设函数y=y(x)由方程2y^3-2y^2+2xy-x^2=1所确定，求函数的极值点并求极值

显示全部楼层 · 2010-11-17 13:26:29

2y^3-2y^2+2xy-x^2=12y^3-y^2-(y^2-2xy+x^2)-1=02y^3-y^2-1=(x-y)^2(y-1)(2y^2+y+1)=(x-y)^2因(x-y)^2>=0,2y^2+y+1=2(y+1/4)^2+7/8>0则y-1>=0,y>=1即y有最小值1,此时x-y=0,x=1所以函数y=y(x)当x=1时有极小值1

千问 · 2010-11-17 13:26:29

两边对x求导：6y^2*y'-4y*y'+2y+2xy'-2x=0即y'=(x-y)/(3y^2-2y+x)令y'=0,得：x=y再将x=y代入原方程，得：2x^3-2x^2+2x^2-x^2=1,得：2x^3-x^2-1=02x^3-2x^2+x^2-1=02x^2(x-1)+(x-1)(x+1)=0(x-1)(2x^2