Question

Найдите точку минимума y=x^2+1000/x

Answer 1

Y = x^2 + 1000/x
y ' = 2x - 1000/x^2 = (2x^3 - 1000)/x^2 = 2(x^3 - 500)/x^2 = 0
x^3 = 500
x0 = ∛500 ~ 8 (8^3 = 512)
y(x0) = ∛(500^2) + 1000/(∛500) = (500 + 1000)/(∛500) = 1500/(∛500) =
= 1500*∛(500^2)/500 = 3∛(250000) = 30∛250
Ответ: (∛500; 30∛250)

Answer 2

$y= \frac{x^2+1000}{x} \; ,\; \; x\ne 0\\\\y'= \frac{2x\cdot x-(x^2+1000)\cdot 1}{x^2} = \frac{x^2-1000}{x^2} =0\; ,\; \; \; x\ne 0\\\\x^2-1000=0\; ,\; \; (x-\sqrt{1000})(x-\sqrt{1000})=0\; ,\\\\x=\pm \sqrt{1000}=\pm 10\sqrt{10}\\\\Znaki\; y'\; :\; \; +++(-10\sqrt{10})---(10\sqrt{10})+++\\\\x_{max}=-10\sqrt{10}\; ,\; \; \underline {x_{min}=10\sqrt{10}}$

$y_{min}=20\sqrt{10}\\\\A(10\sqrt{10}\, ;\, 20\sqrt{10})$