Answer
$f(g(x))$ is continuous for all values of $x$ such that $x\gt1.$
Work Step by Step
Using Theorem $1.12:$
$g(x)$ is continuous for all values of $x$ but $f(x)$ is continuous for $x\gt0\to$
$g(x)\gt0\to x-1\gt0\to x\gt1.$
$f(g(x))$ is continuous for all values of $x$ such that $x\gt1.$
Note:
The restrictions on $f(x)$ come from the presence of a square root in the denominator; the radicand cannot be negative and the denominator cannot be $0$, hence $x\gt0.$
