Answer
$${f^{ - 1}}\left( x \right) = \frac{{\sqrt 7 x}}{{\sqrt {1 - {x^2}} }}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{x}{{\sqrt {{x^2} + 7} }} \cr
& \left( {\text{a}} \right) \cr
& {\text{Let }}y = f\left( x \right) \cr
& y = \frac{x}{{\sqrt {{x^2} + 7} }} \cr
& {\text{Interchange }}x{\text{ and }}y \cr
& x = \frac{y}{{\sqrt {{y^2} + 7} }} \cr
& {\text{Solve for }}y \cr
& {x^2} = \frac{{{y^2}}}{{{y^2} + 7}} \cr
& {x^2}{y^2} + 7{x^2} = {y^2} \cr
& {y^2} - {x^2}{y^2} = 7{x^2} \cr
& {y^2} = \frac{{7{x^2}}}{{1 - {x^2}}} \cr
& y = \sqrt {\frac{{7{x^2}}}{{1 - {x^2}}}} \cr
& y = \frac{{\sqrt 7 x}}{{\sqrt {1 - {x^2}} }} \cr
& {\text{Let }}y = {f^{ - 1}}\left( x \right) \cr
& {f^{ - 1}}\left( x \right) = \frac{{\sqrt 7 x}}{{\sqrt {1 - {x^2}} }} \cr
& \left( {\text{b}} \right){\text{ Graph shown below}} \cr
& \left( {\text{c}} \right){\text{ The graphs of }}f\left( x \right){\text{ and }}{f^{ - 1}}\left( x \right){\text{ are reflections of each other}}{\text{.}} \cr
& \left( {\text{d}} \right){\text{Domain of }}f\left( x \right) = {\text{Range }}{f^{ - 1}}\left( x \right) = \left( { - \infty ,\infty } \right) \cr
& {\text{Domain of }}{f^{ - 1}}\left( x \right) = {\text{Range }}f\left( x \right) = \left( {1,1} \right) \cr
& \cr
& {\text{Graph}} \cr} $$
