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I've encountered this problem:

Let $f(x)=|x-p|+|x-15|+|x-p-15|$, where $0 < p < 15$. Determine the minimum value taken by $f(x)$ for $x$ in the interval $p \leq x\leq15$.

I'm not sure what the question is asking for, because at least the way I understand it, there is no minimum value of x! You can sub any real number x into the function, and it will work. At most, given the that $p \leq x\leq15$, the minimum value of x is that 1>x>0!

Can someone tell me what the question wants me to do, without telling me how to solve the question?

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  • $\begingroup$ for example |x-1|+|x-2| have minimum value at $x \in [1,2]$you have to find that x for which f(x) is minimum $\endgroup$ – Deepesh Meena Aug 18 '18 at 3:37
  • $\begingroup$ For $x$ in the given interval, you can simplify a couple of the absolute values expressions. $\endgroup$ – John Wayland Bales Aug 18 '18 at 3:54
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We are not asked for the value of $x$ that makes a minimum, we are asked for the minimum value of $f(x)$. As $f(x)$ is a sum of absolute values it is always nonnegative. Therefore $0$ is a lower bound, so the set of possible values has a greatest lower bound, which is what we are asked for.

As the question is asked it implies that the answer is independent of $p$ when $p$ is in the given range. I would try $p=1$ and $p=14$ and see what I find.

When I did that the minimum was $15$ at $x=15$ in both cases. You can justify that.

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  • $\begingroup$ I would not say that the statement of the problem implies that the minimum is independent of $p$. For example: Let $g(x)=x^2+(r-x)^2$ where $r\in \Bbb R.$ Determine $\min \{g(x):x\in \Bbb R\}.$ Wouldn't you answer $r^2/2$? $\endgroup$ – DanielWainfleet Aug 18 '18 at 4:54
  • $\begingroup$ @DanielWainfleet: I believe we were asked for and answer independent of $p$. Your question is much harder. I would be torn between answering $r^2/2$ and punching you in the nose for asking a question without an answer because it shouldn't depend on $r$. This is English, not math, and not a part that many have experience with. I don't think there is a correct answer. $\endgroup$ – Ross Millikan Aug 18 '18 at 5:02
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Given $f(x)=|x-p|+|x-15|+|x-p-15|$, where $0 < p < 15$, the minimum of $f(x)$ in $p \leq x\leq15$ would be: $$x\ge p \Rightarrow x-p\ge 0 \Rightarrow |x-p|=x-p; \\ x\le 15 \Rightarrow x-15\le 0 \Rightarrow |x-15|=15-x;\\ x\le 15 \ \text{and} \ p>0 \Rightarrow x-15-p<0 \Rightarrow |x-p-15|=15+p-x;\\ f(x)=x-p+15-x+15+p-x=30-x;\\ f_{min}(15)=15.$$

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