+0  
 
0
1107
2
avatar

what are the points of discontinuity? are they all removable?  y= (x-5) / x^2 - 6x + 5

 Apr 26, 2015

Best Answer 

 #1
avatar+94619 
+10

Note......

(x - 5) / [ x^2 - 6x + 5] =

(x - 5) / [ (x - 1) (x -5) ] =

1 / (x - 1)

There will be a "hole" at x = 5......but the discontinuity at x = 1  isn't "removable".....there's a vertical asymptote at that point......

Here's the graph.......

https://www.desmos.com/calculator/m598ibdrql

 

{Believe me....there is a "hole" at x = 5  !!!}

 

  

 Apr 26, 2015
 #1
avatar+94619 
+10
Best Answer

Note......

(x - 5) / [ x^2 - 6x + 5] =

(x - 5) / [ (x - 1) (x -5) ] =

1 / (x - 1)

There will be a "hole" at x = 5......but the discontinuity at x = 1  isn't "removable".....there's a vertical asymptote at that point......

Here's the graph.......

https://www.desmos.com/calculator/m598ibdrql

 

{Believe me....there is a "hole" at x = 5  !!!}

 

  

CPhill Apr 26, 2015
 #2
avatar+95369 
0

I have never heard of the term "removable" in relation to discontinuities  

 Apr 26, 2015

16 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.