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Math Help - Absolute min and max

  1. #1
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    Absolute min and max

    Find the absolute minimum and absolute maximum values of f on the given interval.

    f(t) = 3t + 3 cot(t/2),
    [π/4, 7π/4]

    Hey guys, I'm stuck on this and on my last attempt, please help me out here,

    so
    the derivative is -3/2cos(t)csc^2(t/2)

    Then I make it = 0, etc,etc.

    My answers I got were pi/2, 3pi/2 but they were both wrong. Please help me out here, PLEASE!
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  2. #2
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    Re: Absolute min and max

    Hello, Oldspice1212!

    Your derivative is wrong . . .


    \text{Find the absolute minimum and absolute maximum values of }f\text{ on the given interval.}

    . . f(t) \:=\:3t + 3\cot(\tfrac{t}{2}),\;\;[\tfrac{\pi}{4},\:\tfrac{7\pi}{4}]

    We have: . f'(x) \;=\;3 - 3\csc^2(\tfrac{t}{2})\!\cdot\!\tfrac{1}{2}

    . . . 3 - \tfrac{3}{2}\csc^2(\tfrac{t}{2}) \;=\;0

    . . . . 2 - \csc^2(\tfrac{t}{2}) \;=\;0

    n . . . . . \csc^2(\tfrac{t}{2}) \;=\;2

    . . . . . . . \csc(\tfrac{t}{2}) \;=\;\pm\sqrt{2}

    . . . . . . . . . . \frac{t}{2} \;=\;\pm\frac{\pi}{4},\:\pm\frac{3\pi}{4},\:\pm \frac{5\pi}{4}\:\hdots

    . . . . . . . . . . t \;=\;\pm\frac{\pi}{2},\:\pm\frac{3\pi}{2},\:\pm \frac{5\pi}{2}\:\hdots

    Therefore: . t \;=\;\left\{\frac{\pi}{2},\:\frac{3\pi}{2}\right\}


    I'll let you calculate the maximum and minimum.
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  3. #3
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    Re: Absolute min and max

    Ok so for pi/2 I got 3+3pi/2, for pi/4 I got 3pi/4+3cot(pi/8), 3pi/2 I got 9pi/2-3, and for 7pi/4 I got 21pi/4-3cot(pi/8)? So is the 3+3pi/2 min and 9pi/2-3 the max? Thanks a lot btw.
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  4. #4
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    Re: Absolute min and max

    Note that the absolute min and max on a closed bounded interval may be at the endpoints of the interval even if the derivative is not zero there.
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