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

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

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

Aylin
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0I see, thanks for the detailed answer!

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

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

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

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

Aylin
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0ok let me try the next problem! thank you so much!!

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

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

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

Aylin
 2 years 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
 2 years 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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