Let $f(x)=|x-p|+|x-15|+|x-p-15|$, where $0 < p < 15$. Determine the [minimum](https://artofproblemsolving.com/wiki/index.php/Minimum) value taken by $f(x)$ for $x$ in the [interval](https://artofproblemsolving.com/wiki/index.php/Interval) $p \leq x\leq15$.
Solution 1
It is best to get rid of the [absolute values](https://artofproblemsolving.com/wiki/index.php/Absolute_value) first. 
Under the given circumstances, we notice that $|x-p|=x-p$, $|x-15|=15-x$, and $|x-p-15|=15+p-x$.
Adding these together, we find that the sum is equal to $30-x$, which attains its minimum value (on the given interval $p \leq x \leq 15$) when $x=15$, giving a minimum of $\boxed{015}$.

Solution 2
Let $p$ be equal to $15 - \varepsilon$, where $\varepsilon$ is an almost neglectable value. Because of the small value $\varepsilon$, the [domain](https://artofproblemsolving.com/wiki/index.php/Domain) of $f(x)$ is basically the [set](https://artofproblemsolving.com/wiki/index.php/Set) ${15}$. plugging in $15$ gives $\varepsilon + 0 + 15 - \varepsilon$, or $15$, so the answer is $\boxed{15}$