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Math Help - Mean Value Thereom

  1. #1
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    Mean Value Thereom

    Hi, I want to show that this function satisfies the Mean Value Thereom however, I am stuck when I try to obtain c. The function is f (x) = x/(x+2) on the interval [1,4].

    So far I have:

    That because it is a rational fuction, f is continuous on [a,b] and differentiable on (a,b).

    f(4) - f(1) = f'(c)(4-1)
    1/3 = f ' (c)*3
    f '(c) = 1/9


    f'(x) = 2/[(x+2)^2]

    however when i try to set f'(x) = 1/9, i do not obtain a value within [1,4]

    Thank you so much for your help!
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  2. #2
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    Quote Originally Posted by aquaglass88 View Post
    f(4) - f(1) = f'(c)(4-1)
    1/3 = f ' (c)*3 f '(c) = 1/9
    f'(x) = 2/[(x+2)^2]
    however when i try to set f'(x) = 1/9, i do not obtain a value within [1,4]
    I found a solution in [1,4].
    Square the term and solve. You should find the x is about 2.2
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  3. #3
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    you mean square the equation 2/[(x+2)^2]?
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  4. #4
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    \frac{2}{{x^2  + 4x + 4}} = \frac{1}{9}
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  5. #5
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    but i get negative 2.2, and that isn't in [1,4]
    [4-sq(72)]/2
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  6. #6
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    oh wait nm it's -4 thats why i messed up sorry!

    thanks so much!
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