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