Question
Given the function of $f(x)=2-\frac{10}{7 x}$ on the interval $[-8,-2]$, find a point $c$ between -8 and -2 so that it satisfies the Mean Value Theorem.

Provide your answer below:
\[
c=\square
\]
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Step 1 ：The Mean Value Theorem states that if a function f is continuous on the closed 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 that point is equal to the average rate of change of the function over the interval [a, b].

Step 2 ：The average rate of change of the function over the interval [a, b] is given by the formula: \[f'(c) = \frac{f(b) - f(a)}{b - a}\]

Step 3 ：So, we need to find a point c in the interval (-8, -2) such that the derivative of the function at that point is equal to the average rate of change of the function over the interval [-8, -2].

Step 4 ：First, we need to find the derivative of the function f(x) = 2 - 10/(7x).

Step 5 ：Then, we need to find the average rate of change of the function over the interval [-8, -2].

Step 6 ：Finally, we need to solve the equation f'(c) = (f(-2) - f(-8))/(-2 - (-8)) for c.

Step 7 ：The solution to the equation f'(c) = avg_rate_of_change for c is [-4, 4]. However, only -4 is in the interval (-8, -2). Therefore, the point c that satisfies the Mean Value Theorem is -4.

Step 8 ：Final Answer: \[c=\boxed{-4}\]