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moongazer

  • 3 years ago

how do you get the domain of this:

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  1. moongazer
    • 3 years ago
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    how do you find it? step-by-step please :)

  2. moongazer
    • 3 years ago
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    √(x^2 - 9x +20)/√(x^2 - 5x + 6) I already know the domain. But, how do you get it?

  3. ganeshie8
    • 3 years ago
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    first workout contraints for squareroot : the thing inside squareroot cannot be negative

  4. ganeshie8
    • 3 years ago
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    next work out constraints for denominator : denominator cannot be equal to 0

  5. ganeshie8
    • 3 years ago
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    consider numerator first : \(\sqrt{x^2 - 9x +20}\)

  6. ganeshie8
    • 3 years ago
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    can u work out domain for that ?

  7. moongazer
    • 3 years ago
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    {x|x>=5} is that right?

  8. ganeshie8
    • 3 years ago
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    somewhat right :) lets see how to solve

  9. ganeshie8
    • 3 years ago
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    x^2-9x+20 >= 0 x^2-4x-5x+20 >=0 x(x-4) -5(x-4) >= 0 (x-4)(x-5) >= 0 x <= 4 or x >= 5

  10. ganeshie8
    • 3 years ago
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    thats one constraint for numerator, lets find out the constraints for denominator

  11. moongazer
    • 3 years ago
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    ohhhh

  12. moongazer
    • 3 years ago
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    I think I am starting to understand it :)

  13. ganeshie8
    • 3 years ago
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    good :) since, denominator is also under radical, it must be >=0 : x^2 - 5x + 6 >= 0 x^2-3x-2x + 6 >= 0 x(x-3) -2(x-3) >= 0 (x-2)(x-3) >= 0 x <= 2 or x >= 3

  14. ganeshie8
    • 3 years ago
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    with those two things, we finished with the first step of finding constraints for radicals

  15. ganeshie8
    • 3 years ago
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    so far we have this : x <= 4 or x >= 5 x <= 2 or x >= 3

  16. moongazer
    • 3 years ago
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    yup, I understand it :)

  17. ganeshie8
    • 3 years ago
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    next one is denominator can never equal to 0. so, x^2 - 5x + 6 \(\ne\) 0 x^2-3x-2x + 6 \(\ne\) 0 x(x-3) -2(x-3) \(\ne\) 0 (x-2)(x-3) \(\ne\) 0 x \(\ne\) 2 or x \(\ne\) 3

  18. moongazer
    • 3 years ago
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    w8 Isn't it that it should be: x >= 4 or x >= 5 x >= 2 or x >= 3 ???

  19. ganeshie8
    • 3 years ago
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    lets combine all constraints and make a meaningful constraint

  20. ganeshie8
    • 3 years ago
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    which one ?

  21. moongazer
    • 3 years ago
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    (x-4)(x-5) >= 0 (x-2)(x-3) >= 0

  22. ganeshie8
    • 3 years ago
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    oh i got ur question, il give quick explanation

  23. ganeshie8
    • 3 years ago
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    you comfortable with parabola graph ?

  24. moongazer
    • 3 years ago
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    I think so

  25. ganeshie8
    • 3 years ago
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    il show u in graph why its x <=4 , x >=5

  26. moongazer
    • 3 years ago
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    I know few things about parabola

  27. moongazer
    • 3 years ago
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    ok

  28. ganeshie8
    • 3 years ago
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    im sketching this parabola : (x-4)(x-5) >= 0

  29. ganeshie8
    • 3 years ago
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    it intersect x axis at 4, 5 right ?

  30. ganeshie8
    • 3 years ago
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    |dw:1348496600444:dw|

  31. ganeshie8
    • 3 years ago
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    between 4 and 5, it is sinking down into x axis eh ?

  32. ganeshie8
    • 3 years ago
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    its becoming NEGATIVE

  33. ganeshie8
    • 3 years ago
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    |dw:1348496705657:dw|

  34. moongazer
    • 3 years ago
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    why did it intersect in x axis at 4, 5?

  35. ganeshie8
    • 3 years ago
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    when u factored you got : (x-4)(x-5)

  36. ganeshie8
    • 3 years ago
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    that means it intersects x-axis at 4 and 5

  37. moongazer
    • 3 years ago
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    ohh, ok what if it says (x-4)(x+5) does it intersect at 4 and -5 ?

  38. ganeshie8
    • 3 years ago
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    thats right, when it says, (x-4)(x+5) = 0 then it intersects at 4 and -5

  39. ganeshie8
    • 3 years ago
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    cuz for it to equal to 0, one of both of the factors must equal to 0

  40. moongazer
    • 3 years ago
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    |dw:1348498226449:dw|

  41. moongazer
    • 3 years ago
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    so the graph is like that?

  42. ganeshie8
    • 3 years ago
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    perfect, thats (x-4)(x+5)

  43. moongazer
    • 3 years ago
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    and the domain is x>=4,x<=-5 is it correct?

  44. ganeshie8
    • 3 years ago
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    thats right ! you became expert ;p

  45. moongazer
    • 3 years ago
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    for (x-4)(x+5)

  46. moongazer
    • 3 years ago
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    Thanks! I know understand it. :) let's now continue with our previous discussion :)

  47. ganeshie8
    • 3 years ago
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    great ! so far what we have

  48. moongazer
    • 3 years ago
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    next one is denominator can never equal to 0. so, x^2 - 5x + 6 ≠ 0 x^2-3x-2x + 6 ≠ 0 x(x-3) -2(x-3) ≠ 0 (x-2)(x-3) ≠ 0 x ≠ 2 or x ≠ 3

  49. ganeshie8
    • 3 years ago
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    yeah we almost done

  50. ganeshie8
    • 3 years ago
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    lets put down all the 3 constraints we got in one place :

  51. ganeshie8
    • 3 years ago
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    1) numerator radical : x <= 4 or x >= 5 2) denominator radical : x <= 2 or x >= 3 3) denominator cant be 0 : x ≠ 2 or x ≠ 3

  52. moongazer
    • 3 years ago
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    I understand it. What's next?

  53. ganeshie8
    • 3 years ago
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    see 2) and 3)

  54. ganeshie8
    • 3 years ago
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    can we combine them as one, like this : 2) 3) : x < 2 or x > 3

  55. moongazer
    • 3 years ago
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    I'll be back just continue.

  56. moongazer
    • 3 years ago
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    someone is calling me

  57. ganeshie8
    • 3 years ago
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    ok after combining 2) and 3) we are left with : 1) numerator radical : x <= 4 or x >= 5 2) x < 2 or x > 3 this is the last step. see if u can figure out

  58. ganeshie8
    • 3 years ago
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    out of x < 2 & x <= 4, x < 2 is more restrictive, so we have this in our domain : \(\color{green}{x < 2}\)

  59. ganeshie8
    • 3 years ago
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    out of x > 3 & x <= 4 x > 3 , x <= 4 , which is same as : \(\color{green}{3<x\le 4}\)

  60. ganeshie8
    • 3 years ago
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    and of course the last one, \(\color{green}{x >= 5}\)

  61. moongazer
    • 3 years ago
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    sorry for being away, I'm here again :)

  62. ganeshie8
    • 3 years ago
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    np :) see if above stuff makes sense

  63. moongazer
    • 3 years ago
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    it makes sense but why in the final answer they used "or" not "and" "," "then" or anything else dom K: {x|x<2 or 3<x<=4 or x>=5}

  64. ganeshie8
    • 3 years ago
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    yeah we always combine domain with "or" why ? cuz its meaningful. let me ask u a q

  65. ganeshie8
    • 3 years ago
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    suppose we write : x < 2 and x >=5

  66. moongazer
    • 3 years ago
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    for the domain it is ALWAYS "or" ?

  67. ganeshie8
    • 3 years ago
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    is that possible ? can x be BOTH less than 2 and greater than 5 at the SAME time ha ?

  68. ganeshie8
    • 3 years ago
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    yes always "or"

  69. moongazer
    • 3 years ago
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    yes, it is not possible

  70. ganeshie8
    • 3 years ago
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    is "1" less than 2 and greater than "5" ?

  71. ganeshie8
    • 3 years ago
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    yeah, but if we put it like this : either "1" is less than 2 or greater than "5"

  72. ganeshie8
    • 3 years ago
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    how meaningful it is, so while phrasing domain also, we use commonsense aswell :)

  73. moongazer
    • 3 years ago
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    yes, Thank you very much for helping me :)

  74. ganeshie8
    • 3 years ago
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    np i learned as well. i forgot these long back in college... i enjoyed going thru these again.. thnks you too :)

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