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Shido88

  • 2 years ago

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

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  1. Shido88
    • 2 years ago
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    anybody ?

  2. amoodarya
    • 2 years ago
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  3. Shido88
    • 2 years ago
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    @amoodarya thanks for your help! but I really can't read what's going on :(

  4. Aylin
    • 2 years ago
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    \[y=\frac{\sqrt{x-2}}{2x-5}\]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
    • 2 years ago
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    give me a moment plz

  6. Shido88
    • 2 years ago
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    the first part, how do you know x is less than 2 ?

  7. Shido88
    • 2 years ago
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    i get the second part but no the first part

  8. Shido88
    • 2 years ago
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    not*

  9. Shido88
    • 2 years ago
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    can't you set up sqrt x-2 =0 and solve ?

  10. Aylin
    • 2 years ago
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    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
    • 2 years ago
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    I see, thanks for the detailed answer!

  12. Aylin
    • 2 years ago
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    You're welcome! I hope my ramblings made sense. :)

  13. Shido88
    • 2 years ago
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    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
    • 2 years ago
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    It does and thank you ! :D

  15. Aylin
    • 2 years ago
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    Well, what do you mean by which goes first? Do you mean out of [2,5/2) and (5/2, infinity)?

  16. Shido88
    • 2 years ago
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    yes

  17. Shido88
    • 2 years ago
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    that and which one is x and y ?

  18. Shido88
    • 2 years ago
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    i thought the x that we found is 5/2 shouldn't that go first before the 2 ?

  19. Shido88
    • 2 years ago
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    I'm sorry if I confuse you lol :[

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

  22. Shido88
    • 2 years ago
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    oh yes!

  23. Shido88
    • 2 years ago
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    ok let me try the next problem! thank you so much!!

  24. Aylin
    • 2 years ago
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    You're very welcome! :)

  25. Shido88
    • 2 years ago
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    hey, are you still here?

  26. Shido88
    • 2 years ago
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    I am stuck... lol

  27. Aylin
    • 2 years ago
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    Yep

  28. Aylin
    • 2 years ago
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    What's the problem?

  29. Shido88
    • 2 years ago
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    is there to determine the domain and range using the graphing calculator ?

  30. Shido88
    • 2 years ago
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    is there a way* excuse my typing..

  31. Aylin
    • 2 years ago
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    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
    • 2 years ago
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    yea

  33. Aylin
    • 2 years ago
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    You 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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