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Thread: 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 $\displaystyle [\frac{pi}{4}, \frac{7pi}{4}]$

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

    I found the derivative:

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

    Now need to find the critical numbers:

    $\displaystyle f'(t) = 1 -csc^2(t)*\frac{1}{2} = 0$
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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 $\displaystyle [\frac{pi}{4}, \frac{7pi}{4}]$

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

    I found the derivative:

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

    Now need to find the critical numbers:

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

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

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

    unit circle values, right? proceed ...


    ... edited for $\displaystyle \frac{t}{2}$ as the trig function argument.
    Last edited by skeeter; May 20th 2012 at 04:36 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 $\displaystyle f'(t) = 1-\frac{\csc^2(t/2)}{2}$.
    Thanks from skeeter
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