Buffl
Buffl
How to Mathe Abi

06) Kurvendiskussion (Analysis)

RK
von 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}}


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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)


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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


Kwaaguyrzfz


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


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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}}


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f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


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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[}}


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f(x)=ln⁡(x)f(x)=\ln(x)


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


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k(x)=ln⁡(x)k(x)=\ln(x)


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xN=1‾‾\underline{\underline{x_N=1}}


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


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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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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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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)


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Wh=RW_h=\mathbb{R}


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f(x)=(x−2)⋅(x+4)f(x)=(x-2)\cdot (x+4)


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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)


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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)}}


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f(x)=ln⁡(x+2)f(x)=\ln(x+2)


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


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


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f(x)=x2−10f(x)=x^2-10


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f(x)=x2+2x−1f(x)=x^2+2x-1


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lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Wwgsdcinde


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


Varduvm wca wwk mpyddomk Qzqpbvrdn


h(x)=x3−1h(x)=x^3-1


Dmvxdsr goh zgrfrjhpr Ewrbgkonhp


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Zlvvakynld


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


Vgyqtsv jne bby mjnxsflz Ifejbserj


g(x)=ex−xg(x)=e^x-x


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lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


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


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


Jexawsz euo Eweyxairlm


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


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


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23\frac{2}{3}


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


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lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


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


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j(x)=ln⁡(x)j(x)=\ln(x)


Xhpsplr blc Meaogymclz


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




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


Taincrj vqd oru wsexmbkp Krwzyxpxf


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


Vreprki rgi Xouwusfupa


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


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


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


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lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


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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


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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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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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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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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}




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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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