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Theorem: Let $I\subset \mathbb R$ be an interval and a function $f:I \to \mathbb R$ be strictly monotone and continuous on $I$.Let $g:=f^{-1}:f(I)\to \mathbb R$.If $f$ is differentiable at $c\in I$ and $f'(c)\neq0$ then $g$ is differentiable at $f(c)$ and $g'(f(c))=\frac{1}{f'(c)}$ .

In this theorem can we replace 'strictly monotone and continuous' by the assumptions that $f$ is 'invertible and continuous' or by 'injective and continuous' or by 'injective and having IVP on $I$'.

I think any of these conditions will act the same i.e. they will all provide me with the facts I need to prove the original version of the theorem. Another question is how to generalize this theorem for arbitrary domain $X$.

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As an example I'll consider an invertible and continuous function and will show that it has to be strictly monotonic.

Assume $x<y$ are points in $I$ and, w.l.o.g, let's assume $f(x) < f(y)$. ($f(x)=f(y)$ would contradict the fact that $f$ is invertible).

Now assume $x < r < y$. Claim: $f(x) < f(r) < f(y)$. Assume that this is not true. The $f(x)\ge f(r) $ of $f(r)\ge f(y)$. Since both cases are similar, I'll only look at the first one. $f(x)=f(r)$ is impossible, so $f(x) > f(r) $. But then be the mean value theorem there is $s \in (r,y)$ sucht that $f(s) = f(x)$, contradicting the invertibility of $f$.

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  • $\begingroup$ So,as I am saying any of these condition is fine,right? $\endgroup$ – Kishalay Sarkar Dec 23 '19 at 9:05
  • $\begingroup$ @KishalaySarkar yep. But you need to work on connected sets, i.e. intervals. $\endgroup$ – Thomas Dec 23 '19 at 9:05
  • $\begingroup$ And the fact $f'(c)\neq 0$ is essential because we want to avoid the situation of infinite derivative of inverse(although it is a bit informal to say,because we do not consider the differential quotient's infinite limit as a derivative).e.g.$ f(x)=x^3$ with $f^{-1}(x)=x^{\frac{1}{3}}$. $\endgroup$ – Kishalay Sarkar Dec 23 '19 at 9:07
  • $\begingroup$ @KishalaySarkar Right. That condition is, of course, still needed. $\endgroup$ – Thomas Dec 23 '19 at 9:09
  • $\begingroup$ How to generalize this theorem for an arbitrary domain $X$ in place of interval domain $I$. $\endgroup$ – Kishalay Sarkar Dec 23 '19 at 9:48

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