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Study23

  • 2 years ago

Range on this function!!!!!

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  1. Study23
    • 2 years ago
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    |dw:1347947263239:dw|

  2. mukushla
    • 2 years ago
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    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 \(0<\theta<\pi\)

  3. mukushla
    • 2 years ago
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    because i know domain is |x|>2

  4. experimentX
    • 2 years ago
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    is it -2 to inf?

  5. experimentX
    • 2 years ago
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    looks like there a hole.

  6. mukushla
    • 2 years ago
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    it becomes\[y=4\frac{\frac{1}{\cos \theta}+1}{|\tan \theta|}\]

  7. Study23
    • 2 years ago
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    \(\ cos \theta? \)

  8. mukushla
    • 2 years ago
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    range is\[(-2,0)\cup (2,\infty)\]

  9. mukushla
    • 2 years ago
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    yeah \(\cos \theta\) do u know why ?

  10. Study23
    • 2 years ago
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    No, I don't

  11. experimentX
    • 2 years ago
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    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

  12. Study23
    • 2 years ago
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    ? I don't get the whole cos thing... and where did the 2 come from?

  13. mukushla
    • 2 years ago
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    wait forgot cos thing...

  14. mukushla
    • 2 years ago
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    exper got it

  15. experimentX
    • 2 years ago
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    check for these behaviour on these intervals (-inf, -2), (-2, 2), (2, inf)

  16. Study23
    • 2 years ago
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    ? How?

  17. experimentX
    • 2 years ago
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    the - infinity part is always confusing, don't cancel .... for that. rest is just same as you did.

  18. experimentX
    • 2 years ago
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    the - infinity part is always confusing, don't cancel to for that. rest is just same as you did.

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