Answer
$\mathrm{See\:the\:table\:below.}$
Work Step by Step
$b.\quad$ Let's say $\ g(x)=y$. Then our composite function is given as:
$(f\circ g)(x)=f(g(x))$
$f(y)=\frac{y-1}{y}$
$\frac{x}{x+1}=1+\frac{1}{y}$
Solve this equation for $\ y\ $ which will give us the value of $\ g(x).$
$yx+y-x-1=xy$
$y=x+1$
$c.\ \ $ It is interesting to note that $\ \sqrt{x^2}=|x|$
$d.\ \ $ We know that $\ (\sqrt{x})^2=x\ $. Since the domain of $\ g(x)\ $ is $\ [0,\infty)\ ,$ we can have $\ x=|x|.$
