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- #1

- Thread starter shamieh
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- Thread starter
- #1

- Feb 29, 2012

- 342

Discontinuity does not mean it is not defined.

Cheers.

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- #3

- Feb 7, 2012

- 2,807

The convention in diagrams like this is that a hollow circle $\circ$ is intended to indicate that the graph does not include that endpoint of the line segment. A filled circle $\bullet$ shows that the graph does include that endpoint of the line segment. So for example $g(3)$ is not defined, because of the hollow circle at the point $(3,0)$ in the diagram, but $g(-3) = -0.5$ because of the filled circle at the point $(-3,-0.5).$Find g(-1) =

So wouldn't g(-1) = Does not exist since the point is discontinuous or would it be 1.5? I'm confused. I feel like it doesn't exist. Can someone elaborate?

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because g(2) = Does not exist, because it is discontinuous, so g(-1) has to be Does not exist right??

You can see that when $x=-1$ there is a hollow circle at $(-1,0.5)$ and a filled circle at $(-1,1.5)$. So what does that tell you about $g(-1)$?