Lösung von Teilaufgabe c: Unterschied zwischen den Versionen

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(Berechnung derjenigen Punkte, für welche die Tangente an den Graphen von f2 durch den Ursprung verläuft)
(Berechnung derjenigen Punkte, für welche die Tangente an den Graphen von f2 durch den Ursprung verläuft)
 
Zeile 136: Zeile 136:
 
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Aktuelle Version vom 24. Januar 2010, 02:23 Uhr

Inhaltsverzeichnis

Tangente im Punkt Wa( a + 2 / 2 ) an Gfa mit dem Schnittpunkt A (0 / 2012 )

mit:\;

x = 0\;
y = 2012\;
x_0 = a + 2\;
f_a( x_0 ) = f_a( a + 2 ) = 2\;
f^{'}_a( x_0 ) = f^{'}_a( a + 2 ) = m = -1\;


  f^{'}_a( a + 2 ) = e^{a + 2 - ( a + 2 )}\cdot ( 1 + a - ( a + 2 ) )
 = e^{a + 2 - a - 2 }\cdot ( 1 + a -  a - 2 ) )
 = e^{0}\cdot ( -1 ) )
= -1\;


Lösung; Tangentengleichung

Tangentengleichung: siehe Formelsammlung Seite 58

y = f^{'}( x_0 )\cdot ( x - x_0 ) + f ( x_0)


y = f^{'}_a( a + 2 )\cdot ( x - ( a + 2 )) + f ( a + 2 )
y = (-1)\cdot ( x - a - 2 ) + 2
y = -x + a + 2 + 2\;
y = -x + a + 4\;
2012 = 0 + a + 4\;\;\;\;\;\;\;          | -4
a = 2008\;

Lösung; Fußweg

  y = m\cdot x + t
f_a( x_0 ) = f^{'}_a( x_0 )\cdot x_0 + t
 f_a( a + 2 ) = f^{'}_a( a + 2 )\cdot x_0 + t
2 = -1\cdot x_1 + t \;\;\;\;\;\;           | - ( -1\cdot x_0)
t = 2 - ( -1\cdot x_0 )
t = 2 - ( -1\cdot ( a + 2 ))
t = 2 - ( -a - 2)\;
t = 2 + a + 2 \;
t = a + 4 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;        |\;einsetzen\; in\; y = m\cdot x + t


y = m\cdot x + a + 4
 2012 = -1\cdot 0 + a + 4
2012 = a + 4 \;
a = 2008\;

Lösung; Clever

\frac{y_2 - y_1}{x_2 - x_1} = f{'}_a ( x )
\frac{2012 - 2}{0 - ( a + 2 )} = -1
\frac{2010}{(-a - 2 )} = -1 \;\;\;\;\;\;\;\;\;\;| \cdot( -a - 2 )
2010 = a + 2\;
2008 = a\;

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)


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)