## anonymous 3 years ago Determine the Domain: y= sqrt (x-2)/2x-5 HELPPPP!!

1. anonymous

anybody ?

2. amoodarya

3. anonymous

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

4. 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)$

5. anonymous

give me a moment plz

6. anonymous

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

7. anonymous

i get the second part but no the first part

8. anonymous

not*

9. anonymous

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

10. 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.

11. anonymous

I see, thanks for the detailed answer!

12. anonymous

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

13. 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 ?

14. anonymous

It does and thank you ! :D

15. anonymous

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

16. anonymous

yes

17. anonymous

that and which one is x and y ?

18. anonymous

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

19. anonymous

I'm sorry if I confuse you lol :[

20. 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.

21. anonymous

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

22. anonymous

oh yes!

23. anonymous

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

24. anonymous

You're very welcome! :)

25. anonymous

hey, are you still here?

26. anonymous

I am stuck... lol

27. anonymous

Yep

28. anonymous

What's the problem?

29. anonymous

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

30. anonymous

is there a way* excuse my typing..

31. 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.

32. anonymous

yea

33. anonymous

You could probably also write a program on one to tell you the domain and range, but I'm terrible with coding.