(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]

The given linear polynomial y = f(x) has(a) 2 zeros(b) 1 zero and the zero is ‘3’(c) 1 zero and the zero is ‘4’(d) No zero1

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 =

Write a polynomial f (x) that meets the given conditions.Polynomial of lowest degree with zeros of -2 (multiplicity 2) and 3 (multiplicity 2) and with f(0)=−108𝑓(0)=−108Select one:a. f (x) = -3x4 + 6x3 + 33x2 - 18x - 108b. f (x) = -3x4 + 6x3 + 33x2 - 36x - 108c. f (x) = -3x4 + 6x3 - 111x2 - 36x - 108d. f (x) = x4 - 2x3 - 11x2 + 12x - 108

Seja a função h(x) = x³ - 2x² - 3x + 2 no intervalo [0, 2]. De acordo com o Teorema do Valor Intermediário, qual das alternativas é verdadeira?h(0) > h(2). h(0) < h(2).h(0) = h(2). A função h(x) não é contínua.