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4.2 Mean Value Theorem & Rolle’s Theorem. Mean Value Theorem. If f is a function that satisfies the following conditions: f is continuous on a closed interval [a, b] f is differentiable on the open interval (a, b) then there exists a number c in ( a, b) such that .

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## 4.2 Mean Value Theorem & Rolle’s Theorem

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**Mean Value Theorem**• If f is a function that satisfies the following conditions: • f is continuous on a closed interval [a, b] • f is differentiable on the open interval (a, b) then there exists a number c in (a, b) such that In other words, there exists a point somewhere in (a, b) such that the instantaneous rate of change is equal to the average rate of change.**Mean Value Theorem**(c, f(c)) (b, f(b)) (a, f(a)) c**Mean Value Theorem**• Example: The points P(1, 1) and Q(3, 27) are on the curve f(x) = x3. Using the Mean Value Theorem, find c in the interval (1, 3) such that f’(c) is equal to the slope of the secant of PQ. Since f(x) is defined for all real numbers, f is continuous on [1, 3]. Also, f’(x) = 3x2 is defined for all real numbers; thus, f’(x) is differentiable on (1, 3). Therefore, by MVT, (use only positive root since it is in the interval (1, 3))**Rolle’s Theorem**• If f is a function that satisfies the following three conditions: • f is continuous on a closed interval [a, b] • f is differentiable on the open interval (a, b) • f(a) = f(b) then there exists a number c in (a, b) such that f’(c) = 0.**Rolle’s Theorem**(c, f(c)) f‘(c) = 0 a b c**Rolle’s Theorem**• Example: If f(x) = x2 + 4x – 5, show that the hypotheses of Rolle’s Theorem are satisfied on the interval [–4, 0] and find all values of c that satisfy the conclusion of the theorem. 1.) f(x) = x2 + 4x – 5 is continuous everywhere since it is a polynomial. 2.) f’(x) = 2x + 4 is defined for all numbers and thus is differentiable on (–4, 0). 3.) f(0) = f(–4) = –5. Therefore, there exists a c in (–4, 0) such that f’(c) = 0 f‘(x) = 2x + 4 = 0 x = –2**Increasing/Decreasing Functions**• A function is increasing on (x1, x2) if: • f(x1) < f(x2) OR • f‘(x) > 0 at each point of (x1, x2) • A function is decreasing on (x1, x2) if: • f(x1) > f(x2) OR • f‘(x) < 0 at each point of (x1, x2)**Antiderivatives**• A function F(x) is an antiderivative of a function f(x) if F’(x) = f(x) for all x in the domain of f.

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