The problem asks to find the inverse of the given function $f(x) = \left( \frac{x}{10} \right)^5$.

Step-by-Step Solution:

1. Start with the definition of the function: $f(x) = \left( \frac{x}{10} \right)^5$

   To find the inverse $f^{-1}(x)$, we solve for $x$ in terms of $y$, where $y = f(x)$.

   $y = \left( \frac{x}{10} \right)^5$

2. Rewrite the equation to isolate $\frac{x}{10}$:

   Raise both sides of the equation to the power $\frac{1}{5}$ to undo the exponent: $y^{1/5} = \frac{x}{10}$

3. Solve for $x$:

   Multiply both sides by 10 to isolate $x$: $x = 10 \cdot y^{1/5}$

4. Rewrite the inverse function:

   Replace $y$ with $x$ to express the inverse: $f^{-1}(x) = 10 \cdot x^{1/5}$

Final Answer:

$f^{-1}(x) = 10 \cdot x^{1/5}$

Let me know if you'd like further explanation!

Here are 5 related questions to explore:

1. What is the domain and range of $f(x) = \left( \frac{x}{10} \right)^5$ and its inverse?
2. How do you verify that $f^{-1}(x)$ is indeed the inverse of $f(x)$?
3. What happens to the inverse if the function is modified to $f(x) = \left( \frac{x}{k} \right)^n$, where $k$ and $n$ are constants?
4. What is the derivative of the inverse function $f^{-1}(x)$?
5. How does the behavior of $f(x)$ change if it is defined as $f(x) = \left( \frac{x}{10} \right)^{-5}$?

Tip: To verify inverses, remember that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.