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Math Help - Another Continuity Problem

  1. #1
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    Another Continuity Problem

    Another continuity problem I'm having trouble on

    On the interval [0,2pi] where is the function f(x) = (sqrt 2cosx - (sqrt 3)) continuous?

    Thanks again guys



    Edit : Sorry about that Plato. Forgot to put the X after Cos. Thanks
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  2. #2
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    That function is meanless as posted.
    Please review what you have posted and correct the function.
    There is no x in the agrument.
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  3. #3
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    is this your function?

    f(x) = \sqrt{2\cos{x} - \sqrt{3}}
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  4. #4
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    In either case f(x) = \sqrt {2\cos (x)}  - \sqrt 3 \,\mbox{or} \,f(x) = \sqrt {2\cos (x) - \sqrt 3 } the function is not even define.
    Can you have the square root of a negative number?
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  5. #5
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    Well there is an answer for it and it's

    [0, pi/6] U [11pi/6, 2pi]
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  6. #6
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    I'm assuming that the function is

     f(x) = \sqrt{2\cos{x} - \sqrt{3}}

    if that is the case, then

    2\cos{x} - \sqrt{3} \geq 0

    you know why, correct?

    simplifying the inequality ...

    \cos{x} \geq \frac{\sqrt{3}}{2}

    your "answer" is the set of x-values for which the above inequality is true.
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