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anonymous

  • 5 years ago

I'm trying to find removable discontinuities. I'm studying Algebra II for dummies. The problem is: x^3-3x^2-16x-12/x^3-4x^2-11x-6. This factors to (x+2)(x-6)(x+1)/(x+1)(x+1)(x-6). Reducing to x+2/X+1. I thought the only removable discontinuity would be 6. The text lists both 6 and -1. But, since -1is a vertical asymptote even though it was factored out...it hasn't been removed. Am I thinking wrong or is the text wrong?

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  1. anonymous
    • 5 years ago
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    the tect is right. a graph is discontinuous when you have to lift your pencil to continue graphing it, and at a verticle asymptote you have to lift your pencil, therefore it is discontinuous

  2. anonymous
    • 5 years ago
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    Okay, but the key word is removable discontinuity. By factoring out 6 it becomes a removable discontinuity. -1 has been factored out, but since it is still in the denominator(it was squared) the discontinuity has not been removed. That is what threw me. I can see the asymptote would be there and the graph would be discontinuous.

  3. anonymous
    • 5 years ago
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    well removable discontinuity is when a point or section of a graph is removed, aka a hole, that is when 2 factors are crossed out during simplification

  4. anonymous
    • 5 years ago
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    Right. To simplify my confusion, if the equation were (x+1)/(x+1)^2 the function would reduce to 1/x+1. There would be an asymptote at -1. But, following the apparent logic of the problem in the text...-1 would be considered a removable asymptote because I factored it out in part. I don't see how something can be both a removable discontinuity and an asymptote at the same time.

  5. anonymous
    • 5 years ago
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    it is just saying that the point lies on the vertical asymptote. the vertical asymptote means the function wont cross that value, yet a point may lie on it.

  6. anonymous
    • 5 years ago
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    Okay, I really appreciate your time. It clears things up a great deal for me.

  7. anonymous
    • 5 years ago
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    oo your welcome. i didnt actually learn that till i was in calculus, and thats where i learned that i really like math

  8. anonymous
    • 5 years ago
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    Now your just making me feel a bit guilty. I had calc ten years ago and haven't really used math very often. I like it a great deal but I'm out of practice and am working to push my knowledge further. Again, I greatly appreciate your help.

  9. anonymous
    • 5 years ago
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    oo your welcome.

  10. anonymous
    • 5 years ago
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    I'm out for the night. I'm sure I'll be back in the future.

  11. anonymous
    • 5 years ago
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    Is this the expression? \[\frac{x{}^{\wedge}3-3x{}^{\wedge}2-16x-12}{x{}^{\wedge}3-4x{}^{\wedge}2-11x-6}\]

  12. anonymous
    • 5 years ago
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    Okay, I'm back. Yes, that is the expression.

  13. anonymous
    • 5 years ago
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    The expression simplifies to: \[\frac{2+x}{1+x}\] When x=-1 in the denominator, you have division by zero which is a no no. There seems to be a root at x=-2 Refer to the plot attachment.

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  14. anonymous
    • 5 years ago
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    Right, the imaginary line is an asymptote. An asymptote creates a discontinuity. A removable discontinuity is one that can be removed by factoring. My solution agrees with the texts solution until the final answer which lists both 6 and -1 as being removable. My point is that -1 is not removable. I think that is what you are also indicating. According to nctarheelsbball, it is an asymptote but listed as removable because that is where the point lies.

  15. anonymous
    • 5 years ago
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    Sorry. I guess I'll have to google "removable discontinuity math"

  16. anonymous
    • 5 years ago
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    From: http://www.math-magic.com/algebra/asymptote.htm E. Removable Discontinuities "A removable discontinuity is when you can factor out a term in the numerator and factor out the same term in the denominator, thus canceling each other out." \[ \frac{x{}^{\wedge}3-3x{}^{\wedge}2-16x-12}{x{}^{\wedge}3-4x{}^{\wedge}2-11x-6}==\frac{(x+1) (x+2)(x-6)}{(x+1)(x+1)(x-6)}\] The factors (x+1) and (x-6) cancel out. Perhaps the confusion is that (x+1) is the simplified version of ( x - (-1) ) which equals ( x + 1 )

  17. anonymous
    • 5 years ago
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    Well, I don't know if anyone will read this now, and I don't mean to be pedantic. I am very well aware of what a removable discontinuity is, but because of this problem I also googled the definition for greater understanding. I also understand what an asymptote is...the problem is the conflict between names, definitions and this particular rational function: Mainly, that I'm very bloody well aware that (x-6) cancels out: thus fulfilling the very definition of a removable discontinuity. (x+1) cancels out in the numerator and the denominator and thus is removed, however: because it is squared in the denominator (x+1) remains a part of the function and because it remains, it creates an asymptote particularly a verticle asymptote which is the very definition of a non-removable discontinuity. The best answer I've seen so far is nctarheelsball who seems to suggest, in a condensed form: yes, those terms conflict, but both are true.

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