Determine the Domain:
y= sqrt (x-2)/2x-5
HELPPPP!!

- anonymous

Determine the Domain:
y= sqrt (x-2)/2x-5
HELPPPP!!

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- jamiebookeater

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- anonymous

anybody ?

- amoodarya

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- anonymous

@amoodarya thanks for your help! but I really can't read what's going on :(

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## More answers

- 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

give me a moment plz

- anonymous

the first part, how do you know x is less than 2 ?

- anonymous

i get the second part but no the first part

- anonymous

not*

- anonymous

can't you set up sqrt x-2 =0 and solve ?

- 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

I see, thanks for the detailed answer!

- anonymous

You're welcome! I hope my ramblings made sense. :)

- 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

It does and thank you ! :D

- anonymous

Well, what do you mean by which goes first? Do you mean out of [2,5/2) and (5/2, infinity)?

- anonymous

yes

- anonymous

that and which one is x and y ?

- anonymous

i thought the x that we found is 5/2 shouldn't that go first before the 2 ?

- anonymous

I'm sorry if I confuse you lol :[

- 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

It's set notation. I think\[2 \le x < \frac{5}{2}\]OR\[\frac{5}{2} < x < \infty\]would make more sense?

- anonymous

oh yes!

- anonymous

ok let me try the next problem!
thank you so much!!

- anonymous

You're very welcome! :)

- anonymous

hey, are you still here?

- anonymous

I am stuck... lol

- anonymous

Yep

- anonymous

What's the problem?

- anonymous

is there to determine the domain and range using the graphing calculator ?

- anonymous

is there a way*
excuse my typing..

- 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

yea

- 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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