Cho hàm sốy=f(x)có đạo hàm f'(x) Hàm số

Đáp án D

Ta có ngay \(\frac{{13}}{4} \le f\left( x \right) \le 6,\forall x \in \left[ { - 1;2} \right]\).

Ta có \(g'\left( x \right) = 3{f^2}\left( x \right).f'\left( x \right) - 3f'\left( x \right) = 0 = 3f'\left( x \right).f\left( x \right).\left[ {f\left( x \right) - 1} \right]\).

Với \(\forall x \in \left( { - 1;2} \right) \Rightarrow \left\{ \begin{array}{l}f'\left( x \right) > 0\\f\left( x \right).\left[ {f\left( x \right) - 1} \right] > 0\end{array} \right. \Rightarrow g'\left( x \right) > 0,\forall x \in \left( { - 1;2} \right)\)

\( \Rightarrow \left\{ \begin{array}{l}M = g\left( 2 \right) = {f^3}\left( 2 \right) - 3f\left( 2 \right) = 198\\m = f\left( { - 1} \right) = {f^3}\left( { - 1} \right) - 3f\left( { - 1} \right) = \frac{{1573}}{{64}}\end{array} \right. \Rightarrow M + m = \frac{{14245}}{{64}}\).