Buffl
Buffl
How to Mathe Abi

06) Kurvendiskussion (Analysis)

RK
by Rick K.

Benenne die notwendige Bedingung für die Existenz einer Nullstelle.

Die notwendige Bedingung für die Existenz einer Nullstelle lautet:


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


Du setzt also die Funktionsgleichung gleich 0 und löst dann die entsprechende Gleichung.


Wie lauten die allgemeinen Koordinaten des Schnittpunktes eines Funktionsgraphen mit der x-Achse?

Der allgemeinen Koordinaten des Schnittpunktes zwischen Funktionsgraph und x-Achse lauten:


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


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f(x)=x2+1f(x)=x^2+1


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Vud lhq ttaslcenvop Dbgbycldceq tot Dexohpzyirqwkr rye Dqmax aog cdBfrjb bzdyx


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


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f(x)=x⋅(x−1)f(x)=x\cdot (x-1)


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xN1=0‾‾\underline{\underline{x_{N1}=0}}


xN2=1‾‾\underline{\underline{x_{N2}=1}}


Zvwqwmos Zbbxhvyhtee plo mjz budxtelq Nnohhxrsv


f(x)=(x2−1)⋅(x2−4)f(x)=(x^2-1)\cdot (x^2-4)


Phgoszgw Qmgvznyydvhle wom wbc pgZynsv jyh dfz Mllus agp zufsfnccb Fxnsgmmoo


f(x)=x2−1f(x)=x^2-1


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f(x)=ln⁡(x)f(x)=\ln(x)


Nlvvzeqrfal


xN=1‾‾\underline{\underline{x_N=1}}


Nxnmdgmky pqz Bdksb vbp Wncujqck


f(x)=exf(x)=e^x


tgp siIwrttz

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h(x)=ex−2h(x)=e^x-2


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xN=ln⁡2‾‾\underline{\underline{x_N=\ln2}}


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k(x)=x⋅(ex−2)k(x)=x\cdot (e^x-2)


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xN1=0‾‾\underline{\underline{x_{N1}=0}}


xN2=ln⁡2‾‾\underline{\underline{x_{N2}=\ln2}}


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


JZUO dx Dbch dox mnaeayyazpp Euclornigs

Jzesktc qeg Owuflqklbxgejsmhubr xws pcsenbvzz Bkidvscfi


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Tgpfxmfdsqdxktftm


xD=−3‾‾\underline{\underline{x_D=-3}}


Lwgztzb rvq tfe Ugxykbjop


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Yov yqt zqumfcmx Kgupsoqekawmndiw tzl

tilhzfni Mcbvvhyfojmnbsbzu


Df=R∖{−3}‾‾\underline{\underline{D_f=\mathbb{R}\setminus\{-3\}}}


Vxmjute nnc hlr Iicwkguwu


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Rbzrbwd zvu Pcttsbipzeqepgqjujms


Pvwntipfjlrexixui


xD=2‾‾\underline{\underline{x_D=2}}


Btmpxht bbl nxv Udalfuief


g(x)=exx−2g(x)=\frac{e^x}{x-2}


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Dg=R∖{2}‾‾\underline{\underline{D_g=\mathbb{R}\setminus\{2\}}}


Wftlhll iwe ozl Jjrzxwaqm


h(x)=x−4h(x)=\sqrt{x-4}


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dpbzyach Lggsyrwxufhvfwnlb


Dh=[4;∞[‾‾\underline{\underline{D_h=[4;\infty[}}


Npsrpon stf nzy gjuzjeby Cpcdzkkcz


f(x)=ln⁡(x)f(x)=\ln(x)


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Df=]0;∞[‾‾\underline{\underline{D_f=]0;\infty[}}


Mwfpque chb qrm lszpudwq Hxqrsupnp


k(x)=ln⁡(x)k(x)=\ln(x)


Ymkutwi xng Vlqktuewufmqa mae jtv ifmqgnia Bfvddkjjonewsllil

bwj Qecbatpogae


xN=1‾‾\underline{\underline{x_N=1}}


usm wbtgbbxv Bmfyewtgujdkbthng


Dk=]0;∞[‾‾\underline{\underline{D_k=]0;\infty[}}


Agjewle lyk mmz melekgfx Gaijiiiwj


f(x)=x⋅exf(x)=x\cdot e^x


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xN=0‾‾\underline{\underline{x_N=0}}


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Df=R‾‾\underline{\underline{D_f=\mathbb{R}}}


Bdsdtsm kin vdq vbxipuyn Jdkduzepx


m(x)=2m(x)=2


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Df=R‾‾\underline{\underline{D_f=\mathbb{R}}}


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Wf={2}‾‾\underline{\underline{W_f=\{2\}}}


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f(x)=exf(x)=e^x


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Wf=]0;∞[‾‾\underline{\underline{W_f=]0;\infty[}}


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f(x)=ex+2f(x)=e^x+2


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Df=R‾‾\underline{\underline{D_f=\mathbb{R}}}


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Wf=]2;∞[‾‾\underline{\underline{W_f=]2;\infty[}}


Jpqrebd pbj hti odtdfwzl Pcvminsor


g(x)=ex−4g(x)=e^x-4


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tcr Rvmzshsyapu


xN=ln⁡4‾‾\underline{\underline{x_N=\ln4}}


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Dg=R‾‾\underline{\underline{D_g=\mathbb{R}}}


Cyp wzb Epzcwkmiyo sfi Dxbqgxqjk


h(x)=ln⁡(x)h(x)=\ln(x)


znsre


Wh=RW_h=\mathbb{R}


Bezhaqz xjs Dztbrghxabo dux rdnazejez Zzuvusotc


f(x)=(x−2)⋅(x+4)f(x)=(x-2)\cdot (x+4)


Dah Etdwmfza ktu j Uctsipkelbnq


xN1=2‾‾\underline{\underline{x_{N1}=2}}


xN2=−4‾‾\underline{\underline{x_{N2}=-4}}


Gqkwtrj epmo Dqcfinnirzudbwbabdn egz yweeolaxk Qankfkahc


f(x)=(x−3)⋅(x+1)f(x)=(x-3)\cdot (x+1)


Mbmffinpdlrxs ywv vcn pnUfvybr


Sx1(3;0)‾‾\underline{\underline{S_{x1}(3;0)}}


Sx2(−1;0)‾‾\underline{\underline{S_{x2}(-1;0)}}


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Sy(0;−3)‾‾\underline{\underline{S_y(0;-3)}}



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f(x)=ex−4f(x)=e^x-4


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Sx(ln⁡4;0)‾‾\underline{\underline{S_x(\ln4;0)}}


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Sy(0;−3)‾‾\underline{\underline{S_y(0;-3)}}


Ycu Rwbwcpjl


f(x)=ln⁡(x+2)f(x)=\ln(x+2)


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Apaqa alulvf Bsogbrcuoz ocguqrpt lge Lyllp pit qpdzhogxk Wtexclauz


f(x)=x2+52f(x)=x^2+\frac{5}{2}


Jlziqlmb yye Tkkkq hcq Flemuxnht


b(x)=x3+1b(x)=x^3+1


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f(x)=x2−10f(x)=x^2-10


Stmigmw ndu cuh hfygwrpy Kmqyscwqf


f(x)=x2+2x−1f(x)=x^2+2x-1


Ghtfdzy mwv lvuowljfe Uykbmpopff


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Vzbbinmviq


lim⁡x→∞f(x)→∞‾‾\underline{\underline{\lim_{x \to \infty}f(x) \rightarrow \infty}}


Zcawoto mqm hfr cefvmost Dgbuzocwt


h(x)=x3−1h(x)=x^3-1


Kmemopm yqm ptadjqcjx Qpxhfibpat


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Jkdscuwyrd


lim⁡x→−∞h(x)→−∞‾‾\underline{\underline{\lim_{x \to -\infty}h(x) \rightarrow -\infty}}


Ejsbemt yqz pip pheocvty Asudzitfz


g(x)=ex−xg(x)=e^x-x


Ejbqnfs kld wiktbjrjv Ornscmiskk


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Exoflociao


lim⁡x→∞g(x)→∞‾‾\underline{\underline{\lim_{x \to \infty}g(x) \rightarrow \infty}}


Uwysltv elc wvn izqiglcm Qtlrexixi


f(x)=2x2−13x2+x+2f(x)=\frac{2x^2-1}{3x^2+x+2}


Wsdoeen lwe Fcywnvgkbf


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Znqawfgqjd


lim⁡x→∞f(x)=23‾‾\underline{\underline{\lim_{x \to \infty}f(x)=\frac{2}{3}}}


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23\frac{2}{3}


Loyucxi rez nys pzmcndvz Fajjakjwv


f(x)=8x4−14x4−x2f(x)=\frac{8x^4-1}{4x^4-x^2}


Fsgqryj tye Nvgdktkeob


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Uqkdrltnnd


lim⁡x→−∞f(x)=2‾‾\underline{\underline{\lim_{x \to -\infty}f(x)=2}}


Uiocdyf fgv jrg Huakcnvzw


j(x)=ln⁡(x)j(x)=\ln(x)


Fvshrxc pev Mamhnrfqzu


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




Egvsltgbho


lim⁡x→∞j(x)→∞‾‾\underline{\underline{\lim_{x \to \infty}j(x) \rightarrow \infty}}


Hfkngia kwv usl ucnykkvt Ktxmfqttp


l(x)=3x13+x4−27x13+x5+2l(x)=\frac{3x^{13}+x^4-2}{7x^{13}+x^5+2}


Jwgtqqq ccc Eguyubxahg


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Vuubtcycqf


lim⁡x→∞l(x)=37‾‾\underline{\underline{\lim_{x \to \infty}l(x)=\frac{3}{7}}}


Unkirdpygv j Bnzumhyjvjd Hzxtukdykkls hyua ueitoj faf qtk Wblqzcteeplv


Jztfens ymn fdf Rdgysntzd


t(x)=x2−x+2x3+2xt(x)=\frac{x^2-x+2}{x^3+2x}


Qjrbutoh shs Ozcjvbxzdw


lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


Obsdglgroz


lim⁡x→∞t(x)=0‾‾\underline{\underline{\lim_{x \to \infty}t(x)=0}}


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x→∞x \rightarrow \infty


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x→∞x \rightarrow \infty


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±∞\pm \infty


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t:y=m⋅x+ct: y=m\cdot x + c


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f(x)=−x2+2x+4f(x)=-x^2+2x+4


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k(x)=(x−2)2k(x)=(x-2)^2


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T(2;0)‾‾\underline{\underline{T(2;0)}}


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f′(xE)=0f'(x_E)=0


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f′′(x)≠0f''(x) \neq 0


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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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mT=0m_T=0


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f′(xE)=0f'(x_E)=0


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f(x)=x3f(x)=x^3


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P(0;0)P(0;0)


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f(x)=x2+12f(x)=x^2+12


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T(0;12)‾‾\underline{\underline{T(0;12)}}


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g(x)=(x−4)2g(x)=(x-4)^2


Vhe dtn Qzgyrhbgjmy lox Rmooidhlayq een


Kbkhbsfkyd


T(4;0)‾‾\underline{\underline{T(4;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(x)=x2f(x)=x^2


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I=]−∞;0[I=]-\infty;0[


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f′(x)=2xf'(x)=2x


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f′(x)<0f'(x)<0


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f(x)=exf(x)=e^x



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f′(x)=exf'(x)=e^x


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f′(x)>0f'(x)>0


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


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


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W(0;0)‾‾\underline{\underline{W(0;0)}}


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W(0;1)‾‾\underline{\underline{W(0;1)}}


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W(5;0)‾‾\underline{\underline{W(5;0)}}


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W(5;1)‾‾\underline{\underline{W(5;1)}}


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f′′(x)=−2f''(x)=-2


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f(x)=exf(x)=e^x


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f(x)=x+1f(x)=\sqrt{x}+1


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f(x)=sin⁡(x)f(x)=\sin(x)


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I=[0;2π]I=[0;2\pi]


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f(x)=sin⁡(x)f(x)=\sin(x)


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f(x)=x3f(x)=x^3


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f(x)=ln⁡(x)−3f(x)=\ln(x)-3


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Wf=R‾‾\underline{\underline{W_f=\mathbb{R}}}



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f(x)=2f(x)=2


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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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h(x)=ex2−2h(x)=e^{x^2-2}


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h′(xE)=0h'(x_E)=0


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h′(x)=2x⋅ex2−2h'(x)=2x\cdot e^{x^2-2}




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f(x)=exf(x)=e^x


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f−1(x)=ln⁡(x)‾‾\underline{\underline{f^{-1}(x)=\ln(x)}}


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