Domain help

What is domain of [x]2x\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
4 years, 5 months ago

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

Brian Charlesworth - 4 years, 5 months ago

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Updated

Kushal Patankar - 4 years, 5 months ago

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O.k., great. Assuming that we are working in the domain of real numbers, we require that g(x)=[x]2x0.g(x) = [x] - 2x \ge 0.

Now clearly for x>0x \gt 0 we have g(x)<0g(x) \lt 0 since it can be at most x2x=x.x - 2x = -x. Since g(0)=0g(0) = 0 we have that 00 is part of the domain of f(x)=g(x).f(x) = \sqrt{g(x)}.

Next, for 12<x<0-\frac{1}{2} \lt x \lt 0 we have that g(x)=12x<0,g(x) = -1 - 2x \lt 0, and hence this interval is not part of the domain of f(x).f(x). However, for 1x12-1 \le x \le -\frac{1}{2} we have that g(x)=12x0.g(x) = -1 - 2x \ge 0.

For 2x1-2 \le x \le -1 we have that g(x)=22x=2(1+x)0g(x) = -2 - 2x = -2(1 + x) \ge 0 since 1+x0.1 + x \le 0.

In general, for kx(k+1)k \le x \le (k + 1) for integers k2k \le -2 we see that g(x)=k2xk2(k+1)=(k+2)0.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)f(x) is (,12]{0}.(-\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.

Brian Charlesworth - 4 years, 5 months ago

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@Brian Charlesworth I am getting the same.

Sudeep Salgia - 4 years, 5 months ago

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@Sudeep Salgia Great. Thanks for the confirmation. :)

Brian Charlesworth - 4 years, 5 months ago

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@Brian Charlesworth Thanks for the answer. But I didn't got the reason behind taking x(12,0) x \in (\frac{-1}{2} , 0) as a case.

Kushal Patankar - 4 years, 5 months ago

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@Kushal Patankar In that interval we have g(x)=12x<12(12)=0.g(x) = -1 - 2x \lt -1 - 2(\frac{1}{2}) = 0. Thus f(x)=g(x)f(x) = \sqrt{g(x)} is not defined (over the reals) in that interval. In the interval [1,12][-1, -\frac{1}{2}] we have g(x)=12x12(12)=0,g(x) = -1 - 2x \ge -1 - 2(\frac{1}{2}) = 0, and so f(x)f(x) is defined in this interval. This is why we had to isolate the interval (12,0).(-\frac{1}{2}, 0).

Brian Charlesworth - 4 years, 5 months ago

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@Brian Charlesworth OK. got you now. Thanks 4 help

Kushal Patankar - 4 years, 4 months ago

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@Kushal Patankar You're welcome. :)

Brian Charlesworth - 4 years, 4 months ago

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@Brian Charlesworth We could easily visualise it graphically.

Indraneel Mukhopadhyaya - 3 years, 4 months ago

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