Cho hàm số $f(x)=\left| 3{{x}^{4}}-4{{x}^{3}}-12{{x}^{2}}+m...

Đặt $g\left( x \right)=3{{x}^{4}}-4{{x}^{3}}-12{{x}^{2}}+m$ trên $\left[ -1;3 \right]$.
Ta có: ${g}'\left( x \right)=12{{x}^{3}}-12{{x}^{2}}-24x=12x\left( {{x}^{2}}-x-2 \right)$.
${g}'\left( x \right)=0\Leftrightarrow \left[ \begin{matrix}
x=-1\in \left[ -1;3 \right] \\
x=0\in \left[ -1;3 \right] \\
x=2\in \left[ -1;3 \right] \\
\end{matrix} \right.$
$g\left( -1 \right)=m-5$ ; $g\left( 0 \right)=m$ ; $g\left( 2 \right)=m-32$ ; $g\left( 3 \right)=m+27$
Thấy: $m-32<m-5<m<m+27,\forall m\in \mathbb{R}$. Vậy $\underset{\left[ -1;3 \right]}{\mathop{\max }} =\max \left\{ \left| m-32 \right|;\left| m+27 \right| \right\}$.
TH1: $\left| m+27 \right|\le \left| m-32 \right|\Leftrightarrow m\le \dfrac{5}{2}$. Khi đó $M=\left| m-32 \right|\ge \dfrac{59}{2},\forall m\le \dfrac{5}{2}\Rightarrow \min M=\dfrac{59}{2}$.
TH2: $\left| m+27 \right|\ge \left| m-32 \right|\Leftrightarrow m\ge \dfrac{5}{2}$. Khi đó $M=\left| m+27 \right|\ge \dfrac{59}{2},\forall m\ge \dfrac{5}{2}\Rightarrow \min M=\dfrac{59}{2}$.
Vậy giá trị nhỏ nhất của $M$ là $\dfrac{59}{2}$.