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October 22nd, 2017, 03:44 AM  #1 
Newbie Joined: Oct 2017 From: canada Posts: 3 Thanks: 0  Why 3,5 can't be a part of domain? Untitled.png Within given equation, you can change this to 4/(x/31). which I think the answer should be All Real Numbers? since X isn't taking any part from denominators 
October 22nd, 2017, 07:53 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,745 Thanks: 1001 Math Focus: Elementary mathematics and beyond 
What about x = 0? What about x = 3?

October 22nd, 2017, 10:47 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 18,595 Thanks: 1493 
For $x \ne 0$, $h(x) = \dfrac{4x}{3  x}$, so the discontinuity at $x = 0$ is removable.

November 8th, 2017, 05:38 AM  #4  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,959 Thanks: 802  Quote: The domain of the initial $\displaystyle \frac{4}{\frac{3}{x} 1}$ is "all real numbers except x= 0 or x= 3". The domain of $\displaystyle \frac{4}{\frac{x}{3} 1}$ is "all real numbers except 3" but that is not equivalent to the original fraction. Quote:
 
November 8th, 2017, 05:59 PM  #5 
Senior Member Joined: May 2016 From: USA Posts: 904 Thanks: 359 
Yes, you can say that $h(x) = \dfrac{4}{\dfrac{3}{x}  1} = \dfrac{4x}{3  x} = g(x) \text { if } x \ne 0 \text { and } x \ne 3.$ But if x = 0, you do not have a definition for h(x) so you cannot say that it equals g(0). And of course neither g(x) nor h(x) is defined if x = 3 so saying h(3) = g(3) is meaningless for two distinct reasons. 
November 8th, 2017, 07:17 PM  #6  
Senior Member Joined: Aug 2012 Posts: 1,709 Thanks: 458  Quote:  

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