Natural Domain

 $f\left(x\right)=\frac{1}{x},\ x\ne0$  

 $f\left(x\right)=\frac{x-9}{\left(x-2\right)\left(x+3\right)},\ x\ne-3,\ 2$  

 $f\left(x\right)=\sqrt{x}\ ,\ x\ge0$  

 $f\left(x\right)=3\sqrt{x-4}+5\ ,\ x\ge4$  

However;  $f\left(x\right)=\sqrt{x^2+3}$  has a domain of All Real Numbers because the radicand cannot be negative (no matter what we substitute for x.)

 $f\left(x\right)=\frac{1}{\sqrt{x}},\ x>0$  notice the  $\ge$  is not just >

 $f\left(x\right)=\frac{3x}{\sqrt{x-2}},\ x>2$