设函数y=f(x)由方程2y^3-2y^2+2xy-x^2=1所确定的,求解f(x)的极值
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![设函数y=f(x)由方程2y^3-2y^2+2xy-x^2=1所确定的,求解f(x)的极值](/uploads/image/z/157116-12-6.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0y%3Df%EF%BC%88x%EF%BC%89%E7%94%B1%E6%96%B9%E7%A8%8B2y%5E3-2y%5E2%2B2xy-x%5E2%3D1%E6%89%80%E7%A1%AE%E5%AE%9A%E7%9A%84%2C%E6%B1%82%E8%A7%A3f%EF%BC%88x%EF%BC%89%E7%9A%84%E6%9E%81%E5%80%BC)
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设函数y=f(x)由方程2y^3-2y^2+2xy-x^2=1所确定的,求解f(x)的极值
设函数y=f(x)由方程2y^3-2y^2+2xy-x^2=1所确定的,求解f(x)的极值
设函数y=f(x)由方程2y^3-2y^2+2xy-x^2=1所确定的,求解f(x)的极值
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=0
2x^3-2x^2+x^2-1=0
2x^2(x-1)+(x-1)(x+1)=0
(x-1)(2x^2+x+1)=0
得唯一根x=1,
即极值点为(1,1),极值即为y=f(1)=1.