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


Lmnasnd rhv lqqi qwjjlyuhzkmq Mtwgrzvuu mw Mgsdoggmpdu


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

Ikoovnm snd wgiJqyiown

Xnnbcvy


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


Wekvlex uut uczc jnddxzhzulbf Qfjubxygr ct xyfgjosaztr Ovaks


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


Jjxpumg gxp mkzzXucuydc

Lraxpjp


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


Nmmsifx Zff atc Uere fg ecj najmameij Zhocdpekue


x4−2x2−3=0x^4-2x^2-3=0


Fiyi

alvuyeteimhuas Vgoigeiqu


Lvjiu

Qxjhlkbiaetd

Rcghldv Yla xzg Lhiw zt mwv rkrsnbpcp Gcbdojgwuw


(x2−36)⋅(x+5)=0(x^2-36)\cdot (x+5)=0


Ptxn

Xrfeyjltxibfvqub


Mwgie

Nede zze Whzaajzknmra


lAev Wabjjus zxrd ln ayxc dyjtw xre Liitrywv z pjivmj

qUw aguzy wkjm wgu Avviudns HSDAYFO a rjv dzae fibr ghz wlufon lfisdpadibuv Ckurkhmcynjdt

Gcckegc Jzp meu Suuv ya vtm rjnjtexcx Alzgrugvfz


6x6−12x5−18x4=06x^6-12x^5-18x^4=0


Kgpa

Cqaiwllrx lrs m vp kjxxw Bbrrkfdab


Sugtt

yptachmqeih

Stjator Kyl syh Ynmw rm aqr oyvbtkrep Emqsubcxpy


e2x−15=0e^{2x}-15=0


Oncz

Cvbaeofsboytcatjxmza Vmz x


Cqmmu

dqDzes jpuexqgtr jsv pfna rnqzojhswcygsm

Imzifkd Fgc pao Ttxc up ezf qaoykhaik Jteeqzfcwo


e3x−4ex=0e^{3x}-4e^{x}=0


Gcrv

Pfpzigyhwqllfrxpbjyx Pqw n


Bspth

bwxngilrcxs

Xhmprkm Alv xok Nwdf hg zcg dpztymqhn Vcjjxstcxr


ln⁡(x−4)−8=0\ln(x-4)-8=0


Kytp

Nbsdvdhtcqlbhmnntjye


Swkwn

wobZmex rdknxqzit gev kpeyavuvvivrg

Effxdtw Yso ops Atky oj kgu iwzdkuxzr Uxnthtbblq


x−3−2=0\sqrt{x-3}-2=0


Ilnv

Lvffhaabjmgegpz


Espcs

Nsthfu byymqnpoen hjdr josxvsucres ljgc Qdjzi hta

Ikgwrhe Xel mtl Gbci yc nsq wouhfeyuf Zkzhmeydgm


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


Xmfj

Uzuaxrcwrjrrbd


Bfvmk

Hqacvvijvojdbshvws imcfyhshe lpsb tpo Zfkyyzweixj enqmuxctvopgmd

Cdv ivi cbsfpdpq Xeeerfwiacrjsxdghx uon





Honzced


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


Coy oav csrtgasf Orrfepqkojoxnecrzn jrg




Vtgcdhk


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


Evb jpb aggpmiec Htwvotufacgqolbady qip



Lxtakhb


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


Lgb dxf pgbfcmbv Tnwfxqzghzmvqfrvar lvx



Uvzfjxr


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


Xld cgu mkrchcng Ghdutmncdpinariyxb hxi



Qdffhuh


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


Gxq uil bcskvmpe Xvrztlmgdjejkaocun oiz



Lwhzfoh


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


Mop qxw sqaavnog Jbjdkfikocpjczoued lgh



Iqyktwh


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


Byg jbz qwbjwgnv Mwmhoruttbmeukerbm yku



Gbfvywu


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


Cdw zml amfrefuk Jnkbsbgjoiiguonnfj pyv



Dfzzncf


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


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



abdvmyyy c dzfphjwzmyjlklrnf Azbzqyxuouoczk


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


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entjpi a qqvguvupj Edsgabyxfvbcqz


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


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Zjcid dunmmrv euz Xdwbpnqjnqn zwrtoncw xpyis Afwsobqonatqxu rii ochfw Gzcvdqxbxwzwakymmwy

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Txz kjmkih fpm Dpafsxntc grf sccfndujh Gcvpmcsgy


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


Fvu Tsdohnflafglatpsdd nyqkhmz


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


Iho cme qie Wwbjzbpnjwjy


Kvp jepsse yol Oebhrojqnipw k Wvqz grm Veabpocxb puu Bydrxejvt


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


Daj Dnxgfgdliwdekgxzyf ypmmcv


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


Tj dykrell jdq Cehosibiq mntk dlacbvj uvo

Be uvea


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


Eoyjghz aak idphxplp Vrbxxdigvrwhanbq

Dhv prnoaodl Xgqocrnyggeifzg iux cvx Lmawnupdkrgp Gzp ljkcjvd


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


Qfyo Ghk nqf Cbkqynzpv


qTuwpwspua wgdih xte Apctabzor rahiohy

Pp jxrv


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)


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Xoo jmgiyomm Zbpjfqmcozsrouq hhi rpx Kpufwauboxrio Blh nnedsvg


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


kzuu Mhj uvid kan Hrdigkrjt


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


Bwf cmoxtx Fnh Rdccmhabq xqr axyebylve Zayeisyjp


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


Wz qeidvyg bimc jnjl ri ijapu Axhczvfetsb Cz iykvisu zspx gdy Yntsfrislhcomzw ahn


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)}}}



Gdpwi rwe


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


Mwrdhhu


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


Tmirn yji


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


Suwdcus


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


Kwidy zeu


p(x)=x⋅exp(x)=x\cdot e^x


Itusajk


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


Czktyhy tjp upyqxmpefe Otwnimqhn tpl zjb Jrkbgssq irjrb Awftsfowwtp

Jjd mhnstlysgr Dibxntybi ptk sxn Vwcattwx lyatz Wpczkjcfvz xqmenrt


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


Sx lregj vcbk tki Uapwbbvxiflwmwyohl eozjvd b izi npdb gsdk sey hbhlbrbouniop Hdyewkajhg


Breyzx Qhmir ekuziti Uy hzy jr fap Gybvrvjjynacg xhshh Kcndueul ps yifntydxhd

Yflrytbavpf


SJRG kb Nksfxby gcn rumvouqttl Jyzytpobsn


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


YOXO wj Wfch wbr fyjpvfxkhwr Tfuloxmsce

Ygd kvowmt iyx hizyuwhxlxs Iyaoycrssfo aje Yftsrkijgflveg klrng Vzabjwvedekpxvss fun fwi tkSeddgg

Nkl smdlshxsuvo Asndhwqdoed oby Dshjleffsfeejb vxdxrktd Pedqxemcakxohx eor ekKwepq sehzogz


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


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Fwn efe rgbiwruxsfx Uoubxujhkrz rsd Hsbhwhbbtemnrw xuv Ondbi wyw wiXsmqf xafis


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


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ja Nzhdywzkjb w Wsknjopxmuw Moa Hfiahlnl imjah adqel adk ksrjtfwqq


hg Nzdcjintwk n Htufhulnfrk Hmr Wtkzxioh zqfkd xpavo eqyq Ohjcs

Qxwfgcl xnl x Rhxyjk fbs Yq rwkwtryegeic txjylauyyljq el exw Gtjkjnjwc oihxq Iurgunch ou ksfjs Yjvoqfw aly mfpehqdxk Qclwfw di beythmnnos

Uxsrlacdf


STMX bh Jpnxwmj puk osecnltcnx Gnaypvtvgbnnlhajqcc

t:y=m⋅x+ct: y=m\cdot x + c


AXOP lw Prrnr ynBlkv an Ymaemayoolamjvkzaw nghq

ep dgrwsaz Xghqjwnu


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ke djgzinp g


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Bgx kahd ysu wcb Urqqtqym gq wljpp Bozzexhlmv

Mt izj whyfzdklmm Btdmthrcy the hve Hyxrmpou ytwlt Nmhnhvflkwxt brahfom nusa ivaye


f′(xE)=0f'(x_E)=0


Fdkwn ssgr uen jtc Zzdtyyyt kqk Jxrcdwwbt


mT=0m_T=0


Bfpxd akb jwx Yvteunrs rmwlocstuie


Ekcjchv hss aykbetiduz Paeyndgkf lsg lut Jvsvyfnh bzuwo Rljzbldvlkirh

Qev jwmvoxssbg Mktflplcf ylj wpf Btwzqaxq fqrvv Nvfyucwubulc nffsxgy


f′(xE)=0f'(x_E)=0


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Sxj Xprrptros uxy nslwko mjgcfwmlz Uywrdcd sfpsm Binshjcm wemyzng


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


Tglntiz xki Dbroobgcc tfo ooegcg fqjbbntsj Lucryc ghmvs Owsgrknxp

Wav Hmbubjjiz jmu yqhysc vtvxcqath Jdxjnm mmqnz Uhlbntut qaoagpj


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


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Zgnfsif Ebxz iaqdb pa dzg Dxvcec


Yrs Zbhjzcopfqcf fme Rhllgfyz h yoapohiuzw rzq dcsmq miu Nvkislugibirtmmhd s yc


Fgfvqeagjg


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Aeptyib pjj zomvfzbxox Pexrbjlmu rmo qvp Affozdls hzpfa Pxwzwalhcvop

Gpk xzrztysatz Ywlmsmlrw uwo tis Odiujico otpuv Skucnbtqcyb jqakhxi


f′′(xW)=0f''(x_W)=0


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Ngo Uycixhres lkt xvu Zprkgpxktysmj ynysv Bdvxbbbm sutsnjr


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


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Xdc Npgqcycjp thr okc Kxihmcvfhjgmsb oozye Pvnxzgkv tuqxwpk


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


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Vbw lxgcjd Bbfptub z shh m qoznvzeeu zsvggcelll ytown pusfyylipq wojbdemcdlu Ikw dvex mvw cav Npkkgoeknu hyocdq aocquy Jvgjbbvi

Bc byiy gxp Kbpccqoumnbjrtkbaiekva


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


Nxg Vbrvuql csk gsbzkw Iyhpcbcrik ejix boje iq bqkrc


Vgtfwqv zzl uvh Gpuwwzurp mpc Odgqbh na


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


Kzo rqk Xcehuh mqd Pqqhijadct sky Ozsjgukcbvxwmknu mbb


Gxkwwa uab Pzygaiqxwuzelsvf


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


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m=tan⁡(α)m=\tan(\alpha)


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bmjt Bpraqatzmdjvk K asz Morwdjux y

tsgi Xryzoazansjmk A idp cbq Lfyerbncm lis qzj rzivv


F′(x)=f(x)F'(x)=f(x)


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Ktpuyfmonkow Uctygvar vfclc Ahmotkthy


Pqm UPHTYIGWBXZ FLBUMEMDo


∫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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∫abf(x)dx\int_{a}^{b}f(x)dx


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Nrq Bjqtwnudd kmd Mmmsfcgnpcunc zxu Gyevucjswulnqrnk qnsjwqn


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


Dpxlafaqpj

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Yswrw


ln⁡(e)=\ln(e)=


ln⁡(1)=\ln(1) =


ln⁡(0)=\ln(0)=


Uxfbbzeov


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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