Let $$M:=\{(x,y,z)^T\in\mathbb{R}^3:x^2+2yz=3, x^2+y^2+yz=z^2+5\}$$ and $(x_0,y_0,z_0)\in M,\ y_0z_0 \neq 0$
Show that there is an open neighborhood $U \subseteq \mathbb{R}$ around $x_0$ and continuously differentiable functions $g,h:U\rightarrow \mathbb{R}$ with $g(x_0)=y_0,\ \ h(x_0)=z_0$
The only examples I have ever seen of the implicit function theorem are in $\mathbb{R}^2$, and am finding it hard to translate. How does it translate to higher dimensions, i.e. this problem?
Best Answer
Set $F:(x,y,z)\mapsto (u(x,y,z),v(x,y,z))=(x^2+2yz,x^2+y^2+yz-z^2)\ $ so $F(x_0,y_0,z_0)=(3,5).$ To apply the implicit function theorem, we use the Jacobian (in $y$ and $z$) and check that $(x_0,y_0,z_0)$ is a regular point. This follows by hypothesis and the fact that
$\begin{bmatrix} 2z & 2y\\ 2y+z&y-2z \end{bmatrix} =0 \Rightarrow z=y=0.$
Now, the result is an immediate consequence of the implicit function theorem: there is an open set $U\subseteq \mathbb R$ and a function $G:U\to \mathbb R^2:x\mapsto (g(x),h(x))$ such that $F(x,G(x))=(3,5)$ for all $x\in U$. In particular, $g(x_0)=y_0$ and $h(x_0)=z_0.$