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Thread: 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)$\displaystyle f(\theta) = 2cos\theta - cos2\theta\,,\,[0,2\pi]$
    So I find the derivative to find the critical numbers:

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

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

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

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

    Add $\displaystyle sin 2\theta$: $\displaystyle sin\theta = sin 2\theta$

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

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

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

    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 $\displaystyle cos (\,-\,x) = cosx$

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

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

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

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

    $\displaystyle \sin{\theta} = 0$

    $\displaystyle \theta = 0 , \pi , 2\pi$

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

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

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

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

    $\displaystyle \sin \theta - \sin 2\theta = 0$

    $\displaystyle \sin \theta \1 - 2 \cos \theta) = 0$ which leads to

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

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

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

    EDIT : too late !
    Last edited by running-gag; Mar 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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