To define the function \( p(x) \) for the interval \( 2 \leq x \leq 10 \) and explore its properties, we first identify the prime and non-prime integers in the floor values within this interval:
- Prime numbers between 2 and 10: 2, 3, 5, 7
- Non-prime numbers between 2 and 10: 4, 6, 8, 9, 10
Now let's compute \( p(x) \) for \( x \) in the intervals determined by the floor function:
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For \( 2 \leq x < 3 \): \(\lfloor x \rfloor = 2\) (prime) \[ p(x) = x + 1 \] Thus, \( p(x) = x + 1 \) for \( x \in [2, 3) \).
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For \( 3 \leq x < 4 \): \(\lfloor x \rfloor = 3\) (prime) \[ p(x) = x + 1 \] Thus, \( p(x) = x + 1 \) for \( x \in [3, 4) \).
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For \( 4 \leq x < 5 \): \(\lfloor x \rfloor = 4\) (not prime)
- The greatest prime factor of 4 is 2. \[ p(x) = p(2) + (x + 1 - 4) = (2 + 1) + (x - 3) = 3 + x - 3 = x \] Thus, \( p(x) = x \) for \( x \in [4, 5) \).
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For \( 5 \leq x < 6 \): \(\lfloor x \rfloor = 5\) (prime) \[ p(x) = x + 1 \] Thus, \( p(x) = x + 1 \) for \( x \in [5, 6) \).
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For \( 6 \leq x < 7 \): \(\lfloor x \rfloor = 6\) (not prime)
- The greatest prime factor of 6 is 3. \[ p(x) = p(3) + (x + 1 - 6) = (3 + 1) + (x - 5) = 4 + x - 5 = x - 1 \] Thus, \( p(x) = x - 1 \) for \( x \in [6, 7) \).
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For \( 7 \leq x < 8 \): \(\lfloor x \rfloor = 7\) (prime) \[ p(x) = x + 1 \] Thus, \( p(x) = x + 1 \) for \( x \in [7, 8) \).
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For \( 8 \leq x < 9 \): \(\lfloor x \rfloor = 8\) (not prime)
- The greatest prime factor of 8 is 2. \[ p(x) = p(2) + (x + 1 - 8) = (2 + 1) + (x - 7) = 3 + x - 7 = x - 4 \] Thus, \( p(x) = x - 4 \) for \( x \in [8, 9) \).
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For \( 9 \leq x < 10 \): \(\lfloor x \rfloor = 9\) (not prime)
- The greatest prime factor of 9 is 3. \[ p(x) = p(3) + (x + 1 - 9) = (3 + 1) + (x - 8) = 4 + x - 8 = x - 4 \] Thus, \( p(x) = x - 4 \) for \( x \in [9, 10) \).
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For \( x = 10 \): \(\lfloor x \rfloor = 10\) (not prime)
- The greatest prime factor of 10 is 5. \[ p(x) = p(5) + (10 + 1 - 10) = (5 + 1) + 1 = 6 + 1 = 7 \] Thus, \( p(10) = 7 \).
Now, we can summarize \( p(x) \) across the intervals defined:
\[ \begin{align*} p(x) & = x + 1 & \text{for } [2, 3) \ p(x) & = x + 1 & \text{for } [3, 4) \ p(x) & = x & \text{for } [4, 5) \ p(x) & = x + 1 & \text{for } [5, 6) \ p(x) & = x - 1 & \text{for } [6, 7) \ p(x) & = x + 1 & \text{for } [7, 8) \ p(x) & = x - 4 & \text{for } [8, 9) \ p(x) & = x - 4 & \text{for } [9, 10) \ p(10) & = 7 \end{align*} \]
Thus, \( p(x) \) can be expressed in piecewise form.
To find the overall range of \( p(x) \) over the defined interval \( [2, 10] \):
- \( p(x) \) achieves values from \( p(2) = 3 \) up to \( p(10) = 7 \).
- Across the different sections, we establish that various segments (e.g., \( [2, 3) \) leads to \( (3, 4) \)) include 4 and extend to 7.
The function is continuous across the defined intervals, so the range of \( p(x) \) is from 3 to 7, inclusive.
Thus, the final range of \( p(x) \) in interval notation is:
\[ \boxed{[3, 7]} \]
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