Expressing variables in terms of each other re: implicit function theorem

Question

The following was given as a question on a practice midterm for my introductory vector analysis class: True or False? If $f: \mathbb{R^3} \rightarrow \mathbb{R}$ is any function that is continuously differentiable and $D_2f(a, b, c) \neq 0$ and $D_3f(a, b, c) \neq 0$, then there is a function $h$ of $(y, z)$ defined near $(b, c)$ such that $f(h(y, z), y, z)) = 0$. I'm thinking since $D_2f(a, b, c) \neq 0$ and $D_3f(a, b, c) \neq 0$, it is possible to express $a$ in terms of $b$ and $c$ (i.e. the function $h(y, z)$ defined near $(b, c)$ exists), but I don't know how to formalize that, nor do I know how to show that $f(h(y, z), y, z)) = 0$ using the text of the IFT itself.

Answer 1

Upon review, it turns out that this is in fact false. The function $f(a, b, c) = b + c$ is clearly $C^1$ with $D_2f(a, b, c) = 1 \neq 0$ and $D_3f(a, b, c) = 1 \neq 0$, but $D_1f(a, b, c) = 0$, so the Implicit Function Theorem does not guarantee an implicit function defining $a$ in terms of $b, c$ because $0$ is not invertible. Furthermore, $b + c$ does not vanish identically in any neighborhood of $\vec{0}$, so the implicit function actually does not exist.