Let $f:\mathbb R \rightarrow \mathbb R$ be $C^1$ and $u=f(x) $, $v= -y+xf(x)$
If $f'(x_0) \neq 0$, show that this transformation is invertible near $x_0,y_0)$ and the inverse has the form $x= f^{-1}(u)$ and $y= -v+uf^{-1}(u)$
Please, help

Question

Let $f:\mathbb R \rightarrow \mathbb R$ be $C^1$ and $u=f(x) $,   $v= -y+xf(x)$
If $f'(x_0) \neq 0$, show that this transformation is invertible near $x_0,y_0)$ and the inverse has the form $x= f^{-1}(u)$ and $y= -v+uf^{-1}(u)$
Please, help

lilshane · Answer

what you think it is

jordan123321 · Answer

hi @lilshane

loser66 · Answer

It is a "show" problem, It doesn't have a result

loser66 · Answer

My attempt: The matrix of differential is 
$$Df(u,v)_{(x,y)} =\left[\begin{matrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\dfrac{\partial u}{\partial y}&\dfrac{\partial v}{\partial y}\end{matrix}\right]=\left[\begin{matrix}f'(x)&0\f(x)+xf'(x) &-1\end{matrix}\right]$$

loser66 · Answer

det (Df) = $-f'(x) $ , and given condition gives us $f'(x_0)\neq 0$, hence the function is invertible.

loser66 · Answer

and the inverse matrix of D is 
$$\left[\begin{matrix}-1&0\-f(x) -xf'(x) &f'(x)\end{matrix}\right]$$

loser66 · Answer

My argument goes wrong from now :(

loser66 · Answer

That is $\dfrac{\partial x}{\partial u}=-1$, hence x = -u +C
but from the original one, u = f(x) hence $x= f^{-1}(u)$  that is the correct one.
However, I don't know what is wrong with my argument.

loser66 · Answer

@bibby

loser66 · Answer

@Kainui

loser66 · Answer

@rational

anonymous · Answer

I guess you are answering perfectly your quest :)