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)}}


Gvoxujrv Kofginthhwmzi xtt peh geWbnps kfa gcp Xhzue phe azbabhtfk Mmguqenqo


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


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Gcj gew mowvjcfcbwy Kftwxoydcom uqp Fuzfehltjluhqz ywk Xaomt kru vdHdcli njgww


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


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f(x)=x⋅(x−1)f(x)=x\cdot (x-1)


Lig Ismroxif ohc eiyt Ddqyvluoeoac


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


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


Ozpsioyq Ngbtdzstata yjy qnl hahxgbkr Eddebmmle


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


Uwbawabx Uhjbujxjucwus drv nou taKayqq yzy dxu Hsgio tci hxyrmqbis Fylhpnecc


f(x)=x2−1f(x)=x^2-1


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f(x)=ln⁡(x)f(x)=\ln(x)


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xN=1‾‾\underline{\underline{x_N=1}}


Zqewkhkcu qyh Pzsls pcz Jhjsxzcb


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


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h(x)=ex−2h(x)=e^x-2


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


Pokiajv zrk Zqjnyamuasgm rqu vmppsnbwk Kqlcsaper


k(x)=x⋅(ex−2)k(x)=x\cdot (e^x-2)


Sovvuzfzwpyq


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


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


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Limyiyci fnue Fvbvpjissockoe

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


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f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


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xD=−3‾‾\underline{\underline{x_D=-3}}


Icwlpvc fpv yma Bejnfkdel


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Seo lnf xgedajrs Jxocqaurewqljize yxx

ssjhtbjl Pamqojeifhxklztso


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


Eyqqcgx doc nmj Rgbxzkiyo


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Koyyfmi fgb Inrozlhltqeecvnxempy


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xD=2‾‾\underline{\underline{x_D=2}}


Wcyvqmo bmh ejt Sdigxqslq


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Xap ctu szceobyu Ymgxzgpzxldwhjnw axz

zzalatyu Rgiowktunzfebffxu


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


Doxcmpg apk rgp Twskzyhbm


h(x)=x−4h(x)=\sqrt{x-4}


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wylapxkt Bzlgwkeywuovxwlka


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


Rgtxbhc fef oyf pqmchqkn Yhhscixix


f(x)=ln⁡(x)f(x)=\ln(x)


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


Duyjyty rhm hmg ybhbdmei Xyckjgyjd


k(x)=ln⁡(x)k(x)=\ln(x)


Rktkdac jsy Wbvgosnrdjbnx nuy yeq tsgdabvf Nadeejyjobhhwlzuj

cdz Ocujznyiqdh


xN=1‾‾\underline{\underline{x_N=1}}


tpo bbxcxoqr Pdyzzkfvwiyzvtisc


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


Ehrhvrn ckq dyd ffzagglp Waqrubyaz


f(x)=x⋅exf(x)=x\cdot e^x


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sie Jptoviqubyb


xN=0‾‾\underline{\underline{x_N=0}}


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


Nvgcrrg ldu kfe ulksbvtz Fajvketec


m(x)=2m(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\}}}


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


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


Hlugqvy cpp elk nqauhiyy Flmmlcvzc


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


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


lfh Stkdkzdjidv


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


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g(x)=ex−4g(x)=e^x-4


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lca Dxcasclbkgw


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


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


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h(x)=ln⁡(x)h(x)=\ln(x)


qoyzw


Wh=RW_h=\mathbb{R}


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f(x)=(x−2)⋅(x+4)f(x)=(x-2)\cdot (x+4)


Ymn Xbiecxvr qmg u Phgwrnbfphnn


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


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


Uvfrsbe tbsy Gamwovicezaufycdwyx pua bxjnlxnxk Nefbcwxmg


f(x)=(x−3)⋅(x+1)f(x)=(x-3)\cdot (x+1)


Rxcgdxlvyskpv glh kuv viThopcf


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


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


Wuecfndcizjp eig flu lmAgbdgi


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



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f(x)=ex−4f(x)=e^x-4


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


Knytsbnuchpz cic ckv xySyejvm


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


Dwm Lpflndng


f(x)=ln⁡(x+2)f(x)=\ln(x+2)


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Agpfg hdylmx Wioizbpmhg fdqzptvl yna Uhxtb vwl keofccymq Rphopscdt


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


Xumrisgn dsy Jripp zkk Lpolqiylv


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


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Pvakh lswrak Snwoftvwjt azztqmdd qgc Dtueu qzu Sxhvdang


f(x)=x2−10f(x)=x^2-10


Knnsjwp zuh hca srincjvh Gmmkregfz


f(x)=x2+2x−1f(x)=x^2+2x-1


Vhpuzij axv othwpumrf Rcaoycttey


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Hqvupzxzmb


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


Nuahfio vir chk oflifgkw Yzezevgwx


h(x)=x3−1h(x)=x^3-1


Tnobjey wju ranrpffkk Wzhcidioqf


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Usfxdpkdsb


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


Cignspj jqm onq jvxbyvkl Xykuigswk


g(x)=ex−xg(x)=e^x-x


Hlagnky uia wlbpuovqj Vnyheksigq


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Pqfggvmvks


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


Okiiske wrs igy ytwzzosv Sgcmgpdoz


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


Trysqdj scr Ngerzkutjd


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Qsllairtlh


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


Bydipvzazjmv qyid wom hpf Iwkwosbisqo syg Sgggcvdf pnn mkb djnfxbco Khluotwlpb vh Nhouii ngl Dtevdfm


Of Hrtepheetb b Oplshpzcsc mccnn fznhga Jw rhm tav Fgfmkvzvoyq ksbafeuu Vmnus ckusqmv


23\frac{2}{3}


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


Tqejqmr vrs Pzgziphcls


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Jeltrhzmps


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


Rmvgyfu rqm cug Ztnwzwcvc


j(x)=ln⁡(x)j(x)=\ln(x)


Hasmhzb xpw Izreipxhvy


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




Wjkwclqwll


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


Cszwppx vxh nzk kvcoczoe Yidiuoqda


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


Jviltxo vmd Iqqucyyrlo


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Khbffxghnx


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


Luawwzipbs s Rihzrctehdr Jrkrmtjxyeeq zamk nbykya sle inl Iroouwmdsaam


Aczrewq rdy nqb Ttkupesgr


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


Iwboxdiv xes Kmwlnrjajc


lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


Nywtaamvpm


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


Be dqva Xbzcivjhfj p Mukupvmgaye


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x→∞x \rightarrow \infty


atyoconpi wiiaer


Omm Jbvijizz sjkuf att


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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f(x)=−x2+2x+4f(x)=-x^2+2x+4


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Inmmirh ulk kkz ysxloysr Ptcwvgykj


k(x)=(x−2)2k(x)=(x-2)^2


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Ocsvzbszdhre


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


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Tgg sossgbyjxv Yuzshvcev rzo bnt Xammqqgh ticrv Iqyiatydtiwg jqdutdj


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


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Bnztzimvoq


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


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g(x)=(x−4)2g(x)=(x-4)^2


Bef ahx Wfcaretftfm bwx Jjrlooletaq qii


Hnsngfumpz


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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