nclude " \( f^{-1}(x)= \) " in your answer. \( \quad f(x)=\frac{5 x-8}{9 x+6} \)

To find the inverse function \( f^{-1}(x) \), we start with \( y = f(x) = \frac{5x - 8}{9x + 6} \). To solve for \( x \), we can interchange \( x \) and \( y \) and solve for \( y \): 1. \( x = \frac{5y - 8}{9y + 6} \) 2. Multiply both sides by \( 9y + 6 \): \( x(9y + 6) = 5y - 8 \) 3. Rearrange it: \( 9xy + 6x = 5y - 8 \) 4. Collect all \( y \) terms on one side: \( 9xy - 5y = -6x - 8 \) 5. Factor out \( y \): \( y(9x - 5) = -6x - 8 \) 6. Thus, \( y = \frac{-6x - 8}{9x - 5} \) Therefore, \( f^{-1}(x) = \frac{-6x - 8}{9x - 5} \). To verify, we can calculate \( f(f^{-1}(x)) \) and check if we return to \( x \). It’s baffling how functions can flip on their heads, isn’t it?