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Intermediate value theorem

  1. Nov 18, 2013 #1
    1. The problem statement, all variables and given/known data

    Let ##a,b\in \mathbb{R}##, ##a<b## and let ##f## be a continuous real valued function on ##[a,b]##. Prove that if ##f## is one-one then ##f([a,b])## is either ##[f(a),f(b)]## or ##[f(b),f(a)].##

    2. The attempt at a solution

    Suppose ##f(a) < f(b)## then by IVT we have if ##x\in(a,b)## then ##f(x) \in [f(a),f(b)]##. Arguing by contradiction suppose ##f(x) < f(a)## or ##f(x)>f(b)##. If ##f(x) <f(a)## then by IVT ##f([x,b])## contains all points between ##f(x)## and ##f(b)## and contains ##f(a)## contradiction that ##f## is one to one.

    Why does that contradict the assumption that ##f## is one to one?
     
  2. jcsd
  3. Nov 18, 2013 #2

    Dick

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    Draw a picture. Suppose f(a)<=f(b) and x is in (a,b) but f(x)<f(a). Doesn't that mean that there is a y in [x,b] such that f(y)=f(a)?
     
  4. Nov 18, 2013 #3
    Thank you! I see my mistake in my reasoning now.
     
    Last edited: Nov 18, 2013
  5. Nov 18, 2013 #4

    Dick

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    I don't think there was any real mistake. Are you clear how on how f(y)=f(a) would contradict one-to-one?
     
  6. Nov 18, 2013 #5
    Yup, because we already know ##a<x<b## and if ##f(a)\in f([x,b])## we have that ##x\le y\le b## thus ##f(y) = f(a)## which contradicts one to one since ##y\ne a##.
     
    Last edited: Nov 18, 2013
  7. Nov 18, 2013 #6

    Dick

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    Exactly.
     
  8. Nov 18, 2013 #7
    Thanks again!
     
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