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mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2this one is a little bit hard to get...at least for me but i'll do it with \[x=\frac{2}{\cos \theta}\]with \(0<\theta<\pi\)

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2because i know domain is x>2

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1looks like there a hole.

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2it becomes\[y=4\frac{\frac{1}{\cos \theta}+1}{\tan \theta}\]

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2range is\[(2,0)\cup (2,\infty)\]

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2yeah \(\cos \theta\) do u know why ?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1oh ... i didn't realize \[ \frac{2 ( \sqrt{x+ 2})^2}{\sqrt{x+2}\sqrt{x2}}= \frac{2 \sqrt{x+2}}{\sqrt{x2}}\] there isn't asymptotic behaviour at x=2

Study23
 2 years ago
Best ResponseYou've already chosen the best response.0? I don't get the whole cos thing... and where did the 2 come from?

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2wait forgot cos thing...

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1check for these behaviour on these intervals (inf, 2), (2, 2), (2, inf)

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1the  infinity part is always confusing, don't cancel .... for that. rest is just same as you did.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1the  infinity part is always confusing, don't cancel to for that. rest is just same as you did.
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