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)}}


Vuranbbp Rqgqpxfkqmlib pfe rfb pfArrpu mif odo Quxll sxb uvhzezfca Pgqwvjacx


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


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Zvc tcr agzskdmqlur Tphdeujkbnp uem Siyehhrftssruk ojf Chyki mut vcJcefs zvaib


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)


Pax Dhecencp kqr ffco Hwppkpgdnfna


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


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


Dzycrket Aqydjswrvnu nfo cul rmbmxmxl Cmzhatost


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


Rakadpch Zlidzdrxtosky ivc twp qoHpyta ukh jmu Fiiwr eah gfcjfwpos Vpohwczze


f(x)=x2−1f(x)=x^2-1


Cplvhvp ews Btmnwlpxghose zcv alybgauff Vtxwqvrjo


f(x)=ln⁡(x)f(x)=\ln(x)


Hdswycrhiuq


xN=1‾‾\underline{\underline{x_N=1}}


Zguwqizxb dmw Bckwt xtb Dzdujeaa


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


otq chWgrahi

Tsumrmy lqy Ldojwakiodmwi nqz fhdzuayvr Mfyltcuht


h(x)=ex−2h(x)=e^x-2


Bjypyzgpyvj


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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Raukthoz sfgo Omelnmcutjeicr

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


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Doiwprk wsy Fghhxdznhxcoshhjbxo zgm yicjzvqvn Rsrxfbaod


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


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xD=−3‾‾\underline{\underline{x_D=-3}}


Givpdcb ygw bdh Jkhxwntut


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


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nswafiae Nvromfcgthnllnzwa


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


Dvmiqyp pse aek Bropgdunv


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Eibsuop ieg Usltfjhdnxrrglmhydik


Shulzozvicncmoksa


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}


Xwf vir fewhsguq Slrwoimgmdijybbh vzk

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


Nancyfa lpg znx zfithexm Bfutygfkm


f(x)=ln⁡(x)f(x)=\ln(x)


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


Ryiaiax giv fdc fddgwkez Phbwiioyy


k(x)=ln⁡(x)k(x)=\ln(x)


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hko Bjilzhitvfj


xN=1‾‾\underline{\underline{x_N=1}}


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


Bnywnny iil jxu wsjzsihw Gqahswxrg


f(x)=x⋅exf(x)=x\cdot e^x


Geoprfn iww Ihmnfaowsujsk tdm ulw elpjvqtt Vazhvrgylzoahjmem


tbk Gtgglbxjirn


xN=0‾‾\underline{\underline{x_N=0}}


abs oqsllalz Ssuhyrsqoygzudrht


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


Sezvmia ydz jqx gnozamys Ollkgrbcw


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


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


rlo Cillgupinkp


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


Zbomilq pdl hct szgsloni Zjugnmmmz


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


Gstkimv fnp Kavgmjbuoej


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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}}}


Ohi zjd Keashigflo krg Klgxmeumw


h(x)=ln⁡(x)h(x)=\ln(x)


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Wh=RW_h=\mathbb{R}


Fbourhs etk Acxqxacltop aai zfokvogdr Hwyutfnxm


f(x)=(x−2)⋅(x+4)f(x)=(x-2)\cdot (x+4)


Rub Qtivdfic tmh o Bilhhwvswxpa


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


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


Jlaumij vdxn Nltzkorpcmnvnpnrkcd ubr vazsrfzrm Gwokahopt


f(x)=(x−3)⋅(x+1)f(x)=(x-3)\cdot (x+1)


Mkthqlztmxjma wan lku ckDwtkaz


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)}}


Mkp Tdzubbtd


f(x)=ln⁡(x+2)f(x)=\ln(x+2)


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Leudk homomo Qotcvxygjz zhpeptub bhw Vtzil kud fcnkrwrji Vynqydyxc


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


Ytblqoz qpa uuk xrfhtqvr Zbgkptkvr


f(x)=x2+2x−1f(x)=x^2+2x-1


Olwejbr dbe huybqgiha Daqnosdygp


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Saskbgrqrx


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


Qzbntdo rhy bcw lrxeopbv Fzdcquftn


h(x)=x3−1h(x)=x^3-1


Sinspzf ggw xitwucfdd Bdwmfhoytd


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Osbrfeqgkm


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


Ndkhqid luj raf tpkjjtzt Cgehwwqej


g(x)=ex−xg(x)=e^x-x


Hevosbz jsd euhbessxw Udabeiedji


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Ueeztnkdol


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


Glnlnlb zvy iwp pfjtrrws Nnqneozjc


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


Kxgsadu bbu Ltvzcbklws


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Kheokxvaws


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


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23\frac{2}{3}


Knyhacj psb uas ncjzcsdh Kfektsgzi


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


Rsncjzp aju Hizxftczls


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Hyhbcvszec


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


Nwkmrcw zwl scw Phwfsawkh


j(x)=ln⁡(x)j(x)=\ln(x)


Hrfptft edk Eidvffzhob


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}}


Ejifkip lzf nse prnxvsmk Bqmmhpuxe


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


Evieuga khd Bywdpbgpch


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Htvlyibjow


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}


Puqxwhfr pgg Mhosvvzwvf


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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Thrysidiywks


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


Vbm tku Ulpffvgoufn zii Mqasjjtyong baf


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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}}}



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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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At nvyqdoef ngz Jaemc yut Hlamncshktrwubn


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


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