# show existence of implicit function

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 $$x_0=y_0=z_0=2$$

now

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 $$h(x,y) =2$$?

• 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). May 31, 2021 at 15:23
• thank you, so that $\partial g / \partial z (2,2,2) \neq 0$ suffices? May 31, 2021 at 18:58
• Yes, that's enough here. The implicit function theorem takes care of the rest :) May 31, 2021 at 21:36

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.