Answer
$${f^{ - 1}}\left( x \right) = 2x + 6$$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)f\left( x \right) = \frac{1}{2}x - 3 \cr
& {\text{Let }}y = f\left( x \right) \cr
& y = \frac{1}{2}x - 3 \cr
& {\text{Interchange }}x{\text{ and }}y \cr
& x = \frac{1}{2}y - 3 \cr
& {\text{Solve for }}y \cr
& 2x = y - 6 \cr
& y = 2x + 6 \cr
& {\text{Let }}y = {f^{ - 1}}\left( x \right) \cr
& {f^{ - 1}}\left( x \right) = 2x + 6 \cr
& \cr
& \left( {\text{c}} \right) \cr
& *{f^{ - 1}}\left( {f\left( x \right)} \right) = 2\left( {\frac{1}{2}x - 3} \right) + 6 \cr
& = x - 6 + 6 \cr
& = x \cr
& *f\left( {{f^{ - 1}}\left( x \right)} \right) = \frac{1}{2}\left( {2x + 6} \right) - 3 \cr
& = x + 3 - 3 \cr
& = x \cr
& \cr
& \left( {\text{d}} \right){\text{Domain of }}f\left( x \right) = {\text{Range }}{f^{ - 1}}\left( x \right) = \left( { - \infty ,\infty } \right) \cr
& {\text{Domain of }}{f^{ - 1}}\left( x \right) = {\text{Range }}f\left( x \right) = \left( { - \infty ,\infty } \right) \cr
& \cr
& \left( {\text{c}} \right){\text{Graph}} \cr} $$
