The sum of the absolute minimum and the absolute maximum values of the function $f\left(x\right)=\left|3x-x^2+2\right|-x$ in the interval $[-1,2]$ is :

$f\left(x\right)=\left\{\begin{array}{l}{x}^2-4x-2,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{\forall }x\in \left(-1,\frac{3-\sqrt{17}}{2}\right)\\ -x^2+2x+2,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{\forall }x\in \left(\frac{3-\sqrt{17}}{2},2\right)\end{array}\right.$

$\begin{array}{l}{f}^{\mathrm{\prime }}\left(x\right)\text{ when }x\in \left(-1,\frac{3-\sqrt{17}}{2}\right)\\ f^{\mathrm{\prime }}\left(x\right)=2x-4=0\Rightarrow x=2\\ f^{\mathrm{\prime }}\left(x\right)=2(x-2)\Rightarrow f^{\mathrm{\prime }}\left(x\right)\text{ is always }\downarrow \\ f\left(2\right)=2\\ f(-1)=3\\ f\left(\frac{3-\sqrt{17}}{2}\right)=\frac{\sqrt{17}-3}{2}\end{array}$

$$\begin{gathered} f^{\mathrm{\prime }}\left(x\right)\text{ when }x\in \left(\frac{3-\sqrt{17}}{2},2\right) \\ \begin{array}{r}{f}^{\mathrm{\prime }}\left(x\right)=-2x+2\\ f^{\mathrm{\prime }}\left(x\right)=-2(x-1)\\ f^{\mathrm{\prime }}\left(x\right)=0\text{ when }x=1\\ f\left(1\right)=3\end{array} \end{gathered}$$

$$\begin{gathered} \text{ absolute minimum value }=\frac{\sqrt{17}-3}{2} \\ \text{ absolute maximum value }=3 \end{gathered}$$