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

2y^3-2y^2+2xy-x^2=1
2y^3-y^2-(y^2-2xy+x^2)-1=0
2y^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