Undefined function If we have an element of the set of real numbers i.e. '-3' and we have the square-root function i.e. $f(x) = \sqrt x$, then $f(-3) = \sqrt {-3}$ is undefined and the ordered pair $(-3, \sqrt {-3})$ is not a member of $f$. Is it what an undefined function is? 
 A: $f(x) = \sqrt x\quad$ IS defined for all $\;x \geq 0,\;$ but is not defined for real numbers $\;x \lt 0$.
So there is a domain on which it is defined, and an interval on which it is not defined.
If a function is undefined, it is usually "relatively" undefined on an interval or at certain points. whether and/or where a function is undefined depends on the given domain of the function. In your case  $f(x) = \sqrt x$ is perfectly defined on the set of non-negative real numbers, on the set of natural numbers, on the set of non-negative integers, etc. But it is undefined for all real negative numbers.  
So you're correct, $(-3, \sqrt {-3}) \notin f$, but we do have $(3, \sqrt 3) \in f$.

Another example: Let $$f: \mathbb R \to \mathbb R,\;\;f(x) = \dfrac 1x.\;$$ Then, when $\;x = 0,\;\;f(0) = \dfrac 10\;$ is undefined, because division by zero is undefined. But $f(x) = \dfrac 1x$ is defined for all other values of $x$: So the domain on which $f$ is defined, i.e., the values of $x$ for which $f$ is defined, includes all $x\in \mathbb R,\;x\neq 0$.  As suggested in a comment below, one way to put this is that no ordered pair of the form $(0, \cdot)$ is a member of $f$. 
