Answer
$a.\quad(f\circ g)(1) = 22$
$b.\quad(g\circ f)(1) = \sqrt{26}$
$c.\quad (f\circ g)(x) = x+21$
$d.\quad (g\circ f)(x) = \sqrt{x^{2}+25}$
Work Step by Step
$(f\circ g)(x)=f[g(x)]=[g(x)]^{2}+8$
$= |x+17|+4 $
since the domain of g is $[-17,\infty)$ (and of $f\circ g$,)
we can drop the absolute value brackets, because the expression $(x+17)$
is not negative on the domain.
$(f\circ g)(x)=x+21$ $\qquad ... \quad(c)$
$(f\circ g)(1)= 1+21 =22 \qquad ... \quad(a)$
$(g\circ f)(x)=g[f(x)]=\sqrt{f(x)+17}$
$= \sqrt{x^{2}+8+17} $
$= \sqrt{x^{2}+25} \qquad ... \quad(d)$
$(g\circ f)(1)= \sqrt{1^{2}+25} =\sqrt{26} \qquad ... \quad(b)$
