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anonymous

  • 5 years ago

Find the domain of the following functions: f(x)=(6x^2-2x+1)/3

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  1. anonymous
    • 5 years ago
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    (- oo, oo)

  2. anonymous
    • 5 years ago
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    Domain is looking for numbers where the function is discontinuous.

  3. anonymous
    • 5 years ago
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    it is continuoues everywhere so its from negative infinity to infinity

  4. anonymous
    • 5 years ago
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    Thank you!

  5. anonymous
    • 5 years ago
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    Yeah. I wasn't saying you were wrong, just trying to put out the general way of solving the problem without just giving the answer.

  6. anonymous
    • 5 years ago
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    oo yeah gotcha, i was giving a reason to my answer

  7. anonymous
    • 5 years ago
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    Alright, no worries.

  8. anonymous
    • 5 years ago
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    you guys are cool! very helpful i think i have a harder one

  9. anonymous
    • 5 years ago
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    And that is?

  10. anonymous
    • 5 years ago
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    f(x)=x+squartroot x^2+1

  11. anonymous
    • 5 years ago
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    cant figure out how to do the squartroot symbol

  12. anonymous
    • 5 years ago
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    Still looking for areas where the function is discontinuous. This one is a little different. Normally you check for values that make the square root bad. So that would be negative numbers in a square root.

  13. anonymous
    • 5 years ago
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    Here you would check \[\sqrt{x^2+1}\] and well what do you find out?

  14. anonymous
    • 5 years ago
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    since the x value will always be positive because it is squared it will be continuous everywhere aswell

  15. anonymous
    • 5 years ago
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    hmm what if its sqrt root of x^2-1 instead. will it be same

  16. anonymous
    • 5 years ago
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    Nope.

  17. anonymous
    • 5 years ago
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    x cant be 1

  18. anonymous
    • 5 years ago
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    Not quite. If x=1 it's still valid.

  19. anonymous
    • 5 years ago
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    You don't want anything under the square root to be negative. So you would set what is under the square root to > or = to 0. And solve.

  20. anonymous
    • 5 years ago
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    pj has it right

  21. anonymous
    • 5 years ago
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    hmm still not understanding

  22. anonymous
    • 5 years ago
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    If you plug zero in for x in \[\sqrt{x^2-1}\] you get \[\sqrt{-1}\] which is imaginary and not good.

  23. anonymous
    • 5 years ago
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    oh ok that wat you meant

  24. anonymous
    • 5 years ago
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    thanks

  25. anonymous
    • 5 years ago
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    Make sense?

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