anonymous
  • anonymous
the domain of f(x)= 1/ sqrt 1-x^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
I understand that a sqrt needs to be > or = to 0 and that the denominator needs to be 0, but I'm lost on how to combine the two concepts
anonymous
  • anonymous
Do I just have to make sqrt 1-x^2 equal to zero?
anonymous
  • anonymous
since the denominator must not be a zero, you can rule out the equality, thus, leaving you with solving \[1-x^2>0\]

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anonymous
  • anonymous
ok, so is it \[\sqrt{1}>x?\]
anonymous
  • anonymous
nope. since you have not considered other values. try doing \[1-x^2>0 \rightarrow x^2-1<0 \rightarrow (x-1)(x+1)<0\]
anonymous
  • anonymous
how do you get to the domain from that?
anonymous
  • anonymous
[-11]
anonymous
  • anonymous
I meant (-11)
anonymous
  • anonymous
From there, we have -1 < x < 1, and this is the range of values permissible for x in the function. So, the domain is -1 < x < 1. Why -1 < x < 1? here's how... \[(x-1)(x+1)<0\] implies that one of the factors x - 1 or x + 1 is negative, i.e. less than 0, and the other factor is positive, i.e. greater than 0. So we would have 2 cases: Case 1: \[x-1>0 \rightarrow x>1\] and\[x+1 <0 \rightarrow x <-1\]however, this cannot be simultaneously true, so no solution for this case. Case 2: \[x-1<0 \rightarrow x<1\]and\[x+1>0 \rightarrow x>-1\]the two inequalities are true for -1 < x < 1, thus the answer.
anonymous
  • anonymous
I give myself an F for not practicing my algebra skills for a year. (-1
anonymous
  • anonymous
no problem... :D

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