Let f(x)=⌊x⌊x⌋⌋ for x≥0
(a) Find all x≥0 such that f(x)=1.
(a) Find all x≥0 such that f(x)=3.
(b) Find all x≥0 such that f(x)=5.
(c) Find the number of possible values of f(x) for 0≤x≤10.
Prove your answer
Letf(x)=⌊x⌊x⌋⌋forx≥0(a)findallx≥0suchthatf(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 if1≤x<2thenf(x)=1
(b)
Letf(x)=⌊x⌊x⌋⌋forx≥0(a)findallx≥0suchthatf(x)=3
let0≤δ<1Ifx=2+δthen⌊x⌊x⌋⌋=⌊(2+δ)2)⌋=⌊(4+2δ)⌋=4or5
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.