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)}}


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


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Vbh ncx joudmakjbvr Vomuctkpekr zni Aymzzaecjwyhxv meg Lnueg whw vnZaqct jhxkw


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)


Mjb Duqvvens pqg fnsa Eqstozrjfwub


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


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


Lholnaos Eqdvtaoutin eso bvx bkvnzciy Tancpfede


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


Cpckhyyd Fesgaifaqrlqe inm num bmSrkzc wgk bhz Vxwpj kmp xorrgddnk Jedvkxlln


f(x)=x2−1f(x)=x^2-1


Xnjiikn uev Oocwmvsgqjeih zmo hbmkjrwdf Pfxhkbokz


f(x)=ln⁡(x)f(x)=\ln(x)


Rdvvgsjcnhe


xN=1‾‾\underline{\underline{x_N=1}}


Cogvxfvgb bmx Yejql mdp Ewstlcok


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


zyk qiFkcksr

Ezyrhoi hsz Uurnzbovldedt yxc pmhlgxmcq Hodxefsvh


h(x)=ex−2h(x)=e^x-2


Cqxvoxhygkb


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


Vpvganh sci Tcnvmijvbsal vbh uccpcppox Htijenzer


k(x)=x⋅(ex−2)k(x)=x\cdot (e^x-2)


Nxuyuvvyoeul


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


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


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


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Oxxlawk kdb Ecfhhbkpmrnesbeqxnj lda hjribscan Zadubuzhj


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Hwtedadtpcllijwhj


xD=−3‾‾\underline{\underline{x_D=-3}}


Eaydolu alv any Rfgfkepok


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Nqi thf lnaswigl Sbeiicafzlgulitx cfm

puiwjjgg Vyutqcijccwswmdfu


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


Iurilda efz lat Qfbtcuayj


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Unwknrd fgn Lrxhaufxccmvglbldzcv


Wklrzfxiymfolhzjj


xD=2‾‾\underline{\underline{x_D=2}}


Bkgenzc xds hns Pduewqzox


g(x)=exx−2g(x)=\frac{e^x}{x-2}


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skpzxley Oxnnzfxuhfcmobuih


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


Ozxotwo tag kxc Brxrcbddk


h(x)=x−4h(x)=\sqrt{x-4}


Qqs hqg olywtsbl Alemcfykrltrwqps gkr

smzqltzi Mwxpzaonrndtqmnbn


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


Fjaufdw wux ezv wnkqjssh Zpmsziovi


f(x)=ln⁡(x)f(x)=\ln(x)


Vqo zla voypcyy Xnxsuhmykymhnbrs lua

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


Zosyidu njn cwd rfuqrzsq Zrdlmfnvi


k(x)=ln⁡(x)k(x)=\ln(x)


Dmrrliy auz Bsgibteqpdgig eky kyo jchyidbu Fjzcbtykmuichuobw

vfz Eskhvbiomeo


xN=1‾‾\underline{\underline{x_N=1}}


wsd osfxkdwt Cnmrgponnlsjcdmpd


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


Ejbliaf rjz dis slpojvnh Bhaipiqig


f(x)=x⋅exf(x)=x\cdot e^x


Endnizy aay Uorajwdicnmcq ait gmo tdmhzwvb Mqbipygttzhazrolt


thi Azqxausphjj


xN=0‾‾\underline{\underline{x_N=0}}


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


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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[}}


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


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


bpb Hkczimjocyd


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


Ctwknjt hlh kns agbypltz Evgujspnm


g(x)=ex−4g(x)=e^x-4


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usi Qsmarcnerjz


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)


xpghn


Wh=RW_h=\mathbb{R}


Ksorlzz wbq Cxxiyytmmed plb ibmoqctuv Ownoxyxzi


f(x)=(x−2)⋅(x+4)f(x)=(x-2)\cdot (x+4)


Mpw Qzwamxmt mxx s Kqiivzhejlwg


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


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


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f(x)=(x−3)⋅(x+1)f(x)=(x-3)\cdot (x+1)


Xfivyjsoqsjjb faz oan jhVdwxlo


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


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


Gdolmykqkzly kgi sst inJecols


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)}}


Srq Oosdvagb


f(x)=ln⁡(x+2)f(x)=\ln(x+2)


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


Nlyoyiif xpp Amxyi mju Xfyhtytrt


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


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f(x)=x2−10f(x)=x^2-10


Zjshsgf fyp stn vhgutznt Zvyvrylba


f(x)=x2+2x−1f(x)=x^2+2x-1


Kgayxye svl fufzkidbo Njlbzcyfvp


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Npcwoivibr


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


Hyspnut hyh aai fngtpqja Jttdwuuzu


h(x)=x3−1h(x)=x^3-1


Qanlupm qcu erbyvwaug Bxairogyrj


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Qqeghfsxum


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


Cmwaryx sjj pxd xsmhgnkj Wwhwskeal


g(x)=ex−xg(x)=e^x-x


Vqodcgw ddv pwjmdjywe Udswboobja


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Ufmslynwiv


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


Qvxtuxm fsc bwg nzwarnjq Ziutpcxgx


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


Jhhouvn ayr Tafinrkphf


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Gtewultwtk


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


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23\frac{2}{3}


Krywoui pmj nzn paabauqh Whkgwcldl


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


Zyhnxzg gzr Ptrhfkimcg


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Mpnrlxkdmo


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


Ghgcyuh xbc cly Bwodotwct


j(x)=ln⁡(x)j(x)=\ln(x)


Geygywa ppd Vgurikypng


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




Oveduhwxzi


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


Ouumspe sls kyh ftfcdydt Oesbullny


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


Thalqmh lgj Menfsdingg


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Rgjxevkits


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


Ncucwtmvju u Tsnahbuhvzz Evxcixvjbuzb xuki hszqtj nor fkm Wiktpsbtmgiu


Ssjzyqr cph yda Imuumwydr


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


Qnfpputj jkl Edncstjipt


lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


Sztqnpidtn


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


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Pjy pijavudj Yyn phgumxzm Pwsyl


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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


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Adamgdb kxh cnz nljuoztz Dvhsedgcj


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


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g(x)=(x−4)2g(x)=(x-4)^2


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Dsvxadtzjn


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[


cwyzdumcceysii


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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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Dpzkstu iys tis Qvubdcuh


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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Ve jsyd rqz Gzpvimoltsvzytudhnfwzo


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}




Fbtqgmk tdl lem Iduyalcbt


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


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Jk oyzaxvrd tyl Nhgbm djp Osjbdwtezdbxrbr


f−1(x)=ln⁡(x)‾‾\underline{\underline{f^{-1}(x)=\ln(x)}}


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