Let \( f(x) = 5x + 1 \). Determine its inverse function and demonstrate that they are inverses by showing that both compositions \( f(f^{-1}(x)) \) and \( f^{-1}(f(x)) \) return x.

To find the inverse of the function \( f(x) = 5x + 1 \) and demonstrate that they are indeed inverses, follow these steps: ### 1. Finding the Inverse Function \( f^{-1}(x) \) To find the inverse function, we need to solve the equation \( y = 5x + 1 \) for \( x \) in terms of \( y \): \[ y = 5x + 1 \] Subtract 1 from both sides: \[ y - 1 = 5x \] Now, divide both sides by 5 to solve for \( x \): \[ x = \frac{y - 1}{5} \] Therefore, the inverse function is: \[ f^{-1}(x) = \frac{x - 1}{5} \] ### 2. Verifying the Inverses through Composition To confirm that \( f \) and \( f^{-1} \) are indeed inverses, we need to show that: 1. \( f(f^{-1}(x)) = x \) 2. \( f^{-1}(f(x)) = x \) #### a. Composition \( f(f^{-1}(x)) \) \[ f(f^{-1}(x)) = f\left(\frac{x - 1}{5}\right) = 5\left(\frac{x - 1}{5}\right) + 1 \] Simplify the expression: \[ 5 \times \frac{x - 1}{5} = x - 1 \] \[ x - 1 + 1 = x \] So, \[ f(f^{-1}(x)) = x \] #### b. Composition \( f^{-1}(f(x)) \) \[ f^{-1}(f(x)) = f^{-1}(5x + 1) = \frac{(5x + 1) - 1}{5} \] Simplify the expression: \[ \frac{5x + 1 - 1}{5} = \frac{5x}{5} = x \] So, \[ f^{-1}(f(x)) = x \] ### Conclusion Since both compositions \( f(f^{-1}(x)) \) and \( f^{-1}(f(x)) \) return \( x \), we have confirmed that \( f(x) = 5x + 1 \) and its inverse \( f^{-1}(x) = \frac{x - 1}{5} \) are indeed inverse functions of each other. **Final Answer:** The inverse function is \( f^{-1}(x) = \dfrac{x - 1}{5} \). Demonstrating the compositions: \[ f(f^{-1}(x)) = f\left(\dfrac{x - 1}{5}\right) = 5 \cdot \dfrac{x - 1}{5} + 1 = x - 1 + 1 = x \] \[ f^{-1}(f(x)) = f^{-1}(5x + 1) = \dfrac{5x + 1 - 1}{5} = \dfrac{5x}{5} = x \] Thus, \( f \) and \( f^{-1} \) are inverses since both compositions equal \( x \).