设y=y(x)由x3+3x2y-2y3=2确定,求y=y(x)的极值.

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问题 设y=y(x)由x3+3x2y-2y3=2确定,求y=y(x)的极值.

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答案x3+3x2y-2y3=2两边对x求导得3x2+6xy+3x2y-6y2y=0.[*]两边再对x求导得6x+6y+12xy+3x2y’’-12yy’2-6y2y’’=0.x=0时,y’’(0)=﹣1,x=0为极大值点,极大值为y=﹣1;x=﹣2时,y’’(﹣2)=﹣l,为y=y(x)的极小值点,极小值为y=1.

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