Question

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

Answer 1

\[y=\frac{x+5}{\sqrt{x^2-1}}\]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\[D= \mathbb{R}-\left\{ x|-1 \le x \le1 \right\}\]

Answer 2

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

Answer 3

You're other equation is right, btw.

Answer 4

*Your

Answer 5

Okay, thanks for letting me know..