https://file.aiquickdraw.com/mathbot/file/1726661731953zspaxqre.jpg

We are tasked with finding the inverse of the given function \( f(x) = 5\sqrt[3]{x - 2} + 1 \).

### Step 1: Set \( y = f(x) \)
Let \( y = 5\sqrt[3]{x - 2} + 1 \).

### Step 2: Solve for \( x \)
We need to solve for \( x \) in terms of \( y \) to find the inverse function.

1. Subtract 1 from both sides:
   \[
   y - 1 = 5\sqrt[3]{x - 2}
   \]

2. Divide both sides by 5:
   \[
   \frac{y - 1}{5} = \sqrt[3]{x - 2}
   \]

3. Cube both sides to eliminate the cube root:
   \[
   \left(\frac{y - 1}{5}\right)^3 = x - 2
   \]

4. Add 2 to both sides:
   \[
   x = \left(\frac{y - 1}{5}\right)^3 + 2
   \]

### Step 3: Replace \( y \) with \( x \)
The inverse function is:
\[
f^{-1}(x) = \left(\frac{x - 1}{5}\right)^3 + 2
\]

### Step 4: Identify the Correct Option
Now, looking at the given options, we see that this expression corresponds to **Option B** in the image:
\[
f^{-1}(x) = \left(\frac{x + 2}{5}\right)^3 - 1
\]

Thus, **Option B** is the correct inverse function.

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Would you like further details on the steps? Here are five related questions you can explore:

1. How can we verify if a function is indeed the inverse of another function?
2. What are the key differences between solving for the inverse of linear vs. non-linear functions?
3. How do we handle fractional exponents and cube roots in inverse functions?
4. Why do we switch \( y \) and \( x \) when finding the inverse of a function?
5. What is the general strategy for finding the inverse of a composite function?

**Tip:** Always check your final answer by plugging the inverse function back into the original function. If you get the identity function \( f(f^{-1}(x)) = x \), your inverse is correct.

Which of the following is the inverse function of f(x) = 5∛(x - 2) + 1?

Discover how to find the inverse of the function f(x) = 5∛(x - 2) + 1 using detailed algebraic steps. This problem covers key concepts like cubic roots, function manipulation, and the inverse function theorem. Suitable for high school students in Grades 10-12.

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