+9519 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.
