(a) Find all $x \ge 0$ such that $f(x) = 1.$
(b) Find all $x \ge 0$ such that $f(x) = 3.$
(c) Find all $x \ge 0$ such that $f(x) = 5.$
(d) Find the number of possible values of $f(x)$ for $0 \le x \le 10.$
I don't exactly know how I can attempt to do this. First I said (for part a) that $1\leq x \cdot \lfloor{x} \rfloor <2$ then divided by $x$ and split the remaining floor function into $x-[x]$, where $[x]$ is the fractional part of $x$. Simplified a few things out and now I have $\frac{x-1}{x} \geq [x] > \frac{x-2}{x}$.
Not sure how to proceed from here or if I'm even working in the right direction. I think if I can get help for part $a$ then I may be able to do the rest on my own?
Thanks in advance!
Best Answer
As you say, you want $1\le x\lfloor x\rfloor<2$. It’s easy to check that $x\lfloor x\rfloor\ge 2$ if $x\ge 2$, so you can assume that $x<2$. Similarly, it’s clear that $x\lfloor x\rfloor<1$ if $x<1$, so you can assume that $x\ge 1$. And if $1\le x<2$, then $\lfloor x\rfloor=1$, so $x\lfloor x\rfloor=x$, and $f(x)=1$.
Note that $f$ is a non-decreasing function, $f(2)=4$, and $f(3)=9$, so if $f(x)=5$, then $2<x<3$. To find the actual interval, we want to solve $x_0\lfloor x_0\rfloor=5$ and $x_1\lfloor x_1\rfloor=6$ for $x_0$ and $x_1$. We know that $\lfloor x_0\rfloor=\lfloor x_1\rfloor=2$, so the first is easy: $x_0=\frac52$. The second has no solution, but $x\lfloor x\rfloor<6$ for $x<3$, so $f(x)=5$ if and only if $\frac52\le x<3$.
I’ll leave the rest to you.