# Thread: Problem involving implicit function theorem.

1. ## Problem involving implicit function theorem.

Suppose $F(x,y)$ is a $C^{1}$ function such that $F(0,0) = 0$. What conditions on $F$ will guarantee that the equation $F(F(x,y),y) = 0$ can be solved for $y$ as a $C^{1}$ function of $x$ near $(0,0)$.

So obviously $\partial_{2}F(0,0) \ne 0$ is one condition just by what the hypothesis of the implicit function theorem uses (I think this part is obvious, but I could be wrong and if I am, please explain why this is part of a condition that will guarantee what we want). However, another condition that is needed that will guarantee what we want is $\partial_{1}F(0,0) \ne -1$, and I just don't see how to arrive at that conclusion. Any help would be appreciated.

2. Will it have something to with being able to solve the equation $x = F(x,y)$, and if so, how exactly can we arrive to the conclusion that $\partial_{1}F(0,0) \ne 1$ and $\partial_{2}F(0,0) \ne 0$ will guarantee we get want we desire? If not, still the same question, how do we arrive that those two conditions will guarantee what we want?

3. Sorry for the multiple posts but I really want to know how to solve this since it's practice for my upcoming midterm. Here is what I have so far.

If $\partial_{y}F(F(0,0),0) = \partial_{y}F(0,0) \ne 0$, then we can solve the equation $F(F(x,y), y) = 0$ for $y$ as a $C^{1}$ function of $F(x,y)$ near $(F(0,0),0) = (0,0)$. Now, we have to show that we can find a function $g$ that is $C^{1}$ near $(0,0)$ such that $g(x) = F(x,y)$...

...and that is where I get stuck. I think this is the right approach but if it is, how do I proceed, and if not, what do I do?