relative Extrema < Analysis < Hochschule < Mathe < Vorhilfe
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(Frage) beantwortet  | | Datum: | 06:44 So 13.02.2005 | | Autor: | Sue20 |
Gesucht sind die relativen Extrema folgender Funktion:
z = f(x,y) = [mm] (x-3)e^{x-1} [/mm] + [mm] 2e^{2y} [/mm] - 4y
Meine Lösung:
[mm] f_{x} [/mm] = [mm] e^{x-1}(1 [/mm] + (x-3)) = [mm] e^{x-1}(x-2)
[/mm]
[mm] f_{y} [/mm] = [mm] 4(e^{2y}-1)
[/mm]
[mm] f_{xx} [/mm] = [mm] e^{x-1}(2+(x-3)) [/mm] = [mm] e^{x-1}(x-1)
[/mm]
[mm] f_{xy} [/mm] = 0
[mm] f_{yy} [/mm] = [mm] 8e^{2y}
[/mm]
stationäre Punkte:
[mm] f_{x}(x,y) [/mm] = [mm] e^{x-1}(x-2) [/mm] = 0 (I)
=> x = 2
[mm] f_{y}(x,y) [/mm] = [mm] 4(e^{2y}-1) [/mm] = 0
[mm] 4e^{2y} [/mm] - 4 = 0
[mm] e^{2y} [/mm] = 1
2y = ln 1
2y = 0
y = 0
=> P(2,0)
Diskrimante D(x,y) = [mm] f_{xx}*f_{yy}-(f_{xy})²
[/mm]
= [mm] e^{x-1}(x-2)*8e^{2y}
[/mm]
P(2,0): D(2,0) = [mm] e^{2-1}(2-2)*8e^{2*0} [/mm] = 0
=> D=0 => keine Aussage
Stimmt das so oder könnte ich mich irgendwo verrechnet haben?
Vielen Dank!
MfG Sue
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