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Thread: 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 $\displaystyle z^3 + ze^{(x+y)} + 2 = 0$ 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?
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    MHF Contributor FernandoRevilla's Avatar
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    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 $\displaystyle (x,y)\in V$ .
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    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?
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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
    $\displaystyle (ii)\quad \dfrac{\partial F}{\partial z}(x_0,y_0,z_0)\neq 0$
    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.
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