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Math Help - Proof of an Inversion of the Mean Value Theorem

  1. #1
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    Proof of an Inversion of the Mean Value Theorem

    Question:

    Let f:[a,b]-> R be a continuous function on the closed interval [a,b] and differentiable on the open interval (a,b). Show that for any c in (a,b) that is not a point of maximum of minimum for f', there exist x_1,x_2 in (a,b) such that:

    f'(c)=\frac{f(x_2)-f(x_1)}{x_2-x_1}

    This seems so matter-of-fact to me, and I know it's simply a sort of inversion of the Mean value Theorem, but I can't see how I would approach this. It looked really easy at first sight, but I'm lost..

    All I know is that if there's an interval (x1,x2), there is at least one point such that the derivative at that point is equal to \frac{f(x_2)-f(x_1)}{x_2-x_1}. (Mean Value Theorem). But I feel like it's not applicable in this proof..

    Suggestions please? Thank you!
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  2. #2
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    Re: Proof of an Inversion of the Mean Value Theorem

    Quote Originally Posted by limddavid View Post
    Question:

    Let f:[a,b]-> R be a continuous function on the closed interval [a,b] and differentiable on the open interval (a,b). Show that for any c in (a,b) that is not a point of maximum of minimum for f', there exist x_1,x_2 in (a,b) such that:

    f'(c)=\frac{f(x_2)-f(x_1)}{x_2-x_1}

    This seems so matter-of-fact to me, and I know it's simply a sort of inversion of the Mean value Theorem, but I can't see how I would approach this. It looked really easy at first sight, but I'm lost..
    You're right, this seems to be harder than it looks. Here's an outline solution.

    Define a new function g on the interval [a,b] by g(x) = f(x) - f'(c)(x-c). Then g is also continuous on [a,b] and differentiable on (a,b). If there are two distinct points x_1,x_2 in [a,b] with g(x_1) = g(x_2) then \frac{f(x_2)-f(x_1)}{x_2-x_1} = f'(c), as required.

    Suppose that there do not exist two such points, so that g takes each of its values only once. In other words, g is an injective function on the interval [a,b]. Use the intermediate value theorem to deduce that g must be a monotone function on the interval. It then follows that g'(x) must always have the same sign (either g'(x)\geqslant0 for all x in [a,b], or g'(x)\leqslant0 for all x in [a,b]). But g'(c) = 0, which means that g' has either a maximum or a minimum at c, and hence so does f' (because g' and f' only differ by a constant).

    Taking the contrapositive, it follows that if f' does not have a max or a min at c, then points x_1,x_2 with the desired property must exist.
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