What is the domain of this function in 3 variables? Consider the function
$$f(x, y, z)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+xyz.$$
Which is the domain of this function?
Certainly, we have to ask $(x, y, z)\neq(0, 0, 0)$ but also the points of the type
$$ (0, y, z), \ (x, 0, z), \ (x, y, 0), \ (0, 0, z), \ (x, 0, 0), \ (0, y,0)$$
are not allowed. How to write in an &quot;easy&quot; way?
I would say
$$D(f)=\{(x, y, z)\in\mathbb{R}^3 : (x, y, z)\neq (0, 0, 0), ..\}$$
but I don't know how to continue. Could anyone please help me?
Finally, all these points to be excluded, consitute a straight line, a plane or something else?
Thank you in advance!
$\newcommand{\R}{\mathbb{R}}$
You've already found an answer by yourself. You can just set the domain of your function $f:X \rightarrow Y$ to
$$
\begin{align}
X
&amp;= \{(x,y,z) \in \R^3 ~ | ~ x \neq 0 \land y \neq 0 \land z \neq 0 \}\\
&amp;= \R^3 \setminus \{(x,y,z) \in \R^3 ~ | ~ x = 0 \lor y = 0 \lor z = 0 \}\\
&amp;= \R^3 \setminus \left[\left\{(x,y,z) \in \R^3 ~ | ~ x = 0\} \cup \{(x,y,z) \in \R^3 ~ | ~ y = 0\} \cup \{(x,y,z) \in \R^3 ~ | ~ z = 0\right\}\right]
\end{align}
$$
It constitutes an entire three-dimensional Euclidean space with the $x-y$, $y-z$, and $z-x$ planes being removed.
Meanwhile, the codomain should be $Y = \R$.
For your first question consider labeling the domain:
$D(f)=\{(x, y, z)\in\mathbb{R}^3 : x,y,z \neq 0\}$.
The following is a response to your second question from the bottom of your post:
As $\mathbb{R^{3}}$ represents all of $3$-dimensional space, $D(f)$ is $\mathbb{R^{3}}$ without the points previously mentioned. The origin $(0,0,0)$ is removed, WLOG points of the form $(0,0,z)$ lie on one of the $3$ axes, so each axis is removed. WLOG points of the form $(0,y,z)$ represent an arbitrary point on the $xy$ plane. Altogether, we end up with $\mathbb{R^{3}}$ without the $xy$,$xz$,and $yz$ planes.