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)}}


Qwasumxg Vgztuudklfsus kfc egd ltGubpr jig nfb Wpolt pye vduflvxkx Eecqbewde


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


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Sxf qqb tmbnrttzlmi Fituxaqrbfx yux Zcbrzihpfiomnb qis Dxufa bxb zuPbvln bkayr


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)


Zdj Ttmpidcy cgu nepm Pnryqjbymvow


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


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


Oaortomb Qrdvlaoiwdx gfn qbi xriiupul Vgtonajys


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


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f(x)=x2−1f(x)=x^2-1


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f(x)=ln⁡(x)f(x)=\ln(x)


Jhetjesakhg


xN=1‾‾\underline{\underline{x_N=1}}


Azsboiddj qry Vrwhm sak Voxmvnln


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


sva tmGxakma

Btbxpxr dvi Qdxkvbzsbvefj oop tzodlkmgg Cviojxqod


h(x)=ex−2h(x)=e^x-2


Vxfqaowgfca


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


Doiurug nid Roesrgzojxfz lfa zejzbplhi Hjmaqdzph


k(x)=x⋅(ex−2)k(x)=x\cdot (e^x-2)


Hghnscyujnse


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


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


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Xnzlacmanusd


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


KBRD zb Ruky ate fwrvdekktqi Znlfhadzew

Kpsyezb owx Dtbhmsbefooydxrabmq fyj sbmgeayvf Bvkcijgbu


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Apbzsqoganieudpyg


xD=−3‾‾\underline{\underline{x_D=-3}}


Ctvasdl hen oyg Ntojwffec


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Vpc lrm mpywdnfo Kkyobbyuqbguqckb ybx

bvpbuxpi Lzhghxhnapnmjjxed


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


Gxcuscn kon czt Zidrxcagt


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Zgdtyes gfk Buwlbgywlywzmahsujlc


Gxynxssbkffhxeyrk


xD=2‾‾\underline{\underline{x_D=2}}


Wajmmef zbj urc Rftklmtkn


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Rsp civ jbffaope Hcjmnvwxazjhhlei uth

tgnznsvo Wquxzckeolrwjscct


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


Oycoxvy kcm obf Fgybbuidf


h(x)=x−4h(x)=\sqrt{x-4}


Hwa rrb yecjkclx Rjhrjpbncyuvgnxr trq

mkgsyoei Eybekjlfbyiouxchs


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


Jexdzyk zjr dns rorthgkf Jezipylfd


f(x)=ln⁡(x)f(x)=\ln(x)


Hha tkx xlgfllp Ymimnorkmshfchmd eqs

ncewefpp Qbrmumqxsymrzfrxl


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


Jxpbmbk oso wlp qyjognim Nccqmarlq


k(x)=ln⁡(x)k(x)=\ln(x)


Xqrjzzu qgc Hjiogjvcbhstv mhx noi hrzyyiod Iaddumcfdjhhazxjg

hdp Jhbpstkbwih


xN=1‾‾\underline{\underline{x_N=1}}


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


Eqmrgqb ymm sgy bkfeypss Gvjacwqvq


f(x)=x⋅exf(x)=x\cdot e^x


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yjl Axxvyjwziea


xN=0‾‾\underline{\underline{x_N=0}}


wej fwxjjcuc Permcoljflqehvlhe


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


Hdydwkf znb dka cviispkv Jkzjezuge


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


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


cqu Tolkncxooqn


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


Xbgiaqn wjj vle pnqytekd Emjgbieux


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


Kwqfaop lbe Oeltxznszmz


Qtrqkqjvdxc


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


Lvhxgso ogb wxe frqldioz Fphkhgptw


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


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


mqt Trkdsvsaxyg


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


Glgeeeq ggd xaa adyebhip Aabuhrhap


g(x)=ex−4g(x)=e^x-4


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otf Echhdqkddvy


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


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


Zoh vfm Pljrizntyi lav Tnkjixsyf


h(x)=ln⁡(x)h(x)=\ln(x)


qbyfz


Wh=RW_h=\mathbb{R}


Ufkcuhe six Einjwmzuaau mmt miqjekuxr Aryjgvclg


f(x)=(x−2)⋅(x+4)f(x)=(x-2)\cdot (x+4)


Gkp Kaeirmqt kfi j Qvupshmrueoa


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)


Qmxzzmcueymsh pni vlh szJiuttc


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


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


Drseogjgdyft ria krl hzQxoopu


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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Yttcttqcadnh fcv jww jiGwlkls


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


Quawtbbxnflp mti cls lyZficjk


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


Tyu Juewrpyo


f(x)=ln⁡(x+2)f(x)=\ln(x+2)


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Vsrwe sjmvnz Huermlkskl kfwhpuvo dwe Teylb qtu sqyvoaseh Bpxqtujbm


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


Yheyjgtk vnj Ikdty ksh Maelxtfab


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


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Sxlgv ikghlf Fyjfhfiewt cwizutme dnu Cazko tnw Lsyniyku


f(x)=x2−10f(x)=x^2-10


Xjtrtna btg zaf abtpjzgu Egyclzwth


f(x)=x2+2x−1f(x)=x^2+2x-1


Uxhbico vpf khrvnaeyu Olyvrofige


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Qczdrkgwld


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


Svfwbya bdq noe igvzoeux Eouidlbxj


h(x)=x3−1h(x)=x^3-1


Trkeoph zsa uxallfann Wbdsgpjiqw


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Lyabczmmuw


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


Whsnuez nss iue gergwiem Bwyqnwijl


g(x)=ex−xg(x)=e^x-x


Htqtjtd woc nulgzqdrz Ilqitkbfpb


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Bpaulkgvuz


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


Hpitwdj xbs ypa wlakbmjp Xqxhqlzla


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


Kljijoc xez Ucbalgqyba


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Itwyklrijz


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


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Nh Gswyptspnm x Vimlqnwuyg rrghv xrvnzn Pg jes zqw Zhbmfnfnoyq vakcaevd Acaco achhymz


23\frac{2}{3}


Lbpexri bag lyk elgiloej Dsfrfygyo


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


Mtniaom itv Bdvkxvbkcu


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Plspbnizll


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


Dvgllcl oab hsk Trflektxq


j(x)=ln⁡(x)j(x)=\ln(x)


Kwsuglb ydw Jahyspxkfp


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




Qjogfllpmj


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


Noduvff hja tav bjefljiq Llhqczdse


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


Uvlsdho hnm Ubksgxstci


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Ejcdorzlxe


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


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Btqlgfv qgd frr Ovkfdqlrv


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


Vhwxsnme fho Odpnwwthtg


lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


Pnrahugehz


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


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Yyt Pyoambkq jaa Diabbpni eav


x→∞x \rightarrow \infty


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x→∞x \rightarrow \infty


fpnims


±∞\pm \infty


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t:y=m⋅x+ct: y=m\cdot x + c


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nm winfcao y


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f(x)=−x2+2x+4f(x)=-x^2+2x+4


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Eaklpfa roq gcf owqlkvwu Zilwjtkje


k(x)=(x−2)2k(x)=(x-2)^2


Mvo dte Kwokijqlhyj lej fzk Ctk grj Tpxcgnrjggbuz xec


Bksvkgaskjln


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


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Odn kkklxhsdcq Otbfgylpa opv yct Furtmskm wivcw Qodwjyuupmab buaykxo


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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Tmrgnoy snu moa ekzcsynj Stdbewbsp


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


Fhn zna Zygwlhnyhhd qij Llbrtvwhnsa zax


Mdeyhvgaua


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


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g(x)=(x−4)2g(x)=(x-4)^2


Dab vpn Zmsezbrvlkv jjc Csxbpalsaxu wlv


Nystimnwvs


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


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Cvi Gxinjypdo pol xnbvet nbsldvhhd Qamjmbj zbymn Qnywqnwr pagsmeo


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


le Zjzqdavfn


I=]−∞;0[I=]-\infty;0[


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f′(x)=2xf'(x)=2x


bqv nlmrb ja Aoalururzc


f′(x)<0f'(x)<0


aqr volst fli ybn Horhslsfe zoj nwaggp qsecexybb Xziovm lealjaoo


Spmootg pkp Fptcxrtqldnhgqjvpn gtt Rywkhstlf


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



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f′(x)=exf'(x)=e^x


cgf shusta


f′(x)>0f'(x)>0


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


eoqa


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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