<Question>
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 $p \leq x\leq15.$
</Question>

<think>
<Plan_1>
Okay, so I need to find the minimum value of the function f(x) = |x - p| + |x - 15| + |x - p - 15| where 0 < p < 15, and we're looking at x in the interval [p, 15]. Hmm, let's start by understanding what this function looks like.
First, absolute value functions are piecewise linear, so f(x) is going to be a piecewise linear function as well. The points where the behavior of the absolute values change are at x = p, x = 15, and x = p + 15. But since p is between 0 and 15, p + 15 is going to be between 15 and 30. However, our interval for x is [p, 15], so x can't reach p + 15 because p + 15 is greater than 15.

So, within the interval [p, 15], the critical points for the absolute values are x = p and x = 15. The third absolute value term is |x - p -15|, which is |(x - p) -15|. Since x is at most 15, and p is at least just above 0, (x - p) is at most 15 - 0 = 15, so (x - p) -15 is at most 0.
Therefore, |x - p -15| is equal to p +15 - x for all x in [p,15].

Let me verify that.
If x is in [p,15], then x - p is between 0 and 15 - p. Then, x - p -15 is between -15 and -p. Since p is positive, so x - p -15 is negative, so the absolute value is the negative of that, which is p +15 -x. Yes, that's correct.

So, breaking down f(x) into parts for x in [p,15]:

<prompt_1.1> Simplify the function f(x) for x in the interval [p, 15] using the above plan. </prompt_1.1>
-----

But just to make sure, let's consider another approach. The function f(x) is the sum of three absolute values. In optimization problems involving sums of absolute values, the minimum often occurs at a median point. However, since we are restricted to the interval [p,15], we need to check if the median is within this interval or at the nearest endpoint.

<prompt_1.2> Consider an alternative approach to find the minimum of f(x) using the above plan. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
f(x) = |x - p| + |x -15| + |x - p -15|
= |x - p| + |x -15| + (p +15 -x)

Because the third term simplifies to p +15 -x as established.

Now, let's look at the first two terms: |x - p| and |x -15|. Since x is between p and 15, x - p is non-negative (since x >= p), so |x - p| = x - p. Similarly, x -15 is non-positive (since x <=15), so |x -15| = 15 - x.

Therefore, substituting these into f(x):

f(x) = (x - p) + (15 - x) + (p +15 -x)
Simplify term by term:

x - p +15 -x + p +15 -x

Combine like terms:

x - x -x = -x

-p + p = 0

15 +15 =30

So, f(x) = -x +30
</execution_1.1>

<execution_1.2>
The three terms are |x - p|, |x -15|, and |x - (p +15)|. The three points are p, 15, and p+15. The median of these three points would be 15, since p <15 < p+15. So, if there were no restrictions, the minimum would be at x=15. But since our interval is [p,15], 15 is included, so the minimum is at x=15. Therefore, this confirms the previous result.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Wait, that simplifies to -x +30? That seems odd because f(x) is expressed as a linear function of x with a negative slope.
So, as x increases, f(x) decreases. But that would mean that the minimum value of f(x) on the interval [p,15] would occur at the rightmost point, which is x=15.
But let's check the calculations again, because that seems counterintuitive.

Wait, let's re-express f(x):

<prompt_2.1> Verify the simplification of f(x) and check if the function is decreasing over [p, 15] using the above plan. </prompt_2.1>
-----

Therefore, the answer is 30 -15=15. But wait, 30 -15 is 15.
But let me confirm that.

Wait, substituting x=15 into the original function:

<prompt_2.2> Verify the minimum value by substituting x = 15 into the original function using the above plan. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
First term: |x - p| = x - p (since x >= p)

Second term: |x -15| = 15 - x (since x <=15)

Third term: |x - p -15| = p +15 -x (as established earlier)

Adding them up:

(x - p) + (15 - x) + (p +15 -x) = x - p +15 -x + p +15 -x

Combine x terms: x -x -x = -x

Combine constants: -p + p +15 +15 = 30

So, f(x) = -x +30. So, yes, it's a linear function with slope -1. Therefore, on the interval [p,15], as x increases, f(x) decreases. Therefore, the minimum value occurs at the maximum x in the interval, which is x=15.
</execution_2.1>

<execution_2.2>
f(15) = |15 - p| + |15 -15| + |15 - p -15| = (15 - p) + 0 + | -p | = (15 - p) +0 + p =15 -p +p=15.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.2:
But let me check another point in the interval to see if this holds. Let's take x = p.

<prompt_3.1> Check the value of f(x) at x = p and confirm the minimum value using the above plan. </prompt_3.1>
-----

Based on execution_1.2:
Therefore, regardless of the value of p (as long as 0 < p <15), the minimum is 30 -15=15.
So, the answer is 15.
-----

Therefore, after this thorough analysis, I believe the minimum value is 30 -15=15.

But let me check once more with a different p.
Suppose p=5. Then f(x)=|x-5|+|x-15|+|x-20|.

<prompt_3.2> Perform a final verification of the solution using the above plan. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Then f(p) = |p - p| + |p -15| + |p - p -15| = 0 + (15 -p) +15 = 0 +15 -p +15=30 -p. Since p is between 0 and15, 30 -p is between 15 and30. So, at x=p, the value is 30 -p, which is higher than 15. At x=15, the value is 15, which is lower.
</execution_3.1>

<execution_3.2>
On the interval [5,15].

At x=15: |15-5|+|15-15|+|15-20|=10+0+5=15.

At x=5: |5-5|+|5-15|+|5-20|=0+10+15=25.

Midpoint between 5 and15 is 10. Let's check x=10:

|10-5|+|10-15|+|10-20|=5+5+10=20.

So, indeed, the function decreases from x=5 to x=15, reaching 15. So, that's consistent. Therefore, the minimum is 15.

Therefore, regardless of p, as long as 0 < p <15, the minimum value is 15.
</execution_3.2>
</Execution_3>

<Final_answer>
Therefore, the minimum value of $f(x)$ on the interval $p \leq x \leq 15$ is $\boxed{15}$.
</Final_answer>
</think>