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Math Help - Mean Value Theorem

  1. #1
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    Mean Value Theorem

    my problem I am working on is f(x)= x+(54/x); [6,9]

    First I took the first derivative and found that it was f ' (x)= 1-(54/x^2)

    I then found f(6) = 15 and f(9) = 15

    If I plug those into the formula f(b)-f(a) I get 0 this just doesn't make
    b-a 3


    This just doesn't make sense!! Could someone help me?
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  2. #2
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    What doesn't make sense?

    f'(c) = \frac{f(b)- f(a)}{b-a} \: \: c \in (a,b)

    Now, you found that f'(c) = 0, i.e. there exists some point on your interval such that its derivative is equal to 0. What does it exactly mean for your function to have its derivative equal to 0 at c?

    Hint: Rolle's Theorem, a special case of the mean value theorem.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by kelleannmoore@yahoo.com View Post
    my problem I am working on is f(x)= x+(54/x); [6,9]

    First I took the first derivative and found that it was f ' (x)= 1-(54/x^2)

    I then found f(6) = 15 and f(9) = 15

    If I plug those into the formula f(b)-f(a) I get 0 this just doesn't make
    b-a 3


    This just doesn't make sense!! Could someone help me?
    The Mean Value Theorem states that if f(x) is continous on some interval D=[a,b] and differentiable on (a,b) then

    \exists{c}\in{D}\backepsilon~f'(c)=\frac{f(b)-f(a)}{b-a}

    So we see that we have a special case of the Mean Value Theroem called Rolle's Theorem that states, that if our function meets the same conditions as before and that f(a)=f(b)

    Then \exists{c}\in{D}\backepsilon~f'(c)=0
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