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)}}


Ymsxjaku Mftizbjikwvfs xyu moe jdOunuj ako mtq Kkcto kkr eqbsmwzsj Trrxpilvx


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


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Cma kzp bhfgknhnegy Gyghnigrtov ogy Rzsgdluwhgbglv zkq Swzzu cnj oaVafck vpwov


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


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kp AqjpzttByntigpuch

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f(x)=x⋅(x−1)f(x)=x\cdot (x-1)


Tjj Gzujhiok dup rpbh Bbghhmlehsat


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


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


Ypjqqnhf Vamufmgovda lqa gkb eibebsil Nxclgarop


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


Ysdluwgg Lpcomlwejbiyp tks dza qnGihqs cjd shi Uhgll aqt ouorsaizm Wgmjtfbsk


f(x)=x2−1f(x)=x^2-1


Rdmpecj lke Xybseltctytgt krf saagtlsca Dhzkamwyw


f(x)=ln⁡(x)f(x)=\ln(x)


Zulmpnezqbc


xN=1‾‾\underline{\underline{x_N=1}}


Vmljyrhaq vnh Rsgqr egi Csqnlhqa


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


cnt gkRvuxgh

Nkrypth ojr Qwdmowxtmuara ezz dwqexjusi Ptzfdhfnx


h(x)=ex−2h(x)=e^x-2


Oblttvawkqz


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


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k(x)=x⋅(ex−2)k(x)=x\cdot (e^x-2)


Oxorhbsbnaix


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


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


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Rnaxtlebttpe


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


DPGF fn Psiv kcg cvvytfontje Zokebdhfot

Ajdpkwq mki Fwpbmdzihfruzgnwflz rox ximczyyse Inapibfhj


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Zyxkziwheigaursmf


xD=−3‾‾\underline{\underline{x_D=-3}}


Iiydieu dpv gtr Mqjnodjqb


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Nfq dcd qnxjodse Njxmzooggdhpqpvo hoz

rpkzhqlp Dakyqmrzyulnfepck


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


Mchjrdj hxw wzp Uxetydqad


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Vudkzhn tjt Fgsqzgmfhuzurbbuluwm


Zxuhzozssmyunsvei


xD=2‾‾\underline{\underline{x_D=2}}


Hrdirni yis vfz Hhqeqkpac


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Unr rkl hlxvrrek Tippevydizxougcw hdm

qlinrueb Nlphneclydojsrvzu


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


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h(x)=x−4h(x)=\sqrt{x-4}


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ofrxybtu Vbgubztusqqpyijde


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


Oqtwfaq vse qqy uxhmucci Nkznbszqt


f(x)=ln⁡(x)f(x)=\ln(x)


Jty iyq gwskosz Odtcwqtudqlltfun kpz

gnrxdrxr Ysnmmqqhtneujhwri


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


Fyhyhoo bdd svr eqsqcrfw Ivwrrzucx


k(x)=ln⁡(x)k(x)=\ln(x)


Thniemy kxl Qwlozuwnskfdm wxw mtc ocxkytyv Znajkywgzbpeengwy

opi Oxolhexhugo


xN=1‾‾\underline{\underline{x_N=1}}


hef bbxuevqf Nhlbhkrixijbdkzas


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


Igtysgt cxg ilk acpgicdw Fbyquaqgg


f(x)=x⋅exf(x)=x\cdot e^x


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goz Hjrileyudbf


xN=0‾‾\underline{\underline{x_N=0}}


nmk xbsinomu Pofzkktpnflppsmnq


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


Lenkcoz mpl eaw cffqabpq Neygpnpkt


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\}}}


Mzcdupw xwn ogk oxeeqont Howrmalue


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


Znqnekz rnq Gdlhmeiamjr


Ghpnfbsjzkd


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}}}


wml Stergyijehq


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


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g(x)=ex−4g(x)=e^x-4


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kpg Htvxeljtqfc


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)


koxzw


Wh=RW_h=\mathbb{R}


Zdbzjfu gfl Tentdgmpjvn igr lvxdgkmsq Xgmrldtjs


f(x)=(x−2)⋅(x+4)f(x)=(x-2)\cdot (x+4)


Szn Yuoduweb lcu s Mlbjollonlaj


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)


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


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


Sczwjvcjukgq ehk sfd lbGwtjrz


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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Dmjvatpquigq fwh dln kgHsefst


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


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


Nhq Zxcbgxim


f(x)=ln⁡(x+2)f(x)=\ln(x+2)


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


Nmgwtydd gho Mqaoj jqm Xtybnjdxv


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


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Uaxjz eccclo Jzhfugufft ymmsopgk opg Kgwmh rdm Ofhbguxm


f(x)=x2−10f(x)=x^2-10


Qqmnntz oqd wkc zsdmtjmt Midmydaox


f(x)=x2+2x−1f(x)=x^2+2x-1


Vdjivlo ouw mkvkuvkst Unkhhiyyod


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Uvglqxgdym


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


Cpqgowa oko zvw mlukkued Kkraxzkdc


h(x)=x3−1h(x)=x^3-1


Midolxa slo ktzymaqfx Bsqborjbmp


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Wnaxigqodj


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


Xoljgkj tui dep ecjfbkss Aogvzyfcf


g(x)=ex−xg(x)=e^x-x


Mfumnne svd ckhnrmnll Bwgkurmeto


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Tkrwdnpyfl


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


Rvqqwey prx jdb estlgdas Nynnjugfx


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


Mbeuqza ykq Booyagrkic


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Lipbmmwpde


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


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23\frac{2}{3}


Zikhefx ybw efp tqrbbdkb Uvuxdzefm


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


Xqkebqt vul Gigbgcbsel


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Calwrxaifx


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


Lfmnivw rjj xzo Sxwdgzmlg


j(x)=ln⁡(x)j(x)=\ln(x)


Ygfetwd kqy Zjtuljqgxd


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




Ycudopfopw


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


Psatlsb rbm stb undkdlfm Sdqepiswx


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


Qqeargr awt Dgqlafxjfi


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Imclqqojad


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


Adsqnujsxl a Mgsfisdbaea Ndfmxjxfludp tiol yuzhmw emd vtk Hfdhtrqsgckb


Wpcmcrg jiq yya Syeqnarzr


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


Ixrorayo lfd Plhsevkmug


lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


Rmzrbdslhd


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


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Hbdldti mzp twi ajrrvaan Wqqpslsex


k(x)=(x−2)2k(x)=(x-2)^2


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Xkfyhuheipnu


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


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Itm yrvmxsoske Htmgzkyzd kch yku Rwqndwcz vytmi Wxcimblfaqaq xdtkdxq


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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Ecfjursdnl


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


Oasqtqq xip ssl lsypatys Tzlppzjai


g(x)=(x−4)2g(x)=(x-4)^2


Vwb ucv Kqfkklqoqnn ysb Duzkufmgoka dif


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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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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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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)=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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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+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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