Answer
$${f^{ - 1}}\left( x \right) = \sqrt {x + 5} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^2} - 5,{\text{ }}x \geqslant 0 \cr
& {\text{Domain of }}f\left( x \right),{\text{ }}D:\left[ {0,\infty } \right) \cr
& \left( {\text{a}} \right)f\left( x \right) = {x^2} - 5 \cr
& {\text{Let }}y = f\left( x \right) \cr
& y = {x^2} - 5 \cr
& {\text{Interchange }}x{\text{ and }}y \cr
& x = {y^2} - 5 \cr
& {\text{Solve for }}y \cr
& x + 5 = {y^2} \cr
& y = \sqrt {x + 5} \cr
& {\text{Let }}y = {f^{ - 1}}\left( x \right) \cr
& {f^{ - 1}}\left( x \right) = \sqrt {x + 5} \cr
& {\text{Domain }}\left[ { - 5,\infty } \right) \cr
& \cr
& \left( {\text{c}} \right) \cr
& *{f^{ - 1}}\left( {f\left( x \right)} \right) = \sqrt {{x^2} - 5 + 5} \cr
& = \sqrt {{x^2}} \cr
& = x \cr
& *f\left( {{f^{ - 1}}\left( x \right)} \right) = {\left( {\sqrt {x + 5} } \right)^2} - 5 \cr
& = x + 5 - 5 \cr
& = x \cr
& \cr
& \left( {\text{d}} \right){\text{Domain of }}f\left( x \right) = {\text{Range }}{f^{ - 1}}\left( x \right) = \left[ {0,\infty } \right) \cr
& {\text{Domain of }}{f^{ - 1}}\left( x \right) = {\text{Range }}f\left( x \right) = \left[ { - 5,\infty } \right) \cr
& \cr
& \left( {\text{c}} \right){\text{Graph}} \cr} $$
