<s>[INST] Find the minimum value of this expression with absolute values The expression is 
$$|x-3| + |x-1| + |x| + |x+2| + |x+4|$$
I know that the minimum values for this expression is when x = 0 but is there any algebraic way to find this out? I did it on the calculator
 [/INST] Let $f$ the function defined by the equation $f(x)=|x-3|+|x-1|+|x|+|x+2|+|x+4|$ for all $x\in \mathbb{R}$. $$x<-4\Longrightarrow f(x)=-(x-3)-(x-1)-x-(x+2)-(x+4)=-5x-2>18$$
$$-4\le x <-2 \Longrightarrow f(x)=-(x-3)-(x-1)-x-(x+2)+(x+4) = -3x+6>12$$
$$-2\le x <0  \Longrightarrow f(x)=-(x-3)-(x-1)-x+(x+2)+(x+4) = -x+10>10$$
$$0\le x < 1 \Longrightarrow f(x)=-(x-3)-(x-1)+x+(x+2)+(x+4) = x+10\ge 10$$
$$1\le x < 3 \Longrightarrow f(x)=-(x-3)+(x-1)+x+(x+2)+(x+4) = 3x+8\ge 11$$
$$x\ge 3  \Longrightarrow f(x)=(x-3)+(x-1)+x+(x+2)+(x+4) = 5x+2\ge 17$$
It follows $f(x)\ge 10=f(0)$ for all $x\in \mathbb{R}$.
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