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File0017 - ’— 58 59 60 61 62 63 65 67 69 Section 2.6...

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Unformatted text preview: ’—. 58. 59. 60. 61. 62. 63. 65. 67. 69. Section 2.6 Continuity Let f(x) be the new position of point x and let d(x) 2 f(x) — x. The displacement function d is negative if x is the left—hand point of the rubber band and positive if x is the right-hand point of the rubber band. By the Intermediate Value Theorem, d(x) : 0 for some point in between. That is, f(x) : x for some point x, which is then in its original position. If f(0) = 0 or f(1) = 1, we are done (i.e., c = 0 orc : 1 in those cases). Then let f(0) = a > 0 and f(1) = b < 1 because 0 _<_ f(x) S 1. Define g(x) = f(x) — x :> g is continuous on [0, 1]. Moreover, g(0) = f(O) — 0 = a > 0 and g(1) = f(l) — 1 = b — 1 < 0 :> by the Intermediate Value Theorem there is a number c in (0, 1) such that g(c) = 0 :> f(c) —c =00rf(c) = e. Let 6 = ”(Tm > 0. Since fis continuous at x = c there is a 6 > 0 such that ix — cl < 6 :> |f(x) — f(c)| < 6 => f(c) ~ 6 < f(x) < f(c) + 6. If f(c) > 0, then 6 = %f(c) :> %f(c) < f(x) < gm) => f(x) > 0 on the interval (0 — 5, c + 6). If f(c) < 0, then 6 = — %f(c) # %f(c) < f(x) < %f(c) => f(x) < 0 on the interval (c — 6,0 + 6). y f(C)+c .\_--- f(c)—e By Exercises 52 in Section 2.3, we have xlifimc f(x) = L 4: hlimO f(c + h) = L. Thus, f(x) is continuous at x = c 4:) Xli_r9nc f(x) = f(c) <1) hlim0 f(c + h) = f(c). By Exercise 61, it suffices to show that hlim0 sin(c + h) = sin c and hlim0 cos(c + h) = cos c. Now 1112110 srn(c + h) = hlrmo [(sm c)(cos h) + (cos c)(sm h)] 2 (sm c) (hlgnocos h) + (cos c) (hlgnosm h) By Example 6 Section 2.2, hlimocos h = 1 and hlimosin h = 0. So hlim0 sin(c + h) = sin c and thus f(x) = sin x is continuous at x = 0. Similarly, hlgrio cos(c + h) = hInn0 [(cos c)(cos h) — (sm c)(srn h)] = (cos c) (hlgnocos h) — (sm c) (blimosm h) = cos c. Thus, g(x) = cos x is continuous at x : c. x x 1.8794, —1.5321, —0.3473 64. x x 1.4516, —0.8547, 0.4030 x x 1.7549 66. x x 1.5596 x z 3.5156 68. x z #39058, 3.8392, 0.0667 x x 0.7391 70. x z —1.8955, 0. 1.8955 83 ...
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