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# Domain help

What is domain of $$\sqrt{[x]-2x}$$

$$[•]$$ represents greatest integer function.

May sound silly question but I have just started learning functions, so plz help.

Note by Kushal Patankar
1 year, 6 months ago

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Could you please clarify what $$[x]$$ means; is it the greatest integer function? · 1 year, 6 months ago

Updated · 1 year, 6 months ago

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. · 1 year, 6 months ago

I am getting the same. · 1 year, 6 months ago

Great. Thanks for the confirmation. :) · 1 year, 6 months ago

We could easily visualise it graphically. · 6 months, 1 week ago

Thanks for the answer. But I didn't got the reason behind taking $$x \in (\frac{-1}{2} , 0)$$ as a case. · 1 year, 6 months ago

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 $$(-\frac{1}{2}, 0).$$ · 1 year, 6 months ago

OK. got you now. Thanks 4 help · 1 year, 6 months ago