# How do you find the domain and range of f(x)= sqrt(x^2-1)?

Aug 5, 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 expression that's under the square root.

More specifically, for real numbers, the square root is defined exclusively for positive numbers, which means that ${x}^{2} - 1 \ge 0$, $\left(\forall\right) x \in \mathbb{R}$.

This quadratic has two solutions

${x}^{2} - 1 = 0$

${x}^{2} = 1 \implies x = \pm 1$

For numbers outside the $\left(- 1 , 1\right)$ interval, this expression will always be positive. However, for values of $x$ Inside the interval, this expression will be negative, which means that the domain of the function cannot include values in the $\left(- 1 , 1\right)$ interval.

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

The range of the function will be determine by the fact that the square root will always give a positive value for real numbers.

This means that the range of the will be $\left[0 , + \infty\right)$.

graph{sqrt(x^2-1) [-7.9, 7.904, -3.946, 3.956]}