Let p(x) be defined on \(2 \le x \le 10\) such that
{ x + 1 if \(\left \lfloor{x}\right \rfloor \) is prime
p(x) = { p(y) + (x + 1 - \(\left \lfloor{x}\right \rfloor \)) otherwise
where Y is the greatest prime factor of \(\left \lfloor{x}\right \rfloor \). Express the range of \(p\) in interval notation.
I tried [3, 8] because when x = 2, p(x) = 3, and when x = 7, p(x) = 8. I thought that was the range, but it is apparently not.
Please help, thanks in advance!
+775 I tried that for guest, and I am pretty sure it is incorrect. I think p(x) can equal 8 because if x = 7, p(x) = 7+1 = 8.
+9673 If you only consider the integer values, then the range is [3,8].
Consider a number infinitesimally close to 8. (7.9999999999999999999999...)
p(7.9999999999999999999...) = x + 1 = 8.999999999999999999999... = 9
So the range is actually [3,9].
+775 Oh, I see what you are saying. I thought the first condition with x+1 if floor x is prime was saying: If floor x is prime, then p(x) = floor x + 1.
That explains a lot. But again, the number can be infinitely close to 8, such as 7.99999..., but the number can never be 8, so the answer range is \(\boxed{[3,9)}\).
