Let \( f(x)=\frac{1}{x+9} \) \( f^{-1}(x)=\square \)

To find the inverse function \( f^{-1}(x) \) of \( f(x) = \frac{1}{x+9} \), we first replace \( f(x) \) with \( y \): \[ y = \frac{1}{x+9} \] Next, we swap \( x \) and \( y \) to solve for \( y \): \[ x = \frac{1}{y+9} \] Now, we multiply both sides by \( y + 9 \) to eliminate the fraction: \[ x(y + 9) = 1 \] Expanding gives: \[ xy + 9x = 1 \] Now isolate \( y \): \[ xy = 1 - 9x \] \[ y = \frac{1 - 9x}{x} \] Thus the inverse function is: \[ f^{-1}(x) = \frac{1 - 9x}{x} \] But we can also express it more neatly: \[ f^{-1}(x) = \frac{1}{x} - 9 \] And there you have it!