What is the inverse of f(x) = x + [x]?

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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!
 
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There is no inverse to x --> [x] due to its not being injective.
 
The problem is the holes in the range of f:\mathbb R \to \mathbb R with f:x\to x + \lceil x \rceil[/tex]. For example, there is no real x that yields f(x)=1.5.<br /> <br /> That said,<br /> <br /> f^{-1}(y)=&lt;br /&gt; y-\frac{\lfloor y \rfloor}2&lt;br /&gt;<br /> <br /> but only if \lfloor y \rfloor is even. The inverse function is undefined if \lfloor y \rfloor is odd.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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