Asked by Fiona

Let p(x) be defined on 2 <= x <= 10 such that p(x) = x + 1 if floor(x) is prime, p(x) = p(y) + (x + 1 - floor(x)) otherwise, where y is the greatest pirme factor of floor(x). Express the factor of p in interval notation.

Answers

Answered by GPT-4o mini
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:

1. **For \( 2 \leq x < 3 \)**:
\(\lfloor x \rfloor = 2\) (prime)
\[
p(x) = x + 1
\]
Thus, \( p(x) = x + 1 \) for \( x \in [2, 3) \).

2. **For \( 3 \leq x < 4 \)**:
\(\lfloor x \rfloor = 3\) (prime)
\[
p(x) = x + 1
\]
Thus, \( p(x) = x + 1 \) for \( x \in [3, 4) \).

3. **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 \( x \in [4, 5) \).

4. **For \( 5 \leq x < 6 \)**:
\(\lfloor x \rfloor = 5\) (prime)
\[
p(x) = x + 1
\]
Thus, \( p(x) = x + 1 \) for \( x \in [5, 6) \).

5. **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 \( x \in [6, 7) \).

6. **For \( 7 \leq x < 8 \)**:
\(\lfloor x \rfloor = 7\) (prime)
\[
p(x) = x + 1
\]
Thus, \( p(x) = x + 1 \) for \( x \in [7, 8) \).

7. **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 \( x \in [8, 9) \).

8. **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 \( x \in [9, 10) \).

9. **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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