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Consider the graph of y = g(x) given.

shamieh

Active member
Sep 13, 2013
539
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??
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Shamieh, notice that the function is defined at [tex]x=-1[/tex], even though it is not continuous there. The value seems to be 1.5 (or something pretty close to it). However, it isn't defined at [tex]x=2[/tex], as there is a hole in the graph. You can see that there is the black dot when [tex]x=-1[/tex], meaning that it is defined in that point.

Discontinuity does not mean it is not defined. :eek:

Cheers.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
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?

View attachment 1730

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because g(2) = Does not exist, because it is discontinuous, so g(-1) has to be Does not exist right??
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 $(-3,-0.5).$

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 $g(-1)$?