Average Rate of Change/Differentiation Problem

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The average rate of change of the function f over the interval from x = 2 to x = 2 + h is expressed as 7e^h - 4cos(2h). To find f'(2), the Mean-Value Theorem is applicable, which requires the correct formulation of the average rate of change as [f(2 + h) - f(2)]/h. The limit of this expression as h approaches 0 will yield the derivative at x = 2. Clarification on the correct limits and application of the theorem is essential for solving the problem. Understanding these concepts will lead to the correct determination of f'(2).
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If the average rate of change of a function f over the interval from x = 2 to x= 2 + h is given by 7e^h -4cos(2h), then f ' (2) = ?

This is what I got:
[f(2 + h) - f(h)]/h = 7e^h -4cos(2h)
I know we are supposed to used Mean-Value Theorem, I just don't get how to do it.
 
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Are you sure you don't mean [f(2 + h) - f(2)]/h? Take the limit of both sides as h->0.
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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