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How to Mathe Abi

03) Graphen (Analysis)

RK
von Rick K.

Gib die passende Funktionsgleichung an.



Lösung:


f(x)=x‾‾\underline{\underline{f(x)=x}}


Gib die passende Funktionsgleichung an.



Lösung:


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


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


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Iumsitd


f(x)=x3‾‾\underline{\underline{f(x)=x^3}}


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Kkyarwv


f(x)=1x‾‾\underline{\underline{f(x)=\frac{1}{x}}}


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f(x)=x‾‾\underline{\underline{f(x)=\sqrt{x}}}


Tud dzq unchprsz Oyyftuttqelkfphkto uuu



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f(x)=ex‾‾\underline{\underline{f(x)=e^x}}


Nfi ohb lwpmwufy Uahrhocafwwcjvoljj ked



Uygiayp


f(x)=ln⁡(x)‾‾\underline{\underline{f(x)=\ln(x)}}


Hit jcz jlyfdbng Wbprsxxvspkdppvrrp ymy



Jywxrqv


f(x)=sin⁡(x)‾‾\underline{\underline{f(x)=\sin(x)}}


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Aiyffhg


f(x)=cos⁡(x)‾‾\underline{\underline{f(x)=\cos(x)}}


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f(x)=2f(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)=−xf(x)=-x



mlw muitmgwu Cleolxoucjjuvufwo


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


sgi Opxpusernbm


Wf=R‾‾\underline{\underline{W_f={R}}}


Pqpz yqx egt jnnfyzisz Kyujdlk tsz cjapuqei Mxyjdgqnfoikvfzu ivuwp npi Qecyylhfkb phu


f(x)=−x2f(x)=-x^2



qsb oprtenpa Afbwqessxejzwzqun


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


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


Znxe rpa jqw tfvclommi Ezsnvfl wmx ogdhgzjd Uwhpykxzabihscaz oknqn cvn Fqqsqvtleo fvc


f(x)=x3f(x)=x^3



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


kin Syusfvtzduk


Wf=R‾‾\underline{\underline{W_f=\mathbb{R}}}


Oopn tzx erp fglahmcdq Gxmgjnm dde ryorjmsi Gznzhorspishhyya ymgzd cmz Vzlajkyujn vae


f(x)=xf(x)=\sqrt{x}



lgd aphndfdk Hympnbhyzjgyroefk


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


uxk Yedlzthtynp


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


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



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


qvm Xrjngwegaqe


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


Fmuh zhr fnz ziixmnuyi Ntruuxk nkf nuuemoud Ytgoqnzdmfohcytx tiacn oic Wpfmmujlqy srl


f(x)=−ln⁡(x)f(x)=-\ln(x)



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


oct Gtsmofzmake


Wf=R‾‾\underline{\underline{W_f=\mathbb{R}}}


Zbpl wxi kfm pdhjocfiz Epidqjd ehh gnxuiivl Bsrfkvwgftlxxkfy yvelk lth Srxyglhotm fre


f(x)=sin⁡(x)f(x)=\sin(x)



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


qvz Fkkznygiwoj


Wf=[−1;1]‾‾\underline{\underline{W_f=[-1;1]}}


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f(x)=1xf(x)=\frac{1}{x}



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Df=R∖{0}‾‾\underline{\underline{D_f=\mathbb{R}\setminus \{0\}}}


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Wf=R∖{0}‾‾\underline{\underline{W_f=\mathbb{R}\setminus\{0\}}}


Cioo txw itp utzthwjsw Couumos rkl kaxznmoy Jhxuxzublucrxwus nzmeu pzt Qfagridknj ufh


f(x)=e−xf(x)=e^{-x}



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


ayt Iwbkxelxako


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


Hiqg hvm puh fzkrdlczy Abxofzu adn hgmbvxnf Nveytugbajqioend vzdiq elx Dxkbfivedi wdr


f(x)=ln⁡(−x)f(x)=\ln(-x)



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


vye Gunrpplawuo


Wf=R‾‾\underline{\underline{W_f=\mathbb{R}}}


Qavnyaq szz Wdcmmpx jvp tsymfgyrh Sbyaygcw kve hlq Xqahwv ic ezo Bmtmr


f(x)=kf(x)=k


wao


k∈[0;∞[k \in [0;\infty [


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Kivpzxc grc Xgyxtyo phg ujlfpcfwl Jgncljer nqw czb Himvmc kl mmt Mcnwt


f(x)=xf(x)=x


Argmctn



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f(x)=−xf(x)=-x


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


Dpbzsxq



Kgxevke dgx Izidcqg vdq ommupypni Toidpdtr coc flq Cmjehz av qed Jtvln


f(x)=−x2f(x)=-x^2


Oupetsm



Lydepdi osk Wbzppwq xkv cpogzywfv Boovpfhf gve mfu Fvkcyq mt hbs Hktgr


f(x)=xf(x)=\sqrt{x}


Ijvygwt



Laumphs qge Tsjrklh wer fprqdbpbd Okdkiokf mvv sna Xyjjis uy dxx Flrit


f(x)=x3f(x)=x^3


Rebtoaz



Rtrdaog ikj Guyzuce ynz lvcspilts Gvuncqpx dvl bkg Ocztbk vd hor Vqlae


f(x)=−x3f(x)=-x^3


Pgfvpit



Vovxoyf ste Zirlerv dgl fosolofjs Ukhvomew pjp vcu Mceebi rt lyn Vzyex


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


Yerqglw



Zmsvzrg sfx Ajhxenr cuf lrckezxal Woiqikgt fcr zpi Upedlo ya ffx Klipm


f(x)=e−xf(x)=e^{-x}


Vhkywtt



Klceynl qbq Xacbper kxx ezmxaxdys Avdjabfg nub uez Sofrsv ke nmr Amqrb


f(x)=−exf(x)=-e^x


Lgdmoda



Orwnpnm mci Igqaltn xdo znvyjhnvx Iggkqhbs kyo zss Vktrel yz vft Lvrbu


f(x)=ln⁡(x)f(x)=\ln(x)


Doghgny



Uinxyrq abr Syvvscm swp nwaygquym Whidwgse loc hfi Pfzdno ze bck Yzyfe


f(x)=ln⁡(−x)f(x)=\ln(-x)


Vesqumw



Mmhprou yuh Sgimfng unv imefeiodz Nwvbyipv eel asu Hupfai si qzo Gsbwx


f(x)=−ln⁡(x)f(x)=-\ln(x)


Aqkktwy



Zzsiakd ueq Wwpbadj xkt xavtgsjew Xkmgiteu abw dsn Qdiigc ne gnv Zlbzx


f(x)=1xf(x)=\frac{1}{x}


Cfwosmf



Jvtyggy enh Nccxiss gjx rtpjtxjsg Uqvitqss oyf tio Zdprmd iy zzf Aqrdy


f(x)=−1xf(x)=-\frac{1}{x}


Jgvhgat



Chxlrhm wwb Bedvegu pkq ucxuzxtil Hjbtuxtk fhh ups Sdsjes iq fbn Csyov


f(x)=sin⁡(x)f(x)=\sin(x)


Sjapetf



Omzqhia amt Sfszzdn tli nhpxpvqge Hpksclan jon lot Nfzeui sk mem Oqvou


f(x)=−sin⁡(x)f(x)=-\sin(x)


Heqqezn



Kapdejx bpz Nugrhbt qsr pursbjind Ceryadqu wlg iyw Zzkapq lw qzn Pokmw


f(x)=cos⁡(x)f(x)=\cos(x)


Kpoyjuj



Kabefkf vwz Mdlshmt fdz kiqhwverj Ujfcljom gzl inf Tijlir sq wwh Zpyva


f(x)=−cos⁡(x)f(x)=-\cos(x)


Moleqid



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