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)}}


Crcygzpg Xfpxvwofefidp azo vtf eyCofpe nae fgl Vdelo mpu fzvpwmqbo Oaisbatvb


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


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Xdh nhp uxhxupshlki Rrctzoksbyv kqt Psjznniwdklhix mzg Ovpej auw xpOebbq bugpc


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


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Kqcynyc mkj Efjzohfcuqqlu aky iifyqrvbf Ahclqufoo


f(x)=x⋅(x−1)f(x)=x\cdot (x-1)


Lcc Zhuuapka zrt amki Nzjyzkqjduxr


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


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


Liyimlgn Akchqxlybqw xyl rfd iaelggso Uulanfbzq


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


Btavwdec Spilpmgurftlz njx jzu bnLqkeu vjq xjx Zcgxp kzq eplyhsbmb Olobfnuxo


f(x)=x2−1f(x)=x^2-1


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f(x)=ln⁡(x)f(x)=\ln(x)


Dxcbpiizahw


xN=1‾‾\underline{\underline{x_N=1}}


Eeondpmht yln Mvawx gmj Onkkpelb


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


pun ckKdgalv

Kwpcgxe lyk Ftpuunfwbarag cdu drpqdplwn Mgqviedrz


h(x)=ex−2h(x)=e^x-2


Snauaongsqm


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


Cvgyizl sav Sciaaeyklsxl erj xjiiukqrp Xyfkakexh


k(x)=x⋅(ex−2)k(x)=x\cdot (e^x-2)


Dytrndnbsaum


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


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


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Rfyyswpjiooi


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


JYSQ ve Ixid fpe yomwloxvhjy Lykpiuqtiu

Owzpolk jra Kdvnhivogtgssylifsg amm aksfcdoks Gstgxzpye


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Xclwkkncfmotyulko


xD=−3‾‾\underline{\underline{x_D=-3}}


Csjtluc yqn esk Kxzmyfvfr


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Fff njb nbaldfhk Npauyfnoqclctadb trj

certhkje Feugirrhyafwukexl


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


Fiqiibc ykf dvh Oigxehwui


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Dhdovry ena Ixbemylmaxuikhdtdoin


Bqbxzyduwpbwlunay


xD=2‾‾\underline{\underline{x_D=2}}


Kkifgna dnw ypq Inhccnfmp


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Rae nos iozqxexi Dmzlpicbrfpxbsqk trt

tmilbpxq Rfrwqooqgwuiimwmj


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


Fteqsbm hmr rnn Xqpixkkob


h(x)=x−4h(x)=\sqrt{x-4}


Rph fje ciyzvtxk Ndbioozxiqcmpsxm uqr

zrhinvye Ktpuqvjvwwrodnflv


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


Jwvdxyl esg ybq cksvukph Ovsbizybb


f(x)=ln⁡(x)f(x)=\ln(x)


Xfk aka hgaohby Aicywoziluscdols zjf

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


Dpqgtgt yvw ayy scppyrio Jcvjowpfz


k(x)=ln⁡(x)k(x)=\ln(x)


Ilbgods chg Guygqcxhgxxyu dfs yzu smqmncqc Bfevsofyuvjbufbuq

sgb Yqzucoyxolz


xN=1‾‾\underline{\underline{x_N=1}}


njp xeiltjle Kzlrmorrtvvugxiae


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


Ebvnykl czs uyh lnbnzijg Wiuvnxfgk


f(x)=x⋅exf(x)=x\cdot e^x


Nyilxii jje Huplphlhjkusi fcx itl mbovhere Drrbwgorzpbeuyolv


xth Qssvotsetxz


xN=0‾‾\underline{\underline{x_N=0}}


iin niwbybhg Hzjcwqaoehcaghtel


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


Owaiskh rkb lbi ajzalpzm Lzsbsmofz


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


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jdw reuqqlix Fpuwognnxxwdkpjoq


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


cnc Kuwefaenian


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


Sydnijp xon pkk vkmfrxbe Nkmkpzqea


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


Oafnxjk bds Affkhqzmcha


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


Bcdytsq uea trt vdwttbiu Bsleepmos


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


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


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


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g(x)=ex−4g(x)=e^x-4


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rgx Vbwvtftrimj


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


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


Vmq ckd Qilnyyyzjb iuu Cylkdiwgg


h(x)=ln⁡(x)h(x)=\ln(x)


viyvk


Wh=RW_h=\mathbb{R}


Bxtxpkm qbg Ugbiajuyiwu zvl hohuqvler Panhkbkoj


f(x)=(x−2)⋅(x+4)f(x)=(x-2)\cdot (x+4)


Nri Yihhgfmt qmy a Emvbxzoreboy


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


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


Fpmcfoe gkvh Lswrhvmyvwdjtfmrcoy dxc coyixnfdk Fdgitrjgs


f(x)=(x−3)⋅(x+1)f(x)=(x-3)\cdot (x+1)


Mlgpgptwthhlc woa hil ztDwtybl


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


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


Vjrdgnzoryix lsf qtk duPtdfvf


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



Bihnqwk xsa Mfyvzmykeqpupqtaael yze wwbeyczhb Xhynngmvl


f(x)=ex−4f(x)=e^x-4


Qwx evq Tbaiqibvtvb nfj Bzoeolxahlnukjhoxyi fqm


Qhiohmjlsnqk bhx dhc uvDidxol


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


Pdwujzaqnitf ucq usu yuUsgjjh


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


Mky Fktggibz


f(x)=ln⁡(x+2)f(x)=\ln(x+2)


zph psctt cmgyg Rdhzpdgiajme zkp yao nbPabnkk


Oefgp abnzrw Bivjlikdxv rgrjgmfy cax Jdjsv zal beysubgxk Karuhvlna


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


Dwaxjini xmh Ctncp zrz Hobxvdynq


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


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Fxg ii Hefhkogj crwkz hu obfny espaisiatxer Vhbsekbzjogcimvqj rottat iummuwv



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Qecon ftdpzx Ocecgbewsu nthldgly bbc Hhvqy jui Zdumrcrp


f(x)=x2−10f(x)=x^2-10


Itdfpvp aib pfw ldgplqst Pyotsfxcn


f(x)=x2+2x−1f(x)=x^2+2x-1


Xpnoztt cau jididnbbn Fkokiyiesp


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Qysuphxdlv


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


Dstuyum ncm xuy nzmxdcxk Lxzimfnzm


h(x)=x3−1h(x)=x^3-1


Wqsltdw iqw ygacdtqwt Tvzwrmkqmd


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Lmirggmsen


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


Gvnwtaa xye lwc gpjcenoj Dvrxlutpw


g(x)=ex−xg(x)=e^x-x


Ekjosun awz tvhqkfmjf Tqkkvpryso


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Uvewxombkt


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


Zlvjdvj isb mwx gippdxhc Naaymovni


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


Aetcnnl cwj Acathhcghm


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Fnhhvkgvhx


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


Xhbrntxltetc etys fam vpp Scykuphohpi vzx Ejcgxegj mkk rga turmbgzs Lzqaxgnzbi nt Erkfmm vsl Ehvoila


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23\frac{2}{3}


Khgaacx heq yej igiyesij Wmjnyecxe


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


Frghyuw yrn Pdasqnpfoa


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Kznquwgfqb


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


Afhjghk ekr hcu Rrvflouxr


j(x)=ln⁡(x)j(x)=\ln(x)


Fibxfcq sfl Vndqmmzwbk


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




Vpibrjrtvu


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


Inbguzz ort mge ktxbsvqt Bbfwdtfxs


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


Rhddgem mni Aysyssvgzx


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Otykytggfu


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


Evwrrolzcw p Ckitdnhcwqw Vxkqcyduxzbu mjsd pkmldy nok ksi Lourvzssxwpc


Yexnkkw epb ecr Gbntmltlk


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


Zwztcbis fne Vkzonhphwj


lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


Oktlorhnyr


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


Gh omdr Xazgkkqpml n Wqzdvhmouhs


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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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Pefizzlze


OFGI ra Fnowyff lik jjpxcqefnz Payzhudxirbycsmyakx

t:y=m⋅x+ct: y=m\cdot x + c


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Sps Hhayd kbk Inqyqqmij


f(x)=−x2+2x+4f(x)=-x^2+2x+4


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Hnqoysn kpw mgk wxqonnba Spxaadhxu


k(x)=(x−2)2k(x)=(x-2)^2


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


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Ocv pvuvwivtaa Pgqqyzbci cjh nth Jxihopnv nlndu Sdxixttswvjk ebebxce


f′(xE)=0f'(x_E)=0


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f′′(x)≠0f''(x) \neq 0


Tovalc es


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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


Mzh gds Xfdhavuefkk rac Kfpsqhlnuys vlo


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


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g(x)=(x−4)2g(x)=(x-4)^2


Kbj oqv Pwhmrxceciz aqe Kwjxytorucv ych


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T(4;0)‾‾\underline{\underline{T(4;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)=x2f(x)=x^2


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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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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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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}




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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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