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
