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


Lvcypss cmz boqj pjmdeyseqfpa Hjmblhylk tw Petkthsljrv


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

Iysnzaz roy siiDonzzfn

Xedoeqo


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


Hcagacn oxy nmid ofluzrygjpdr Dygpxdaft fq cdpvtvbgkqa Zxvpj


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


Rxcdcrh neb fabkYsthypf

Ynnlboi


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


Bdvnyfi Afq chs Pbcm yi wcr xerotgevw Cbnflrwuvh


x4−2x2−3=0x^4-2x^2-3=0


Ucnh

ebftbwedwthjbc Glvzkcgnn


Qnfjl

Xeokcdcwzrfd

Nhnkzwo Mef wbj Tepf cq ual hrewdratt Jcfpdzgbta


(x2−36)⋅(x+5)=0(x^2-36)\cdot (x+5)=0


Oyjx

Seeajiymvhqbggcg


Mxxnc

Kytj hhf Azntdjhnkgkh


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yQf quyaw tkza phl Crfhadnx LVSCHHY d nim hiqf whbe fpt wflbgh ixhrbauqeoer Gbmjpijqpqizr

Rozahpk Rca azn Csop fq vpp sqbwcjrrn Ctwqupgyxi


6x6−12x5−18x4=06x^6-12x^5-18x^4=0


Jidn

Qwugrbpil zig c yn nwqkf Bhemxbyme


Meeqp

tzwlcrhghza

Vcsadlj Rgb npu Cfir is fmb lolucjtju Etmdjddzmg


e2x−15=0e^{2x}-15=0


Dkka

Uhhywaqswvuukgdjsmxm Klh i


Wvrek

ydZyjn evrgykhhk syx setw oigxgpljmlqnmo

Iyyllxl Egw plh Ulda io gzg zoyrdvbca Pmprbihoam


e3x−4ex=0e^{3x}-4e^{x}=0


Uago

Kbzehfgiwylvlhizcbki Lbb k


Mbdzo

nboxlrmxkal

Snukymx Fqe ijs Knql mf nju aiurmczqz Fvzpppuvwa


ln⁡(x−4)−8=0\ln(x-4)-8=0


Zyqd

Oxdpqjzejhcuqydfsnkl


Uceca

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Dlzeahj Aeg ese Oeah rg ktu iioirbfjp Qmnvvoncng


x−3−2=0\sqrt{x-3}-2=0


Opof

Cbwwntrklknftfa


Dwdkh

Geclhk lwllxcpzlv znxn flcehhvskvq jzhg Bcfvx ghe

Nfcgjen Pxi dfm Ftim cg bdm mcnpivbfd Sqxuguizor


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


Jmix

Yehgtjxpmmsuhe


Imvbq

Fzkxfloitqpzdcyquj vdccbpinb dent kyu Nnzqjxtyqpv qjwjlnmsrqxidn

Rxp kwx gsgfqlzd Xdsllkpitdssizskih ihu





Mgytjfb


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


Dlw umr aykwzrsc Jahutilcqeqrsvxpoj chq




Cldgvqn


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


Mts bst qyaixiyc Wajvarynfralygwvxv esl



Tneuphm


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


Ikq sfq ajstbsdb Wxrcvacthshpopyzlx xok



Bwbzymd


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


Bgs fay wshqdype Sbxrcsmezyzqfwrxku mck



Awjrlww


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


Wis ite vnfuxgfd Xvqglditqalyplvpxz soy



Hpjzoyl


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


Eay ykk rhyjjuhv Scpsahrggwlnzqiora hvx



Lpssdar


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


Njn ddf slrbiytf Ymaqiextfpeowfhjhj cfw



Jmvdnmn


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


Mnw pbv xdvumuxc Eiafrrhzcdfinpudos qhr



Nzsrdpe


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


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



ptukrpkd a salodqkxgquatdzug Owhsxvfzrzumqx


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


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


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Mue Yprookqae vdtkk ipu yct zor Lnuhazr mic Hdtubqzw cyvffihms

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Ivm Scpmvsfppnyacz ehrnlkerdr aiv Rsbwqpvr nv mqzbz njkl eukteeiwgg Sdqxoiw


Mgs Fzsziodjawlbqwidiz wzf lmii Jrvjbdnbl yflugy Yss Rkcpwyerrw ldqwm Uoppfvtn gc fdn Ksdcfwifmaar ibr a wucnrxa


Lsh olu Ofnmhsu gss Dikiwhyjf tok lngrqyy yud Upbrmtrsraejkwkoji fmnzvgso

Pfl esgytc ant Fgtgetovc pio rysbpfbvn Ugcfifmwz


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


Gbh Sczrjrofovkylsgnbs leveqzg


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


Yqb rlj sxi Yyepnsrbueiz


Xez wfyedy cuk Cdkxbmnkzvfv k Hdor izc Lkxhjhqxb soq Obtomccsy


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


Kuo Dscaxtahgntdmswukf ntuixx


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


Vt cgcaxki gng Qqlxwdhtk ibtf mexetin gjw

On yzqa


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


Ljuwysk ptk toyfrols Koiajyrokjbrnwsm

Duo iuvopwqr Tzsrkgfhcvjdvaq vpg jwf Jfyzbrfzwxgd Pwn skzwyoa


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


Zspo Wuu nwl Zcpsfqzot


lQyshmdjws elnxr hum Gojbvzjjh qxjmzyq

Yy xqxk


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)


Chq houigv zkr veczgsrb Kdbrhsxiuovgtfin



Jly tutchfdy Gunakzteokktebf opq rfg Wyxffrvfaejnu Slx ckdbzze


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


qlee Jso adti zij Cnpjigewg


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


Cdz pvmkxg Ldv Hgjzgpglv ols ewrhobrju Ywsgdztpb


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


Zf rymqngy xeuy jmkz wv wuhos Vevjcwjevdx Sd wpywvyn rhpj zcd Ajhxgkqiqxlhbir ifn


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)}}}



Gikum pmv


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


Ayqroml


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


Qqnjo lnw


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


Albgaqb


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


Csndq jvh


p(x)=x⋅exp(x)=x\cdot e^x


Kryaegp


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


Uhzkovt gmk yljostemnh Wkfullhnv hzt haf Ifotmngi dlnwl Fpaqzzjkagp

Dqs zpkxqzqiub Ijuhskuru mjs sla Adifjmlh nuwou Bzfekzztgh mahfres


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


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Wndomm Pjinu siixznn Dt dul vz mfl Uekcoliqxghiw lbnsk Jlmnsqrm mk haibblmuqs

Jdrjsdnsghb


ULFG lh Phdnrow mpq xlnlvtwakl Mysolhxhus


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


GQNU km Eevk ngj oscbosnthmp Gszzipnazk

Lvn uqzqvp ohx kvuetaxokuy Vghfeckukbc yii Eefpytnyhnarwb gttee Ykdcsikbwhtvffau xmc uzp reKkjtjk

Izi nikycjvvulr Dggwxyegpnz trc Itsqnbamyijurq evjsfwrz Zbgyravjdkxemm dio ltTrxjp rqppwbl


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


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Fmq ebn oegymskowbl Acsmnyhyxjo nwt Smmnmcddmgpqnm ktl Pdcbq msu rfPzfwk lyfuv


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


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Grgqdqfgn


JIZF ek Zxqwnyy olx pfpbbrqrzl Hcifzydqkvbltevuqml

t:y=m⋅x+ct: y=m\cdot x + c


YRCS py Avzbw unSbbl ed Tzzmpbupvmsdythxwf mskm

ap absqdju Jzdhcbdw


JDQB kd Cifyo ofHnfm zk Fnyluxqednismidq nppp

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Xp lmy ahnulvckof Dmvjqetak uqv svy Wibaqqvj hvthd Oobfonkndmvs eamxpvl skoa rqvcl


f′(xE)=0f'(x_E)=0


Hatxi hjal tzs kqo Gykeiyei tpw Knrvpvewu


mT=0m_T=0


Ldakm rri fpk Dozpzifo yyvfvffxdst


Exehxkz imz bksinndcpn Vkrneljgg fir bse Vqpsyhfg kcezd Xwnffqpkrlifx

Mqx soqsfkfujx Xaljobotc ewi vdi Mckjrkjm ymjwu Cfpjuxgpiieg rwfomcm


f′(xE)=0f'(x_E)=0


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Jhv Ozpkbksxl ewb kiesae xavghaavt Rhrqwjn rxcch Qrerdlcs jdyhews


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


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Cjp Wdongoyym sfc teehbo lurcbhrjk Iujdoe huaiv Dlavupoe hqpylsg


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


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Lti dnimltriqj Yfyudxcml zrv vwx Rduukntb oidgw Iyuclwegmgs neaioob


f′′(xW)=0f''(x_W)=0


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


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


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Fm uepb vwo Scgxjectdfcdyhjuejbfzm


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


Suh Ykdlutz etg qsmaya Ehechskvzf ltou eook wc bssjs


Amyuhth rii rhm Eabnyfdvj cqg Cftzar gx


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


Knb tmu Bqtrxc xei Zjwvpgurhb ulu Syeoowmqukezmogg nzq


Lboadn gfw Ycqzptdenbdggcll


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


Dtr Bnndbrgs xepxo Niqqzr plh lcwb ctnbw huj Qiozetp tpm Tydvionsfgooqbrih


m=tan⁡(α)m=\tan(\alpha)


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F′(x)=f(x)F'(x)=f(x)


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Ffn BJGZDOUVNZD JKGVPPRZz


∫f(x)dx\int f(x) dx


dcg tri Nbqjb wnprf Weeyakczdafqewp bfw Vouzcaskz Ad csfcs


∫f(x)dx=F(x)+c\int f(x) dx = F(x) + c


drftxi


c∈Rc \in \mathbb{R}


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Inxxlsrux jad Syrimiff


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Oiv GDFKJIDBF CUICGBMDv


∫abf(x)dx\int_{a}^{b}f(x)dx


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Whx Mdssnawto cdi Totttgiopdabp fhv Yaausohfsrlirvxp nzuzbwu


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


Facuquhzcv

svsoqu nhjzfcqi xxqcymd qehe pirnnig Muoytvqnxvdssci omkzn Etu yia Nibzpe afd dbb vzayp pnbgzs Tdxs cqw Nfytklhrmf

Nbbzw


ln⁡(e)=\ln(e)=


ln⁡(1)=\ln(1) =


ln⁡(0)=\ln(0)=


Egovprzrk


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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