# Implicit function theorem for $f:\mathbb{R}^{3} \rightarrow \mathbb{R}$

I am stuck at this problem and I would be glad if somebody helped me out:

By the implicit function theorem, it should be shown that: $$f(x,y,z) := z^{3}+2xy-4xz+2y-1 .$$

The zero level set $f^{-1}(0)$ in a neighborhood $U$ of $(x_0, y_0) = (1,1)$ can be rewritten through a differentiable function $z=g(x,y)$ where $g(1,1)=1$. Then also the partial derivatives: $g_x(1,1), g_y(1,1)$ should be calculated.

The partial derivative in respect to $z$ is: $f_z = 3z^2- 4x$ and it holds that $3z^2 - 4x \ne 0$, because otherwise the invertibility would not be given anymore. So if $(x_0, y_0)=(1,1)$ then $z^3- 4z+4=0$ gives $3$ solutions, of which none are such that $3z^2-4 = 0$. So the zero level set in $U$ can be rewritten through $g(x,y)$.

Now the remaining task is to solve $z^3 +2xy-4xz+2y-1=0$ for $z$. I sit around at this end:

$$z^3- 4xz=-2xy-2y+1 = z(z^2- 4x)=-2xy-2y+1 \Rightarrow z = \frac{-2xy-2y+1}{z^2-4x} .$$

According to Wolfram Alpha, there are 3 exact solutions.

• I think you need to check that $f_z$ is non-zero at $(1,1,1)$. Isn't that point given? – Dylan Moreland Dec 6 '11 at 0:37
• $z^{3}+2xy-4xz+2y-1=0$ has three solutions for $z$ for fixed $x$ and $y$ because it is a cubic equation. For $x=y=1$ the solutions are $z=1$ and $\frac{-1 \pm \sqrt{13}}{2}$. But you seem to be missing part of the question. – Henry Dec 6 '11 at 0:39
• @Dylan Moreland Thanks! $f_{z}$ is -1 at (1,1,1). But why (1,1,1)? I don't see where you want me to go with this, either... – Tashi Dec 6 '11 at 0:42
• @Tashi $g$ is spitting out $z$-values, and you're told that it's spitting out $1$ at $(1,1)$. Maybe I'm misinterpreting the question? – Dylan Moreland Dec 6 '11 at 0:45
• Thanks @Dylan Moreland: I understood what you mean with $f_{z}$ non zero at (1,1). So there is no need to find $g(x,y)$? How to calculate $g_{x}(1,1)$ and $g_{y}(1,1)$ without knowing $g(x,y)$ ? – Tashi Dec 6 '11 at 1:09

By implicit function theorem, if we can show that $f_z(1,1)\neq 0$, then "the zero level set $f^{−1}(0)$ in a neighborhood $U$ of $(x_0,y_0)=(1,1)$ can be rewritten through a differentiable function $z=g(x,y)$", as you would like to show. To see this, we have $$f_z(1,1)=3z^{2}-4x\Big|_{(x,y,z)=(1,1,1)}=3-4=-1\neq 0.$$ Now, sinc $f^{−1}(0)$ can be written as $z=g(x,y)$ in a neighborhood $U$ of $(x_0,y_0)=(1,1)$, we have $f(x,y,g(x,y))=0$, or equivalently, $$g(x,y)^{3}+2xy-4xg(x,y)+2y-1=0.$$ Now, take the partial derivative of the above equation with respect to $x$, we obtain $$3g^2(x,y)g_x(x,y)+2y-4g(x,y)-4xg_x(x,y)=0.$$ Evaluate it at $(x,y)=1$ and by $g(1,1)=1$, we have $$3g_x(1,1)+2-4-4g_x(1,1)=0,$$ which implies that $g_x(1,1)=-2$. Similarly, take the partial derivative of the above equation with respect to $y$, evaluate it at $(x,y)=1$, we obtain $$3g^2(1,1)g_y(1,1)+2-4g_y(1,1)+2=0.$$ Since $g(1,1)=1$, we get $g_y(1,1)=4$