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
$\endgroup$ 32 Answers
$\begingroup$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 $D_f = \mathbb R^2$.
$\endgroup$ $\begingroup$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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