# Confusion with implicit function theorem

According to my understanding of the implicit function theorem, if $$F(y_0,x_0)=c$$ and $$\frac{dF}{dy}$$ at $$(y_0,x_0)$$ is nonzero then there is a unique function $$y=F(x)$$ for $$x \in B$$ where B is an open ball around $$x_0$$. The part that is confusing to me is the uniqueness. As the common example of a circle shows ($$x^2+y^2=c^2$$), for every open ball around $$x_0$$ there are still two functions $$y=\sqrt{x^2-c^2}$$ and $$y=-\sqrt{x^2-c^2}$$. What am I missing here?

• There is a unique function $F$ in $B$ around $x_0$ such that $y_0 = F(x_0)$. Mar 22 at 12:58

What you're missing is the fact that we first fix some $$(x_0,y_0)$$. In particular, if we choose $$(x_0,y_0)$$ to be a point on the upper half-circle, then you'll get the positive solution $$y=\sqrt{x^2-c^2}$$ and if you choose $$(x_0,y_0)$$ on the lower half-circle you get the negative solution $$y=-\sqrt{x^2-c^2}$$. The key thing here is that both of them cannot cross $$(x_0,y_0)$$ at the same time, and so you do in fact get a unique solution as the theorem tells you.
• @ Lorago, Thanks but why isn't said that there is an open ball around $(x_0,y_0)$? it is only said around $x_0$.
• @ASB I mean your statement of the theorem is somewhat incomplete. It should have that $f$ is the unique function satisfying $F(f(x),x)=c$ for all $x$ in some ball around $x_0$ and which is such that $f(x_0)=y_0$ Mar 22 at 13:15
For all the points $$(x_0, y_0)$$ on the circle and in the strict upper plan, your circle can be approximated by $$y = \sqrt{x^2 - c^2}$$ and for all the points $$(x_0, y_0)$$ on the circle and in the strict lower plan, your circle can be approximated by $$y = -\sqrt{x^2 - c^2}$$.
Your confusion comes from the fact that the development depends on your chosen point. In the first case, all $$y$$ are negative, and are therefore around a negative $$y_0$$. Same for the second one.