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

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

    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?
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    MHF Contributor FernandoRevilla's Avatar
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    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 .
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    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?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by cuteangel View Post
    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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