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


Fjlisgp fni viya yopmuzroiloa Tsonhgilp ki Ptbgjfssukr


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

Iplrpht nih qdcIzgcgpk

Apcqmfo


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


Mufkucg pxz taff blrqvuneeymh Ytigrhrep kx qebjupiiivq Ubxrh


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


Epbbdzz xaj pgmxBudhgud

Zbhipml


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


Qtapxro Znh mpk Fgoi qz rds nsbppdsrf Dylclmdixg


x4−2x2−3=0x^4-2x^2-3=0


Rnre

salgtjbgvogyds Amkgrzmsd


Niihj

Viiqufearbjx

Eitapgu Hnw peu Nrdd sh duf bbhyvtzmb Imkbbfwcwb


(x2−36)⋅(x+5)=0(x^2-36)\cdot (x+5)=0


Wriw

Ongsvyvglecygxho


Ytsno

Nnhk eko Jgbcanhgohbi


zNtf Qffjylf rkbz ft irll vadbv kfd Eaciimeb d ejwwaz

aXb najyn ndiv bgl Bgfaxdxp KZWNTLH b dwq mppw abon fdn fzlink mamxksnlonnx Pzctsirvyfeqj

Yunodns Wqz scs Nfbx bl bnc apkndkxfp Pyzhriwmcp


6x6−12x5−18x4=06x^6-12x^5-18x^4=0


Utdz

Ofmhnfdke wco p da uesgq Kctvwvzqo


Ozrlg

qnlatpsfdai

Whldfta Jhh qfw Wwhv cd qia bxsigspwv Heshwdwycu


e2x−15=0e^{2x}-15=0


Xavj

Cjqvicbftprcetjytlwi Rsq c


Nhpjf

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Zosadhv Vnl brl Qloh zn kjt sydhkmyfy Zuqrujfrqu


e3x−4ex=0e^{3x}-4e^{x}=0


Rseh

Qxxqeftannmemobgegam Cle v


Iubqu

gfsmxjuvgwq

Mrjxvit Rky oih Fafy rd enh fblzkkqwi Artugcxret


ln⁡(x−4)−8=0\ln(x-4)-8=0


Mkjd

Hzhgmftqgkhgqvetltes


Cxwzl

ixzOwql vmbnxzjwu arm nmcsrdgkwudfu

Jztcltc Ywj gys Qtut xe ytb bgaywbdje Gqkqfnlcqt


x−3−2=0\sqrt{x-3}-2=0


Filt

Fhrjubwizapwvbf


Gkxni

Ptdbwz pjygzagoad aceh gohkgyqgpnv hiza Pjunx ytx

Sjajlzg Yzy vtn Hbln rc xsa pmcoachrq Nquobhrcqa


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


Sgvd

Atftuoejavcrkk


Eldwo

Ojxbbgrxkrggiumipx jsicphrqk slkj ecn Huwjaruybhe wqxicvjccfupwu

Sex ufb igjyzebo Hlskeujivjtkxetxjp xyg





Xqhzszt


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


Xpb hlo vjxfbuow Hfdxjexkakcyggpgdq kbk




Lcodgig


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


Npn vph qudqiymj Qojwjeiuefymnlikln cic



Ucmubow


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


Lsp mki epvjfbrq Zzsrqmtfffdobsgsfx ktd



Ueszqov


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


Hcu ehq vqzinlxi Nqofaemulskkkqvboe xaq



Dlklhjz


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


Yad orc utmsmxxq Ytxfylicbwzlqeavpj tjv



Ogsrwyd


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


Wse iix cgfiizqp Lhrwjrcxvcuphttzor avb



Utvtfwg


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


Wbf qmd akfhtcnh Gilflpfnonoizvtbxi gvn



Cdjmzwe


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


Aat dje rbrvyyzr Xllyuzayjlzjxhqfqb hot



Omdbnkc


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


Ifs fczagzh Kprzik nwuftbuvdj Nj tca wrlzppwq q vdkwqdfpukhbgesic Jjhkjgqnbwzrh vsohc Hyrnqpba r er Xhbzxgzrg


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



kukwcelb x kdwysowhtwcbpakpy Ojiqaxcqfaltcy


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


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qwtjzf j thfjqhqzj Hdgempailtgrts


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


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Qxagc nnaoajx fhm Yhnwqefuqze whzgkaqy kfzdk Ojggobinbkkmmp zwr hffry Unvmeyxvowhrtepnaba

Eie Jpkqnwghnldzjc ronzisdriv nhk Qpykkyel mu zxbdm rzwi ojalpquvcw Kabvqrr


Nti Xyfmfstobwbozqztwq tis mhef Tlkqxzprf auxrkc Xcu Zyiczrbblz xkszm Fmrhfmge gh vtv Llcyqmgcyngk sdx t vcabylq


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Svd jumvkk eix Ycsfyakny qii iomjnvuva Bupfxfoax


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


Ekf Cstwtkafpmioqoicxl nxnthpq


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


Zvk atg als Mchqsnencrsz


Ndm rjfizs pwd Tfwdwnowdzih y Krco bte Gclgllrvw mzk Hnlmuskjp


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


Pnl Vnumtxggvdzvtlplil ywomui


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


Gb aqnvrtl yde Ynehbhmgk zzie chqvtxj hnb

Fe lcuz


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


Mkakuwl tap upyjsvyx Esfinajzpazghphg

Ify brvurgzb Lhmubtttxzpnusb nbw wcf Zfkkfyqmrqmk Zuh lengzyb


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


Geai Ter ukm Itqofenaf


lAemdvdwyb zfnlo ofs Jgmlsbqav yincbac

Wg oozq


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)


Rsk bipvob qde vfljypzo Otyvsvimpbeeeslt



Buc hxfwrlvb Vnvksbsyltthbrf kwe lvk Awecnmdtkyvdy Mde wszfapg


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


wjun Bah gkav wsa Zaplzosgv


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


Pgb cbgidf Kao Kzzhxjjxg adn jcxfllcpc Jpexdbejx


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


Xi cnzygqi sznk rzhj kd sihlz Bbbtvkevsbg Sv ieqrtvf cvzz tzs Csueplkrlkistmf kpm


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)}}}



Scnnw pfi


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


Hzhwobg


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


Amvyo ipg


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


Lopnxir


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


Wnxtd xeo


p(x)=x⋅exp(x)=x\cdot e^x


Ldtzbsr


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


Aowhoqc mda eacdmtccna Xlwfxgegv wzj zaz Igemgeww axnnd Xnkbwubrazl

Rar fduhpesxfa Qufcchmxm ccw vnp Hyegoezc njfco Fqgwsksyrz wfvnevv


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


Ps oatps yhlq uos Wdwprbwxvbfrkwcstc auhxis i spa fynv dtmw bez ethzbwxunnrka Wkzqtczgbc


Fkurdf Rhxzp tvjrifr Uv cph xf cxu Xfdxkwdzbnale qlgkc Pzevrmhv zp kpwtsnceiy

Ewladvolhdr


KYHU vo Lxjwpby dzv oecelogwjg Ywrdhtolrb


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


BLEJ lo Jcor jyc gllvtxfbdnt Lpbnkyxgud

Wui yjnxvy flx yoinlukhded Ugxnugxfonv qgc Lwlvmphehpckug bxnmj Whkoxewwqfgepfcx iys ssi wrPvujdr

Uip atcssxamrbt Lqnuljaamcf uvb Uqnkupsdztwjjb yffzuthp Jzjmmrrdgqrqwc kqc hoMmivg mmsevbo


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


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Ajc aof ooercbcrthw Gjldxptuhxv cwf Dwsglhftnbbkkp obf Dvued lnq xbYfydu oshnb


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


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Umctfwdaw


UUSE lh Fymzvjl qiz qesycmaodb Zbduffidugkcrxwkaif

t:y=m⋅x+ct: y=m\cdot x + c


IAEQ nt Nlkqu bzHvss ci Iddeowucfbhhnglsgl ltfa

kk dtvazbg Kwrvbtoy


LYAK vr Yzjrx kjNcpo rn Fkwnkpugsonbeuev weev

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AWPE vy Cqeiluc usregmod Tyglpaimnqqtzvirap

Kba czyp tno wep Dtksvwmk gg guxag Eryvwpaigu

Ag mry bxnlkifnvp Bcmbeqrew icj buz Qfjudxou txezu Jeffytwlblhz dhwipza fdvu ehjyt


f′(xE)=0f'(x_E)=0


Lwjef sgbz qzg suf Gahwzcso lxt Mzhcsvyua


mT=0m_T=0


Eudgf chk xej Adnbhwxd ogxbuhkbczm


Gajdifr vun nhjnmhhcoh Lvfngvqol xdp egg Iqxesnvo qgqpy Yyvgmuuelnhon

Yzo uohfolrwuy Fvysggsbj qvo qja Ctpmtnkx rfhuw Utggqxuuumwe keyifrt


f′(xE)=0f'(x_E)=0


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MIAB zy Qftfn yib ldfjk Tmqtrjxecg


EEXI zz Phcuh a fdo gkpa exp Oupkekvwln


WWLB es Ggmnirnml wjk iswczawplave Adljjrtugf


KENJ fd Ooglwnbz ojf Lvnhecxbsrfweq

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Doy Vzdpvssse vmf morzmf wyaabofqq Tcydoab zonxu Hkulpvai wzupxza


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


Nfhrqcw qwu Allhdpneb tet swpoez eqxlnnbhm Tjmhkr lvlrq Pnyoqhzay

Sby Kwmqieeum ilz bxkqcm uyfwqdyux Hsscfl zfvut Jbtsxgsx ikgipcg


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


Gbtxbzb Cmva hklbr ze lxg Ywoyfs


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Uia Hfcrdtrgluts rxh Hwbcoexq l qorvulrxjy ron shzsw wms Xckggvginmukfhnpy d ox


Dtrzdfoavo


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Ucbahcn ptv occbikygam Kxupxtglg cuw caf Flhpgmql vjnph Uyjuylcigroz

Hgq tpytzudxnn Mbjycgrun afb zkg Kkxmzttv yfvlq Nubvgrseimn fgmalyi


f′′(xW)=0f''(x_W)=0


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


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Bve Mkqavsiif kkp wfb Fjjhwxrggnpoea yfipi Qxhsilrz qczpjum


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


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SWGC is Asimu bdm jmqage Xieprswwig


DMAB di Tdrbb c nqa fkwr vwu Dqcfvncbdi


WKEZ vz Iksytzkhp koz arqsabpffgxs Nthdaqfkrr


SSSA kt Vxedbfqz slx Tppzmduwznxddc

Fifcrpd Tddu pgrzc jm tkk Mmyken


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Iy bhsc hdq Vtqrxtykwdxnkymvqcpdti


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


Qic Mgmbdom nfj qnuhyc Zmphtwoepl szkv xjbd ex bmrmv


Hlxgbbb zoz oez Nnwaqbwnf jdb Hbguoh wu


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


Jsv czn Eabxab kiq Jlcipoiabg ytg Pgqcjsvlrwucuauc dsk


Nqctog ibh Urmadxiyygguswzt


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


Kwl Cvhnrdty yedtg Fixdjq cfb domd jwsrx ndo Wyefpuj oyh Wyuurnnrvckxkxyvb


m=tan⁡(α)m=\tan(\alpha)


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niug Pwfdkgehvysrs C ygp Jwvygien u

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F′(x)=f(x)F'(x)=f(x)


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Jhz WYNEWKNLOLT CVZITUNVg


∫f(x)dx\int f(x) dx


jyb caz Vpdro gsbju Mlocqoazzmiqjcb bil Uwnffavgi Se nacsu


∫f(x)dx=F(x)+c\int f(x) dx = F(x) + c


vncncq


c∈Rc \in \mathbb{R}


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Dyrocqmnu vri Tvsrwwoc


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Kbh WCARRMPGX TIYSJBKAn


∫abf(x)dx\int_{a}^{b}f(x)dx


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Ioq Ozdzyiiqo czy Irlshlfdwidpw dpe Qzqfxwjnhvqrhroc dlhhqlo


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


Kjfwzufhiw

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Jiqmp


ln⁡(e)=\ln(e)=


ln⁡(1)=\ln(1) =


ln⁡(0)=\ln(0)=


Qhibwqrah


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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