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Denebel

How do you find the domain and range for a) sqrt ( (1/x)+1 ) b) 1/ sqrt (x+1) ?

  • one year ago
  • one year ago

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  1. SNSDYoona
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    for (a) x cannot be 0 if not its undefined

    • one year ago
  2. SNSDYoona
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    range is y cannot be 1

    • one year ago
  3. SNSDYoona
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    (b) is x cannot be -1 if not denominator is 0 n 1/0 is undefined

    • one year ago
  4. SNSDYoona
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    range is y cannot be 0

    • one year ago
  5. Denebel
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    Can you show me how did you get that?

    • one year ago
  6. Denebel
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    for a) I got (1/x) + 1 >= 0 ? Did I set it up correctly?

    • one year ago
  7. TheViper
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    http://cs.gmu.edu/cne/modules/dau/calculus/domains_ranges/dom2_frm.html

    • one year ago
  8. SNSDYoona
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    Umm u dont even actually have to set up.. just apply this few rules (1) anything divided by 0 ( a fraction) is undefined so for (a) square root of (1/x + 1) u know that 1 over 0 is undefined when x = 0 so ur domain would be x cannot be equal to 0.. as simple as that

    • one year ago
  9. panlac01
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    the denominator cannot be zero and function is only 1 to 1

    • one year ago
  10. Denebel
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    Do I need to think about the negatives as well?

    • one year ago
  11. SNSDYoona
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    yes.. for example 1/(x+1).. ur x cannot be negative one cause -1 +1 = 0

    • one year ago
  12. Denebel
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    For a) if the denominator was negative, it would make the equation inside the sqrt negative

    • one year ago
  13. SNSDYoona
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    umm brother... its 1 OVER x...

    • one year ago
  14. Denebel
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    -2, -3, etc...

    • one year ago
  15. SNSDYoona
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    even if x is -100000000000 its 1 OVER it.. thats like 0.00000000000 bla bla bla

    • one year ago
  16. Denebel
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    Oh nvm

    • one year ago
  17. SNSDYoona
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    it doesnt matter a fraction is always smaller than a whole number when the numerator is smaller than the denominator

    • one year ago
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