Question

Найдите точку максимума функции у= -х^2+49/x

Answer 1

$y=- \frac{x^2+49}{x} \\\\y'=- \frac{2x\cdot x-(x^2+49)\cdot 1}{x^2}= -\frac{x^2-49}{x^2} =- \frac{(x-7)(x+7)}{x^2}=0\\\\x_1=-7,\; x_2=7\\\\Znaki\; y':\; \; +++(-7)---(7)+++ \\\\.\qqiad\qquad \qquad \qquad \nearrow \; (-7)\; \; \searrow \; \; \; \; (7)\; \; \nearrow \\\\x_{max}=-7\\\\y_{max}=- \frac{(-7)^2+49}{-7} =14\\\\Tochka\; \; \; (-7,14)\; .$