Find the minimum value of this expression with absolute values

Question

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

Answer 1

First consider the extremes. If $x$ gets very positive or negative, you would nkow the expression becomes very large as the absolute value takes away the negative sign. Now, evaluate the expression at each of your critical values, in this case $3,1,0,-2,-4$. Then pick the minimum of these values.

Answer 2

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 $x\in \mathbb{R}$.

Answer 3

A statistical way to solve the problem:

We know that mean absolute deviation is minimum about median.

So consider median of $\{3,1,0,-2,-4\}$

From here conclude that $x=0$ is minimum.

Answer 4

Applying the properties of the module, we find easily $$|x-3| + |x-1| + |x| + |x+2| + |x+4|=|3-x| + |1-x| + |x+2| + |x+4|+ |x| \geq |3-x+1-x+x+2+x+4|+ |x| =10+ |x| \geq 10. $$ It seems that this minimum is attained for $x = 0.$