{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

File0017

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}