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Math Help - Min and Max of Sec & Csc

  1. #1
    Senior Member sakonpure6's Avatar
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    Min and Max of Sec & Csc

    "Find the equation for the min , max and the asymptotes for y = csc x"

    I can manage finding the equation for the recurrence of the asymptotes, however when I check in my text book it says that csc x has no min or max values? I completely disagree. Am I right?
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  2. #2
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    Re: Min and Max of Sec & Csc

    \lim_{x \to 0^+} \csc x = \infty, \lim_{x \to 0^-} \csc x = -\infty. I agree that it has no minimum or maximum value.
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    Senior Member sakonpure6's Avatar
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    Re: Min and Max of Sec & Csc

    I do not understand limits :$ ( have not learned them yet )
    . To me the graph of y = csc x has parabola like cuves. So I concluded that there would be corresponding x values for a min / max.
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    Re: Min and Max of Sec & Csc

    Pl go through this site, it will clarify many doubts
    Graphs of the Secant and Cosecant Functions
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    Re: Min and Max of Sec & Csc

    Quote Originally Posted by sakonpure6 View Post
    I do not understand limits :$ ( have not learned them yet )
    . To me the graph of y = csc x has parabola like cuves. So I concluded that there would be corresponding x values for a min / max.
    Suppose you claim n\le 1 is the maximum value for \csc x. Then, \csc\left(\dfrac{\pi}{4}\right) = \sqrt{2} > n. Suppose you claim that n>1 is the maximum value for \csc x. Then, 0 <\dfrac{1}{n+1} < \dfrac{1}{n} < 1, so there is some value of x so that \sin(x) = \dfrac{1}{n+1}. Then \csc x = \dfrac{1}{\sin x} = \dfrac{1}{\left(\tfrac{1}{n+1}\right)} = n+1>n. This contradicts the assumption that n is the maximum value for \csc x.

    I can do the same thing with negative numbers. Suppose n< -1. Then 1 < -n and I can find a value for x so that \csc x = 1-n > -n. Then, \csc (-x) = -\csc x = -(1-n) = n-1<n, so n cannot be the minimum value for \csc x. This is true for any value of n you choose.
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