Question

Найдите точку максимума функции:
$y=- \frac{x^2+4}{x}$

Answer 1

$y=- \frac{x^2+4}{x} \\ \\ x \neq 0 \\ \\ y= \frac{-x^2-4}{x} \\ \\ y'= \frac{(-x^2-4)'(x)-(-x^2-4)(x)'}{x^2} = \frac{-2x^2+x^2+4}{x^2} = \frac{-x^2+4}{x^2} \\ \\ y'(x) =0 \\ \\ \frac{-x^2+4}{x^2}=0 \\ \\ -x^2+4=0 \\ \\ (x+2)(x-2)=0 \\ \\ x=-2; x=-2 \\ \\ -----[-2]+++++(0)+++++[2]----- \\ \\ x_{min} =-2; \\ \\ x_{max}=2$