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Easy to understand, hard to solve

  1. Jan 8, 2004 #1
    that is my first time, and i have questions for my exam.

    * Show that if a function "f" is continuous and has no zeros on an interval, then either f(x)>0 or f(x)<0 for every "x" in the interval
    (Intermediate Value Theorem)

    I solved it in that way:
    since no zeros in the interval then the interval will be
    so f(x) will not be 0.... but that can be possible, and didn't know how to do it???

    please i need the answer in maximum 2 days :frown:
    any efforts will be appreciated
    thank you
  2. jcsd
  3. Jan 8, 2004 #2


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    I always had trouble with these. I always would just think, "that's true by definition. How can I prove it?"

    First, assume that it is not true. Then, for some value of x on the interval, f(x) and f(x+dx) will have opposite signs, so that:

    f(x)f(x+dx)<0 (positive * negative is always <0)

    using the basic definition of a derivitive, f'(x)=[f(x+dx)-f(x)]/dx you can solve for f(x+dx)=dx*f'(x)+f(x)

    substituting, you get:


    Take the limit as dx->0 and:


    The left side of the equation is a square, it can not be <0

    So assuming the thing you are trying to prove is false yields an impossibility.

    I think this is good.

    PS- I know I used sloppy notation, dx instead of delta x. I forgot how to make deltas.
  4. Jan 8, 2004 #3


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    Homework Helper

    Nice approach, but just because the function is continuous, doesn't mean that it's differentiable.

    Lets say we have f(x) > 0 and f(y) < 0 with x,y on the interval.
    let a = min (x,y) and b=max(x,y)
    Clearly (a,b) is contained in the interval. Therefore f is continuous on [a,b].
    It's also clear that 0 is between f(a) and f(a)

    Then by applying the intermediate value theorem we get hat there must be some point x on the interval [a,b] such that f(x)=0, but this contradicts the hypothesis that [tex]f(x) \neq 0[/tex] on the interval.
  5. Jan 9, 2004 #4


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    How you would do this depends upon what prior theorems you have to work with. Have you had the "intermediate value theorem"? That says that if f(x) is continuous on [a,b] then, on [a,b], f takes on all values between f(a) and f(b).
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