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Intermediate Value Theorem : Is this version correct?

  1. Jul 17, 2009 #1
    I recently came across a version of the Intermediate Value Theorem in "Cracking the GRE" by Princeton Review.


    Intermediate Value Theorem :

    f is a real function continuous on a closed interval [a,b].

    Let m be the absolute minimum of f on [a,b] and M be the absolute maximum.

    Then for all Y s.y. m < or = Y < or = M, there exists a c in [a,b] s.t. f(c) = Y.



    The one that I am familiar with says Y lies between f(a) and f(b) instead of the absolute min and absolute max. Is this version also correct?
     
  2. jcsd
  3. Jul 17, 2009 #2

    HallsofIvy

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    Assuming that they are the min and max of f on that interval, yes, it is correct. and it is a bit more general than your version.

    If m is a minimum of f on [a, b] then there exist c in [a, b] such that f(c)= m.
    If M is a minimum of f on [a, b] then there exist d in [a, b] such that f(d)= M.

    Apply your version of the intermediate value theorem to [c, d] or [d, c] (depending on whether c< d or d< c), which are subintervals of [a, b].
     
  4. Jul 17, 2009 #3
    Yes. Perhaps you can prove this version from yours?
     
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