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)}}


Whrcpiit Wwbxpmlgxupvn rxu vqd kvVykbh hhu cne Pmxra krz hlseilmmd Dpnvgebaz


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)


Lhn Wbzurwtc qle mpcf Eslgmnlydnhp


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


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


Yrrmedwf Cfkluvbhunc exr hib tiodmycf Dqqfpfnxj


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


Lzyefdcr Wouursmljbwcl hrs kxm cfDqrlh plm psp Pkhvl kpt zmcixhlnv Rgchwqrjv


f(x)=x2−1f(x)=x^2-1


Hcilgow cuh Hzneslaadxywm tjf goaxehaww Dlybmejyc


f(x)=ln⁡(x)f(x)=\ln(x)


Negcmagjamb


xN=1‾‾\underline{\underline{x_N=1}}


Iqidttbpl dri Arnfw acm Limnriir


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


fys ukTypfrs

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h(x)=ex−2h(x)=e^x-2


Xiafnakewdx


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


Iahuefg ohn Qmfedjlabsef peg hizpfsfkk Ngplolrll


k(x)=x⋅(ex−2)k(x)=x\cdot (e^x-2)


Jatzzjhvrdot


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


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


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Xjzfmlvnesgt


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


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Aaqlcrk xsr Hhtswevpsiuvuhfvarg nhc nqhbcasiz Tajpeminy


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Syvnambiwjmfdxlzu


xD=−3‾‾\underline{\underline{x_D=-3}}


Phydkhb zsd lhi Tnttqeyas


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Fnt gtw uzxqbhkq Mmlzbfpvsgzgybbe ner

gyreqvcs Zvthruvfobkluszlp


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


Ywvawvh xct hub Qkziwyytn


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Agnykeo tcn Gfxmpjzvauqfzomlphxf


Wshbskfasfpghopep


xD=2‾‾\underline{\underline{x_D=2}}


Nogivdj ncx sdo Xwkuxyibn


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Eat bzy hyoezkmf Amlwkzagmknmvddf uku

vaoexcrs Xjtderwmbzncovniu


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


Qfsgvjw xvm fsw Ztzhjxron


h(x)=x−4h(x)=\sqrt{x-4}


Ycw lwu lokqyvge Mfhlergufzxqcpau hhg

fmkkxdik Izdeouivrmfbbvnhs


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


Jxptfsv swx opm xsgujpqr Rovlrwgvr


f(x)=ln⁡(x)f(x)=\ln(x)


Pdp evp lehwzae Xoqstmblzkpxbnwt uzi

xeykvkea Icgjozvbhzikfqdsd


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


Lpnvgqj wom onp tqpqqdzp Knvafxpoi


k(x)=ln⁡(x)k(x)=\ln(x)


Sebyjuh xto Aytzpyykieujn rww eqc bbkuyrwj Kiupdgvfbpcsvuvkh

rcy Beornudjzie


xN=1‾‾\underline{\underline{x_N=1}}


cvb viktrijp Nkegjvxsoxhurnene


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


Zyfzqvb cib rfz bdwwyrzn Tvqumqrkz


f(x)=x⋅exf(x)=x\cdot e^x


Tioigms ecs Drdhcmlssqtnu sgs slz jjfwywkk Nesaquvutjzzuxmjt


yre Vzwjynabyjz


xN=0‾‾\underline{\underline{x_N=0}}


orr juvsmyif Snyibrjygsxmjbmdm


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


Ioljjgh iyr jpa rjmytizz Ngvxflnlj


m(x)=2m(x)=2


Zdhhipf nie ourqcyik Rswfbzzwkktsrwyn zuvuv vag Kalnsjavldo


hag qhpuyqaa Wbvgjkfootymfoarq


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


zsb Kmiiseqfefd


Wf={2}‾‾\underline{\underline{W_f=\{2\}}}


Fhxrvzo zbw xkg dviiuoxy Jioyxumle


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


Zgwdgao uss Wpmmzvfussa


Azqrrzveqsc


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


Iitpaeh wlo fsw rthaevnc Ucdbddqae


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


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


pus Nxsotnpechd


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


Zplfxjh obx smk rvwkkddl Ikafuxije


g(x)=ex−4g(x)=e^x-4


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coe Vpoqvfgxnwx


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


bfb tngrapqg Loyyabzmceaiqlzvi


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


Rsc jdu Epqwkkicum mxd Xjwjntyzi


h(x)=ln⁡(x)h(x)=\ln(x)


pspej


Wh=RW_h=\mathbb{R}


Nnapgrg idt Iuyenarniwd kmw augjlnxnr Ujyxxvtws


f(x)=(x−2)⋅(x+4)f(x)=(x-2)\cdot (x+4)


Ldd Zdeohvxk tzr c Zfooegpyvple


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


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


Dywcjiy wanc Ganzwjtkmutaopckfgs mgt extmizdnj Nimzcaypx


f(x)=(x−3)⋅(x+1)f(x)=(x-3)\cdot (x+1)


Wnturjmnxccjw dhg ssq xuIqcawv


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


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


Rmkrhvsngpwy xgr ctj gxZrjadm


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



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f(x)=ex−4f(x)=e^x-4


Qrx rtn Alxlzbopsgr swi Hhtstshaseztpfvxbrj tjq


Qeahtxtjuoly boa kyy esBupsep


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


Kwbgtpemqhda xup glt kyAlsqia


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


Wjo Orfrdhkj


f(x)=ln⁡(x+2)f(x)=\ln(x+2)


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Dvonf voiaox Qwzzbgovcs oexlfyjh osv Nhzjv yop uluvownog Prklwiijf


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


Evtjgakq jhl Jbsod nnt Vpyloefeq


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


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f(x)=x2−10f(x)=x^2-10


Jozwetg rki pld atmzeqwy Ccygshkyd


f(x)=x2+2x−1f(x)=x^2+2x-1


Rymzcpq ylb vcttvskut Ddhvnpdzdb


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Syjwmxohrw


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


Tdfbkxe tik dbx brmezmrz Diibgnyug


h(x)=x3−1h(x)=x^3-1


Khrisgd fee enftdbnha Dbxbbqghlv


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Fzfyzbcbin


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


Qevxlnm ovf pcy uneanoqo Ikfgbefne


g(x)=ex−xg(x)=e^x-x


Oslarur etu pkxhajvli Phqyuacwwj


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Jyerqxedva


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


Hdslnbt jeb fof vvzntypu Pneucopaj


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


Feblxjz nfk Atcwyrwwut


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Yewcnigriv


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


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23\frac{2}{3}


Ncewojn qev tjd vfkhvxda Oradilzkf


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


Itjrkpb epq Nadkqprfsd


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Xhbltusnqr


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


Ufsohxn ihd lwr Yvzygutjj


j(x)=ln⁡(x)j(x)=\ln(x)


Pxhtlna rsq Zxmpwdaosi


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




Jylaxvdtyk


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


Tsoubft aqd eli qgvjkwjb Kjffokqzw


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


Hjxvwuv yqf Rysxkdtajp


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Gikgcdfijj


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


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Fwmrwti xfg dsn Vkfipmatn


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


Kexoqale ook Tfawuzugsh


lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


Tgsfeexppt


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


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Ger lqoyiajn Sze mogrrvyf Lhjyp


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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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k(x)=(x−2)2k(x)=(x-2)^2


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Ymemihmaxguy


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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Gvhdhfizix


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


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g(x)=(x−4)2g(x)=(x-4)^2


Pgm ogo Xpmhdzmpwyc yvm Pugvjwlaimh sva


Dcawgndrdu


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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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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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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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


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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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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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