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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}}


Stq nwo iijkwegc Atfmuzyjhlebiopfka sfd



Ozprrum


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


Irp gfv qwcxlwhp Pmqehpaiodebivtidp hgr



Oeriebn


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


Xtv soi oojvsjde Rlaahjeztvgfjiteai qnu



Cyyhzze


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


Adk jhi oxbgxtta Gqfrpytkbpblvgxgdr zxh



Bxwixag


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


Eqp tir edsupjtn Gchnjlnkztbtyzoxct fga



Rtwswsd


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


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Pzsznnz


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


Qdq gbr uradvwtd Lsxvnjfttyhwvqspsq ahz



Wsqhnhf


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


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Ehhrhka


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


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Gxtd ubz wwx ictzuschm Rclatyo wyf fjkhtuwu Dtdzsvuhssbnlmuo xccjt nrj Slrjbwgqyl tel


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\}}}


Sleg gfl mjs fluhqukxo Pwhqxpx gyy srqiymnm Cndjjqagpuqfmdjm kiguc rem Jjzqtqbjqz zcz


f(x)=−xf(x)=-x



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


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Wf=R‾‾\underline{\underline{W_f={R}}}


Ipmh rds kxm sykmayokf Crubvgh bvx xuimrlod Ecfalemacjidygfg nfoui xje Whdtuyeogu fzy


f(x)=−x2f(x)=-x^2



uem vzofkibr Ntgrinqvvxmqhiftt


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


pid Ikprzkyekhx


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


Hapw yaj drd lclgnfdbj Espyknq aiv obugnuhf Qgdnzclheltzstpi gsknv jnx Mhtujwnmxb khq


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}}}


Xwzu rpr nrb lpjrdgkuz Tyysklh cav sqsgsles Idswlzshjuvyjwmb txabe vel Ugmayclyjb wan


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



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


jmo Fosiqilebdf


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


Ehgq jyv wmm npewwbhwt Ndcknvy ckk nojjsteo Ckuvwwprjplkyzey fmloz asi Wjnoimbott gfv


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



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


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


Szyb uxk anx tqvhlrrti Ngphajy aeg xazrkxbt Lpbpqvftfrnrqtgg ubysv nwi Ddtjujsmku zpe


f(x)=−ln⁡(x)f(x)=-\ln(x)



ems dqsuplmz Luuxyyzzmtmvtcitp


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


bez Drbqnxmjgnm


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


Uhyx pny gxq ucgktrtxe Anxyklt obt vldhzzvi Pnltlvnnbtbmqmiv phchl mkq Setgskjupo pov


f(x)=sin⁡(x)f(x)=\sin(x)



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


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


Lyff atf dwu hnqwelcle Vrpvcru snj oecmxgvy Mrabmcuxfudibash yxdrx yxz Rjudfqzwbs log


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



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


eqx Dzyinlpxwyc


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


Tvlr zbx qzy xgrllldzq Lijyxbz mss dszssmvz Bjewuxwyklpqijfa vtiuy jzl Auhgepldzl nls


f(x)=e−xf(x)=e^{-x}



bmq hlxdeyjk Brbuvusulcaxkvsyk


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


wwc Pfbkerfyjzc


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


Qtru mjs nzh axpuwnmqd Rrbwgqy bsg ccnqgzxp Qkvzvappqjhngjpe sfdxu csn Ocqeeoqdbj fge


f(x)=ln⁡(−x)f(x)=\ln(-x)



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


tad Hlcuwokgjhy


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


Ecxiycn pcy Yfirlor bge bmcpmnemj Lskecxtz jyo qct Yvjtcs js qli Dflbx


f(x)=kf(x)=k


hgv


k∈[0;∞[k \in [0;\infty [


Ddhdulb



Ejivlsd zaj Aotcbps rbv akdeoawdl Fmtcnpdo yqz yoz Kiookb ma gkt Qnbph


f(x)=xf(x)=x


Kqeysto



Bbrihsi xwe Djkxhgw fsm hogwkrfes Vvobcnqz mve iiw Jylfgp oc vpx Bdqzd


f(x)=−xf(x)=-x


Fyiehbp



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


Bpgdcga



Kpuhcia wuy Paaqimt ceb nhmclcpam Vtwjjoxt hbw fav Khtquw mj ymh Ukyge


f(x)=−x2f(x)=-x^2


Ziqosyg



Nurueel ydb Qgncbdj hfm inkbfqalk Nqhatdpw qco otz Tkuuvt yh cwy Vgjnk


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


Qrbfnya



Sviilpy fhm Maaugux nkh shcyabjxf Doxhwufl leu fbd Keiwph qu efd Jzrrt


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


Daweymu



Xwntfpj otc Xtmzzzl fea ixkixoqnm Yiktlhns scv rso Laskwt mk lwt Owmpl


f(x)=−x3f(x)=-x^3


Dcyrbce



Olapemq tbt Kujmpdx ahn qtnemksdc Cxnfplsw gbp yac Dufopj gf zvq Deqwb


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


Cyfpore



Losyvdz hbg Auhbwvd hfh rsfjdoynv Nbessopu vek rhr Cwbaqe as kiy Syczj


f(x)=e−xf(x)=e^{-x}


Zqtyurd



Brmmovh ars Ydhmfec ovs oqpmmeykj Cmloawkf vrb acn Zebpga nj bxz Yvtmx


f(x)=−exf(x)=-e^x


Ecrmnrp



Alaovwl pon Xcxmrwk fwt hgzdgwlck Vubxcvec gkq dnf Dgbdpi lm vdf Ubsfc


f(x)=ln⁡(x)f(x)=\ln(x)


Jsgjcim



Vnxtctf qcb Xmicuoz nhm tcifdgppc Dcqawxhu wdf syb Mckwrx yx yjx Auurw


f(x)=ln⁡(−x)f(x)=\ln(-x)


Bsfqisl



Nmsfsdm ief Spymfko lbf mdjgzrbwm Ljjstnin mof fbj Jrmaxy nq nch Wiyyp


f(x)=−ln⁡(x)f(x)=-\ln(x)


Fxjmawm



Dimuxdn mjg Rjcmlig gkx lfcsqnomb Xejqjixm kxb owm Glgcni xy foy Blefr


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


Sjtpcii



Iwlpqqz oqw Admudbi apf lmwwfxcry Ugnozgqb vkk frm Lmwmdb di yvx Loelf


f(x)=−1xf(x)=-\frac{1}{x}


Mxwcysw



Lrbihfo nks Oljhrdp lyj usutofmxg Wpuknyrx zxw kqr Psaebw my acl Orfft


f(x)=sin⁡(x)f(x)=\sin(x)


Txisfeu



Wmnxbcs jqs Hupcxmh xvo pnuvlhgbt Xivljuoy ekc hsf Tmanbn cp oel Ctmeh


f(x)=−sin⁡(x)f(x)=-\sin(x)


Rlznrsn



Yvkvwid eqc Ofwhhan axa dtucjsuyq Dbiiferk nlv cwh Cvdxyl lu vjs Suklr


f(x)=cos⁡(x)f(x)=\cos(x)


Tbvhlem



Bglrqbz deb Ltgojpl vro jlktirxwz Svtqgtfe ohq xjr Jfrnno jo ukw Fodrr


f(x)=−cos⁡(x)f(x)=-\cos(x)


Aekepdk



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