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# Mean Value Theorem

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### Transcript

• 1. 4.2 Mean Value Theorem
• 2. Mean Value Theorem for Derivatives
• 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
• 3. Using Mean Value Theorem
• Show that f(x) = 2x 2 satisfies the mean value theorem on the interval [0,2]. Then find the solution to the equation on the interval.
• Find f’(x)
• f’(x) = 4x
• 4 = 4x
• X= 1
• 4. Using Mean Value Theorem
• f(x) = l x – 1 l on [0, 4]
• f(a) = -1
• f(b) = 3
• 1 = l x -1 l
The function does not satisfy the mean value theorem because there is a cusp so the function is not continuous on [0,4]
• 5. Mean Value Theorem
• f(x) = -2x 3 + 6x – 2 , [-2 , 2]
• f(-2) = -2(-2) 3 + 6(-2) - 2 = 2 f(2) = -2(2) 3 + 6(2) - 2 = - 6
• f(x) = -6 - 2 = -2
• 2 - (-2)
• f '(x) = -6x 2 + 6
• -2 = -6x 2 + 6
• X = 2 -2 √3, √3
Mean value is satisfied because the function is continuous on [-2, 2]
• 6. Using Mean Value Theorem
• f(x) = x 3 + 3x – 1, [0,1]
• f(b) = 3
• f(a) = -1
• f’(x) = 3x 2 + 3
• 4 = 3x 2 +3
• X =
• 7. Mean Value with Trig. Functions
• 8. More mean value Solve.