## Precalculus (10th Edition)

$f(g(x))=(\frac{2}{x-3})^2+2$. The domain is all real numbers except $-3$. $f(g(5))=3$.
$f(g(x))=f(\frac{2}{x-3})=(\frac{2}{x-3})^2+2$. The denominator of a function cannot be $0$, hence the domain is all real numbers except $-3$. $f(g(5))=(\frac{2}{5-3})^2+2=(\frac{2}{2})^2+2=1^2+2=3$.