Domain of $f(x)=1/\sqrt{\left\{x+1\right\}-x^2+2x}$ where {.} denotes the fractional part of $x$. Normally, in finding domain for square root function, what I do is that what ever is under the root should be equal to or greater than $0$. Here the square root is in denominator so $\left\{x+1\right\}-x^2+2x > 0$. But due to the fractional part function I don't know how to proceed. I know $\left\{x+1\right\} = x+1 -[x+1]$ but this isn't helping much either. Any help will be appreciated. Thank you.
 A: Note that $0\leq\{x+1\}<1$ so consider the boundary values of $0$ and $1$.  at $0$ we have $-x^2+2x=x(-x+2)$ which is nonnegative positive between $0$ and $2$, and adding a nonnegative number will keep it nonnegative, so we have that entire part in.  Now,  moving away from $[0,2]$ makes $-x^2+2x$ negative, to counteract that we need the fractional part of ${1+x}$ to be greater than or equal to it.  Obviously if $x\geq 3$ we have gone far too negative, likewise if $x\leq -1$ it is too negative for the fractional part to make up for it, so we only have to worry about the values in $(-1,0)$ and $(2,3)$.
In $(2,3)$ the fractional part of $x+1$ is $x-2$,  so we can plug that into our polynomial and :  $$-x^2+2x+x-2=-x^2+3x-2=-1(x^2-3x+2)=-(x-1)(x-2)$$, which is positive between $1$ and $2$,  so we don't pick up anything here.
In $(-1,0)$ the fractional part of $x+1$ is $x+1$,  so plugging that in we get
$$-x^2+2x+x+1=-x^2+3x+1$$ which has roots at
$$\frac {3\pm \sqrt {13}}{2}$$
Intersecting that with (-1,0) gets us the lower half of the domain intersects with $(-1,0)$,  so putting that together with $[0,2]$
gets to a final domain of
$(\frac{3 - \sqrt{13}}{2}, 2)$ $-$ {$0$}
This graph shows the possible domain, same as the @Alan has mentioned above, just in a graph: Look at function 2's max and min domain as function 2 is a subset of function 1 (asymptote at $\frac {3 - \sqrt {13}}{2}$).
Btw, desmos only likes decimal apparently, so $\frac {3 - \sqrt {13}}{2}$ = -0.303 (3 decimal rounding).

A: To proceed further,
Notice that we can take out integer from the floor function,
i.e. 1 from the last equation you wrote.
The full solution is attached below
![Solution](https://i.stack.imgur.com/prTs5.jpg
