I have a function of two real variables which is given by the transformation rule $$f(x,y)=\frac{y}{1+x^2+y^2}.$$ I have to find the domain of $f$ which consists of all points $(x,y)$.

When I examine the function I would say the domain is $$|x,y \in \Bbb{R}^2:y\neq0, x \text{ are real numbers|}$$, but looking at the results-list it says that both $x$ and $y$ are real numbers. How come that is?

This might be straightforward for some of you, but I can't seem to wrap my head around this on my own and hope some of you can help. Thanks in advance

  • 1
    $\begingroup$ The domain of a function is the set of values which the function can take as input. You're probably tasked with finding the maximal domain in $\mathbb{R}^2$. So you have to ask yourself, for which $(x,y)\in\mathbb{R}^2$ is the function $f$ you stated (not) defined. $\endgroup$ – blub Aug 10 '18 at 10:03
  • $\begingroup$ Why do you think $f$ should not take the value $0$. As long as the denominator in non-zero the functions is well defined. $\endgroup$ – Kavi Rama Murthy Aug 10 '18 at 10:03
  • $\begingroup$ However, if $x,y \in \Bbb{C}$, things will be a bit different. $\endgroup$ – twalberg Aug 10 '18 at 15:52

For the domain of the given function, the denominator must be different than zero, but :

$$1+x^2+y^2 \neq 0 \Leftrightarrow 1 \neq -x^2 - y^2$$

Note that $-x^2 -y^2 \leq 0 \; \forall \; x,y \; \in \mathbb R$ and since $1$ is a positive number, this can never equal it. There are no other constraints to check. Thus, the domain is $D_f = \mathbb R^2$.


We have $1+x^2+y^2 \ge 1 >0$ for all $(x,y) \in \mathbb R^2$. Hence $1+x^2+y^2 \ne 0$ for all $(x,y) \in \mathbb R^2$. This shows that $f$ is defined for all $(x,y) \in \mathbb R^2$.


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