Calculus

# Calculus - Chapter 3 Applications of Differentiation...

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Chapter 3 Applications of Differentiation 81 Chapter 3 Section 3.1 5. 0 1/2 0 () ( 1 ) 11 1 () 2 1 ( 1 ) 2 fx f x f == = = ′′ So 00 0 ( ) ( ) 1 1( 1 ) 2 11 22 Lx f x f x x x x x =+ 7. 0 0 0 2 9 , 0 0 ) 209 3 1 ( 2 9 ) 2 ( 2 9 ) 2 1 ( 0 )( 209 ) 3 x x f x x f −− = + = = + + = So 0 ( ) 1 3(0 ) 3 1 3 3 x x 9. 0 0 0 () s i n3, 0 0 )s i n 30s i n () 3 c o s3 ( 0 )3 c o s 303 xx f x f = = = = 0 ( ) 03 ( 0 ) 3 x x = 11. 2 0 20 0 0 2 0 0 , 0 0 ) 1 () 2 ( 0 )2 2 2 x x fx e x f e e e f e e = = = === = 0 ( ) 12 ) 21 x x 13. 2 () t a n (0) tan0 0 e c x f x = = 2 (0) sec 0 1 ( ) (0) (0)( 0) 0 1( 0) (0.01) .01 (0.01) tan 0.01 .0100003 (0.1) 0.1 (0.1) tan(0.1) .1003 (1) 1 tan1 1.557 f f f x x x L f L f L f = + = = =≈ = = 15. 4 (0) 4 0 2 1 ( 4 ) 2 x f x = (0) (4 0) 24 1 ( 0 ) ( 0 ) ( 0 ) 2 4 1 (0.01) 2 (0.01) 2.0025 4 (0.01) 4 0.01 2.002498 1 (0.1) 2 (0.1) 2.025 4 (0.1) 4 0.1 2.0248 1 2 2.25 4 4 1 2.2361 f f f x x L f L f L f = = + = = =+≈ = 17. i n, 0 3 3 32 () c o s 1 cos 33 2 31 3 2 2 3 1 .842 3 f x f f f x x L π π  =   = ππ π π = + π −≈

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Chapter 3 Applications of Differentiation 82 19. 4 0 4 3/4 () 1 6 , 0 (0) 16 0 2 1 ( 1 6 ) 4 11 (0) (16 0) 43 2 1 ( 0 ) ( 0 ) ( 0 ) 2 32 1 (0.04) 2 (0.04) 2.00125 32 fx xx f x f Lx f f x x L =+ = = = = + = 21. 4 0 1 6 , 0 1 ( ) 2 (from exercise 19) 32 1 (.16) 2 (.16) 2.005 32 x L = = 23. 19) 4 16.04 2.0012488 (0.04) 2.00125 2.0012488 2.00125 .00000117 L = = −= 20) 4 16.08 2.0024953 (.08) 2.0025 2.0024953 2.0025 .00000467 L = = 21) 4 16.16 2.0049814 (.16) 2.005 2.0049814 2.005 .0000186 L = = 25. The interval is about ( 0.307, 0.307). 27. a. 4 (24) (20) ( (30) (20)) 10 18 0.4(14 18) 16.4 ff ≈+ = b. 6 (36) (30) ( (40) (30)) 10 14 0.6(12 14) 12.8 = 29. a. 8 (208) (200) ( (220) (200)) 20 128 0.4(142 128) 133.6 = b. 12 (232) (220) ( (240) (220)) 20 142 0.6(136 142) 138.4 = 31. 2 22 2 lim lim 2 4 4 x x x →→ == 33. 32 00 0 0 3 lim lim sin cos 1 6 lim sin 6 lim 6 cos x x x x x x = −− = 35. lim lim 1 ln 1/ x 37. 1 lim lim ; undefined cos 1 sin ee = 39. 2 0 lim 0 cos x x = (L’Hôpital’s rule does not apply.) 41. 1 ln ln lim ; undefined ln x x x 43. 2 sin 2 cos lim lim 1 2 x x x The result matches Example 1.5. 45. 33 36 sin 1 cos 1 lim 1; lim 2 47. The limit of cos x as x approaches 0 is not 0, so L'Hôpital's rule does not apply and the second limit is not equal to the first (which in fact is ). 49. The first step is plausible, since for x close to 0, sin . The second step (dividing out the x ’s) is actually correct. 51. 1 lim lim 1 cx x ec e c x
Chapter 3 Applications of Differentiation 83 53. 2 0 0 2 0 2.5(4 sin 4 ) lim 4 2.5(4 4 cos4 ) lim 8 2.5( 16 sin 4 ) lim 0 8 tt t ω ωω = == 55. For small x we approximate b y1 . x ex + 2/ 22 11 4 2 2 dL Le e ee dd L LL d L L d ππ π + ªº §· +− ¨¸ «» ©¹ ¬¼ ++ ≈= 4.9 () 2 9 . 8 fd d d ≈⋅ = Section 3.2 5. For 0 0 x = the method fails. 7. 32 0 2 3 1 0 , 1 () 3 6 fx x x x fx x x =+ = = a. 0 10 0 2 13 93 31 61 xx =− +⋅ − =− = ⋅+ 1 21 1 33 2 3 1 2 3 36 79 .5486 144 + =≈ b. .53209 9. 42 0 3 3 , 1 () 4 6 x x x x + = = a. 0 3 0 11 1 1 2 41 61 −⋅ + = = ⋅− 1 1 3 15 28 46 −+ = b. .61803 11. 2 ( )4 3 1 8 3 x x x x y 4 x 0 1 0 1 5 for 0, 1, 2, 3, ... n nn n x n + = n n x 1 –4.718750 2 –4.686202 3 –4.6857796 4 –4.6857795 4.685780 x ≈− 13. 53 ( )3 1 ( )5 9 1 x x x x + =++ y 8 x 0 1 0 1 .5 for 0, 1, 2, 3, ... n n x n + = = n n x 1 .526316 2 .525262 3 .525261 4 .525261 .525261 x

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Chapter 3 Applications of Differentiation 84 15.
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## This homework help was uploaded on 04/19/2008 for the course MATH 105 taught by Professor Schroder during the Spring '08 term at Millersville.

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Calculus - Chapter 3 Applications of Differentiation...

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