Smooth mapping on $ \mathbb R$ I am going through differentiable manifolds and come across a problem:
How to construct a smooth mapping $f: \mathbb R \rightarrow  \mathbb R$ such that

*

*$f^{-1}(0)=0$


*$f^{'}(0) \neq 0$


*$\forall \epsilon >0,f^{-1}(-\epsilon ,\epsilon )$ is not homeomorphic to $(-\epsilon ,\epsilon )$
Any help?
 A: SUGGESTION (not a full answer): Something that oscillates a lot would be a good choice. You might want to start looking at
$$
f(x) = x^2 \sin \frac1x
$$
(where $f(0)$ is defined to be 0, but I didn't want to write out cases).
Now $f'(0) = 0$, so it doesnt quite work. But if
$$
g(x) = f(x) + x
$$
you might have something to work with...
A: $$f(x)=x\left(\sin^2(x)+\frac1{1+x^2}\right)$$

*

*There is only one value of $x$ such that $f(x)=0$

*$f'(0)=1\ne0$

*$f$ oscillates between values arbitrarily close to $0$ and arbitrarily large values.

The last property guarantees that $f^{-1}((-\varepsilon,\varepsilon))$ is made of a union of infinitely many disjoint open intervals, hence it's not homeomorphic to $(-\varepsilon,\varepsilon)$, for any $\varepsilon>0$.

With the example in my comment above, $f(x)=\dfrac{x}{1+x^2}$, the constraints where only partially fulfilled: for a large enough $\varepsilon>0$, $f^{-1}((-\varepsilon,\varepsilon))=\Bbb R$, because that $f$ is bounded. Here it doesn't happen.
