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monroe17 Group Title

Find the domain and range of: f(x)=18sqrt(x)-17

  • one year ago
  • one year ago

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  1. TheMind Group Title
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    D={x: where x belong to R and x >=17} R={ y>=0}

    • one year ago
  2. monroe17 Group Title
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    so, would it be 18x-17>=0? and solve?

    • one year ago
  3. dpaInc Group Title
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    is this your function: \(\large f(x)=18\sqrt{x}-17 \)

    • one year ago
  4. monroe17 Group Title
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    yes

    • one year ago
  5. monroe17 Group Title
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    |dw:1346012919057:dw|

    • one year ago
  6. dpaInc Group Title
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    the domain of f is all the x values that give f(x) a real value.... so looking at the function, the function will not give you a real number when x is less than 0....

    • one year ago
  7. dpaInc Group Title
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    so the actual question in finding the domain is "what x values is \(\large \sqrt x =real\) number

    • one year ago
  8. monroe17 Group Title
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    would I do.. 18x-17<0 and solve?

    • one year ago
  9. love_math Group Title
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    x>=-17

    • one year ago
  10. love_math Group Title
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    root of x is always larger than 0

    • one year ago
  11. qpHalcy0n Group Title
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    You're just looking to be sure that what's under the root does not go negative. So x cannot be negative. Since the smallest value of x is 0 (because of the domain), then the range's minimum value will happen when x is 0. (So plug in 0 for x, and see what y is). The upper limit of the function will go on forever.

    • one year ago
  12. monroe17 Group Title
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    -17?

    • one year ago
  13. qpHalcy0n Group Title
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    You got it, that's the minimum of the range. As x gets bigger and bigger, there is no limit, it just grows and grows forever.

    • one year ago
  14. monroe17 Group Title
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    so.. [-17,infinity) ?

    • one year ago
  15. qpHalcy0n Group Title
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    That is your range, yes.

    • one year ago
  16. monroe17 Group Title
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    how do I find the domain with this function?

    • one year ago
  17. monroe17 Group Title
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    (-inf,inf) ?

    • one year ago
  18. qpHalcy0n Group Title
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    Well, as we talked about earlier, the domain would be anything that is a "no no". You have a square root, and here, you cannot take the square root of a negative number, as it has no real solutions.

    • one year ago
  19. qpHalcy0n Group Title
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    anything that is NOT a "no-no", sorry ;]

    • one year ago
  20. monroe17 Group Title
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    1?

    • one year ago
  21. qpHalcy0n Group Title
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    No, you simply have sqrt(x), since you cannot take the square root of a negative number, we know that x MUST be greater than or equal to 0.

    • one year ago
  22. monroe17 Group Title
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    so.. 17?

    • one year ago
  23. qpHalcy0n Group Title
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    |dw:1346014535907:dw|

    • one year ago
  24. qpHalcy0n Group Title
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    We're looking for x's that might make the equation undefined. That's what it means to find the domain. The only thing in this equation that can become undefined (or have no real solutions) is the square root. Which can't be negative.

    • one year ago
  25. monroe17 Group Title
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    so x is all real numbers that are positive then?

    • one year ago
  26. qpHalcy0n Group Title
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    Yep, or in other words x >= 0

    • one year ago
  27. monroe17 Group Title
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    how do I put that into interval notation though? [0,inf)

    • one year ago
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