Lösung von Teilaufgabe c) 2.: Unterschied zwischen den Versionen

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(Verwendung der Tangentialgleichung)
(Berechnung derjenigen Punkte, für welche die Tangente an den Graphen von f2 durch den Ursprung verläuft)
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:                      <math> \Rightarrow B_1(1 + \sqrt{3} / 2{,}601)</math>
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:                      <math> \Rightarrow B_1(1 + \sqrt{3} / 2{,}601)</math>   [[Bild:TANGENTE_b1.png|400px]]
  
  
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:                      <math> \Rightarrow B_2(1 - \sqrt{3} / -310{,}164)</math>
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:                      <math> \Rightarrow B_2(1 - \sqrt{3} / -310{,}164)</math>   [[Bild:TANGENTE_b2.png|400px]]
  
  
  
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=== Bilder als Hilfen ===
<br />
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[[Bild:TANGENTE_b1b2.png|400px]]
 
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Version vom 26. Januar 2010, 04:22 Uhr

Berechnung derjenigen Punkte, für welche die Tangente an den Graphen von f2 durch den Ursprung verläuft

Verwendung der Tangentialgleichung

y = f^{'}( x_0)\cdot ( x - x_0 ) + f ( x_0 )
 y = ( x_0 - a - 1 )\cdot ( -e^{a + 2 - x_0})\cdot ( x - x_0 ) + ( x_0 - a )\cdot e^{a + 2 - x_0})


mit:\;

y = 0\;
x = 0\;
a = 2\;


0 = ( x_0 - 3 )\cdot ( -e^{4 - x_0} )\cdot ( -x_0 ) + ( x_0 - 2 )\cdot ( e^{4 - x_0} )
 0 = ( x_0 - 3 )\cdot ( e^{4 - x_0} )\cdot ( x_0 ) + ( x_0 - 2 )\cdot ( e^{4 - x_0} )
 0 = ( x_0^{2} - x_0\cdot 3 )\cdot ( e^{4 - x_0} ) + ( x_0 - 2 )\cdot ( e^{4 - x_0} )
 0 = e^{4 - x_0}\cdot ( x_0^{2} - 3\cdot x_0 +  x_0 - 2 )
 0 = e^{4 - x_0}\cdot ( x_0^{2} - 2\cdot x_0 - 2 )\;\;\;\;\;\;\;\;|e^{4 - x_0}>0
\Rightarrow ( x_0^{2} - 2\cdot x_0 - 2 ) = 0

Lösen quadratischer Gleichungen mit Hilfe der Mitternachtsformel  x_{1,2} = \frac{-b\pm\sqrt{b^{2}-4\cdot a\cdot c}}{2a}

 x_{1,2} = \frac{2\pm\sqrt{4--8}}{2}
 x_{1,2} = \frac{2\pm\sqrt{4+8}}{2}
 x_{1,2} = \frac{2\pm\sqrt{12}}{2}
 x_{1,2} = \frac{2\pm\sqrt{4\cdot 3}}{2}
 x_{1,2} = \frac{2\pm2\cdot\sqrt{3}}{2}
 x_{1,2} = {1\pm\sqrt{3}}


\Rightarrow x_{1} = {1 + \sqrt{3}}
\Rightarrow x_{2} = {1 - \sqrt{3}}


f_a(x_1)=\;
= f_a(1 + \sqrt{3})\;
= ( 1 + \sqrt{3} - a )\cdot e^{a + 2 - ( 1 + \sqrt{3})}
 = ( 1 + \sqrt{3} - 2 )\cdot e^{2 + 2 - ( 1 + \sqrt{3})}
 = ( \sqrt{3} - 1 )\cdot e^{4 - 1 - \sqrt{3})}
 = ( \sqrt{3} - 1 )\cdot e^{3 - \sqrt{3})}
\approx 2{,}601


 \Rightarrow B_1(1 + \sqrt{3} / 2{,}601) TANGENTE b1.png


f_a(x_2) =\;
= f_a(1 - \sqrt{3})\;
 = ( 1 - \sqrt{3} - a )\cdot e^{a + 2 - ( 1 - \sqrt{3})}
 = ( 1 - \sqrt{3} - 2 )\cdot e^{2 + 2 - ( 1 - \sqrt{3})}
 = ( -\sqrt{3} - 1 )\cdot e^{4 - 1 + \sqrt{3})}
 = ( -\sqrt{3} - 1 )\cdot e^{3 + \sqrt{3})}
\approx -310{,}164


 \Rightarrow B_2(1 - \sqrt{3} / -310{,}164)
Fehler beim Erstellen des Vorschaubildes: Ungültige Thumbnail-Parameter


Bilder als Hilfen

TANGENTE b1b2.png