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Shido88
Determine the Domain: y= sqrt (x-2)/2x-5 HELPPPP!!
@amoodarya thanks for your help! but I really can't read what's going on :(
\[y=\frac{\sqrt{x-2}}{2x-5}\]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\[[2,\frac{5}{2}) \cup (\frac{5}{2}, \infty)\]
the first part, how do you know x is less than 2 ?
i get the second part but no the first part
can't you set up sqrt x-2 =0 and solve ?
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.
I see, thanks for the detailed answer!
You're welcome! I hope my ramblings made sense. :)
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 ?
It does and thank you ! :D
Well, what do you mean by which goes first? Do you mean out of [2,5/2) and (5/2, infinity)?
that and which one is x and y ?
i thought the x that we found is 5/2 shouldn't that go first before the 2 ?
I'm sorry if I confuse you lol :[
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.
It's set notation. I think\[2 \le x < \frac{5}{2}\]OR\[\frac{5}{2} < x < \infty\]would make more sense?
ok let me try the next problem! thank you so much!!
hey, are you still here?
is there to determine the domain and range using the graphing calculator ?
is there a way* excuse my typing..
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.
You could probably also write a program on one to tell you the domain and range, but I'm terrible with coding.