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anonymous

  • one year ago

Find the domain of the function. g(x)=^4sq.rt x^2+2x

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  1. anonymous
    • one year ago
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    yep that's what it's supposed to be

  2. IrishBoy123
    • one year ago
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    scan or link the question pls

  3. IrishBoy123
    • one year ago
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    sorry, been here done that! just scan or link it :p

  4. anonymous
    • one year ago
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    the one you typed is the correct one

  5. IrishBoy123
    • one year ago
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    scan or link your question no point is answering to ghosts

  6. anonymous
    • one year ago
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    i don't know how to scan it and the link wont work since it's under my account name.. you already know what the problem looks like, you sent it yourself, so it should be good enough?

  7. anonymous
    • one year ago
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    \[\sqrt[4]{x^2+2x}\]

  8. anonymous
    • one year ago
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    ?

  9. freckles
    • one year ago
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    what happens if you have this\[\sqrt[4]{-1}\]? does this exist over the real numbers?

  10. anonymous
    • one year ago
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    I don't think so since you shouldnt have -'s under a radical

  11. freckles
    • one year ago
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    great so you want x^2+2x to be positive or zero right?

  12. anonymous
    • one year ago
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    yeah

  13. freckles
    • one year ago
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    for a number to be positive or equal it must be greater than or equal to 0

  14. freckles
    • one year ago
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    so you basically need to solve x^2+2x>=0

  15. freckles
    • one year ago
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    to find the domain

  16. anonymous
    • one year ago
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    do we just get rid of the radical then?

  17. freckles
    • one year ago
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    you told me just a sec ago that we didn't want the thing under the radical to be negative which means we want it to be zero or positive

  18. freckles
    • one year ago
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    the thing under the radical was x^2+2x

  19. freckles
    • one year ago
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    so you want x^2+2x>=0

  20. anonymous
    • one year ago
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    it has to be in interval notation

  21. anonymous
    • one year ago
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    so would I start with (0,_)

  22. freckles
    • one year ago
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    you would solve x^2+2x>=0

  23. freckles
    • one year ago
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    then we can put an interval notation

  24. anonymous
    • one year ago
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    x^2>or= -2x

  25. anonymous
    • one year ago
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    and then would we square both sides?? because then there'd be a negative on the right

  26. freckles
    • one year ago
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    \[x^2+2x \ge 0 \\ x(x+2) \ge 0\] draw a number line x(x+2) is zero when x=0 or x+2=0 |dw:1441829182423:dw| test the intervals

  27. freckles
    • one year ago
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    |dw:1441829210139:dw|

  28. anonymous
    • one year ago
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    OHH so x can equal 0 or -2?

  29. anonymous
    • one year ago
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    can;t*

  30. freckles
    • one year ago
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    let me know when you have tested the intervals and yes x=-2 and x=0 will be included in the solution for the domain but we have to also consider the numbers around them

  31. anonymous
    • one year ago
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    3,-1, and 3?

  32. freckles
    • one year ago
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    |dw:1441829448997:dw|

  33. freckles
    • one year ago
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    so this tells us x^2+2x is positive on the first and last intervals

  34. freckles
    • one year ago
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    in this number line

  35. freckles
    • one year ago
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    |dw:1441829505021:dw|

  36. anonymous
    • one year ago
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    (-infinity,-2)U(-2,0)U(-2,0)U(0,infinity)?

  37. freckles
    • one year ago
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    you need to include both -2 and 0... also why are you include (-2,0) x^2+2x is negative on (-2,0)

  38. freckles
    • one year ago
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    remember we were trying to find x such that x^2+2x was positive or zero

  39. freckles
    • one year ago
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    we found x^2+2x was zero at x=-2, x=0 we found x^2+2x was positive for values to the left of -2, also values to right of 0

  40. freckles
    • one year ago
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    you should not include (-2,0) do you understand why? you should include -2 and 0 in your solution do you understand this also?

  41. anonymous
    • one year ago
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    no I don't :-(

  42. freckles
    • one year ago
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    hmm... so do you remember when we tested the intervals earlier and for the member that represented the set of numbers on (-2,0) we got a negative number?

  43. freckles
    • one year ago
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    |dw:1441829916519:dw|

  44. freckles
    • one year ago
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    |dw:1441829938691:dw| I put check marks on the intervals we want because we were looking for intervals where x^2+2x was positive I put an X mark on intervals we don't want because we were not looking for when x^2+2x was negative.

  45. freckles
    • one year ago
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    the 4 root of 0 exists and it is 0 so x^2+2x=0 will satisfy our domain x(x+2)=0 when x=0 and x=-2 these numbers have to be a part of the domain

  46. freckles
    • one year ago
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    |dw:1441830041033:dw|

  47. anonymous
    • one year ago
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    oh, so we dont include -2,0 because it would result in a negative

  48. freckles
    • one year ago
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    no we don't include (-2,0) because any number from that set would result in a negative we do include -2 and 0 though because pluggin in those numbers result in zero

  49. anonymous
    • one year ago
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    okay so we include those but not (-2,0)

  50. freckles
    • one year ago
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    yes any number from (-2,0) when pluggin in to x^2+2x will give you a negative number you have already said the 4th root of a negative number doesn't exist over the real numbers

  51. anonymous
    • one year ago
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    so how would I write that in interval notationn without including (-2,0)

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