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- 1. 4.2 Mean Value Theorem
- 2. Mean Value Theorem for Derivatives <ul><li>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 </li></ul>
- 3. Using Mean Value Theorem <ul><li>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. </li></ul><ul><li>Find f’(x) </li></ul><ul><li>f’(x) = 4x </li></ul><ul><li>4 = 4x </li></ul><ul><li>X= 1 </li></ul>
- 4. Using Mean Value Theorem <ul><li>f(x) = l x – 1 l on [0, 4] </li></ul><ul><li>f(a) = -1 </li></ul><ul><li>f(b) = 3 </li></ul><ul><li>1 = l x -1 l </li></ul>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 <ul><li>f(x) = -2x 3 + 6x – 2 , [-2 , 2] </li></ul><ul><li>f(-2) = -2(-2) 3 + 6(-2) - 2 = 2 f(2) = -2(2) 3 + 6(2) - 2 = - 6 </li></ul><ul><li>f(x) = -6 - 2 = -2 </li></ul><ul><li> 2 - (-2) </li></ul><ul><li>f '(x) = -6x 2 + 6 </li></ul><ul><li>-2 = -6x 2 + 6 </li></ul><ul><li>X = 2 -2 √3, √3 </li></ul>Mean value is satisfied because the function is continuous on [-2, 2]
- 6. Using Mean Value Theorem <ul><li>f(x) = x 3 + 3x – 1, [0,1] </li></ul><ul><li>f(b) = 3 </li></ul><ul><li>f(a) = -1 </li></ul><ul><li>f’(x) = 3x 2 + 3 </li></ul><ul><li>4 = 3x 2 +3 </li></ul><ul><li>X = </li></ul>
- 7. Mean Value with Trig. Functions
- 8. More mean value Solve.

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