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monroe17 Group Title

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

  • 2 years ago
  • 2 years 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}

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

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

    • 2 years ago
  4. monroe17 Group Title
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    yes

    • 2 years ago
  5. monroe17 Group Title
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    |dw:1346012919057:dw|

    • 2 years 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....

    • 2 years 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

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

    • 2 years ago
  9. love_math Group Title
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    x>=-17

    • 2 years ago
  10. love_math Group Title
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    root of x is always larger than 0

    • 2 years 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.

    • 2 years ago
  12. monroe17 Group Title
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    -17?

    • 2 years 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.

    • 2 years ago
  14. monroe17 Group Title
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    so.. [-17,infinity) ?

    • 2 years ago
  15. qpHalcy0n Group Title
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    That is your range, yes.

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

    • 2 years ago
  17. monroe17 Group Title
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    (-inf,inf) ?

    • 2 years 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.

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

    • 2 years ago
  20. monroe17 Group Title
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    1?

    • 2 years 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.

    • 2 years ago
  22. monroe17 Group Title
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    so.. 17?

    • 2 years ago
  23. qpHalcy0n Group Title
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    |dw:1346014535907:dw|

    • 2 years 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.

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

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

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

    • 2 years ago
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