Is the function $ \varphi:[-1,+1]\to [-1,+1] $ continuous? Let $g:\Omega\to \mathbb{R}$, $\Omega\subseteq \mathbb{R}^2$ open, a polynomial functions in the variables $x$ and $y$, i.e.
$$
g(x,y)= \sum_{\ell=0}^{m}\sum_{\mu+\nu=\ell} b_{\mu\nu}x^{\mu}y^{\nu}.
$$
Suppose $[-1,+1]\times [-1,+1]\subset \Omega$ and that the function $g$ satisfy the conditions:
$$
\begin{array}{ccc}
 g(x,-1)<0 \mbox{ for all } x\in[-1,1]
&\quad \mbox{ and }\quad &
g(x,1)>0 \mbox{ for all } x\in[-1,1]
\\
\end{array}
$$
For each $x\in [-1,+1]$ fix the set
$$
\Lambda(x)=\left\{y\in [-1,1] \left|\begin{array}{rl} g(x,y)&=0\\ (\partial/\partial y)g(x,y)&>0 \end{array}\right.\right\}
$$
Set the function $\varphi:[-1,+1]\to [-1,+1]$ by
$$
\varphi(x)= \min\overline{\Lambda(x)},
$$
that is, $\varphi(x)$ is the minimum of the set of adherent points of set $\Lambda(x)$. 

Question. Is the function  $ \varphi:[-1,+1]\to [-1,+1] $ continuous?

My attempt. In fact, let $x_n\to \tilde{x}\in [-1,+1]$. I'm trying to prove that $\lim_{n\to\infty}\varphi(x_n)=\varphi(\tilde{x})$.For each $x_n\in [-1,+1]$ there is a $y_n\in\overline{\Lambda(x_n)}$ such that 
$$
y_n=\min \overline{\Lambda(x_n)}
$$
I'm trying to use the fact that both $\Lambda(x_n)$ and the range $[-1,+1]$ are sequentially compact. But I'm not getting any success. The problem is that when $ y_n $ varies for $ y_{n + 1} $ the set $ \Lambda (x_n)$ varies for $ \Lambda (x_{n + 1}) $.

First of all, $\varphi(x)$ may not even be defined, since $\Lambda(x)$ can be empty.  For instance, if $g(x,y)=y^3$ then $\Lambda(x)$ is empty for all $x$.
Even when $\varphi$ is defined, it may not be continuous.  For instance, let $g(x,y)=(y^2-x/4)(y-2/3)$.  Then $2/3\in\Lambda(x)$ for all $x$ so $\varphi(x)$ is always defined.  Moreover, $2/3$ is the only point in $\Lambda(x)$ if $x\leq 0$, so $\varphi(x)=2/3$ for $x\leq 0$.  But when $x>0$, $\varphi(x)=-\sqrt{x}/2$, and in particular $\varphi(x)$ approaches $0$ as $x\to 0$ from above.
