anonymous
  • anonymous
Determine the Domain: y= sqrt (x-2)/2x-5 HELPPPP!!
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
anybody ?
amoodarya
  • amoodarya
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anonymous
  • anonymous
@amoodarya thanks for your help! but I really can't read what's going on :(

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anonymous
  • anonymous
\[y=\frac{\sqrt{x-2}}{2x-5}\]We'll start with all real numbers. We know that the square root of (x-2) will not give real results when x is less than 2, so let's restrict the range to\[2 \le x < \infty\]Now, we also know that x cannot be able value which will make the demonimator equal 0. So we'll say\[2x-5=0\]\[2x=5\]\[x=\frac{5}{2}\]So we know that x cannot be 5/2. Since that is greater than 2, we have to exclude it. The domain is then\[[2,\frac{5}{2}) \cup (\frac{5}{2}, \infty)\]
anonymous
  • anonymous
give me a moment plz
anonymous
  • anonymous
the first part, how do you know x is less than 2 ?
anonymous
  • anonymous
i get the second part but no the first part
anonymous
  • anonymous
not*
anonymous
  • anonymous
can't you set up sqrt x-2 =0 and solve ?
anonymous
  • anonymous
Well, if x >2 then (x-2) is a positive number so the square root of (x-2) is a real number. Likewise, if x=2, then (x-2)=0, and sqrt(0)=0. However, if we take any number slightly smaller than 2, and let x equal that, say x=1.9999, (x-2) would be:\[(1.9999-2)=-0.0001\]And\[\sqrt{-0.0001}=\sqrt{-1 \times 0.0001}=\sqrt{-1} \times \sqrt{0.0001}=i \times 0.01\]and this is clearly not a real number. So the absolutely smallest value of x can be 2.
anonymous
  • anonymous
I see, thanks for the detailed answer!
anonymous
  • anonymous
You're welcome! I hope my ramblings made sense. :)
anonymous
  • anonymous
one more thing about this " [2,5/2)∪(5/2,∞)" how do you know which goes first and which goes last ? I mean the way to set it up ?
anonymous
  • anonymous
It does and thank you ! :D
anonymous
  • anonymous
Well, what do you mean by which goes first? Do you mean out of [2,5/2) and (5/2, infinity)?
anonymous
  • anonymous
yes
anonymous
  • anonymous
that and which one is x and y ?
anonymous
  • anonymous
i thought the x that we found is 5/2 shouldn't that go first before the 2 ?
anonymous
  • anonymous
I'm sorry if I confuse you lol :[
anonymous
  • anonymous
I think I see where the confusion is. \[[2,\frac{5}{2}) \cup (\frac{5}{2},\infty)\]is the same thing as\[x \in [2,\frac{5}{2})\]OR\[x \in (\frac{5}{2},\infty )\]That is, everything I wrote down represents the values that x is allowed to take. Neither of them refer to y.
anonymous
  • anonymous
It's set notation. I think\[2 \le x < \frac{5}{2}\]OR\[\frac{5}{2} < x < \infty\]would make more sense?
anonymous
  • anonymous
oh yes!
anonymous
  • anonymous
ok let me try the next problem! thank you so much!!
anonymous
  • anonymous
You're very welcome! :)
anonymous
  • anonymous
hey, are you still here?
anonymous
  • anonymous
I am stuck... lol
anonymous
  • anonymous
Yep
anonymous
  • anonymous
What's the problem?
anonymous
  • anonymous
is there to determine the domain and range using the graphing calculator ?
anonymous
  • anonymous
is there a way* excuse my typing..
anonymous
  • anonymous
Hmm, well I suppose you could graph the function and then look at the graph. Though sometimes it can be misleading if the window is set incorrectly.
anonymous
  • anonymous
yea
anonymous
  • anonymous
You could probably also write a program on one to tell you the domain and range, but I'm terrible with coding.

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