Question

Determine the Domain:
y= sqrt (x-2)/2x-5
HELPPPP!!

Answer 1

anybody ?

Answer 2

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

Answer 3

\[y=\frac{\sqrt{x-2}}{2x-5}\]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\[[2,\frac{5}{2}) \cup (\frac{5}{2}, \infty)\]

Answer 4

give me a moment plz

Answer 5

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

Answer 6

i get the second part but no the first part

Answer 7

not*

Answer 8

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

Answer 9

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.

Answer 10

I see, thanks for the detailed answer!

Answer 11

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

Answer 12

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 ?

Answer 13

It does and thank you ! :D

Answer 14

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

Answer 15

yes

Answer 16

that and which one is x and y ?

Answer 17

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

Answer 18

I'm sorry if I confuse you lol :[

Answer 19

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.

Answer 20

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

Answer 21

oh yes!

Answer 22

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

Answer 23

You're very welcome! :)

Answer 24

hey, are you still here?

Answer 25

I am stuck... lol

Answer 26

Yep

Answer 27

What's the problem?

Answer 28

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

Answer 29

is there a way*
excuse my typing..

Answer 30

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.

Answer 31

yea

Answer 32

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