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 the function.

First, f(x) is the sum of three absolute value expressions. Absolute value functions can have different linear pieces depending on the value of x relative to the points where each absolute value expression changes its behavior. These points are where the expressions inside the absolute values are zero, so for each term:

1. |x - p| changes at x = p
2. |x - 15| changes at x = 15
3. |x - p - 15| changes at x = p + 15

Given that p is between 0 and 15, p + 15 would be between 15 and 30. Since we're considering x in [p, 15], which is entirely to the left of p + 15 (since 15 < p + 15), the point x = p + 15 is outside our interval of interest. Therefore, within [p, 15], the only critical points where the function might change its slope are x = p and x = 15.

Wait, but let me confirm that. Let's see, for x in [p, 15], when x is between p and 15, then:

- |x - p| is equal to x - p, since x ≥ p
- |x - 15| is equal to 15 - x, since x ≤ 15
- |x - p - 15| is equal to (p + 15) - x, since x ≤ p + 15 (which is always true here because x ≤15 and p +15 ≥15)

Therefore, within the interval [p, 15], all three absolute value expressions can be rewritten without the absolute value signs as linear functions. Let's write out f(x) in this interval:

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

Simplify this expression:

First term: x - p
Second term: 15 - x
Third term: p + 15 - x

Add them together:

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

Simplify term by term:

x cancels with -x in the second term, then there's another -x: so x - x -x = -x

-p + p cancels out: -p + p = 0

Then 15 + 15 = 30

So overall, f(x) simplifies to -x + 30 in the interval [p, 15]

Wait, that's interesting. So within [p, 15], f(x) is a linear function with slope -1, starting at x = p and ending at x =15. Therefore, the function is decreasing in this interval. So its minimum would be at the right endpoint, x =15, and the maximum at the left endpoint, x = p.

But wait, let me check the calculation again to make sure I didn't make a mistake.

Original f(x):

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

In [p,15]:

|x - p| = x - p

|x -15| = 15 - x

|x - p -15| = p +15 -x (since x ≤ p +15, which is always true as x ≤15 and p +15 ≥15)

So adding them up:

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

Combine like terms:

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

x cancels with -x, leaving -x

-p + p cancels, leaving 0

15 +15 =30

So yes, -x +30. Therefore, f(x) = -x +30 in [p,15], which is a straight line with slope -1.

Therefore, the function is decreasing on [p,15], so the minimum occurs at x =15, and the maximum at x =p.

But the question is asking for the minimum value of f(x) on [p,15]. So according to this, the minimum is f(15) = -15 +30 =15.

Wait, but that seems too straightforward. Let me check if I considered all the cases correctly.

Alternatively, maybe I should consider the entire function over all x, but since we're restricted to [p,15], and in that interval, the function is linear and decreasing, then yes, the minimum is at x=15.

But just to be thorough, let me consider the function outside the interval [p,15] to see if there's a lower value. However, since the problem restricts x to [p,15], we don't need to consider other intervals. Therefore, the answer is 15.

But let me verify once again by evaluating f(15):

f(15) = |15 - p| + |15 -15| + |15 - p -15| = |15 - p| +0 + |-p| = (15 - p) +0 + p =15 -p +p =15. So yes, that's correct.

Wait, but 15 is the result. However, wait, if p is between 0 and15, then 15 - p is positive, so |15 - p| is 15 - p. Then |-p| is p. So 15 - p + p =15. So yes, f(15)=15 regardless of p. But then, if the function is decreasing on [p,15], then the minimum is 15. But that seems to ignore the other terms. Wait, but let me check another point.

Suppose p=7.5. Then f(x)= |x-7.5| + |x-15| + |x-22.5|

But in [7.5,15], f(x) = (x -7.5) + (15 -x) + (22.5 -x) = x -7.5 +15 -x +22.5 -x = (-x) + ( -7.5 +15 +22.5) = -x +30. So yes, same as before. So f(15) = -15 +30=15, and f(7.5)= -7.5 +30=22.5. So the function is decreasing, as x increases, f(x) decreases.

But if we consider the entire function, maybe there's a different behavior. For example, if x is less than p, then |x - p| = p -x, |x -15|=15 -x, |x - p -15|=p +15 -x. So f(x)=p -x +15 -x +p +15 -x= 2p +30 -3x. That's a linear function with slope -3. So decreasing with steeper slope.

If x is between p and15, slope -1.

If x is between15 and p+15, |x -p|=x -p, |x -15|=x -15, |x -p -15|=x -p -15. So f(x)=x -p +x -15 +x -p -15=3x -2p -30. Slope +3, increasing.

If x > p+15, |x -p|=x -p, |x -15|=x -15, |x -p -15|=x -p -15. So f(x)=x -p +x -15 +x -p -15=3x -2p -30. Same as above.

But since our interval is [p,15], and p+15 >=15, so in [p,15], the function is decreasing with slope -1, so the minimum is at x=15, which is 15.

But wait, if p is very small, say p approaches 0, then f(x) would be |x| + |x -15| + |x -15|. Wait, but x is in [0,15]. Then f(x)=x + (15 -x) + (15 -x) =x +15 -x +15 -x=30 -x. So slope -1, decreasing. So minimum at x=15, which is 15.

If p is approaching 15, say p=14.9, then f(x)=|x -14.9| + |x -15| + |x -29.9|

In [14.9,15], |x -14.9|=x -14.9, |x -15|=15 -x, |x -29.9|=29.9 -x. So f(x)=x -14.9 +15 -x +29.9 -x= (-x) + ( -14.9 +15 +29.9)= -x +30. So same as before. So f(15)=15, which is the minimum.

Therefore, regardless of p in (0,15), the function f(x) in [p,15] is linear with slope -1, decreasing, so the minimum is at x=15, which is 15. Therefore, the answer is 15.

But let me think again. Suppose p is 0. Then f(x)=|x| + |x -15| + |x -15|. For x in [0,15], f(x)=x +15 -x +15 -x=30 -x. So minimum at x=15, which is 15. But p must be greater than 0, so p approaches 0, but not equal to 0. So even when p approaches 0, the minimum is still 15.

Alternatively, if p is exactly 0, then x is in [0,15], and f(x)=|x| + |x -15| + |x -15|. But since p>0, we don't need to consider p=0.

Alternatively, maybe there's a different approach. Let's think about the median of the points. The function f(x) is the sum of absolute deviations from three points: p, 15, and p+15. The minimum of the sum of absolute deviations occurs at the median of the points. However, since we're restricted to x in [p,15], which is between p and15, and p+15 is outside this interval. So the median of the three points is 15, since the points are p,15, p+15. The median is the middle value, which is 15. Therefore, the minimum occurs at x=15, which is within [p,15], so the minimum is 15.

Therefore, regardless of the value of p (as long as 0 < p <15), the minimum value is 15.

Wait, but let me check p=7.5 again. Then the three points are 7.5,15,22.5. The median is 15. So the sum of absolute deviations is minimized at x=15, which gives |15 -7.5| + |15 -15| + |15 -22.5| =7.5 +0 +7.5=15. So yes, that's correct. Therefore, this approach also confirms that the minimum is 15.

Therefore, the answer is 15.

**Final Answer**
The minimum value of \( f(x) \) is \boxed{15}.
Therefore, the minimum value of $f(x)$ for $x$ in the interval $\left[p, 15\right]$ is $\boxed{15}$.
</think>