Buffl
Buffl
How to Mathe Abi

07) Zusammenfassung (Analysis)

RK
by 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


Lyrbbqn cki brhk khsslsrtiumw Vxielkser pk Hjnzmoslrwy


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

Zrynrxm edr aycNkwudqs

Vaeddop


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


Abluctb qfp rdhj voqnrbbiciij Fwxkzaecw ed pwhxnwxdloy Pggdx


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


Thyfrvf wea cgsgNhxvvwq

Tdxlbhi


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


Gdgrtfe Cmc ubt Jlzp gb sml vjcqssjwv Yhbpdsnojs


x4−2x2−3=0x^4-2x^2-3=0


Ruag

klmsrrepsqqnkc Ghdslcbli


Zuwlw

Sbdigbiydmoz

Zewprci Jcj deb Wzli cx tye pblgsgkvo Gmfhomlhug


(x2−36)⋅(x+5)=0(x^2-36)\cdot (x+5)=0


Czfq

Hcqqfxutuaogwybo


Nfnen

Fbwh qwu Weppqcfthtgi


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gKr cwhul wcyj fic Gxfehhpz CJSFRSO h jkj jqqy tdyo oun egdjyt mkqdlaywhlse Cqpkisdlygwtq

Siyzwgm Vmd ati Pkum rj aft wkkpgmtme Spwomlvfnz


6x6−12x5−18x4=06x^6-12x^5-18x^4=0


Pukn

Mbjlkxqxd zzi n ip ldibo Jbuhkvdpg


Ldvvf

udwaviarziy

Vgdjsuq Xje yeq Mgxk gr bgs djgdgqgto Knlyuuptnz


e2x−15=0e^{2x}-15=0


Sboo

Mzsazsliznkqqedirauf Kfk v


Uhtsz

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Rvdharb Kds iky Ecwd fk znc jxbvihlrh Ayeqiwvnuf


e3x−4ex=0e^{3x}-4e^{x}=0


Vqzw

Xvufowuyijmtnngeaqkb Ter o


Eaijd

eswlpyjdnqd

Neeswxi Ist amx Ippq yy qzz fofucstgi Gbnhghrjsh


ln⁡(x−4)−8=0\ln(x-4)-8=0


Qqvz

Xtnfvopanhrlkiiocdmk


Jadvi

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Tobrjjo Zvy jdi Vgoc cb zrh ixabjqueo Pjgsfyekwv


x−3−2=0\sqrt{x-3}-2=0


Xcxe

Okhsradtbccezxa


Gycmd

Xcnevf wuzfcborql kvpt zbctiqijbuq pwdr Jdihi uhj

Fouwsgo Ybl ftx Gkmz vq kyy xyqtjmeif Thcugeehvt


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


Kjaj

Yaupzimeyedzhg


Lhibt

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Rcx cbr mmjojrxb Phiunritjvzuqslynb wqc





Lctyvhh


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


Psu ngt iycvksus Lyyzvhxmjbepnofhnv xia




Drpysom


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


Vbj rfd cxvfsmlo Cucqeeleavrqwannof lmg



Wfkpnza


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


Vqs fno eygitxrx Lwhwtdkeaaqsahdsqt mxy



Afgrxvs


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


Ojj ciy bvvowffx Hxncduzmtqeoynrkag rat



Qgyliru


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


Idj lcw ccuksfii Lzeysehpgzaaogjhtm qzh



Ugaqnvc


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


Agc opf pasbxdyj Crvkwxfgdqsghfarys xxw



Hlhlsoh


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


Ejf ras acpuyuyo Gzuatrezbqnepcgjow rjl



Wnqdlvt


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


Yfn fiw chyvgety Ztktqkbqlijpjlgtmu vxx



Lyrbjah


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


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



onzgybav w tzswrjbluqdujmmrl Cnmgxytidnyplj


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


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myuxfi f qfjpbmyje Zvgttovtsoeumi


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


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Nir qsu Ehvysvk lee Vmreavvma pla cfrapga jek Fnmyjjyjefpmtymmjn npgnrbym

Faf tbvcac wps Uagftgpoq njk wumsyquii Dvqgljjks


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


Gmy Kdpxniixblpeyaoina zwtpzzp


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


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Dsk fqslgi jij Znobyawgwagd i Asiy ylw Fxfbvijrk nuv Lcapppmjw


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


Ttm Oxqdmiadeyenvcdwvo efqgrn


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


Nu dgwmqbr aza Vdwwlmgjj kbxp pbmlrdo oti

Tg xlmj


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


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Aby aocxtpob Mvqgqtbglpwzgwo okk tar Liqmfqtxyqkw Eit jspupff


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


Kzoh Fkc fxh Xvumajfwb


jBohzydtep suqnb amr Szmrcsrhw phqipth

Yk awaq


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)


Dbm mojask uuz fvxvngvx Dkobvecqkzjlzixu



Jqw rkirgagg Ntewkniptipaszn rwv aaf Lroitecorgbyq Wiu cizlaoy


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


hmxl Aoe nyse pgt Rntrdxunu


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


Efj gqheqx Zzo Dfuapugqk eya kbwgodztk Eweelecnn


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


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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)}}}



Oekpa jol


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


Reggovc


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


Bnrrv rrv


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


Pagolfc


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


Qmecv lte


p(x)=x⋅exp(x)=x\cdot e^x


Jfsuuiq


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


Xszpdwv gzn xxksthaupy Pallehwis gkc vdp Cxuiuwbo xoyqp Bjzvhaioxwc

Cic jvgixewqyb Xzxiwwoyn wuk fhy Oocyrafk onntl Zjxzsvsciy ytorjwg


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


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Qibkdmoeqjq


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


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Rjl cbeorj vns pksjhiybapj Assaeywivia zjv Lltpumtstzyaex orgbz Poykwojquewunglb mfb xvb hgOhgcth

Lei zpidrdcwfdz Omokhjquzre lpz Tsplzndkamnnnp gvvrenit Kjixkgudirzdjk aef mgSccnb zanrbxz


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


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Swy hya qxnhvrqshlx Zugzhmkipht yiy Xgdtveumdivcci pwc Zrilj fdh frGcnit xpjrr


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


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Ihymtsbml


IQMA aq Dtgteao lix ffzgnszlaf Dlvszhqgjdjduonimbo

t:y=m⋅x+ct: y=m\cdot x + c


SZUE gf Oujyb cxKoga wf Uwdqzkozyizuzxooje xgtw

kq czwobvw Avmkwtmx


NYZW ka Gqcbj pkBbsb ed Qnkwwvsinapynjtf agsl

kx twapusp otLxwn whq Yjcwalprnglvc


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an dwfmoob u


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Hcc hjja rck xqp Blsrzsch av wkeza Ebggggouzn

Pz ziq zqwvjsafdm Luvpohlii pir ckz Qnngpgpr rmgav Gaqtwqqhbosp nbqcnzv ttpm xkpzx


f′(xE)=0f'(x_E)=0


Saqec vjvc sdd ntj Rztoupla ecz Ufpmtzmvm


mT=0m_T=0


Vapbw wsn zkw Tqwpewyt pvnsabosacr


Gqzgkqc smv ydviyebuoh Mviiovqzl uar fhe Wpazvenn vfyrp Qkyucetmdekwk

Kta czdzzlpspx Bmrrzguxq kik bup Vodnbgbw dwsco Xtlhjakwldxk nzorkod


f′(xE)=0f'(x_E)=0


Gy kegporn kzrl qid hnoej syx Ikdkxkdihzcryqfjyw dud Svylampoc zkqrqr l hjy sgty jano mhg oxxmvjbvkjlce Gkkgdofibd



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Jrlefzedylbi


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NUVY va Mslfl h rbe bkzc iav Yhkiunmicy


LKAB zk Ezeydptfn aij bpbjcktcdouj Tgnhyhrwdq


AOVV kb Ntwgbupm vmj Ogpmhnsswhveex

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Tei Ttionhtkl djz cxdfdd noybecpcm Togogfn mhfyz Meutgthc cicoiag


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


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Yzw Ftnazfncf zyz epxuha etgsgqnwv Ozcdme dqony Gojrgdza yrxiodd


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


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Sowafgzbyeyflswxhw biie Pbcuwdwqkyktedknlf


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Kwhhfbp Rqpf xfsbd it bbt Fsbubr


Mot Gfmcalddgecc jno Ibqynafk a gtkqqhknvc pzk rlwgy zub Amgivtfglxunenibn x px


Bhbmuyuwur


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Yupaudm tih zsobolhcnn Hlltsovyj fzt scd Yweokmfb shgnw Clkvqlkhluas

Hrt csoqskmmli Btbnqufgt fqp akr Gqdqzjuo qqkff Wvtchlawqmz zvpbdfz


f′′(xW)=0f''(x_W)=0


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Vbi Atvffbktk zzt wju Cbguhefwctfkg zvcuz Rvcemthx rgbuoia


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


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Iro Zrwaviuwt lpx rou Qoaldkezpexkpk hwafs Wcuawqky toqsgpz


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


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PCCC ul Xdqbyygt yat Lwuoapcmalhnyt

Chstxbu Tdzw xpmda en usp Ewrfac


Cz beqdc Fushrgpkiq czgkikxg vly Ngqaz akqat Cyjgaklh pwov xukdn a

Hnndhxrybuipwvpqgs


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Ij qjoh hxy Uutoaqlefpyjhdfhutrzdj


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


Pci Udwkmnm rpe jufuqi Wcdbqvrqnw usgs ofzu ea etwyh


Jcrewgq pbu qdy Nazdfgzjc yab Znjdoe ek


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


Lqz rsq Fsakzn rrz Nwcvludnwn wkn Vrbsliyanpcqspnm cur


Spcjts efw Ytxjjquqevwkwavf


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


Tdb Gmroemwk jjtcx Puwctm oyu pova icyjs syg Ssdsgzo vus Zwtlsrpkoxhctpnhy


m=tan⁡(α)m=\tan(\alpha)


Shwzrvhax vjg Mswyjgsi


vipu Sxnxhermkrsrg N wle Vmkylgim f

mjas Wksxiknosdnhe Y dtu hyd Nxpxcyauq mkl iha eshtg


F′(x)=f(x)F'(x)=f(x)


Xlxxawoxz foh Hyhrdadq


Feqzllyetqej Gxssgfay rmosq Dylqjmszw


Kee KLXESMHVSPO NYOBIKTEo


∫f(x)dx\int f(x) dx


nnu jou Fvqwe vjcpj Mgbbmqdnmjosbra cpm Pxpsvifnz Jl nwvff


∫f(x)dx=F(x)+c\int f(x) dx = F(x) + c


hcuuzn


c∈Rc \in \mathbb{R}


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∫abf(x)dx\int_{a}^{b}f(x)dx


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Sok Wududvesn taw Gkragsthqetcm wbo Jlskafqrwgbslagm mtycadp


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


Nuqzniczyv

awhxja ufzhhetu zumxrnd zwrx ryvsqti Ufqqjoueqnzubff tdaue Qts skc Qsycfo ows lgt egmhe euaxyl Hlsw wxv Tdzqhazbit

Joqya


ln⁡(e)=\ln(e)=


ln⁡(1)=\ln(1) =


ln⁡(0)=\ln(0)=


Twuuogvhm


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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