The problem: We have a function $f: \mathbb{R}^3 \rightarrow \mathbb{R}^2$ such that $f(x,y,z) = (x-xy, x+2y+z^2)$. For a point $(a,b,c)$ such that $f(a,b,c) = 0$, find a condition on $(a,b,c)$ such that there exists an open interval V containing $a$ and an open set W containing $(b,c)$ such that $g: V \rightarrow W$ exists and is differentiable, and $f(x,g(x)) = 0$ for all $x \in V$.

My work:

We need to use the implicit function theorem, so create the matrix of partial derivatives. The required condition is given when the determinant is non-zero. Work out the entries individually:

For notational convenience, let $f^1 = x-xy, f^2 = x+2y+z^2$

$$a_{11} = \frac {\partial f^1}{\partial y}(a) = -x|_a = -a\\ a_{12} = \frac {\partial f^1}{\partial z}(a) = 0\\ a_{21} = \frac {\partial f^2}{\partial y}(a) = 2\\ a_{11} = \frac {\partial f^2}{\partial z}(a) = 2z|_a = ?$$

My question is how to evaluate this last equation. Do I just substitute $z=a$? It seems like I shouldn't. I did that for the first equation since $a$ is the first entry of the ordered triple $(a,b,c)$ in $\mathbb{R}^3$, but the value corresponding to $z$ would be $c$, so do I sub that in instead?

  • $\begingroup$ Do you still need help with this? $\endgroup$ – Git Gud Apr 21 '14 at 0:02

In this answer I stated the implicit function theorem. I'll be using the notation I used in that answer which is pretty much the same as your notation.

Your first and only mistake was looking at $f_1$ and $f_2$ as one variable functions. This is wrong. One has $f_1\colon \mathbb R^3\to \mathbb R, (x,y,z)\mapsto x-xy$ and $f_2\colon \mathbb R^3\to \mathbb R, (x,y,z)\mapsto x+2y+z^2$.

Thus $$\begin{cases} \dfrac{\partial f_1}{\partial y}(a,b,c)=-a ,& \dfrac{\partial f_1}{\partial z}(a,b,c)=0\\ \dfrac{\partial f_2}{\partial y}(a,b,c)=2, & \dfrac{\partial f_2}{\partial z}(a,b,c)=2c \end{cases}$$

And so you need to require, among other things, that $ac\neq 0$. Proceed.

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