Calculus 10th Edition

$f(g(x))$ is continuous for all values of $x$ such that $x\ne1$ and $x\ne-1.$
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 $x\ne-1.$