Question

cho hàm số \(y=\dfrac{1}{x^2-2x+5}\). có bao nhiêu số nguyen \(x\in\left(-9;9\right)\) để \(y'\left(x\right)\ge0\)

Answer 1

\(y'=-\dfrac{2x-2}{\left(x^2-2x+5\right)^2}=\dfrac{2-2x}{\left(x^2-2x+5\right)^2}\)

\(y'\ge0\Leftrightarrow\dfrac{2-2x}{\left(x^2-2x+5\right)^2}\ge0\Rightarrow x\le1\)

Có \(1-\left(-8\right)+1=10\) số nguyên