Step 1 :\(\left(f^{-1} \circ f\right)(x) = f^{-1}(f(x)) = f^{-1}\left(\frac{9-x}{x}\right)\)
Step 2 :Substitute \(f^{-1}(x)=\frac{9}{x+1}\) into the equation: \(f^{-1}\left(\frac{9-x}{x}\right) = \frac{9}{\frac{9-x}{x}+1}\)
Step 3 :Simplify the denominator: \(\frac{9}{\frac{9-x+ x}{x}} = \frac{9}{\frac{9}{x}}\)
Step 4 :Simplify the fraction: \(9 \cdot \frac{x}{9} = x\)
Step 5 :So, \(\left(f^{-1} \circ f\right)(x) = x\)
Step 6 :\(\left(f \circ f^{-1}\right)(x) = f(f^{-1}(x)) = f\left(\frac{9}{x+1}\right)\)
Step 7 :Substitute \(f(x)=\frac{9-x}{x}\) into the equation: \(f\left(\frac{9}{x+1}\right) = \frac{9-\frac{9}{x+1}}{\frac{9}{x+1}}\)
Step 8 :Simplify the numerator: \(\frac{9x+9-9}{9} = \frac{9x}{9}\)
Step 9 :Simplify the fraction: \(\frac{9x}{9} = x\)
Step 10 :So, \(\left(f \circ f^{-1}\right)(x) = x\)
Step 11 :Since both \(\left(f^{-1} \circ f\right)(x) = x\) and \(\left(f \circ f^{-1}\right)(x) = x\), we can conclude that \(f^{-1}(x)=\frac{9}{x+1}\) is indeed the inverse of \(f(x)=\frac{9-x}{x}\)
Step 12 :\(\boxed{f^{-1}(x)=\frac{9}{x+1}}\) is the inverse of \(\boxed{f(x)=\frac{9-x}{x}}\)