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anonymous
 one year ago
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anonymous
 one year ago
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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0C is easy, simply find where the derivative is zero, But i cannot for the life of me figure out a and b.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I thought right/left end behaviors were only for polynomials

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1rewrite y in terms of x by switching the sec to cos then it should look more intuitive

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0arccos(1/x)= x= 1/cos(y)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Still dont see how this results in an end model... :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433356872213:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0somethung like that (cant draw with a mouse)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1good x= 1/cos(y) then get y = f(x) and look at: \(\lim_{x > \pm \infty}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0.... is it positive and negative infinity

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1if x= 1/cos(y) and y = f(x) then f(x) = ???

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1mmm x= 1/cos(y) cos(y) = ??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The course gave me the formula arccos(1/x)=arcsec(x) ... does that qualify?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It would make sense... given the lesson didnt talk about end behaviors at all

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1"The course gave me the formula arccos(1/x)=arcsec(x) ... does that qualify?" not interested! my belief is that if you can work it out from the simple stuff, there is virtually nothing to remember. and no reliance on millions of formulae. now, you managed: \(x= \frac{1}{cos(y)}\). doesn't it follow that: \(cos(y)= \frac{1}{x}\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No I mean the lesson wasnt talking about end behaviors, but it did bring up the fact that arccos(1/x)=arcsec(x). Yeah obviously the math which makes them equal is simple, I was just justifying this random question. Sorry I worded that badly, Thanks :) Because it is true for all values, this makes that the correct answer for both a and b doenst it?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1OK, arccos(1/x)=arcsec(x) ?? we can test that: let y = arccos(1/x) cos y = (1/x) secy = x y = arcsec(x) so arccos(1/x)=arcsec(x) that's what i mean. ie, IMHO, it 's bad enough having to remember that sec = 1/cos. so are we finished? do you see the end behaviours? we have: y = arccos(1/x), and: x > \( \pm \infty\),

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lim x>+inifinty (arcsec(x)/arccos(1/x) ) = 1 so i believe that makes arccos(1/x) the end behavior model

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1brill! we're done :p
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