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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

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    $\begingroup$ 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! $\endgroup$ – Macavity Apr 26 '14 at 3:42
  • $\begingroup$ See also math.stackexchange.com/questions/113270 $\endgroup$ – punctured dusk Apr 4 '15 at 9:22
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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 $x\in \mathbb{R}$.

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  • $\begingroup$ 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 $\endgroup$ – user140484 Apr 26 '14 at 3:28
  • $\begingroup$ 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 $ac>bc$. $\endgroup$ – Ángel Mario Gallegos Apr 26 '14 at 3:32
  • $\begingroup$ 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 $\endgroup$ – user140484 Apr 26 '14 at 3:38
  • $\begingroup$ By just plugging in zero into the expression? So wouldn't it just be f(x) = 10 for f(0) $\endgroup$ – user140484 Apr 26 '14 at 3:39
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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.

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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.

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  • $\begingroup$ Why is the minimum of each absolute value term 0 $\endgroup$ – user140484 Apr 26 '14 at 3:23
  • $\begingroup$ i am so sorry i read the question wrongly. here is my new proposed solution. $\endgroup$ – bryanblackbee Apr 26 '14 at 3:26
  • $\begingroup$ Thank you that helps so much! $\endgroup$ – user140484 Apr 26 '14 at 3:30
  • $\begingroup$ if you want to be very sure, you can consider evaluating at the midpoints of each critical value like 2,0.5,-1,-3. you will still see the minimum at a critical point, not any other points. $\endgroup$ – bryanblackbee Apr 26 '14 at 3:42
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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 $x = 0.$

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