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Dodo1 Group Title

Please help me. Domain of sqrt ((x-8)(x+4)) thanks

  • one year ago
  • one year ago

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  1. mikoza Group Title
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    is (-infinity ,-4]U[8,+infinity)

    • one year ago
  2. Dodo1 Group Title
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    i entered it and it told me that it was wrong

    • one year ago
  3. davisla Group Title
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    Answers without solutions don't help understanding so then you can't do the next one or the one on your exam. The value we are taking the square root of must be greater than or equal to 0, so in the expression \[\sqrt{(x-8)(x+4}\] the radicand is (x-8)(x+4) set up the inequality \[(x-8)(x+4)\ge0\]

    • one year ago
  4. davisla Group Title
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    Now, the only way we can go from positive to negative is to pass through 0 and the only way we can go from negative to positive is to pass through 0, so find the zeros. For what values of x is (x - 4)(x+8) = 0?

    • one year ago
  5. mikoza Group Title
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    dear davisla (x-4)(x+8)is not equal to (x-8)(x+4)

    • one year ago
  6. davisla Group Title
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    you are correct ... so for what values of x is (x - 8)(x + 4) = 0?

    • one year ago
  7. Dodo1 Group Title
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    I thought that X=-8. and X=+4. I am not sure how the infinity blacket works

    • one year ago
  8. Dodo1 Group Title
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    sorry I mean X=+8 and X=-4

    • one year ago
  9. davisla Group Title
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    once you find the zeros, you know where it COULD change signs. That does not mean it does. So, from neg infinity to -4 it NEVER changes signs, from -4 to 8 it NEVER changes signs and from 8 to infinity it NEVER changes signs Plug in a number in each of those 3 intervals to determine the sign of the expression in that interval. Try -5: (-5 - 4)(-5 + 4) = -9 (-1) which is +9 ... but all we care about is that it is positive. So in the interval (-infinity, -4) it is positive - that is okay so that is part of the domain. Now try a number in the second interval and see if it is positive or negative.

    • one year ago
  10. Dodo1 Group Title
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    so the second interval should be (8,infinity)?

    • one year ago
  11. davisla Group Title
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    the second is (-4, 8) the third is (8, infinity)

    • one year ago
  12. davisla Group Title
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    in the second interval, I would choose 0 to plug in since it is easy

    • one year ago
  13. Dodo1 Group Title
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    Ok so, the answer should be (-infinity,-4)U(-4,0)U(0,8)U(8,infinity)?

    • one year ago
  14. davisla Group Title
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    no, when you plug 0 in you get: (0-8)(0+4) = -8(4) and that is a negative number - so that interval is no good and is not in your domain what about the last interval? do you get a positive or a negative?

    • one year ago
  15. Dodo1 Group Title
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    I got positive.

    • one year ago
  16. davisla Group Title
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    so that interval is in our domain ... now, what about -4 and 8? Can the expression = 0?

    • one year ago
  17. davisla Group Title
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    we can take the sq root of zero, so we do want to include those endpoints in our intervals

    • one year ago
  18. davisla Group Title
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    we wind up with the intervals given by someone else in the beginning and that is correct. \[(-\infty,-4]\cup[8,\infty)\]

    • one year ago
  19. davisla Group Title
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    I am wondering if it didn't like the way you input ... did you put a + in front of the second infinity?

    • one year ago
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