# What is the domain and range of  y=sqrt(x^2-1)?

Aug 26, 2015

Domain: $\left(- \infty , - 1\right] \cup \left[1 , + \infty\right)$
Range: $\left[0 , + \infty\right)$

#### Explanation:

The domain of the function will be determined by the fact that the expression that's under the radical must be positive for real numbers.

Since ${x}^{2}$ will always be positive regardless of the sign of $x$, you need to find the values of $x$ that will make ${x}^{2}$ smaller than $1$, since those are the only values that will make the expression negative.

So, you need to have

${x}^{2} - 1 \ge 0$

${x}^{2} \ge 1$

Take the square root of both sides to get

$| x | \ge 1$

This of course means that you have

$x \ge 1 \text{ }$ and $\text{ } x \le - 1$

The domain of the function will thus be $\left(- \infty , - 1\right] \cup \left[1 , + \infty\right)$.

The range of the function will be determined by the fact that the square root of a real number must always be positive. The smallest value the function can take will happen for $x = - 1$ and for $x = 1$, since those values of $x$ will make the radical term equal to zero.

$\sqrt{{\left(- 1\right)}^{2} - 1} = 0 \text{ }$ and $\text{ } \sqrt{{\left(1\right)}^{2} - 1} = 0$

The range of the function will thus be $\left[0 , + \infty\right)$.

graph{sqrt(x^2-1) [-10, 10, -5, 5]}