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The question I'm trying to solve is ${\frac {2x+1}{x \sqrt{x^2-1}}}$

The denominator should not be zero. And the radicand should be non-negative. If there was no $x$ behind the radical, then the domain could have been calculated with $x^2-1 > 0$, but that $x$ behind the radical makes problem for me.

Also, could you please help me with find the domain of ${\frac {2x+1}{x- \sqrt{x^2-1}}}$

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    $\begingroup$ What did you try?Do you know how to find the domain? $\endgroup$ – kingW3 Jan 7 '17 at 9:46
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You want $x\sqrt{x^2-1} \neq 0$, thus $x \neq 0$, and $x^2 - 1 > 0\implies x < -1 $ or $x > 1$. Thus $|x| > 1$ is what gives the domain of the function. For the second function, since $x - \sqrt{x^2-1} \neq 0$, we require $x^2 - 1 \ge 0\implies |x| \ge 1$ gives the domain.

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The domain of a function is the set of $x$-values for which the function is defined.

Example: the domain of $f(x) = \frac{1}{x}$ is equal to $(-\infty,0)\cup(0,\infty).$

Example (2): the domain of $g(x) = \sqrt{x}$ is equal to $[0,\infty)$.

Now evaluate the domain of $\frac{2x+1}{x\sqrt{x^2-1}}.$

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  • $\begingroup$ This is not a very thorough answer as it does not help the OP. $\endgroup$ – DeepSea Jan 7 '17 at 11:04
  • $\begingroup$ it does. I provides him with a structure how to tackle problems like these. I prefer that over giving the explicit answer like you. $\endgroup$ – user394255 Jan 7 '17 at 11:12
  • $\begingroup$ It is like you answer to no one ! $\endgroup$ – DeepSea Jan 7 '17 at 11:13

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