The function provided is:

$f(x) = \sqrt[3]{3x - 2}$

Step 1: Find the inverse function

To find the inverse of a function $f(x)$, follow these steps:

1. Replace $f(x)$ with $y$: $y = \sqrt[3]{3x - 2}$

2. Swap $x$ and $y$ (this reflects finding the inverse): $x = \sqrt[3]{3y - 2}$

3. Now, solve for $y$:

   • Cube both sides to remove the cube root: $x^3 = 3y - 2$
   • Isolate $y$: $3y = x^3 + 2$ $y = \frac{x^3 + 2}{3}$

So, the inverse function is:

$f^{-1}(x) = \frac{x^3 + 2}{3}$

Step 2: Prove the inverse by composition

To verify that $f^{-1}(x)$ is the correct inverse, we need to check two conditions:

1. $f(f^{-1}(x)) = x$
2. $f^{-1}(f(x)) = x$

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

Substitute $f^{-1}(x) = \frac{x^3 + 2}{3}$ into $f(x) = \sqrt[3]{3x - 2}$: $f(f^{-1}(x)) = f\left( \frac{x^3 + 2}{3} \right) = \sqrt[3]{3 \left( \frac{x^3 + 2}{3} \right) - 2}$ Simplify the expression inside the cube root: [ f(f^{-1}(x)) = \sqrt[3]{x^3 + 2 - 2} = \sqrt[3]{x^3} = x ] This shows that $f(f^{-1}(x)) = x$.

2. Prove $f^{-1}(f(x)) = x$:

Substitute $f(x) = \sqrt[3]{3x - 2}$ into $f^{-1}(x) = \frac{x^3 + 2}{3}$: [ f^{-1}(f(x)) = f^{-1}\left( \sqrt[3]{3x - 2} \right) = \frac{\left( \sqrt[3]{3x - 2} \right)^3 + 2}{3} ] Simplify the expression: $f^{-1}(f(x)) = \frac{3x - 2 + 2}{3} = \frac{3x}{3} = x$ This shows that $f^{-1}(f(x)) = x$.

Conclusion:

Since both compositions result in $x$, we have verified that the inverse function is correct:

$f^{-1}(x) = \frac{x^3 + 2}{3}$

Let me know if you would like further clarifications or details. Here are some related questions:

1. How do you find the inverse of more complex functions?
2. What is the domain and range of the inverse function?
3. Can you explain why the domain of $f(x)$ is all real numbers?
4. How does composition help in verifying inverse functions?
5. What other methods exist for verifying inverse functions?

Tip: When solving for the inverse of a function, always check if the function is one-to-one, as this ensures that it has an inverse.