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Proving a function is not bijective

  1. Nov 19, 2013 #1
    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. jcsd
  3. Nov 19, 2013 #2

    Dick

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    Homework Helper

    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
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