Hàm số \(y = {{x^2} + 2x + 3} }}{x}\) có đạo hàm \(y' = \frac{{ax + {{x^2} + 2x + 3} }}\). Tìm \(\max \left\{ {a,b}

* Hướng dẫn giải

Ta có:

\(\begin{array}{l}y' = \frac{{\frac{{2x + 2}}{{2\sqrt {{x^2} + 2x + 3} }}.x - \sqrt {{x^2} + 2x + 3} }}{{{x^2}}}\\y' = \frac{{{x^2} + x - {x^2} - 2x - 3}}{{{x^2}\sqrt {{x^2} + 2x + 3} }}\\ = \frac{{ - x - 3}}{{{x^2}\sqrt {{x^2} + 2x + 3} }}\\ \Rightarrow \left\{ \begin{array}{l}a =  - 1\\b =  - 3\end{array} \right.\\ \Rightarrow \max \left\{ {a;b} \right\} = \max \left\{ { - 1; - 3} \right\}\\ =  - 1\end{array}\)