# Implicit Function Theorem -Confusion-

I am trying to figure out what the implicit function theorem is. Can anyone explain it to me? I was reading the wiki: http://en.wikipedia.org/wiki/Implicit_function_theorem but still don't understand it too well.

Why does the function $f: R^{n+m} \rightarrow R^m$ have to be continuously differentiable?

• – Git Gud Jan 2 '14 at 0:18
• You can find many documents about IFT, for instance you can look the notes rutherglen.science.mq.edu.au/wchen/lnmvafolder/mva03.pdf – daulomb Jan 2 '14 at 0:20
• It doesn't (as @Git Gud said). It depends on what properties you want for the implicit function. Read the Generalizations section in en.wikipedia.org/wiki/Implicit_function_theorem – ir7 Jan 2 '14 at 0:29
• @user40615 THanks, that article is good. – asd Jan 2 '14 at 0:44
• You're welcome. Good luck. – daulomb Jan 2 '14 at 0:53

It might be easiest to look at an example from $R^1$ to $R^1$ that's differentiable but not continuously differentiable. My favorite is $$f(x) = \begin{cases} x^2 \sin (\frac{1}{x}) & x \ne 0 \\ 0 & x = 0 \end{cases}.$$
This function has a discontinuous derivative, but the derivative is zero at many many points arbitrarily near the origin, so the preimage of the origin ends up not being an embedded submanifold (of dimension 0) in $R^1$.
• But $f'(0)=0$ violates the maximal rank condition. – Ted Shifrin Jan 2 '14 at 1:02
• Right. Good point, Ted. Add $x$ to the function, i.e., $x + x^2 \sin(\frac{1}{x})$ has rank 1 at the origin, is differentiable, but not continuously differentiable. – John Hughes Jan 2 '14 at 1:18
• But now $f^{-1}(0)$ is just $0$, certainly locally. – Ted Shifrin Jan 2 '14 at 1:23