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

mukushlaBest ResponseYou've already chosen the best response.2
because i know domain is x>2
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
looks like there a hole.
 one year ago

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

mukushlaBest ResponseYou've already chosen the best response.2
range is\[(2,0)\cup (2,\infty)\]
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
yeah \(\cos \theta\) do u know why ?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
oh ... 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
 one year ago

Study23Best ResponseYou've already chosen the best response.0
? I don't get the whole cos thing... and where did the 2 come from?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
wait forgot cos thing...
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
check for these behaviour on these intervals (inf, 2), (2, 2), (2, inf)
 one year ago

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

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