Implicitly Defined Real-Valued Function

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Theorem

Let $F: \left({\mathbf X' \subseteq \R^{n+1}}\right) \to \left({\mathbb I\,' \subseteq \R}\right)$ have continuous partial derivatives.



Let $\left({\mathbf x, z}\right)$ denote an element of $\R^{n+1}$, where $\mathbf x \in \R^n$ and $z \in \R$.

Suppose $\exists (\mathbf x_0,z_0) \in \mathbf X'$ such that:

$F\left({\mathbf x_0, z_0}\right) = 0$
$\dfrac{\partial}{\partial z}F\left({\mathbf x_0, z_0}\right) \ne 0$


Then there exists a unique mapping of the form:

$g: \mathbf X \to \mathbb I$

where $\mathbf X \subseteq \R^n$ contains $\mathbf x_0$ and $\mathbb I$ is an open real interval containing $z_0$, such that:

$\forall \mathbf x \in \mathbf X, z \in \mathbb I: F\left({\mathbf x, z}\right) = 0 \iff z = g\left({\mathbf x}\right)$

and $g$ itself has continuous partial derivatives.


Proof

This is a special case of the Implicit Function Theorem.

$\blacksquare$


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