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@rizwan_uet can you help?
@RadEn can u help?
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.
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`.
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
okay how about for graph number two i'll look it up for the word at x = 4
Hmm just like the first graph, it looks like we have a `jump` at \(x=-2\) yes? :)
yes we do
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`.
Then at \(x=4\) we have a discontinuity that we saw in the first graph. Can you recognize it? :) hmm
yes the cusp thingee lol
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.
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.
So like in the first graph, when we see an asymptote, we have an `infinite discontinuity`.
ohh okay got it
|dw:1359671410605:dw|Here is another form of the `infinite discontinuity` that you might see come up.
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`.
okay thank you