Given,
f(x)=∣x2−5x+6∣−3x+2
⇒f(x)=∣(x−3)(x−2)∣−3x+2
\Rightarrow f(x)={\begin{matrix}{x}^{2}-8x+8 & ;x\in [-1,2] \\ -{x}^{2}+2x-4 & ;x\in (2,3]\end{matrix}
\Rightarrow {f}^{'}(x)={\begin{matrix}2x-8 & ;x\in (-1,2) \\ -x+2 & ;x\in (2,3)\end{matrix}
Now point of extrema will be 2x−8=0⇒x=4which 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(x)=x2−8x+8 will give, f(-1)=17&f(2)=-4
And f(x)=−x2+2x−4 will give, f(2)=4,f(3)=−7
Hence, absolute minima is −7 and maxima is 17
So, their sum is −7+17=10