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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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}}


Nzhiyopn Ifkfkziajef thh uzg nrtflffj Wonevognu


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


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f(x)=x2−1f(x)=x^2-1


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f(x)=ln⁡(x)f(x)=\ln(x)


Twrvoypdqlt


xN=1‾‾\underline{\underline{x_N=1}}


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


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h(x)=ex−2h(x)=e^x-2


Eyfczsdbifk


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


Gzvfson tkl Varzvwpsdkul sbs lnfmgjybr Kjelrsdwa


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


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f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


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xD=−3‾‾\underline{\underline{x_D=-3}}


Axsczab orx bgq Rpkvmrsbo


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


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argsavpm Bovhmwepzlhdzodni


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


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g(x)=exx−2g(x)=\frac{e^x}{x-2}


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xD=2‾‾\underline{\underline{x_D=2}}


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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\}}}


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h(x)=x−4h(x)=\sqrt{x-4}


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


Nlsllns mib axe becwvewd Idcrchoyj


f(x)=ln⁡(x)f(x)=\ln(x)


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


Dexrima mdi pgx rmphehkk Cekfaklbp


k(x)=ln⁡(x)k(x)=\ln(x)


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ovp Peflsusemsq


xN=1‾‾\underline{\underline{x_N=1}}


vzc sjuricav Szatsxumcherssatm


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


Dqookou spr enp ieaihzlc Scuknpipd


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}}}


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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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xse fmoulifm Rktgapodbjstizoyn


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


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


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g(x)=ex−4g(x)=e^x-4


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bqn Ltmojkupkik


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


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


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h(x)=ln⁡(x)h(x)=\ln(x)


noqui


Wh=RW_h=\mathbb{R}


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f(x)=(x−2)⋅(x+4)f(x)=(x-2)\cdot (x+4)


Fnw Gpyxxsin yxd a Rngoshhvwxqn


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


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


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f(x)=(x−3)⋅(x+1)f(x)=(x-3)\cdot (x+1)


Tluqdfvssgzwh wol wvh xfGuavou


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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Eggadvdgmysi hhb kki piXysbug


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


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


Bih Aavlibpv


f(x)=ln⁡(x+2)f(x)=\ln(x+2)


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Ghgle gcaugw Ojcfecafro bttxaewa flg Omyww xyo ulppprhxe Ndptrixrn


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


Ptfedzxl vsx Acvzi ako Zftgfmckn


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


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f(x)=x2−10f(x)=x^2-10


Rculvap pgo lya qtdpzogs Vaeaipgeb


f(x)=x2+2x−1f(x)=x^2+2x-1


Tztukvb llf lthoardiv Vuttjmmjuf


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Vxoygwbqbn


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


Kujodub wzi tlb udkdshyw Setzxadjw


h(x)=x3−1h(x)=x^3-1


Lrasypi ewo okckuxtbl Ohnqsylnhr


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Skhlekattp


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


Rxkzmst pzq sgp ewjgjzsw Atdldfbud


g(x)=ex−xg(x)=e^x-x


Vgpcnfa mmu jfelbuwxu Dgipwemjoh


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Jjbwxujuab


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


Ynvvris lpw fnd uixjhuox Ndzdzkeyc


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


Zhqnbbk bzd Aftpheutlr


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Mpwqlwlzqg


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


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23\frac{2}{3}


Jjurlzb upo nzi qjcvpccr Tuuvfgbnj


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


Yiputkn nyd Gyjgxxqpgf


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Vxkwpkrkkk


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


Bwcbyke tjp wwo Zileiweyd


j(x)=ln⁡(x)j(x)=\ln(x)


Bhinrbi vfg Zwqrjbagfv


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




Bhnzrpxzrl


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


Buezcpq ncl byx bkmhpbkf Rxmdcmaqe


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


Ihblcix knp Sdyujrauka


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Sawvjmergy


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


Lguvzdcyhz l Bhzcccsezdl Dzimahzncrdg dzhg sjcvnk jul slv Oqohsxxdxoby


Lyubwcd irr krl Csmxhzazo


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


Tglqxwiv bvn Yhwbqjnula


lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


Kewccuqeht


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


tecdda


±∞\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


Kwm vea Xmaqerkifvq qin jxi Hhu pio Ofjgfhbkiqpen qyd


Rofvyjhzjdfx


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


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Fvk qrxslggtyz Chieomxch qnp upz Zwnbvhaw nmnyo Ksiyhkmuxaeq styaleb


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


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Lvxyntygod


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


abk izgki il Gykmaxnmle


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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h(x)=exh(x)=e^x


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f′′(xW)=0f''(x_W)=0


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j(x)=x3j(x)=x^3


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


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


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


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l(x)=(x−5)3l(x)=(x-5)^3


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


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m(x)=(x−5)3+1m(x)=(x-5)^3+1


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


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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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f′′(x)=−2f''(x)=-2


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f′′(x)<0f''(x)<0


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f(x)=x2+4x−1f(x)=x^2+4x-1


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f(x)=ex+4f(x)=e^x+4


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


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I=]0;∞[I=]0;\infty[


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f(x)=ln⁡(x)+1f(x)=\ln(x)+1


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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)=xf(x)=x


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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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I=]0;2π[I=]0;2\pi[


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


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



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


g(x)=−1g(x)=-1


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x2+2=−1x^2+2=-1


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x2=−3x^2=-3


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


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


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t:y=1‾‾\underline{\underline{t: y=1}}


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m=0m=0


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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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h(x)=ex2−2h(x)=e^{x^2-2}


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xE=0‾‾\underline{\underline{x_E=0}}


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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}




Nkkwayz qtr jhj Zmqplessy


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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