Question

Найти наибольшее значение функции y=x^2+25/x на отрезке [-10;-1]

Answer 1

$y=\frac{x^2+25}{x}\; ,\; \; x\in [-10,-1\, ]\\\\ODZ:\; \; x\ne 0\\\\y'=\frac{2x\cdot x-(x^2+25)\cdot 1}{x^2}=\frac{x^2-25}{x^2}=\frac{(x-5)(x+5)}{x^2}=0\; \; \to \; \; \left \{ {{(x-5)(x+5)=0} \atop {x\ne 0}} \right.\\\\x_1=-5\; ,\; \; x_2=5\\\\Znaki\; y':\; \; +++(-5)---(0)---(5)+++\\\\.\qquad\qquad \qquad \nearrow \; \; (-5)\; \; \searrow \; \; (0)\; \; \searrow \quad (5)\; \; \; \nearrow \\\\x_{max}=-5\; ,\; \; x_{min}=5\\\\x\in [-10,-1\, ]:\\\\y(-10)=\frac{100+25}{-10}=-12,5\\\\y(-5)=\frac{25+25}{-5}=-10\\\\y(-1)=\frac{1+25}{-1}=-26\\\\y_{naiboishee}=y(-5)=-10$