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monroe17

Given that f(x)=sqrt{2 + x} and g(x)=sqrt{2 - x}, find formulas for the following functions, and their domains. In each case, enter the domain using interval notation. (a) f+g= and its domain is

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  1. AndreaYoung1
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    Have you searched the web? I havn't studied this yet /:

    • one year ago
  2. monroe17
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    I have.. I can't find anything similar :/

    • one year ago
  3. monroe17
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    @qpHalcy0n can you help me?

    • one year ago
  4. monroe17
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    @jim_thompson5910 can you please help? :)

    • one year ago
  5. jim_thompson5910
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    f(x) + g(x) literally means "add the functions f(x) and g(x)"

    • one year ago
  6. jim_thompson5910
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    but since f(x) = sqrt(2+x) and g(x) = sqrt(2-x), we know that f(x) + g(x) = sqrt(2+x) + sqrt(2-x)

    • one year ago
  7. monroe17
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    that's what I got.. But, I wasn't sure if f(x) + g(x) = sqrt(2+x) + sqrt(2-x) was correct or not.. how would I find the domain?

    • one year ago
  8. jim_thompson5910
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    yeah it seems too simple that it looks like a trick question, but it's not

    • one year ago
  9. jim_thompson5910
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    the domain is the set of allowable inputs, or x values

    • one year ago
  10. jim_thompson5910
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    remember that you can't take the square root of a negative number, so 2+x >= 0 and 2-x >= 0

    • one year ago
  11. jim_thompson5910
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    those two inequalities become x >= -2 and x <= 2 which combine to -2 <= x <= 2

    • one year ago
  12. monroe17
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    so, in interval notation.. it would be (-inf,-2)U(2,inf) ?

    • one year ago
  13. jim_thompson5910
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    no, [-2, 2]

    • one year ago
  14. jim_thompson5910
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    that represents \(\Large -2 \le x \le 2\)

    • one year ago
  15. monroe17
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    oh okay... and would this be correct? f(x) - g(x) = sqrt(2+x) - sqrt(2-x)

    • one year ago
  16. jim_thompson5910
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    yes that's 100% correct

    • one year ago
  17. jim_thompson5910
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    the domain is the same because the radicands (the stuff in the square roots) are the same

    • one year ago
  18. monroe17
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    f(x) * g(x) = sqrt(2+x) * sqrt(2-x) ?

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  19. jim_thompson5910
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    yes

    • one year ago
  20. jim_thompson5910
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    domains are the same (for the same reason explained above)

    • one year ago
  21. jim_thompson5910
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    domain is*

    • one year ago
  22. monroe17
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    f(x)/g(x) = sqrt(2+x)/sqrt(2-x) and same domain?

    • one year ago
  23. jim_thompson5910
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    the first part is correct, but the domain will be slightly different

    • one year ago
  24. jim_thompson5910
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    now we have to avoid dividing by zero, so 2-x can't be zero 2-x = 0 x = 2 so if x = 2, then 2-x is zero. So we toss out x = 2 from the domain So the domain is now [-2, 2)

    • one year ago
  25. monroe17
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    so now what is the difference between f+g and fog(x)?

    • one year ago
  26. jim_thompson5910
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    f+g means "add f and g" fog(x) means "start with f(x), and plug in g(x) as the input"

    • one year ago
  27. jim_thompson5910
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    fog(x) is the same as f(g(x))

    • one year ago
  28. jim_thompson5910
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    \[\Large f(x) = \sqrt{2+x}\] \[\Large f(g(x)) = \sqrt{2+g(x)}\] \[\Large f(g(x)) = \sqrt{2+\sqrt{2-x}}\]

    • one year ago
  29. monroe17
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    well it's with different numbers, I just wanted to know the difference.

    • one year ago
  30. jim_thompson5910
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    i gotcha, do you see how I'm getting that?

    • one year ago
  31. monroe17
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    f(g(x))=2+sqrt(2-x)?

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  32. jim_thompson5910
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    f(x) = sqrt(2+x) and g(x) = sqrt(2-x) correct?

    • one year ago
  33. monroe17
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    yes.

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  34. jim_thompson5910
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    Then \[\Large f(g(x)) = \sqrt{2+\sqrt{2-x}}\] or \[\Large f\circ g(x) = \sqrt{2+\sqrt{2-x}}\]

    • one year ago
  35. monroe17
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    can you draw that? cause its not showing up correct.,. and im a lil confused

    • one year ago
  36. jim_thompson5910
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    |dw:1346027228896:dw|

    • one year ago
  37. jim_thompson5910
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    oh my bad, there should be a closing parenthesis after g(x)

    • one year ago
  38. jim_thompson5910
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    but everything else is correct

    • one year ago
  39. monroe17
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    okay so if f(x)=1/x and g(x)= 9x+1 fog(x) would be... 1/(9x+1)?

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  40. jim_thompson5910
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    yes perfect

    • one year ago
  41. monroe17
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    and its domain would be all real numbers except ...?

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  42. jim_thompson5910
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    what makes 9x+1 equal to zero?

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  43. monroe17
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    anything less than zero?

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  44. jim_thompson5910
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    9x +1 = 0 x = ???

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  45. monroe17
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    -1/9

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  46. jim_thompson5910
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    good, so x can be any number but -1/9

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  47. monroe17
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    [-1/9,inf] ?

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  48. jim_thompson5910
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    (-inf, -1/9) U (-1/9, inf)

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  49. jim_thompson5910
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    start with (-inf, inf), which is the entire number line then poke a hole at -1/9 to get (-inf, -1/9) U (-1/9, inf)

    • one year ago
  50. monroe17
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    for the domain it it says.. its domain is all real numbers except __ so would I just put in -1/9?

    • one year ago
  51. jim_thompson5910
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    yes

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  52. jim_thompson5910
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    that's in a more understandable way (in my opinion)

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  53. jim_thompson5910
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    interval notation can be a little tricky to grasp

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  54. monroe17
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    haha ya! gof(x) would be 9(1/x)+1?

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  55. jim_thompson5910
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    yes, which becomes 9/x + 1

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  56. jim_thompson5910
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    the 1 isn't part of the fraction

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  57. monroe17
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    then to find the domain would I do x+1=0 x=-1?

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  58. jim_thompson5910
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    no, the denominator is just x

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  59. jim_thompson5910
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    \[\Large f \circ g(x) = \frac{9}{x}+1\]

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  60. monroe17
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    oh its (9/x)+1?

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  61. jim_thompson5910
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    yes

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  62. monroe17
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    so how would you do that domain then? would it be.. x cannot be zero?

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  63. jim_thompson5910
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    yes, x can be any number but 0

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  64. monroe17
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    would fof(x) be 1/(1/x).. so 1?

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  65. jim_thompson5910
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    1/(1/x) = (1/1)/(1/x) 1/(1/x) = (1/1)*(x/1) 1/(1/x) = x/1 1/(1/x) = x

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  66. jim_thompson5910
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    So 1/(1/x) simplifies to x

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  67. jim_thompson5910
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    We have to add the restriction that x can't be zero to make sure that the two expressions are completely equivalent

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  68. monroe17
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    fof(x)=x and its domain is all real numbers except 0?

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  69. jim_thompson5910
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    yes

    • one year ago
  70. jim_thompson5910
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    you nailed it

    • one year ago
  71. monroe17
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    gog(x)=9(9x+1)+1=81x+9+1=81x+10?

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  72. jim_thompson5910
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    perfect

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  73. jim_thompson5910
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    domain: all real numbers

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  74. jim_thompson5910
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    since the result is a polynomial

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  75. monroe17
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    it says... its domain is (__,__)

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  76. monroe17
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    -inf, inf?

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  77. jim_thompson5910
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    yes it is

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  78. monroe17
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    yyayyy! thank you!

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  79. jim_thompson5910
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    you're welcome

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  80. monroe17
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    one more thing :/ ?

    • one year ago
  81. jim_thompson5910
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    what's that

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  82. monroe17
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    How do I find the corresponding function values. 1. f(g(-2)) 2. f(g(0))

    • one year ago
  83. jim_thompson5910
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    1) what is g(-2)??

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  84. monroe17
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    that's the point at (2,-2)?

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  85. jim_thompson5910
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    look at g(x) and not f(x)

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  86. monroe17
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    (-2,2)?

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  87. jim_thompson5910
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    you are correct, the point (-2,2) is on g(x)

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  88. jim_thompson5910
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    so f(g(-2)) = f(2)

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  89. monroe17
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    so what would be the corresponding function value? how do I put that together?

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  90. monroe17
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    cause I tried f(2) and f(1) for the next one and they're both wrong.

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  91. jim_thompson5910
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    notice how (2,-2) is on f(x)

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  92. monroe17
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    yes

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  93. jim_thompson5910
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    so f(2) = -2

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  94. jim_thompson5910
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    which means overall f(g(-2)) = -2

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  95. jim_thompson5910
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    this example may be a bit confusing with all the 2's...

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  96. jim_thompson5910
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    but the idea is you start with g(x), plug in x = -2 then you get some result, which you plug into f(x) to get your answer

    • one year ago
  97. monroe17
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    so for the second one.. f(g(0)) (0,1)?

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  98. jim_thompson5910
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    close g(0) = 1 f(g(0)) = f(1) f(g(0)) = -1

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  99. monroe17
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    all right thank you!

    • one year ago
  100. jim_thompson5910
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    sure thing

    • one year ago
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