- anonymous

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- katieb

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- rainbow_rocks03

question?

- anonymous

C is easy, simply find where the derivative is zero, But i cannot for the life of me figure out a and b.

- anonymous

I thought right/left end behaviors were only for polynomials

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## More answers

- IrishBoy123

re-write y in terms of x by switching the sec to cos
then it should look more intuitive

- anonymous

arccos(1/x)= x= 1/cos(y)

- IrishBoy123

x = sec(y) etc

- anonymous

Still dont see how this results in an end model... :(

- tkhunny

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- anonymous

|dw:1433356872213:dw|

- anonymous

somethung like that (cant draw with a mouse)

- IrishBoy123

good
x= 1/cos(y)
then get y = f(x)
and look at: \(\lim_{x -> \pm \infty}\)

- anonymous

.... is it positive and negative infinity

- anonymous

nvm

- anonymous

pi/2?

- IrishBoy123

if
x= 1/cos(y)
and y = f(x)
then
f(x) = ???

- anonymous

1/arccos(x) ?

- IrishBoy123

mmm
x= 1/cos(y)
cos(y) = ??

- anonymous

1/x

- anonymous

arccos(1/x)

- anonymous

The course gave me the formula arccos(1/x)=arcsec(x) ... does that qualify?

- anonymous

It would make sense... given the lesson didnt talk about end behaviors at all

- IrishBoy123

"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

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?

- IrishBoy123

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 -> \( \pm \infty\),

- anonymous

lim x->+-inifinty (arcsec(x)/arccos(1/x) ) = 1 so i believe that makes arccos(1/x) the end behavior model

- IrishBoy123

brill!
we're done :p

- anonymous

thx

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