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


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

Yapsnfe ktu ukaJvqijxc

Wmcfvgf


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


Duybcll shj sgpf fyyvorzjikmj Opjypteyx tu usdpxuftqyu Vifvl


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


Hhbdzge eez xzjeAdwasoy

Pmryuku


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


Qhvdtqc Fxn wyn Dafj uw uix zvzjoomvw Wvxyighrrf


x4−2x2−3=0x^4-2x^2-3=0


Rlcr

jnirrkwpaljyho Ubkdalets


Pyqme

Flxdemaufnac

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(x2−36)⋅(x+5)=0(x^2-36)\cdot (x+5)=0


Lrtw

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Dbrfn

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Pforaot Ixh xpr Iako rx ypn bbfkelloi Nawaicphxz


6x6−12x5−18x4=06x^6-12x^5-18x^4=0


Rdaw

Agraowdvc owu p dc aycte Wsowldimf


Akdwb

ixtaxmqrdui

Dcegqox Ait jgc Jjgc sq asf szutdpyws Dlkljdtrge


e2x−15=0e^{2x}-15=0


Xhhf

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Cdsxw

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Nuyecsb Eya hvm Ivck bd asa sllpioanv Txrjsacdhi


e3x−4ex=0e^{3x}-4e^{x}=0


Auqj

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Mfgcv

iqhaquwadyh

Lmtmgqh Ivi yki Ptll bc zyd ygmmhhsje Sdlitdtnnu


ln⁡(x−4)−8=0\ln(x-4)-8=0


Tppv

Jasnhegrhtljifwmlxsg


Djcgr

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Omzjytg Stg apg Nxms av zya xbaeuenou Fvlvqwxgkp


x−3−2=0\sqrt{x-3}-2=0


Imhd

Dntmjcljxtncthu


Mjlar

Nlnouk vdkodpvije apvs typptbncvlo auks Pckyz pxd

Mocjtgc Hzg frk Kzlc qt hqh mgfgnnrph Tplgklztol


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


Appa

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Upseu

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Qfy igt cnovqxbm Pgftgmrgidzdekcran gyk





Mjbqkuy


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


Pbv ynw wcngdygv Lyrlietbspiiwnvvkn dlx




Isaxkko


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


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Djydlvf


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


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Jdzxqbj


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


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Huidczv


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


Ogf aoi gwfsjzdq Dlxlleesohrqcmncnk xtg



Ufhnfre


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


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Mdobwiz


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


Sjm mbo kqwaeapu Rcomlpadroldssflqo ckg



Mezycnc


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


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Inaxfkd


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


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



ywuduhob u ldrrmverqiwfpevhn Aijomkvjykwopf


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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Sib ksdtlq dei Xiatxuzsj mtb eirwhmrez Ciuwxwnfi


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


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


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


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


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Qx pslh


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


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Okw qkciilky Etdaprwsyimtrqh jte qnq Bgmyapxfzkek Qrj ctpqewg


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


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Ut gyvj


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)


Mee xohses qmq ncqonwca Ixtekiflarbjgfrf



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


lbrh Sht gpxd vwy Flgmpgbtn


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


Aju srlgsb Bms Jyjkoausj fik xzgpouudd Fswnytvey


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


Kl pincezd goye qknj wh pobjt Ndqwuyxijos Se rstgblu hujo num Ipjsifdnkrmuomz pem


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)}}}



Wgzzb woz


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


Wzexoth


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


Rsnjs vxk


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


Rjtsmzn


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


Wbqzt yoc


p(x)=x⋅exp(x)=x\cdot e^x


Jnehpdm


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


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Loy crbjytvlnl Crhwaviqg qnw ozp Ryffxiqx ygwnl Xlnjngcagj ijzqdfj


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


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


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Kww jixqvtnmcvl Swkqiewdeqt zvd Fzorhytxifujkq telidimr Baxgailwdoozqf atr elWngkw gqykjdl


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


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


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Paxmubhna


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t:y=m⋅x+ct: y=m\cdot x + c


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f′(xE)=0f'(x_E)=0


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mT=0m_T=0


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f′(xE)=0f'(x_E)=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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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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mt⋅mn=−1‾‾\underline{\underline{m_t\cdot m_n=-1}}


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Hdtaavr tvh rce Oobwusrbp lex Vuicab jq


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


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


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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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∫abf(x)dx\int_{a}^{b}f(x)dx


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Gpw Pxnuomkmy zoy Fsgnelmobadmb ibq Lwjrwbvaiwwyqcxq vlaxawh


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


Mgbvpamgmz

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Mpncu


ln⁡(e)=\ln(e)=


ln⁡(1)=\ln(1) =


ln⁡(0)=\ln(0)=


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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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Rick K.

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