August 12th, 2017, 08:15 PM  #1 
Newbie Joined: Aug 2017 From: Davao Posts: 9 Thanks: 0  Guyss pleasee help me
How to find the points of discontinuity of these functions? 1. x²+3/x²16 2. 3x+2/x²6x+9 3.3(x+2)³/x³+4x²+x6 4.x²3x/x²+2x²+5x 5.x²+2x1/4x⁴+4x³+3x²x1 
August 12th, 2017, 08:27 PM  #2 
Senior Member Joined: May 2016 From: USA Posts: 889 Thanks: 357 
Hmm I thought I had answered this. Look at item number 1. Are there one or more values of x where the function is not defined? Why is it not defined there? What does that mean? By the way, you may need parentheses in one or more of your questions. For example, is the first one $x^2  \dfrac{3}{x^2}  16$? That is what you wrote 
August 12th, 2017, 09:05 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 18,568 Thanks: 1483 
(2). x²  6x + 9 = (x  3)² (3). x³ + 4x² + x  6 = (x + 3)(x + 2)(x  1) (4). x² + 2x² + 5x = x(3x + 5) (5). 4x^4 + 4x³ + 3x²  x  1 = (2x + 1)(2x  1)(x² + x + 1) In each case, for what values of x is the denominator zero? 
August 12th, 2017, 09:17 PM  #4 
Newbie Joined: Aug 2017 From: Davao Posts: 9 Thanks: 0 
The given for #1 is x²+3 divided by x²16

August 12th, 2017, 10:01 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 18,568 Thanks: 1483 
Is there a typing error in #4?

August 12th, 2017, 10:29 PM  #6 
Newbie Joined: Aug 2017 From: Davao Posts: 9 Thanks: 0 
No, the given for #4 is correct. Btw how to know the points of discontinuity?

August 13th, 2017, 05:02 AM  #7 
Math Team Joined: Jul 2011 From: Texas Posts: 2,700 Thanks: 1358  
August 13th, 2017, 05:11 AM  #8 
Senior Member Joined: Oct 2009 Posts: 232 Thanks: 82  Actually none of these functions have any discontinuities, if you use the definition that 99.999% of the mathematical world uses. I don't doubt your teacher doesn't know this however.
Last edited by Micrm@ss; August 13th, 2017 at 05:17 AM. 
August 13th, 2017, 05:55 AM  #9 
Newbie Joined: Aug 2017 From: Davao Posts: 9 Thanks: 0 
thank you so much guys big help

August 14th, 2017, 03:52 PM  #10  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,944 Thanks: 797  Quote:
However the definition I have always used is "A function is continuous at x= a if and only if: a) f(a) exists b) $\displaystyle \lim_{x\to a} f(x)= f(a)$ I would say that a function is discontinuous at points where the function value does not exist. Perhaps you would say it is neither continuous nor discontinuous there. Last edited by Country Boy; August 14th, 2017 at 03:55 PM.  

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