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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 \(0 \le x \le 10\).

 

Prove your answer

 May 22, 2022
 #1
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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

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