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Relative extrema confusion

  1. Sep 26, 2010 #1
    Hello All,

    My problem is to find the relative extrema of abs(sin 2x); 0< x < 2pi.

    I got the derivative right, I think: 2 cos 2x if x > 0, and -2 cos 2x if x < 0.

    However, the solution says that f ' (x) does not exist at x = pi/2, x = pi, and x = (3pi)/2.

    Glancing at my trigonometry charts, I see that sin and cos are never undefined,

    so why does 2 cos 2x or -2 cos 2x not exist at pi/2, pi, or 3pi/2?

    Thanks,

    Nick
     
  2. jcsd
  3. Sep 26, 2010 #2

    LCKurtz

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    But your interval only includes x > 0. And it isn't the sign of x that is important, it is the sign of sin(x) because of the absolute value signs.

    Draw the graph of |sin(2x)| and you will answer your own questions.
     
  4. Sep 26, 2010 #3
    Thank you. I had a brain freeze on that one.
     
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