Question

Найдите наибольшее значение функции y= (x^2+49)/x на отрезке [-19;-1]

Answer 1

$y= \frac{x^2+49}{x}\; ,\; \; ODZ:\; \; x\ne 0\\\\y=x+\frac{49}{x}\\\\y'=1-\frac{49}{x^2}=0\; ,\; \; \; \frac{49}{x^2} =1\; ,\; \; \; x^2=49\; ,\; \; x=\pm 7\\\\+++(-7)---(0)---(7)+++\\\\.\quad \nearrow(-7)\quad \searrow \; \; (0)\quad \searrow \quad (7)\quad \nearrow \\\\x_{max}=-7\; ,\; \; x_{min}=7\\\\x\in [\, -19,-1\, ]\; ,\; \; \; -7\in [-19,-1\, ]\\\\y(-19)\approx -21,58\\\\y(-7)=-14\\\\y(-1)=-50\\\\y_{naibolshee}=y(-7)=-14$