Đáp án B
Ta có: \(f\left( x \right) = {x^3} - 3x + 2 \Rightarrow f'\left( x \right) = 3{x^2} - 3\)
Có: \(f'\left( x \right) = 0 \Leftrightarrow 3{x^2} - 3 = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = 1}\\{x = - 1}\end{array}} \right.\)
Mặt khác : \(f\left( { - 3} \right) = - 16,f\left( { - 1} \right) = 4,f\left( 1 \right) = 0,f\left( 3 \right) = 20\).
Vậy \(\mathop {\max }\limits_{\left[ { - 3;3} \right]} f\left( x \right) = 20\).
