\(\Leftrightarrow\max\limits_{\left[0;3\right]}f\left(x\right)\le16\)
Trên \(\left[0;3\right]\) xét hàm \(g\left(x\right)=x^3-3x+m\Rightarrow g'\left(x\right)=3x^2-3=0\Rightarrow x=1\)
Ta có: \(g\left(0\right)=m;\) \(g\left(1\right)=m-2\) ; \(g\left(3\right)=m+21\)
\(\Rightarrow\max\limits_{\left[0;3\right]}f\left(x\right)=max\left\{\left|m-2\right|;\left|m+18\right|\right\}\)
TH1: \(\left\{{}\begin{matrix}\left|m-2\right|\ge\left|m+18\right|\\\left|m-2\right|\le16\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m\le-8\\-14\le m\le18\end{matrix}\right.\)
\(\Rightarrow-14\le m\le-8\)
TH2: \(\left\{{}\begin{matrix}\left|m+18\right|\ge\left|m-2\right|\\\left|m+18\right|\le16\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m\ge-8\\-34\le m\le-2\end{matrix}\right.\)
\(\Rightarrow-8\le m\le-2\)
Vậy \(-14\le m\le-2\)
