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

Question

$f: \mathbb{R^2} \rightarrow \mathbb{R}$, $f(x,y) = x^2+y^2-1$

$X:= f^{-1} (\{0\})=\{(x,y) \in \mathbb{R^2} | f(x,y)=0\}$

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

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1. $[df(x,y)]=(\frac{\delta f}{\delta x}, \frac{\delta f}{\delta y})=(2x, 2y)$

2. $0=x^2+y^2-1$ $\Rightarrow$ $y=\sqrt{-x^2+1}$, it follows that $x \in [-1,1]$ and $y \in [0,1]$.

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

Answer 1

1. $f$ is continuously differentiable means that partial derivatives of $f$ are continuous. Clearly $\frac{\partial f}{\partial x}=2x$ and $\frac{\partial f}{\partial y}=2y$ are continuous, so $f$ is continuously differentiable.

2. The Implicit Function Theorem says that if $f(x,y)=0$ and $\frac{\partial f(x,y)}{\partial y}\neq0$, then we can express $y$ as a function of $x$ in a some neighborhood of the point $(x,y)$. Since $\frac{\partial f}{\partial y}=2y\neq0$ for $y\neq0$ we have that we can express y as a function of $x$ for any $y\neq0$. Note that your guess with $y\in[0,1]$ is wrong since for $y=0$ we have $f(x,0)=x^2-1=0$ for $x=\pm1$.

3. It's easy to see that the equation $f(x,y)=x^2+y^2-1=0$ defines a circle of unit radius centered at the origin and from (2) such function $g(x)$ exists everywhere except such point $(x,y)$ that $\frac{\partial f(x,y)}{\partial y}\neq0$, i.e. except $(-1,0)$ and $(1,0)$, then trivially $(a,b)=(0,1)$ and $g(x)=\sqrt{1-x^2}$.