What will be the domain of the function $f(x)=\sqrt{2\{x\}^2-3\{x\}+1}$? I need to find the domain of the function

$$f(x)=\sqrt{2\{x\}^2-3\{x\}+1}$$

where $x \in [-1,1]$ and $\{.\}$ represents the fractional part of $x$

So here's what I tried:

Clearly the part inside the square root has to be greater than or equal to zero for it to exist, and by factoring the quadratic in terms of $\{x\}$ I get,

$$(\{x\}-1)(2\{x\}-1)\geq0$$

So, from here I got $$\{x\} \in \bigg(-\infty,\frac{1}{2}\bigg] \cup \bigg[1,\infty\bigg)$$

But we know, that the fractional part of x can only vary between 0 and 1, ie,

$\{x\} \in[0,1)$

So finally, I get $\{x\} \in \bigg(-\infty,\frac{1}{2}\bigg]$, which further reduces to the following

$$\{x\} \in \bigg[0,\frac{1}{2}\bigg]$$

And, from the question, $x \in[-1,1]$

But how do I finally get the domain for $x$ here? I'm seemingly stuck at the last step
Hint:
Write $X = \left \lfloor X \right \rfloor + \left \{ X \right \}$
$\\ And  \left \lfloor X \right \rfloor = 0 \  or \ -1$
Hint:
Write $X = \left \lfloor X \right \rfloor + \left \{ X \right \}$
$\\ And  \left \lfloor X \right \rfloor = 0 \  or \ -1$