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Math Help - Mean value theorem

  1. #1
    Senior Member
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    Mean value theorem

    I'm having some trouble with this question

    Use the mean value theorem to show that:

    <br />
1)\ \frac{4}{27}<3-23^ \frac{1}{3}< \frac{3}{16}<br />

    Would you start of like this:

    \frac{f(27)-f(23)}{27-23}= f'(c),\ c \in (23,27)

    Let\ f(x)=x^ \frac{1}{3}

    \frac{27^ \frac{1}{3} - 23^ \frac{1}{3}}{27-23} = \frac{1}{3c^ \frac{2}{3}}

    \frac{3- 23^ \frac{1}{3}}{4}= \frac{1}{3c \frac{2}{3}}

    23<c<27

    23^ \frac{2}{3}<c^ \frac{2}{3}< 27^ \frac{2}{3}

    3* 23^ \frac{2}{3}<3c^ \frac{2}{3}< 3*9

    \frac{1}{3* 23^ \frac{2}{3}}> \frac{1}{3c^ \frac{2}{3}}> \frac{1}{27}

    From here on, I don't know what to do?
    Last edited by acevipa; April 26th 2010 at 03:22 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by acevipa View Post
    I'm having some trouble with these two questions:

    1)\ \frac{4}{27}<3-23^ \frac{1}{3}< \frac{3}{16}

    For 1) would you start off like this. I don't really know what to do after.

    \frac{27^ \frac{1}{3}-23^ \frac{1}{3}}{27-23}= \frac{1}{3c \frac{2}{3}}
    That is meaningless, try posting the original question as set.

    CB
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  3. #3
    Senior Member
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    Quote Originally Posted by CaptainBlack View Post
    That is meaningless, try posting the original question as set.

    CB

    Sorry, this is question I'm having trouble with:

    Use the mean value theorem to show that:

    <br />
1)\ \frac{4}{27}<3-23^ \frac{1}{3}< \frac{3}{16}<br />

    Would you start of like this:

    \frac{f(27)-f(23)}{27-23}= f'(c),\ c \in (23,27)

    Let\ f(x)=x^ \frac{1}{3}

    \frac{27^ \frac{1}{3} - 23^ \frac{1}{3}}{27-23} = \frac{1}{3c^ \frac{2}{3}}

    \frac{3- 23^ \frac{1}{3}}{4}= \frac{1}{3c \frac{2}{3}}

    23<c<27

    23^ \frac{2}{3}<c^ \frac{2}{3}< 27^ \frac{2}{3}

    3* 23^ \frac{2}{3}<3c^ \frac{2}{3}< 3*9

    \frac{1}{3* 23^ \frac{2}{3}}> \frac{1}{3c^ \frac{2}{3}}> \frac{1}{27}

    From here on, I don't know what to do?
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