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 \(0 \le x \le 10\).
Prove your answer
+118608 \(Let\;\;f(x)=\lfloor x\lfloor x \rfloor \rfloor \quad for \;\;x\ge 0\\ (a)\;\; find\;all\;x\ge0\;\;such\;\;that\;\; f(x)=1\)
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.
