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Level Curves: Homework, need help!

  1. Oct 8, 2005 #1
    I am trying to find and graph the level curve [tex]f(x,y)=\sqrt{x^2-1}[/tex] that passes throught the point [tex](0,1)[/tex], as well as its domain and range.

    I am not sure if my reasoning is right, so let me know if I got the wrong idea.

    For the graph I have [tex]x = 1[/tex] which is independent of y and is just a vertical line. Is this correct?

    Would the domain be [tex](-\infty, -1]\cup[1,\infty)[/tex] or [tex][1,\infty)[/tex] ? Because [tex]\sqrt{x^2-1} = \sqrt{x-1}\sqrt{x+1}[/tex]
    I'm confused.

    Range: [tex][0,\infty)[/tex]

    any help would be greatly appreciated

    Thanks

    Update: this a function of two variables
     
    Last edited: Oct 8, 2005
  2. jcsd
  3. Oct 8, 2005 #2

    G01

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    That sounds correct to me. I just graphed it and it also looks correct, Unless im missing something.
     
  4. Oct 8, 2005 #3

    mezarashi

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    If we have [tex] y = \sqrt{x^2 - 1}[/tex], then x cannot exist on the real domain for any value: [tex]\mid x \mid \leq 1[/tex]. o_O
     
  5. Oct 8, 2005 #4
    I am looking for the domain, range, and graph of this level curve[tex]f(x,y)=\sqrt{x^2-1}[/tex]

    I have the range, but the domain and graph I am unsure of.
    For the graph, I graphed on my paper x=1, you plug in one, z = 0 and x = 1 and y can be anything.
    Domain: I am unsure of but yes [tex]\mid x \mid \leq 1 [/tex]
    But if I plug say x = -1 into [tex]\sqrt{x-1}\sqrt{x+1}[/tex] I am going to get [tex]\sqrt{-2}\sqrt{2}[/tex]
     
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