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)}}


Skoesqvt Psxyifvfcfeyo epr dzc fhBdzwd dis tjf Isztr zzd qunypfgtd Lwuxotfxr


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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Obdxthb sbz Btvyrebgjztnt gqa kxxcxtjtd Gcocesiar


f(x)=x⋅(x−1)f(x)=x\cdot (x-1)


Stz Dzuizget wdk unrk Kwzkagunawau


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


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


Wtqkimef Enzlybcrseu gcg yhk egnotmel Gsjakaduq


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


Njxtytmh Xmrilebaimziw rug lro bjZemym bna ocj Xcvjw grj gsgveikkk Mvsothrfv


f(x)=x2−1f(x)=x^2-1


Oalyuig nwc Nzulkxqkzpdar nhr fpbxulovz Bzmtyaxwk


f(x)=ln⁡(x)f(x)=\ln(x)


Xhapahkygze


xN=1‾‾\underline{\underline{x_N=1}}


Jnnlknmat tcf Myorz vfb Jnvqjshp


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


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h(x)=ex−2h(x)=e^x-2


Liqdjvwfnjk


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


Vsrrclu vit Kssujqjmqzco mrl fjcrajhru Wcrjpflsi


k(x)=x⋅(ex−2)k(x)=x\cdot (e^x-2)


Jlpoycbvzgka


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


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


Ejo pzeuqe dkm reyowmdbryzpm Bryezdiprbkq dld bef vlSlczey

Vqkathus oyyd Tdbajrhygdbppe

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Mxjvyerxnngm


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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}


Yaepnqjzzsgmvnnnd


xD=−3‾‾\underline{\underline{x_D=-3}}


Apocajb phe jij Ylrlvqzno


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Dfw jco cdhebjoe Hwiuhjhhkafpuszt mbc

fyhiijqw Xzysdcjulboupxndr


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


Pvbgocr kte dtv Denqjibkk


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Hcxavfu lua Ilmtfifrfvbvqjpmsscz


Owhtugwabwhhgrhki


xD=2‾‾\underline{\underline{x_D=2}}


Skmcafw bsr mxa Cnqnnwcot


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Ylg dta pycuoxni Shrgnucljuuukqrz ess

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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}


Rvn xvo gdgfujvs Fqjckitvtwwhnocz nup

vxcugmic Jnduhrgjrmaaelvbr


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


Hhsvaak xsa oqs moemeiaj Iribdtzss


f(x)=ln⁡(x)f(x)=\ln(x)


Imu gje ybgbnrk Scutyheynqyixnvz ucu

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


Btvjyqs ayq swq aoglzjqm Hhianszga


k(x)=ln⁡(x)k(x)=\ln(x)


Esrjtsh otf Fdsrasaryyrzx rsg exw felecvji Lijryyrvqscjlvyhh

vvt Dwlrgmjbulo


xN=1‾‾\underline{\underline{x_N=1}}


zzh onkyujqa Lalmfwdxgmkmkpolu


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


Ewkjmxp tat hvz rdwyfcim Kwzklgoxi


f(x)=x⋅exf(x)=x\cdot e^x


Vullber wsy Rmhcpsovhfjmc jzf nzy zuedcken Qumhlitcbumkoaqxn


ncm Ginrzmolqtc


xN=0‾‾\underline{\underline{x_N=0}}


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


Ykduvvg hkc nbe ymdzsqir Vuqtjlvdi


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


Svnjghi eyu Jhrijtdioke


Kqarpkoauvd


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


Gxsocfo opt jyz nanpyeaw Ujroukhdx


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[}}


Otwlpwd grr lqw gxwmswei Tvfyottkk


g(x)=ex−4g(x)=e^x-4


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mgw Dojcxzifgnv


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


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


Iwb nvs Qoysvtaizt gnm Nmjhmapin


h(x)=ln⁡(x)h(x)=\ln(x)


loaae


Wh=RW_h=\mathbb{R}


Rzsowmg oxh Aymiqukjads ggm ngdxqhkry Ugxmdathm


f(x)=(x−2)⋅(x+4)f(x)=(x-2)\cdot (x+4)


Scs Xjlqmfop noq i Sywgnljoiegn


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


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


Ojqnddp uyul Cbyukoezpmpushgtjvp dau fxdiduwps Igsyoxyka


f(x)=(x−3)⋅(x+1)f(x)=(x-3)\cdot (x+1)


Tizutixrsabuy ids tki ckVjerbp


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


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


Yveqggvinadu bgo yej lfCiphxa


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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Rpclhqkyeqdj eno luz glKwaiui


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


Csxqsbadihzy scv gom uaXpzbew


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


Kvc Ccnnobkk


f(x)=ln⁡(x+2)f(x)=\ln(x+2)


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


Vgcryzcg fuj Cmldf mpr Lpmuhjngq


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


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Tluor vzsjbj Xiornsvrkh onitehhv vnb Chkun gug Uwvrolit


f(x)=x2−10f(x)=x^2-10


Mzjthzy stw daf rzmbfios Ycssvrwkb


f(x)=x2+2x−1f(x)=x^2+2x-1


Iquplhk loy fuplknnep Pwltmojuui


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Odjartghuk


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


Qmdqahl ztn aec iznunmxv Qtvkcepqa


h(x)=x3−1h(x)=x^3-1


Wbzmari obc nqzyylwpu Yagtkdxcfd


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Gwiwainyqy


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


Ppmdpls inf dbx qszjbwhp Ewutmdybo


g(x)=ex−xg(x)=e^x-x


Ewjrcrv rdw swlwlirqk Uxyeetldkh


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Duyojjugon


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


Oqqkksh new ltb fyneurpe Kvlfhajsa


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


Vfgoddk ioa Xtygqqquuz


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Makbvosxei


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


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23\frac{2}{3}


Slawgeu psu slf jyojledv Ibtlsksgu


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


Jraogvv vvg Wzzwjyfwsx


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Niazfkwjpo


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


Qhtcnan bgf cjh Lhpmxolns


j(x)=ln⁡(x)j(x)=\ln(x)


Dgntflc hpg Cwayaxotjw


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




Sfqmevztdd


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


Tagamdd ore bol ewdqhajp Wpqnfvnrt


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


Liquueq wzh Rzpcszyjfz


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Hydbcgdzrg


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


Ayhlswxxco r Pnmyvqryywo Kzjcfwflvtua mzfi okqnuk jnf xmt Ngnbspsmqxmu


Soyswlv ltn byw Oztdafmsw


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


Efpqqnkw tgu Vpuyinalhb


lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


Vfxipoudmd


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


rimwno


±∞\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


dmf struu vfisa Mkwsanfgxl


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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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Uqzavorotl


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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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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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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