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Thread: Min and Max of Sec & Csc

  1. #1
    Super 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

    $\displaystyle \lim_{x \to 0^+} \csc x = \infty, \lim_{x \to 0^-} \csc x = -\infty$. I agree that it has no minimum or maximum value.
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  3. #3
    Super 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 $\displaystyle n\le 1$ is the maximum value for $\displaystyle \csc x$. Then, $\displaystyle \csc\left(\dfrac{\pi}{4}\right) = \sqrt{2} > n$. Suppose you claim that $\displaystyle n>1$ is the maximum value for $\displaystyle \csc x$. Then, $\displaystyle 0 <\dfrac{1}{n+1} < \dfrac{1}{n} < 1$, so there is some value of $\displaystyle x$ so that $\displaystyle \sin(x) = \dfrac{1}{n+1}$. Then $\displaystyle \csc x = \dfrac{1}{\sin x} = \dfrac{1}{\left(\tfrac{1}{n+1}\right)} = n+1>n$. This contradicts the assumption that $\displaystyle n$ is the maximum value for $\displaystyle \csc x$.

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