Предмет: Математика, автор: pavlodar311p40qc1

Найти промежутки возрастания функции у=11+12x-3х2

Ответы

Автор ответа: Miroslava227

1

Ответ:

$y = 11 + 12x - 3 {x}^{2}$

$y' = 12 - 6x \\ \\ y' = 0 \\ 12 - 6x = 0 \\ x = 2 \\ \\ + \: \: \: \: \: \: \: \: \: - \\ - - 2 - - > \\ \\ x\in( - \infty; 2 )$

- на этом промежутке функция возрастает

Автор ответа: NNNLLL54

1

Ответ:

$y=11+12x-3x^2\\\\y'=12-6x=0\ \ ,\ \ \ 6x=12\ \ ,\ \ x=2\\\\znaki\ y'(x):\ \ \ +++(2)---\\{}\qquad \qquad \qquad \ \ \ \ \nearrow \, \, \ (2)\ \ \searrow \\{}\qquad \qquad \qquad \qquad \ \ \ \, max\\\\y(x)\ vozrastaet\ \ pri\ \ x\in (-\infty ;\ 2\ ]$

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