gebrochen rationale < Rationale Funktionen < Analysis < Oberstufe < Schule < Mathe < Vorhilfe
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| Status: |
(Frage) beantwortet  | | Datum: | 09:23 Do 17.09.2009 | | Autor: | itil |
| Aufgabe | | f(x) = [mm] \bruch{x^3}{x^2-16} [/mm] |
1) Polstellen:
[mm] x^2 [/mm] - 16 = 0
[mm] x^2 [/mm] = 16
x1 = +4
x2 = -4
______________________________________
Lücken:
f(4) = [mm] \bruch{4^3}{4^2-16} [/mm] = [mm] \bruch{64}{0} [/mm] = keine Lücke
f(-4) = [mm] \bruch{-4^3}{-4^2-16} [/mm] = [mm] \bruch{-64}{0} [/mm] = keine Lücke
______________________________________
Asymtote: Z>N --> Polynomdivision
[mm] (x^3+0x^2+0x+0):(x^2-16)=x [/mm] -16
______________________________________
Nullstellen:
f(x) = [mm] \bruch{x^3}{x^2-16} [/mm] = 0
[mm] x^3 [/mm] = 0
x1 = 0
x2 = 0
x3 = 0
______________________________________
Extremwerte: f'(x) = 0
f(x) = [mm] \bruch{x^3}{x^2-16}
[/mm]
f'(x) = u/v = [mm] \bruch{u'v - uv'}{v²}
[/mm]
= [mm] \bruch{3x^2 * (x^2-16) - ((x^3)(2x))}{(x^2-16)²}
[/mm]
= [mm] \bruch{3x^4 -48x^2 - 2x^4}{(x^2-16)²}
[/mm]
= [mm] \bruch{x^4 - 48x^2 }{(x^2-16)²}
[/mm]
[mm] \bruch{x^4 - 48x^2 }{(x^2-16)²} [/mm] = 0
[mm] x^4 [/mm] - [mm] 48x^2 [/mm] = 0
E1 = 0
E2 = 0
E3 = -48
E4 = 48
f(x) = [mm] \bruch{x^3}{x^2-16}
[/mm]
f(48) = 48,33566
f(-48) = 47,6689
__________________________________________
Wendepunkt: f''(x) = 0
f'(x) = [mm] \bruch{x^4 - 48x^2 }{(x^2-16)²}
[/mm]
f''(x) = [mm] [(4x^3-96x )*(x^4-32x^2+16)]-[(x^4 [/mm] - [mm] 48x^2 [/mm] ) * [mm] 2*(x^2-16) [/mm] *2x)]
[mm] [4x^7 [/mm] - [mm] 128x^5 [/mm] + [mm] 64x^3 -96x^4 [/mm] + [mm] 3072x^3 [/mm] + 1536x] - [ [mm] (2x^4 [/mm] - [mm] 96x^2)*(2x^3 [/mm] - 32x)]
[mm] [4x^7 [/mm] - [mm] 128x^5 [/mm] + [mm] 64x^3 -96x^4 [/mm] + [mm] 3072x^3 [/mm] + 1536x] - [mm] [4x^7 [/mm] - [mm] 64x^5 [/mm] - [mm] 192x^5 [/mm] + [mm] 3072x^3]
[/mm]
[mm] 4x^7 [/mm] - [mm] 128x^5 [/mm] + [mm] 64x^3 -96x^4 [/mm] + [mm] 3072x^3 [/mm] + 1536x - [mm] 4x^7 [/mm] + [mm] 64x^5 [/mm] + [mm] 192x^5 [/mm] - [mm] 3072x^3
[/mm]
f''(x) = [mm] \bruch{+ 128x^5 -96x^4 + 64x^3 + 1536x}{[(x^2-16)²]²}
[/mm]
[mm] \bruch{+ 128x^5 -96x^4 + 64x^3 + 1536x}{[(x^2-16)²]²} [/mm] = 0
+ [mm] 128x^5 -96x^4 [/mm] + [mm] 64x^3 [/mm] + 1536x = 0
Raten: 0
Rest im unreellen Zahlenbereich.
einsetzen:
f(0) = 0
________________________________________
Wendetangente
y = kx + d
0 = 0k + d
k = 0
d = 0
keine Wendetangente
________________________________________
korrekt?
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| Status: |
(Antwort) fertig  | | Datum: | 09:35 Do 17.09.2009 | | Autor: | fred97 |
> f(x) = [mm]\bruch{x^3}{x^2-16}[/mm]
> 1) Polstellen:
>
> [mm]x^2[/mm] - 16 = 0
> [mm]x^2[/mm] = 16
>
> x1 = +4
> x2 = -4
Korrekt !
> ______________________________________
>
> Lücken:
>
> f(4) = [mm]\bruch{4^3}{4^2-16}[/mm] = [mm]\bruch{64}{0}[/mm] = keine Lücke
Das ist Unfug ! 4 ist doch Polstelle ! Du teilst durch 0 ?!
>
> f(-4) = [mm]\bruch{-4^3}{-4^2-16}[/mm] = [mm]\bruch{-64}{0}[/mm] = keine
> Lücke
Wie oben: Unfug.
>
> ______________________________________
>
> Asymtote: Z>N --> Polynomdivision
>
> [mm](x^3+0x^2+0x+0):(x^2-16)=x[/mm] -16
Das ist falsch ! Richtig:
[mm]\bruch{x^3}{x^2-16}[/mm]= [mm] $x+\bruch{16x}{x^2-16}$
[/mm]
>
> ______________________________________
>
> Nullstellen:
>
>
> f(x) = [mm]\bruch{x^3}{x^2-16}[/mm] = 0
>
> [mm]x^3[/mm] = 0
> x1 = 0
> x2 = 0
> x3 = 0
Korrekt
> ______________________________________
>
> Extremwerte: f'(x) = 0
>
> f(x) = [mm]\bruch{x^3}{x^2-16}[/mm]
>
> f'(x) = u/v = [mm]\bruch{u'v - uv'}{v²}[/mm]
Im Quelltext hast Du es richtig:
f'(x) = u/v = [mm]\bruch{u'v - uv'}{v^2}[/mm]
> = [mm]\bruch{3x^2 * (x^2-16) - ((x^3)(2x))}{(x^2-16)²}[/mm]
Ebenso:
= [mm]\bruch{3x^2 * (x^2-16) - ((x^3)(2x))}{(x^2-16)^2}[/mm]
>
> = [mm]\bruch{3x^4 -48x^2 - 2x^4}{(x^2-16)²}[/mm]
>
> = [mm]\bruch{x^4 - 48x^2 }{(x^2-16)²}[/mm]
>
> [mm]\bruch{x^4 - 48x^2 }{(x^2-16)²}[/mm] = 0
Nochmal:
[mm]\bruch{x^4 - 48x^2 }{(x^2-16)^2}[/mm] = 0
>
> [mm]x^4[/mm] - [mm]48x^2[/mm] = 0
>
> E1 = 0
> E2 = 0
Korrekt
> E3 = -48
> E4 = 48
Nein: [mm] E_3 [/mm] = [mm] \wurzel{48}
[/mm]
[mm] E_4 [/mm] = [mm] -\wurzel{48}
[/mm]
>
> f(x) = [mm]\bruch{x^3}{x^2-16}[/mm]
> f(48) = 48,33566
> f(-48) = 47,6689
Siehe oben
FRED
> __________________________________________
>
> Wendepunkt: f''(x) = 0
>
> f'(x) = [mm]\bruch{x^4 - 48x^2 }{(x^2-16)²}[/mm]
>
> f''(x) = [mm][(4x^3-96x )*(x^4-32x^2+16)]-[(x^4[/mm] - [mm]48x^2[/mm] ) *
> [mm]2*(x^2-16)[/mm] *2x)]
>
> [mm][4x^7[/mm] - [mm]128x^5[/mm] + [mm]64x^3 -96x^4[/mm] + [mm]3072x^3[/mm] + 1536x] - [ [mm](2x^4[/mm]
> - [mm]96x^2)*(2x^3[/mm] - 32x)]
>
> [mm][4x^7[/mm] - [mm]128x^5[/mm] + [mm]64x^3 -96x^4[/mm] + [mm]3072x^3[/mm] + 1536x] - [mm][4x^7[/mm] -
> [mm]64x^5[/mm] - [mm]192x^5[/mm] + [mm]3072x^3][/mm]
>
> [mm]4x^7[/mm] - [mm]128x^5[/mm] + [mm]64x^3 -96x^4[/mm] + [mm]3072x^3[/mm] + 1536x - [mm]4x^7[/mm] +
> [mm]64x^5[/mm] + [mm]192x^5[/mm] - [mm]3072x^3[/mm]
>
>
> f''(x) = [mm]\bruch{+ 128x^5 -96x^4 + 64x^3 + 1536x}{[(x^2-16)²]²}[/mm]
>
>
> [mm]\bruch{+ 128x^5 -96x^4 + 64x^3 + 1536x}{[(x^2-16)²]²}[/mm] =
> 0
>
> + [mm]128x^5 -96x^4[/mm] + [mm]64x^3[/mm] + 1536x = 0
>
> Raten: 0
> Rest im unreellen Zahlenbereich.
>
> einsetzen:
>
> f(0) = 0
> ________________________________________
>
> Wendetangente
>
> y = kx + d
> 0 = 0k + d
> k = 0
> d = 0
>
> keine Wendetangente
> ________________________________________
>
>
> korrekt?
>
>
>
>
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| Status: |
(Frage) beantwortet | | Datum: | 10:02 Do 17.09.2009 | | Autor: | itil |
stimmt mein wendepunkt? das wollte ich eigentlich wissen 
vielen dank für den rest!
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| Status: |
(Antwort) fertig | | Datum: | 10:28 Do 17.09.2009 | | Autor: | M.Rex |
Hallo
Die Notwendige Bedingung für den Wendepunkt f''(0)=0 ist erfüllt, aber du hast die hinreichende Bedingung nicht "abgeklopft". Hättest du das getan, hättest du auch festgestellt dass 0 eine Sattelstelle ist.
Herauszufinden, was das dann für Auswirkungen auf die Satteltangente hat, überlasse ich dann dir.
Marius
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