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Mean Value Theorem inequality

  1. Dec 26, 2012 #1
    1. The problem statement, all variables and given/known data

    For every x in the interval [0,1] show that:j

    [tex]\frac{1}{4}x+1\leq\sqrt[3]{1+x}\leq\frac{1}{3}x+1[/tex]



    3. The attempt at a solution

    Well i subtracted 1 from all sides and divided by x and I got:

    [tex]\frac{1}{4}\leq\frac{\sqrt[3]{1+x}-1}{x}\leq\frac{1}{3}[/tex]

    But now I need to find a function that has a derivative with a max value of 1/3 and a min value of 1/4 there where i'm stuck. Any help would be very much appreciated.
     
    Last edited: Dec 26, 2012
  2. jcsd
  3. Dec 26, 2012 #2

    MarneMath

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    Education Advisor

    Is the inequality even true? Let x = 0, you have -1 < 1 < -1
     
  4. Dec 26, 2012 #3
    Sorry my mistake i edited it.
     
  5. Dec 26, 2012 #4

    Dick

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    Homework Helper

    Not sure quite how this works in with the mean value theorem but f(x)=(x+1)^(1/3) is concave down. x/3+1 is the tangent line at x=0, so it lies above f(x). x/4+1 is below the secant line connecting x=0 and x=1. So it lies below f(x).
     
    Last edited: Dec 26, 2012
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