let $ F: {\mathbb{R}}^{3} \rightarrow {\mathbb{R}}, (x_0, y_0, z_0)^{T}=(2,2,2)^{T}$ and $F(x,y,z)=x^3-2xyz+3xz^2+3z^3$

I first had to find $(x_0, y_0, z_0)^{T}$ such that $F(x_0, y_0, z_0)=40$, I came to the solution of $x_0=y_0=z_0=2$


Prove that in a neighborhood T of $(x_0, y_0)^{T}$ there exists an implicit function $h(x,y)$, such that $F(x,y,h(x,y))=40$ $\forall (x,y) \in T$

I calculated all the partial derivatives of $F$ and showed that they are continuous in $(x_0, y_0)^{T}$, but I do not know how to continue from here? is $h(x,y) =2$?

  • 1
    $\begingroup$ You don't need to find $h$, just show that it exists, so you should apply the Implicit Function Theorem. All you have to show is that its conditions are satisfied here, i.e. that $g$ is continuously differentiable and $\frac{\partial g}{\partial z}\neq 0$ at $(x_0,y_0,z_0)$, where $g(x,y,z) = F(x,y,z) - 40$ (so that $F(x_0,y_0,z_0)=40\implies g(x_0,y_0,z_0) = 0$, as required by most statements of the IFT). $\endgroup$ May 31, 2021 at 15:23
  • $\begingroup$ thank you, so that $\partial g / \partial z (2,2,2) \neq 0$ suffices? $\endgroup$
    – Inocenciaa
    May 31, 2021 at 18:58
  • 1
    $\begingroup$ Yes, that's enough here. The implicit function theorem takes care of the rest :) $\endgroup$ May 31, 2021 at 21:36

1 Answer 1


To apply the implicit function theorem here, the last thing you should verify is that the partial derivative of $F$ with respect to $z$ at the particular solution you found (i.e. at the point $(2,2,2)$) is not zero. If that is verified, then the implicit function theorem assures you that such a function $h$ exists, but in general there isn't a way to give the function $h$ explicitly.


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