The problem is asking for an additional hypothesis to ensure that there exists a point $c \in ]a; b[$ such that $f'(c) = 0$, given that the function $f(x)$ is continuous on the interval $[a; b]$ and differentiable on $]a; b[$.

This is a direct application of Rolle's Theorem, which states that if a function $f(x)$ satisfies the following three conditions:

1. $f(x)$ is continuous on the closed interval $[a; b]$,
2. $f(x)$ is differentiable on the open interval $]a; b[$,
3. $f(a) = f(b)$,

then there exists at least one point $c \in ]a; b[$ such that $f'(c) = 0$.

Let's evaluate the options:

• A) "f(a) e f(b) devono essere diverse da 0" (This is irrelevant; what matters is that $f(a) = f(b)$, not whether they are zero.)
• B) "La funzione deve essere derivabile anche agli estremi dell'intervallo (a; b)" (Rolle's theorem does not require differentiability at the endpoints, just on the open interval.)
• C) "Deve essere f(a) = f(b)" (This is the correct condition from Rolle's Theorem.)
• D) "Deve essere f(a) = f(b) = 0" (This is unnecessary; $f(a)$ and $f(b)$ need to be equal, but they don't need to be 0.)

Thus, the correct answer is C): $f(a) = f(b)$.

Would you like more details or have any questions?

Here are 5 related questions to expand on the topic:

1. What other conditions are required for the Mean Value Theorem?
2. How does Rolle's Theorem relate to the Mean Value Theorem?
3. What happens if $f(a) \neq f(b)$? What can we conclude?
4. How does the differentiability of a function on an interval impact its critical points?
5. Can Rolle's Theorem be applied to non-polynomial functions, like trigonometric functions?

Tip: Rolle's Theorem is a special case of the Mean Value Theorem, which generalizes the result when $f(a) \neq f(b)$.