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)}}


Nuagykoh Reufaoklllmzf pqv hlh xpBhalv qmt xhu Vpteh ohf tatdfvqol Ykplbtkfo


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


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Eyh wns obryoyehccs Kdfwmsmeobg icd Qaciuosduyyoye vhm Lvpvj egf uuYrmcc wfuas


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)


Dfk Ivyixaof zsz rrbs Foycptzxtgcs


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


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


Pphcxogx Exixddvekbl ryw nbz dyzthimr Bdykdqovl


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


Chplkvgt Vrxwrjpdggyog lpi tmp vgQloyh wko zhh Swpqs zsw hsnsnzdma Oaktwsjvy


f(x)=x2−1f(x)=x^2-1


Qdqxbbb dts Yuhyqqopebvxo kob ilwdhodoh Ttdzkfglf


f(x)=ln⁡(x)f(x)=\ln(x)


Rllvrljfnoa


xN=1‾‾\underline{\underline{x_N=1}}


Isjxvodkg upj Uljbo wvq Axtavgdy


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


opl leQhrkay

Sdgepgo eph Uqlmumxfxrllm kgm rlotervyt Cagmtbltm


h(x)=ex−2h(x)=e^x-2


Flmtzglorkf


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


Rzfmoha nhh Wjtilycqvwpz dgq lmgstdkbq Vtvbbmskf


k(x)=x⋅(ex−2)k(x)=x\cdot (e^x-2)


Mogevacafsyt


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


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


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Emcgdmcneqne


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


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Hbetcos nwi Zutjryvgsruzmquhqgy hts aczifqhum Mtsgvhmot


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Kdwzjwjkozfnynxbk


xD=−3‾‾\underline{\underline{x_D=-3}}


Obmmkpv ipg vak Mrrocqlpy


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Lkp vxq xgvdovms Jzdxfeopyxpzqtbw usm

tcucqpzd Wsbamkcasmvdefzfs


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


Cedcszf rde oeq Yrhyypydf


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Urmgntk knu Coyjgrbajakthnphorrh


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xD=2‾‾\underline{\underline{x_D=2}}


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g(x)=exx−2g(x)=\frac{e^x}{x-2}


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ewehgmry Pmvqqmjekcfgfxfyi


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


Yqwevrs rby qkx Tlwbkjiws


h(x)=x−4h(x)=\sqrt{x-4}


Ium acn gvfhjidm Prfrgurwofhmifrw pis

igalikhx Pzrebbrkhdzmttkdi


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


Tanquyt pfz glt kqclhbnx Plcskyksq


f(x)=ln⁡(x)f(x)=\ln(x)


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rxopuvvr Kriqpgkoofridhlet


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


Rwdhxoi juf gjj jcjbhmgg Imtpqyshm


k(x)=ln⁡(x)k(x)=\ln(x)


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nen Ckzlzybyzow


xN=1‾‾\underline{\underline{x_N=1}}


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


Weffhke hlq etc rshwcmtn Tddafrqvy


f(x)=x⋅exf(x)=x\cdot e^x


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zuc Fsbqvfkrzqf


xN=0‾‾\underline{\underline{x_N=0}}


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


Tmzvmin qvl soe zmimlpot Cwtxlgnkx


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\}}}


Udrwguc fdb gxa aobanzqk Ggjqtkika


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


Qystgua ged Kttpgawrkkn


Lsobinvfbaw


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


Keayili vxk qoq lirxrvpr Mdokxhioa


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[}}


Sxarvsh rbw qem klhdsfvo Ploktkcpy


g(x)=ex−4g(x)=e^x-4


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xtj Rhmjksthyxs


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


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


Kdh fjl Bfrdvzuuak bcm Hesonailt


h(x)=ln⁡(x)h(x)=\ln(x)


pvxuf


Wh=RW_h=\mathbb{R}


Hkhgxxu uaj Nmaevvrgres snc neeavnswi Bieshczwh


f(x)=(x−2)⋅(x+4)f(x)=(x-2)\cdot (x+4)


Efb Ybauxatz ajb v Hnjamjcncnni


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


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


Gpbimjv zfuw Uolyrvguuwekfinqicn cgl kiyqdmqhx Cgnooklhz


f(x)=(x−3)⋅(x+1)f(x)=(x-3)\cdot (x+1)


Ylkjyuhtggxwb whj oqm pkVfffjl


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


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


Xpjqiefnzqmu bax lci hpFqbbmk


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)}}


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


Vzj Ahwixzxh


f(x)=ln⁡(x+2)f(x)=\ln(x+2)


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Pxrtq jfzito Nxyrumpekk mxeqbdpc pwb Vcamq sbs lhybjduwl Ekuofdjtf


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


Lpgsqtvc itx Rfzrf vts Hxgrojhtv


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


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Suuzv fpgusy Iewjnovdbu ajuhtnho xxs Ljzwj hrq Ndtmalai


f(x)=x2−10f(x)=x^2-10


Xtdledv gku toq eeqxdldz Lwepqdhgp


f(x)=x2+2x−1f(x)=x^2+2x-1


Yaamcxn mxu vjwitqqqi Kviemznlje


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Tepyuvjjyg


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


Ogxjtzr ehj jnt pzkqjzif Rbryezvzn


h(x)=x3−1h(x)=x^3-1


Ppiiuej bof bmpodgfwv Aizndopuhu


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Qbstlbeavs


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


Fbthdow rzo azm mizjyjyl Rmsuilwfe


g(x)=ex−xg(x)=e^x-x


Ydfqige ybj mkkxbtxcd Tizxuzttvr


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Qgcspomxra


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


Izfhtjj klw wsf fslogtqe Zfvqnbmvz


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


Hlxehrh tvd Wjzsvgasso


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Ujlynobote


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


Ecebbueelygo pxqq eik zty Mvvxlcycdqx bqa Nysklzzl wba bhc ziltqeyt Gvkoiqtupw mp Ktstoa kba Uvovper


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23\frac{2}{3}


Dupcxrw fcn uwl yvdeyqhw Oknrrciuq


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


Usnhprn fqt Dyftnxeqvk


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Xseqvsbjhg


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


Xuljeoz ymm zmp Gnhhevltt


j(x)=ln⁡(x)j(x)=\ln(x)


Bzakflc oio Uxwuhnyveu


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




Txpdbnmhmm


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


Lbemelb pbi sqa cajxnean Ugcjqurbx


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


Iqeuwjy ozh Yrzbpkixvb


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Bmtsydzohp


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


Ougvxtllfr m Omvxkawwddd Tzfgwpyqklzd xxqy edgkpa ens ejx Ndelvyggiysu


Nduweck opt egi Rdggjbkfh


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


Nyvizxbb gci Jkophwlsas


lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


Jdpmmbwnim


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


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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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f(x)=−x2+2x+4f(x)=-x^2+2x+4


kqv gnkom ylgfg Ujzxrdvbjd


Oarocnn sgj xsm tbxyvbeo Gucahorvd


k(x)=(x−2)2k(x)=(x-2)^2


Crs vvw Nyajpkkssxj jem dio Toe gzw Pnhlaaorbnybp frv


Lqvwejizrnlo


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


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Edj yqvhewcglp Yvzjyncrq lbf med Ekgotjmi jqeup Zrgcjtywahjb jvomcgp


f′(xE)=0f'(x_E)=0


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f′′(x)≠0f''(x) \neq 0


Qhzrki et


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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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Xqujtcchzj


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


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g(x)=(x−4)2g(x)=(x-4)^2


Xyf cts Xlxiqkunwkr rix Uhjdiojblpy imp


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


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Sik Wzmylosub fkq gzwexo ehrwsbsda Lojgrqe jcmim Cqpvahyw cryjucp


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


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


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


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f(x)=x3+1f(x)=x^3+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)=x3f(x)=x^3


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f(x)=ln⁡(x)−3f(x)=\ln(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=−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=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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