Question

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

Answer 1

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