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
3 years ago

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

Brian Charlesworth - 3 years ago

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Updated

Kushal Patankar - 3 years ago

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

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.

Brian Charlesworth - 3 years ago

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@Brian Charlesworth I am getting the same.

Sudeep Salgia - 3 years ago

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

Brian Charlesworth - 3 years ago

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@Brian Charlesworth We could easily visualise it graphically.

Indraneel Mukhopadhyaya - 1 year, 12 months ago

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

Kushal Patankar - 3 years ago

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@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 \((-\frac{1}{2}, 0).\)

Brian Charlesworth - 3 years ago

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

Kushal Patankar - 3 years ago

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@Kushal Patankar You're welcome. :)

Brian Charlesworth - 3 years ago

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