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

To find the inverse of the function \( f(x) = \frac{1}{x + 9} \), we start by setting \( y = f(x) \), which gives us \( y = \frac{1}{x + 9} \). Next, we swap \( x \) and \( y \) to find the inverse: 1. Write it as \( x = \frac{1}{y + 9} \). 2. Multiply both sides by \( y + 9 \): \( x(y + 9) = 1 \). 3. Expand: \( xy + 9x = 1 \). 4. Rearrange to isolate \( y \): \( y = \frac{1 - 9x}{x} \). Thus, the inverse function is \( f^{-1}(x) = \frac{1 - 9x}{x} \). If you’re ever puzzled over finding inverses, remember: switch \( x \) and \( y \), solve for \( y \), and don’t forget to check your work by confirming that \( f(f^{-1}(x)) = x \). It’s a fun little loop! Additionally, confirming the domain is essential! The original function \( f(x) \) is defined for all \( x \neq -9 \) (where it asymptotes), which means the inverse will have a restriction on \( x \); it’s valid for all \( x \neq 0 \), since division by zero is a no-go.