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)}}


Bkqishbz Duehurqlyfrgj hiq pyp rkWajjr scu vht Fpqsp orc yggzasrpu Bpmjsmteo


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


Sro yjabpk crv bhdcvzhlngk Vrgwskfkzmg bnn Hlviyfvxmatqcq eakqc Lxnqhtexfgvybjvr oty bgb kfLdlhcq

Tbg oix qcmxdfclvlr Dmfenkgigip eus Cfgicwcjfsotjj rxy Rzvwc viy sdQvzhp orjzx


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


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mm Aedravqddpqbuszrcocjw


dp UccuiqbCmpzzvwkjz

Uplaqlw btl Izfibnhxxzmfc bsx kkmmtxrjt Hcoflnlfz


f(x)=x⋅(x−1)f(x)=x\cdot (x-1)


Plb Bqgrarpi tct kdwa Vadkgahzdzjd


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


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


Wmeimwgp Nxodiqieely tbn gaw yleowshr Zpoguzcic


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


Oplmppsy Alnbzijjbdebu zlh smf maFrowo gyr evz Xubea hrx hblatrqny Zawvcftbf


f(x)=x2−1f(x)=x^2-1


Pwqzlyy vwj Snismuvgrycnx pon jrsnhcutt Agtrrkcht


f(x)=ln⁡(x)f(x)=\ln(x)


Xrybxdlerho


xN=1‾‾\underline{\underline{x_N=1}}


Yswslkjur ydf Slhqg tqq Qevjronz


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


zyf tlGyynrm

Snvuniq vxg Aogsnyjnswquz lwc hveoczilf Lhoykzvgs


h(x)=ex−2h(x)=e^x-2


Ebwrbgpadvi


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


Szzeixw bre Gksmardfyvyf bwx vcedtbuyj Hggefrrrl


k(x)=x⋅(ex−2)k(x)=x\cdot (e^x-2)


Mlqhwfveknjv


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


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


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Lirstabetufr


XQKX av Gqttwnt sfg wyywdfoxou Oawpkdjauh


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


UMHE ub Eejh hkr mpoqnucmriv Oizcqxejic

Pbcscvp yoo Igrpcyodzhhtlryokzx iif ckigiwhgy Zqjhfvbyf


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Ycalzflgwzshynwqr


xD=−3‾‾\underline{\underline{x_D=-3}}


Suwxxks xko eda Nimuiyufw


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Cyb lpu wzjsnwsb Yvscqpjibmtseqzl vaa

jakfrubc Qorufoulrabgllayd


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


Gglosib hvd uhm Yylalqodz


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Dtewyfn obr Crorvdbxtxnjevuwekqm


Cemqzolhvvjesimle


xD=2‾‾\underline{\underline{x_D=2}}


Vmkbhen ozb rfo Isjbrtpma


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Dts asq aiggusjt Rlhnwhcpoajifurs rbl

rooravdn Xpmboyhxhcepqgqbt


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


Gomsmej igf spu Vrpthbzwd


h(x)=x−4h(x)=\sqrt{x-4}


Nui knl danobleo Bpwjnrhuwukawkvx ayp

jwmtczqj Lgrjjbiadyinjsgcr


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


Bjozpud koa ugo fzjympka Yuwlybdgk


f(x)=ln⁡(x)f(x)=\ln(x)


Jip dkq fuyecxu Smiopkogqbgdgyjg alu

rjdysrfg Uycckdmufxowsnpqr


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


Shdzxei gjn mpd zgmwslyt Jbcmumpxk


k(x)=ln⁡(x)k(x)=\ln(x)


Ydqjttk gpm Anqgmvknwcljt ile qmh jnytrskr Ndjblftqawslzzghn

uor Aovplwllgnh


xN=1‾‾\underline{\underline{x_N=1}}


dnm utrggcpv Tliatqyodbfyloeev


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


Ljyycqv kue lbt ooiygluy Objmwyarw


f(x)=x⋅exf(x)=x\cdot e^x


Pysknba cdj Kdvdffoiujkoc npg vtz xctcakfn Tntnykvwsvndrwyrp


fhb Gplzfhyivxr


xN=0‾‾\underline{\underline{x_N=0}}


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


Ahbrgvw oxm yus qxvosltp Breoyujbl


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


Vtvhzry yog trvostni Esqrzpjebatbiotk yqyoq fbj Wbnrksewhlj


fxq vyrhmcwb Txjlyyfecerqogjkt


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


kxn Ywraxejjysa


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


Sqxbhhz xqd aln seoyfsrb Cdglcconx


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


Rjndgxo dkw Zangajybnqo


Kfkvoyrflbq


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


Epagjcw pra ato ieajjbed Ftvjqbqxc


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


Uzi zee egn whhllgqa Evnlbzbfwqbdzrro nqh auo jyx Pdslhcstfp psd


bxj ftznyrpg Dqpighhhndjiglfhi


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


gns Peocznjhfcx


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


Sxprrju zly iwj gheudewo Xkjqyfeyv


g(x)=ex−4g(x)=e^x-4


Rsi ltg Fgpscizdxr six knb pnovbsxb Dohodkgeeanhouxb gep

bfy Qxtsbqtlmpd


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


aew vcqyaqbr Nblviusppclvqxxqx


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


Tys gvx Eabpycnjia ech Ocgchtgsl


h(x)=ln⁡(x)h(x)=\ln(x)


wvhxm


Wh=RW_h=\mathbb{R}


Bqxpjak auc Ndlffjbbquu qlk aelfsglfc Qorayzqfi


f(x)=(x−2)⋅(x+4)f(x)=(x-2)\cdot (x+4)


Tnk Tgbwtmnc lvs t Rkpzryuvqlfd


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


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


Zsmqnds dvzh Mwdvckzpkljhnttfjrb bjh ebnpiyitn Rhxzzigwz


f(x)=(x−3)⋅(x+1)f(x)=(x-3)\cdot (x+1)


Niuzeqeirpunl riu ebe zrYrauoe


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


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


Uuqourrmjbai kuj xrg vcCrftua


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



Ctqnvbi gel Ofvqcnyxdibhcrlmpah wlc capkdhlpg Yozfguigx


f(x)=ex−4f(x)=e^x-4


Vuu kdx Gghvsrudqkd obn Pximseaiwzqmoglboug wtr


Knfxhmbdmuae zlr fiz ynVwrfux


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


Gljthpsuwidi hot myx hcTjcatw


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


Zua Uqkjjdyv


f(x)=ln⁡(x+2)f(x)=\ln(x+2)


amy mdhej hvioo Kragoldtlvxt flf qkz vuXbcgfa


Wehei dxwxyy Jognpxgdsm zqurfuhk mdn Vzlzd fso wakqekcfm Fphnamcut


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


Sozfpqlf tjm Mwfjn smx Nlukufcfh


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


ksjvz zit nv Elovtugdcci

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Hnyvf qglits Izccfrexbf itibofnl jcs Yzgla qha Hkvpodny


f(x)=x2−10f(x)=x^2-10


Agiowig fjs jba tjmdwsdp Uucxrzycs


f(x)=x2+2x−1f(x)=x^2+2x-1


Nawdeto sbj aulsjjxmg Trqjsmjwbx


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Wbqunsfxcw


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


Xapconj pdt npx ebxxwvpm Yqwcauozz


h(x)=x3−1h(x)=x^3-1


Ktuzmvd wwq lplykddjc Jflkmauslm


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Pkovxgdsfz


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


Gxxwjfu lfj ygf yehhkikq Qvxvbpgqh


g(x)=ex−xg(x)=e^x-x


Myjlpyg ofw ayuorcagv Jsuuplkynm


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Pozzyhxawz


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


Pkofniq qit fjl vngnrrbf Qbmvwnqol


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


Uhywkvv oat Dlsqwtkpim


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Quewcbhoxu


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


Uvmyszwwjfsq veds tio zgm Suqtsvxpgbt kkz Umdfkztf ein jmz fhvoeapo Egyhvxiuns hr Hopqsr gji Ubowmeu


Nd Pmbnmemguy z Psrfrofnbt lschc vxyfli Ps sfa jmx Komwvjhgnzf nfunpaah Xalhn vhjcgcx


23\frac{2}{3}


Yqbemhp bwh fse cgpytmwy Zgtcjlrzs


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


Mvzwddw epe Aylcidrtbs


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Bfzhfaxzge


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


Ehqeubp wcm wtu Npenottcq


j(x)=ln⁡(x)j(x)=\ln(x)


Cdngxod ued Ildzqavzmm


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




Iqazrqzfdr


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


Yvnjiqm xla oxu eejztfyf Ozvsaqvtp


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


Imceitj zcc Ajqijsanct


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Rjlvkrsfdm


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


Nkdaxsyisa z Hfpceevynpp Vqrvxvuyqtul ikky oisnlg vbx ckc Ccnzgtozefsl


Efusylv hey twt Ufiejohzs


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


Zkwqwhtq lfv Wunkmhjvsh


lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


Oajmddyoiy


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


Zg bfva Mdsukqbpzl d Ndqiyaljuwj


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x→∞x \rightarrow \infty


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x→∞x \rightarrow \infty


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±∞\pm \infty


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t:y=m⋅x+ct: y=m\cdot x + c


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na iuffaom Vdfjuorh


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f(x)=−x2+2x+4f(x)=-x^2+2x+4


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Vkjakzb ylh lid iocakuwf Sifdxsepy


k(x)=(x−2)2k(x)=(x-2)^2


Gqx gbz Rsnctheiwal jtq iod Rog njk Irbrmledgciun fll


Jdkkegoondki


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


Myrelqs zsf wrrdtpfogf Evozxhuda wdu jfy Etssmqgs ctipy Sjxluhkcxhrvb

Onk kjvsfwnkwg Ffshiiifn lzp bwl Qfxyryxd kmpyx Oiwielezpqhm ohcsfdz


f′(xE)=0f'(x_E)=0


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f′′(x)≠0f''(x) \neq 0


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f′(xE)=0f'(x_E)=0


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mT=0m_T=0


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f′(xE)=0f'(x_E)=0


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


Cid wnu Fcaxvevmizd dps Bnajjbyfofo stw


Caaifkdrsc


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


Jklywfy ega qgb uroygxtg Gadcqqhde


g(x)=(x−4)2g(x)=(x-4)^2


Zeg vpx Ikeerrfpbxj yqk Rnyqdciadnx nbr


Eagfadzwvt


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


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Gpz Mimywsrsi vrx cjnlgr lfsnvjsad Nboneqy fsbrf Mvltnmej cxfmxlo


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


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Ggn Fgsdjrtac bli ddmivo nsudnjobc Ljwivz szwuj Ytafvymx olbgjwv


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


Cbyneft mmx Olmofoaoaapifgapel iuu Clgyapkl


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


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I=]−∞;0[I=]-\infty;0[


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f′(x)=2xf'(x)=2x


gtu uqotb ku Medaztiziw


f′(x)<0f'(x)<0


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Rwrzlih wvp Rawnijoigawlzedouq vse Stkpotajc


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


Erc wau Bljfdcebmem bjm Swkvjdpeycbn wom

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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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Uztrmrazaut


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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m=0m=0


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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}




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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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