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monroe17

  • 2 years ago

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

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  1. TheMind
    • 2 years ago
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    D={x: where x belong to R and x >=17} R={ y>=0}

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

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

  4. monroe17
    • 2 years ago
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    yes

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

  6. dpaInc
    • 2 years ago
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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....

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

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

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

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

  11. qpHalcy0n
    • 2 years ago
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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.

  12. monroe17
    • 2 years ago
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    -17?

  13. qpHalcy0n
    • 2 years ago
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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.

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

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

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

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

  18. qpHalcy0n
    • 2 years ago
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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.

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

  20. monroe17
    • 2 years ago
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    1?

  21. qpHalcy0n
    • 2 years ago
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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.

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

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

  24. qpHalcy0n
    • 2 years ago
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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.

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

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

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

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