Implicit functions: some remarks

$\bullet$ Let us consider the equation
\begin{equation} \label{eq1}
F(x,y)=0
\end{equation}
and the set
\begin{equation}
A=\{(x,y)/ x,y \in \mathbb{R} \ \ and \ \ F(x,y)=0 \} \subset  \mathbb{R}^{2}
\end{equation}
of real solutions of \eqref{eq1}.
   The set of points of the Cartesian plane represented by the ordered pairs of $A$ is called the graph of  \eqref{eq1}.
$\bullet$ If there is a set $E \subset  \mathbb{R}$ such that
$$
\begin{array}{ccc}
 \forall x \in E,  & \exists \ \ y  \in \mathbb{R}, &   F(x,y)=0
\end{array}
$$
then we can map for any $x \in E$ a single $y=f(x)$ such that
$$
F(x,f(x))=0
$$
In this way a function $f:E \rightarrow \mathbb{R}$ with formula $y=f(x)$ is defined and this function will be called an implicit function defined by \eqref{eq1}.
$\bullet$ If this is the case, the graph of the implicit function $y=f(x)$, $f:E \rightarrow \mathbb{R}$ defined by \eqref{eq1} will be a part of the graph of \eqref{eq1}.
Remarks: 
1. It is interesting to note that an implicit function may be defined through an equation even if there is no way to obtain an analytic expression for the function's formula by solving the equation. This is the case (why?) for example for the equation
\begin{equation}
2y-2x-siny=0
\end{equation}
2. Even in cases in which the formula is easy to handle, many "unexpected" functions may be defined implicitly through an equation; most of them may not even be continuous: this is the case for example with the equation of a circle
\begin{equation}
x^{2}+y^{2}=1
\end{equation}
when it is solved with respect to $y$ a host of functions arise; some of them are continuous:
$$
\begin{array}{ccc}
 \begin{array}{l}
y=f_{1}(x)=\sqrt{1-x^{2}}  \\
[-1,1] \rightarrow \mathbb{R}
 \end{array}   &  \begin{array}{l}
                            y=f_{2}(x)=-\sqrt{1-x^{2}}  \\
                            [-1,1] \rightarrow \mathbb{R}
                            \end{array}
\end{array}
$$

while others are not, either at a point:



$
 y=f_{3}(x)= \left\{%
                               \begin{array}{l}
                               \sqrt{1-x^{2}}, \ \ x \in [-1,0) \\
                              -\sqrt{1-x^{2}}, \ \ x \in [0,1]
                               \end{array}
                                \right.$

or nowhere in their domain:
$$
 y=f_{4}(x)=\left\{%
                            \begin{array}{l}
                            \sqrt{1-x^{2}}, \ \ x \in [-1,1],  x: \ rational \\
                           -\sqrt{1-x^{2}}, \ \ x \in [-1,1], x: \ irrational
                            \end{array}
                            \right.
$$
Note that for all the above functions the domain is common $D_{f_{i}}=[-1,1]$, $i=1,2,3,4$.