## Shido88 Group Title Determine the Domain: y= sqrt (x-2)/2x-5 HELPPPP!! one year ago one year ago

1. Shido88 Group Title

anybody ?

2. amoodarya Group Title

3. Shido88 Group Title

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

4. Aylin Group Title

$y=\frac{\sqrt{x-2}}{2x-5}$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. Shido88 Group Title

give me a moment plz

6. Shido88 Group Title

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

7. Shido88 Group Title

i get the second part but no the first part

8. Shido88 Group Title

not*

9. Shido88 Group Title

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

10. Aylin Group Title

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. Shido88 Group Title

I see, thanks for the detailed answer!

12. Aylin Group Title

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

13. Shido88 Group Title

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. Shido88 Group Title

It does and thank you ! :D

15. Aylin Group Title

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

16. Shido88 Group Title

yes

17. Shido88 Group Title

that and which one is x and y ?

18. Shido88 Group Title

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

19. Shido88 Group Title

I'm sorry if I confuse you lol :[

20. Aylin Group Title

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. Aylin Group Title

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

22. Shido88 Group Title

oh yes!

23. Shido88 Group Title

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

24. Aylin Group Title

You're very welcome! :)

25. Shido88 Group Title

hey, are you still here?

26. Shido88 Group Title

I am stuck... lol

27. Aylin Group Title

Yep

28. Aylin Group Title

What's the problem?

29. Shido88 Group Title

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

30. Shido88 Group Title

is there a way* excuse my typing..

31. Aylin Group Title

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. Shido88 Group Title

yea

33. Aylin Group Title

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