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# Floor function

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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 $$f(x) = 1$$.

(a) Find all $$x \ge 0$$ such that $$f(x) = 3$$.

(b) Find all $$x \ge 0$$ such that $$f(x) = 5$$.

(c) Find the number of possible values of $$f(x)$$ for $$0 \le x \le 10$$.

May 22, 2022

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$$Let\;\;f(x)=\lfloor x\lfloor x \rfloor \rfloor \quad for \;\;x\ge 0\\ (a)\;\; find\;all\;x\ge0\;\;such\;\;that\;\; f(x)=1$$

If  0<=x<1 then  f(x)=0

If  1<=x<2 then   f(x)= floor of (x*1) = 1

If x>=2      then    f(x)>1

so      $$if\;\; 1\le x<2 \;\;then \;\;f(x)=1$$

(b)

$$Let\;\;f(x)=\lfloor x\lfloor x \rfloor \rfloor \quad for \;\;x\ge 0\\ (a)\;\; find\;all\;x\ge0\;\;such\;\;that\;\; f(x)=3$$

$$let\;\; 0\le\delta <1\\ If\;\;x=2+\delta \;\;then\\ \lfloor x\lfloor x \rfloor \rfloor=\lfloor (2+\delta) 2) \rfloor=\lfloor (4+2\delta) \rfloor=4\;\;or\;\;5$$

So if  x is 1 and a bit then f(x)=1

If x=2 and a bit then f(x)=4 or 5

f(x) cannot equal 3

You can think about the rest yourself.

May 23, 2022