Okay. The derivative of |x-a| is:
1 if x-a>0
0 if x=0
-1 if x-a<0
Since we have a sum we the derivative of f(x) is the sum of the derivatives of the modulus:f'(x)=
If f'(x) is to be 0 we can't have some of the elements
to be zero because then all the other elements can't be zero. We will have 15 elements that can be 1 or -1 and one which is 0 and their sum cannot be 0.
Suppose we have a number i=n such as for i>n
. This is easy enough to prove
. We will also have
for i<=n since there can't be a zero term. From f'(x)=0 we can conclude that n=8 (8 times 1 and 8 times -1=0).
So
for n>8
and
for n<=8
All that is left is to prove that this is a minimum but I'll leave that to you.
Don't know if someone has suggested this. I didn't read all the posts.