# applying the implicit function theorem and finding a unique solution

I think I roughly understand what the implicit function theorem states, but I'm having trouble so far actually applying it in the exercises. I believe I understand how it's applied at the most common example $x^2 + y^2 = 1$, but I'm struggling with more complex applications.

Take for example the function

$$f(x, y) = x^2 + x^3 + y^3 + e^y - 1$$

with $x, y \in \mathbb{R}$. I now want to show that for each $x \in \mathbb{R}$ there's one but only one $y \in \mathbb{R}$ with $f(x, y) = 0$. Using this I want to find a solution $g(x) = y$ that's defined for all $x \in \mathbb{R}$.

Now as far as I understand, I first have to find any $(x, y) \in \mathbb{R}^2$ that satisfy $f(x, y) = 0$. So let's take $x = y = 0$. Next, I need to determine the derivative with respect to y: $f_y(x, y) = 3y^2 + e^y$. Which gives us $f_y(0, 0) = 1$, which is invertible, therefore, the implicit function theorem tells us that there is an open intervall $0 \in I \subseteq \mathbb{R}$, so that there exists a continuously differentiable function $g: I \to \mathbb{R}$ with $f(x, y) = 0 <=> g(x) = y$, in particular with $g(0) = 0$.

So... this is where I don't know to continue. The implicit function theorem only gives me one solution for an open intervall in $\mathbb{R}$. But I want to show that there's a unique solution for the entirety of $\mathbb{R}$, and want to find that solution. How would I do that? I suppose that by rearranging $f(x, y) = 0$ to the form $y = ...$ would be sufficient to prove the existence of a unique solution for $x \in \mathbb{R}$, and would also give me the desired solution $y = g(x)$. But so far, I've had trouble rearranging $f(x, y) = 0$ into that form.

Any help would be appreciated.

In this case we don't need the implicit function theorem. Define $g:\mathbb {R}\to \mathbb {R}$ by $g(y)= y^3 +e^y.$ Verify that $g$ is a bijection. Setting $f(x,y)=0$ amounts to the equation $g(y) = 1-x^2-x^3.$ It follows that for any $x\in \mathbb {R},$ there is one and only one $y\in \mathbb {R}$ that solves the equation, namely $g^{-1}(1-x^2-x^3).$
• Thanks, I didn't expect it to be that easy. I wonder though; what's the general idea for finding a function $g$ with $g(x) = y$? Can we always basically put all the summands that contain a $y$ at the LHS of the equation and the ones that contain an $x$ at the RHS, and define $g$ as the inverse function of the LHS to eventually get a $g$ that's defined on at least an open intervall $I \subseteq \mathbb{R}$? – moran Sep 6 '15 at 18:38