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 October 22nd, 2017, 02:44 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/3-1). which I think the answer should be All Real Numbers? since X isn't taking any part from denominators
 October 22nd, 2017, 06:53 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,842 Thanks: 1068 Math Focus: Elementary mathematics and beyond What about x = 0? What about x = 3? Thanks from topsquark
 October 22nd, 2017, 09:47 AM #3 Global Moderator   Joined: Dec 2006 Posts: 19,299 Thanks: 1688 For $x \ne 0$, $h(x) = \dfrac{4x}{3 - x}$, so the discontinuity at $x = 0$ is removable. Thanks from topsquark
November 8th, 2017, 04:38 AM   #4
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Quote:
 Originally Posted by ninjaapple Attachment 9246 Within given equation, you can change this to 4/(x/3-1).
What do you mean by this? How can you just flip the "3/x" to make it "x/3"?? I do see that you could multiply both numerator and denominator by x to get $\displaystyle \frac{4x}{3- x}$ but that is not the same thing.

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:
 which I think the answer should be All Real Numbers? since X isn't taking any part from denominators

 November 8th, 2017, 04:59 PM #5 Senior Member   Joined: May 2016 From: USA Posts: 1,084 Thanks: 446 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, 06:17 PM   #6
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Quote:
 Originally Posted by ninjaapple Attachment 9246 Within given equation, you can change this to 4/(x/3-1).
The trick here is to carefully write out the derivation from the original equation to the new form, step by step ... and ask yourself, under what circumstances is each step valid?

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