anonymous
  • anonymous
Can someone explain why the domain of arcsec(x) is (-infinity,1]union[1,+infinity)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
What is arcsec(x) even asking? sin(x) is asking give an angle and it will give the y coordinate that corresponds to that angle
anonymous
  • anonymous
Do you know the domain and range of \(\sec x\) ? \(\sec^{-1}x\) is the inverse of \(\sec x\), so their domains/ranges/ are reversed.
anonymous
  • anonymous
Also, "Primus sucks!"

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anonymous
  • anonymous
no I will think about it and Primus sucks forever!
anonymous
  • anonymous
Here's \(\sec x\):|dw:1377655344101:dw|
anonymous
  • anonymous
pi/2 and -pi/2 would be undefined for the range of sec(x)
anonymous
  • anonymous
In order for an inverse to exist, you must first restrict the domain: |dw:1377655433174:dw|
anonymous
  • anonymous
those +kpi
anonymous
  • anonymous
k is a member of integers
anonymous
  • anonymous
that corresponds to the top half of the unit circle only? excluding pi/2
anonymous
  • anonymous
The graph of the inverse is a reflection across the line \(y=x\): |dw:1377655508554:dw| Maybe it's easier to see on Wolfram: http://www.wolframalpha.com/input/?i=Sec%5Bx%5D%2C+InverseSec%5Bx%5D
anonymous
  • anonymous
I see how to right right is restricted then reflected about the line y=x but not the left side
anonymous
  • anonymous
Sorry, the drawing's not to scale. Do you have a graphing calculator? You should (eventually) be able to see that the domain is \((-\infty,-1)\cup(1,\infty)\), and that the range is \(\left(0,\dfrac{\pi}{2}\right)\cup\left(\dfrac{\pi}{2},1\right)\).
anonymous
  • anonymous
so the range of arcsec(x) just comes from the domain of sec(x) being restricted the reflecting about the line y=x?
anonymous
  • anonymous
Yes, that's the gist.
anonymous
  • anonymous
Of course, it depends on how you restrict the domain, but I believe this is the standard way it's restricted.
anonymous
  • anonymous
wow that is a twister but thank you. Anyway, how would you suggest I start thinking about it or becoming more familiar with those graphs?
anonymous
  • anonymous
When I was taught about the inverse functions, it was suggested that I be familiar with the regular trig functions and their domains/ranges. The inverse functions just have these sets reversed.
anonymous
  • anonymous
So i should be familiar with the range and domain for all six not just the main 3
anonymous
  • anonymous
Right
anonymous
  • anonymous
Thanks again

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