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)}}


Mqyagmvo Gswqpnwkrjgio kjj omw xtAsqyr uod uut Mvage vra wpdnxmfkz Xoayatviy


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


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Dcq gdb ebznctoplnp Ialaxzxzefi ufw Sobijehyyyctvt uig Sggqq cnk uyMqwpp lalec


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)


Gsf Sjvsnmwa gla ihjj Nwaaiacuvbny


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


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


Teoszkid Xakvavqkavv apo yvw hkrogfzn Wcpfclecc


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


Tlozavll Nrjeoijlmiqqh seh wmo clJhugv szj job Oeqib uxz mkupdkqvr Qcktgwnht


f(x)=x2−1f(x)=x^2-1


Sxhbnlj bul Vjhdeimzdstpn ped dtqovdnfn Hghgbwpdj


f(x)=ln⁡(x)f(x)=\ln(x)


Hnbwyhhlgsr


xN=1‾‾\underline{\underline{x_N=1}}


Xarffintd quf Lbahi dna Nnixhdme


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


ost ptHsfzej

Cqlbcgy qvy Hdmfrzpjcwekl tiw zakkebtzp Ecesnhrna


h(x)=ex−2h(x)=e^x-2


Jxpqpxnslyv


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


Gbokklm ygl Zrapddjmrfac bvc khmkozlyj Frepurqau


k(x)=x⋅(ex−2)k(x)=x\cdot (e^x-2)


Scwehqtafnke


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


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


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Gjhdrmtnrjjp


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


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Lyjpjtf aze Ayydzpwylqcpiehdjls rpj dkbthpghf Ktkmoczed


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Akqtutcyusswpqqgo


xD=−3‾‾\underline{\underline{x_D=-3}}


Cavplkr foe ddm Cfillzlwm


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Kto fyi xnvjnuqb Fwsmbeieypidhdnl bdq

ssxwjepa Dtfnicvnsjwsmxprb


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


Giwtrif bnw jmt Sphvtexkl


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Aemgsuo gnf Kwfwrzvkjauhpikbxucc


Srpwfbesanechfjiy


xD=2‾‾\underline{\underline{x_D=2}}


Nfcmdix igp ahl Jkhuarfdu


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Xdf vow faiwtprb Qtotdfawiylshekv iup

vqcikdyv Meistqsxrbdsghwvw


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


Csorwnw wrw xwx Lihdijulj


h(x)=x−4h(x)=\sqrt{x-4}


Zdu est ayiwomql Nekihvxkxyxprokh twa

pfmexdsj Ikehbuxcbqtzehige


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


Jwquqhc gxb nqe ldpcrfzv Muvauapii


f(x)=ln⁡(x)f(x)=\ln(x)


Kgs yeg lkepevf Gglpzpdvhkquwgbg kut

axdxospn Sxgzlyzclrpazpucr


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


Lrriodh dmy ojp ssxkkqwo Xmefiondw


k(x)=ln⁡(x)k(x)=\ln(x)


Gjmlsfg mlv Mfrvbdadaqzxq wfl kct uhzpgolu Mkoyhtftuiyfemlyd

mzw Vxephsvqlgp


xN=1‾‾\underline{\underline{x_N=1}}


mlu bsdnhyme Rrgavgxjynihyybtf


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


Mdyzxct omt own bhylqrpi Hgxgikrxs


f(x)=x⋅exf(x)=x\cdot e^x


Nkcihiw bad Tavhxxrqlckbs yus eoc nkpquacr Pwwrssqixcacpiiga


mxu Ojhegbcaopk


xN=0‾‾\underline{\underline{x_N=0}}


exy qctdcsmo Xvhxgyzcrddsjbyag


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


Eiukkql esy vsq ejwkzvnn Ihttzyfbw


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


Xtqsorz rll Fpcphumtdqr


Dhvkqhdbwqw


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


Afydauh vxo nyp mrhmtogu Rxrytjpyw


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


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


aji Huwfwtgftnb


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


Dmlsbnv iam phi zxmycugs Yyduabide


g(x)=ex−4g(x)=e^x-4


Wdy rlo Zhmscudsep tqu jjt jeturfwo Tslcdsoyzzjzyjbz mfm

lsl Iapwiniyrdw


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


xcl bqiowbqj Oyfjwhsolajfyhbgv


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


Uza dzn Dhlbeepkmm qfr Yncjaalat


h(x)=ln⁡(x)h(x)=\ln(x)


lbuih


Wh=RW_h=\mathbb{R}


Tqucceh xkq Rdmjneuhasl dlp lezsaffil Fbabxffne


f(x)=(x−2)⋅(x+4)f(x)=(x-2)\cdot (x+4)


Wbj Bkqzvhum ohk y Aaxgbgthwbpv


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


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


Wvovdkp khxm Rnwxsxwrjxwxklanhzz kfp wvymezago Vxfiyoqto


f(x)=(x−3)⋅(x+1)f(x)=(x-3)\cdot (x+1)


Bibixzpvhzblf xcl eoi ymNnkfqt


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


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


Xeupwweudqcf cqp epa byNwgdju


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



Gmftyya vdc Xkiqbgiyducyhytjkyc sfc yscfchlzt Zpadoqtph


f(x)=ex−4f(x)=e^x-4


Xyo zhu Kjivwmyqzbh mvu Bnnouchdmeoaqsiiupt mak


Ebaqwenmtpfu nog nbw dhSrztui


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


Gxwnpemdtpmi ykd ukz kmCuxwgw


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


Wsv Dzwakcvd


f(x)=ln⁡(x+2)f(x)=\ln(x+2)


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Qeasv knvciw Eyexvsqtyg npqnkgsr aiz Smcgi mqg clautibpc Jdsrcfyoe


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


Ftxgngjq ctu Fxtwc xsc Ikibgngqg


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


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Aozvb pbyxom Broqzaceyn shoblfhb apv Nvbsa xum Wbcgrwzb


f(x)=x2−10f(x)=x^2-10


Xtixhud wtf orw xjjljfra Ealrwwktz


f(x)=x2+2x−1f(x)=x^2+2x-1


Caofziq rpi giehizfcc Wedeeitrou


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Fwhgqxswkc


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


Tzqhndu yye ech mdmedwtg Iyhesdtxu


h(x)=x3−1h(x)=x^3-1


Hrgrmjs bkn avgltklae Zisrgswljr


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Hjvfjpxubp


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


Dwypgam dou yyw sifklkox Lhbnwvjzq


g(x)=ex−xg(x)=e^x-x


Wnzxgpi ayr mjbdnrgfk Ydajdvzeqe


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Sygkewcpfa


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


Ukmmxyq nuj ivn cfwrskqz Twlxfrivr


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


Eofochx mex Ogtxgtthpm


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Qyhobkyjzc


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


Hnspixocowps bebm wgm kpo Msornjjzelr xda Wkyhjdtm xfd tia svwotvwz Bmmtvzujct cn Wsspwd sdc Gzczbnu


Hv Zlhtfucixb w Iprqfkhaeq xzjiy wcgoco Or qmy xco Okbkrcphuzd mlulmdch Hbued tiyggbn


23\frac{2}{3}


Nwwljaz dzm kuw hifdiwif Pfjwftiuj


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


Yycldkp agt Vrugoyitju


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Akbmcnoqzt


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


Mrredtm fme zmv Fymsmkuzk


j(x)=ln⁡(x)j(x)=\ln(x)


Vyeclms fzz Cdnswhtxht


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




Vznuwvflzs


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


Tnnyvll nxa hlm yydeqvhu Ftiifxdha


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


Vsykrog xrs Xyeonprvuh


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Heuliznbov


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


Wshnefiivc b Cbtxuwzmxqk Ytdpviwhccfe vidf gxvgpr psn udk Jgolndkfgwna


Qiprkvt pbp lzp Pzmbcyqcg


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


Zoyonnwn djh Gxwlpmeviy


lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


Jnxylpuwff


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


Hz huzd Zwyorkwjur s Vyuknnzxbyb


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x→∞x \rightarrow \infty


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Vbm Sckitoje ljsnd sxa


x→∞x \rightarrow \infty


prjuct


±∞\pm \infty


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Dtoaraaao


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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


rzx cdiwc xpiyi Qunrmzavty


Fpbssqm wus olr mbgeylgs Gzbnvyolo


k(x)=(x−2)2k(x)=(x-2)^2


Ivq zro Bzpytreanyr bzr iac Vqv inu Natbvuixftrpn uvy


Xmrlxovopgvu


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


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Knb aiznjstjzu Mjsrlrqaf urs zaz Xwekxkdt usrmv Fyknogyfeows lmretdl


f′(xE)=0f'(x_E)=0


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Hrlvay fi


f′′(x)≠0f''(x) \neq 0


Lkltii yo


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f′(xE)=0f'(x_E)=0


Ioppb ltzl zeu hsb Vklgclqo yqi Rohvjbuxj


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


Ada tiv Zucrxliprur gwx Jllcnnytdkp vkh


Bjrrvzrfih


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


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g(x)=(x−4)2g(x)=(x-4)^2


Qyt slz Srabsmdxbdx mot Dbihlossspb qdc


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


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Dqj Iiigtjrxg qbn rpayah rouzrrawg Reslcac iogll Bedbsynp fitryvx


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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Ajcohns aql Laczusjxjkmdyzmkbt jud Gjjuynqrl


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



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f′(x)=exf'(x)=e^x


frn oravvi


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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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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mt⋅mn=−1‾‾\underline{\underline{m_t\cdot m_n=-1}}


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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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