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Homework Help: Finding Relative Extrema & Points Of Inflection

  1. Mar 24, 2004 #1


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    For the function below, I have to find the exact values of x for which relative extreme exist and the exact values of x for which points of inflection exist.

    [tex]f(x) = 1x/2 - sin(x)[/tex] when x is in the interval [tex](0,2pi)[/tex]

    Here's what I have:

    [tex]f'x = 1/2 - cos(x) = 0[/tex] (I'm not sure how to solve for x in this spot)

    Also, for the finding the inflection points, I have this:

    [tex]f"x = sin(x) = 0[/tex]

    [tex]x = 0/sin = 0[/tex] (This right? Or did I screw up before this?)

    I guess my troubles are based around the basic trig functions as well as algebra. I'm in the process of moving, so my algebra book will be hard to find; however, I'm about to start looking.

    Any help to point me in the right direction would be fantastic. Thanks.
    Last edited: Mar 24, 2004
  2. jcsd
  3. Mar 24, 2004 #2


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    First of all, wtf is up with your post? It there's a big black area and I can see some of the page's source code.

    [tex]f'(x) = \frac{1}{2} -cos(x) = 0[/tex]

    [tex]cos(x) = \frac{1}{2}[/tex]

    You can either cheap out and use your calculator or you can draw those goofy triangles. http://myfiles.dyndns.org/math/value_triangle1.jpg is the one that applies in this case.
    Turn 60 degrees into radians by dividing by 180 then multiplying by pi.

    [tex]x = \frac{\pi}{3}[/tex]

    I can't even read what you have for inflection stuff because of the black area I mentioned.

    [tex]f''(x) = sin(x) = 0[/tex]

    [tex]x = 0, \pi , 2\pi[/tex]

    That is NOT the answer though! The interval is between 0 and 2[tex]\pi[/tex] but those are rounded brackets. Round brackets mean you do NOT include the limits, that means 0 and 2[tex]\pi[/tex] are not answers.

    [tex]x = \pi[/tex]
    Last edited by a moderator: Apr 20, 2017
  4. Mar 24, 2004 #3
    I have a program that finds roots. First extremum x is a little more than 1 radian and second about 5.20
  5. Mar 24, 2004 #4


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    Excellent point.

    cos(x) = 1/2 is a referance angle. The first correct answer is pi/3. The second answer is mirrored around the x axis (sine is mirrored around the y axis). The second answer should be 2pi - pi/3 which is 5pi/3 (about 5.23 as beyond had said).
  6. Mar 25, 2004 #5


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    Thanks for the help fellas. I don't know why my post looks like that. I checked it about 20 times looking for a mistake in the Latex coding, but there aren't any.
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