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Let π‘“(π‘₯)=⌊π‘₯⌊π‘₯βŒ‹βŒ‹f(x)=⌊x⌊xβŒ‹βŒ‹ for π‘₯β‰₯0.

 

(a) Find all π‘₯β‰₯0xβ‰₯0 such that π‘“(π‘₯)=1.f(x)=1.
(b) Find all π‘₯β‰₯0xβ‰₯0 such that π‘“(π‘₯)=3.f(x)=3.
(c) Find all π‘₯β‰₯0xβ‰₯0 such that π‘“(π‘₯)=5.f(x)=5.
(d) Find the number of possible values of π‘“(π‘₯)f(x) for 0≀π‘₯≀10.

Thanks so much!

 Jan 29, 2022
 #1
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Sorry,

It doesn't make much sense to me either.

 Jan 29, 2022

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