What is the inverse of \( f(x)=2 \sqrt[3]{x-3} \) ? (1 point) \( f^{-1}(x)=\left(\frac{x}{2}-3\right)^{3} \) \( f^{-1}(x)=\left(\frac{x}{2}\right)^{3}+3 \) \( f^{-1}(x)=\left(\frac{x}{2}+3\right)^{3} \) \( f^{-1}(x)=\left(\frac{x}{2}\right)^{3}-3 \)

To find the inverse of the function \( f(x) = 2 \sqrt[3]{x - 3} \), we start by substituting \( f(x) \) with \( y \): \[ y = 2 \sqrt[3]{x - 3} \] Now, solving for \( x \) in terms of \( y \): 1. Divide both sides by 2: \[ \frac{y}{2} = \sqrt[3]{x - 3} \] 2. Cube both sides to eliminate the cube root: \[ \left(\frac{y}{2}\right)^3 = x - 3 \] 3. Finally, solve for \( x \): \[ x = \left(\frac{y}{2}\right)^3 + 3 \] Thus, the inverse function is: \[ f^{-1}(x) = \left(\frac{x}{2}\right)^3 + 3 \] This matches the second option given: \( f^{-1}(x) = \left(\frac{x}{2}\right)^3 + 3 \). Double-checking options: - \( f^{-1}(x) = \left(\frac{x}{2} - 3\right)^{3} \) - This one's incorrect; we want to keep the \( +3 \) in our final answer. - \( f^{-1}(x) = \left(\frac{x}{2}\right)^{3} + 3 \) - Bingo! - \( f^{-1}(x) = \left(\frac{x}{2} + 3\right)^{3} \) - Wrong form; the addition must not be cubed. - \( f^{-1}(x) = \left(\frac{x}{2}\right)^{3} - 3 \) - Not correct since we added 3. So, the correct inverse function is indeed \( f^{-1}(x) = \left(\frac{x}{2}\right)^{3} + 3 \).