Bài 2: \(y=\dfrac{x^2-8x+7}{x^2+1}\)
TXĐ: \(D=\mathbb Z\)
\(y'=\dfrac{2x-8}{(x^2+1)^2}=0\Leftrightarrow x=4\)
\(y(4)=\dfrac{-9}{17}\)
\(\lim\limits_{x \to -\infty} \dfrac{x^2-8x+7}{x^2+1}=\lim\limits_{x \to -\infty} \dfrac{1-\dfrac{8}{x}+\dfrac{7}{x^2}}{1+\dfrac{1}{x^2}}=1\)
\(\lim\limits_{x \to +\infty} \dfrac{x^2-8x+7}{x^2+1}=\lim\limits_{x \to -\infty} \dfrac{1-\dfrac{8}{x}+\dfrac{7}{x^2}}{1+\dfrac{1}{x^2}}=1\)
BBT như hình vẽ
Vậy hàm số đạt GTNN bằng \(\dfrac{-9}{17}\) tại \(x=4\).
