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Shido88Best ResponseYou've already chosen the best response.0
@amoodarya thanks for your help! but I really can't read what's going on :(
 one year ago

AylinBest ResponseYou've already chosen the best response.1
\[y=\frac{\sqrt{x2}}{2x5}\]We'll start with all real numbers. We know that the square root of (x2) 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\[2x5=0\]\[2x=5\]\[x=\frac{5}{2}\]So we know that x cannot be 5/2. Since that is greater than 2, we have to exclude it. The domain is then\[[2,\frac{5}{2}) \cup (\frac{5}{2}, \infty)\]
 one year ago

Shido88Best ResponseYou've already chosen the best response.0
the first part, how do you know x is less than 2 ?
 one year ago

Shido88Best ResponseYou've already chosen the best response.0
i get the second part but no the first part
 one year ago

Shido88Best ResponseYou've already chosen the best response.0
can't you set up sqrt x2 =0 and solve ?
 one year ago

AylinBest ResponseYou've already chosen the best response.1
Well, if x >2 then (x2) is a positive number so the square root of (x2) is a real number. Likewise, if x=2, then (x2)=0, and sqrt(0)=0. However, if we take any number slightly smaller than 2, and let x equal that, say x=1.9999, (x2) would be:\[(1.99992)=0.0001\]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.
 one year ago

Shido88Best ResponseYou've already chosen the best response.0
I see, thanks for the detailed answer!
 one year ago

AylinBest ResponseYou've already chosen the best response.1
You're welcome! I hope my ramblings made sense. :)
 one year ago

Shido88Best ResponseYou've already chosen the best response.0
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 ?
 one year ago

Shido88Best ResponseYou've already chosen the best response.0
It does and thank you ! :D
 one year ago

AylinBest ResponseYou've already chosen the best response.1
Well, what do you mean by which goes first? Do you mean out of [2,5/2) and (5/2, infinity)?
 one year ago

Shido88Best ResponseYou've already chosen the best response.0
that and which one is x and y ?
 one year ago

Shido88Best ResponseYou've already chosen the best response.0
i thought the x that we found is 5/2 shouldn't that go first before the 2 ?
 one year ago

Shido88Best ResponseYou've already chosen the best response.0
I'm sorry if I confuse you lol :[
 one year ago

AylinBest ResponseYou've already chosen the best response.1
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.
 one year ago

AylinBest ResponseYou've already chosen the best response.1
It's set notation. I think\[2 \le x < \frac{5}{2}\]OR\[\frac{5}{2} < x < \infty\]would make more sense?
 one year ago

Shido88Best ResponseYou've already chosen the best response.0
ok let me try the next problem! thank you so much!!
 one year ago

Shido88Best ResponseYou've already chosen the best response.0
hey, are you still here?
 one year ago

Shido88Best ResponseYou've already chosen the best response.0
is there to determine the domain and range using the graphing calculator ?
 one year ago

Shido88Best ResponseYou've already chosen the best response.0
is there a way* excuse my typing..
 one year ago

AylinBest ResponseYou've already chosen the best response.1
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.
 one year ago

AylinBest ResponseYou've already chosen the best response.1
You could probably also write a program on one to tell you the domain and range, but I'm terrible with coding.
 one year ago
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