Let \( f(x)=8+\sqrt{x-7} \). Find \( f^{-1}(x) \) \( f^{-1}(x)=\square \) Now for fun, verify that \( \left(f \circ f^{-1}\right)(x)=\left(f^{-1} \circ f\right)(x)=x \)

To find the inverse function \( f^{-1}(x) \) for \( f(x) = 8 + \sqrt{x-7} \), we start by setting \( y = f(x) \): \[ y = 8 + \sqrt{x-7} \] Next, we isolate \( \sqrt{x-7} \): \[ \sqrt{x-7} = y - 8 \] Now, we square both sides: \[ x - 7 = (y - 8)^2 \] Then, we solve for \( x \): \[ x = (y - 8)^2 + 7 \] Now we swap \( x \) and \( y \) to get the inverse: \[ f^{-1}(x) = (x - 8)^2 + 7 \] In conclusion: \[ f^{-1}(x) = (x - 8)^2 + 7 \] Now let’s verify that \( (f \circ f^{-1})(x) = x \) and \( (f^{-1} \circ f)(x) = x \). Starting with \( f(f^{-1}(x)) \): \[ f(f^{-1}(x)) = f((x - 8)^2 + 7) \] \[ = 8 + \sqrt{((x-8)^2 + 7) - 7} \] \[ = 8 + \sqrt{(x - 8)^2} \] \[ = 8 + |x - 8| \] Since \( |x - 8| = x - 8 \) when \( x \geq 8 \) and \( |x - 8| = 8 - x \) when \( x < 8 \), we can see that: - For \( x \geq 8 \): \( f(f^{-1}(x)) = 8 + (x - 8) = x \) - For \( x < 8 \): \( f(f^{-1}(x)) = 8 + (8 - x) = x \) So \( f(f^{-1}(x)) = x \) for all \( x \). Now, for \( f^{-1}(f(x)) \): \[ f^{-1}(f(x)) = f^{-1}(8 + \sqrt{x - 7}) \] \[ = ((8 + \sqrt{x - 7}) - 8)^2 + 7 \] \[ = (\sqrt{x - 7})^2 + 7 \] \[ = x - 7 + 7 = x \] Thus, \( f^{-1}(f(x)) = x \) as well. We have verified both compositions, confirming they are indeed the identity function. Feel free to ask for more details or any other math help!