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If f has the IVP, prove kf does too

  1. Sep 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose f has the intermediate value property on an interval I and let k be a constant. Prove that kf has the intermediate value property.


    2. Relevant equations



    3. The attempt at a solution

    Since f has IVP on I, then there is distinct a and b in I and f(c)=v exists and is between f(a) and f(b).

    Let k be a constant real number.

    Then kf(a), kf(b) and kf(c) are all real numbers where kf(c) is between kf(a) and kf(b) by properties of real numbers.

    So then kf has IVP on I?
     
  2. jcsd
  3. Sep 19, 2012 #2

    HallsofIvy

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    Staff Emeritus
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    No, that is not the intermediate vaue theorem. Rather, for any a and b, if v is any number between f(a) and f(b), then there exists c, between a and b, such that f(c)= b.
    You are "going the wrong way", assuming from the beginning that your number is of the form kf(x). That is, basically, what you want to prove.

    Let a and b be any two numbers and v be a number between kf(a) and kf(b). Then, if k> 0, v/k lies between f(a) and f(b) so, because f has the intermediate value property, there exist c such that f(c)= v/k. Then kf(c)= k(v/k)= v and we are done.

    You do the cases (a) k< 0 and (b) k= 0.
     
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