is f continuous

Jul 2009
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0
\(\displaystyle |x+1|/(x^2-1)\)

i graphed it and i think it has a jump discontinuity at -1 can someone tell me if this is right
 
Jan 2010
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138
The way I learned it, "jump discontinuity" means that there is a jump in function values that makes the gap impossible to plug with a single point. I don't think this is what you mean. Rather, at -1, the graph has a "removable discontinuity." The graph is continuous everywhere except for the "hole" at x = -1. If we "plug" the hole we would have a continuous graph at -1.

By the way, there is another place on the graph where we have a discontinuity that you didn't mention. Where is it?
 
Oct 2009
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\(\displaystyle |x+1|/(x^2-1)\)

i graphed it and i think it has a jump discontinuity at -1 can someone tell me if this is right


\(\displaystyle \frac{|x+1|}{x^2-1}=\left\{\begin{array}{ll}\frac{-x-1}{x^2-1}=-\frac{1}{x-1}&\,if\,\,x<-1\\{}\\\frac{x+1}{x^2-1}=\frac{1}{x-1}&\,if\,\,x>-1\\{}\\unde{f}ined&\,if\,\,x=\pm 1\end{array}\right.\)

Tonio
 
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Apr 2010
384
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Canada
what would the graph be?
Has your professor not told you how to graph rational functions?

Find: Points of Inflection, Vertical/Horizontal Asymtotes, intervals of decreasing/increasing functions via first derivative test.

I forget, but I think that's all we need to graph.
 
Dec 2009
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This is what the graph looks like.

If you look at Tonio's response, you can evaluate the limit the graph approaches,
as x approaches -1 from the left and from the right.
 

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Jul 2009
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so its not continuous and has a removable discontinuity at -1 and a infinite discontinuity at x=1