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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

bryan.blackbee
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user140484
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    Yet another way is geometric - you seek an $x$ on the real line that minimises the absolute distances from some given points (here $-4, -2, 0, 1, 3$). This is when you choose the median of those points! – Macavity Apr 26 '14 at 03:42
  • See also http://math.stackexchange.com/questions/113270 – Bart Michels Apr 04 '15 at 09:22

4 Answers4

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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 $x\in \mathbb{R}$.

Ángel Mario Gallegos
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  • Where are you getting −5x−2>18 and −3x+6>12? I'm sorry I just don't see how you get all of these expressions – user140484 Apr 26 '14 at 03:28
  • If $x<-4$ then $-5x-2>-5(-4)-2=20-2=18$. In a similar fashion, if $-4\le x < -2$ then $-3x+6>-3(-2)+6=6+6=12$. Here I am using a property about inequalities, namely, when $a<b$ and $c<0$ we have $ac>bc$. – Ángel Mario Gallegos Apr 26 '14 at 03:32
  • Oh okay, I see thank you! But then how do you get to the conclusion that f(x)≥10=f(0) for all x∈ℝ – user140484 Apr 26 '14 at 03:38
  • By just plugging in zero into the expression? So wouldn't it just be f(x) = 10 for f(0) – user140484 Apr 26 '14 at 03:39
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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.

Argha
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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.

bryan.blackbee
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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 $x = 0.$

medicu
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