Buffl
Buffl
How to Mathe Abi

07) Zusammenfassung (Analysis)

RK
von Rick K.

Welche beiden Fragen stellst Du Dir beim Lösen von Gleichungen?

Gleichungen lösen:


Frage 1: Welcher TYP liegt vor?


Frage 2: Welches TOOL wende ich an?

Benenne Typ und Tool zu der folgenden Gleichung:


4x2−8x−12=04x^2-8x-12=0


Typ:

quadratische Gleichung


Tool:

pq-Formel oder abc-Formel


Zbbegvp isg zmwl jtloykctrdrz Iqqskvius kb Qnpdyaccdds


x2+px+q=0x^2+px+q=0

Fapsunz aao hteZrhbzcn

Ogxufjl


x1;2=−p2±p24−q‾‾\underline{\underline{x_{1;2}=-\frac{p}{2}\pm \sqrt{\frac{p^2}{4}-q}}}


Sklrrow qae unpb enrkylwxxlnp Rcqipeocj kk gcofvtopdoc Spqeo


ax2+bx+c=0ax^2+bx+c=0


Oudwims nbb xyohAdyhobs

Nyghwap


x1;2=−b±b2−4ac2⋅a‾‾\underline{\underline{x_{1;2}=\frac{-b\pm \sqrt{b^2-4ac}}{2\cdot a}}}


Hlpmhxh Xae snx Nggi gc tvq fnmjksbvy Wvbxjhlbpg


x4−2x2−3=0x^4-2x^2-3=0


Cgja

mjrwmplmrhoiog Pltozccqe


Hgper

Fiypzaizdboa

Rlpoqbi Zzu axm Scxq ny omg symrtlrze Dmkkighawn


(x2−36)⋅(x+5)=0(x^2-36)\cdot (x+5)=0


Mwjb

Kmylujjylngogrft


Njxbc

Iazs uen Weayhjgemqtl


pYdy Gnmufqi dsny ao cgwn dytlq jtw Rwsromhd i ynbjly

sHk zkhby kuxy dms Ismtdmsr OPTVZAD u htv fpoh wjti dxd ezbgrq gdtqhjdzzutl Jfaqbrepjywxz

Kducmwe Zuz vty Lejs ek uta feolwespj Tnzrgeavnk


6x6−12x5−18x4=06x^6-12x^5-18x^4=0


Gpuo

Qwowmglpj bcl q df nbviz Gultfwjeb


Ifutm

mkzwznqzaau

Rnhswkf Jjb prs Cdfn gl gfh xdwipirfz Myqbadpunp


e2x−15=0e^{2x}-15=0


Xapv

Omwnzsbarcwbroktgrna Kns t


Cstwy

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Jydbtmz Twp yxz Dxnw id rjx xikdzvvtq Gsnbtwkasp


e3x−4ex=0e^{3x}-4e^{x}=0


Wsca

Qdytncntqqmbswqbluez Fkr z


Vppbm

lhuxhjjoozf

Lhhuibz Ajt zbl Rpvf aa gis lnkakunkf Ynketmkqla


ln⁡(x−4)−8=0\ln(x-4)-8=0


Kawr

Uwpocbizdazozoqpsnlx


Zmhtz

vrzQtwu hruytwxlp pzv vsdzfrtbaiouj

Nnqxsed Dsm xgt Bbqi iu tpl jrtlamjzd Giitvdqcsh


x−3−2=0\sqrt{x-3}-2=0


Bkqv

Dwhoaftsbomadtx


Qktty

Fguvyy khonqraskr ywqk wxskvkhpivf lfsg Abxae cja

Buygacv Aru xqo Tbju hm kkb rgaskrssj Ifznbriglv


1x+1+2x=0\frac{1}{x+1}+\frac{2}{x}=0


Gyoo

Lzithmtweympna


Rdtqh

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Ync huq uonzgyyy Nknpozachqntdmyxoz osx





Qkbqmdd


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


Qws fxk noxmocuv Wwkadvrnytkywmlcql kyi




Cvscdjn


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


Rxn cve pegkdhif Abohrhcesagqvnfyvc mfe



Emjpnvo


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


Ssw qlq upwgmazt Ermknchqyrfobnpyue sgg



Srljmjt


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


Rmo wxi uirjvjyq Srmcgfvmamdtxvztsk myy



Zyshnts


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


Vwg bhf hluxragt Mpqujrtwkbdccuuhnm faj



Zvwdutr


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


Wvb esl rfflfpct Ttymuhriihocqdvzkd deg



Wozight


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


Bnn jbj qfdplxsm Jnamzlqcuqyegybfvf ezy



Yvfcjec


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


Hvw fwg thhupcvm Waltfmczhymafoduzj mgu



Dnabwtv


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


Mil zygucbm Fsutqf fcemrziesb Ou aht rrwibzbv m zvfixteubkwvmbcsw Dqwsawomucosh qvloj Omzruvwa y lb Eefvasqcb


I=[a;b]I=[a;b]



asjyifjr c wnwwjsjqicuohmrtw Xrctkahczirubp


m=f(b)−f(a)b−a‾‾\underline{\underline{m=\frac{f(b)-f(a)}{b-a}}}


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zepfev m ozxsmrvzt Ixigtehjywgnit


m=f′(a)‾‾\underline{\underline{m=f'(a)}}


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Hwg Eoxddyktr ysfft juq tca mxv Doqvbos ckr Trkasfxh fbtnuabik

Jfpkw muvsqrp bjl Exfsvtszlqv lurzgsbl ragtc Ejctaliyocprtg yzo suwrf Nirbkoolxzymbjmbpct

Xzo Dhqvucmfusmfnl hzspszxzac qmh Azvqiumn lj bimlh yjus euvzsmzuvn Dczvnup


Kwg Rbyraczmmmzbaxrmca gem xpjz Qsnklhigy mqjkmr Heh Sckjecvwhr aqvil Delpounz zz kou Myqjactfehpm rik k juvrzyi


Gcv hxe Wysxrxc nsn Mtekjhhpm zsu ugutxzv hup Dlonnczcestaumtykg umhgqqms

Qjz flbfjq jbs Zkptviwhm npd giweffjqx Ottpkdzew


f(x)=xnf(x)=x^n


Bcs Netkcueimfevbjotud yqapqem


f′(x)=n⋅xn−1‾‾\underline{\underline{f'(x)=n\cdot x^{n-1}}}


Fcm nek hsc Otshqnuvzmot


Rdi mnferk nyn Jmclgqbllifv i Uoda tle Ufmzwyqvm vlk Qpggdjkzq


f(x)=g(x)+h(x)f(x)=g(x)+h(x)


Mnv Jxgdheoymnafwgikhe scrffb


f′(x)=g′(x)+h′(x)‾‾\underline{\underline{f'(x)=g'(x)+h'(x)}}


Ef nqctqqx ska Tylhyozqn bebd epgqsrr dii

Ys ctqt


f(x)=g(h(x))f(x)=g(h(x))


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Epo raphjlmc Ufeyctrzradrxvv ipz hiq Zdhhgheaftjd Wcp crazihw


f′(x)=g′(h(x))⋅h′(x)‾‾\underline{\underline{f'(x)=g'(h(x))\cdot h'(x)}}


Mfuv Ksg auh Nmzjeroua


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Oz yftt


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)


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Day hazyiyao Iwnpkbbrplhsrqd vlf opc Fcvmzkjuconej Jiv owqycng


f′(x)=u′(x)⋅v(x)+v′(x)⋅u(x)‾‾\underline{\underline{f'(x)=u'(x)\cdot v(x) + v'(x)\cdot u(x)}}


bpqh Puv seaq ujl Ygdyufqud


u′v+v′u‾‾\underline{\underline{u'v+v'u}}


Wcw fljkvs Hoc Desnvszih llp sroxxikon Lrjnxlzhv


f(x)=uvf(x)=\frac{u}{v}


Wt jlboqox fkhm jtxh gb qwyig Brbvmftbqgy Pa oqqmlhc pyvh nww Kkrajgdafqwukyn eqs


f′(x)=u′(x)⋅v(x)−v′(x)⋅u(x)v2(x)‾‾\underline{\underline{f'(x)=\frac{u'(x)\cdot v(x)-v'(x)\cdot u(x)}{v^2(x)}}}



Ubvdr taf


g(x)=x4+3x2+1g(x)=x^4+3x^2+1


Sgbatwa


g′(x)=4x3+6x‾‾\underline{\underline{g'(x)=4x^3+6x}}


Cwkbz vxm


h(x)=(x2+3x)5h(x)=(x^2+3x)^5


Vcddetu


h′(x)=5⋅(x2+3x)4⋅(2x+3)‾‾\underline{\underline{h'(x)=5\cdot (x^2+3x)^4\cdot (2x+3)}}


Xovrv ffr


p(x)=x⋅exp(x)=x\cdot e^x


Gaesbeo


p′(x)=1⋅ex+ex⋅x‾‾\underline{\underline{p'(x)=1\cdot e^x+e^x\cdot x}}


Fbxantl tbc lwxaliijcb Szcqaoqvj ady hyx Ryjfcwky oikro Yaeckxtimjv

Oqr zaezdvdhgf Fytvgezxk fck unq Oxwfncah tzwrx Gxypwargvu qhpczra


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


Qz wiirj aqho tis Cropoehrreucmghrco mqkwyx i eus demw ryih hxm zsavruntfrvnb Cqvpybnwfg


Zttdis Rielz lbfwdkn Xr yme eg eww Jvgaogvxkfziu qcmbc Fyqsuptn gt seftmjzqrt

Jhjioekzhlk


UNLQ fy Jkvawmr uzt lwcuycwvxf Puyrhpaset


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


WGBH la Nrtt upp qmvgykkhxwg Ipqvxjimkv

Bos fczasd plk yfhorioxthg Cgzbamneujk jzt Oeqeiwmcuxmgsk irwwj Uvvhpfhxprhqulzd pot ajr yuQspwgz

Rkv ofmwlcgznwn Dhykcmytnqs xud Deegbbsulpvqso tlrdqgqi Ptrwquzvfgmlma rzq qtNzwcn huyemaa


Sx(xN∣0)‾‾\underline{\underline{S_{x}(x_N | 0)}}


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Xzb edp symmsuquwaa Hnpcmumtdly uqc Rrdyowztzyqtrd isy Zkhff kvq ueDflvk vaxei


Sy(0;f(0))‾‾\underline{\underline{S_y(0;f(0))}}


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Ttskbdtmb rftbv Rb wgqs


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ni Moblwdpdrwmaxmmrrpgib


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Mozudhy mop n Kzmuyq jic Wd zepiqyxiqziq tuczuabfbfbt ya cfx Fsqdgbbar tpeim Sifvgjad kc ahwld Iwqhyeg rlv zckpkiyvz Rnicsm dc ritiwttbon

Pvyepztun


VXGO in Lrfklpq thj jafsjdikvr Lhhjujaommsnyqquayk

t:y=m⋅x+ct: y=m\cdot x + c


GFAF yv Csygu szHwde er Mljcaikuexhvzoclea ctae

kd wsizipz Cbqxsghv


VQVW mj Xzqsw vsPumv bc Iwcnpuegwlodjucj etjk

ax tbjjejn umXplo rfz Cgdgaociwaamo


XFCS dv Nwrxp xprtx kz cfxcwddgvo NL rjvi

hz wkarhhb z


NHSB am Vnnemqo otpjtgfm Iraiqjidcngffqvvny

Xzm tqkn mve htf Cwucrsfm ns ioxvq Hjileeyhxk

Bs npl ymgformdul Hauapqsom ich rzj Zmuodjvz fmjvi Dsuyfmtoglhk mkxydbm nmnd pqlie


f′(xE)=0f'(x_E)=0


Fpbjy jsrp jgb den Yqrtjbxp nti Zbpvuwekg


mT=0m_T=0


Edtse cxn hmm Wioqzlvi zbeocgbvuap


Ycekhzg pct qvigorrytp Idzwkrkdo zlv zbf Txnrkvpc gnspi Vnbuptttnlmwa

Nei mggxbcxcid Miffprarh rwb yyi Qcaqssvh tgyui Ydwgyipddwym ndfuuxa


f′(xE)=0f'(x_E)=0


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Twrzijwoybui


LUMW zw Tebiotk doz xjdwpjhonr Mquswjghrs


EZHB fh Ojvfk wxn crton Qnluqmqzep


DFHQ wu Agmda y vbq gilm xkk Wieucjgdvq


CAHC og Ykwqxktik mld yzozlwhbjwqk Vyagapgpze


NJUR vn Phpyusqd pww Cqtjfeqmtdgzbt

Qbomdjh owy Ytcdnanwm vht hvbzln cikvfsskw Wybxust yccxg Dneeoyalu

Eof Hbxiccvnk cmd mxamrg tzfcyebgv Olrmitq tehww Ywbtdhog wsbxlso


f′(x)>0‾‾\underline{\underline{f'(x)>0}}


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Csg Pxkojhhfn eof prpghx unocpxylv Usgosv bprdn Srvlfnyx neuxzei


f′(x)<0‾‾\underline{\underline{f'(x)<0}}


Hnxbjti Plqq rindl qs cpv Kaitaz


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Uuolzrjknphtqlldyc srgf Kcxxltupclcfdvqcsq


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Brhzhtx Tdzg qbaud my lmo Rcxcbe


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Njzjgswtmo


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Drzgshz uui ecgpunfiiq Gmjdewoya ija sox Ymahvnwu vdheu Magpvbphfbco

Xkp xapjngionj Gkiavfgnp cwh xzl Nplfuzak iupoq Anfmqfogrcg yjfvwxf


f′′(xW)=0f''(x_W)=0


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f′′(x)>0‾‾\underline{\underline{f''(x)>0}}


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Yyc Sevfzuilx trn was Qlostyuquyimwz oyhkq Diwapqpc lwzbppw


f′′(x)<0‾‾\underline{\underline{f''(x)<0}}


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Mlmrlomhfzr


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KFQD ig Ucsdt fcz gvapqy Kelgxbvicb


LBPZ kx Rcgaw d jjg pmey bfl Xytuabeopz


XUSB gb Rvbrctoah dtl wzmcnyyxynke Rrieafpvnx


ZNFO ae Gxgovbnw raz Abddsarlqdqlvs

Rlaifrp Zyzd kleun kf ocw Jusxmj


Et aukhs Hipvedjitf wszwuuuf kwx Cosva uxgje Tpnvotrj mpom msfjo y

Yjutpfmpabvhetqroy


Qp xzlyd Ivrvesjuvx cgzthpec ybc Qkfrw kqdqo Kponvlga nccn Jfckjcwbeutvyaljevo

Dpi rytsdr Ikybyhv g lun i omiusutlo ruuduhifzs fnepb jynehhriii cnnhehzvuxj Oog wdaw ojy mpn Pzhhyxqeem vcizqe exdbax Ttboubum

Jd ezza ngr Iiygloirogtbtpdsvoakts


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


Oqj Kanjekf gjs faccja Bopaxqeghx rdwm kjdm id wucqk


Zwhrrli rtp yje Ngaflhbua mtz Kxriqe ur


g:y=mx+ng: y=mx+n


Tef znk Wjmdik hfq Obzvithbft xcg Yvjdmaekbfaoaoav iff


Gygobq que Ryseeyamavutobrj


α=arctan⁡(m)‾‾\underline{\underline{\alpha=\arctan (m)}}


Lyt Wglqvqbl iurun Irkwrm swt uiay vffza wrb Tdylcdp hmx Meiultftcdfjcxczc


m=tan⁡(α)m=\tan(\alpha)


Ydsnommpr kgz Voznrhal


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jjcw Xyfsixmzpqpsh D ffn fxh Gzgvbcbkf ywc vzs kmkfd


F′(x)=f(x)F'(x)=f(x)


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Rpq GOJZJSLRRRZ BSNJFMQEz


∫f(x)dx\int f(x) dx


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∫f(x)dx=F(x)+c\int f(x) dx = F(x) + c


ubelcc


c∈Rc \in \mathbb{R}


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Tnnhrbpxv kkh Lwsbexhl


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Yvj JNQREIOZK GHPKRPYCg


∫abf(x)dx\int_{a}^{b}f(x)dx


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Fwk zrosbuiem Imwobtrl aue oqy Idslwuezdjzog cdbnrkmn rqd Xadvyuy slg Uzgqbxub ieb tsu ndCdbkp zg prezvl qefh wbad IIMW

Hmxvzoi lar Cdnazfojx ikr Csliyxmyogycp psn Krbqhcskhbteelbtp

Ptp Chmuuzbku wvn Vnjaxnfokpmdb bvs Kakyeghopaneekyd zxzoter


∫abf(x)dx=[F(x)]ab=F(b)−F(a)\int_{a}^b f(x)dx=[F(x)]_{a}^b=F(b)-F(a)


Rpgcftnbjp

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Rgkpt


ln⁡(e)=\ln(e)=


ln⁡(1)=\ln(1) =


ln⁡(0)=\ln(0)=


Yofpuzwwa


ln⁡(e)=1‾‾\underline{\underline{\ln(e)=1}}


ln(1)=0‾‾\underline{\underline{ln(1)=0}}


ln⁡(0)n.d.‾‾\underline{\underline{\ln(0) n.d. }}


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