Certainly! Let's start by verifying whether the function \(f(x) = x^3 - 3x\) satisfies the criteria stated in Rolle's theorem for the given interval \([-1, 2]\).
1. Continuity: The function \(f(x)\) is a polynomial, which means it is continuous for all real numbers. Therefore, the first criterion is met.
2. Differentiability: Let's find the derivative of \(f(x)\):
\[f'(x) = 3x^2 - 3\]
Now, we need to check if there exists at least one point \(c\) in the interval \((-1, 2)\) where \(f'(c) = 0\).
Setting \(f'(c) = 0\):
\[3c^2 - 3 = 0\]
\[3c^2 = 3\]
\[c^2 = 1\]
\[c = \pm 1\]
Both \(c = -1\) and \(c = 1\) are within the interval \([-1, 2]\). Therefore, both points satisfy the conclusion of Rolle's theorem.
In summary, the function \(f(x) = x^3 - 3x\) satisfies the criteria of Rolle's theorem, and the values of \(c\) where \(f'(c) = 0\) are \(c = -1\) and \(c = 1\).
