Cho hàm số $f\left( x \right)=\left|...

Đặt $g\left( x \right)=3{{\text{x}}^{4}}-4{{\text{x}}^{3}}-12{{\text{x}}^{2}}+m$.
Có ${g}'\left( x \right)=12{{\text{x}}^{3}}-12{{\text{x}}^{2}}-24\text{x};\text{ {g}'}\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& x=-1 \\
& x=0 \\
& x=2 \\
\end{aligned} \right.$.
Ta có: $g\left( -1 \right)=m-5;\text{ g}\left( 0 \right)=m;\text{ g}\left( 2 \right)=m-32;\text{ g}\left( 3 \right)=m+27$.
Ta thấy: $m-32<m-5<m<m+27,\forall m$.
TH1: Nếu $m-32<m+27\le 0\Leftrightarrow m\le -27$ thì $M=\left| m-32 \right|$ và $\min M=59$.
TH2: $\left\{ \begin{array}{*{35}{l}}
m-32<0<m+27 \\
\left| m-32 \right|\le \left| m+27 \right| \\
\end{array} \right.\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
-27<m<32 \\
-m-27\le m-32\le m+27 \\
\end{array} \right.\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
-27<m<32 \\
m\ge \dfrac{5}{2} \\
\end{array} \right.\Leftrightarrow \dfrac{5}{2}\le m<32 $ thì $ M=\left| m+27 \right| $ và $ \min M=\dfrac{59}{2}$.
TH3: $\left\{ \begin{array}{*{35}{l}}
m-32<0<m+27 \\
\left| m+27 \right|\le \left| m-32 \right| \\
\end{array} \right.\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
-27<m<32 \\
m-32\le m+27\le -m+32 \\
\end{array} \right.\Leftrightarrow \left\{ \begin{array}{*{35}{l}}
-27<m<32 \\
m\le \dfrac{5}{2} \\
\end{array} \right.\Leftrightarrow -27<m\le \dfrac{5}{2} $ thì $ M=\left| m-32 \right| $ và $ \min M=\dfrac{59}{2}$.
TH4: Nếu $0\le m-32<m+27\Leftrightarrow m\ge 32$ thì $M=\left| m+27 \right|$ và $\min M=59$.
Vậy $\min M=\dfrac{59}{2}$ khi $m=\dfrac{5}{2}$.