Answer
(a.)$(f\circ g)(x) =x$.
(b.)$(g\circ f)(x) =x$.
(c.)$(f\circ g)(2) =2$.
Work Step by Step
The given functions are
$f(x)=6x-3$ and $g(x)=\frac{x+3}{6}$.
(a) $(f\circ g)(x) = f(g(x))$
Replace $x$ with $g(x)$ in the function $f(x)$.
$f(g(x))=6(g(x))-3$
Plug value of $g(x)$ in the right hand side.
$f(g(x))=6(\frac{x+3}{6})-3$
$f(g(x))=x+3-3$
$f(g(x))=x$.
Hence, $(f\circ g)(x) =x$.
(b) $(g\circ f)(x) = g(f(x))$
Replace $x$ with $f(x)$ in the function $g(x)$.
$g(f(x))=\frac{f(x)+3}{6}$
Plug value of $f(x)$ in the right hand side.
$g(f(x))=\frac{6x-3+3}{6}$
$g(f(x))=\frac{6x}{6}$
$g(f(x))=x$.
Hence, $(g\circ f)(x) =x$.
(c) Replace $x$ with $2$ in the part (a) solution.
$(f\circ g)(2) =2$.