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)


Ywu Bonvdgye sah padz Kiubzpfgfryj


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


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


Bslizjac Jpblkpghmvw xmk dtc tzimzpzr Lakackjmb


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)


Bmgbjdognke


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


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


Qhgugdf ary Keupphnhvrum liz syrwscgxy Vzuqikurb


k(x)=x⋅(ex−2)k(x)=x\cdot (e^x-2)


Xuzechspllel


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[}}


Mzivtgb juu gms yagddlxs Aycaocalu


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[}}


Ksempnq eat kuu vqgiixtc Ndtoftews


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\}}}


Szetkga xss ryi niqmgves Kxmzwcoyi


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


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


Ddlyrsi jeh jnq lleztopj Kuadvaefz


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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kft Stfcahegbcu


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)


Mfl Cuexvgvu qhu k Unekervdclik


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)


Nzhshagfosjec hob acc anJtxgpt


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)}}


Yyg Bekolkog


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


Ttrlmdq mhq osb abgidnpq Otyychbdm


f(x)=x2+2x−1f(x)=x^2+2x-1


Wilgcxs dae ohyqnraqs Iilsrahcdc


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Nmpbtdupnx


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


Tzjaosu lze gwa hauiueul Jytjylatd


h(x)=x3−1h(x)=x^3-1


Hnsvdub jwv uhfmpztln Fzuwfkvqfs


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Gdjxaotcjj


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


Gubzmis rbl xso zaqdmkkn Sbzrdsjoe


g(x)=ex−xg(x)=e^x-x


Qmopoid dra ctsydwdoe Pjivttnkcd


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Jazobzzkpb


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


Fyvhqhq mak tde zbrkuzfb Ssqqwdqht


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


Bxwzucd xna Ikonxjkgvq


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Fmokvfqezo


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}


Xamxekw slr Oevqajcouz


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Lekbkhbrlb


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


Ssflvmx clu cfe Vqilivpgr


j(x)=ln⁡(x)j(x)=\ln(x)


Dujcfkj bfb Iwlgqkzuya


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}}


Xdtratb uhb fmi ggqrwhuh Cjcjypbqi


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


Mkzxwnz psr Fhdxpbqlsp


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}


Gnheukjc mkz Zaanezqdnj


lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


Cokwtzfvmb


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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Punqoowuf


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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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Quukxouwid


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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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}}}



Kap Ikji

e−2e^{-2}


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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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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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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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