If f(1) = 3 and f(5) = 11, there exists c in (1,5) with f'(c) equal to (f(5) - f(1))/(5 - 1). What is that value?

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Multiple Choice

If f(1) = 3 and f(5) = 11, there exists c in (1,5) with f'(c) equal to (f(5) - f(1))/(5 - 1). What is that value?

Explanation:
Mean Value Theorem gives that, for a differentiable function on (1,5) and continuous on [1,5], there is a c in (1,5) with f'(c) equal to the average rate of change over the interval. Compute that average rate: (f(5) − f(1)) / (5 − 1) = (11 − 3) / 4 = 8/4 = 2. So f'(c) = 2 for some c in (1,5). The value is 2.

Mean Value Theorem gives that, for a differentiable function on (1,5) and continuous on [1,5], there is a c in (1,5) with f'(c) equal to the average rate of change over the interval. Compute that average rate: (f(5) − f(1)) / (5 − 1) = (11 − 3) / 4 = 8/4 = 2. So f'(c) = 2 for some c in (1,5). The value is 2.

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