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Math Help - Mean Value Theorem - Inequality Problem

  1. #1
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    Mean Value Theorem - Inequality Proof

    Quite an interesting problem:

    Use the inequality sqrt(3) < 1.8 to prove that 1.7 < sqrt(3) < 1.75.

    The book where I found this says to use the square root function in the MVT for derivatives and use that function on the interval [3,4].

    Still confused how to do this.... need help!!!
    Last edited by zachd77; December 27th 2012 at 02:49 PM.
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  2. #2
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    Re: Mean Value Theorem - Inequality Problem

    Let f(x)=\sqrt{x}, x>0.

    Then, f^{\prime}(x) = \frac{1}{2\sqrt{x}} , x>0.

    Since f is continuous on [3,4] and differentiable on (3,4), there exists c\in(3,4) such that

    f^{\prime}(c)=\frac{f(4)-f(3)}{4-3}=f(4)-f(3).

    So, there exists c\in(3,4) such that

    \sqrt{4}-\sqrt{3}=2-\sqrt{3}=\frac{1}{2\sqrt{c}}.


    Now we prove the boundaries:

    3<c<4 \implies \sqrt{3}<\sqrt{c}<2 \implies \frac{1}{2}<\frac{1}{\sqrt{c}}<\frac{1}{\sqrt{3}} \implies \frac{1}{4}<\frac{1}{2\sqrt{c}}<\frac{\sqrt{3}}{6}

    Since \frac{1}{2\sqrt{c}}=2-\sqrt{3} and the problem gives us that \sqrt{3}<1.8 , the last inequality implies

    \tiny{\frac{1}{4}<2-\sqrt{3}<\frac{1.8}{6} \implies 0.25-2<-\sqrt{3}<0.3-2 \implies -1.75<-\sqrt{3}<-1.7 \implies 1.7<\sqrt{3}<1.75}.
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  3. #3
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    Re: Mean Value Theorem - Inequality Problem

    Thanks!!!! ..... I realized that I did the MVT part wrong.. I solved for c even though I didn't have to...
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  4. #4
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    Re: Mean Value Theorem - Inequality Problem

    Quote Originally Posted by abender View Post

    Now we prove the boundaries:

    3<c<4 \implies \sqrt{3}<\sqrt{c}<2 \implies \frac{1}{2}<\frac{1}{\sqrt{c}}<\frac{1}{\sqrt{3}} \implies \frac{1}{4}<\frac{1}{2\sqrt{c}}<\frac{\sqrt{3}}{6}

    Since \frac{1}{2\sqrt{c}}=2-\sqrt{3}.....
    note that the information √3 < 1.8 is not needed:

    from

    2 - \sqrt{3} < \frac{\sqrt{3}}{6}

    we have:

    12 - 6\sqrt{3} < \sqrt{3} \implies 12 < 7\sqrt{3} \implies \frac{12}{7} < \sqrt{3}

    and 12/7 is approx. 1.7142 which is certainly larger than 1.7 (or 17/10 < 12/7 since: 7*17 = 119 < 120 = 10*12).

    we also have:

    \frac{1}{4} < 2 - \sqrt{3} \implies 1 < 8 - 4\sqrt{3} \implies 4\sqrt{3} < 7 \implies \sqrt{3} < \frac{7}{4} = 1.75

    (as a practical matter, this means that 97/56 is a VERY good rational approximation of √3).
    Last edited by Deveno; December 27th 2012 at 11:36 PM.
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