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:
- 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.
- That the hypotheses given above do not imply that $f$ is continuous on an open interval containing $a$, but the theorem holds anyway.
- 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.
Best Answer
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 $x\in\mathbb Q$.