To verify Rolle's theorem for $f(x) = x^2 - 5x + 6$ on the interval $[2, 3]$, follow these steps:

1. Check the conditions of Rolle's theorem:

   • Continuity: $f(x)$ is a polynomial, so it is continuous on $[2, 3]$.
   • Differentiability: $f(x)$ is a polynomial, so it is differentiable on $(2, 3)$.
   • Equal values at endpoints: Calculate $f(2)$ and $f(3)$: $f(2) = 2^2 - 5(2) + 6 = 4 - 10 + 6 = 0$ $f(3) = 3^2 - 5(3) + 6 = 9 - 15 + 6 = 0$ Since $f(2) = f(3) = 0$, the conditions are satisfied.

2. Find $c$ in $(2, 3)$ such that $f'(c) = 0$:

   • Differentiate $f(x)$: $f'(x) = 2x - 5$
   • Set $f'(x) = 0$: $2x - 5 = 0 \quad \implies \quad x = \frac{5}{2} = 2.5$

3. Verify $c \in (2, 3)$: $c = 2.5$ is indeed in the interval $(2, 3)$.

Conclusion:

Rolle's theorem is satisfied, and $c = 2.5$ is the point where $f'(c) = 0$.

Would you like more details or have any questions?

Here are 5 related questions:

1. Can you generalize Rolle's theorem to other polynomial functions?
2. How does the Mean Value Theorem differ from Rolle's theorem?
3. Could there be multiple values of $c$ in $(2, 3)$ where $f'(c) = 0$?
4. What happens if $f(a) \neq f(b)$ for some interval $[a, b]$?
5. Can Rolle's theorem apply to non-polynomial functions?

Tip: Always ensure the conditions of Rolle's theorem are thoroughly verified before solving for $c$.