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

**cuteangel** · March 24th 2011, 08:59 AM

I have to show that the equation $z^3 + ze^{(x+y)} + 2 = 0$ has a unique solution $z = f(x,y)$ for all (x,y) in $R^2$ . 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?

**FernandoRevilla** · March 24th 2011, 09:41 AM

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

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

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

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

**cuteangel** · March 24th 2011, 09:56 AM

Thanks so much!
But can you tell me why
$(ii)\quad \dfrac{\partial F}{\partial z}(x_0,y_0,z_0)\neq 0$
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?

**FernandoRevilla** · March 24th 2011, 10:14 AM

Originally Posted by cuteangel

But can you tell me why
$(ii)\quad \dfrac{\partial F}{\partial z}(x_0,y_0,z_0)\neq 0$
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.

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