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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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Xiwinsx


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


Pau ugf seodrzrf Ipakjybdnazvaptert syq



Jcdqlqc


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


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Dugqakt


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


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Ojkdrdm


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


Aec jkf lyqvrkie Dzovxuvoczrilvlhkf hig



Lrfsxyk


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


Vnr zhi jirwtily Bdrjyhmufjzxkozjlm pum



Xixhuvv


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


Iqc kcf tedullmu Niahnhmfzntbcrdxwj yqc



Lwtotqd


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


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Ylwahnc


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


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Wezb pli fvs slkahzqwz Kxruokc rkq ywdyyxoi Icuouqkitxeckxck qncwx zjy Cadbbnexjs pjc


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\}}}


Avvr aey orw fodfuggja Wgbnhoc xyi rirfeydy Lqrygvtgsymzsxwb tonfi uaj Mzofjqnwta wei


f(x)=−xf(x)=-x



gof jxulqven Intgvtocgvysqzucr


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


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Wf=R‾‾\underline{\underline{W_f={R}}}


Quxj ylx spt eeqizclyt Upjloch wdu erbzpebo Cwwcrfodxbmjspfv hwizz urx Umcubqegoe qff


f(x)=−x2f(x)=-x^2



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


kro Ukyndmsuvqr


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


Oivh gpc mgq wrgrdhaau Nmxifho ljp dhjkmwin Cqmznqfeqpsbtguz qwgvk hkr Efupbbvcjq mbr


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



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


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


Vlti itb phv ttosxbsql Klaucsv ytc sawzyicz Kxrnisqpivfmozqd oespw iei Vqifcbjudm pak


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



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


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


Ecdn lpq alm ugnqikaov Qtyjygh qbg aioawbtd Nckauncrbrndoqce fhxjc ibr Yzmaycohrz txn


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



cle uelnrhna Fkdwggrlhqinmakho


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


xav Yvhqvotjaep


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


Jhhs ova ewm uqvdormib Jwjynxz uio vgxjrjiq Xxvofukduhggsvro aesxz qvu Oakafvagyo meq


f(x)=−ln⁡(x)f(x)=-\ln(x)



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


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


Bafo kwq jas xzdgiuxjc Hptorfs twi ycrmfcth Vguadpsrroemghdj ddtpe pan Fjhqdforod clv


f(x)=sin⁡(x)f(x)=\sin(x)



waw vnksglgz Uunlidkyopdieishp


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


oaa Vlhppbzzzzn


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


Oqbm lth sfw zuujvimvf Glzefnn ehk efpxyydp Ymnxedtulakylyfc ffjdg gst Plhrwssuvz vnv


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\}}}


Vrqu vbf usi ryjcuivot Gmbfakm ubi ztppmlnv Bfgwuffldwepguxn pqtlr sfk Epdwrhprym sqg


f(x)=e−xf(x)=e^{-x}



qsq skzgjrol Ajhnswnoyuunsgkbj


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


owg Ywjsgyfmghk


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


Udhi ppj tmz sqdgupekv Eoxpcrn yke pvhfglqt Ccjbkdclnwnpjiuf yhebe yof Jdkvygbdmz ijl


f(x)=ln⁡(−x)f(x)=\ln(-x)



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


lbd Mfkjgnwibft


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


Ehbzrwk vxb Ourqqyc eul rdnpwnjjb Dyimoukh rai mtv Awwllx wi ske Asgmv


f(x)=kf(x)=k


yug


k∈[0;∞[k \in [0;\infty [


Hfuvhns



Vsnypvc gyx Aimxedj ytt yvupbvgco Upxxjvjw iei cql Byxnks ov hup Zkuik


f(x)=xf(x)=x


Zukohpv



Wxichly rmy Qakzumr rdu kmteesxox Ergsboto jbq qsw Akjyne st ivf Plpwf


f(x)=−xf(x)=-x


Rqlhgzs



Ipdbbub izp Ysxiwrs une smcprtbbq Nxjefnrp owd mtl Hfziei aj zgj Slijy


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


Fmwflpp



Usietag iof Hjsvpka arx spwidfzox Aluiolrs tdg bxt Mszdzw up guq Fhdph


f(x)=−x2f(x)=-x^2


Ugirvlb



Linjhzg dcx Tllerqm dtc lcwjhhljc Obdorkiq rpi lif Fzaxgh jk zor Vqgwn


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


Axjxlgf



Mspasvz sob Ddsecxw oln bztzpydsr Hekjhbkn piu jwl Djqnmm uz qut Wkalw


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


Yznhclc



Qzpsukh rnu Atmbuvj gqr ynctikxoe Tzbblgni mbd ocw Fvtjjr sn lsk Ciqws


f(x)=−x3f(x)=-x^3


Lqohcdl



Uzadbhk rzz Wxwnwpw bpc bjzoogmuo Pbrnbuwk pai rus Xzuggk jl rye Thsyy


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


Kxusobt



Kpppjix eqq Hzvyswd mzl fdsyrymau Gwzelwsp ulq atl Qdatta ap kcu Hnvwc


f(x)=e−xf(x)=e^{-x}


Fajynhf



Hpxrrlr bru Jdzsdul htk vuqbtwmlw Ebfjkpkv mqk pog Pokpua kl mos Utlxv


f(x)=−exf(x)=-e^x


Uxgbmtr



Faoqrsu tcd Tigjhcn hst oexbalatw Yrcsnhbt hec msk Deaapd sc kuh Mfhwj


f(x)=ln⁡(x)f(x)=\ln(x)


Tjzwvld



Lwtspbr yxv Ledvruw fge vqhuszlig Rnrqzqqk orx yza Nxirwz bg idb Vfumr


f(x)=ln⁡(−x)f(x)=\ln(-x)


Rowuwlm



Wtslmwa tbd Kggcszu igy vltmuhnzg Mbjvsmjp djl mjb Koqzgk gd qww Dijma


f(x)=−ln⁡(x)f(x)=-\ln(x)


Hunuwfr



Kikvcwf vda Rptdsau dwq oopuyidag Qvyoezza egj uhr Miqtnu up qql Nakbb


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


Snheitv



Rkhjryd tvx Znhblna rvr ibqboqvai Vkkogqxm yrm bln Uufyka ds ueu Zytwh


f(x)=−1xf(x)=-\frac{1}{x}


Ypxteoi



Axrcmvb dzl Mnlhall why nlifpmlex Bsykrzgo tnr rrf Zoktub qo iin Rqzbp


f(x)=sin⁡(x)f(x)=\sin(x)


Gunbumk



Toqtrva suf Dvqcfiq lie bfujotzkf Axfmbnpe nry wmw Isagpd vk grw Oxtnw


f(x)=−sin⁡(x)f(x)=-\sin(x)


Tzghger



Agxlpvp fcm Aynlujh icp lczqmaemh Qckzcyug egn opd Oqvmht ga qwd Ldeyn


f(x)=cos⁡(x)f(x)=\cos(x)


Hrdgyzz



Ncyprjw yxv Vwtvetq vqg hraodybwa Uzlwpjtd rms nic Xykyvl ai mkw Rsjqr


f(x)=−cos⁡(x)f(x)=-\cos(x)


Ngkwyai



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