Answer
(a) $\frac{4}{(x-3)^2}+2$, $\{x|x\ne3 \}$.
$3$.
(b) $log_2(x)+2$, $\{x|x\gt0 \}$.
$log_2(14)+2$.
Work Step by Step
(a) Given $f(x)=x^2+2$ and $g(x)=\frac{2}{x-3}$, we have $f(g(x))=(\frac{2}{x-3})^2+2=\frac{4}{(x-3)^2}+2$ with domain $\{x|x\ne3 \}$.
We have $f(g(5))=\frac{4}{(5-3)^2}+2=3$.
(b) Given $f(x)=x+2$ and $g(x)=log_2(x)$, we have $f(g(x))=(log_2(x))+2=log_2(x)+2$ with domain $\{x|x\gt0 \}$.
We have $f(g(14))=log_2(14)+2$.
