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
    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. – zzuussee Aug 10 at 10:03
  • Why do you think $f$ should not take the value $0$. As long as the denominator in non-zero the functions is well defined. – Kavi Rama Murthy Aug 10 at 10:03
  • However, if $x,y \in \Bbb{C}$, things will be a bit different. – twalberg Aug 10 at 15:52
up vote 8 down vote accepted

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 $(x,y) \in \mathbb R^2$.

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.