The sum of the absolute minimum and the absolute maximum values of the function f x = 3 x - x 2 + 2 - x in…

Given $f\left(x\right)=\left\{\begin{array}{ll}{x}^2-4x-2,& \forall x\in \left(-1,\frac{3-\sqrt{17}}{2}\right)\\ -x^2+2x+2,& \forall x\in [\frac{3-\sqrt{17}}{2},2)\end{array}\right.$

When $x\in \left(-1,\frac{3-\sqrt{17}}{2}\right)$

$f^{\text{'}}\left(x\right)=2x-4=0\Rightarrow x=2$

Now $f\left(2\right)=2$, $f\left(-1\right)=3$

$f\left(\frac{3-\sqrt{17}}{2}\right)=\frac{\sqrt{17}-3}{2}$

When $x\in [\frac{3-\sqrt{17}}{2},2)$

$f^{\text{'}}\left(x\right)=-2x+2$

$f^{\text{'}}\left(x\right)=-2\left(x-1\right)$

$f^{\text{'}}\left(x\right)=0$ when $x=1$

$f\left(1\right)=3$

Hece, absolute minimum value $=\frac{\sqrt{17}-3}{2}$

and absolute maximum value $=3$

Sum $=\frac{\sqrt{17}-3}{2}+3=\frac{\sqrt{17}+3}{2}$