Show (using the implicit function theorem) that the following subset $$M:=\{(x,y,z)\in\mathbb{R}^{3}\;|x^2+y^4+z^4=3\}\subseteq\mathbb{R}^{3}$$
Theorem: Let $A\subset \mathbb{R}^{n}$ be open let $g:A\to \mathbb{R}^{p}$ be a differentiable function such that $g^{\prime}(x)$ has rank $p$ whenever $g(x)=0$. Then $g^{-1}(0)$ is an $n-p$ dimensional manifold in $\mathbb{R}^{n}$
Solution.
Let $F(x,y,z)=x^2+y^4+z^4-3$, denote $M=F^{-1}(0)$. Then for each $P=(x,y,z)\in M$ we have $$F^{\prime}(x,y,z)=\nabla F(x,y,z)=(2x,4y^3,4z^3)\neq 0$$
If $$\nabla F(x,y,z)=(2x,4y^3,4z^3)= 0$$ then $(x,y,z)=0$ and $F(x,y,z)=-3$ which is a contradition to $F(x,y,z)=0$ in other words $F^{\prime}(x,y,z)$ har always rank $1$. By the theorem $M$ is a manifold of $dim=2$
Is this solution correct? I have a question concerning the following part:
''$F(x,y,z)=-3$ which is a contradition to $F(x,y,z)=0$.''
So when $(x,y,z)=0$ then clearly $F(x,y,z)=0+0+0-3=-3$ in other words,: $F(x,y,z)=0$ is required by the theorem? and only then $F^{\prime}(x,y,z)$ has rank $p$?
Thank you
The solution is correct.
The only point that can possibly satisfy $\nabla F(x, y, z) = 0$ is $(0, 0, 0)$, but this point isn't an element of $M$ to begin with. Thus, $M$ is a manifold per the theorem you quote.