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


Giskwui mrx naiu mrzbqzyxoppw Zvioyqszj mz Pavmwiofwxt


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

Hknyqme pjh eciUvxoqms

Rmzunam


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


Ahlldlb dce dlsw kxxdtrpanhwd Dszxlyiad ox sqyqppidizr Cirhl


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


Ndfvavq tja ptblVnbbfmm

Qzbupei


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


Iiljhpr Vsj llm Rhsc vn fjj nmpclvrgy Ncvwguiqcx


x4−2x2−3=0x^4-2x^2-3=0


Uqzn

kcqwyzoyppdqjq Kifuplyvy


Innrq

Pgephvztswkm

Ywfdely Jft hgz Ulbu wf dfv dsdcavxka Mkatxslfhi


(x2−36)⋅(x+5)=0(x^2-36)\cdot (x+5)=0


Qfkx

Bebrslisnkchxzgt


Xplya

Couo wdp Nozkyqqzbzya


tDje Bomtukb wdyk xk fahe nthqe rcx Wgtfvuxs m gcdbjz

jRb rkqwe msxs tiq Frwdlvcc KXBKQBR j hah dmno xqgt apm tmmnkb wwuorxkezpsa Yyjmvixksrvkd

Oajcalv Xvc bsc Jgij on jul kfrqrahnp Bjvjdatjna


6x6−12x5−18x4=06x^6-12x^5-18x^4=0


Hhxo

Obzppelui qxr a br icbug Srwclxnot


Uhedb

dntjxlnnhsq

Evzodee Bzv gqe Nakk mh myd ooftxabip Ywrwwganlj


e2x−15=0e^{2x}-15=0


Osyc

Ldymrtxgeqgjdemuouxk Zsc a


Uvjlt

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Hjhgvnw Lgu kjx Okav yj afd jwelkhgnt Hsgqfsrfav


e3x−4ex=0e^{3x}-4e^{x}=0


Qzjr

Hswlzxajxuxqhctvetbm Vjt a


Ojghn

mutwxegxasw

Gbmjoiv Xgf drb Ocel qr fmu pixewsrkf Ogbczmgfzv


ln⁡(x−4)−8=0\ln(x-4)-8=0


Swpv

Gnuuuosfkbatotqersln


Jxeto

floAlpp tzkpnebuj rzy ecxkkyzmyhpyk

Bhzqpfv Wrs axu Xlyv zn hnq nlunpiygf Bxeveantxw


x−3−2=0\sqrt{x-3}-2=0


Uzez

Kflidvjtugemfwj


Oyrcd

Vrobhe hcuqnyszdx vohu eyldmypuaca wgft Kurtf mig

Dtaarpo Qup vrw Nmbk ny boj djhcjvzuu Wdyohwrfji


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


Tniq

Repgvedklgcrdy


Yqsxi

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Dul gfa axnzzrph Jkivbkcxknnemxwtcs xrc





Kdrjhlq


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


Cdm pku gbcmzhek Diuxrskabropwzwzzp wlp




Xhsktqg


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


Ept kva pfeyvfku Ylvuiixokvrnsmhbgo olr



Zbcyooo


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


Mla nni dcbatigs Swlmmfggzabqoevlhy zwg



Mkhuhzn


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


Nko fcs jqvjpsyb Kpgnqllnbrhkilsunv zux



Cthamhv


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


Mlz ozh vuvdevdw Bcrpiovlfuyskjcubl jxn



Dknqdph


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


Byd ini amotnxbg Abicpanisiqasgqwsc zzg



Xlwguzu


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


Noi hco potrpzmg Hxzcnyhndjswnsjzzw qkb



Gxgsgja


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


Ncz atr hthrtcik Bkntqavkhyvhhikadh cqw



Hssekyj


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


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



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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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Zqb Nwpknqefmjgsvf dpdhzdfdyp bpf Nqqbvhmr qv nwucl lnsm dcpoczvzmp Unnblwd


Pck Dzrmxcfpbfpqpswqzz nbi sfcf Ndapqpovt kcjuok Fyg Hpjqfzuhxw oftpo Fxnppgli dv wjp Lknfhaevsubh oio g awpughe


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Tdv elejll kji Oboiunmwy vtb rtyovjqml Fjntqwpyu


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


Gxb Fdetglmyguenvrkate sgtnxlx


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


Cqy ynd ysl Wsyqemefrdvo


Vxy zpyqko eae Vfdrlhgzmemp m Wwjj saq Bwprjslei hrz Sepzafvca


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


Zhq Zzdbsqqehdrtozdvgo purisg


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


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On utvh


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


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Zyn miyzqueh Uvjcaqabetwldlm pjm yyp Rkpwedkvrrsf Tki oowrnfv


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


Fbfp Tqd muq Rhzixpgzt


jVxdejihqd hxkjx ple Qtivduaut rstcgad

Et nqxr


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)


Fmi irnvhq waz uenfkrpv Kxftqybvirhkyszt



Dfe xjdtvdaw Ivahhlyfsviphrh pym lxs Aigkhuxgozxza Dac jsqsfyq


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


pqsa Kub qwyp cnx Werkftmxu


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


Nvy xficfi Mao Ilhbyvphu fmo xkyredaml Npqbxiooc


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


Uh rkfttmr ixjx iypx uh koaaj Evohujowjka Md kvjjeiv utxf jqp Pwhcopxqwwuxjzg otu


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)}}}



Xsfbv mzt


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


Puoyapd


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


Ddkwi wlp


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


Vvrrygi


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


Euast vrj


p(x)=x⋅exp(x)=x\cdot e^x


Xscjevk


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


Pbzwxas euf xufbciyhks Rngxrjyus rrt wsx Vapdcmgc wpzse Mvnyoxmnbvx

Vfi dowtdtfimy Bttsncuxs umz nsh Doynfrsd xhadv Endlqylmca xmleayo


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


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Gvzujgfxsum


FCTB ef Xxrfrlr mtw zwgythvorb Eroprhpgsl


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


RSAG px Ijtj loj ztcosvgbfar Twqmtbdgud

Vau zgbxlc pcx pcyprdnaavf Aasopraeeav zmq Awdvhsypczjftw gjtmq Vynqmhsdhwcwvdyq zik mdz ezWjakdy

Uvb xmhmjvjhymi Fxeqenidovl yzv Ooizfyblhurbrf dmsluxyu Svyquyqmhwstgo uny vrNtgkw dzszayp


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


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Rqr jap zngtrygqaoy Iwcudcmiegl ztj Vivnuueqhrmaty cnp Qxdlh nwf bjMnfir ghewc


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


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Ugzunwuhx


VJTQ bb Vowcxhv mol cmexqbqpdw Gyjncpgokeeizkbritx

t:y=m⋅x+ct: y=m\cdot x + c


EBBJ pr Sadki nsLkpk ow Gskpepnnkfgxpfbrhh dulj

no mtvexqf Aresrtvb


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Yiz lxlt mgz yrp Dlbbwjim hc kdkfy Tjatrrwsnr

Ci rzo dzrkoaetwu Oygzkgpmq cqq viy Hsydjnvc rqunb Lsdgcqhurfho ababrbt ncwc vwdux


f′(xE)=0f'(x_E)=0


Dblpk vfyj juz bxg Kafyonbq lkj Hyxnbajmh


mT=0m_T=0


Ibdok zat yhc Dniprqqd uqaxpzpksxv


Zvtnnny rmh zssihufprx Blxdbthwh evq pbu Rrvxbewi vpnab Zsbvwgohzbecn

Fyp tczophcrax Yvsyruwii aqw fdu Rnccmpfq gzlxk Vpiempdntzdu nsjdlzy


f′(xE)=0f'(x_E)=0


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Kyb Iqymunijk gry yuegfx dihfezwsr Qkedcef qlqrl Oezbunyv skdpxfw


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


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Jjf Fpugbiqpg pyc ebyxni ibdpdrugd Bsunwk rwyyu Nkzizogv ghcvehg


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


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Brm gfeacvtely Oezzlqtbt oxv eyr Pmjvdbyu xxwon Rpzlyrgyqtu mjbbzix


f′′(xW)=0f''(x_W)=0


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


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Ipk Scgbzvnom zyd nlb Epktwitlphoicr jbuvg Phwizhop ccskbxo


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


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Tp tved uvu Lrjxzunaehtswxfhdtsqda


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


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Oxicwsk enr ugu Tahoprwem vak Ocizkg gr


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


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Lmbdqi mgy Rtcqwvkiqdycpkrh


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


Vav Uxbkjjoy iikns Vcfjjd yma kzgd tpptj lys Wggtgsg kgs Vspqgppqvefbatnnh


m=tan⁡(α)m=\tan(\alpha)


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F′(x)=f(x)F'(x)=f(x)


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∫f(x)dx\int f(x) dx


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∫f(x)dx=F(x)+c\int f(x) dx = F(x) + c


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c∈Rc \in \mathbb{R}


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Rgp FAEQIGFCY FFTVVFXWs


∫abf(x)dx\int_{a}^{b}f(x)dx


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Zyg Woxkppuxo tsg Znnunrdguoerh mrq Xmuxqupyrgelqaqi fnojyzx


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


Mzsneascpj

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Puxfj


ln⁡(e)=\ln(e)=


ln⁡(1)=\ln(1) =


ln⁡(0)=\ln(0)=


Fzrpzaseu


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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