Problem Breakdown:

The function $f(x) = x^3 - 9$ is one-to-one. We need to:

1. Find the inverse function, $f^{-1}(x)$.
2. Verify that the inverse satisfies the conditions $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.

Step 1: Finding the Inverse

To find the inverse of $f(x)$:

1. Replace $f(x)$ with $y$: $y = x^3 - 9$.
2. Swap $x$ and $y$ to find $x$ in terms of $y$: $x = y^3 - 9$.
3. Solve for $y$: $y^3 = x + 9 \implies y = \sqrt[3]{x + 9}.$
4. Thus, the inverse function is $f^{-1}(x) = \sqrt[3]{x + 9}$.

Step 2: Verification

Verifying $f(f^{-1}(x)) = x$:

1. Substitute $f^{-1}(x)$ into $f(x)$: $f(f^{-1}(x)) = f(\sqrt[3]{x + 9}).$
2. Using the definition of $f(x)$: [ f(\sqrt[3]{x + 9}) = (\sqrt[3]{x + 9})^3 - 9 = x + 9 - 9 = x. ]

Verifying $f^{-1}(f(x)) = x$:

1. Substitute $f(x)$ into $f^{-1}(x)$: $f^{-1}(f(x)) = f^{-1}(x^3 - 9).$
2. Using the definition of $f^{-1}(x)$: [ f^{-1}(x^3 - 9) = \sqrt[3]{(x^3 - 9) + 9} = \sqrt[3]{x^3} = x. ]

Both conditions are satisfied.

Final Answer

• The inverse function is $\boxed{f^{-1}(x) = \sqrt[3]{x + 9}}$.
• Correct option: B.

Would you like further clarification? Below are related questions to deepen understanding.

5 Related Questions

1. What are the conditions for a function to have an inverse?
2. How do you algebraically verify an inverse function for all $x$?
3. How does the domain and range of $f(x) = x^3 - 9$ relate to $f^{-1}(x)$?
4. What is the geometric interpretation of a function and its inverse?
5. How does the process change for functions that are not one-to-one?

Tip:

For cube root functions like $\sqrt[3]{x}$, the domain is all real numbers, and there is no restriction on $x$ in $f^{-1}(x)$.