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I’m preparing a lesson on the domain of functions for my high school students and came across an interesting problem. The function I’m analyzing is:

$$ f(x) = \frac{x}{\sqrt{x-1}} $$

This function was given in a textbook, without the domain being explicitly stated. Normally, to determine the domain, we must impose the condition that the denominator is non-zero and the square root is real. So, we would require:

$$ x > 1 $$

However, I wanted to introduce my students to a more nuanced situation involving complex numbers. For example, at $x = 0$, if we allow $f(x)$ to take values in the complex plane, we get:

$$ f(0) = \frac{0}{\sqrt{0-1}} = \frac{0}{i} = 0 $$

As high school students should know, dividing complex zero by a non-zero complex number still results in the complex zero. This might lead students to incorrectly think that $x = 0$ could belong to the domain of $f(x)$, since we can assign a value to it when working in $\mathbf{C}$.

Of course, to compute $f(0)$, we have to interpret the square root of $-1$ in the complex plane ($\mathbf{C}$), and all of this reasoning suggests that $x = 0$ should not be part of the domain of $f(x)$.

Given this context, my question is: Should the function’s domain be restricted solely to the reals where $x > 1$, or could we consider extending it to include complex values (such as $x = 0$) to allow for this broader interpretation?

I’m particularly interested in how to present this in a way that avoids ambiguity for high school students.

Thanks for your insights!

Updates

Obviously, I have my own opinion on this issue, which is as follows. I will provide the following definitions of the domain of a function:

Definition 1. The set $I$ of values that can be attributed to $x$, such that the corresponding value of $y$ exists, is called the domain of existence or the domain of definition of the function $y$.

It is implied that $I \subseteq \mathbb{R}$ and that 'exist' means being a real number.

Definition 2. The domain of existence of a function, given by an analytical relation, depends on the relationship between the two variables. In general, it is determined by examining for which values of the independent variable $x$ the operations that need to be performed on it in order to obtain the value of the function $y$ are meaningful.

At first glance, the two 'definitions' seem equivalent, but they are not. The first one considers the function 'globally' and is applicable even if an analytical expression is not known. Thus, in the case of my function, the number 0 is an element of the domain, since $\frac{0}{\sqrt{-1}} = 0 \in \mathbb{R}$. The second one favors a morphological analysis: the function is given by an equation that includes a term. This term is first broken down into sub-terms, following the operations and the composition; each of these sub-terms is used to define a function, for which the domain is determined. From these domains, using typical intersection operations, we finally answer the question. With this analysis, my function breaks down into the functions $y = x$ and $y = 1/\sqrt{x-1}$, from which the function is obtained by multiplication. If we intersect the domains of these functions, we get $x > 1$. What do you think?

Mark
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1 Answers1

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I would be clear in the distinction between the two cases.

For reals functions we need in this case $x>1$ since by definition $$y=\sqrt x\ge 0 \iff y^2 =x\ge 0$$ then we can also extend $f(x)$ to the complex version $f(z)$ with $f: \mathbb C \to \mathbb C$ which is in general a multivalued function (square root for complex needs to be properly defined) and for the special case $z=0+i0$ it returns $0+i0$ as result.

user
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  • Thank you! I updated my question, adding my opinion on the problem... what do you think? – Mark Oct 13 '24 at 17:03
  • It seems reasonable and maybe the definitions $1$ and $2$ can be given together, the first one more general and the second one for special cases, when function is expressed by an analytical expression. In the definition $1$, it could be useful explain that the domain $D$ can also be a chosen as a suitable subset of $I$, e.g. for $y=x^2$ in order to define the inverse $y=\sqrt x$. – user Oct 13 '24 at 19:22
  • @Mark Here there is a discussion on the concept of function, hope this can also be useful. – user Oct 13 '24 at 19:24
  • thank you for your additional comments. I would like to kindly ask if any university textbooks contain the second definition I provided, or if you would recommend that I ask a new question requesting specific bibliographic references. Thank you for your help! – Mark Oct 14 '24 at 04:47
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    In these wiki articles there are useful discussion and many references function, partial functions I, partial functions II. – user Oct 14 '24 at 09:05