Vtyjkyuk

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 $p leq xleq15$.

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 $p leq xleq15$, 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?

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.

Given $f(x)=|x-p|+|x-15|+|x-p-15|$, where $0 < p < 15$, the minimum of $f(x)$ in $p leq xleq15$ would be:
$$xge p Rightarrow x-pge 0 Rightarrow |x-p|=x-p; \
xle 15 Rightarrow x-15le 0 Rightarrow |x-15|=15-x;\
xle 15 textand p>0 Rightarrow x-15-p<0 Rightarrow |x-p-15|=15+p-x;\
f(x)=x-p+15-x+15+p-x=30-x;\
f_min(15)=15.$$