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anonymous

  • one year ago

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  1. rainbow_rocks03
    • one year ago
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    question?

  2. anonymous
    • one year ago
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    C is easy, simply find where the derivative is zero, But i cannot for the life of me figure out a and b.

  3. anonymous
    • one year ago
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    I thought right/left end behaviors were only for polynomials

  4. IrishBoy123
    • one year ago
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    re-write y in terms of x by switching the sec to cos then it should look more intuitive

  5. anonymous
    • one year ago
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    arccos(1/x)= x= 1/cos(y)

  6. IrishBoy123
    • one year ago
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    x = sec(y) etc

  7. anonymous
    • one year ago
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    Still dont see how this results in an end model... :(

  8. tkhunny
    • one year ago
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    |dw:1433356700597:dw|

  9. anonymous
    • one year ago
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    |dw:1433356872213:dw|

  10. anonymous
    • one year ago
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    somethung like that (cant draw with a mouse)

  11. IrishBoy123
    • one year ago
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    good x= 1/cos(y) then get y = f(x) and look at: \(\lim_{x -> \pm \infty}\)

  12. anonymous
    • one year ago
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    .... is it positive and negative infinity

  13. anonymous
    • one year ago
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    nvm

  14. anonymous
    • one year ago
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    pi/2?

  15. IrishBoy123
    • one year ago
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    if x= 1/cos(y) and y = f(x) then f(x) = ???

  16. anonymous
    • one year ago
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    1/arccos(x) ?

  17. IrishBoy123
    • one year ago
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    mmm x= 1/cos(y) cos(y) = ??

  18. anonymous
    • one year ago
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    1/x

  19. anonymous
    • one year ago
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    arccos(1/x)

  20. anonymous
    • one year ago
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    The course gave me the formula arccos(1/x)=arcsec(x) ... does that qualify?

  21. anonymous
    • one year ago
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    It would make sense... given the lesson didnt talk about end behaviors at all

  22. IrishBoy123
    • one year ago
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    "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: \(cos(y)= \frac{1}{x}\)?

  23. anonymous
    • one year ago
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    No 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?

  24. IrishBoy123
    • one year ago
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    OK, 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 -> \( \pm \infty\),

  25. anonymous
    • one year ago
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    lim x->+-inifinty (arcsec(x)/arccos(1/x) ) = 1 so i believe that makes arccos(1/x) the end behavior model

  26. IrishBoy123
    • one year ago
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    brill! we're done :p

  27. anonymous
    • one year ago
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    thx

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