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Sketching the Curves of a Function W/In an Interval - Simple (1st Year Calcu

  1. Feb 19, 2008 #1
    [SOLVED] Sketching the Curves of a Function W/In an Interval - Simple (1st Year Calcu

    1. The problem statement, all variables and given/known data

    Sketch the graph of the function on the interval [0, 2pi].

    y = cosx - 1/2(cos2x)

    2. Relevant equations

    [​IMG]

    3. The attempt at a solution

    so the problems that i have been practicing like this have been pretty simple. i determine all of the following:

    1. domain and range
    2. x and y intercepts
    3. whether or not there is discontinuity
    4. whether or not there is symmetry
    5. intervals of increasing and decreasing order
    6. extrema
    7. points of inflection
    8. concavity
    9. whether or not there exist asymptotes

    the problem is that mathematically i located a single extreme value at (pi, -3/2), but graphically, there appears to be more.

    [​IMG]

    is not each change in direction a relative minimum/maximum value? i can only seem to locate the absolute minimum value within the interval, but none of the other relative extreme values. any idea as to where i am going wrong?

    so, the first derivative of the given function is -sinx + sin2x. solving this equation is where i find possible extrema at 0, pi, and 2pi. any ideas why i can't seem to locate the other extrema algebraically?
     
  2. jcsd
  3. Feb 19, 2008 #2
    How did you solve this equation?
     
  4. Feb 19, 2008 #3
    ah. figured it out. i had worked out the values mentally real quick and hadn't realized that i needed to use the double angle trig formula. sheesh. yet another simple algebra mistake. is there a way i could mark this for deletion or something? by the way, thank you. sometimes it can be difficult to see a mistake of your own you know.
     
  5. Feb 19, 2008 #4
    You are welcome! :smile:

    Just edit the title and marked it [SOLVED]
     
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