The function \( f(x)=(x+8)^{5} \) 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 \( f^{-1}(f(x))=x \). B. \( f^{-1}(x)= \), for \( x \geq \) C. \( f^{-1}(x)=\sqrt[5]{x}-8 \), for all \( x \) b. Verify that the equation is correct. \( f\left(f^{-1}(x)\right)=f(\square) \) for \( x \leq \) \( =\square \) and \( \quad f^{-1}(f(x))=f^{-1}(\square) \) Substitute. Simplify.

To find the inverse function \( f^{-1}(x) \) of \( f(x) = (x + 8)^{5} \), we start by letting \( y = f(x) \). This gives us the equation \( y = (x + 8)^{5} \). To find the inverse, we solve for \( x \) in terms of \( y \): 1. Rewrite the equation: \( y = (x + 8)^{5} \). 2. Take the fifth root of both sides: \( \sqrt[5]{y} = x + 8 \). 3. Isolate \( x \): \( x = \sqrt[5]{y} - 8 \). Thus, the inverse function is \( f^{-1}(x) = \sqrt[5]{x} - 8 \) for all \( x \). Next, we must verify that our inverse function is correct by proving that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \). To start with \( f(f^{-1}(x)) \): 1. Substitute \( f^{-1}(x) \) into \( f \): \[ f(f^{-1}(x)) = f(\sqrt[5]{x} - 8) = ((\sqrt[5]{x} - 8) + 8)^{5} \] Simplifying this, we get: \[ = (\sqrt[5]{x})^{5} = x. \] Now, let's check \( f^{-1}(f(x)) \): 1. Substitute \( f(x) \) into \( f^{-1} \): \[ f^{-1}(f(x)) = f^{-1}((x + 8)^{5}) = \sqrt[5]{(x + 8)^{5}} - 8. \] Simplifying gives us: \[ = (x + 8) - 8 = x. \] Both verifications show that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \), confirming that we have the correct inverse function. To fill in the blanks: B. \( f^{-1}(x)= \sqrt[5]{x}-8 \), for \( x \geq (8^5) \) C. \( f^{-1}(x)=\sqrt[5]{x}-8 \), for all \( x \) b. Verify that the equation is correct: \( f\left(f^{-1}(x)\right)=f(\sqrt[5]{x}-8) \) for \( x \leq (8^5) \) \( = x \) and \( \quad f^{-1}(f(x))=f^{-1}((x+8)^5) \) Substitute. Simplify \( = x \).