How to find minimum value of given multiple modulus function

Question

So the question is to find the minimum value of given modulus function. $$f(x)=|x-1|+|3x-1|+|5x-1|+|7x-1|+|9x-1|+|11x-1|$$

So my first approach was to just make a graph of it On making the graph the minimum value was at $x=1/9$. And also rechecked the value in Wolfram alpha and it was indeed $1/9$

But now let's do this by triangle inequality: $$we\;know\; |a|+|b|\ge|a+b|\;so\;minimum\;|a|+|b|=|a+b|$$ Now in our Ques let's group first and last term, second and second last and third and third last $$(|x-1|+|11x-1|)+(|3x-1|)+|9x-1|)+(|5x-1|+|7x-1|)$$

Now using traingle inequality all grouped term min value will be $|12x-2|$

So now

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

Now we can see that at $x=1/6$ the function has its minimal value

$$f\left(\frac{1}{6}\right)\geq3\left|\frac{12}{6}-2\right|=0$$

So what's wrong in this method as the minimum value is at $x=1/9$ not $x=1/6$?

Answer 1

The triangle inequality indeed shows that $$|x-1|+|3x-1|+|5x-1|+|7x-1|+|9x-1|+|11x-1| \geq 3|12x-2|$$ for all $x$. But there is no reason that the minimum of the right hand side should also be a minimum of the left hand side. I suggest plotting both functions to see what happens.

Answer 2

Your solution is wrong because you don't save the case of the equality occurring.

Try to substitute $x=\frac{1}{6}$.

The result is not $0$.

My solution.

The graph of $f(x)=|x-1|+|3x-1|+|5x-1|+|7x-1|+|9x-1|+|11x-1|$ is an union of segments and rays. Also, $f(x)\geq0.$

Thus, $$\min f=\min\left\{f(1),f\left(\frac{1}{3}\right),f\left(\frac{1}{5}\right),f\left(\frac{1}{7}\right),f\left(\frac{1}{9}\right),f\left(\frac{1}{11}\right)\right\}=f\left(\frac{1}{9}\right)=\frac{22}{9}.$$

Answer 3

A sum of absolute value expressions will result in a piecewise linear function. The most straightforward way to solve this particular problem is to (as already mentioned) graph it. With some work, you could also rewrite each of the portions in their piecewise form; in doing so, the problem would reduce to that of finding the minimum of line segments.

The changes in the piecewise function representing the initial expression occur at the various zeros, so it is reasonable to guess that the minimum (if one exists, and it does since the end behavior is $+\infty$ on both sides) will be at one of the corresponding zeros. Indeed, you can verify that the global minimum of this function is when $|9x - 1| = 0$ i.e. when $x = 1/9$.

I endeavored to graph the given function and then construct part of its piecewise representation; specifically, the component around the minimum. You can find this in the Desmos graph linked here. It is also depicted in the screenshot below:

As you can see, the minimum occurs when the functions $f$ and $h$ meet, which is at $x = 1/9$.

Answer 4

The derivative never vanishes because $1+3+5+7<9+11$ and $1+3+5+7+9>11$. So the minimum must be at the point where there is no derivative. The minimum it remains to choose of 6 values.