The function \( f(x)=\frac{4 x+7}{x-2} \) 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 \( f^{-1}(f(x))=x \). (Simplify your answers. Use integers or fractions for any numbers in the expression.) \[ \begin{array}{rlrlr} f\left(f^{-1}(x)\right) & =f(x) & \text { and } & f^{-1}(f(x)) & =f^{-1}(x) \\ & =x & & \text { Substitute. } \\ & =x & & \text { Simplify. } \end{array} \] The equation is verified.

```markdown We are given \[ f(x)=\frac{4x+7}{x-2}, \] and we need to find its inverse \( f^{-1}(x) \) and verify that the compositions \( f\left(f^{-1}(x)\right) \) and \( f^{-1}(f(x)) \) yield \( x \). ### (a) Finding the Inverse Function 1. **Write the function in terms of \( y \):** \[ y = \frac{4x+7}{x-2}. \] 2. **Interchange \( x \) and \( y \) to solve for \( y \):** \[ x = \frac{4y+7}{y-2}. \] 3. **Clear the fraction by multiplying both sides by \( y-2 \):** \[ x(y-2) = 4y+7. \] 4. **Distribute \( x \):** \[ xy - 2x = 4y + 7. \] 5. **Gather like terms (group the terms with \( y \) on one side):** \[ xy - 4y = 2x + 7. \] 6. **Factor \( y \) from the left-hand side:** \[ y(x-4) = 2x + 7. \] 7. **Solve for \( y \):** \[ y=\frac{2x+7}{x-4}. \] 8. **Rewrite using the inverse function notation:** \[ f^{-1}(x)=\frac{2x+7}{x-4}. \] ### (b) Verifying the Inverse Function We now verify by showing both \( f\left(f^{-1}(x)\right)=x \) and \( f^{-1}(f(x))=x \). #### 1. Verify \( f\left(f^{-1}(x)\right)=x \) - **Substitute \( f^{-1}(x) \) into \( f \):** \[ f\left(f^{-1}(x)\right)=f\left(\frac{2x+7}{x-4}\right)=\frac{4\left(\frac{2x+7}{x-4}\right)+7}{\frac{2x+7}{x-4}-2}. \] - **Simplify the numerator:** \[ 4\left(\frac{2x+7}{x-4}\right)+7=\frac{8x+28}{x-4}+\frac{7(x-4)}{x-4}=\frac{8x+28+7x-28}{x-4}=\frac{15x}{x-4}. \] - **Simplify the denominator:** \[ \frac{2x+7}{x-4}-2=\frac{2x+7-2(x-4)}{x-4}=\frac{2x+7-2x+8}{x-4}=\frac{15}{x-4}. \] - **Combine the results:** \[ f\left(f^{-1}(x)\right)=\frac{\frac{15x}{x-4}}{\frac{15}{x-4}}=x. \] #### 2. Verify \( f^{-1}(f(x))=x \) - **Substitute \( f(x) \) into \( f^{-1} \):** \[ f^{-1}(f(x))=f^{-1}\left(\frac{4x+7}{x-2}\right)=\frac{2\left(\frac{4x+7}{x-2}\right)+7}{\frac{4x+7}{x-2}-4}. \] - **Simplify the numerator:** \[ 2\left(\frac{4x+7}{x-2}\right)+7=\frac{8x+14}{x-2}+\frac{7(x-2)}{x-2}=\frac{8x+14+7x-14}{x-2}=\frac{15x}{x-2}. \] - **Simplify the denominator:** \[ \frac{4x+7}{x-2}-4=\frac{4x+7-4(x-2)}{x-2}=\frac{4x+7-4x+8}{x-2}=\frac{15}{x-2}. \] - **Combine the results:** \[ f^{-1}(f(x))=\frac{\frac{15x}{x-2}}{\frac{15}{x-2}}=x. \] ### Conclusion The inverse function is correctly found and verified: \[ f^{-1}(x)=\frac{2x+7}{x-4}, \] and we have demonstrated that \[ f\left(f^{-1}(x)\right)=x \quad \text{and} \quad f^{-1}(f(x))=x. \] ```