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The domain of $f(x)= (x-1)^{1/2}$ are all the real numbers. Now I can sub in x=0 into the expression, and I get the square root of -1 which is a complex number. So is this okay?

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  • $\begingroup$ Assuming you are talking about the real numbers ($\Bbb R$) only for the domain, you can't then have a complex number including in that domain. $\endgroup$ – frog1944 Mar 13 '16 at 2:02
  • $\begingroup$ @frog1944 If $f(x) = (x - 1)^{1/2}$ is a real-valued function of a real variable, we need to restrict the values of $x$ so that we do not include a complex number in the range, which is why the domain of $f$ is $[1, \infty)$. $\endgroup$ – N. F. Taussig Mar 13 '16 at 11:37
  • $\begingroup$ @N.F.Taussig I completely agree with you, but wasn't the question asking if it's okay to include a complex number in the domain. I was stating that we can't include a complex number in the domain. $\endgroup$ – frog1944 Mar 13 '16 at 19:49
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You state immediately that "The domain of $f(x)=(x−1)^{1/2}$ are all the real numbers". How did you arrive at that? If we require that f(x) be a real number, then the domain is "all real number larger than or equal to 1".

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The domain of $f(x)= (x-1)^{1/2}$ is all real $x\ge 1$. If you are talking about the domain of a function over the real numbers, you have restricted it to the values of $x$ for which the function is real.

You are correct in saying that $f(0)$ is not a real number. However, this means that $x=0$ is not apart of the domain, and thus it cannot be all real numbers.

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Domain of a function is defined as the set of input arguments for which the function is defined.

The function is not defined for $x<1$ and so, the domain of the given function is $$x\geq 1$$

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  • $\begingroup$ Consider using \geq vs =>. $\endgroup$ – YoTengoUnLCD Mar 13 '16 at 1:43
  • $\begingroup$ @YoTengoUnLCD Thank you for editing $\endgroup$ – Win Vineeth Mar 13 '16 at 1:44
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Assuming you are finding the domain over the real numbers ($\Bbb R$), then $$\sqrt {x-1} \geq 0$$ As you can't have the square root of a number less than 0 in the real numbers. Solve the expression by squaring it out and then putting the $1$ on the right hand side and you find the domain is:

All real $x$, where $x \geq 1$

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