Section 4.2 Mean Value Theorem.

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Section 4.2 Mean Value Theorem

Mean Value Theorem 𝑓 ′ 𝑐 = 𝑓 𝑏 −𝑓(𝑎) 𝑏 −𝑎 . If y = f(x) is continuous at every point of the closed interval [a,b] and differentiable at every point of its interior (a,b), then there is at least one point c in (a,b) at which 𝑓 ′ 𝑐 = 𝑓 𝑏 −𝑓(𝑎) 𝑏 −𝑎 . In other words, if the hypothesis is true, then there is some point at which the instantaneous rate of change is equal to the average rate of change for the interval.

Find the point at which the function satisfies the Mean Value Theorem 1. f(x) = x3 – 3x + 5 -1 < x < 1 (Be sure to verify that the conditions are met for using the MVT) 2. f(x) = xex on the interval [0, 1]

Implications of the MVT If f’ > 0, then f is….. Increasing. If f’ < 0, then f is….. Decreasing If f’ = 0, then f is….. Constant.

Example Determine where the following function is increasing, decreasing, constant. f(x) = 2x3 – 9x2 + 12x - 5

Another Application of MVT If f’(x) and g’(x) are equal at each point of an interval, then f(x) and g(x) are separated only by a constant. Example: Find a function whose derivative is 6x2 + 5