Let $J\subseteq\mathbb{R}$ be an interval and $f: J\to\mathbb{R}$, $f$ differentiable in $x_0\in J$ with $f'(x_0)\neq 0$. Does there exist a neighborhood of $f(x_0)$ such that $f$ has a continuous and in $f(x_0)$ differentiable inverse function?
I tried to apply the inverse function theorem, but this has continuously differentiable as an assumption.
Is there a work around?
Set $$ f(x)=x+2x^{2}\sin(1/x) ~ (x \not=0), ~~ f(0)=0. $$ Then $f'(0)=1$ and $$ f'(x)=1+4x\sin(1/x) - 2\cos(1/x) ~~ (x \not=0). $$ Thus $(-1,3)\subseteq f'((-\varepsilon,\varepsilon))$ for each $\varepsilon > 0$, and so there is no local inverse near $0$.
Edit: Assume for some $\varepsilon >0$ that $f$ restricted to $I:=(-\varepsilon,\varepsilon)$ has an inverse. Then $f$ has to be injective (thus strictly monotone) on $I$. But $f'(I)$ contains positive and negative values.