Rolle's Theorem Objectives:

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Presentation transcript:

Rolle's Theorem Objectives: Be able to determine if Rolle’s Theorem can be applied to a function on a closed interval. 2. Be able to apply Rolle’s Theorem to various functions. Critical Vocabulary: Rolle’s Theorem

If f is continuous on a closed interval [a, b], then f has both a minimum and maximum on the interval. However, the minimum and/or maximum may occur at the endpoints Example 1: Absolute Max: (3, 9/23) Absolute Min: (0, 0)

Rolle’s Theorem (named after French mathematician Michel Rolle) gives us conditions that guarantees the existence of an extreme value on the interior of the closed interval. Let f be continuous on a closed interval [a, b] and differentiable on the open interval (a, b). If f(a) = f(b), then there is at least one number c in (a, b) such that f’(c) = 0 (horizontal tangent line). Example 2: Determine if Rolle’s Theorem can be applied to the function f(x) = x2 + 6x + 13; [-7, 1] f(-7) = (-7)2 + 6(-7) + 13 f(1) = (1)2 + 6(1) + 13 f(-7) = 20 f(1) = 20 f’(x) = 2x + 6 Rolle’s Theorem can be applied 2x + 6 = 0 x = -3; (-3,4)

Example 3: Determine if Rolle’s Theorem can be applied to the function Let f be continuous on a closed interval [a, b] and differentiable on the open interval (a, b). If f(a) = f(b), then there is at least one number c in (a, b) such that f’(c) = 0 (horizontal tangent line). Example 3: Determine if Rolle’s Theorem can be applied to the function Rolle’s Theorem can’t be applied to this function. This function is discontinuous at x = 0

Example 4: Find the two x-intercepts of the function f and show that f’(x) = 0 at some point between the two intercepts. 1st: Find the x-intercepts (y = 0): 2nd: Find the derivative

Example 5 (Problem 10 on page 326): Determine whether Rolle’s Theorem can be applied to f on the closed interval [a, b]. If Rolle’s Theorem can be applied, find all the values of c in the open interval (a,b) such that f’(c) = 0. f(x) = (x - 3)(x + 1)2; [-1, 3] f(-1) = (-1 - 3)(-1 + 1)2 f(3) = (3 - 3)(3 + 1)2 f(-1) = 0 f(3) = 0 Yes, Rolle’s Theorem can be applied!!!! g(x) = x - 3 g’(x) = 1 h(x) = (x + 1)2 h’(x) = 2(x + 1) f’(x) = (x + 1)2 + 2(x + 1)(x - 3) 0 = (x + 1)(3x - 5) x + 1 = 0 3x - 5 = 0 f’(x) = (x + 1)[(x + 1) + 2(x - 3)] x = -1 x = 5/3 f’(x) = (x + 1)(3x - 5)

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