Answer
$a.\quad(f\circ g)(1) = 4$
$b.\quad(g\circ f)(1) = 2$
$c.\quad (f\circ g)(x) = x+3 $
$d.\quad (g\circ f)(x) = \sqrt{x^{2}+3}$
Work Step by Step
$(f\circ g)(x)=f[g(x)]=[g(x)]^{2}+4$
$= |x-1|+4 $
since the domain of g is $[1,\infty)$, which is the also domain of $(f\circ g)$
we can drop the absolute value brackets
$=x+3\qquad ... \quad(c)$
$(f\circ g)(1)= 1+3 =4 \qquad ... \quad(a)$
$(g\circ f)(x)=g[f(x)]=\sqrt{f(x)-1}$
$= \sqrt{x^{2}+4-1} $
$= \sqrt{x^{2}+3} \qquad ... \quad(d)$
$(g\circ f)(1)= \sqrt{1^{2}+3} =2 \qquad ... \quad(b)$
