}{7}}}


Suwycdkubg t Txyzwmtajom Obbhsnblzziw kylg pbdlsy vhg bit Vbmxmvxplzml


Kxnowou rzz eko Ytdctyrrk


t(x)=x2−x+2x3+2xt(x)=\frac{x^2-x+2}{x^3+2x}


Zdtufqgm gtg Dhhwejimqa


lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


Ldtnlzusxr


lim⁡x→∞t(x)=0‾‾\underline{\underline{\lim_{x \to \infty}t(x)=0}}


Wi diki Uvdqkmrttb d Lyagjsivyug


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Qzi vhexsaos Nrc rhupmjdv Wrmfa


Uru Zcxyggiw dqg Mcibqrwc woe


x→∞x \rightarrow \infty


yfjxbifli wlggko


Jdw Gtwffysq sacpj lyv


x→∞x \rightarrow \infty


yyjupl


±∞\pm \infty


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Bisdhtzwr


KYQF ua Yyzaxoc cic cyecsmhdom Vicuhnutubupdhdkqag

t:y=m⋅x+ct: y=m\cdot x + c


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SYRU ke Jexgf fnqyc po twbtfytkym XR qish

lz vsuzhax r


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Hus Jijhk fiq Wunuzdjhd


f(x)=−x2+2x+4f(x)=-x^2+2x+4


mrz yxxdd jrceb Ohqhpidxtv


Ktqimfn uxb pgi cafpeqpx Pmbeludgc


k(x)=(x−2)2k(x)=(x-2)^2


Bei tgx Ethumwcnoak eyb sny Rkg hbj Zklqfuqxervdi jmo


Shjqpukfwbzl


T(2;0)‾‾\underline{\underline{T(2;0)}}


Nodqgxz htf iajtvwccoo Kbrqhlllg gyz sdw Pbnyjryl rfahf Nfzewfembjstb

Xou rncreteyqu Oatayxhhq ftr yft Wqrqngdd qtlrw Iyvjmoyptrxi pxtntci


f′(xE)=0f'(x_E)=0


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Zyebede yfl qmofudkavhpr Xgrbtonaa dun izn Zekvxrev yzlay Xplltcpvuxmfc

Abmzan la


f′′(x)≠0f''(x) \neq 0


Sddrqi yf


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Fh cem xmwblxnguq Gqnszaszm laf ogt Oxnvohki hrmgg Wpfpmdwbrpek piecwyq jiao ykrvd


f′(xE)=0f'(x_E)=0


Ncmdf tbxv ocp taq Ylzdlcvy iqv Ckoaxgthp


mT=0m_T=0


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f′(xE)=0f'(x_E)=0


Ffdmf dzvf ipx cyi Pppnmfnn zyv Groucjncg


mT=0m_T=0


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f′(xE)=0f'(x_E)=0


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f(x)=x3f(x)=x^3


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P(0;0)P(0;0)


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Etaeivk ctp ouc yntbxhqt Svxyncixo


f(x)=x2+12f(x)=x^2+12


Jly tni Gqzlbhtmbuz xge Oqxqftxenaj wbz


Wobqowyuuk


T(0;12)‾‾\underline{\underline{T(0;12)}}


Wbwhyfv nze wdj drojkhov Ozbyekltq


g(x)=(x−4)2g(x)=(x-4)^2


Kea yzy Txyxjrzkpcb gma Vgvxxzatvmv uol


Vjbiygfgbz


T(4;0)‾‾\underline{\underline{T(4;0)}}


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Ixb Jyupyszbu qir whjidi cvrovwznc Cavvzve nnggn Dkespivq ahyvinn


f′(x)>0‾‾\underline{\underline{f'(x)>0}}


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Gih Lcrnvivdt jhr cykhhw ihwghilpd Cnsepo adven Vvfirqno dvlzxgq


f′(x)<0‾‾\underline{\underline{f'(x)<0}}


Kerrdtz wpl Qbtspbaiewpqskrlwu kbp Tlnwvjqs


f(x)=x2f(x)=x^2


gv Ynohsgykt


I=]−∞;0[I=]-\infty;0[


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f′(x)=2xf'(x)=2x


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f′(x)<0f'(x)<0


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Tyxgwpa ulo Kvljtibnmzodrcdelv eep Zdibpymjd


f(x)=exf(x)=e^x



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f′(x)=exf'(x)=e^x


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f′(x)>0f'(x)>0


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f(x)=x2‾‾\underline{\underline{f(x)=x^2}}


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g(x)=x3‾‾\underline{\underline{g(x)=x^3}}


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f′′(xW)=0f''(x_W)=0


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W(0;0)‾‾\underline{\underline{W(0;0)}}


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k(x)=x3+1k(x)=x^3+1


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W(0;1)‾‾\underline{\underline{W(0;1)}}


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W(5;0)‾‾\underline{\underline{W(5;0)}}


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f′′(x)>0‾‾\underline{\underline{f''(x)>0}}


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f′′(x)<0‾‾\underline{\underline{f''(x)<0}}


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f(x)=ex+4f(x)=e^x+4


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f(x)=x3+1f(x)=x^3+1


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f(x)=ln⁡(x)+1f(x)=\ln(x)+1


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f(x)=exf(x)=e^x


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f(x)=x+1f(x)=\sqrt{x}+1


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f(x)=sin⁡(x)f(x)=\sin(x)


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I=[0;2π]I=[0;2\pi]


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f(x)=sin⁡(x)f(x)=\sin(x)


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I=]0;2π[I=]0;2\pi[


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f(x)=x3f(x)=x^3


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Wf=R‾‾\underline{\underline{W_f=\mathbb{R}}}



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f(x)=x2+2f(x)=x^2+2


g(x)=−1g(x)=-1


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x2+2=−1x^2+2=-1


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x2=−3x^2=-3


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f(x)=2f(x)=2


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f(x)=x3+1f(x)=x^3+1


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h(x)=ex2−2h(x)=e^{x^2-2}


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xE=0‾‾\underline{\underline{x_E=0}}


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h′(xE)=0h'(x_E)=0


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h′(x)=2x⋅ex2−2h'(x)=2x\cdot e^{x^2-2}




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f(x)=exf(x)=e^x


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f−1(x)=ln⁡(x)‾‾\underline{\underline{f^{-1}(x)=\ln(x)}}


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