$$
\begin{array}{l}
\text { 2. Let } f(x)=x-p^{!}+\left|x-15^{\prime}+x-p-15\right|, \text { where } \\
\text { } 0<p<15 \text {. }
\end{array}
$$

Find: For $x$ in the interval $p \leqslant x \leqslant 15$, what is the minimum value of $f(x)$?
\begin{array}{l}\text { Solve } \because p \leqslant x \leqslant 15 \text{ and } p>0, \therefore p+15 \geqslant x . \\ \therefore f(x)=x-p+15-x+p+15-x=30-x, \\ \text { Also } \because x \leqslant 15, \\ \therefore f(x) \geqslant 30-15=15 . \\ \therefore \text { the minimum value of } f(x) \text { is } 15 .\end{array}