Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Assume that $f:I: \rightarrow J$ is a bijection of class $C^1$ and $f'(x)>0$ for all $x \in I$, where $I$, $J$ are intervals in $\mathbb R$. Then by the known theorem $f^{-1}$ is of class $C^1$ and $(f^{-1})'(y)=\frac{1}{f'(f^{-1}(y))}$ for $y\in J$.

Assume now that $f$ is of class $C^n$, $n \in \mathbb N$. Why $f^{-1}$ is also of class $C^n$?

share|cite|improve this question
It seems to me that proof is by induction. If $f^{-1}$ is of class $C^n$ for some $n$. Then by formula on $(f^{-1})'$ we have that $f'\circ f^{-1}$ is of class $C^n$ whence $f^{-1}$ is of class $C^{n+1}$. – L.T Oct 16 '12 at 16:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.