Implicit function theorem problem I have the function
$$(x-2)^3y+xe^{y-1}=0$$
And I have to see if $y$ can be described as a function of $x$ around (1,1). The implicit function theorem can't be applied in this case. What should I do?
 A: Let's write your equation in the following form:
$$
\frac{y}{e^{y-1}}=\frac{x}{{(2-x)^3}}.
$$
Without lose of generality we can assume that $y\ge0$. So $\frac{y}{e^{y-1}}\ge0$. By differencing it's easy to see that $1$ is a maximum point of $\frac{y}{e^{y-1}}$. So it must be $0\le\frac{x}{{(2-x)^3}}\le1$. But for every $0\lt a \lt 1$ we have two $y$ such that $\frac{y}{e^{y-1}}=a$, because $1$ is also local maximum point of $\frac{y}{e^{y-1}}$. Therefore $y$ can't be described as a function of $x$ around $(1,1)$.
A: You can just find two polynomial approximation of $e^z$ in a neighbourhood of $z=0$, the first being a lower bound and the latter being an upper bound, in order to have $g_1(x)\leq y\leq g_2(x)$ in a left neighbourhood of $x=1$, with $g_1$ and $g_2$ being algebraic functions.
Anyway, you can express $x$ as a function of $y$ by solving a third-degree polynomial. By doing so, it is easy so check that $x=h(y)$ has a stationary point in $y=1$, hence we cannot express $y$ as a function of $x$ in a neighbourhood of $x=1$.
