# Implicit Function Thm

How do I apply the implicit function theorm to

$$f(x,y,z)= \begin{bmatrix}(x-1)^2+y^2+z^2\\ (x+1)^2+y^2+z^2\end{bmatrix}$$

at the point $$(0,0,1)$$??? I'm not sure if I'm doing it correctly. I've tried to do $$\begin{bmatrix}(f_1)_y &(f_1)_z\\ (f_2)_y& (f_2)_z\end{bmatrix}$$ and got $$\begin{bmatrix}2y &2z\\2y &2z\end{bmatrix}$$ but that doesn't seem right and even if it is right, I don't know what to do next.

• Hi and welcome to MSE! You can find math formatting requirements here Commented Apr 6, 2021 at 21:19
• So what do you know about the Implicit function theorem? What is it for, and what exactly does it say? Commented Apr 6, 2021 at 21:21
• just for my own knowledge @RobertTheTutor is this where $z$ is made to be the chart? we take $z=1$? Commented Apr 6, 2021 at 21:22

Given $$f(x,y,z) = \begin{bmatrix}{(x-1)}^2 + y^2 + z^2 \\ {(x+1)}^2 + y^2 + z^2 \end{bmatrix}$$, we need to expand the definition to get a square matrix, thus:
$$f(x,y,z) = \begin{bmatrix}{(x-1)}^2 + y^2 + z^2 \\ {(x+1)}^2 + y^2 + z^2 \\ z \end{bmatrix}$$ and $$Df = \begin{bmatrix} 2x-2 & 2y & 2z \\ 2x+2 & 2y & 2z \\ 0 & 0 & 1\end{bmatrix}$$ and $$Df(0,0,1) = \begin{bmatrix} -2 & 0 & 2 \\ 2 & 0 & 2 \\ 0 & 0 & 1 \end{bmatrix}$$
The implicit function theorem in regular calculus says that $${(f^{-1})}^{'}(b) = \frac{1}{f^{'}(a)}$$ where $$f(a) = b$$. I was going to extend this definition to a $$3x3$$ matrix, but unfortunately that matrix is not invertible!