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Thread: Mean Value Theorem

  1. #1
    Nov 2008

    Mean Value Theorem

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

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

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

    so is $\displaystyle g(1)$ infinity ?

    and then..

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

    I got confused here...

    will $\displaystyle f(b)-f(a)=f"(c)(b-a)$ ???

    and then I tried another way...

    $\displaystyle g(2)-g(1)=g'(c)(2-1)$
    $\displaystyle g(-2)-g(1)=g'(c)(-2-1)$
    $\displaystyle 1-g(1)=g'(c)$
    $\displaystyle 5-g(1)=-3g'(c)$
    $\displaystyle g(1)=2 g'(c)=-1$

    Am I wrong again???
    Last edited by Fosite; Nov 18th 2008 at 06:42 PM.
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  2. #2
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK
    Think geometrically! The condition $\displaystyle g''(x)\geqslant4$ 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 $\displaystyle g(x) = 2x^2+ax+b$. 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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