The sum of the abosolute maximum and minimum values of the function...

Given,

$f\left(x\right) = \left|{x}^{2} - 5x + 6\right| - 3x + 2$

$⇒ f\left(x\right) = \left|\left(x - 3\right)\left(x - 2\right)\right| - 3x + 2$

⇒ f\left(x\right) = \left\{\begin{cases}{x}^{2} - 8x + 8 & ; x \in \left[- 1, 2\right] \\ - {x}^{2} + 2x - 4 & ; x \in ( 2, 3 ]\end{cases}\right)</p><p>⇒ {f}^{'}\left(x\right) = \left\{$\begin{cases}2x - 8 & ; x \in \left(- 1, 2\right) \\ - x + 2 & ; x \in \left(2, 3\right)\end{cases}$\right)

Now point of extrema will be $2x - 8 = 0 ⇒ x = 4$which is not in domain and $- x + 2 = 0 ⇒ x = 2$

Now for finding the value of global minima and maxima we will check the value of function at extrema points and boundary points,

So, $f\left(x\right) = {x}^{2} - 8x + 8$ will give, $f\left(- 1\right) = 17 \text{ and } f\left(2\right) = - 4$

And $f\left(x\right) = - {x}^{2} + 2x - 4$ will give, $f\left(2\right) = 4, f\left(3\right) = - 7$

Hence, absolute minima is $- 7$ and maxima is $17$

So, their sum is $- 7 + 17 = 10$