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October 22nd, 2017, 03:44 AM   #1
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Why -3,5 can't be a part of domain?

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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
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October 22nd, 2017, 07:53 AM   #2
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What about x = 0? What about x = 3?
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October 22nd, 2017, 10:47 AM   #3
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For $x \ne 0$, $h(x) = \dfrac{4x}{3 - x}$, so the discontinuity at $x = 0$ is removable.
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November 8th, 2017, 05:38 AM   #4
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Quote:
Originally Posted by ninjaapple View Post
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
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November 8th, 2017, 05:59 PM   #5
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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.
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November 8th, 2017, 07:17 PM   #6
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Quote:
Originally Posted by ninjaapple View Post
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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