# Proving a function is not bijective

1. Nov 19, 2013

### mahler1

The problem statement, all variables and given/known data.
Let $f:\mathbb R \to \mathbb R$ be the function defined as $f(t)=t+2t^2sin(\frac{1}{t})$ if $t≠0$ and $f(0)=0$. Prove that $f$ is not bijective in any neighbourhood of $0$.

The attempt at a solution.

I've tried to prove that is not injective. The only idea I had in mind (later I realized it was completely wrong) was to suppose there is a neighbourhood of $0$ where the function is bijective. Then, there is $\epsilon>0$ such that $B(0,\epsilon) \subset U$. I considered $t_1=(2k\pi)^{-1}$ and $t_2=(-2k\pi)^{-1}$ with $k \in \mathbb N$ such that $k>(2\pi\epsilon)^{-1}$ (I can find such $k$ by the archimidean property of the natural numbers), then $t_1$ and $t_2$ are in $U$.I did all that without considering that $f=t+2t^2sin(\frac{1}{t})$, I thought of the function $f=2t^2sin(\frac{1}{t})$, the term $t$ ruins me everything because it takes into account the sign. Can anyone suggest me a different approach?

Last edited: Nov 19, 2013
2. Nov 19, 2013

### Dick

Your function is differentiable. If a function is bijective and continuous on an interval, then clearly it has be monotone increasing or decreasing. Can you think of a way to use the derivative to show that isn't true? You can use parts of your previous argument here.

Last edited: Nov 19, 2013