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Mean Value Teorem

  1. Aug 5, 2005 #1
    How do I prove that (x-1)/x < lnx < (x–1), for x > 1 by using the mean value teorem?
     
  2. jcsd
  3. Aug 5, 2005 #2

    TD

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    Homework Helper

    Well, since [itex]\ln \left( 1 \right) = 0[/itex], we can write:

    [tex]\begin{array}{l}
    \frac{{x - 1}}{x} < \ln \left( x \right) - \ln \left( 1 \right) < x - 1 \\ \\
    \frac{{x - 1}}{{x\left( {x - 1} \right)}} < \frac{{\ln \left( x \right) - \ln \left( 1 \right)}}{{x - 1}} < \frac{{x - 1}}{{x - 1}} \\ \\
    \frac{1}{x} < \frac{{\ln \left( x \right) - \ln \left( 1 \right)}}{{x - 1}} < 1 \\
    \end{array}[/tex]

    Now you can apply the Mean Value Theorem to the middle expression, which states that, if [itex]f\left( x \right)[/itex] is differentiable over (a,b) and continuous over [a,b], there exists a [itex]c \in \left( {a,b} \right)[/itex] so that:
    [tex]f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}[/tex]

    Does that help?
     
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