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) |
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− | == Tangente im Punkt W<sub>a</sub> ( a + 2 / 2 ) an G<sub>f<sub>a</sub></sub> mit dem Schnittpunkt A (0 / 2012 ) == | + | == Tangente im Punkt W<sub>a</sub>( a + 2 / 2 ) an G<sub>f<sub>a</sub></sub> mit dem Schnittpunkt A (0 / 2012 ) == |
+ | <math>mit:\;</math><br /> | ||
+ | |||
+ | :<math>x = 0\;</math><br /> | ||
+ | :<math>y = 2012\;</math><br /> | ||
+ | :<math>x_0 = a + 2\;</math><br /> | ||
+ | :<math>f_a( x_0 ) = f_a( a + 2 ) = 2\;</math><br /> | ||
+ | :<math>f^{'}_a( x_0 ) = f^{'}_a( a + 2 ) = m = -1\;</math><br /> | ||
− | === | + | |
+ | : <math> f^{'}_a( a + 2 ) = e^{a + 2 - ( a + 2 )}\cdot ( 1 + a - ( a + 2 ) )</math><br /> | ||
+ | :::: <math> = e^{a + 2 - a - 2 }\cdot ( 1 + a - a - 2 ) )</math><br /> | ||
+ | :::: <math> = e^{0}\cdot ( -1 ) )</math><br /> | ||
+ | ::::<math>= -1\;</math><br /> | ||
− | |||
− | + | === Lösung; Tangentengleichung === | |
+ | Tangentengleichung: siehe Formelsammlung Seite 58<br /> | ||
+ | <math>y = f^{'}( x_0 )\cdot ( x - x_0 ) + f ( x_0)</math><br /> | ||
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− | + | :: <math>y = f^{'}_a( a + 2 )\cdot ( x - ( a + 2 )) + f ( a + 2 )</math><br /> | |
− | + | :: <math>y = (-1)\cdot ( x - a - 2 ) + 2</math><br /> | |
− | + | :: <math>y = -x + a + 2 + 2\;</math> <br /> | |
− | + | :: <math>y = -x + a + 4\;</math> <br /> | |
− | + | : <math>2012 = 0 + a + 4\;\;\;\;\;\;\; | -4</math><br /> | |
− | + | :: <math>a = 2008\;</math><br /> | |
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=== Lösung; Fußweg === | === Lösung; Fußweg === | ||
− | + | ::: <math> y = m\cdot x + t </math><br /> | |
− | + | ||
− | + | :: <math>f_a( x_0 ) = f^{'}_a( x_0 )\cdot x_0 + t </math><br /> | |
− | + | : <math> f_a( a + 2 ) = f^{'}_a( a + 2 )\cdot x_0 + t</math><br /> | |
− | + | :::: <math>2 = -1\cdot x_1 + t \;\;\;\;\;\; | - ( -1\cdot x_0)</math><br /> | |
− | + | :::: <math>t = 2 - ( -1\cdot x_0 )</math><br /> | |
− | + | :::: <math>t = 2 - ( -1\cdot ( a + 2 ))</math><br /> | |
− | + | :::: <math>t = 2 - ( -a - 2)\;</math><br /> | |
− | + | :::: <math>t = 2 + a + 2 \;</math><br /> | |
+ | :::: <math>t = a + 4 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; |\;einsetzen\; in\; y = m\cdot x + t</math><br /> | ||
− | + | ||
− | + | :: <math>y = m\cdot x + a + 4</math><br /> | |
− | + | : <math> 2012 = -1\cdot 0 + a + 4</math><br /> | |
− | + | : <math>2012 = a + 4 \;</math><br /> | |
+ | :: <math>a = 2008\;</math><br /> | ||
=== Lösung; Clever === | === Lösung; Clever === | ||
− | <math>\frac{ | + | :: <math>\frac{y_2 - y_1}{x_2 - x_1} = f{'}_a ( x )</math><br /> |
+ | |||
+ | : <math>\frac{2012 - 2}{0 - ( a + 2 )} = -1 </math><br /> | ||
+ | |||
+ | :: <math>\frac{2010}{(-a - 2 )} = -1 \;\;\;\;\;\;\;\;\;\;| \cdot( -a - 2 )</math><br /> | ||
+ | |||
+ | :::: <math>2010 = a + 2\;</math><br /> | ||
+ | :::: <math>2008 = a\;</math><br /> | ||
− | + | == Berechnung derjenigen Punkte, für welche die Tangente an den Graphen von f<sub>2</sub> durch den Ursprung verläuft == | |
− | |||
− | + | === Verwendung der Tangentialgleichung === | |
− | + | ||
+ | : <math>y = f^{'}( x_0)\cdot ( x - x_0 ) + f ( x_0 )</math><br /> | ||
− | + | :<math> y = ( x_0 - a - 1 )\cdot ( -e^{a + 2 - x_0})\cdot ( x - x_0 ) + ( x_0 - a )\cdot e^{a + 2 - x_0})</math><br /> | |
− | === | + | <math>mit:\;</math><br /> |
+ | : <math>y = 0\;</math><br /> | ||
+ | : <math>x = 0\;</math><br /> | ||
+ | : <math>a = 2\;</math><br /> | ||
+ | |||
+ | |||
+ | |||
+ | :: <math>0 = ( x_0 - 3 )\cdot ( -e^{4 - x_0} )\cdot ( -x_0 ) + ( x_0 - 2 )\cdot ( e^{4 - x_0} )</math> | ||
+ | |||
+ | :: <math> 0 = ( x_0 - 3 )\cdot ( e^{4 - x_0} )\cdot ( x_0 ) + ( x_0 - 2 )\cdot ( e^{4 - x_0} )</math> | ||
+ | |||
+ | :: <math> 0 = ( x_0^{2} - x_0\cdot 3 )\cdot ( e^{4 - x_0} ) + ( x_0 - 2 )\cdot ( e^{4 - x_0} )</math> | ||
+ | |||
+ | :: <math> 0 = e^{4 - x_0}\cdot ( x_0^{2} - 3\cdot x_0 + x_0 - 2 )</math> | ||
+ | |||
+ | :: <math> 0 = e^{4 - x_0}\cdot ( x_0^{2} - 2\cdot x_0 - 2 )\;\;\;\;\;\;\;\;|e^{4 - x_0}>0</math> | ||
+ | |||
+ | :: <math>\Rightarrow ( x_0^{2} - 2\cdot x_0 - 2 ) = 0</math> | ||
+ | |||
+ | Lösen quadratischer Gleichungen mit Hilfe der [http://de.wikipedia.org/wiki/Mitternachtsformel?title=Mitternachtsformel&redirect=no Mitternachtsformel] <math> x_{1,2} = \frac{-b\pm\sqrt{b^{2}-4\cdot a\cdot c}}{2a}</math> | ||
+ | |||
+ | : <math> x_{1,2} = \frac{2\pm\sqrt{4--8}}{2}</math><br /> | ||
+ | : <math> x_{1,2} = \frac{2\pm\sqrt{4+8}}{2}</math><br /> | ||
+ | : <math> x_{1,2} = \frac{2\pm\sqrt{12}}{2}</math><br /> | ||
+ | : <math> x_{1,2} = \frac{2\pm\sqrt{4\cdot 3}}{2}</math><br /> | ||
+ | : <math> x_{1,2} = \frac{2\pm2\cdot\sqrt{3}}{2}</math><br /> | ||
+ | : <math> x_{1,2} = {1\pm\sqrt{3}}</math><br /> | ||
+ | |||
+ | |||
+ | : <math>\Rightarrow x_{1} = {1 + \sqrt{3}}</math><br /> | ||
+ | |||
+ | : <math>\Rightarrow x_{2} = {1 - \sqrt{3}}</math><br /> | ||
+ | |||
+ | |||
+ | |||
+ | : <math>f_a(x_1)=\;</math><br /> | ||
+ | : <math>= f_a(1 + \sqrt{3})\;</math> | ||
+ | : <math>= ( 1 + \sqrt{3} - a )\cdot e^{a + 2 - ( 1 + \sqrt{3})}</math><br /> | ||
+ | : <math> = ( 1 + \sqrt{3} - 2 )\cdot e^{2 + 2 - ( 1 + \sqrt{3})}</math><br /> | ||
+ | : <math> = ( \sqrt{3} - 1 )\cdot e^{4 - 1 - \sqrt{3})}</math><br /> | ||
+ | : <math> = ( \sqrt{3} - 1 )\cdot e^{3 - \sqrt{3})}</math><br /> | ||
+ | : <math>\approx 2{,}601</math><br /> | ||
+ | |||
+ | |||
+ | : <math> \Rightarrow B_1(1 + \sqrt{3} / 2{,}601)</math> | ||
+ | |||
− | |||
− | + | : <math>f_a(x_2) =\;</math><br /> | |
+ | : <math>= f_a(1 - \sqrt{3})\;</math> | ||
+ | : <math> = ( 1 - \sqrt{3} - a )\cdot e^{a + 2 - ( 1 - \sqrt{3})}</math><br /> | ||
+ | : <math> = ( 1 - \sqrt{3} - 2 )\cdot e^{2 + 2 - ( 1 - \sqrt{3})}</math><br /> | ||
+ | : <math> = ( -\sqrt{3} - 1 )\cdot e^{4 - 1 + \sqrt{3})}</math><br /> | ||
+ | : <math> = ( -\sqrt{3} - 1 )\cdot e^{3 + \sqrt{3})}</math><br /> | ||
+ | : <math>\approx -310{,}164</math><br /> | ||
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− | + | : <math> \Rightarrow B_2(1 - \sqrt{3} / -310{,}164)</math> | |
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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 )
Lösung; Tangentengleichung
Tangentengleichung: siehe Formelsammlung Seite 58
Lösung; Fußweg
Lösung; Clever
Berechnung derjenigen Punkte, für welche die Tangente an den Graphen von f2 durch den Ursprung verläuft
Verwendung der Tangentialgleichung
Lösen quadratischer Gleichungen mit Hilfe der Mitternachtsformel