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Math Help - max min value of function [pi/4, 7pi/4]

  1. #1
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    max min value of function [pi/4, 7pi/4]

    Find absolute max and absolute min of the function on the interval [\frac{pi}{4}, \frac{7pi}{4}]

    f(t) = t +cot(\frac{t}{2})

    I found the derivative:

    f'(t) = 1 -csc^2(t)*\frac{1}{2}

    Now need to find the critical numbers:

    f'(t) = 1 -csc^2(t)*\frac{1}{2} = 0
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  2. #2
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    Re: max min value of function [pi/4, 7pi/4]

    Quote Originally Posted by rabert1 View Post
    Find absolute max and absolute min of the function on the interval [\frac{pi}{4}, \frac{7pi}{4}]

    f(t) = t +cot(\frac{t}{2})

    I found the derivative:

    f'(t) = 1 -csc^2(t)*\frac{1}{2}

    Now need to find the critical numbers:

    f'(t) = 1 -csc^2(t)*\frac{1}{2} = 0
    \csc^2\left(\frac{t}{2}\right) = 2

    \csc\left(\frac{t}{2}\right) = \pm \sqrt{2}

    \sin\left(\frac{t}{2}\right) = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}

    unit circle values, right? proceed ...


    ... edited for \frac{t}{2} as the trig function argument.
    Last edited by skeeter; May 20th 2012 at 05:36 AM.
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  3. #3
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    Re: max min value of function [pi/4, 7pi/4]

    Just wanted to add that f'(t) = 1-\frac{\csc^2(t/2)}{2}.
    Thanks from skeeter
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