Question

Suppose that the functions \(f: \mathbb{R}^{3} \rightarrow \mathbb{R}, g: \mathbb{R}^{3} \rightarrow \mathbb{R}\), and \(h: \mathbb{R}^{3} \rightarrow \mathbb{R}\) are continuously differentiable and let \(\left(x_{0}, y_{0}, z_{0}\right)\) be a point in \(\mathbb{R}^{3}\) at which

$$ f\left(x_{0}, y_{0}, z_{0}\right)=g\left(x_{0}, y_{0}, z_{0}\right)=h\left(x_{0}, y_{0}, z_{0}\right)=0 $$

and

$$ \left\langle\nabla f\left(x_{0}, y_{0}, z_{0}\right), \nabla g\left(x_{0}, y_{0}, z_{0}\right) \times \nabla h\left(x_{0}, y_{0}, z_{0}\right)\right\rangle \neq 0 $$

By considering the set of solutions of this system as consisting of the intersection of a surface with a path, explain why that in a neighborhood of the point \(\left(x_{0}, y_{0}, z_{0}\right)\)

the system of equations

$$ \begin{aligned} &f(x, y, z)=0 \\ &g(x, y, z)=0 \\ &h(x, y, z)=0, \quad(x, y, z) \text { in } \mathbb{R}^{3} \end{aligned} $$

has exactly one solution. Also explain this by using the Inverse Function Theorem.

Answer 1