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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

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  • $\begingroup$ For which $x$ values do you get a fractional part between $0$ and $\frac{1}{2}$ ? $\endgroup$ May 9 '20 at 18:38
  • $\begingroup$ @NinadMunshi I know that for $-1\le x < 0$, $\{x\} = x+1$, so here for negative values of x, how would I get the answers? In between $0$ to $1$, I only can have values between $0$ to $\frac{1}{2}$ and also include $1$ as that yields zero, but what about for negative values of $x$? Like between $-1$ and $0$? $\endgroup$
    – Techie5879
    May 9 '20 at 18:41
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Hint: Write $X = \left \lfloor X \right \rfloor + \left \{ X \right \}$ $\\ And \left \lfloor X \right \rfloor = 0 \ or \ -1$

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