Rolle Problems - Contents 5 Rolles Theorem the Mean Value...

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Contents 5 Rolle’s Theorem, the Mean Value Theorem, and L’Hˆopital’s Rule 80 5.1 Rolle’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 5.2 Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 5.3 L’Hospital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 Close the Window to Return to the Table of Contents
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Chapter 5 Rolle’s Theorem, the Mean Value Theorem, and L’Hˆopital’s Rule 5.1 Rolle’s Theorem In the following problems (a) Verify that the three conditions of Rolle’s theorem have been met. (b) Find all values z that satisfy the conclusion of the theorem. 1. (a) f ( x ) = x 2 - 7 x + 10 , on [2 , 5] (b) f ( x ) = x 2 - 7 x + 10 , on [0 , 7] (c) f ( x ) = x 2 - 7 x + 10 , on [ - 1 , 8] 2. (a) f ( x ) = x 2 - 4 x, on [0 , 4] (b) f ( x ) = x 2 - 4 x, on [ - 1 , 5] (c) f ( x ) = x 2 - 4 x, on [ - 4 , 8] 3. (a) f ( x ) = x 3 - 5 x 2 - 17 x + 21 , on - 3 x 7 (b) f ( x ) = x 3 - 16 x, on - 4 x 4 (c) f ( x ) = x 3 + 2 x 2 - x - 2 , on - 2 x 1 4. (a) f ( x ) = x 2 - 6 x - 7 , on [ - 1 , 7] (b) f ( x ) = x 3 + x 2 - 6 x, on [0 , 2] (c) f ( x ) = x 2 - x - 2 , on [ - 1 , 2] (d) g ( x ) = x 3 + 5 x 2 + 6 , on [ - 3 , 0] 5. (a) f ( x ) = sin 2 x, on [0 , π ] (b) f ( x ) = cos ( x 2 ) , on [ π, 3 π ] (c) f ( x ) = sin x + cos x on £ - π 4 , 3 π 4 / ( Hint: 3 π 4 = π 4 + π 2 ) Close the Window to Return to the Table of Contents
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81 5.2 Mean Value Theorem In problems 6 to 11 determine in what way the function fails to meet the conditions of Rolle’s theorem. 6. f ( x ) = 4 x 2 - 8 x for 0 x 3. 7. f ( x ) = 1 - cos 4 x + 2 cos x for 3 π 2 x 7 π 2 . £ f ( 3 π 2 ) 6 = 0 / 8. f ( x ) = 3 x 2 - 2 x +4 x - 2 for 1 x 5. 9. f ( x ) = sin 2 x + cos 2 x for 0 x 3 π 8 . 10. f ( x ) = 3 x +2 ( x - 1) 2 for 1 2 x 2. 11. f ( x ) = 1 - x 2 3 for - 1 x 1. [ f 0 (0) is undefined] 12. Find a function f ( x ) on the interval [ - 1 , 1] such that f ( x ) fails to meet the conditions of Rolle’s theorem, but [ f ( x )] 2 does meet them. In what way does f ( x ) fails to meet the conditions? 13. Use Rolle’s theorem to prove that, regardless of the value of b , there is at most one point x in the interval - 1 x 1 for which x 3 - 3 x + b = 0. 5.2 Mean Value Theorem For problems numbered 14 to 22, (a) Verify that the conditions of the mean value theorem have been met. (b) Find all numbers z that satisfy the theorem. Example : f ( x ) = x 3 - 2 x 2 for 1 x 3. Solution. (a) f ( x ) is continuous (all polynomials are continuous) f 0 ( x ) = 3 x 2 - 4 x exists for x (1 , 3). Thus the conditions of the mean value theorem have been met. (b) f (3) = 27 - 18 = 9. f (1) = 1 - 2 = - 1. Therefore, f ( b ) - f ( a ) b - a = f (3) - f (1) 3 - 1 = 9 - ( - 1) 3 - 1 = 10 2 = 5 f 0 ( z ) = f ( b ) - f ( a ) b - a (M.V.T.) 3 z 2 - 4 z = 5 3 z 2 - 4 z - 5 = 0 z = 4 ± p 16 - 4( - 5)(3) 6 = 4 ± 76 6 2 . 1 or - 0 . 78 Since - 0 . 78 is not in the interval discard it, but 2.1 is in the interval so that z = 4+ 76 6 is the answer.
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