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Math Help - Mean Value Theorem follow on question

  1. #1
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    Mean Value Theorem follow on question

    Hi, please could someone help me with the following

    f(x) = square root (x), x>0
    Apply the Mean Value Theorem to f(x) on the interval [100,102] to prove that

    111/11 < sqrt (102) < 101/10

    I've proved that sqrt (102) < 101/10 by saying that there exists c such that the gradient at c = ( f(102) - f(100) ) / (102-100), then by saying that the gradient at c is less than the gradient at x=100

    Annoyingly, if i try saying the gradient at c is greater than the gradient at x=102 then I end up with exactly the same inequality. How do i prove the left hand inequality, 111/11 < sqrt (102)?

    Thanks.
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  2. #2
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    let  f(x) = \sqrt{x}

     \sqrt{102} - \sqrt{100} = \sqrt{102} - 10 = \frac {1}{2 \sqrt{c}} (102-100) = \frac {1}{\sqrt{c}}

    where  c \in [100,102]


     100 < c < 102

     10 < \sqrt{c} < \sqrt{102} < 11

     \frac {1}{11} < \frac {1}{\sqrt{c}} < \frac {1}{10}

    which implies  \frac {1}{11} < \sqrt{102} - 10 < \frac {1}{10}

     \frac {1}{11} + 10 < \sqrt{102} < \frac {1}{10} + 10

     \frac {111}{11} < \sqrt{102} < \frac {101}{10}
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  3. #3
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    Thanks so much, I would never have thought of introducing root (102) <11.

    So I can use my original method as well, by saying the gradient at c is greater than at x=121.

    Thanks again!
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