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Inverse of f(x) = x + [x]

  1. Dec 11, 2008 #1
    I'm having trouble finding the inverse of f(x) = x + [x]. I think it comes back to what is the inverse of the greatest integer function, [x]. I have graphed [x], and its inverse is the reflection along the y = x line, which appears to be similar, although the inverse graph is "vertical". Is there a name for this inverse graph? I have tried 1/[x], and [1/x], but those are not it. Also, I can't solve for the inverse by factoring out x.

    If it isn't already too much, I can't seem to find the inverse of f(x) = x/(1-x^2)
    for -1 <= x <=1 either, since I can't factor out the x. I have checked that this function is one-to-one on the interval, so it should be possible. Are there any suggestions?

    Thanks in advance!
  2. jcsd
  3. Dec 11, 2008 #2
    There is no inverse to x --> [x] due to its not being injective.
  4. Dec 11, 2008 #3

    D H

    Staff: Mentor

    The problem is the holes in the range of [itex]f:\mathbb R \to \mathbb R[/itex] with [itex]f:x\to x + \lceil x \rceil[/tex]. For example, there is no real x that yields f(x)=1.5.

    That said,

    y-\frac{\lfloor y \rfloor}2

    but only if [itex]\lfloor y \rfloor[/itex] is even. The inverse function is undefined if [itex]\lfloor y \rfloor[/itex] is odd.
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