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lgbasallote

  • 2 years ago

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

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  1. lgbasallote
    • 2 years ago
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    i smell a race....

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

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

  4. lgbasallote
    • 2 years ago
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    ^@Jonask

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

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

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

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

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

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

  11. swissgirl
    • 2 years ago
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    yes x=0

  12. lgbasallote
    • 2 years ago
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    hmm oh yeah

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

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

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

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

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

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

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

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

  21. lgbasallote
    • 2 years ago
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    ^?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  39. lgbasallote
    • 2 years ago
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    i did

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

  41. ParthKohli
    • 2 years ago
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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

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

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

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

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

  46. ParthKohli
    • 2 years ago
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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 ?

  47. lgbasallote
    • 2 years ago
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    just state your point

  48. ParthKohli
    • 2 years ago
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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.

  49. lgbasallote
    • 2 years ago
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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

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

  51. lgbasallote
    • 2 years ago
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    ^?

  52. ParthKohli
    • 2 years ago
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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.

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

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

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

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

  57. lgbasallote
    • 2 years ago
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    when???

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

  59. ParthKohli
    • 2 years ago
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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?

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

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

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

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

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

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

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