I found domains and ranges for when $\lfloor x \rfloor$ is prime and otherwise. [hide=My work]We want $\lfloor x \rfloor$ to be prime. The only intervals that work for this are $[2,4)$, $[5,6)$, and $[7,8)$. Evaluating $x+1$ for each interval, we find there is a range of $[3,5)$, $[6,7)$, and $[8,9)$, respectively.
We use the rest of the intervals, $[4,5)$, $[6,7)$, and $[8,10]$ for the other piece, which is everything else. We need to evaluate $p(y)+(x+1-\lfloor x \rfloor)$ for each interval. We know $y$ must be the greatest prime factor of $\lfloor x\rfloor$. Everything in the first interval, $[4,5)$, has a floor of $4$. The greatest prime factor of $4$ is $2$. So, we have $p(y)=p(2)$ for the first interval. We know what $p(2)$ is so we can evaluate it for our expression.
Our equation is now $3+(x+1-\lfloor x \rfloor)$. Evaluating $x+1-\lfloor x \rfloor$, we find if we use anything in our first interval, we have $4.m-4$, because the floor is always going to be $4$, but $x$ might not be. Adding one gives us $1.m$. We don’t know what $m$ is, but we find that $1.m$ has a range of $[1,2)$. Adding on three gives us the range $[4,5)$.
We use the same steps for the next interval, because we only have one floor, $6$! Our interval is $[5,6)$. But for the last interval, we have $3$ different cases$-$when the floor is $8$, $9$, and $10$. The largest prime factors of $8$ and $9$ are $2$ and $3$, respectively. Since we already evaluated those, we don’t need to do them again. But the last case, where the floor is $10$, we get $p(5)$, which we haven’t evaluated yet! Solving for $p(5)$ gives us $6$, in turn giving us the interval $[7,8)$.
Combining our intervals we got for the range, $[3,5)$, $[6,7)$, $[8,9)$, $[4,5)$, $[5,6)$, and $[7,8)$. We combine these all to get the interval $\boxed{[3,9)}$.
