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O.k., great. Assuming that we are working in the domain of real numbers, we require that $g(x) = [x] - 2x \ge 0.$

Now clearly for $x \gt 0$ we have $g(x) \lt 0$ since it can be at most $x - 2x = -x.$ Since $g(0) = 0$ we have that $0$ is part of the domain of $f(x) = \sqrt{g(x)}.$

Next, for $-\frac{1}{2} \lt x \lt 0$ we have that $g(x) = -1 - 2x \lt 0,$ and hence this interval is not part of the domain of $f(x).$ However, for $-1 \le x \le -\frac{1}{2}$ we have that $g(x) = -1 - 2x \ge 0.$

For $-2 \le x \le -1$ we have that $g(x) = -2 - 2x = -2(1 + x) \ge 0$ since $1 + x \le 0.$

In general, for $k \le x \le (k + 1)$ for integers $k \le -2$ we see that $g(x) = k - 2x \ge k - 2(k + 1) = -(k + 2) \ge 0.$

So putting the pieces together, we conclude that the domain of $f(x)$ is $(-\infty, -\frac{1}{2}] \cup \{0\}.$

Edit: I just checked WolframAlpha; oddly, it responded with "unable to determine domain". I think I've got it right, though; it's an unusual, discontinuous function, but breaking down the intervals as I have reveals the general form of the function.

@Kushal Patankar
–
In that interval we have $g(x) = -1 - 2x \lt -1 - 2(\frac{1}{2}) = 0.$ Thus $f(x) = \sqrt{g(x)}$ is not defined (over the reals) in that interval. In the interval $[-1, -\frac{1}{2}]$ we have $g(x) = -1 - 2x \ge -1 - 2(\frac{1}{2}) = 0,$ and so $f(x)$ is defined in this interval. This is why we had to isolate the interval $(-\frac{1}{2}, 0).$

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## Comments

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TopNewestCould you please clarify what $[x]$ means; is it the greatest integer function?

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Updated

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O.k., great. Assuming that we are working in the domain of real numbers, we require that $g(x) = [x] - 2x \ge 0.$

Now clearly for $x \gt 0$ we have $g(x) \lt 0$ since it can be at most $x - 2x = -x.$ Since $g(0) = 0$ we have that $0$ is part of the domain of $f(x) = \sqrt{g(x)}.$

Next, for $-\frac{1}{2} \lt x \lt 0$ we have that $g(x) = -1 - 2x \lt 0,$ and hence this interval is not part of the domain of $f(x).$ However, for $-1 \le x \le -\frac{1}{2}$ we have that $g(x) = -1 - 2x \ge 0.$

For $-2 \le x \le -1$ we have that $g(x) = -2 - 2x = -2(1 + x) \ge 0$ since $1 + x \le 0.$

In general, for $k \le x \le (k + 1)$ for integers $k \le -2$ we see that $g(x) = k - 2x \ge k - 2(k + 1) = -(k + 2) \ge 0.$

So putting the pieces together, we conclude that the domain of $f(x)$ is $(-\infty, -\frac{1}{2}] \cup \{0\}.$

Edit: I just checked WolframAlpha; oddly, it responded with "unable to determine domain". I think I've got it right, though; it's an unusual, discontinuous function, but breaking down the intervals as I have reveals the general form of the function.

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$x \in (\frac{-1}{2} , 0)$ as a case.

Thanks for the answer. But I didn't got the reason behind takingLog in to reply

$g(x) = -1 - 2x \lt -1 - 2(\frac{1}{2}) = 0.$ Thus $f(x) = \sqrt{g(x)}$ is not defined (over the reals) in that interval. In the interval $[-1, -\frac{1}{2}]$ we have $g(x) = -1 - 2x \ge -1 - 2(\frac{1}{2}) = 0,$ and so $f(x)$ is defined in this interval. This is why we had to isolate the interval $(-\frac{1}{2}, 0).$

In that interval we haveLog in to reply

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