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Theloshua

  • one year ago

Find the domain of the following functions

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  1. Theloshua
    • one year ago
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    \[f(x)=\frac{ \sqrt{x+2} }{ x-5 }\]

  2. Theloshua
    • one year ago
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    \[g(x)=\frac{ x-3 }{ \sqrt{x-1}}\]

  3. jim_thompson5910
    • one year ago
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    how far did you get with any of these?

  4. Theloshua
    • one year ago
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    not far at all

  5. Theloshua
    • one year ago
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    like i need helpppp i haves so much work to do and so lil time, i dont wanna be up all nite...

  6. jim_thompson5910
    • one year ago
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    well first off, hopefully you see that x-5 in the denominator cannot be zero, so x-5 = 0 leads to x = 5 if x = 5, then the denominator x-5 is 0 that means x = 5 is NOT part of the domain

  7. jim_thompson5910
    • one year ago
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    agreed so far?

  8. Theloshua
    • one year ago
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    yes

  9. jim_thompson5910
    • one year ago
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    in the numerator of f(x), we have a radical you cannot take the square root of a negative number, so x+2 cannot be negative x+2 is either 0 or positive solve \[\Large x+2 \ge 0\] for x. Tell me what you get

  10. Theloshua
    • one year ago
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    x greater or equal to -2

  11. jim_thompson5910
    • one year ago
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    yes

  12. Theloshua
    • one year ago
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    oh wait, i already solved for this one. i need help with the second

  13. jim_thompson5910
    • one year ago
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    so x cannot be 5 AND x has to be greater than or equal to -2

  14. jim_thompson5910
    • one year ago
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    same idea with g(x) you cannot divide by 0 and you cannot take the square root of a negative number

  15. jim_thompson5910
    • one year ago
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    what makes the denominator of g(x) equal to 0 ?

  16. Theloshua
    • one year ago
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    3 cant be an option

  17. jim_thompson5910
    • one year ago
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    why not?

  18. Theloshua
    • one year ago
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    because the top

  19. Theloshua
    • one year ago
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    you subtract -3

  20. Theloshua
    • one year ago
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    i mean add

  21. Theloshua
    • one year ago
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    right?

  22. jim_thompson5910
    • one year ago
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    you can have 0 in the numerator though

  23. jim_thompson5910
    • one year ago
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    what do you get when you solve sqrt(x-1) = 0 for x?

  24. Theloshua
    • one year ago
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    1

  25. jim_thompson5910
    • one year ago
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    so if x = 1, then the whole denominator turns to 0 which is why x = 1 is NOT part of the domain

  26. jim_thompson5910
    • one year ago
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    also, you need to ensure that x-1 is never negative

  27. Theloshua
    • one year ago
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    okay

  28. Theloshua
    • one year ago
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    and the other part?

  29. jim_thompson5910
    • one year ago
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    solve x-1 >= 0 for x to get what?

  30. Theloshua
    • one year ago
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    1

  31. jim_thompson5910
    • one year ago
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    x = 1 ? or x <= 1 ? or x >= 1 ?

  32. Theloshua
    • one year ago
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    x greater than 1

  33. jim_thompson5910
    • one year ago
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    greater than or equal to 1, yes

  34. jim_thompson5910
    • one year ago
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    oh wait, nvm the "or equal to" part

  35. jim_thompson5910
    • one year ago
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    x cannot equal 1 because it leads to a division by zero error

  36. Theloshua
    • one year ago
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    okay

  37. Theloshua
    • one year ago
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    so? explaiiiin im confused tell me how to solve

  38. jim_thompson5910
    • one year ago
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    you just said it: the domain is x > 1

  39. jim_thompson5910
    • one year ago
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    any number larger than 1 will work as an input if x = 1 or smaller, then you run into problems

  40. Theloshua
    • one year ago
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    okay so the answer is x is not one and x is greater than 1?

  41. jim_thompson5910
    • one year ago
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    saying x > 1 implies that x is not equal to 1

  42. jim_thompson5910
    • one year ago
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    you don't need to say them both if you just say x > 1

  43. Theloshua
    • one year ago
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    okay, so thats the answer tho

  44. jim_thompson5910
    • one year ago
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    yes

  45. Theloshua
    • one year ago
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    thnxxxx

  46. jim_thompson5910
    • one year ago
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    no problem

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