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This question is about the inverse function theorem for real-valued functions.

Suppose $f$ is a one-to-one, that $a$ is in the domain of $f$, and that $f$ is defined on an open interval containing $a$. Suppose further that $f$ is differentiable at $a$, and $f'(a)\neq0$. Does it follow that $f^{-1}$ is differentiable at $f(a)$, and $$ \bigl(f^{-1}\bigr)'\bigl(f(a)\bigr)=\frac{1}{f'(a)} \, ? $$ I ask this question because some presentations of the inverse function theorem (e.g. in Spivak's Calculus) seem to additionally require that $f$ is continuous on an open interval containing $a$. I see three possibilities:

  1. That the hypotheses given above imply that $f$ is continuous on an open interval containing $a$, and so it is redundant to state this as a hypothesis.
  2. That the hypotheses given above do not imply that $f$ is continuous on an open interval containing $a$, but the theorem holds anyway.
  3. That the hypothesis that $f$ is continuous on an open interval containing $a$ is in fact necessary, and so there is a counter-example to the "theorem" stated above.
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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Commented Sep 29, 2022 at 13:02

1 Answer 1

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As it turns out, there is a relatively simple counter-example to the "theorem" given above. Let $f$ be given by $$ f(x)=\begin{cases} x^3+x & \text{if $x$ is rational,} \\ x &\text{if $x$ is irrational.} \end{cases} $$ We have $f'(0)=1$, but $f^{-1}$ is not differentiable at $f(0)=0$ as it is not defined in a neighbourhood of $0$. This is because we can find arbitrarily small rational numbers $y$ which are not of the form $x^3+x$ for some $x\in\mathbb Q$.

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    $\begingroup$ It is an interesting exercise to verify the last sentence of my post. I used the rational root theorem to prove that $x^3+x=1/2^n$ does not have rational solutions for $n\in\mathbb N$. $\endgroup$
    – Joe
    Commented Sep 28, 2022 at 9:36

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