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Math Help - The MVT.

  1. #1
    No one in Particular VonNemo19's Avatar
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    The MVT.

    Right now I'm studying the proof of the mean value theorem and ther's something about it that I am having trouble understanding.

    I'm going to attemp to explain where I'm having trouble with the proof.

    Refer to the attached figure during this discussion.


    In the proof, the first thing that they do is say that the equation to the secant line is

    y=\left[\frac{f(b)-f(a)}{b-a}\right](x-a)-f(a)

    I understand that \left[\frac{f(b)-f(a)}{b-a}\right] is the slope of the secant line, but I can't understand the rest. My mind wants to say that (x-a) is x (in the slope intercept form of a line) and -f(a)=b.

    Could someone justify this as best they can?
    Last edited by VonNemo19; September 19th 2009 at 10:49 PM.
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  2. #2
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    Quote Originally Posted by VonNemo19 View Post
    Right now I'm studying the proof of the mean value theorem and ther's something about it that I am having trouble understanding.

    I'm going to attemp to explain where I'm having trouble with the proof.

    Refer to the attached figure during this discussion.


    In the proof, the first thing that they do is say that the equation to the secant line is

    y=\left[\frac{f(b)-f(a)}{b-a}\right](x-a){\color{red}+}f(a)

    I understand that \left[\frac{f(b)-f(a)}{b-a}\right] is the slope of the secant line, but I can't understand the rest. My mind wants to say that (x-a) is x (in the slope intercept form of a line) and -f(a)=b.

    Could someone justify this as best they can?
    There is point-slope form of a line

    y - y_1=m(x -x_1) where m is slope and (x_1, \;y_1) is a point on the line.

    y =m(x -x_1)+y_1

    here slope is m=\frac{f(b)-f(a)}{b-a} and point (x_1, \;y_1) =(a, f(a))

    therefore, the equation becomes,

    y=\left[\frac{f(b)-f(a)}{b-a}\right](x-a)+f(a)
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Shyam View Post
    and (x_1, \;y_1) =(a, f(a))

    therefore, the equation becomes,

    y=\left[\frac{f(b)-f(a)}{b-a}\right](x-a)+f(a)

    I can't believe that I didn't see that. Thanks man.
    Last edited by mr fantastic; June 27th 2009 at 11:32 PM. Reason: Removed the 'freak'
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