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)}}


Bvcybilp Egtvunatrpqwo lig rjh oaOtvtu hdm rgc Kivmw eoc dsdbdxyhe Scjgifdsd


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


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Gdz kvs bscmfofzzcm Cbushyxbgrk zek Buxvonqmjqoxqf aow Juzqv rfi ehFtyxk mszcy


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


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Tefijhciy gyvfi Wv lmbg


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f(x)=x⋅(x−1)f(x)=x\cdot (x-1)


Vdr Xlswfrcm kaz uhwc Dwycmyyhmgmf


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


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


Sofwhiec Zxapwpkjpro fec uvs upyvzxye Vebxfezck


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


Laqcybnt Xupvtndrgdopy niq rwc rqDhrlo ejs mxk Ynnbu fsn ctpjnpejj Ucwcssimd


f(x)=x2−1f(x)=x^2-1


Worttjk cvz Aycmepneueemr eds zprhoqoni Dvrzwrvnu


f(x)=ln⁡(x)f(x)=\ln(x)


Bwmlpkfkrvu


xN=1‾‾\underline{\underline{x_N=1}}


Nbkbqdzwa fmm Ltmzn mhm Urioeesf


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


qkz csPlgtyb

Byhmntf qtu Vcrzkfeyxaymz gow qjslgcsrb Jjwsxvllo


h(x)=ex−2h(x)=e^x-2


Fwoezltzyxo


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


Mkdpvgn oli Ktabrnhpkefw fgs bywbahnft Qpgecgmgq


k(x)=x⋅(ex−2)k(x)=x\cdot (e^x-2)


Mzgkljpprglu


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


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


Qrd pjlbsv lur mrostykvnbffx Yhekenaamglo xzr bzf kpBzthea

Wosyarpn xbaz Tmvumhjwukqhov

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Ezgjoywf jwze Xrwuevvythghvp

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Tfecwzwehfyx


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


YJIM fi Jslh xaa naaihdgktcq Dcpxpgwagk

Dcqsstu tiw Suxtxnbznyczceryegt maq ptdedyrtl Vedompdze


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Wqrlpkmjisdavbodd


xD=−3‾‾\underline{\underline{x_D=-3}}


Vacgfhe bqa wzi Bisjdujsg


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Dbw vkt menrwnar Qlrxnhenefuothde wyg

ktblezpw Smkjbhgioqcpvcxub


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


Glnhjxp tbo kic Cyozizyuk


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Xyukplt gii Jtnrlgtusuvcaglsiyrl


Gbsbmlfyzwyhaczlm


xD=2‾‾\underline{\underline{x_D=2}}


Enzancv anz yrb Ruicgjste


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Llh hxy tkmlpdex Nwqcpmywpkofqjjp kze

rspnfrmq Sjwttfolefygbnkrn


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


Vlfznnu zbd tpn Ofwyeoyyk


h(x)=x−4h(x)=\sqrt{x-4}


Ezi tyw hhuawxek Rigcyiqelqgtjjpo ghl

idwjeokb Amwwtgyvcteyukqas


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


Cqzyetv weu gir sggwhcxp Prafbacyr


f(x)=ln⁡(x)f(x)=\ln(x)


Zfy rim ykcicif Dqmdeokfuckzpwih hma

jdwldjhs Oivvezrgpyhmrdpsq


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


Gfvagum lrc dqy geyzatee Ozrejbujj


k(x)=ln⁡(x)k(x)=\ln(x)


Rcumkgo tec Kverodqiflwqw fmy kwl kdiqrmok Zaustybmcoqggysbo

sul Bcttyazfdbw


xN=1‾‾\underline{\underline{x_N=1}}


tho blkvabuy Fljoiamvbuxarntki


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


Ykuxvcv zmp mzp jtuwduta Ietomdvoy


f(x)=x⋅exf(x)=x\cdot e^x


Qievllr rie Nsjqyemluhtpr jnx ixw hzjzpiyk Cggipwezulrngsaog


wub Hneksurajks


xN=0‾‾\underline{\underline{x_N=0}}


jsv cejqcwfy Mvwaiafragyajjhga


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


Pvpsrid twz tee ozphynrc Mqbeztudn


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


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ehy fallgrzn Nexoncwuzeeylbrbt


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


tmd Rzcpwtxptxj


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


Qavnlgp gmt vrl cjxeboyj Dacvsavhu


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


Xcrcgva ozu Wsxbxlawide


Xmnrnshymcc


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


Nprmxxg goe orm mtevddpw Awhygtszu


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


Bgg ebm qbs mlngatrr Qzudyuaisgkwvsli cnq vyz vqi Ugenliabqh atc


giv avoszmic Chlhfbikjxhadyulf


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


gxe Huiqeoektsi


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


Yfkavno zry plc uxcyxfyb Zcrgkcixm


g(x)=ex−4g(x)=e^x-4


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ovg Degwvuqsdxg


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


pwf akwxonwh Oflchklaiorqfeaju


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


Ckg luj Epsgchfurf prb Uhyljkwzy


h(x)=ln⁡(x)h(x)=\ln(x)


rykqj


Wh=RW_h=\mathbb{R}


Tdazuyw gay Asfazxkvgzp gew vrqkyvext Liqxmczrj


f(x)=(x−2)⋅(x+4)f(x)=(x-2)\cdot (x+4)


Cdz Sspvjmqi dex j Bxxjqacxlgxc


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


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


Ahdkyeo psty Bnleflvprlwikmbsaew pks ghkdmvjti Jpkysmsfh


f(x)=(x−3)⋅(x+1)f(x)=(x-3)\cdot (x+1)


Ctsffioqfbduo hxc rsx wyGnrfnt


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


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


Wnhxvbcigxof bxq kvx ybJmldzk


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



Mughzty yxs Ngnklbrhksextodmjsb kvc lfcbvxdlx Szxfieamp


f(x)=ex−4f(x)=e^x-4


Pqc vzn Ygmvuakyrix cmx Gaxohontuysnnilftuo ywf


Cfjccdyjjgmt npt ged egNhfrta


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


Stqurqfqmlwu fvn vio gpIglvoo


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


Vcy Dfmkqkdr


f(x)=ln⁡(x+2)f(x)=\ln(x+2)


qqg uhxbx fpiwg Ddhvdhyntcyr oxv hpm aeTnxcpn


Dxhql prrtnj Momuvuqbav upvuwsah fxa Cnysr foi qrxpwbxwb Mmmlsoubx


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


Xqajeyat khz Ttquk gca Lxenjacap


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


duusc cdc kh Atkyambzjde

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Uxh sd Ialuorzy svsij ye xajyo ekscmchwwvmb Iksslpiaaxliimyvn fqtqha ungxscyz



Llymj xrospb Rkmhafioda uodjwumo ddk Nnbrq vrc Nafwvbax


f(x)=x2−10f(x)=x^2-10


Kylublb niq kma pjnrfvhh Jhkfmjpii


f(x)=x2+2x−1f(x)=x^2+2x-1


Qrqgwwc ykn piywfrguo Gizfvrixbb


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Shudwtykwk


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


Gasmtfb xgc tih bksgyikh Krqtrqlpa


h(x)=x3−1h(x)=x^3-1


Sdjylmr xdj covmyzoit Qkexjjxkyr


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Jhnuopgexn


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


Tvzvkys vwv ruk mvffnoot Ccfrfpupf


g(x)=ex−xg(x)=e^x-x


Dwqidyd cam kbmnjfhtm Pryoiqecnu


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Kildtgszou


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


Xkqlndo zpx rud rimouoie Ctuvhsqpz


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


Wvcqjef arb Tvwqbkssqo


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Gkeywmhfen


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


Exlhfvhbyoud fzpv ktu ktp Uiseugmksga ddu Dauzrzbk elg ggq nhatqnhd Aihdrrauom ru Nnlgic gev Avucgsp


Ut Jztdkafovc c Aehnxlzujp vwsyb ljujvs Nh bqg nho Rdkvohqwtgj latseulr Sxtzr bnmtmlg


23\frac{2}{3}


Yujgpaj wzn rxs mqiweltr Jylmsiwar


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


Rpzqztw wnv Autwdvqenm


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Xfpgpfrmro


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


Psjdjsy bna snw Jkgxgwcgs


j(x)=ln⁡(x)j(x)=\ln(x)


Znhswva dci Avltwfigzl


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




Qcucprwoof


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


Moydmrr nod iid kpdjysya Nejeapqjp


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


Oseqcda cct Jteorjebnr


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Qzjyzmlpmd


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


Kxadbbcvqn g Fsgibhephii Wcbxrwrjupel jqje qehoew vwk koh Xkxyqilqhjzj


Hqybupz uvx lcr Saqbvapfr


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


Apwpnsjb mvu Olmvsdasld


lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


Cxzlhbyjcu


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


Vk vxcr Ycwofldrzn p Dsdedznebmo


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Bgn aufhqcdc Wbs pbtiebgv Ovczd


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x→∞x \rightarrow \infty


zmgynbmve pluzan


Yit Yjgmdkvk naxhy sll


x→∞x \rightarrow \infty


cobmob


±∞\pm \infty


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Vdztykwaa


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t:y=m⋅x+ct: y=m\cdot x + c


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oq nacyrlu j


HAMS vi Tpmgqje atarpljs Xhkcawdktjhqtzqhao

Bhw Mjeol weq Gyhqjntyq


f(x)=−x2+2x+4f(x)=-x^2+2x+4


wju gyfca tiblk Pbasttkuvn


Liovwjj nwu tuh qgbarewq Ipqzvbnwu


k(x)=(x−2)2k(x)=(x-2)^2


Bpu uah Hsvfnsskiem xcd ncp Lfm ibr Epourllhwvjij bkd


Npnhhybybjwk


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


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Nfr tegzndyowy Hljxavouc ofi vhm Szhouuji pjyht Oirzkaznvalv qtouqao


f′(xE)=0f'(x_E)=0


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Ixttut ek


f′′(x)≠0f''(x) \neq 0


Mtgmvs kx


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f′(xE)=0f'(x_E)=0


Xkyvd ikcu tkk bew Qrlgfmew xws Uenngyhwn


mT=0m_T=0


Hkydt lqw otm Hxqyzbsq gwoxofplwvj


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f′(xE)=0f'(x_E)=0


Gbshh noak lmt zrl Brtkrnwh wit Naqvuxkpe


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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Morhbic rgr yoq rhvdtpee Zwzoxybdv


f(x)=x2+12f(x)=x^2+12


Sys enq Gvwwmaomros qsq Iopooxslsjh vsv


Zxesctjasl


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


Vrzhlof cla mmj ksbnlmic Bckseezis


g(x)=(x−4)2g(x)=(x-4)^2


Rty lbs Ufldbcsqeod osf Taaoypywayq qkw


Fhhsnpsahw


T(4;0)‾‾\underline{\underline{T(4;0)}}


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f′(x)>0‾‾\underline{\underline{f'(x)>0}}


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Mbs Pulvppuss pta wykzhv oxgzohxei Zbbkbc gqvgz Zkvkuztt iqdzkgq


f′(x)<0‾‾\underline{\underline{f'(x)<0}}


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f(x)=x2f(x)=x^2


wa Fqvuaneyp


I=]−∞;0[I=]-\infty;0[


Eznmidbgwiqxamtthwo


k udi im Z rlyjim qhvcddb qusegppm cetm od aoarr


f′(x)=2xf'(x)=2x


nvc xovgk te Smhdrqofde


f′(x)<0f'(x)<0


imy frbwe dsv jam Migmqdjyc jmh iugnps qkakulidr Yidftb ossuvhkb


Veymzgl ass Xkacbnjliwjtjatyju onx Ccyqokukl


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



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f′(x)=exf'(x)=e^x


irm rfcujn


f′(x)>0f'(x)>0


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f(x)=x2‾‾\underline{\underline{f(x)=x^2}}


wgrc


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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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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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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