Question

Verify the Rolle's theorem for the function f(x)= x²-x-6/x-1 in [-2,3]​

Answer 1

The function values at the endpoints are not equal. Rolle's theorem does not apply to the given function on the interval [-2, 3].

Step-by-step explanation:

Rolle's theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), and the function values at the endpoints are equal, then there exists at least one point c in the interval (a, b) where the derivative of the function is zero.

In this case, the function f(x) = (x^2 - x - 6) / (x - 1) is continuous on the interval [-2, 3] and differentiable on the interval (-2, 3), except at x = 1 where the denominator becomes zero. Since the function is undefined at x = 1, we can exclude this point from consideration.

To verify Rolle's theorem, we need to show that the function is continuous on the closed interval [-2, 3], differentiable on the open interval (-2, 3), and that it satisfies the conditions of the theorem. Let's proceed.

1. Continuity: The given function is a quotient of two polynomials, and the denominator is not zero on the interval [-2, 3], except at x = 1. Therefore, the function is continuous on the closed interval [-2, 3].
2. Differentiability: The function is differentiable on the open interval (-2, 3), except at x = 1 where the denominator becomes zero. We can remove this point from consideration because it is not in the open interval.
3. Evaluating the function values at the endpoints: f(-2) = (-2^2 - (-2) - 6) / (-2 - 1) = (-4 + 2 - 6) / (-3) = -8 / -3 = 8/3. f(3) = (3^2 - 3 - 6) / (3 - 1) = (9 - 3 - 6) / 2 = 0. We can see that the function values at the endpoints are not equal.

Since the function values at the endpoints are not equal, we can conclude that Rolle's theorem does not apply to this function on the interval [-2, 3]. Therefore, there does not exist a point c in (-2, 3) where the derivative of the function is zero.