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Let $f:\mathbb{R^3} \to \mathbb{R}.$

Let $ (x_0,y_0,z_0)$ be a solution of the equation $f(x-y,y-z,z-x)=0$.

Find sufficient conditions such that extracting $z$ with respect to $x,y$ is possible.

I study about implicit function theorem and I find out this theorem but the equation $f(x-y,y-z,z-x)=0$ gets me a little bit confused.

My attempt:

For applying implicit function theorem in $(a,b,c)$ I have to demand:

1.$f(a,b,c)=0$.

2.There is a neigborhood $U$ of $(a,b,c)$ such that $f$ is $C^1$ in $U$ and

3.$\frac{\partial f}{\partial z}(a,b,c)\neq 0$

Denote $a:=x_0-y_0,b:=y_0-z_0,c:=z_0-x_0 \implies f(a,b,c)=0$.

$f\in C^1$.

I have an issue finding the derivative of $f$.

Is my solution correct?

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  • $\begingroup$ It is not $\partial f/\partial z$ which must be non-zero at $(a,b,c).$ It is $\partial g/\partial z,$ where $g(x,y,z)=f(x-y,y-z,z-x).$ To compute $\partial g/\partial z,$ use the chain rule: $\partial g/\partial z(x,y,z)=(\partial_3f-\partial_2f)((x-y,y-z,z-x).$ $\endgroup$ Feb 11, 2023 at 15:53
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    $\begingroup$ @AnneBauval Sorry , you are right , fixed $\endgroup$
    – Algo
    Feb 11, 2023 at 15:55

2 Answers 2

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You can define $F(x,y,z) = f(x-y,y-z,z-x)$ and $\phi(x,y,z)= (x-y, y-z,z-x)$. You have $F(x,y,z) = (f \circ \phi)(x,y,z)$. From there, you just have to apply the chain rule:

$$\frac{\partial F}{\partial z}(x,y,z) = -\frac{\partial f}{\partial y}(x-y,y-z,z-x) + \frac{\partial f}{\partial z}(x-y,y-z,z-x).$$ Finally, the condition you're looking for is

$$\frac{\partial f}{\partial y}(x_0-y_0,y_0-z_0,z_0-x_0) \neq \frac{\partial f}{\partial z}(x_0-y_,y_0-z_0,z_0-x_0)$$ applying implicit function theorem.

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  • $\begingroup$ thanks , did you take derivative of $y$ since there is $y-z$ is the second coordinate ? I meant , in case $y$ doesn't appear in the second coordinate of $f$, let's say its only $z$ $\endgroup$
    – Algo
    Feb 13, 2023 at 8:16
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It is not $\frac{\partial f}{\partial z}(a,b,c)$ which must be non-zero. It is $\frac{\partial g}{\partial z}(x_0,y_0,z_0),$ where $$g(x,y,z)=f(x-y,y-z,z-x).$$ To compute $\frac{\partial g}{\partial z},$ use the chain rule: $$\frac{\partial g}{\partial z}(x,y,z)=(\partial_3f-\partial_2f)(x-y,y-z,z-x).$$

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