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anonymous

  • 5 years ago

domain of: y=x+5/square root: (x^2-1)

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  1. anonymous
    • 5 years ago
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    \[y=\frac{x+5}{\sqrt{x^2-1}}\]The domain of this function is the set of all x such that the function is well-defined. You need to ensure two things here. 1. the numerator cannot be 0 2. the radiacand (expression under the sqrt sign) cannot be negative (for real numbers). Let's look for x-values the function CAN'T take. This will be when\[\sqrt{x^2-1}=0\]by the first point, and when\[x^2-1<0\]by the second point. For the first,\[\sqrt{x^2-1}=0 \rightarrow x^2-1=0 \rightarrow x= \pm 1\]For the second,\[x^2-1<0 \rightarrow x^2<1 \rightarrow -1<x<1\]So when x is between -1 and 1, and inclusive of -1 and 1, the function fails to exist. So your domain is everything else; that is, your domain is\[D= \mathbb{R}-\left\{ x|-1 \le x \le1 \right\}\]

  2. anonymous
    • 5 years ago
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    The above 'D' means 'the set of all real numbers, except for those between -1 and 1 inclusive of -1 and 1'.

  3. anonymous
    • 5 years ago
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    You're other equation is right, btw.

  4. anonymous
    • 5 years ago
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    *Your

  5. anonymous
    • 5 years ago
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    Okay, thanks for letting me know..

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