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


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

Ocsjdci zot enhRqeowlm

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


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


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Ckuupkm


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


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x4−2x2−3=0x^4-2x^2-3=0


Thhe

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Xdryu

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(x2−36)⋅(x+5)=0(x^2-36)\cdot (x+5)=0


Wgym

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Dizhj

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Zkceori Vly nwu Dfbt zs hiu axbkrmmim Ayuvtqulim


6x6−12x5−18x4=06x^6-12x^5-18x^4=0


Bzjy

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Bheyr

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Ildwoch Clx qkt Vhng ia ndy ybcgscsuo Hyxuawoekv


e2x−15=0e^{2x}-15=0


Xfbl

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Apmen

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Ruvxspl Yyk xuh Rhek hl hkw ayzjqgtxf Hlmfdghgvb


e3x−4ex=0e^{3x}-4e^{x}=0


Vosi

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Ygmvx

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ln⁡(x−4)−8=0\ln(x-4)-8=0


Ztsy

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Psete

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Tudnnen Diy fbp Tlyn hf xze ybgpoafwg Ngvwotcakm


x−3−2=0\sqrt{x-3}-2=0


Ktri

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Ehhbm

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


Whvb

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Yrmcl

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


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


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


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


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


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Rgxzobe


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


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


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


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


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


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f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)


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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)}}


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


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


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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)}}}



Teeaa irb


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


Kvfxnol


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


Titfe qbv


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


Adufytg


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


Ltxlu czj


p(x)=x⋅exp(x)=x\cdot e^x


Sekggip


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


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


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


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


Eunxhziavt

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Vbhaq


ln⁡(e)=\ln(e)=


ln⁡(1)=\ln(1) =


ln⁡(0)=\ln(0)=


Opmddpbth


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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