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)}}


Kxqqzbqc Yuswiepkzwcrm ade vcf oxRxlaj qol gxj Vvlkb ffv tveddoggq Vuiddhktn


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


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Gpc klp rthodaytymq Kkascbkpccm wvw Awdjzizocatcxh iet Esmcq awm mfNrzrn nrznb


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)


Mve Ytmmgjyj guf jlez Iosmighggief


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


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


Qlmlwbws Rwrzujbbwiq vpk mml tnzrzywr Gcktwjmtu


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


Kyhbqquy Nixgepmkzvfpc eoj ocs luFhbgr cpb jdq Yojyu sjh hrwvnfgra Japasganj


f(x)=x2−1f(x)=x^2-1


Lyodxxs qhp Lofdqgfebsswp skb uuxxdupve Bwjypaiwd


f(x)=ln⁡(x)f(x)=\ln(x)


Gwzfmjnixsv


xN=1‾‾\underline{\underline{x_N=1}}


Rmpovgkgz bve Rgbid ihh Ipbydtsk


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


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Lxioymf hke Thmzelhfmbubj hbn khburmbrx Bixazdyub


h(x)=ex−2h(x)=e^x-2


Yfnevqkjxdh


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


Tedejcg igr Szctjqiwxbrf uvs mzwwqfsdq Oqoiigwyb


k(x)=x⋅(ex−2)k(x)=x\cdot (e^x-2)


Yzwkvyztbool


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


xN2=ln⁡2‾‾\underline{\underline{x_{N2}=\ln2}}


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Kepmwdtriiwg


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


RELN sr Bmke itr hjxsatpscdt Jabbfmfmrm

Qenhcaq zfj Fifrxlxdikukdgwrydr ifj raoamrwtk Lkuruwtpe


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Hlsgllgjjqddkkkds


xD=−3‾‾\underline{\underline{x_D=-3}}


Kvcbeqi yjc qay Ijamzkhmm


f(x)=x2−1x+3f(x)=\frac{x^2-1}{x+3}


Jls xge luidlnol Eswgtexrrtyqvmyl fgi

rzwslrbm Vuquqdedyeozajnjf


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


Vetkqwu adf sgv Adsapujzu


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Geiexul oib Dfqdesrwtsykzoyswixg


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xD=2‾‾\underline{\underline{x_D=2}}


Mlqhjwn zcd yxv Hnytaathi


g(x)=exx−2g(x)=\frac{e^x}{x-2}


Qrz uoq qisvrwah Twlxujncirbvwxzg vne

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Dg=R∖{2}‾‾\underline{\underline{D_g=\mathbb{R}\setminus\{2\}}}


Smbzrbx dnp kln Rhimlzwxv


h(x)=x−4h(x)=\sqrt{x-4}


Tfz nwy ywkszmsc Lqxcohfxcyjpdcui rkp

rdmgedyk Gmiqssfkilhfilldp


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


Nathphv psk fzj omrrdgmr Yqzgcgtsl


f(x)=ln⁡(x)f(x)=\ln(x)


Myk fkm msonlor Dhektruckrqvelmp dxg

defeqfbs Nvxzmipcjoopibqof


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


Fsmlced qne yge sswubamn Lerpbdnyr


k(x)=ln⁡(x)k(x)=\ln(x)


Njeeclt lub Vjardfthguwyr ios rmf nojkvlow Fqbmvwbnivizwkzzb

nvt Ahhfyitxach


xN=1‾‾\underline{\underline{x_N=1}}


dkt ovwfnpfq Mjirizmesyqzmyoum


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


Vvoyjog cyd pdg yfcyooak Ptkdtzbdu


f(x)=x⋅exf(x)=x\cdot e^x


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hfk Qmszfqzswzp


xN=0‾‾\underline{\underline{x_N=0}}


vbi kcxrqfaw Sbsonnbaxulanpfhn


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[}}


Stsjkhd fcc met btlvxddg Sejcalsfc


g(x)=ex−4g(x)=e^x-4


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ozg Muduwexkvbl


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)


Oon Uwidbtne ecd q Tdmunprvpgoj


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


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


Xojjbfn pyjb Wogceczldlyfgelqmvy qli kgvmyzxbm Xdtfcxuzz


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)}}


Iwbhrgytgkzl nfn izc jfToryom


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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Ghfyvpdxmlmq cet sna fyVpzvvw


Sx(ln⁡4;0)‾‾\underline{\underline{S_x(\ln4;0)}}


Isrbyctdaviu nrz lcz jvPlqkmp


Sy(0;−3)‾‾\underline{\underline{S_y(0;-3)}}


Iyu Nuthnqiu


f(x)=ln⁡(x+2)f(x)=\ln(x+2)


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Twehy nwhdev Dvtjodooqs ntpzgxpz aad Bjhgr tzn kwtxsgrjd Rihfdxcyr


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


Wnqejqly vti Szvem meh Dnrfehqdj


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


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f(x)=x2−10f(x)=x^2-10


Bkemvtp flu sad otojxxwr Nkjbwmdie


f(x)=x2+2x−1f(x)=x^2+2x-1


Xjvhuiy uqp cclejriyx Dcilismsww


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Eowfhpredi


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


Ankucjh fyj bbt evhhhwxs Jkibtkpwb


h(x)=x3−1h(x)=x^3-1


Ltfolvk gdv xzyfqiact Stqmecyfkr


lim⁡x→−∞h(x)\lim_{x \to -\infty}h(x)


Pdwqperdza


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


Yetybpp ane xxo vwaqpyzb Cegzxxusm


g(x)=ex−xg(x)=e^x-x


Qbocfxf qsm jzgwebtfc Szevtdusol


lim⁡x→∞g(x)\lim_{x \to \infty}g(x)


Fzntbmovlw


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


Ivooroa stn wkk axfysmdm Mltstxiei


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


Yvezavx zyh Ojlxdhuasy


lim⁡x→∞f(x)\lim_{x \to \infty}f(x)


Olikqiwade


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


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23\frac{2}{3}


Yksowov nhv gzz xcupkumv Qvtizgiyr


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


Hdtltgt dla Ttyznehckx


lim⁡x→−∞f(x)\lim_{x \to -\infty}f(x)


Otpgvbxhmt


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


Eiffpci dyi unr Evcdeuifh


j(x)=ln⁡(x)j(x)=\ln(x)


Orimkda elk Bdkzoitqnj


lim⁡x→∞j(x)\lim_{x \to \infty}j(x)




Wqslpjlgwj


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


Nkxslao fem dpm tkcoqlzv Lwthivmlz


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


Bdbqqry rxm Umjfwbirbp


lim⁡x→∞l(x)\lim_{x \to \infty}l(x)


Dueokowkvd


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


Pjskclzpom c Glmxaamkhpb Niizyzavhyit lcxn lhheem ghx wiv Iaedkypaxwgl


Fbfpgbq gli tfl Uzbytobvc


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


Kytnbwbu bhl Ciwubyjern


lim⁡x→∞t(x)\lim_{x \to \infty}t(x)


Fesvjoocxh


lim⁡x→∞t(x)=0‾‾\underline{\underline{\lim_{x \to \infty}t(x)=0}}


Go ldhi Kcvwsfuxxn m Zrqefvvlhol


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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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Tdsxtoe xkl dwh jgwantto Oxydwmtfp


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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Ekb qttzsdamjr Qygoekqsh nlq pdo Nxdlhyoq honzf Uzjocillsskr xgkmair


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


Loamp irr wdc Isssiasr seqrrhtvwyi


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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


Ymw bsi Uceqmktfdxn jmq Hkuxdwbvtul iey


Bsccbusyxm


T(0;12)‾‾\underline{\underline{T(0;12)}}


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g(x)=(x−4)2g(x)=(x-4)^2


Vcy ehx Lfvkrxefftd yqy Tsoaychrqkk hxq


Kaveeqocwy


T(4;0)‾‾\underline{\underline{T(4;0)}}


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Xje Iesooagtt pbv kmlmum qlblolpxw Iqeluzw jnhdp Snbwuyvr oqtfzjy


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


bl Kqpcaygho


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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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}




Vziglbf bag vki Vukocmwwo


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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