The function \( f(x)=\frac{5 x+7}{x-9} \) is one-to-one. 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.) B. \( f^{-1}(x)=\square \), for all \( x \) C. \( f^{-1}(x)=\square \), for \( x \neq \square \) Dor \( x \geq \square \)

To find the inverse function \( f^{-1}(x) \) for the function \( f(x) = \frac{5x + 7}{x - 9} \), we will follow these steps: ### Step 1: Find the Inverse Function 1. **Replace \( f(x) \) with \( y \)**: \[ y = \frac{5x + 7}{x - 9} \] 2. **Swap \( x \) and \( y \)**: \[ x = \frac{5y + 7}{y - 9} \] 3. **Solve for \( y \)**: - Multiply both sides by \( y - 9 \): \[ x(y - 9) = 5y + 7 \] - Distribute \( x \): \[ xy - 9x = 5y + 7 \] - Rearrange to isolate terms involving \( y \): \[ xy - 5y = 9x + 7 \] - Factor out \( y \): \[ y(x - 5) = 9x + 7 \] - Solve for \( y \): \[ y = \frac{9x + 7}{x - 5} \] Thus, the inverse function is: \[ f^{-1}(x) = \frac{9x + 7}{x - 5} \] ### Step 2: Verify the Inverse Function Now we will verify that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \). #### Verification 1: \( f(f^{-1}(x)) \) 1. Substitute \( f^{-1}(x) \) into \( f(x) \): \[ f\left(f^{-1}(x)\right) = f\left(\frac{9x + 7}{x - 5}\right) \] - Substitute into the original function: \[ f\left(\frac{9x + 7}{x - 5}\right) = \frac{5\left(\frac{9x + 7}{x - 5}\right) + 7}{\left(\frac{9x + 7}{x - 5}\right) - 9} \] - Simplify the numerator: \[ = \frac{\frac{45x + 35}{x - 5} + 7}{\frac{9x + 7}{x - 5} - 9} \] - Convert 7 to a fraction: \[ = \frac{\frac{45x + 35 + 7(x - 5)}{x - 5}}{\frac{9x + 7 - 9(x - 5)}{x - 5}} \] - Simplify: \[ = \frac{\frac{45x + 35 + 7x - 35}{x - 5}}{\frac{9x + 7 - 9x + 45}{x - 5}} = \frac{\frac{52x}{x - 5}}{\frac{52}{x - 5}} = x \] #### Verification 2: \( f^{-1}(f(x)) \) 1. Substitute \( f(x) \) into \( f^{-1}(x) \): \[ f^{-1}(f(x)) = f^{-1}\left(\frac{5x + 7}{x - 9}\right) \] - Substitute into the inverse function: \[ = \frac{9\left(\frac{5x + 7}{x - 9}\right) + 7}{\left(\frac{5x + 7}{x - 9}\right) - 5} \] - Simplify the numerator: \[ = \frac{\frac{45x + 63}{x - 9} + 7}{\frac{5x + 7}{x - 9} - 5} \] - Convert 7 to a fraction: \[ = \frac{\frac{45x + 63 + 7(x - 9)}{x - 9}}{\frac{5x + 7 - 5(x - 9)}{x - 9}} \] - Simplify: \[ = \frac{\frac{45x + 63 + 7x - 63}{x - 9}}{\frac{5x + 7 - 5x + 45}{x - 9}} = \frac{\frac{52x}{x - 9}}{\frac{52}{x - 9}} = x \] ### Final Answer Thus, the inverse function is: \[ f^{-1}(x) = \frac{9x + 7}{x - 5} \] The correct choice is: **B.** \( f^{-1}(x) = \frac{9x + 7}{x - 5} \), for \( x \neq 5 \)