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Shido88
 one year ago
Best ResponseYou've already chosen the best response.0@amoodarya thanks for your help! but I really can't read what's going on :(

Aylin
 one year ago
Best 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\[[2,\frac{5}{2}) \cup (\frac{5}{2}, \infty)\]

Shido88
 one year ago
Best ResponseYou've already chosen the best response.0the first part, how do you know x is less than 2 ?

Shido88
 one year ago
Best ResponseYou've already chosen the best response.0i get the second part but no the first part

Shido88
 one year ago
Best ResponseYou've already chosen the best response.0can't you set up sqrt x2 =0 and solve ?

Aylin
 one year ago
Best ResponseYou've already chosen the best response.1Well, 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.

Shido88
 one year ago
Best ResponseYou've already chosen the best response.0I see, thanks for the detailed answer!

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

Shido88
 one year ago
Best ResponseYou've already chosen the best response.0one 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 ?

Shido88
 one year ago
Best ResponseYou've already chosen the best response.0It does and thank you ! :D

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

Shido88
 one year ago
Best ResponseYou've already chosen the best response.0that and which one is x and y ?

Shido88
 one year ago
Best ResponseYou've already chosen the best response.0i thought the x that we found is 5/2 shouldn't that go first before the 2 ?

Shido88
 one year ago
Best ResponseYou've already chosen the best response.0I'm sorry if I confuse you lol :[

Aylin
 one year ago
Best ResponseYou've already chosen the best response.1I 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.

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

Shido88
 one year ago
Best ResponseYou've already chosen the best response.0ok let me try the next problem! thank you so much!!

Shido88
 one year ago
Best ResponseYou've already chosen the best response.0hey, are you still here?

Shido88
 one year ago
Best ResponseYou've already chosen the best response.0is there to determine the domain and range using the graphing calculator ?

Shido88
 one year ago
Best ResponseYou've already chosen the best response.0is there a way* excuse my typing..

Aylin
 one year ago
Best ResponseYou've already chosen the best response.1Hmm, 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.

Aylin
 one year ago
Best ResponseYou've already chosen the best response.1You 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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