This week in calculus appeared to be easier than the first two weeks. The beginning of the week was just review for the test. We reviewed derivatives again which I think I finally got the hang of them (most of them anyway), and we reviewed finite and infinite limits from last year and how to find points of discontinuity and if the graph is continuous or not. To find points of discontinuity you have to factor the top and the and make cancellations if possible. Whatever cancels, you set equal to zero and that is a removable at the number you get. An example is
x^2-x-12/ (x^2-6x=9) = (x-4) (x+3)/(x+3) (x+3)
The (x+3)s cancel and you are left with (x-4)/(x+3)
To find if there are any vertical asymptotes, you have to set what is left in the bottom, after you factored equal to zero.
(x+3) = 0
Therefore, there is a vertical asymptote at x= -3
And a removable at x= -3
And I think if they come out to the same number you only have to list the vertical asymptote but I’m not exactly sure so it would be great if someone could confirm that.
If there is a point of discontinuity, then the graph is not continuous.
On Thursday and Friday this week we learned something new and the first day, I was completely lost but I think I started catching on Friday. I think the most confusing part for me is the new vocabulary words and which ones go with what derivative and I get confused on how to find if the first or the second derivative is above or below the axis.
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a point of discontinuity does not necessarily mean that the graph is not continuous
ReplyDeleteif you have an asymptote at x=2 but if the limit of x->(2-) and x->(2+) are the same then it is continuous
Nope. If there is an infinite it is not continuous. The limit exists but the graph is not continuous there. There are some texts that will argue that point but AP perspective is that if the graph contains anything that is considered a point of discontinuity then the graph is not continuous.
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