Minimizing a function involving several modulus terms

Question

$f(x)=3|x+1|+|x|+3|x-1|+2|x-3|$

finding minimum value of this function

One conventional way would be to make five cases regarding the various inequalities and hence simplifying to find minimum value in a graph as in

I was trying another way

method : Aim is to minimise each term .Since difference of x and 1 would matter thrice as much as that between x and 3 would , the value of x should be closer to 1 .

Applying this thought,x={3(1)+3(-1)+2(3)+1(0)}\9 which gives as x=2/3 . This way comes to be incorrect.

I wanted to know if there were shorter solutions to this problem rather than using the graphical approach each time as when the modulus terms increase to 10 terms , it would be unreasonable to make 11 cases.

Answer 1

$f$ is a convex function because a sum of convex functions is a convex function

and since the graph of $f$ is an union of segments and rays,

$\lim\limits_{x\rightarrow+\infty}f(x)=+\infty$ and $\lim\limits_{x\rightarrow-\infty}f(x)=+\infty$, we obtain:

$$\min{f}=\min\{f(-1),f(0),f(1),f(3)\}=f(1)=11.$$

Answer 2

We want a point where the derivative goes from negative to positive. For |x-a|, the derivative is negative if x < a and positive if x > a. If f(x) = $\sum|x-a_i|$, then the derivative at x is thus going to be (sum of all $a_i$ greater than x) minus (sum of all $a_i$ less than x). So we want a point where those are equal. That is, we want a number where as many numbers are less than it as greater than. That's called the median. Note that if you have an even number of terms, then there will be a region between the two "middle" terms where the function is constant and minimal. Also note that when calculating the median, you should take repetition into account; that is, if an absolute value term has a coefficient of n, take n copies of the associated $a_i$. For example, in this case you should take the median of -1,-1,-1,0,1,1,1,3,3. And finally, note that it's a bit more complicated if some of the absolute value terms have negative coefficients.

Answer 3

By triangle inequality we have for the function $$g(x) = 3|x+1|+3|x-1| = |3x+3|+|3-3x|\geq |3x+3+3-3x| =6$$ with equality exactly when $x-1$ and $x+1$ are of equal sign and this is exactly when $-1\leq x\leq 1$. But in this case we have $2|x-3| = 6-2x$. So we have

$$f(x)=g(x)+|x|+2|x-3|\geq 12-2x+|x|$$

So, if $x\geq 0$ we have $f(x) =12-x \geq 11$ and if $x\leq 0$ we have $f(x) =12-3x \geq 12$.

So $f$ achieve minimum at $x=1$ and it is $11$.

Answer 4

Hint: For the (convex) function $f(x) = \sum \lambda_n|x-a_n|$, ($a_n$ are distinct numbers). we always have a minimum point where derivative does not exist (it is not hard to prove this).

Hence, $$\min{f}=\min\{f( a_1),f(a_2),...,f(a_n)\}$$

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I added this part to prove my claim! (Although it is somehow trivial! I am writing this exclusively for down-voter)

Proof : We want to show that there is a minimizer which happens to be nondifferentiable point. we assume $\lambda_n > 0$ (actually we don't need assume this as long as we are sure minimum exist!) Let $c \in \Bbb R$ be the minimum point then it has to be critical point too which means either $f'(c) =0$ or $f'(c) = \text{does not exist}.$ In second case we are done! So assume first case has been happened. So we have $f'(c) =0$. Since $f$ is pieacewise linear, there exist a maximal interval say $[a,b]$ containing $c$ such that $f$ is constant on it. Also note $ -\infty < a< b < + \infty $ because $\lim_{x \to \pm \infty} f(x) = + \infty$. This shows $f(a) = f(b)= f(c)$ and so $a,b$ are minimizer too and moreover (it is easy to show that) $a,b$ are nondifferentiable points.

Conclusion: In order to find a minimum point, just evaluate $f$ at Absolutvalues' roots and pick the smallest one! Then you don't need think about graphical method which may lead to $11$ cases.