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
Best Answer
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 $D_f = \mathbb R^2$.