# How do you find the domain of h(x)=sqrt(4-x)+sqrt(x^2-1)?

Jan 6, 2016

Domain of $h \left(x\right)$ $= \left(- \infty , - 1\right] \cup \left[1 , 4\right]$

#### Explanation:

Domain is all possible allowable $x$ value inputs.

There are generally 2 cases that are not allowed in real functions :

1. Even square roots of negative numbers. (Not defined in $\mathbb{R}$).
2. Division by zero. (Not defined at all).

In this case there are no divisions (denominators) in $h \left(x\right)$ so we only consider case 1 of the square roots not allowed to be negative.
Doing so, it becomes clear that the first term is only negative if $x > 4$ and the second term is only negative if $x < 1$.

So #

Hence the domain is $\left\{x \in \mathbb{R} | - \infty < x \le 1 \mathmr{and} 1 \le x \le 4\right\}$

$= \left(- \infty , - 1\right] \cup \left[1 , 4\right]$.

The graph makes it clear :

graph{sqrt(4-x)+sqrt(x^2-1) [-7.28, 10.5, -1.93, 6.95]}