The question you provided is written in Greek and states:

"Given the continuous function $f : [0, 3] \rightarrow (0, +\infty)$. Prove that there exists $\xi \in (0, 3)$ such that:
$f(\xi) = \sqrt{f(1) \cdot f(2)}.$"

This is a typical problem where we need to apply the Intermediate Value Theorem (IVT), since the function $f$ is continuous and maps the interval $[0, 3]$ to positive real values $(0, +\infty)$. Let's go step by step through the proof:

Proof:

1. Define a new function:
   Define a new function $g(x) = f(x) - \sqrt{f(1) \cdot f(2)}$. Notice that $g(x)$ is continuous because $f(x)$ is continuous.

2. Evaluate $g(x)$ at the endpoints:
   Let's check the values of $g(1)$ and $g(2)$: $g(1) = f(1) - \sqrt{f(1) \cdot f(2)} = f(1) - \sqrt{f(1) \cdot f(2)}$ $g(2) = f(2) - \sqrt{f(1) \cdot f(2)} = f(2) - \sqrt{f(1) \cdot f(2)}$

3. Analyze the signs:
   Since $f(1)$ and $f(2)$ are both positive (as $f$ maps to $(0, +\infty)$), the values of $g(1)$ and $g(2)$ will have opposite signs. This is because:

   • If $f(1) > \sqrt{f(1) \cdot f(2)}$, then $g(1) > 0$.
   • If $f(2) < \sqrt{f(1) \cdot f(2)}$, then $g(2) < 0$.

4. Apply the Intermediate Value Theorem:
   By the IVT, since $g(x)$ is continuous and $g(1)$ and $g(2)$ have opposite signs, there must exist some $\xi \in (1, 2)$ such that: $g(\xi) = 0.$ This implies: $f(\xi) = \sqrt{f(1) \cdot f(2)}.$

Therefore, we have proven that such a $\xi \in (0, 3)$ exists.

Would you like further clarification on any of the steps?

Here are 5 follow-up questions to explore the concept further:

1. What conditions are necessary for the Intermediate Value Theorem to apply?
2. How would the proof change if the function $f$ were not continuous?
3. Can the function $f$ ever take negative values in this scenario?
4. What would happen if $f(1) = f(2)$?
5. How does the range $(0, +\infty)$ of the function affect the problem?

Tip: The Intermediate Value Theorem is a powerful tool for proving the existence of solutions, but it requires the function to be continuous on the given interval.