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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?

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  • $\begingroup$ depends on what you mean by "around" (1,1). If you only need a first order approximation, use implicit differentiation to find the slope at the point (1, 1). Then solve a linear approximation, $y = mx + b$, where $m$ is the slope, and the line must pass through (1,1). $\endgroup$ – cnick May 25 '14 at 21:33
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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 $(1,1)$.

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    $\begingroup$ It should be $$\frac y{e^{y-1}}\ =\ \frac x{(2-x)^3}\,.$$ $\endgroup$ – Berci May 25 '14 at 22:46
  • $\begingroup$ I'm sorry. But does it change things? $\endgroup$ – pointer May 26 '14 at 3:42
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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 $x=1$.

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