The sum of the absolute minimum and the absolute maximum values of...

Given f\left(x\right) = \left\{\begin{cases}{x}^{2} - 4x - 2, & \forall x \in \left(- 1, \dfrac{3 - \sqrt{17}}{2}\right) \\ - {x}^{2} + 2x + 2, & \forall x \in [ \dfrac{3 - \sqrt{17}}{2}, 2 )\end{cases}\right)</p><p>Whenx \in \left(- 1, \dfrac{3 - \sqrt{17}}{2}\right)$</p><p>${f}^{'}\left(x\right) = 2x - 4 = 0 ⇒ x = 2$</p><p>Now$f\left(2\right) = 2$,$f\left(- 1\right) = 3$</p><p>$f\left(\dfrac{3 - \sqrt{17}}{2}\right) = \dfrac{\sqrt{17} - 3}{2}$</p><p>When$x \in [ \dfrac{3 - \sqrt{17}}{2}, 2 )$</p><p>${f}^{'}\left(x\right) = - 2x + 2$</p><p>${f}^{'}\left(x\right) = - 2\left(x - 1\right)$</p><p>${f}^{'}\left(x\right) = 0$when$x = 1$</p><p>$f\left(1\right) = 3$</p><p>Hece, absolute minimum value$= \dfrac{\sqrt{17} - 3}{2}$</p><p>and absolute maximum value$= 3$</p><p>Sum$= \dfrac{\sqrt{17} - 3}{2} + 3 = \dfrac{\sqrt{17} + 3}{2}$