Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval?f(x) = x3 − 3x + 4,   [−2, 2]Yes, it does not matter if f is continuous or differentiable; every function satisfies the Mean Value Theorem.Yes, f is continuous on [−2, 2] and differentiable on (−2, 2) since polynomials are continuous and differentiable on .    No, f is not continuous on [−2, 2].No, f is continuous on [−2, 2] but not differentiable on (−2, 2).There is not enough information to verify if this function satisfies the Mean Value Theorem.If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem. (Enter your answers as a comma-separated list. If it does not satisfy the hypotheses, enter DNE).c =

et f(x) = (x − 3)−2. Find all values of c in (2, 5) such that f(5) − f(2) = f '(c)(5 − 2). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)c = 52​ Based off of this information, what conclusions can be made about the Mean Value Theorem?This contradicts the Mean Value Theorem since f satisfies the hypotheses on the given interval but there does not exist any c on (2, 5) such that f '(c) = f(5) − f(2)5 − 2.This does not contradict the Mean Value Theorem since f is not continuous at x = 3.    This does not contradict the Mean Value Theorem since f is continuous on (2, 5), and there exists a c on (2, 5) such that f '(c) = f(5) − f(2)5 − 2.This contradicts the Mean Value Theorem since there exists a c on (2, 5) such that f '(c) = f(5) − f(2)5 − 2, but f is not continuous at x = 3.Nothing can be concluded.

EXAMPLE 2 Since f(x) = 4 + x2 is continuous on the interval [−4, 2], the Mean Value Theorem for Integrals says there is a number c in [−4, 2] such that2−4(4 + x2) dx = f(c)[2 − (−4)].In this particular case we can find c explicitly. Using this example, we find that fave = 8, so the value of c satisfiesf(c) = fave = 8.Therefore4 + c2 = soc2 = .So in this case there happen to be two numbers c = ± 2 in the interval [−4, 2] that work in the Mean Value Theorem for Integrals.

(1) f (x) = x2 − 4x + 3; [1, 3](2) f (x) = x3 − x; [−1, 1](3) f (x) = x3 − 9x; [−3, 3]Solution(1) f (x) = x2 − 4x + 3; [1, 3].f (1) = 1 − 4 + 3 = 0f (3) = 9 − 12 + 3 = 0The polynomial f (x) is continuous and dierentiable on [1, 3].Therefore the hypotheses of Rolle's Theorem are satised.

Verify that the function satisfies the three hypotheses of Rolle's theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle's theorem. (Enter your answers as a comma-separated list.)f(x) = 3x2 − 6x + 4,   [−1, 3]

Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)f(x) = x,    [0, 16]