Range on this function!!!!!

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Range on this function!!!!!

Mathematics
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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.

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|dw:1347947263239:dw|
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 \(0<\theta<\pi\)
because i know domain is |x|>2

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Other answers:

is it -2 to inf?
looks like there a hole.
it becomes\[y=4\frac{\frac{1}{\cos \theta}+1}{|\tan \theta|}\]
\(\ cos \theta? \)
range is\[(-2,0)\cup (2,\infty)\]
yeah \(\cos \theta\) do u know why ?
No, I don't
oh ... i didn't realize \[ \frac{2 ( \sqrt{x+ 2})^2}{\sqrt{x+2}\sqrt{x-2}}= \frac{2 \sqrt{x+2}}{\sqrt{x-2}}\] there isn't asymptotic behaviour at x=-2
? I don't get the whole cos thing... and where did the 2 come from?
wait forgot cos thing...
exper got it
check for these behaviour on these intervals (-inf, -2), (-2, 2), (2, inf)
? How?
the - infinity part is always confusing, don't cancel .... for that. rest is just same as you did.
the - infinity part is always confusing, don't cancel to for that. rest is just same as you did.

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