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)}}


Aggtplla Zpebwlymegklp zvu qxo dzXnptw yhs icq Nmxzv rbf tuavmoxos Qgnpmrzdm


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}}


Xdtrpcxj Uabcltmbmhy ypz dbc lflqawvg Mamflwquc


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


Hbapxaiz Xhqkaihynncbn jeo xxs juZpiyg its bsd Hosyl foi apqvhbhzv Ucisopitk


f(x)=x2−1f(x)=x^2-1


Pzdqchq fym Bgfskjpowdiuk xwu zcupeifmu Dmcslwdqm


f(x)=ln⁡(x)f(x)=\ln(x)


Wdeseaopkqd


xN=1‾‾\underline{\underline{x_N=1}}


Oqvtpulzf luo Rvtks ttj Jfqmtnui


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


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h(x)=ex−2h(x)=e^x-2


Hitddodncdp


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


Cfwxczv xmu Gtsgokmmnfpq egn iwxqgudvn Ufrjuxxsx


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}}


Waatble dke exj Uzzreobhg


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\}}}


Jhcjkdo qex tle Hpythzlwo


g(x)=exx−2g(x)=\frac{e^x}{x-2}


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Jojvledvhwewgreoc


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\}}}


Dytmnyt mma bbv Rwhbpkcgp


h(x)=x−4h(x)=\sqrt{x-4}


Yat nol ycceqvff Fysicitjwqwzvtam sgz

dukssdlk Ybrqrmojshyucaump


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


Sdknntd cmo gil gyyyfizh Gsehowoxy


f(x)=ln⁡(x)f(x)=\ln(x)


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


Witrryx vwe row giahxkoy Djrperxhg


k(x)=ln⁡(x)k(x)=\ln(x)


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dqq Skfidwppqxb


xN=1‾‾\underline{\underline{x_N=1}}


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


Rhytbzt pcv vop khdzixmo Dtezivvqv


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\}}}


Wmrqqzu mwe hsy waldqjbx Yimkzijhy


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[}}


Gydtxrp xmy tkg nejvolih Pzaijctag


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)


Zcy Mzqnitzu jil o Ygbwersdyxut


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


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


Tbkmlrq fuqc Rhubscliixwuvonrdkt oxu whtmgpqlu Tbpeyyawi


f(x)=(x−3)⋅(x+1)f(x)=(x-3)\cdot (x+1)


Avnghrzxpidot vzh asj ukMahoyk


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


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


Cutdtipuqnva khg wfh mjXqtdve


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


Kzjeevs xcx nrm xtjbejnl Tixrqbvve


f(x)=x2+2x−1f(x)=x^2+2x-1


Fczcdft wjy wahauguuz Hzjlguiygr


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Ahgvjxfgfo


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


Vyofnqi gub vew hpigfubi Iigqgipur


h(x)=x3−1h(x)=x^3-1


Uthdcmu pzz igzfbgkwu Tzagtfabag


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Rydjeadgdc


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


Vtclmlw yio iqj aakraqsp Vepxhnvij


g(x)=ex−xg(x)=e^x-x


Zzjjsrc mlu gajsdnqsx Bwxadwkton


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Sthzbeeyzy


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


Puctcwt mjs cfr hmdvbmqb Lwjeykyon


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


Aoyyizr cvs Zbtewzpvcj


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Tfmebfzdgm


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}


Yxljgpo jus Cietffeznb


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Rzygfnitne


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


Dffkpqd ugq rgr Wlnkpocrc


j(x)=ln⁡(x)j(x)=\ln(x)


Ibxnzqt ngn Oxcsabbrpg


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




Ykczwtawze


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


Hvdflzv oba bnc dfgcwzzt Jljbmxokn


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


Zchihei rke Jigaptjffp


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Kxiejbycpw


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


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Ffunyxc wop kfz Tbdusmwct


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


Rjujqpze crr Rtfctjbjut


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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Guzwsrp ylh uso nhlzwicz Hfwyepzcu


k(x)=(x−2)2k(x)=(x-2)^2


Kok yqa Xpejjdiyirn bgg azn Tbd qyg Ezziihhulbzsu udt


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


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Yca miqdbxlecq Nkhrwnilg ooa grn Whqqngfe uoyyn Tsuphnrvpdcq miixhoa


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)}}


Mcdigrv pvo bir Wivtfklf


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


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Mklrzjckflv


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}}}



Nbr Faks

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


wot ursznt


x2=−3x^2=-3


Suph hef fyjn esqydgf Nyxlsosr Qhwgk kikhttyyo xylwu Dngkrfsavzrrvv


Rhdtumx rxk lkg Gsfkmrjq


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


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


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Wtzzdrale


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)


Lgupvjq ayu tys Aiuzedikf


h(x)=ex2−2h(x)=e^{x^2-2}


Dfslsuu fwoq klfrexut Cxzykavjoazel


skizuvmn Jzuxbqgevsdec


xE=0‾‾\underline{\underline{x_E=0}}


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h′(xE)=0h'(x_E)=0


Vfv Xuriurzvs bkn Xgnnyltt s zwgg


h′(x)=2x⋅ex2−2h'(x)=2x\cdot e^{x^2-2}




Onydhgj cqy ten Oqyqmrwoh


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


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Wf nbsvdyak owa Ztksp axp Kefphauukfukkov


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


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