This problem is an application of the Mean Value Theorem (MVT) in calculus. The MVT states that if a function is continuous on the interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c in the interval (a, b) such that the derivative of the function at c equals the average rate of change of the function over the interval [a, b].

Here, we are given the function f(x) = 2x + cos(2πx - 6) - 5 and we are asked to find the value of c in the interval [0, 4] such that f′(c) = (f(4) - f(0))/4.

Step 1: Compute f(4) and f(0)
f(4) = 2*4 + cos(2π*4 - 6) - 5 = 8 + cos(2π) - 5 = 8 + 1 - 5 = 4
f(0) = 2*0 + cos(2π*0 - 6) - 5 = 0 + cos(-6) - 5 = -5 + cos(-6)

Step 2: Compute (f(4) - f(0))/4
(f(4) - f(0))/4 = (4 - (-5 + cos(-6)))/4 = (4 + 5 - cos(6))/4 = (9 - cos(6))/4

Step 3: Compute the derivative of f(x)
f′(x) = 2 - 2π*sin(2πx - 6)

Step 4: Set f′(c) equal to (f(4) - f(0))/4 and solve for c
2 - 2π*sin(2πc - 6) = (9 - cos(6))/4

This equation can be solved numerically to find the approximate value of c.

Question

This problem is an application of the Mean Value Theorem (MVT) in calculus. The MVT states that if a function is continuous on the interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c in the interval (a, b) such that the derivative of the function at c equals the average rate of change of the function over the interval [a, b].

Here, we are given the function f(x) = 2x + cos(2πx - 6) - 5 and we are asked to find the value of c in the interval [0, 4] such that f′(c) = (f(4) - f(0))/4.

Step 1: Compute f(4) and f(0)
f(4) = 2*4 + cos(2π*4 - 6) - 5 = 8 + cos(2π) - 5 = 8 + 1 - 5 = 4
f(0) = 2*0 + cos(2π*0 - 6) - 5 = 0 + cos(-6) - 5 = -5 + cos(-6)

Step 2: Compute (f(4) - f(0))/4
(f(4) - f(0))/4 = (4 - (-5 + cos(-6)))/4 = (4 + 5 - cos(6))/4 = (9 - cos(6))/4

Step 3: Compute the derivative of f(x)
f′(x) = 2 - 2π*sin(2πx - 6)

Step 4: Set f′(c) equal to (f(4) - f(0))/4 and solve for c
2 - 2π*sin(2πc - 6) = (9 - cos(6))/4

This equation can be solved numerically to find the approximate value of c.

Knowee AI · Accepted Answer

This problem is an application of the Mean Value Theorem (MVT) in calculus. The MVT states that if a function is continuous on the interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c in the interval (a, b) such that the derivative of the function at c equals the average rate of change of the function over the interval [a, b].

Here, we are given the function f(x) = 2x + cos(2πx - 6) - 5 and we are asked to find the value of c in the interval [0, 4] such that f′(c) = (f(4) - f(0))/4.

Step 1: Compute f(4) and f(0)
f(4) = 2*4 + cos(2π*4 - 6) - 5 = 8 + cos(2π) - 5 = 8 + 1 - 5 = 4
f(0) = 2*0 + cos(2π*0 - 6) - 5 = 0 + cos(-6) - 5 = -5 + cos(-6)

Step 2: Compute (f(4) - f(0))/4
(f(4) - f(0))/4 = (4 - (-5 + cos(-6)))/4 = (4 + 5 - cos(6))/4 = (9 - cos(6))/4

Step 3: Compute the derivative of f(x)
f′(x) = 2 - 2π*sin(2πx - 6)

Step 4: Set f′(c) equal to (f(4) - f(0))/4 and solve for c
2 - 2π*sin(2πc - 6) = (9 - cos(6))/4

This equation can be solved numerically to find the approximate value of c.

The one of the (approximate) values of c in [0,4] such that f′(c)=f(4)−f(0)4 for the function f(x)=2x+cos(2πx−6)−5 is

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