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 $X = \left \lfloor X \right \rfloor + \left \{ X \right \}$ $\\ And \left \lfloor X \right \rfloor = 0 \ or \ -1$