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.

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.

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TopNewestCould you please clarify what \([x]\) means; is it the greatest integer function? – Brian Charlesworth · 2 years, 1 month ago

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– Kushal Patankar · 2 years, 1 month ago

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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 · 2 years, 1 month ago

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– Sudeep Salgia · 2 years, 1 month ago

I am getting the same.Log in to reply

– Brian Charlesworth · 2 years, 1 month ago

Great. Thanks for the confirmation. :)Log in to reply

– Indraneel Mukhopadhyaya · 1 year, 1 month ago

We could easily visualise it graphically.Log in to reply

– Kushal Patankar · 2 years, 1 month ago

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

– Brian Charlesworth · 2 years, 1 month 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 \((-\frac{1}{2}, 0).\)Log in to reply

– Kushal Patankar · 2 years, 1 month ago

OK. got you now. Thanks 4 helpLog in to reply

– Brian Charlesworth · 2 years, 1 month ago

You're welcome. :)Log in to reply