Найдите промежутки, на которых функция = х^2 / x+1 убывает.. Created by divaella. algebra-ru

x+1 убывает...

divaella @divaella

July 2022 1 16 Report

Найдите промежутки, на которых функция = х^2 / x+1 убывает.

Answers & Comments

zzhash

Функция y убывает там, где y' < 0

$y'=\frac{(x^2)'\cdot (x-1) - x^2 \cdot (x-1)' }{(x-1)^2} = \frac {2x(x-1) - x^2\cdot 1}{(x-1)^2} = \frac{2x^2-2x-x^2}{(x-1)^2} = \frac{x^2-2x}{(x-1)^2}\\\frac{x^2-2x}{(x-1)^2} < 0\\x \in (0;1) \cup (1;2)$

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