Verified answer

In order for f(x) to be a real number, the expression (4x^2 - 5) must be 0 or greater. If that number equals less than zero, its square root will be imaginary.

So x can only be values such that (4x^2 - 5) >= 0.

4x^2 - 5 >= 0

4x^2 >= 5

x^2 >= 5/4

x >= sqrt(5/4), or sqrt(5) / 2

That is the domain of the function.

The range will be real numbers.

Domain is:

4x^2 - 5 >= 0

4x^2 >= 5

x^2 >= 5/4

x^2 >= 1.25

x >= sqrt(1.25) or x <= -sqrt(1.25)

Approximately:

x >= 1.1180339887498948482045868343656 or x <= -1.1180339887498948482045868343656

The range is: f(x) >= 0, since sqrt() is assumed to yield a non-negative number unless otherwise specified.