Answer
$f(g(x))=\frac{1}{\sqrt{x+2}}$; Domain: $(-2,\infty)$
$g(f(x))=\sqrt{\frac{1}{x}+2}$; Domain: $(-\infty,-\frac{1}{2}]∪(0,\infty)$
The two composite functions are not equal.
Work Step by Step
$f(g(x))$
$f(\sqrt{x+2})$
$\frac{1}{\sqrt{x+2}}$
The expression under the square root must be greater than or equal to 0:
$x+2\geq0$
$x\geq-2$
Also the denominator must not equal 0 so:
$\sqrt{x+2}\ne0$
$x\ne-2$
Therefore the domain is $x>-2$
$g(f(x))$
$g(\frac{1}{x})$
$\sqrt{\frac{1}{x}+2}$
The expression under the square root must be greater than or equal to 0:
$\frac{1}{x}+2\geq0$
$\frac{1}{x}\geq-2$
$x\leq-\frac{1}{2},x>0$
The two composite functions are not equal since they do not have the same output or domain.
