A quick way to tell local minimum given a sum of absolute value functions

As mentioned by @Aniruddha Deb, it becomes tricky but thankfully, we can use the concept of mean deviation (MD) about $x$ of the data points $x_i$ with respective frequency as $f_i$: $$MD_x=frac{1}{N}sum_{i+1}^{n} f_i ~|x_i -x|,~~ N=sum_{i=1}^n f_i.$$ The MD is known to be the least if measured about $M$ the median of the data. Now, we can see the given function $$F(x)=2|x-1+k|x-2|+3|x-3|+|x-5|.$$ So $x_i$ are $1,2,4,5$ their frequencies are $2, k,3,1$; respectively.

For $k=1$: The median of this data: $1,1,2,3,3,3,5$ is a single number $M=3$ so the data has single minimum $F_{min}=F(3)=7$, this can be checked to be a local(relative) min as $F(2.9)=7.1, F(3.1)=7.5.$

For $k=2$: The data is $1,1,2,2,3,3,3,5$, the medians are $M=2,3$, so $F(2le x le 3)=8$. Hence, the function is bounded from below and the least value is $8$ ($F(x)ge 8$) and $F(x)$ doesn't have local min.

For $k=3:$

The data is $1,1,2,2,2,3,3,3,5$ the median $M=2$, so $F_{min}=F(2)=8$ and it is a local min as $f(1.9)=8.5, F(2.1)=8.1$

For $k=4$:

There are two but equal medians $2,2$ so $F(x)$ will again have a local nib at $x=2$.