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Continuity (intermediate value theorem)

  1. Sep 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Let f be a continuous function on the interval I=[a,b] such that for every x in [a,b] f(x)≠0.

    Show that the function f(x) doesn't change its sign.( like increasing or decreasing)



    3. The attempt at a solution

    Well for this to be true, we need to have f(a)>0 and f(b)>0 and f(x) is increasing so then it won't change the monotony. If we have f(a)<0 and f(b)<0, then f(x) is decreasing, hence we will not find any x in the interval I such that f(x)=0. Therefore for every x in the interval I f(x)≠0. Am I correct? Do I need to explain a bit more or what?
     
    Last edited: Sep 28, 2012
  2. jcsd
  3. Sep 28, 2012 #2

    jgens

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    Gold Member

    This is nonsense. If [itex]f[/itex] does change sign, then there exist real numbers [itex]c,d \in [a,b][/itex] such that [itex]f(c) < 0[/itex] and [itex]0 < f(d)[/itex]. Now use the intermediate value theorem to derive a contradiction.
     
  4. Sep 28, 2012 #3

    HallsofIvy

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    Did you add the parenthetical statement? I believe you are misunderstanding the question! A function is "increasing or decreasing" if and only if the derivative of f doesn't change sign. [itex]f(x)= x^2+ 1[/itex], for example is continuous and never 0 on [-1, 1] but it certainly is NOT always increasing or always decreasing.

    Use a proof by contradicction. If f(x1)> 0 and f(x2)< 0 (f changes sign) and f is continuous, what does the intermediate value theorem tell you?



     
  5. Sep 28, 2012 #4
    If f(x1)>0 and f(x2)<0 that entails f(x1)*f(x2)<0 so that means that there at least exists a number c such that f(x2)<f(c)=0<f(x1). Correct right?
     
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