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Continuity ( monotony table )

  1. Sep 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Let g be a function defined as [tex]g(x)=(\frac{1}{4})x^{2}-sin(x)[/tex]

    Give a monotony table for g in the domain [tex][-\frac{\pi}{2},\frac{\pi}{2}][/tex]



    3. The attempt at a solution

    I calculated the first derivative of g and i got g'(x)=(1/2)x-cos(x)

    and then when I wanted to solve the equation (1/2)x-cos(x)=0 I found that there was no solution so we can't know if it increases or decreases.
     
  2. jcsd
  3. Sep 25, 2012 #2

    jbunniii

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    There is a solution, because g'(0) = -cos(0) = -1, and g'(pi/2) = pi/4. As g' is continuous and has different signs at 0 and pi/2, it must be true that g'(x) = 0 for some x with 0 < x < pi/2. (Intermediate value theorem.)
     
  4. Sep 25, 2012 #3
    I don't understand what you said because (1/2)x-cos(x)=0 has no solution.
     
  5. Sep 25, 2012 #4

    SammyS

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    Well, (1/2)x-cos(x)=0 has a solution at about x = 1.02987 .

    What is a monotony table ?
     
  6. Sep 25, 2012 #5

    http://www.fmaths.com/studyoffunction/lesson.php

    Take a look at this link it shows a monotony table.
     
  7. Sep 25, 2012 #6

    I like Serena

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    Hi mtayab1994!

    There is a set of solutions.
    Not an exact one, but an approximate one.

    There are different ways to deal with problems like those.
    The appropriate method here is probably by drawing a graph.

    From the graph of g'(x), you can see where it increases monotonously, and where it decreases monotonously.
    The exact boundary will be approximate.
     
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