Answer
$f\circ g(x)=3x-5$; domain: all reals, $(-\infty,\infty)$
$g\displaystyle \circ f(x)=\frac{6x-5}{2}$; domain: all reals, $(-\infty,\infty)$
$f\circ f(x)=36x-35$; domain: all reals, $(-\infty,\infty)$
$g\displaystyle \circ g(x)=\frac{x}{4}$; domain: all reals, $(-\infty,\infty)$
Work Step by Step
f(x) is defined for all x,
g(x) is defined for all x
$f\circ g(x)=f[g(x)]=6g(x)-5$
$=6(\displaystyle \frac{x}{2})-5$
$=3x-5$; domain: all reals, $(-\infty,\infty)$
$g\displaystyle \circ f(x)=g[f(x)]=\frac{f(x)}{2}$
$=\displaystyle \frac{6x-5}{2}$; domain: all reals, $(-\infty,\infty)$
$f\circ f(x)=f[f(x)]=6f(x)-5$
$=6(6x-5)-5$
$=36x-30-5$
$=36x-35$; domain: all reals, $(-\infty,\infty)$
$g\displaystyle \circ g(x)=g[g(x)]=\frac{g(x)}{2}$
$=\displaystyle \frac{\frac{x}{2}}{2}$
$=\displaystyle \frac{x}{4}$; domain: all reals, $(-\infty,\infty)$
