Answer
$f(g(x))$ is continuous for all values of $x$ such that $x\ne1$ and $x\ne-1.$
Work Step by Step
Using Theorem $1.12:$
$g(x)$ is continuous for any value of $x$ but $f(x)$ has restrictions:
For $f(x)\to x\ne6.\to g(x)\ne6\to$
$ x^2+5\ne6\to x\ne1$ and $x\ne-1.$
$f(g(x))$ is continuous for all values of $x$ such that $x\ne1$ and $x\ne-1.$
