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2阶偏导判别法,可否举出实例?

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dodonaomik Posted at 2016-7-8 22:20:35 |Read mode
Last edited by hbghlyj at 2025-3-31 00:14:33二阶偏导检验法
函数f在(x,y)附近,乃是连续的二阶偏导数, 且△f(x
现令D=fxx(x,y)fyy(x,y)-fxy²(x,y)
假使D>0,且fxx(x,y)<0⇒f(x,y)为极大值
假使D>0,且fxx(x,y)>0⇒f(x,y)为极小值
假使D<0, f(x,y)就不是极值,(x,y)称为鞍点
那么,D=0的话,检验法失效,

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川木 Posted at 2016-7-9 20:40:30
你可以考虑各种平面二次和三次曲线在三维空间里平移出的曲面, 然后你可以搜一下椭球面, 鞍面, 自己算一下.

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hbghlyj Posted at 2023-1-6 03:15:01

实例

$f = x^2 + k y^2$
$f_{xx}(0,0)=2$
$f_{yy}(0,0)=2k$
$f_{xy}(0,0)=0$
所以$D=4k$
当$k>0$时$D>0$, 因为$f_{xx}(0,0)>0$, 所以$(0,0)$为极小值
当$k<0$时$D<0$, 则$(0,0)$为鞍点.

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