Câu hỏi 29

Ta có \(y' = {x^2} + x - 2 = 0 \Leftrightarrow \left[ \begin{array}{l}x = 1\,\,\,\,\,\,\,\,\left( {tm} \right)\\x =  - 2\,\,\,\left( {loai} \right)\end{array} \right.\).

Ta có \(y\left( 1 \right) =  - \dfrac{{13}}{6};\,\,\,y\left( 0 \right) =  - 1;\,\,y\left( 2 \right) =  - \dfrac{1}{3}\).

\( \Rightarrow {y_{\min }} = y\left( 1 \right) =  - \dfrac{{13}}{6}\).

Chọn B