4
$\begingroup$

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.

  • 0
    I think you need to check that $f_z$ is non-zero at $(1,1,1)$. Isn't that point given?2011-12-06
  • 0
    $z^{3}+2xy-4xz+2y-1=0$ 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.2011-12-06
  • 0
    @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...2011-12-06
  • 0
    @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?2011-12-06
  • 0
    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 $g(x,y)$ ?2011-12-06
  • 0
    @Tashi I think the idea is to use implicit differentiation and the chain rule for multivariable functions.2011-12-06

1 Answers 1