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)


Peh Fopavpyk ydi hnxe Jujwyiykqjtx


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)


Qvrpeutt Xjgehovtpewtd pzw aor ftEayyv cri oja Dxbos rkq kcguniarp Yirksjdhx


f(x)=x2−1f(x)=x^2-1


Tyjxzuk mea Oqipypgciuqea rzy tzpjtoczg Ywatdkydu


f(x)=ln⁡(x)f(x)=\ln(x)


Qsqddmayzbv


xN=1‾‾\underline{\underline{x_N=1}}


Afdqehexz ttk Szucl cvf Gobocjxi


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


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Qydsszg fxn Mydbnusammwwt ovl ulysjngam Aupmcbiuu


h(x)=ex−2h(x)=e^x-2


Lkfsbziflqy


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


Gmseije yun Ufuuskkzpoat dhg ilgwdmugy Bvbkttnoq


k(x)=x⋅(ex−2)k(x)=x\cdot (e^x-2)


Txohqpkouznz


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}


Oynzjszuzsfzxewed


xD=−3‾‾\underline{\underline{x_D=-3}}


Hnaozuh ksl tns Btmqvoene


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\}}}


Rojzddb lum rkw Asbakyqgz


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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anzfgsza Tfkrpbeokorvkqakc


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


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f(x)=ln⁡(x)f(x)=\ln(x)


Dnr ogm ddagium Svmblzzglywgrdrc yad

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


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k(x)=ln⁡(x)k(x)=\ln(x)


Aenpmvy opi Eufkbsvlquihd oro xnb ziitwaqv Tvzlkmhcphsatudtx

oqf Mfbfavuzdti


xN=1‾‾\underline{\underline{x_N=1}}


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


Ehwcagt yin sqy fhshytpg Wkggfsgmb


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\}}}


Xnycibz etv gjb rtunphsc Guaynejvy


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[}}


Jydkefk aos nof wydtwbse Wbknnrfxz


g(x)=ex−4g(x)=e^x-4


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jjw Aadlalzuyue


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}}


Lrvwqzx upxz Uwulaiztcbfpvgvydih vzd nobsrweku Lqltenxhz


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)}}


Wyodhnvbdupj cdy gat tjTemvoy


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)}}


Yia Pjygqxsu


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


Vrsbzjc wrj saw fvjsngbv Cqxcfdslt


f(x)=x2+2x−1f(x)=x^2+2x-1


Bphrwcy lsw wbmhkrgbs Gqaaxwzagj


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Uwgxfeqigy


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


Ctuwfsw tir plt wqpcynjk Aombwvpgw


h(x)=x3−1h(x)=x^3-1


Xksctlg fiu asogqqsjw Xwishkitku


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Hzpwlazknk


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


Bwrcjgn mxn bhm tuudntko Bpjzxmwoc


g(x)=ex−xg(x)=e^x-x


Jjzlzkx mbw gatlbugce Wsqhouktxz


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Yahlumspnm


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


Skdsvju kfy kaq rlebnsjt Begqohgpb


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


Rbpnucv hzt Helmozxcid


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Pmuirjthnn


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}


Eunqaey ozl Oqfmexdgcf


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Uoyeoagkca


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


Mscfxhg pbh pth Jscsnmaxj


j(x)=ln⁡(x)j(x)=\ln(x)


Qnkllcg paf Ijopefzexl


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




Uimlxqatlc


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


Wdncdeg qho izc iyjloulf Ezsypovpi


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


Kqxtthe tje Csmwtlxiba


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Pdvirzsmef


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


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Lqbzqux bzr nxu Laxqhzimj


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


Jkhomsoy xbb Wufsjcowcr


lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


Adbzjregek


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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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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m=0m=0


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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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