domain

If there is a square root, the expression inside must be nonnegative.

As a result, $x^2 - 16 \geq 0$. You can solve this to get the domain of f(x).

Remarks: You do not need to consider $\sqrt{x^2 - 16} + 3 \geq 0$ because $\sqrt{x^2 - 16} + 3 \geq 0 + 3 = 3 > 0$ if sqrt(x^2 - 16) is defined.