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Let $f: \mathbb{R}^3 \to \mathbb{R}^2$ be of class $C^1$; write $f$ in the form $f(x,y_1,y_2)$. Assume that $f(3,-1,2) = \mathbf{0}$ and $$ Df(3,-1,2) = \begin{pmatrix} 1 & 2 & 1 \\ 1 & -1 & 1 \end{pmatrix}.$$

(a) Show that there is a function $g: B \to \mathbb{R}^2$ of class $C^1$ defined on an open set $B$ in $\mathbb{R}$ such that $f(x,g_1(x),g_2(x)) = \mathbf{0}$ for $x \in B$ and $g(3) = (-1,2)$.

(b) Find $Dg(3)$.

I have done a. I have no idea how to solve b. Please help.

user159691
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  • My earlier comment was a little misleading. It is a direct application of the implicit function theorem, but I had accidentally written $x$ instead of $y$. I have atoned for my sin below :-). – copper.hat Jun 29 '14 at 20:33

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Write $f$ (which is $C^1$) as $f:\mathbb{R} \times \mathbb{R}^2 \to \mathbb{R}^2$.

Note that ${ \partial f(3,(-1,2)^T) \over \partial y}$ is invertible, and $f(3,(-1,2)^T)=0$. The implicit function theorem gives the existence of a $C^1$ function $g: \mathbb{R} \to \mathbb{R}^2$ such that $f(x,g(x)) = 0$ for $x$ in a neighbourhood of $3$ and $g(3) = (-1,2)^T$.

The implicit function theorem also gives a formula for the derivative of $g$ as ${ \partial g(3) \over \partial x} = - { \partial f(3,(-1,2)^T) \over \partial y}^{-1} { \partial f(3,(-1,2)^T) \over \partial x} = (0,-1)^T $.

copper.hat
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  • Why are you writing $(-1,2)^T$? I read the question from the munkres text and am trying to understand your notation – john fowles Aug 03 '18 at 10:58
  • I was being pedantic. (i) Instead of $f(x,y_1,y_2)$ I use $f(x,y)$, where $y \in \mathbb{R}^2$. (ii) For elements of $\mathbb{R}^2$ I write a column vector, hence the $^T$. I had a minor error as well, $g$ is a function of $\mathbb{R}$. – copper.hat Aug 03 '18 at 14:57