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Math Help - proof using mean value theorem

  1. #1
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    proof using mean value theorem

    I need to prove this using the mean value theorem.

    1/9 < sqrt(66) - 8 < 1/8

    I assumed a = 64 and b = 66 and let a function f(x) = sqrt(x) such that x lies in the interval [a, b]

    Then by the mean value theorem,

    f'(x) = 1/2sqrt(x) = (sqrt (66) - sqrt (64)) / 66-64 = (sqrt(66) - 8)/2

    But at this point, I'm beautifully stuck, and dont know what else to do.

    Any ideas?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dgmath View Post
    I need to prove this using the mean value theorem.

    1/9 < sqrt(66) - 8 < 1/8

    I assumed a = 64 and b = 66 and let a function f(x) = sqrt(x) such that x lies in the interval [a, b]

    Then by the mean value theorem,

    f'(x) = 1/2sqrt(x) = (sqrt (66) - sqrt (64)) / 66-64 = (sqrt(66) - 8)/2

    But at this point, I'm beautifully stuck, and dont know what else to do.

    Any ideas?
    Problem: Using the MVT prove that \frac{1}{9}<\sqrt{66}-8<\frac{1}{8}

    Proof: Since \sqrt{x} satisfies all the conditions of the mean value theorem on [64,66] there exists some c\in(64,66) such that \frac{\sqrt{66}-\sqrt{64}}{64-66}=\frac{1}{2\sqrt{c}}. Or equivalently, \sqrt{66}-8=\frac{1}{\sqrt{c}}\quad\color{red}\star. Since \frac{1}{\sqrt{x}} is a decreasing function on [64,66] we know that for any k\in(64,66) it's true that \frac{1}{\sqrt{66}}<\frac{1}{\sqrt{k}}<\frac{1}{\s  qrt{64}}=8\quad\color{red}\star\star. Combining \color{red}\star,\star\star we see that \frac{1}{\sqrt{66}}<\frac{1}{\sqrt{c}}=\sqrt{66}-8<8. Lastly noting that \frac{1}{\sqrt{66}}>\frac{1}{\sqrt{81}}=\frac{1}{9  } finishes the proof.
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  3. #3
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    oh wow, that makes perfect sense. thanks bunch!
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