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The function f(x)=x^3+5x^2-12x-56 is graphed below. Plot a line segment connecting the points on f where x=-6 and x=2. Afterwards, determine all values of c which satisfy the conclusion of the Mean Value Theorem for f on the closed interval -6≤x≤2.​

The function f(x)=x^3+5x^2-12x-56 is graphed below. Plot a line segment connecting-example-1

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Final answer:

To apply the Mean Value Theorem to the function f(x) = x^3 + 5x^2 - 12x - 56, find the average rate of change over the interval [-6, 2] and then find the f'(x) values that match this average rate.

Step-by-step explanation:

The student has been asked to apply the Mean Value Theorem (MVT) to a specific function (f(x) = x^3 + 5x^2 - 12x - 56) and find all values of c in the closed interval [-6, 2] that satisfy the theorem's conclusion.

The Mean Value Theorem states that if a function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one value c in the interval (a, b) where the instantaneous rate of change (slope of the tangent) equals the average rate of change (slope of the secant) on [a, b]. The average rate of change of the function on the interval [-6, 2] can be calculated as (f(2) - f(-6))/(2 - (-6)).

Firstly, by evaluating f at the endpoints of the interval, we get f(-6) and f(2). Afterward, we find the average rate of change by dividing the change in function values by the length of the interval. To find values of c that satisfy the theorem, we take the derivative of the function, f'(x), and set it equal to the average rate of change. Solving for x gives us the values of c that meet the theorem's condition. These values of c represent the x-coordinates of the points where the tangent to the curve has the same slope as the secant line connecting f(-6) and f(2).

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