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Regarding the previusmean theorim question

  1. Nov 26, 2008 #1
    i got a different solution to my problem :

    f(x) continues in [a,b] interval,and differentiable (a,b) b>a>0
    alpha differs 0
    proove that there is b>c>a

    in that formula:

    i was told that in this question i dont use "mean theorim"
    but the "couchy mean theorim"

    F(x)=f(x)/[x^&] G(x)=1/(x^&)

    [F(b)-F(a)]/[G(b)-G(a)] = [f(b)/(b^&) - f(a)/(a^&)]/[1/(b^&) - 1/(a^&)] =

    = [f(b)*(a^&) - f(a)*( b^&)]/[a^& - b^&]

    F'(c) / G'(c)=[f'(c)* (1/c^&) -&*c^(-&-1)*f(c)]/[-&*c^(-&-1)]=

    =[&*f(c)-c*f'(c)]/& =f(c) - c*f'(c)/&

    my proffesor solved it like this after about 50 minutes when
    he tried to use the mean theorim
    and then he said that we need to use couchy mean theorim
    i dont know how he desided that??

    and the second most important problem is picking the variables.
    how he picked
    F(x)=f(x)/[x^&] G(x)=1/(x^&)
    what proccess do i need to do in order to deside
    that the way of solving such a problem is picking

    F(x)=f(x)/[x^&] G(x)=1/(x^&)

  2. jcsd
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