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# Algebra - Floor Functions

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Let $$f(x) = \lfloor x \lfloor x \rfloor \rfloor$$ for $$x \ge 0$$.

(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 $$x \ge 0$$ for $$0 \le x \le 10.$$

I've solved part (a), and here's my answer:

In order for $$f(x)$$ to equal $$1$$, we must have the inequality $$1 \leq x \lfloor x \rfloor <2$$. From this we can easily see that if $$x<1$$, then $$\lfloor x \rfloor < 1$$.

Similarly, if $$x\geq2$$, then $$\lfloor x \rfloor\geq2$$.

However, if $$1\le x<2$$, then $$x\lfloor x \rfloor=x$$, and so $$f(x)=1$$. Hence, we have our answer.

Jun 22, 2022