Học lớp hướng dẫn giải
\(\\ y' = 3{x^2} + 6x - 9,y' = 0 \Leftrightarrow \left[ \begin{array}{l} x = 1 \in \left[ {0;3} \right]\\ x = - 3 \notin \left[ {0;3} \right] \end{array} \right. \\ \\ \begin{array}{l} f\left( 0 \right) = 1,f\left( 1 \right) = - 4,f\left( 3 \right) = 28\\ \Rightarrow \mathop {\max }\limits_{\left[ {0;3} \right]} f\left( x \right) = 28,\mathop {\min }\limits_{\left[ {0;3} \right]} f\left( x \right) = - 4 \end{array}\)