For the function $\mathrm{f}$, use composition of functions to show that $\mathrm{f}^{-1}$ is as given.
\[
f(x)=\frac{9-x}{x}, f^{-1}(x)=\frac{9}{x+1}
\]

To show that $f^{-1}(x)=\frac{9}{x+1}$, find $\left(f^{-1} \circ f\right)(x)$ and $\left(f \circ f^{-1}\right)(x)$ and check to see that

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 \(\boxed{f(x)=\frac{9-x}{x}}\)