A quick way to tell local minimum given a sum of absolute value functions Given a function $$f(k,x)=2|x-1|+k|x-2|+3|x-3|+|x-5|, ~~k=1,2,3$$ which is a five piece continuous function. The question is: can  we can tell if there will exist a local minimum without plotting this five-piece function for these three values of $k$?
 A: 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 $F(2\le x \le 3)=8$.
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 $x=2$.
