If $f(x)=\sqrt{x-3}$, what is the smallest real number $x$ in the domain of $f(f(x))$?
I know that $f(x)$ is only defined for $x \geq 3$, since I need to have a non-negative radicand under the square root. So, for $f(f(x))$ to be defined, I need to have $f(x) \geq 3$ as well, since it will be the new radicand. This means that $\sqrt{x-3} \geq 3$, or equivalently, $x-3 \geq 9$, after squaring both sides. Adding $3$ to both sides, I get $x \geq 12$. Therefore, the smallest real number $x$ in the domain of $f(f(x))$ is $12$. 