Show $f(t)$ is not injective in any neighbourhood of $0$ This comes from Baby Rudin chapter 9's exercises: define 
$$f(t)=t+2t^2\sin\frac1t,\,t\ne 0;\quad f(0)=0$$ show $f$ 
fails to be injective in any neighbourhood of $0$. 
My thought was to find points where $f'(t)=0$ but $f''(t)\ne 0$. Letting $t_n=1/(2n\pi)$ and $t'_n=1/(2n\pi+\pi)$ it's easy to see that $f'(t_n)\to -1$ while $f'(t'_n)\to 3$. So by the continuity of $f'(t),t\ne 0$ there exists   some $t_0$  between $t_n$ and $t'_n$ for each $n$ such that $f'(t_0)=0$. But it's much more troubling to show the second order derivative doesn't vanish. 
Is there any slicker way? 
 A: Hint.
You have $$f^\prime(x)=1 - 2\cos \left( \frac{1}{x}\right) + 4x \sin \left( \frac{1}{x}\right)$$ Hence for $\vert x \vert \le \frac{1}{8}$ we get $$-\frac{1}{2} \le f^\prime(x) -(1 - 2\cos \left( \frac{1}{x}\right)) \le \frac{1}{2}$$ and $$\frac{1}{2}- 2\cos \left( \frac{1}{x}\right) \le f^\prime(x) \le \frac{3}{2} - 2\cos \left( \frac{1}{x}\right)$$
From there you can find intervals as closed as you want from $0$ on which $0 < \frac{1}{2}- 2\cos \left( \frac{1}{x}\right) \le f^\prime(x)$ and other intervals as closed as you want from $0$ on which $f^\prime(x) \le \frac{3}{2} - 2\cos \left( \frac{1}{x}\right) < 0$.
Consequently you get a sequence of points converging to $0$ which are local maxima. Proving that $f$ is not injective in any neighborhood of $0$.
For a similar case with additional details, you can have a look at "A function whose derivative at 0 is one but which is not increasing near 0"
A: A continuous injective function on an interval $I$ must be either strictly increasing or strictly decreasing on $I.$ It follows that if $f$ is in addition differentiable on $I,$ then either $f'\ge 0$ on $I$ or $f'\le 0 $ on $I.$ In your problem you have shown that $f'$ takes on positive and negative values in any neighborhood of $0.$ Therefore $f$ is not injective in any neighborhood of $0.$
