Cho hàm số \(y = f\left( x \right) = \frac{{2x + 1}}{{1 - x}}\).

* Hướng dẫn giải

\(f'\left( x \right) = \frac{3}{{{{\left( {x - 1} \right)}^2}}} \Rightarrow f''\left( x \right) =  - \frac{6}{{{{\left( {x - 1} \right)}^3}}}\)

\(f'\left( x \right) + f''\left( x \right) = 0 \Leftrightarrow \frac{3}{{{{\left( {x - 1} \right)}^2}}} - \frac{6}{{{{\left( {x - 1} \right)}^3}}} = 0 \Leftrightarrow 1 = \frac{2}{{x - 1}} \Leftrightarrow x = 3.\)