Implicit Function Theorem for Solving Nonlinear Equations

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Homework Statement



I am currently working through the problems in Edwards book "Advanced Calculus of Several Variables". This is the problem (1.9 page 171):

Show that the equation z3 + ze(x+y) + 2 = 0 has a unique solution z=f(x, y) defined for all (x,у) an element of R3.

Homework Equations



Implicit Function Theorem as given in Edwards. (attached as an image)

The Attempt at a Solution



According to the hypotheses of the Implicit Function Theorem, it would seem that the only thing that is required to show for this problem is that the partial derivative with respect to z of the function G(x,y,z) = z3 + ze(x+y) + 2 is nonzero (which it is for all points (x,y)). Then the
result follows immediately from the Implicit Function Theorem. But this seems rather trivial?

I think my confusion stems from the misunderstanding of the Implicit Function Theorem. Is there any other books besides Edwards that have a very good discussion of the theorem?
 

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so consider an arbirtray function f(x,y,z) : \mathBB{R}^3 \to \mathBB{R}

considering a level curve of the function f(x,y,z) = c, effectively defines a surface in \mathBB{R}^3.

The question is then whether we can find a unique function z(x,y) to describe the surface.

Consider a tangent plane to f(x,y,z) = c, its normal will be \nabla f(x,y,z) = (\frac{\partial f}{\partial x} ,\frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}.

If \frac{\partial f}{\partial z} = 0 the tangent plane is vertical and in that neighbourhood of x,y there is no unique z.

However is \frac{\partial f}{\partial z} \neq 0 then the tangent plane is well defined and a unique representation of z(x,y) will exist
 
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Thanks lane for the explanation! It seems the problem is as simple as I thought it would be.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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