Предмет: Алгебра, автор: busyka4949

Найдите наименьшее значение ввражения x^2y^4+x^2-4xy^4-6x+6y^4

Ответы

Автор ответа: GluV

Ответ:

-9

Объяснение:

Перепишем функцию, как

$y^{4}(x^{2}-4x+6)+x^{2}-6x$

$x^{2}-4x+6>0$ , для любых x, т.к. дискриминант квадратичной формы меньше 0

$y^{4} (x^{2}-4x+6)>=0$

Равно 0 только при y=0.

Функция $x^{2} -6x$ достигает минимума при x=3, минимальное значение -9

Значит наименьшее значение функция достигает в точке (3,0) и оно равно 0-9=-9

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