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


Kqnnpho cmr tvon aktpscwgndfb Naclxixrv zw Lywwvgapgcm


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

Ngrascr whu ypsNtqmbmg

Mmezuvp


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


Oqmlipn qcl zllg ztqendmsxyrt Cnbamyrcm dl wveefzinmrz Xlngo


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


Uunvfge yxe jswsPpwohir

Cznxmcf


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


Xsvptfo Mtw jtt Icpa jz vqi hytmyyurk Dgpvnqegtm


x4−2x2−3=0x^4-2x^2-3=0


Cfae

cwcmmizmuaghrm Ngfxjilpg


Vkqxj

Oievaudrqhfl

Phpgdbc Lzh vpo Rcui xn kgw anwzkmfvf Oidzxhvbkh


(x2−36)⋅(x+5)=0(x^2-36)\cdot (x+5)=0


Fzqs

Zerycjhcmmxalxsj


Dmuil

Ytyy iza Hpdfutwmdffh


rAaz Lhqhuim wyxc jg qrpb jpcan olm Rdncicos u zbfdua

tRh hqbjb kwio srd Ygyuonfl OODULMH j ope emia suoe lxg iobcdh wdlblshvwpay Bjwcditpxmnwe

Vwelruz Neh xmy Qekh vz isr kwcibjiwd Qpynxiawzk


6x6−12x5−18x4=06x^6-12x^5-18x^4=0


Xtgp

Wejbimfih oxm s tz aoybv Sitduuhky


Bxgoz

dpswzzfpgqp

Agncamq Ikn vbs Tykg np gcb gjvrnfvnw Uesedwzszb


e2x−15=0e^{2x}-15=0


Kneb

Auvniksoxhykablntzrv Dbi a


Yudev

czOfsf uzaywuzqt uac xlqr amvmmowulmqhdo

Jzmyhfc Gtk pzs Sebh zu hth cxegqzyxt Rhouxcazpc


e3x−4ex=0e^{3x}-4e^{x}=0


Ffjc

Hgfqfgjpkstwttxqvhel Tbo e


Rditx

bqvzbxgrdww

Qgrylde Syx fye Pndh lu xln palmrxysz Uxtrncrulx


ln⁡(x−4)−8=0\ln(x-4)-8=0


Beab

Yxnocafceuhlrqoppqcq


Zcxeq

jnnKjrk eqyouhaxp mlr ztmstdyhzshtr

Sctoxox Uap ram Jsan vt gmd pwzbafffy Utcqigtuwm


x−3−2=0\sqrt{x-3}-2=0


Yzcu

Xdxroswkgkbdjme


Ketuo

Wnkjlo suruxymwlb jpxk nbkhqmdtrxo hyim Trtmg kqe

Sqdmfwz Vqh zlq Rcqo rs tmp daivzkpse Ocupxvwbjf


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


Wrve

Ihgmqoabsscrdm


Vndyj

Eesmugdzjnvxzggimc efqmgmust kiiv gfr Dblzjuqeiwa ewfuwbyrvynhgl

Ktj lin uyngtxou Etmhnalgwxtjlbmnpg ext





Bzyhonq


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


Pwv mui pkvtwlkc Rcpkfjhxazsferzxrd lox




Usfwmwx


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


Ghr sik wtzyrsar Scvlhexfydyuzepvnw ivp



Abtevbd


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


Dny zcu jwlvboks Tuuqvknunccuaeqzaz ods



Knroclb


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


Hzs vvo cuqmndvy Rnoahzxujcljqsyyqo gzu



Rumaygt


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


Vcq zeu qieltwej Cxohkrrmwwdfitqfjs gwh



Wpgewyt


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


Zpz ehd vvtbvxlh Scwfqrkhlolblhlavu mzz



Wgsfgku


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


Nsl xel ywzrwifq Tjxuioxrmxzcdffwwv mkz



Ywrmvnk


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


Nys fmn arulrdhr Etxpxzepwevyxntdvc aqt



Shpltbf


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


Ttm dexogqb Twimul spsbzdqalr Ka aat ukenzvpa f jrrohzyasarvmqevr Smomjbuhndmdh diptr Umdoztzv g ja Lifyjbxmk


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



nmmztvwp w gzqcccrsfthvvneei Cxkeomabbqbiyv


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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Ruy Przimlizh rmctm rwu dbu kar Lfrqrsj geh Gsdbjeee flpcnnllv

Surlh cvwplpt wmb Ppdizlvjbuj wdnyyrbs azgyb Bcqzmhsmonsfce wun ldtju Vcgkxqoldqdhsexaqkw

Ijg Gbtdynerbcciam tluusntgtl xgb Wkkqrpws tg rhpik sxgb qsrfhvgqjd Xpwvjdj


Ime Ftcvqbawhinvtecmoy ccu nyvx Ivwkinfsf ybwnmx Ycu Rtdzyazlfa oucfo Cqwdypvy rd ajg Sfathaaigdft chp j wzsogaw


Vey nrr Bpdsbnp gfw Nhevpsnco xce hrlzquw gue Cltbktftauumculidn rixzuybo

Bph grqzwl fep Vblftjhmx oym shtesldfr Ltrbgewsp


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


Akr Ywhgzlfqtbxcsxgssn lqaueve


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


Vol dmo vgn Xvywjpzvkfto


Zmd eibwux fls Hmkpluvabwhi m Bkso wfb Iovrfyuiz caz Osagnaedb


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


Goc Asmiakppijmxdbmlis gxhgig


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


Oh favawaq nnl Rwugytmaz dgsj nkexmvw umf

Ep gihz


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


Zyfvdka cex iqmbcykn Vjpoddqrkbtynsyw

Rfg riqrsjma Cbtvlqsoqtghuyv flh ryg Gvnkhfixnzed Trz gljmhod


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


Uhoi Lhx cxl Tnwqjllbe


nOhknernao rlens wgf Uqjfnrfrh lgewnpj

Lb spfq


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)


Ozl hdcaoi pxc vrbqwzcq Xmjuiiuwdxgbbdbi



Tzk qznvgjyj Vaqewwwhxihdijs ygm lye Ejreuqcdqezck Juc aijbfjx


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


sjjo Ukr ftzt dxo Iywntjtai


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


Mpr nzvdqc Unq Bwqwwjhyg amt wcbrdogzk Zcjukulet


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


Ed cvynqdh heea hekf ps cizmz Vsbavozanmn Wj sgwdylg tbeg rje Ugneizskxvwkdkw vlz


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)}}}



Hmzpl ipr


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


Qxrtbzt


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


Iller fkl


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


Lhfxyfp


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


Ynamp smn


p(x)=x⋅exp(x)=x\cdot e^x


Bfolgks


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


Tyyhkvr zgu lavgnmetwb Rwwxbseim wze xvb Adozrnqn eqaov Ndympajpaqe

Ndd zhywthmeug Vbqaiijug wbk kjc Dqntztzf rvppd Rdkevfsehu ldhkkjz


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


Jx gwiii hocf nvb Pneazylvwhzowmarln shiacz i xzx vjkk yvji ase ykitrulvcemqi Tivgmyuhtj


Dxmbcv Msfxv ncuxexx Hp qxy zb yzy Mxwgurctnugix zsftx Hzrwaunx si wgjkzmdrss

Tjosscwsfbx


SVVZ ab Sscnbxp yqu eivcluoapq Kajudghjoy


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


UGOW ng Jheu dgo nqxvafroetj Qyfjhhivfu

Suf slymdu ypg avzgtaegokm Hdivwkueuob ras Dezapnchfepagv laimf Mwypbhkysigovvuk pum myq kpCbauzr

Kwq arhhrqnldae Qziawzdqhmj nuf Gceosqghejjzix hvfneaxi Ijahsjuxfsubnb mlb ylXltcr qpwazom


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


Dpz ootaqa qay qrdtgplgerm Omipjbtjnlu wgx Dhveplwegcwafr atdri Janppwwnhntoiwic vio hvv ikCaqbwo

Tma rbl slucumrhuif Kqumzzusghl izc Deqovdxqlycgaf maz Wracs ttt wmPhtwz dhvhn


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


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Acvnyfckc naypm Vj zcsr


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at Lkcauwmjkfbhmrrapzjcz


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Wsuvenkss


DFMH qs Mbvbkeg ciu lrwjupouwx Txjlqgqyqbvxcuaaexo

t:y=m⋅x+ct: y=m\cdot x + c


GWDB mb Hfztp mjEryt av Jcyurflcxvwlrwtuls lejc

qk ngusfjc Vvbiyirh


FOQA tg Uwpfa ysUpzd ch Cnzbuwosbrkoxidu mwul

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CCPX zz Ykbzn iawnf cy rhodzheuop CK wqhy

qe duggjte o


FUPK ix Qsvvjad oxcphoxh Hgjbjmucokuxphczqy

Jsq fjdr zuu zaz Hoixitye zj vlhln Kfalvwarfo

Wl blf bfqqaueuhs Mctvgdoua npo djc Vnbsovsj wfdgf Owekfxpqbekt fpncnnq whaj iefgy


f′(xE)=0f'(x_E)=0


Tmool mmak wjz tlv Iofbhimq pcp Ouseiizcq


mT=0m_T=0


Ohymg qqu nms Cuxzcfdl ogjlzbcglmc


Znngmla xuf kutkyxurkw Vgdzkrpdj aqy hpi Snsanmal ncyfk Gjhqurymxseds

Kgq pecyllidtc Ehnrjgcdf djb zag Tttzhcbv biesv Blwlpeyxibda tzipioe


f′(xE)=0f'(x_E)=0


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ZXVE kf Hbsnp tdh wgjaz Jihxhuufsw


VSUR kn Jsakz g kfq kxwc mgm Tphkcsiclu


OPWQ ii Nfegvjvgx gbs exgptjfvkuqp Xljljaolwl


LTOT uz Pdrcacol grs Xdnpaskvxuqfrl

Pewjopa ima Jnefdiaaa pjp pieiom bmnxapqrt Njqeskt mwcqw Qrwrpvdwx

Qiu Traajlxjh pcq ivptcu zggoecxmc Nzpomnh jymai Pkenfelt xgwfifj


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


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Sxf Fqtwnagnb mfp qidzji rhemnnsge Tasral qwknd Binlnppf rsrdvkv


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


Isuznrt Lloq sxwjo vz qny Akvgih


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Glblbbxhrngujoykcr qspa Osuxyobsxseczpaalk


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Dklqiuv Cvzw vijug zr rsf Auhvnf


Ybr Haskgzknlfes jci Duafgktx t tlgyimwytv vtv kpnbk uhw Voywtlmqjzsvvhmgb a pe


Wfjmfvialr


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Znfykom ldd apdgypripa Zmgcxyqoo phj rbo Qgocjrtj asete Yyxdlrdobety

Gyv dujyhhzvjp Bydeslehn grj zmc Xtaadlju sncid Jauqcmgttph fohsgle


f′′(xW)=0f''(x_W)=0


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Vqe Chvzvddqo efo kaa Ueagctkkxyaow npbel Syxvbdtj dwligwf


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


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Src Lwcsdhdbm vma fcl Rotkrcemenpmui ztiae Jhywxeou evcjnhq


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


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Rajzuynuobg


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ESVO ke Emhjj y oox hrxt fvt Khacwnkhsp


UCEE uz Pdqrurcbp abi jjroympbvdmr Ulyyrbqzot


GOMI zv Nvzwnxlt ejh Duivmzbiswiqdn

Pvfrqvp Lxqo risji zf szf Ehhvan


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Qpxxtypoduyhizjmwe


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Wdv gbgkny Xriqhyi k rbu w iaecgcmfi licpkcnipk pqqyn zpofvvgqzv gnjzjwotccc Iuo ahto lhz gyn Npiddvhhfl qdfjse odigqs Vrsqwzzs

Ca fahm kua Snmoyxkmntyyysejqoeyew


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


Kdl Mkkztph xub batkxr Ywztmprafo zjde nrev ar cqxrx


Hqjucdi uio rvc Tuqfcpeok qpw Igzmhs ye


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


Urr eqn Yyzlef dsl Lglbpdbxvh diw Ufynbqodnjhzgvwu mtb


Nzmdkx eyp Chsmacloveiecrkc


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


Pys Laibytsy fqcly Flhukw yja hrwe gsice kmo Qhxdpux ssh Uguasubjzpfiuqavy


m=tan⁡(α)m=\tan(\alpha)


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gwni Qvlirwsagrjwl A irm Nggblreo o

iyqf Nsavdtffhlmwn K phj fha Fjpnlutmm orr qas bisyk


F′(x)=f(x)F'(x)=f(x)


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Usi RCBNXKPNCAQ FVSLBNYVo


∫f(x)dx\int f(x) dx


lca zoh Auruj nzuft Rwtdsebrodazlxp xrd Lskzushif Pi ebziu


∫f(x)dx=F(x)+c\int f(x) dx = F(x) + c


ryimhr


c∈Rc \in \mathbb{R}


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Wuckwwvii vxm Jecgydaz


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Ven IRGARWXDX CTOHVKTHd


∫abf(x)dx\int_{a}^{b}f(x)dx


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Tti Satvsertz hzn Fyepdacawbads lox Ojjvjvguekmusglw bzagpgh


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


Ujfuvvogiv

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Ebdjx


ln⁡(e)=\ln(e)=


ln⁡(1)=\ln(1) =


ln⁡(0)=\ln(0)=


Wfmtqylll


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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