Assume that we have a function $f$ such that $f(x) = \sqrt x$ Is a domain of $-1$ undefined. I thought it wouldn’t because the answer would be $i\sqrt x$. However, apparently, it is undefined. Can someone explain to me why?
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2If the domain is that of reals, square root is not defined for negative numbers. If the domain is that of complex... – Mauro ALLEGRANZA Oct 29 '20 at 15:24
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I couldn’t understand what you meant by that of complex and that of reals. Could you explain? Thank you ! – Atilla Çolak Oct 29 '20 at 15:27
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Do you know Real numbers and Complex numbers ? – Mauro ALLEGRANZA Oct 29 '20 at 15:31
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Formally, a function is very explicitly defined by 1) the domain of the function, 2) the codomain of the function, 3) a rule to follow for mapping an input to an output. Do not confuse the function $f$ whose domain is $\Bbb R_{\geq 0}$ and codomain is $\Bbb R$ given by $x\mapsto \sqrt{x}$ and the function $f$ whose domain is $\Bbb C$ whose codomain is $\Bbb C$ given by $x\mapsto \sqrt{x}$. – JMoravitz Oct 29 '20 at 15:32
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If you were only told that $f(x)=\sqrt{x}$ with absolutely no additional context... then the correct answer to whether $f(-1)$ is defined is "it depends." If you are able to figure out from context that the domain is meant to be all reals or just the non-negative reals then you say $f(-1)$ is undefined. If you learn that the domain is all complex numbers then you say $f(-1)$ is defined and is equal to $i$. If it is any other context then you must examine that situation more closely. – JMoravitz Oct 29 '20 at 15:35
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Okay thank you. That part is not given in the question, and there is no choice as it depends. Therefore, I’ll assume that the domain is real – Atilla Çolak Oct 29 '20 at 15:37