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lgbasallote Group Title

Find the domain: \[h(x) = \sqrt{4-x} + \sqrt{x^2 - 1}\]

  • 2 years ago
  • 2 years ago

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  1. lgbasallote Group Title
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    i smell a race....

    • 2 years ago
  2. ParthKohli Group Title
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    Hint: you can't use positive numbers more than 4.

    • 2 years ago
  3. lgbasallote Group Title
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    ^i wish you would have let me try first....

    • 2 years ago
  4. lgbasallote Group Title
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    ^@Jonask

    • 2 years ago
  5. lgbasallote Group Title
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    ^and then you did it again @Jonask

    • 2 years ago
  6. Jonask Group Title
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    sorry,like you said its a race

    • 2 years ago
  7. lgbasallote Group Title
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    i hate races.....

    • 2 years ago
  8. swissgirl Group Title
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    Ok just try to see when each square root is negative

    • 2 years ago
  9. lgbasallote Group Title
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    anyway....

    • 2 years ago
  10. lgbasallote Group Title
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    i suppose \(\sqrt{x^2 - 1}\) will have no restrictions?

    • 2 years ago
  11. swissgirl Group Title
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    yes x=0

    • 2 years ago
  12. lgbasallote Group Title
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    hmm oh yeah

    • 2 years ago
  13. lgbasallote Group Title
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    didn't think of that

    • 2 years ago
  14. lgbasallote Group Title
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    then \(\sqrt{4-x}\) would be 4 above?

    • 2 years ago
  15. swissgirl Group Title
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    I would say greater than 4

    • 2 years ago
  16. sauravshakya Group Title
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    @lgbasallote was that the question of 100 level???

    • 2 years ago
  17. lgbasallote Group Title
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    greater than 4...4 above...same shiz

    • 2 years ago
  18. Yahoo! Group Title
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    Take Seperately and Connect it by union (U)

    • 2 years ago
  19. lgbasallote Group Title
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    to be the best, one needs to master the basics @sauravshakya

    • 2 years ago
  20. Jonask Group Title
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    \[f(x)=x^2,g(x)=\sqrt{1-x}\] domain of\[gof,fog\] domain

    • 2 years ago
  21. lgbasallote Group Title
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    ^?

    • 2 years ago
  22. Jonask Group Title
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    just a question that i wont interupt you can try

    • 2 years ago
  23. swissgirl Group Title
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    Domain is just 1 and 0

    • 2 years ago
  24. lgbasallote Group Title
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    g o f would be sqrt(1 - x^2) it would only be non-negative if 0 <= x <= 1

    • 2 years ago
  25. lgbasallote Group Title
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    so the domain is 0<=x<=1

    • 2 years ago
  26. swissgirl Group Title
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    for f(g(x))

    • 2 years ago
  27. lgbasallote Group Title
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    f o g would be 1- x so all real numbers

    • 2 years ago
  28. Yahoo! Group Title
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    \[Domain=(-\infty,0) U (0,4]\]

    • 2 years ago
  29. ParthKohli Group Title
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    0 isn't included in the domain since \(\sqrt{0^2 - 1} = \sqrt{-1}\) and this is a real function.

    • 2 years ago
  30. Yahoo! Group Title
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    Lol...@ParthKohli u see it is a open Bracket.....

    • 2 years ago
  31. lgbasallote Group Title
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    "real" function?

    • 2 years ago
  32. ParthKohli Group Title
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    @Yahoo! I wasn't talking to you.

    • 2 years ago
  33. ParthKohli Group Title
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    Yeah, which has all ranges as real numbers.

    • 2 years ago
  34. Yahoo! Group Title
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    oh.....Sorry...) @ParthKohli

    • 2 years ago
  35. lgbasallote Group Title
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    now you confuse me...you're telling me they're wrong?

    • 2 years ago
  36. ParthKohli Group Title
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    swissgirl Best Response 1 Domain is just 1 and 0 that ^^

    • 2 years ago
  37. lgbasallote Group Title
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    hmm seems one of the answers here is wrong then...wonder which

    • 2 years ago
  38. ParthKohli Group Title
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    Okay, let me ask you a question. Why didn't you include all real numbers in the domain? @lgbasallote

    • 2 years ago
  39. lgbasallote Group Title
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    i did

    • 2 years ago
  40. lgbasallote Group Title
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    when didn't i?

    • 2 years ago
  41. ParthKohli Group Title
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    lgbasallote Best Response 0 i suppose sqrt(x62−1) will have no restrictions? swissgirl Best Response 1 yes x=0 lgbasallote Best Response 0 hmm oh yeah

    • 2 years ago
  42. ParthKohli Group Title
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    0 is included in all real numbers.

    • 2 years ago
  43. lgbasallote Group Title
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    since when was \(\sqrt{-1}\) real @ParthKohli ?

    • 2 years ago
  44. ParthKohli Group Title
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    You answered your own question @lgbasallote

    • 2 years ago
  45. lgbasallote Group Title
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    ....what exactly are you saying?

    • 2 years ago
  46. ParthKohli Group Title
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    ParthKohli 0 0 isn't included in the domain since sqrt(0^2 - 1) = sqrt(-1) and this is a real function. <argument happens> ParthKohli 0 Okay, let me ask you a question. Why didn't you include all real numbers in the domain? @lgbasallote lgbasallote Best Response 0 i did lgbasallote Best Response 0 when didn't i? <some more convo> lgbasallote Best Response 0 since when was √−1 real @ParthKohli ?

    • 2 years ago
  47. lgbasallote Group Title
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    just state your point

    • 2 years ago
  48. ParthKohli Group Title
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    You are contradicting yourself. First, you say that you have included all real numbers in the domain. Then, by "since when was √−1 real @ParthKohli ?," you mean that 0 is not in the domain.

    • 2 years ago
  49. lgbasallote Group Title
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    by the way 1) none of us said 0 is included in the domain (we were talking about restrictions) 2) sqrt -1 is not a real function

    • 2 years ago
  50. Jonask Group Title
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    \[D_g=(-\infty,1),D_f=\mathbb{R}\] \[fog=\left| 1-x \right|\]\[gof=\sqrt{1-x^2}\]

    • 2 years ago
  51. lgbasallote Group Title
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    ^?

    • 2 years ago
  52. ParthKohli Group Title
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    No, real function means that for all domains, the range will be real. We have to get a domain which keeps all ranges real.

    • 2 years ago
  53. lgbasallote Group Title
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    and your point is?

    • 2 years ago
  54. ParthKohli Group Title
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    That 0 is not in the domain.

    • 2 years ago
  55. lgbasallote Group Title
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    and isn't that what we said?

    • 2 years ago
  56. ParthKohli Group Title
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    You said that you included all real numbers in the domain. -.-

    • 2 years ago
  57. lgbasallote Group Title
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    when???

    • 2 years ago
  58. lgbasallote Group Title
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    we were talking about RESTRICTIONS

    • 2 years ago
  59. ParthKohli Group Title
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    ParthKohli 0 Okay, let me ask you a question. Why didn't you include all real numbers in the domain? @lgbasallote lgbasallote i did 10 minutes ago lgbasallote Best Response 0 when didn't i?

    • 2 years ago
  60. ParthKohli Group Title
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    I was talking about the domain then.

    • 2 years ago
  61. Jonask Group Title
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    \[D_{gof}=[-1,1]\] \[Dfog=x \in (-\infty,1]\]

    • 2 years ago
  62. lgbasallote Group Title
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    you really tend to make things big when you misread, don;t you @ParthKohli

    • 2 years ago
  63. Jonask Group Title
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    guys i think we can close the question i did not mean to create a big issue

    • 2 years ago
  64. lgbasallote Group Title
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    it has been closed a long time ago @jonask

    • 2 years ago
  65. Jonask Group Title
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    oh i dint notice,cos commentsn are still running

    • 2 years ago
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