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.

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 =

Find the average value and the value(s) of c guaranteed by the mean value theorem for integrals if f left parenthesis x right parenthesis equals 3 x squared plus 2 x over the interval left square bracket 0 comma 2 right square bracket.A. f subscript a v g end subscript equals 12c equals fraction numerator short dash 1 plus square root of 37 over denominator 3 end fractionB. f subscript a v g end subscript equals 6c equals fraction numerator short dash 1 minus square root of 19 over denominator 3 end fraction space a n d space c equals fraction numerator short dash 1 plus square root of 19 over denominator 3 end fractionC. f subscript a v g end subscript equals 12c equals fraction numerator short dash 1 minus square root of 37 over denominator 3 end fraction space a n d space c equals fraction numerator short dash 1 plus square root of 37 over denominator 3 end fractionD. f subscript a v g end subscript equals 6c equals fraction numerator short dash 1 plus square root of 19 over denominator 3 end fraction

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.

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]

Let f be a function such that f (1) = − 2 and f (5) = 7. Which of the following conditions ensures that f(c) = 0 for some value c in the open interval (1,5)?Responses∫15𝑓(𝑥)𝑑𝑥 exists. Integration of f of x of d x exists from 1 to 5f is increasing on the closed interval [1, 5].f is increasing on the closed interval [1, 5].f is continuous on the closed interval [1, 5].f is continuous on the closed interval [1, 5].f is defined for all values of x in the closed interval [1, 5].