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onegirl

  • one year ago

Use the given graph to identify all x-values at which the function is discontinuous.

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  1. onegirl
    • one year ago
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    1)

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  2. onegirl
    • one year ago
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    2)

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  3. onegirl
    • one year ago
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    @rizwan_uet can you help?

  4. onegirl
    • one year ago
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    @RadEn can u help?

  5. zepdrix
    • one year ago
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    Ok let's look at the first one. If we slide over to \(x=-2\), the function is defined where the black dot is located. So the function value is -1 ish. But notice that if we try to move any further to the left, we have to `jump` up to grab the other line. This is called a `jump discontinuity`. The curve doesn't continue in one straight path. That's bad.

  6. onegirl
    • one year ago
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    ok

  7. zepdrix
    • one year ago
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    Around \(x=2\), the line curves down to negative infinity on the left, and up towards positive infinity from the right side of 2. We call this an `infinite discontinuity` or sometimes called an `asymptotic discontinuity`.

  8. zepdrix
    • one year ago
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    Then at \(x=4\) we have a `sharp corner`. It would probably be more accurate to describe it as a `cusp`, but whatever :) Sharp corners like that are bad! Discontinuity! I can't remember what type of discontinuity it is though :( hmm

  9. onegirl
    • one year ago
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    okay how about for graph number two i'll look it up for the word at x = 4

  10. zepdrix
    • one year ago
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    Hmm just like the first graph, it looks like we have a `jump` at \(x=-2\) yes? :)

  11. onegirl
    • one year ago
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    yes we do

  12. zepdrix
    • one year ago
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    Then we move over to \(x=2\). There are no problems are the point, only AT that particular point. It's missing!!! We call this a `removable discontinuity`.

  13. onegirl
    • one year ago
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    ok

  14. zepdrix
    • one year ago
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    Then at \(x=4\) we have a discontinuity that we saw in the first graph. Can you recognize it? :) hmm

  15. onegirl
    • one year ago
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    yes the cusp thingee lol

  16. zepdrix
    • one year ago
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    no no no, not a cusp! :) There is no sharp corner. What's happening is, the line is going up up up on the right. and down down down on the left.

  17. zepdrix
    • one year ago
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    They draw a `pink line` to show that there is an asymptote located there. Meaning that our function does not exist in the pink area.

  18. onegirl
    • one year ago
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    okay

  19. zepdrix
    • one year ago
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    So like in the first graph, when we see an asymptote, we have an `infinite discontinuity`.

  20. onegirl
    • one year ago
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    ohh okay got it

  21. zepdrix
    • one year ago
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    |dw:1359671410605:dw|Here is another form of the `infinite discontinuity` that you might see come up.

  22. zepdrix
    • one year ago
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    Sometimes the branches go the same direction. But that dotted line let's us know that the function isn't defined there. The branches go up up up getting closer and closer to one another as they approach infinity. But they will not connect, since the dotted line will always separate them. So it's an `infinite discontinuity`.

  23. onegirl
    • one year ago
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    okay thank you

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