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korosh23

  • one year ago

Calculus 12 Question! Please explain the part I highlighted. Thank you. Wait for the attachment to load

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  1. korosh23
    • one year ago
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  2. korosh23
    • one year ago
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    There is the solution to the answer, but I don't understand it. Could you please explain?

  3. anonymous
    • one year ago
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    When the x makes the denominator 0, it is not included in the domain. \[\frac{ 1 }{ x(x-1) }\]

  4. anonymous
    • one year ago
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    x=? makes it zero.

  5. anonymous
    • one year ago
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    and x=1

  6. anonymous
    • one year ago
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    i think the explanation you need might be just that \[(-\infty, 0)\cup (0,1)\cup (1,\infty)\] is just a really long winded way of saying all numbers except \(0\) and\(1\)

  7. korosh23
    • one year ago
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    yes x-1=0 x=1

  8. anonymous
    • one year ago
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    1(1-1)=0

  9. korosh23
    • one year ago
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    @satellite73 right, but I don't understand how? the way it is written is confusing

  10. korosh23
    • one year ago
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    @shalante I am following you. Go ahead

  11. anonymous
    • one year ago
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    \[(-\infty,0)\] means \[x <0\]

  12. anonymous
    • one year ago
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    meaning x can 0.0001 or -0.0001 but not at 0

  13. anonymous
    • one year ago
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    (0,1) means it can be from 0.0001 to 0.9999 but not at 1 or 0

  14. anonymous
    • one year ago
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    \[(1,\infty)\] means it can be from 1.0001 to infinity.

  15. korosh23
    • one year ago
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    Ok and what does the U shape represent?

  16. anonymous
    • one year ago
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    only 0 and 1 makes it undefined So we wont include \[x \le0\] or \[x \le 1\] U is just a set of interval

  17. anonymous
    • one year ago
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    A gap between undefined values.

  18. korosh23
    • one year ago
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    But in this case it has to be x cannot equal those numbers. Since any number less or more than those would be defined. Am I right?

  19. korosh23
    • one year ago
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    @Shalante, one more question. If something is like x cannot equal 1, can we say the npvs or non permissible values are 1

  20. anonymous
    • one year ago
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    For the domain?

  21. korosh23
    • one year ago
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    Yes I believe. Like saying x cannot equal 1. So npvs of x are 1. IS it correct?

  22. anonymous
    • one year ago
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    Domain has to be an interval of defined value

  23. anonymous
    • one year ago
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    If it asks you for vertical asymptote, it would work.

  24. zepdrix
    • one year ago
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    npv's? Hmm ive never heard that before :) interesting

  25. korosh23
    • one year ago
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    Yes it was in my pre-cal 11 textbook. lol

  26. korosh23
    • one year ago
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    @zepdrix and @Shalante, I am fine with solving questions, but my main problem now is understnading important concepts. Can I tag you in other questions to explain something like this. If it is ok with you guys?

  27. anonymous
    • one year ago
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    Yep, the name is interesting. Havent heard it in a while.

  28. anonymous
    • one year ago
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    Now I want a break and play. If no one does, I will. (In like 20 minutes)

  29. korosh23
    • one year ago
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    Ok thanks for help.

  30. zepdrix
    • one year ago
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    I don't think it was mentioned earlier, just want to add, U means union. It's like.. the sum of those intervals, is your domain|dw:1443234023659:dw|x can lie anywhere in the ( brackets .

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