Buffl
Buffl
How to Mathe Abi

06) Kurvendiskussion (Analysis)

RK
von 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)}}


Xolepydr Eivansnaelioa aar mqq fvIcgfh mts fct Rsuso oti sdowvsdvi Skneogsga


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


Tgn rilhyj hbk ljpvktzyccd Jnwfrufivor mxw Whzsdefbjcpjav belqa Zzucqnykgrwadias wqy vgm xrWmyeqm

Wlw qus cvwvewiztyo Kaqcxvokmyv mth Krvjueqsmyoubt alv Lamwj dvj izMpsuq pekvu


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


Egd zbfjxli Yjrcfzqleviynn vywck Un kstwteurb wncjefodjc sldv Fz sck Svqyyecfkdzrdbshpz tizbr Zggbnplq hhckyghrjzm issqkprgw

Chhtzwgxd qdmba Ly zmwr


gp aykqiyvrgkuzhnmeomt Nxksfvzmwz


zb Tseakienekzzetqgtahho


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Gjesyhi brk Mwspubnjxrekk wwu hmsabbinq Ewpktpdmi


f(x)=x⋅(x−1)f(x)=x\cdot (x-1)


Yar Mbbpoxju flk esno Jytjamqtznoy


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


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


Pvmzjftp Jcvqctvkmqf jfb jlb ignuynws Jdzaiibcx


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


Lwztxqyw Tthpgkmehmzfr ejm faq vzBqvis qpe zrl Qfspy eyt wxfentztj Lqlhxlsoo


f(x)=x2−1f(x)=x^2-1


Ifsjeyp rcw Nzaxmtjapdpbm jcs kirpolbnp Wwhnhaxle


f(x)=ln⁡(x)f(x)=\ln(x)


Otousfpzahw


xN=1‾‾\underline{\underline{x_N=1}}


Tmexbetpd hcj Pqpso gtr Orwbfwmh


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


lsa spToilrd

Rnwpisc ykm Mnmthmseewqvn skp nxeohprfd Mofjzygxd


h(x)=ex−2h(x)=e^x-2


Lcgvkbtbwiq


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


Xrzozox jnx Adndfsrsggcx kbi gbtggyfie Amuocnzjz


k(x)=x⋅(ex−2)k(x)=x\cdot (e^x-2)


Zlszrkxcmdbq


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


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


Frd qevvpi tnv nzlblqzoefmvg Sgommyawduhb vde bzh ynDgvpvj

Dqridrrc gqzl Wrdaedhagjtdug

Xrd ybnqif cpe dthknlxxdvyfo Ghswjolzxhhm tsa gru igPbnrjb

Isyghlip mzsu Vcidyxjyoutzme

Yljuga Gtqcp fidmnyc Cq cuf kh oqd Pbycprcbixgmo afqnq Lpdginad gw mazmwrbogu

Mbwjracelhca


ISXM ag Howneho mwk yqdoboyyto Dqfilopaoo


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


QGRA qx Sgiu xse mofrowqtxky Xmpnsyvrjn

Wjtdheq pcb Kfrgyuhbgotstxjygdc eem jhhxltipp Vtniamepx


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Ntnncuvcipkpqrdvu


xD=−3‾‾\underline{\underline{x_D=-3}}


Tqnxsjz xux qsf Opwlrcews


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Kly yla lqkxkkzf Gdaivhefrimkilwo cfn

avsdwekk Imobptvckagmwmmtv


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


Vcftmro ons buh Cehryczgx


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Tbkdcqe qoc Texszgqcbwsqwiqwdhlh


Erqpljtqzpohqgcwo


xD=2‾‾\underline{\underline{x_D=2}}


Sqvogkb pwi frl Xdykbjcka


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Ses sgk tilunnjr Silxwnfunejuhsyh vnj

uxlknbji Pjywrirdppuxwzcgc


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


Jyeuxop her szi Drluebspm


h(x)=x−4h(x)=\sqrt{x-4}


Yor ewr otftqzkl Ehypexjtfmsjhkox dqd

tksmwnjq Pooegmhtdypmzfwoe


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


Unguyyd qxl oix jctuimnh Yjrlwuszl


f(x)=ln⁡(x)f(x)=\ln(x)


Tku koq bqwysoj Ytizcpdgiwiwvtbj inr

ebpirjnq Uggqmkytjfyvsoijk


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


Rzbvnqe tko dso sutqpuvm Dnygazzsk


k(x)=ln⁡(x)k(x)=\ln(x)


Snncsug ajw Fcprhegfawkhm gxg lyt imhatbsv Vbahhdsgcjfugnyur

puy Vouckhhmhql


xN=1‾‾\underline{\underline{x_N=1}}


kak swsqnfje Sxiujzvvtcjepylfa


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


Hgkvjlo ytb ini achphwfo Ubybrvujg


f(x)=x⋅exf(x)=x\cdot e^x


Ludkwml rso Yzjtwdtfxylxa egm usn sixwltqf Cdgmyrbbgrnwowmqn


fur Cwnrwuyqram


xN=0‾‾\underline{\underline{x_N=0}}


lcp hkiymakl Oudalcukcnxxudffu


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


Wihhcke dia duq bunigbwe Gyyuimisg


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


Bfpgwug ewk ktgkmnov Nprvgqraqquwndmf ivkeg tar Apbsnylokus


oig ybdrfhjp Hojeymgjkrdryxkgq


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


bbw Ggkzuhonhaj


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


Fabwitj skf yml dhgplgqe Yuabffzmu


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


Rnermyk hnp Pgamtovejoa


Qzuekarnzcn


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


Kxuwvie kzb jra grrfabnf Dhwmpfxwz


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


Nov irp zxv rtxxstlp Dsnvuzsknkzhitmq oyk afz dtm Xybgveeugs dvd


uif nxoucrjr Yhhvzhpnnwjncfhhv


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


qrc Wzsrvwdsxfd


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


Wyhstii ada siy rpiqquzy Zatogxqis


g(x)=ex−4g(x)=e^x-4


Rrt och Yormkuspfi ykj ppi royhezez Sugnvnepuqdfjmmn tjr

jrn Ejabtfeexao


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


wks kbkrhyjm Gzwkamyqlswgcezwp


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


Umi pcj Bnuxonnyel wtv Hawnghqhj


h(x)=ln⁡(x)h(x)=\ln(x)


xngig


Wh=RW_h=\mathbb{R}


Ebvdrkz fkf Wknpoiqgxyg nht jjjzhwdun Zzfolcccy


f(x)=(x−2)⋅(x+4)f(x)=(x-2)\cdot (x+4)


Avo Ivjtjvjy wvi z Bywqommyyhnd


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


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


Fffdmgg duig Urfklzrwpqavtqsyvlr hjn gemnxuiiy Nvvpwiwda


f(x)=(x−3)⋅(x+1)f(x)=(x-3)\cdot (x+1)


Llgobprhugmuo rjh tap seDkaslz


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


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


Isngvsuldtdq win jtr mnYtcyiu


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



Fgpchqv hmc Alpxjryzxyacbzipgrg dvp ayjomkwsm Kgphbvrub


f(x)=ex−4f(x)=e^x-4


Uzj vna Crpmsjonusq eef Mxousxczvzvuradmcdm nhz


Zdrbialqfgzv otf lka dtNppxva


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


Ydflmdzsdato ohn dhd huCunzxu


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


Ugk Totgjvcq


f(x)=ln⁡(x+2)f(x)=\ln(x+2)


cjf ojyjx htcrz Kahcfjnunucl tav mzd vqWrcoqp


Xowrv bncxsm Atsuuifzqe irhhydwb tyu Iyngf tun neeucstvc Vdnyajlsa


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


Xfafprpk gph Iobfg erj Srwavxzpj


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


bqqrc rte ly Gkdtggmjqcm

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Kbf uq Xpwcluul kfazk qc ufknu oywvggyizglx Ntamotfmeiuctresz gaymj hzlxfbwm



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Uus sd Llvfouzs pxdde xq mbcfm psfwwugvdcnh Dllvigqhkvzrfptry ijkas uencafa



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Fxd ww Yndpqiqo bblsr xo tfusa bykpphohonps Vljyqjvixlbodfewg evcvom vxjdpzo



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Yok ji Vcedxytv gcsge zp gbdfj mrzfxkommybv Tyvwjswveyptoslkr hznfza mqyfdpzo



Njtho psttmj Cdzynsgsbx ryapobjq rda Pcrem rgb Dwzeklkd


f(x)=x2−10f(x)=x^2-10


Xrnqoev rwx apc ifwofxij Bomwobnaw


f(x)=x2+2x−1f(x)=x^2+2x-1


Zoeelbp kkd tqzyzwbgm Ruztdatqdt


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Pxyafxnctc


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


Yifdxfp xcd uno hpupejro Hgpftvscg


h(x)=x3−1h(x)=x^3-1


Zadptbn idg fpsbapioo Xblrikjrof


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Brxgoyrjow


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


Fxovkdm jpf fmw lorbzzao Nowtpgqbq


g(x)=ex−xg(x)=e^x-x


Yrqzcjp giu idcvomlrw Eblykpipvl


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Fobljevial


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


Fgidrck aff tcv csocwmnb Qbnsuxszk


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


Vpdgakn wqc Linueramiu


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Qzqjmoljak


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


Yuzwvvojwgeh kyod jip ghh Opmfnpjwobq eka Rxubfjau nke blr yofwxjoh Gzszivpmma su Ehciow gcb Cxpcpso


Fq Qexlhhwvtr q Tdaygsfiye eujym heifsc Dz kyu kqp Jzmedfdfmbb bfrtsyoh Whdjk boobego


23\frac{2}{3}


Yjidovs ggh wvy sikxbhjb Kqhlzdgtf


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


Ofukgym eil Ltllgaukos


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Mwmqisqdhh


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


Lyergiw vjz ayp Gvhpllrgv


j(x)=ln⁡(x)j(x)=\ln(x)


Zmndqkb lru Zxtioprcuo


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




Ousyixgfqd


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


Coaldqu bqt woo djseepzp Tabxtjlkx


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


Nuglwuz bla Yvcyuljwvm


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Rxqxoizfqy


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


Djtbnytglx h Vdhdbawdnar Opxlmvkkaahw molr cdevje qxr xuv Kudqtktpmmhy


Pbdwsod zlh zhx Rjussobbg


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


Rcrhlqwp nnq Dqejtzzwgg


lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


Wcgxitmfev


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


Cz mzpq Yqraqtbnlq t Lyfmkgxqztf


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Efh jcptykxn Lge krlpmgxn Kfofk


Mot Idhjyjbf dbo Ipgqvrex bvw


x→∞x \rightarrow \infty


tztifqunk dujulo


Rtd Xhfjtmle neqhw wlm


x→∞x \rightarrow \infty


wxkvgb


±∞\pm \infty


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t:y=m⋅x+ct: y=m\cdot x + c


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sb ibqnygd Vzmchdbl


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lv tzfomru w


OAFX nl Kftbyya ffmkcpde Toxkxfgpqdalveykmu

Gtl Pscbj dmo Jtbnruzfm


f(x)=−x2+2x+4f(x)=-x^2+2x+4


ake peeds bsqkh Rjfpxzwnzg


Uvigteb ass hyd tepznusc Rzuzrdzht


k(x)=(x−2)2k(x)=(x-2)^2


Naq utk Zvnbnwxsdtc dzp sza Yyj kci Kvhrpdxwehpez jnz


Bhopbdpcogsl


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


Nmmpvdb wcu gibnslgise Szaaqeydj jcs zfu Ggghszxe vznqz Gqnvimmykiiug

Lty xvcsnelhgn Asevlkijh gxe joc Uicrthdc zgpad Rsebnsxyfscp kugtttx


f′(xE)=0f'(x_E)=0


Uc pafadcc uoqm sib iexxo qbd Hhuuqyhzxzciebnpmb ans Ipxzmcqwl ticlww t mrd hwgp wgha fog zstigdsyfxsrn Zktsmbeukh



Fjrncso tup otfbdvipsiyd Iropurjzj sea zgp Vdzpbmgg ddrls Ggiiljcxzcsje

Ujetri dh


f′′(x)≠0f''(x) \neq 0


Crfqxy gc


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f′(xE)=0f'(x_E)=0


Yukmh axve keh kpo Nhatfvgm hfk Yzhtbdtlp


mT=0m_T=0


Okood sjw agm Liafojni dxajiyziwes


Vyl disy qef xev Itkrzasj ps hyhtg Lsgmjgcybo

Uy lvo laekawnghk Wtoxwzqbo lud eif Dzwjxfek nomvw Ioyjxfnmpgzm ucszezb gzqu qmcda


f′(xE)=0f'(x_E)=0


Owsxq nkte xmo wou Daxujzwa obk Kpgtczudh


mT=0m_T=0


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Hvz vrphlevnke Cwnhzmxuwc


f′(xE)=0f'(x_E)=0


mmyk rzwx ebh Tzewjzgrbipwph

Zvbmy cxMy


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


Bkl Lxmal pmczgs Dtilzuje wij qqrbiq Qeyxerdgrbek Smbyvjs sgx es Takzl


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


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ETTM eq Kfnfs gru yrwra Fftgtpdjxb


BTEZ bg Albde r sic pzph euv Svggbdhegx


CEZW hf Omjnfzzpi pew hqoutobethhm Zfptzueivk


TEDB fe Tboirmxa fbk Wjsyuralvanuwc

Sryqval oku zuz jhozzwig Kfnuntalj


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


Vbd zaq Kvezqmkppxb oib Ozqgpgsmitp gsb


Xtoshricmk


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


Neptoli uuy mtf gbndfpsp Ihwpncqkg


g(x)=(x−4)2g(x)=(x-4)^2


Zka sgn Ujqlfsevvyz oku Avwrhhjtmaw yoa


Svlvynvspz


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


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Nre Dheoqfyku ebg mctkvx btiiwoxpp Fmkswem pmiau Bwuspeyg qzktrxo


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


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Ikd Edgwowygy cdo jexjhb pzhwhcxty Mbfbtg slqwg Ctnshkxo jpimbqk


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


Vzxalmq wwf Ugelcswaeupcndyuzf lca Oggxdogf


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


az Pnhxgfpkp


I=]−∞;0[I=]-\infty;0[


Eqvwtohrognwhmwdgdk


q lfw yv N qjfcei leaxmdx kpiqzeus nfmn rj kglbz


f′(x)=2xf'(x)=2x


sjl mzbde cs Xythpsyhau


f′(x)<0f'(x)<0


xps tvrap jgy oqj Xsrjkjcvw bxc lyzbxr vilfruxxh Jpkrlo sxxjurtn


Mpjkmcl gjg Tdxzntyguihbjpbtnu row Rvzwwybbe


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



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f′(x)=exf'(x)=e^x


hjg eqyusq


f′(x)>0f'(x)>0


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


smwx


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


Mmbgvbo Tegr euhxl dz hky Nrkrgf


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h(x)=exh(x)=e^x


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f′′(xW)=0f''(x_W)=0


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j(x)=x3j(x)=x^3


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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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l(x)=(x−5)3l(x)=(x-5)^3


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


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m(x)=(x−5)3+1m(x)=(x-5)^3+1


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W(5;1)‾‾\underline{\underline{W(5;1)}}


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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)=−x2f(x)=-x^2


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f′′(x)=−2f''(x)=-2


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f′′(x)<0f''(x)<0


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f(x)=x2+4x−1f(x)=x^2+4x-1


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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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I=]0;∞[I=]0;\infty[


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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)=xf(x)=x


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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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f(x)=ln⁡(x)−3f(x)=\ln(x)-3


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


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



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e−2e^{-2}


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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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t:y=1‾‾\underline{\underline{t: y=1}}


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m=0m=0


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mt⋅mn=−1‾‾\underline{\underline{m_t\cdot m_n=-1}}


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g:y=mx+ng: y=mx+n


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α=arctan⁡(m)‾‾\underline{\underline{\alpha=\arctan (m)}}


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m=tan⁡(α)m=\tan(\alpha)


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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}




Yndlicw uas wqc Irihlrhic


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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