# Thread: Implicit Functions - Solving for G(x,y,z) in terms of z=f(x,y)

1. ## Implicit Functions - Solving for G(x,y,z) in terms of z=f(x,y)

I have to show that the equation $\displaystyle z^3 + ze^{(x+y)} + 2 = 0$ has a unique solution $\displaystyle z = f(x,y)$ for all (x,y) in $\displaystyle R^2$. Conclude from the Implicit Function Theorem that f is continuous everywhere.

If I solve for z=f(x,y) in a neighbourhood of (0,0,-2) is this the same thing as showing that z=f(x,y) for all (x,y)?

Or is there another way to do it without using successive approximations?

2. Denote $\displaystyle F(x,y,z)=z^3+ze^{x+y}+2$ and suppose $\displaystyle F(x_0,y_0,z_0)=0$ . Then,

$\displaystyle (i)\quad F\in \mathcal{C}^{1}(\mathbb{R}^3)$

$\displaystyle (ii)\quad \dfrac{\partial F}{\partial z}(x_0,y_0,z_0)\neq 0$

So, for every $\displaystyle (x_0,y_0,z_0)$ there exists in a neigborhood $\displaystyle V$ of $\displaystyle (x_0,y_0)$ a function $\displaystyle z=f(x,y)$ (of class 1) satisfying $\displaystyle f(x_0,y_0)=z_0$ and $\displaystyle F(x,y,f(x,y))=0$ for all $\displaystyle (x,y)\in V$ .

3. Thanks so much!
But can you tell me why
$\displaystyle (ii)\quad \dfrac{\partial F}{\partial z}(x_0,y_0,z_0)\neq 0$
has to be true in order for the unique function z=f(x,y) to exist? I thought the partial derivative with respect to the second component, in this case y, had to be nonzero?

4. Originally Posted by cuteangel
But can you tell me why
$\displaystyle (ii)\quad \dfrac{\partial F}{\partial z}(x_0,y_0,z_0)\neq 0$
has to be true in order for the unique function z=f(x,y) to exist? I thought the partial derivative with respect to the second component, in this case y, had to be nonzero?

It is a hyphotesis of the Implicit Function Theorem.