Lời giải:\(f\left( x \right) = 2{x^3} + 3{x^2} - 1 \Rightarrow f'\left( x \right) = 6{x^2} + 6x\); \(f'\left( x \right) = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{x = 0{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( {ktm} \right)}\\{x = {\rm{\;}} - 1{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( {tm} \right)}\end{array}} \right.\)
Hàm số \(f\left( x \right)\) liên tục trên \(\left[ { - 2; - \dfrac{1}{2}} \right]\), có \(f\left( { - 2} \right) = - 5;f\left( { - 1} \right) = 0;f\left( { - \dfrac{1}{2}} \right) = {\rm{\;}} - \dfrac{1}{2}\)
\( \Rightarrow m = \mathop {\min }\limits_{\left[ { - 2; - \dfrac{1}{2}} \right]} f\left( x \right) = {\rm{\;}} - 5;{\mkern 1mu} {\mkern 1mu} M = \mathop {\max }\limits_{\left[ { - 2; - \dfrac{1}{2}} \right]} f\left( x \right) = 0\)\( \Rightarrow P = M - m = 5\).
Chọn C.
