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Math Help - Increasing/Decreasing and Abs Max and Min

  1. #1
    Member Jonboy's Avatar
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    Increasing/Decreasing and Abs Max and Min

    i'd like your thoughts on this one. thank you!

    Find the intervals on which f is increasing/decreasing and state any relative extrema
    1) f(\theta) = 2cos\theta - cos2\theta\,,\,[0,2\pi]
    So I find the derivative to find the critical numbers:

    f\,'\,(\theta) = 2(\,-\,sin\theta) - (\,-\,sin\,2\theta)(2)

    = -2sin\theta + 2sin\,2\theta

    So I set that equal to zero: -2sin\,\theta + 2sin\,2\theta = 0

    Divide by -2: sin\theta - sin2\theta = 0

    Add sin 2\theta: sin\theta = sin 2\theta

    Sub sin2\theta = 2sin\theta cos\theta: sin\theta = 2sin\theta cos\theta

    Divide by sin \theta: 1 = 2cos\theta

    So: cos\theta = \frac{1}{2}

    So I went on the unit circle and found all the places where the x coordinate is one half and included the negative of those angles angles since cos (\,-\,x) = cosx

    Based on that I got \theta = \pm\frac{\pi}{3}\,,\,\pm\frac{5\pi}{3}

    Is this right so far?
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  2. #2
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    skeeter's Avatar
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    you missed solutions when you divided by \sin{\theta} , note that the correct procedure is to factor ...

    \sin{\theta} - 2\sin{\theta}\cos{\theta} = 0

    \sin{\theta}(1 - 2\cos{\theta}) = 0

    \sin{\theta} = 0

    \theta = 0 , \pi , 2\pi

    \cos{\theta} = \frac{1}{2}

    \theta = \frac{\pi}{3} , \frac{5\pi}{3}

    also, note that the solution interval is 0 to 2\pi , so the negative solutions are not needed.
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  3. #3
    MHF Contributor
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    Hi

    You should not divide by \sin \theta but rather factor

    \sin \theta - \sin 2\theta = 0

    1 - 2 \cos \theta) = 0" alt="\sin \theta \1 - 2 \cos \theta) = 0" /> which leads to

    \sin \theta = 0 or \cos \theta = \frac12

    And since the interval is [0,2\pi]

    \theta=0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}, 2\pi

    EDIT : too late !
    Last edited by running-gag; March 28th 2009 at 01:57 PM. Reason: Too late
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  4. #4
    Member Jonboy's Avatar
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    thank you both!
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