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Math Help - Mean Value Theorem

  1. #1
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    Mean Value Theorem

    Suppose that g is a function defined and continuous on [-2, 2] and that g" exists on the
    open interval (-2, 2) . If g(-2) = 5 , g(2) = 1 and g"(x)>= 4 for all x in (-2, 2) , how
    large can g(1) possibly be?


    I did the follow step, but something goes wrong..

    g(1)-g(-2)=g"(c)(1+2)
    g(1)=g(-2)+3g"(c)
    3g"(c)>=12
    g(1)=5+3g"(c)>=17

    so is g(1) infinity ?

    and then..


    g(1)-g(2)=g"(c)(1-2)
    g(1)=g(2)-g"(c)
    -g"(c)=<4
    g(1)=1-g"(c)=<-3

    I got confused here...

    will f(b)-f(a)=f"(c)(b-a) ???


    and then I tried another way...

    g(2)-g(1)=g'(c)(2-1)
    g(-2)-g(1)=g'(c)(-2-1)
    1-g(1)=g'(c)
    5-g(1)=-3g'(c)
    g(1)=2 g'(c)=-1

    Am I wrong again???
    Last edited by Fosite; November 18th 2008 at 06:42 PM.
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  2. #2
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    Think geometrically! The condition g''(x)\geqslant4 tells you that the function g is convex. The maximum value of g(1) will occur when g''(x) is equal to 4 throughout the interval. If g''(x) is ever greater than 4 (with the endpoints at (-2,5) and (2,1) remaining fixed) then the curve will become more convex, forcing g(1) to be smaller.

    So let g''(x) = 4. Integrate twice to get g(x) = 2x^2+ax+b. You can find the values of the constants a and b by using the facts that g(-2)=5 and g(2)=1, and then you will know what g(1) is.
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