Level Curves: Homework, need help!

  • Thread starter gokugreene
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  • #1
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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:

Answers and Replies

  • #2
G01
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That sounds correct to me. I just graphed it and it also looks correct, Unless im missing something.
 
  • #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
 
  • #4
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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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