Xét hàm số $\text{g}\left(\text{x}\right)={\text{x}}^3-2{\text{x}}^2+(\text{m}+2)\text{x}+5$ trên đoạn [-1;2].

$\Rightarrow {\text{g}}^{\text{'}}\left(\text{x}\right)=3{\text{x}}^2-4\text{x}+\text{m}+2{\Delta }^{\text{'}}=2^2-3( \text{m}+2)=-(3 \text{m}+2)<0$ với mọi m dương.

$\Rightarrow {\text{g}}^{\text{'}}\left(\text{x}\right)>0,\forall \text{x}\in ℝ$. Vậy hàm số $\text{g}\left(\text{x}\right)={\text{x}}^3-2{\text{x}}^2+(\text{m}+2)\text{x}+5$ luôn đồng biến trên [-1;2].

Suy ra $\underset{[-1;2]}{\mathrm{Max}}\text{g}\left(\text{x}\right)=\text{g}\left(2\right)=2 \text{m}+9;\underset{[-1;2]}{\mathrm{M}in}\text{g}\left(\text{x}\right)=\text{g}(-1)=-\text{m}$.

Tính $\left\{\begin{array}{l}\text{y}(-1)=\left|\text{m}\right|\\ \text{y}\left(2\right)=\left|2 \text{m}+9\right|\end{array}\right.$. Khi đó $\underset{[-1;2]}{\mathrm{Max}}\text{y}=\{\left|\text{m}\right|;\left|2 \text{m}+9\right|\}$.

TH1. Nếu $|2 \text{m}+9|<\left|\text{m}\right|\iff {(2 \text{m}+9)}^2<{\left(\text{m}\right)}^2\iff (\text{m}+9)(3 \text{m}+9)<0\iff -9<\text{m}<-3$ (L).

TH2. Nếu $\left|\text{m}\right|\le |2 \text{m}+9|\iff {\left(\text{m}\right)}^2\le {(2 \text{m}+9)}^2\iff (-3 \text{m}-9)(\text{m}+9)\le 0\iff \left[\begin{array}{l}\text{m}\le -9\\ \text{m}\ge -3\end{array}\right.$

$\Rightarrow \underset{[-1;2]}{\mathrm{Max}}f\left(x\right)=|2 \text{m}+9|\le 11\iff -11\le 2 \text{m}+9\le 11\iff -10\le \text{m}\le 1$