Question

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

Answer 1

Domain of `h(x)` `=(-oo,-1]uu[1,4]`

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 `RR`).
2. Division by zero. (Not defined at all).

In this case there are no divisions (denominators) in `h(x)` 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 `{x in RR | -oo < x<= 1 and 1 <= x <= 4}`

`=(-oo,-1]uu[1,4]`.

The graph makes it clear :

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