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

1. Shido88

anybody ?

2. amoodarya

3. Shido88

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

4. Aylin

$y=\frac{\sqrt{x-2}}{2x-5}$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

give me a moment plz

6. Shido88

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

7. Shido88

i get the second part but no the first part

8. Shido88

not*

9. Shido88

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

10. Aylin

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

I see, thanks for the detailed answer!

12. Aylin

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

13. Shido88

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

It does and thank you ! :D

15. Aylin

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

16. Shido88

yes

17. Shido88

that and which one is x and y ?

18. Shido88

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

19. Shido88

I'm sorry if I confuse you lol :[

20. Aylin

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

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

22. Shido88

oh yes!

23. Shido88

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

24. Aylin

You're very welcome! :)

25. Shido88

hey, are you still here?

26. Shido88

I am stuck... lol

27. Aylin

Yep

28. Aylin

What's the problem?

29. Shido88

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

30. Shido88

is there a way* excuse my typing..

31. Aylin

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

yea

33. Aylin

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