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Homework Partial Lab - Calculus

Posted on 2013-06-16
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Last Modified: 2013-06-17
Hello Experts,

This is a little different question, but I am hoping someone can help anyway.  I need an answer for the first two questions on my calculus lab.  I understand the other problems, but these on continuity seem to be giving me a headache....  If you can answer, please explain with your response.

Thanks in advance!

aj95
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by:TommySzalapski
ID: 39252224
For problem 1, a closed circle means the function is defined there and an open circle means it is not. So if an x value has a closed and open circle in the y axis, the answer is the value at the closed circle. If there is an open circle only, then the function is not defined there.

For limits, the open circles can be the value, but only if there is just one possible answer. Closed or open makes no difference for the limits.

If that makes sense, great. If you are still not sure, post what you get and we can confirm.
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by:aj85
ID: 39252229
Hello,

Thanks for the quick reply.  So for f(-2)  the answer is 1?
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Expert Comment

by:TommySzalapski
ID: 39252233
You mean f(-1)? f(-2) looks like it's right where it's crossing at 0.
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Author Comment

by:aj85
ID: 39252241
Okay I guess that is where I am confused, as on the graph of f, the point looks like it is (-1).  Can you tell me how you are getting 0?

Thanks.
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by:TommySzalapski
ID: 39252251
Yes. There is a closed circle at x=-1 and y=1. So f(-1) = 1.
But you asked about f(-2). When x is -2, what is y?
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TommySzalapski earned 500 total points
ID: 39252259
Well, I'm off to bed, but remember these points:

Closed circle (or line) means the function has an actual value at that point. So part 1 should be pretty easy.

The limit is what the points on both sides are approaching as x approaches the point.
If the left side is approaching one number and the right side is approaching a different number, the limit does not exist. If they are approaching the same number, that's the limit.
The actual value at the point doesn't matter for limits. At all.

Hope that's all you need.
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Author Closing Comment

by:aj85
ID: 39254716
Thanks for the quick help.
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