Given that
$$f(x)=|x|+2|x−1|+|x−2|+|x−4|+|x−6|+2|x−10|$$, $$x \epsilon (-\infty, \infty)$$
the minimum value of f(x) is _________.

The critical points for $$f(x)$$ are $$x=0,1,2,4,6,10.$$

When $$x=0$$, $$f(x)=0+2\times\ 1+2+4+6+2\times\ 10=0+2+2+4+6+20=34$$

when $$x=1$$, $$f(x)=1+2\times\ 0+1+3+5+2\times\ 9=28$$

when $$x=2$$, $$f(x)=2+2\times1+0+2+4+2\times\ 8=26$$

when $$x=4$$, $$f(x)=4+2\times3+2+0+2+2\times\ 6=26$$

when $$x=6$$, $$f(x)=6+2\times5+4+2+0+2\times\ 4=30$$

when $$x=10$$, $$f(x)=10+2\times9+8+6+4+0=46$$

For, $$x<0$$, value of $$f(x)$$ clearly will be greater than f(0)=34

Similarly, for $$x>0$$, value of $$f(x)$$ will be greater than f(10)=46

So, the minimum value of $$f\left(x\right)=26$$ when $$x$$ lies in the interval [2,4]