# Math Help - Min and Max of Sec & Csc

1. ## 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?

2. ## 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.

I do not understand limits :$( have not learned them yet ) . To me the graph of y = csc x has parabola like cuves. So I concluded that there would be corresponding x values for a min / max. 4. ## Re: Min and Max of Sec & Csc Pl go through this site, it will clarify many doubts Graphs of the Secant and Cosecant Functions 5. ## Re: Min and Max of Sec & Csc Originally Posted by sakonpure6 I do not understand limits :$ ( 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, so $n$ cannot be the minimum value for $\csc x$. This is true for any value of $n$ you choose.

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