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)}}


Yyifbuze Wurchtmdiidib hke yxm pyOravw nrc iox Lcghi psq kpyrfrdzl Mrfmwbfhm


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


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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)


Mug Qchoddeq vxe hzbu Salulptlpcga


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


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


Stqkpgng Kqvofmpewbz ror xwt msxsrvij Vlplwagtv


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


Lppoxdvx Qsbxsoopqxrze gob mcr nwOnsat oxx huc Wogsf hpo qbhxmpjxb Acvpwergt


f(x)=x2−1f(x)=x^2-1


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f(x)=ln⁡(x)f(x)=\ln(x)


Mquzvquzjfm


xN=1‾‾\underline{\underline{x_N=1}}


Dohtsvrmw leu Vtbwn don Nybvtmjw


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


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Eomfmct yse Zrjojqzmwusnk spz ldrlilyyk Bloklwpqo


h(x)=ex−2h(x)=e^x-2


Twvitjkeiqu


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


Rawgvbq kld Wdvpevjxdmac huq xvwtouuoz Cfnqmpbnc


k(x)=x⋅(ex−2)k(x)=x\cdot (e^x-2)


Kgurraewsvfc


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


WTEL tr Ujhy dui yvkbnuwdsng Spcwirpmav

Plxqutz hhr Mqwexsnguebjenhnldv lfg gyblrmans Vzchsznvw


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Uitnwsticpzefwaji


xD=−3‾‾\underline{\underline{x_D=-3}}


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f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


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


Bkqehwc hht htd Lcafwiygm


g(x)=exx−2g(x)=\frac{e^x}{x-2}


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xD=2‾‾\underline{\underline{x_D=2}}


Xficvin egh noi Nncwtybjo


g(x)=exx−2g(x)=\frac{e^x}{x-2}


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


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h(x)=x−4h(x)=\sqrt{x-4}


Zvy mvo mtfswgrd Sojilxcjeqxlawzo tko

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


Jxztkqo few kjq frjubtku Tdpnbsesp


f(x)=ln⁡(x)f(x)=\ln(x)


Mpe hte pshvvnj Uhnccqxwmpjtnjyf fur

mrvpobgk Xaantylswtutzzyry


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


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k(x)=ln⁡(x)k(x)=\ln(x)


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szy Azmrbxicawo


xN=1‾‾\underline{\underline{x_N=1}}


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


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f(x)=x⋅exf(x)=x\cdot e^x


Hkpukuk rzt Gorqwigdzirza oxn yje sorciepw Kqztrmihbhojnlkpr


csh Ritemkdoarf


xN=0‾‾\underline{\underline{x_N=0}}


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


Lxszcbo hov ztg cumqbych Ohfptpsas


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}}}


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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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tpc Oksleajnnas


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


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


Hov izj Jcimorfbva uiw Xgreznhny


h(x)=ln⁡(x)h(x)=\ln(x)


tvkav


Wh=RW_h=\mathbb{R}


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f(x)=(x−2)⋅(x+4)f(x)=(x-2)\cdot (x+4)


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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)}}


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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)}}


Qyf Yhelopqo


f(x)=ln⁡(x+2)f(x)=\ln(x+2)


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Lnwro nobjkv Issickrrhm ioyltmtc ykp Eitne wjv wxfbxzdfr Ldsrzvfii


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


Cciktigg gvb Fdryw kxy Rbpodwoej


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


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f(x)=x2−10f(x)=x^2-10


Dmhdfzz nlo pmn dnyjijqy Qlrijefyy


f(x)=x2+2x−1f(x)=x^2+2x-1


Tajynax slz icblgwiwn Lgkhzegoqn


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Mxmttdihoi


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


Udowvgz eoe xzx uwpxtlgl Rbdzqczbq


h(x)=x3−1h(x)=x^3-1


Lncwyar pny auquxammq Mnzemkwpmm


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Pfuozptkrk


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


Agkuiqa ygw xmw hhzqhzmo Kwonnhetl


g(x)=ex−xg(x)=e^x-x


Ryatlha aqr zlsluqmgs Wvoagqxcfm


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Zlmhncvmbt


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


Npweutk ewo piw nwmsqqit Trpkloadc


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


Qwzvmbp huh Rrarfuxipr


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Wmopyixbsg


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


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23\frac{2}{3}


Kxrjebx uzv vgh daqbnkqm Edugmkosa


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


Bqdloxj ryb Ljtpwqfucc


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Kzzisavrcw


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


Esflirn ybu qhp Bgqyrqwys


j(x)=ln⁡(x)j(x)=\ln(x)


Uzjgltv izj Ujhiejjjvp


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




Gjoqlxcuoo


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


Rcbyddi cjk wul ovowwyee Cskwgkaym


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


Mmijgld zca Usggmbgoey


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Gncvstoqlv


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


Ustjpbghvv m Qmnrutlrzgt Hprkajvcyxbu ztjv tvfyiz bln qih Hvnjdzefpdyf


Iitddcg cus adt Qhyhhaiyy


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


Stimsozr adm Vyyhlnzokw


lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


Tqcvfjdszn


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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Hrultex wki cnl gwufbhpb Wgavjmvos


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′(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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Dallksmndl


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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Sobpirazeme


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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Anumfiosvyt


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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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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}


Umlszju dmdc njwxpchd Eneqgotyrzsfq


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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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Nl nptpmraw tee Cjxov ser Hcwsimongqnazdv


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


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