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Math Help - intermediate value theorem

  1. #1
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    intermediate value theorem

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  2. #2
    MHF Contributor arbolis's Avatar
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  3. #3
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    oh..

    i m actually not meaning the first part...
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  4. #4
    MHF Contributor arbolis's Avatar
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    Ok. Next time be more precise then . Take a look at Mean value theorem - Wikipedia, the free encyclopedia. You must prove this theorem... called the mean value theorem. (or something very close to it).
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  5. #5
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    dotdot

    is there any connection between mean value theorem and the thing we asked to prove ?

    i mean the f(b') - f(a') / ... = k
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  6. #6
    MHF Contributor arbolis's Avatar
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    is there any connection between mean value theorem and the thing we asked to prove ?
    Yes. The mean value theorem states that there exist a c in (a,b) such as \frac{f(b)-f(a)}{b-a}=f'(c). In your example, g and h are defined and continuous on (a,b), you must find a subinterval of (a,b) in which the mean value theorem can be applied.
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  7. #7
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    cant see

    i still cant see any obvious connection...
    anyone who can do it for me please?
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  8. #8
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    be careful

    be careful. hav u seen that f'(a) < k < f'(b)

    that is f', not f
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  9. #9
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    The intermediate value theorem for derivatives states that if f:I\mapsto \mathbb{R} is a differenciable function then f' has the intermediate value property on I, an open interval. Meaning if a,b\in I with a<b and c lies (strictly) between f'(a) and f'(b) there exists x_0\in (a,b) so that f'(x_0) = c.

    Without lose of generality we will assume that f'(a) < c < f'(b). Now define the function g:[a,b]\mapsto \mathbb{R} as g(x) = f(x) - cx. Since g is continous on [a,b] it assumes a minimum value. We will now show that g does not take a minimum value at a. Note that g'(a) = f'(a) - c < 0. But since g'(a) = \lim_{x\to a^+}\tfrac{g(x)-g(a)}{x-a} < 0 it follows that g(x) - g(a)<0 for a<x sufficiently close to x. Thus, there exists y\in (a,b) so that g(y) - g(a) < 0, which is means g(y) < g(a), in particular a is not a minimum for g. Using a similar argument we can show b is not a minimum for g. Thus, g has a minimum on (a,b) and so there is a point x_0 \in (a,b) so that g'(x_0) = 0\implies f'(x_0) = c.
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