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What are the points of discontinuity? Are they all removable? TIA!

 

y = (x+5)(x+3) / x2+8x+15

 
Guest Apr 17, 2018
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 #3
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The defined function is equivalent to

 

\(y=(x+5)(x+3)/[(x+5)(x+3)],\)

 

so the denominator is zero, and therefore \(y\) is undefined, at \(x = -5\) and \(x = -3\). For all other values of \(x \) we have \(y=1\), so we see that \(y\) is continuous everywhere it is defined. The questioner therefore has made a mistake, since continuity is only meaningful where a function is defined (see https://en.wikipedia.org/wiki/Classification_of_discontinuities#Removable_discontinuity). The points \(x = -5\) and \(x = -3\) are actually removable singularities which can simply be removed by defining \(y(-5)=y(-3)=1\) so that \(y(x)=1\) for all \(x\), and \(y(x)\) is then continuous everywhere.

 
Guest Apr 17, 2018
 #4
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thank you so much for replying! so the points are x = -5 and x = -3 and they are removable? just to clarify.

 
Guest Apr 17, 2018
 #5
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I now wonder, did you understand how to solve these equations?
Let me try to explain to you, first you do the multiplication action between the parentheses. after that you have a square equation that you decide with the help of the formula the discriminant of the quadratic polynomial ax+ bx + c is b2-  4ac

This and any other information you can find on the homework help online and there to ask any questions you are interested in, I'm sure that they will help you and explain.

 
AKM17  Apr 17, 2018

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