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BloomLocke367

  • one year ago

Help? it says "Identify the x-values of each discontinuity, and write if it is removable or not. If it is nonremovable then classify the type."

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  1. anonymous
    • one year ago
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    factor the denominator, then set it equal to zero and solve

  2. BloomLocke367
    • one year ago
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    I have a graph D:

  3. anonymous
    • one year ago
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    if one of the factors cancels with a factor in the numerator, then it is "removable" because you can remove it

  4. anonymous
    • one year ago
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    oh then it is even easier, you do it with your eyeballs instead of algebra

  5. anonymous
    • one year ago
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    |dw:1441327182915:dw|

  6. anonymous
    • one year ago
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    course it would be much easier to help if we could see the picture :D

  7. BloomLocke367
    • one year ago
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    I'm trying. give me a moment.

  8. BloomLocke367
    • one year ago
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    |dw:1441327370892:dw|

  9. BloomLocke367
    • one year ago
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    okay sorry it looks bad. the single point is at (-1,1) and the open circle is at (-1,3) and the other line runs between -2 and -3 on the x-axis. @satellite73

  10. anonymous
    • one year ago
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    so the one with the hole is "removable" the other one isn't

  11. BloomLocke367
    • one year ago
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    Can you explain why?

  12. mathmate
    • one year ago
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    |dw:1441327899031:dw| The empty dot could have been a removable discontinuity, because we could have added a definition of the function for f(-1)=3. However, since f(-1) has been defined as -1, this discontinuity becomes non-removable. x=-2 is a vertical asymptote, which is a non-removable discontinuity.

  13. anonymous
    • one year ago
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    |dw:1441328135999:dw|

  14. BloomLocke367
    • one year ago
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    I am so confused

  15. BloomLocke367
    • one year ago
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    @mathmate

  16. BloomLocke367
    • one year ago
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    @freckles

  17. anonymous
    • one year ago
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    the one with the hole is removable cause you can remove it by filling in the hole

  18. anonymous
    • one year ago
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    the one with an "infinite discontinuity" can't be adjusted to make it continuous, so it is not removable

  19. BloomLocke367
    • one year ago
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    which one is that? I'm sorry for so many questions but I don't understand

  20. mathmate
    • one year ago
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    Not everyone may agree with me, but according to the definition, a removable discontinuity is when a point (a hole) is undefined, in which case we can fill the hole and the function becomes continuous. (a hole is where the limit of the function from the left and right are defined and identical) However, in your case, the value of f(-1) has been defined (f(-1)=1), so we can no longer fill the hole (to make it continuous), so it is no longer removable.

  21. BloomLocke367
    • one year ago
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    yes, I learned what it is and I'm looking at my notes right now and it just isn't making any sense at all.

  22. BloomLocke367
    • one year ago
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    @mathmate

  23. BloomLocke367
    • one year ago
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    when it says to identify the x-values, is it asking for the domain?

  24. mathmate
    • one year ago
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    Please explain what bothers you.

  25. BloomLocke367
    • one year ago
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    Just the question in general... what it's asking me to do

  26. mathmate
    • one year ago
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    There are different kinds of discontinuities - removable discontinuities (where a hole interrupts an otherwise continuous curve) - essential discontinuities, which subdivides into - vertical asymptotes, and (as in x=-3) - step discontinuities (a jump in y-value) See: http://www.mathwords.com/e/essential_discontinuity.htm and http://www.mathwords.com/r/removable_discontinuity.htm The question is asking you to locate and identify the type of discontinuity in the given example.

  27. BloomLocke367
    • one year ago
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    I only learned continuous, removable discontinuity, jump discontinuity, and infinite discontinuity.

  28. mathmate
    • one year ago
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    Well these are the same, but with a different name. continuous is not a discontinuity jump => step vertical asymptote =>infinite removable is where you can plug the "hole". In the given example, since f(-1)=1, you can no longer plug the "hole" (-1,3), so in my opinion, it becomes a jump discontinuity.

  29. BloomLocke367
    • one year ago
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    okay

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