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Functions, Mappings and Intervals.

  1. Oct 27, 2007 #1
    1. The problem statement, all variables and given/known data
    This is a problem I am just trying to do myself to work out some other problem.

    I am trying to prove: f:[a, b] → [a, b]
    Given: f is continuous on [a, b], for all x in [a, b] then df/dx < 1 , f(a) ≥ a , f(b) ≤ b.

    2. The attempt at a solution

    First I proved that f(b) - f(a) ≤ b - a. It is simple, but I can give the proof if you wish.
    Now I need to prove that for all x in [a, b] , f(a) ≤ f(x) ≤ f(b), but I have no clue how to do this. I thought of using the Mean Value Theorem somehow, but I don't quite know how. I also thought of showing that f(x) ≤ f(x + dx) for x in [a, b], but I don't know how to do that either. Any help ?
  2. jcsd
  3. Oct 27, 2007 #2


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    [itex]f(a)\ge a[/itex] and [itex]f(b)\le b[/itex] are easy. You are told that f(a) and f(b) are in [a, b]!

    Are you assuming that f is onto [a, b]? Otherwise, neither [itex]f(a)\le f(x)\le f(b)[/itex] not df/dx< 1 is true for all x in [a, b]. Take, for example, a= -1, b= 1, f(x)=x. f(a)= f(b)= 1, but f(x) is less than 1 for all other x. Also df/dx> 1 for x> 1/2.
  4. Oct 27, 2007 #3
    I do not need to prove that, that is one of the assumptions.

    I want to prove that f maps from the interval [a,b] onto [a,b].

    The assumptions are:
    1) f is continuous on [a, b]
    2) for all x in [a, b] then df/dx < 1 ,
    3) f(a) ≥ a , f(b) ≤ b.

    What I have shown:
    f(b) - f(a) ≤ b - a

    Now to prove that f maps from [a,b] to [a,b], then I also need to show that for all x in [a,b] it is true that f(a) ≤ f(x) ≤ f(b), that is, no point in the interval [a,b] can be mapped to a point outside the interval [a,b]. How do I do this?
  5. Oct 29, 2007 #4
    Anyone :(
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