anonymous
  • anonymous
v
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

rainbow_rocks03
  • rainbow_rocks03
question?
anonymous
  • 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
  • anonymous
I thought right/left end behaviors were only for polynomials

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

IrishBoy123
  • IrishBoy123
re-write y in terms of x by switching the sec to cos then it should look more intuitive
anonymous
  • anonymous
arccos(1/x)= x= 1/cos(y)
IrishBoy123
  • IrishBoy123
x = sec(y) etc
anonymous
  • anonymous
Still dont see how this results in an end model... :(
tkhunny
  • tkhunny
|dw:1433356700597:dw|
anonymous
  • anonymous
|dw:1433356872213:dw|
anonymous
  • anonymous
somethung like that (cant draw with a mouse)
IrishBoy123
  • IrishBoy123
good x= 1/cos(y) then get y = f(x) and look at: \(\lim_{x -> \pm \infty}\)
anonymous
  • anonymous
.... is it positive and negative infinity
anonymous
  • anonymous
nvm
anonymous
  • anonymous
pi/2?
IrishBoy123
  • IrishBoy123
if x= 1/cos(y) and y = f(x) then f(x) = ???
anonymous
  • anonymous
1/arccos(x) ?
IrishBoy123
  • IrishBoy123
mmm x= 1/cos(y) cos(y) = ??
anonymous
  • anonymous
1/x
anonymous
  • anonymous
arccos(1/x)
anonymous
  • anonymous
The course gave me the formula arccos(1/x)=arcsec(x) ... does that qualify?
anonymous
  • anonymous
It would make sense... given the lesson didnt talk about end behaviors at all
IrishBoy123
  • 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
  • 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
  • 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
  • anonymous
lim x->+-inifinty (arcsec(x)/arccos(1/x) ) = 1 so i believe that makes arccos(1/x) the end behavior model
IrishBoy123
  • IrishBoy123
brill! we're done :p
anonymous
  • anonymous
thx

Looking for something else?

Not the answer you are looking for? Search for more explanations.