Implicit function theorem for f(x,y) = x^2+y^2-1

[Chris222]

[MATH]f: \mathbb{R^2} \rightarrow \mathbb{R}, f(x,y) = x^2+y^2-1[/MATH]
[MATH]X:= f^{-1} (\{0\})=\{(x,y) \in \mathbb{R^2} | f(x,y)=0\}[/MATH]
1. Show that [MATH]f[/MATH] is continuous differentiable.
2. For which [MATH](x,y) \in \mathbb{R^2}[/MATH] is the implicit function theorem usable to express [MATH]y[/MATH] under the condition [MATH]f(x,y)=0[/MATH] as a function of [MATH]x[/MATH]?
3. Let [MATH](a,b) \in X[/MATH] with [MATH]b>0[/MATH]. Find the largest possible neighbourhood [MATH]V[/MATH] of [MATH]a[/MATH] in [MATH]\mathbb{R}[/MATH] and a continuous differentiable function [MATH]g:V \rightarrow \mathbb{R}[/MATH] such that [MATH]f(x,g(x))=0[/MATH] and [MATH]g(a)=b[/MATH].


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1. [MATH][df(x,y)]=(\frac{\delta f}{\delta x}, \frac{\delta f}{\delta y})=(2x, 2y)[/MATH]2. [MATH]0=x^2+y^2-1 \Rightarrow$[/MATH] [MATH]y=\sqrt{-x^2+1}[/MATH], it follows that [MATH]x \in [-1,1][/MATH] and [MATH]y \in [0,1][/MATH].

Is 1 and 2 correct? How do I do 3?

 

[Singleton]

1. your notation is funny. Anyway, go back to the definition. Say more in words why it's continuously differentiable.
2. Look at the implicit function theorem. There is a condition that needs to be satisfied (a derivative being not zero, which?) That's what part 2 is asking you to check.
3. You already found g(x) in part 2, sort of, when you take square root it's + or - . You choose the right one corresponding to (a,b). The info you need is given. And the largest interval ... you figure out.