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

1. anonymous

$y=\frac{x+5}{\sqrt{x^2-1}}$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

The above 'D' means 'the set of all real numbers, except for those between -1 and 1 inclusive of -1 and 1'.

3. anonymous

You're other equation is right, btw.

4. anonymous

*Your

5. anonymous

Okay, thanks for letting me know..