Absolute minimum

Let

$f(x)=|x+8|+|x+3|+|x-2|+|x-6|$

The points of non differentiability where the function changes are at $x=-8,x=-3,x=2,x=6$ , since

$|x+8|=\begin{cases} -x-8\phantom{cccc}\text{if }x<-8\\ x+8\phantom{cccccc}\text{if }x\geq -8 \end{cases}$

$|x+3|=\begin{cases} -x-3\phantom{cccc}\text{if }x<-3\\ x+3\phantom{cccccc}\text{if }x\geq -3 \end{cases}$

$|x-2|=\begin{cases} -x+2\phantom{cccc}\text{if }x<2\\ x-2\phantom{cccccc}\text{if }x\geq 2 \end{cases}$

$|x-6|=\begin{cases} -x+6\phantom{cccc}\text{if }x<6\\ x-6\phantom{cccccc}\text{if }x\geq 6 \end{cases}$

Hence

$f(x)=\begin{cases} -x-8-x-3-x+2-x+6=-4x-3\phantom{cccc}\text{if }x<-8\\ x+8-x-3-x+2-x+6=-2x+13\phantom{ciccc}\text{if }-8\leq x<3\\ x+8+x+3-x+2-x+6=19\phantom{cccccicccccc}\text{if }-3\leq x<2\\ x+8+x+3+x-2-x+6=2x+15\phantom{ccccicc}\text{if }2\leq x<6\\ x+8+x+3+x-2+x-6=4x+3\phantom{cccicccc}\text{if }x\geq 6 \end{cases}$

$f(x)=\begin{cases} -4x-3\phantom{cccccc}\text{if }x<-8\\ -2x+13\phantom{ciccc}\text{if }-8\leq x<3\\ 19\phantom{cccccicccccc}\text{if }-3\leq x<2\\ 2x+15\phantom{ccccicc}\text{if }2\leq x<6\\ 4x+3\phantom{cccicccc}\text{if }x\geq 6 \end{cases}$

Evaluating the minimum value of each of the parts of the function

$\min\{f(x)\}=\begin{cases} -4(-8)-3=29\\ -2(-3)+13=19\\ 19\\ 2(2)+15=19\\ 4(6)+3=27 \end{cases}$

Thus, the minimum of $f(x)$ is $\boxed{19}$

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A graph of $f(x)$

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