The function \( f(x)=\frac{5 x+7}{x-9} \) is one-to-0ne, a. Find an equation for \( f^{-1}(x) \), the inverse function, b. Verify that your equation is correct by showing that \( f\left(f^{-1}(x)\right)=x \) and \( f^{-1}(f(x))=x \). a. Select the correct choice below and fill in the answer box(es) to complete your choice, (Simplify your answers, Use integers or fractions for any numbers in the expression.) A. \( f^{-1}(x)= \) \( \square \) , for all \( x \) B. \( f^{-1}(x)=\frac{9 x+7}{x-5} \), for \( x \neq 5 \) c. \( f^{-1}(x)= \) \( \square \) , for \( x \geq \) \( \square \) D. \( f^{-1}(x)= \) \( \square \) , for \( x \leq \) \( \square \) b. Verify that the equation is correct. (Simplify your answers, Use integers or fractions for any numbers in the expression.) \[ \begin{array}{rlrlrl} f\left(f^{-1}(x)\right) & \equiv f(\square) & \text { and } & f^{-1}(f(x)) & \equiv f^{1}(\square) & \\ & \equiv \square & & \equiv \square & & \text { Substitute. } \\ & & & \text { Simplify. } \end{array} \]

To solve the problem, we will follow the steps outlined: ### Part a: Find the inverse function \( f^{-1}(x) \) 1. **Start with the function**: \[ f(x) = \frac{5x + 7}{x - 9} \] 2. **Replace \( f(x) \) with \( y \)**: \[ y = \frac{5x + 7}{x - 9} \] 3. **Swap \( x \) and \( y \)** to find the inverse: \[ x = \frac{5y + 7}{y - 9} \] 4. **Multiply both sides by \( y - 9 \)** to eliminate the fraction: \[ x(y - 9) = 5y + 7 \] 5. **Distribute \( x \)**: \[ xy - 9x = 5y + 7 \] 6. **Rearrange the equation to isolate terms involving \( y \)**: \[ xy - 5y = 9x + 7 \] 7. **Factor out \( y \)**: \[ y(x - 5) = 9x + 7 \] 8. **Solve for \( y \)**: \[ y = \frac{9x + 7}{x - 5} \] Thus, the inverse function is: \[ f^{-1}(x) = \frac{9x + 7}{x - 5} \] ### Part b: Verify the inverse function 1. **Verify \( f(f^{-1}(x)) = x \)**: \[ f(f^{-1}(x)) = f\left(\frac{9x + 7}{x - 5}\right) \] Substitute \( \frac{9x + 7}{x - 5} \) into \( f(x) \): \[ f\left(\frac{9x + 7}{x - 5}\right) = \frac{5\left(\frac{9x + 7}{x - 5}\right) + 7}{\left(\frac{9x + 7}{x - 5}\right) - 9} \] Simplifying the numerator: \[ 5\left(\frac{9x + 7}{x - 5}\right) + 7 = \frac{45x + 35}{x - 5} + \frac{7(x - 5)}{x - 5} = \frac{45x + 35 + 7x - 35}{x - 5} = \frac{52x}{x - 5} \] Simplifying the denominator: \[ \left(\frac{9x + 7}{x - 5}\right) - 9 = \frac{9x + 7 - 9(x - 5)}{x - 5} = \frac{9x + 7 - 9x + 45}{x - 5} = \frac{52}{x - 5} \] Therefore: \[ f(f^{-1}(x)) = \frac{\frac{52x}{x - 5}}{\frac{52}{x - 5}} = x \] 2. **Verify \( f^{-1}(f(x)) = x \)**: \[ f^{-1}(f(x)) = f^{-1}\left(\frac{5x + 7}{x - 9}\right) \] Substitute \( \frac{5x + 7}{x - 9} \) into \( f^{-1}(x) \): \[ f^{-1}\left(\frac{5x + 7}{x - 9}\right) = \frac{9\left(\frac{5x + 7}{x - 9}\right) + 7}{\left(\frac{5x + 7}{x - 9}\right) - 5} \] Simplifying the numerator: \[ 9\left(\frac{5x + 7}{x - 9}\right) + 7 = \frac{45x + 63}{x - 9} + \frac{7(x - 9)}{x - 9} = \frac{45x + 63 + 7x - 63}{x - 9} = \frac{52x}{x - 9} \] Simplifying the denominator: \[ \left(\frac{5x + 7}{x - 9}\right) - 5 = \frac{5x + 7 - 5(x - 9)}{x - 9} = \frac{5x + 7 - 5x + 45}{x - 9} = \frac{52}{x - 9} \] Therefore: \[ f^{-1}(f(x)) = \frac{\frac{52x}{x - 9}}{\frac{52}{x - 9}} = x \] ### Final Answers a. The inverse function is: \[ f^{-1}(x) = \frac{9x + 7}{x - 5}, \text{ for } x \neq 5 \] b. Verification: \[ f\left(f^{-1}(x)\right) \equiv x \text{ and } f^{-1}(f(x)) \equiv x \]