# 46 dt dx c1 t c1 x c2 t 25 when x 0 so c2

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Unformatted text preview: near x = 1.3; let f (x) = e−x − ln x, f (x) = −e−x − 1/x, x1 = 1.3, e−xn − ln xn , x2 = 1.309759929, x4 = x5 = 1.309799586 xn+1 = xn + −x e n + 1/xn 2 -2 2 -2 1 0 2 -4 17. √ The graphs of y = x2 and y = 2x + 1 intersect at points near √ x = −0.5 and x = 1; x2 = 2x + 1, x4 − 2x − 1 = 0. Let f (x) = x4 − 2x − 1, then f (x) = 4x3 − 2 so x4 − 2xn − 1 . xn+1 = xn − n 3 4xn − 2 If x1 = −0.5, then x2 = −0.475, x3 = −0.474626695, x4 = x5 = −0.474626618; if x1 = 1, then x2 = 2, x3 = 1.633333333, · · · , x8 = x9 = 1.395336994. 4 -0.5 2 0 Exercise Set 6.4 18. 200 The graphs of y = x3 /8 + 1 and y = cos 2x intersect at x = 0 and at a point near x = −2; x3 /8 + 1 = cos 2x, x3 − 8 cos 2x + 8 = 0. Let f (x) = x3 − 8 cos 2x + 8, x3 − 8 cos 2xn + 8 then f (x) = 3x2 + 16 sin 2x so xn+1 = xn − n 2 . 3xn + 16 sin 2xn x1 = −2, x2 = −2.216897577, x3 = −2.193821581, · · · , x5 = x6 = −2.193618950. 2 -3 2 -2 x2 − a 1 n = 2xn 2 a xn 20. (a) f (x) = x2 − a, f (x) = 2x, xn+1 = xn − (b) 19. a = 10; x1 = 3, x2 = 3.166666667, x3 = 3.162280702, x4 = x5 = 3.162277660 (a) f (x) = (b) xn + 1 1 − a, f (x) = − 2 , xn+1 = xn (2 − axn ) x x a = 17; x1 = 0.05, x2 = 0.0575, x3 = 0.058793750, x5 = x6 = 0.058823529 21. f (x) = x3 + 2x + 5; solve f (x) = 0 to ﬁnd the critical points. Graph y = x3 and y = −2x − 5 to see x3 + 2xn + 5 that they intersect at a point near x = −1; f (x) = 3x2 + 2 so xn+1 = xn − n 2 . 3xn + 2 x1 = −1, x2 = −1.4, x3 = −1.330964467, · · · , x5 = x6 = −1.328268856 so the minimum value of f (x) occurs at x ≈ −1.328268856 because f (x) > 0; its value is approximately −4.098859132. 22. From a rough sketch of y = x sin x we see that the maximum occurs at a point near x = 2, which will be a point where f (x) = x cos x + sin x = 0. f (x) = 2 cos x − x sin x so xn cos xn + sin xn xn + tan xn xn+1 = xn − = xn − . 2 cos xn − xn sin xn 2 − xn tan xn x1 = 2, x2 = 2.029048281, x3 = 2.028757866, x4 = x5 = 2.028757838; the maximum value is approximately 1.819705741. 23. Let f (x) be the square of the distance between (1, 0) and any point (x, x2 ) on the parabola, then f (x) = (x − 1)2 + (x2 − 0)2 = x4 + x2 − 2x + 1 and f (x) = 4x3 + 2x − 2. Solve f (x) = 0 to ﬁnd 4x3 + 2xn − 2 2x3 + xn − 1 the critical points; f (x) = 12x2 + 2 so xn+1 = xn − n 2 = xn − n 2 . x1 = 1, 12xn + 2 6xn + 1 x2 = 0.714285714, x3 = 0.605168701, · · · , x6 = x7 = 0.589754512; the coordinates are approximately (0.589754512, 0.347810385). 24. The area is A = xy = x cos x so dA/dx = cos x − x sin x. Find x so that dA/dx = 0; cos xn − xn sin xn 1 − xn tan xn = xn + . d2 A/dx2 = −2 sin x − x cos x so xn+1 = xn + 2 sin xn + xn cos xn 2 tan xn + xn x1 = 1, x2 = 0.864536397, x3 = 0.860339078, x4 = x5 = 0.860333589; y ≈ 0.652184624. 25. (a) (b) 26. Let s be the arc length, and L the length of the chord, then s = 1.5L. But s = rθ and L = 2r sin(θ/2) so rθ = 3r sin(θ/2), θ − 3 sin(θ/2) = 0. θn − 3 sin(θn /2) . 1 − 1.5 cos(θn /2) θ1 = 3, θ2 = 2.991592920, θ3 = 2.991563137, θ4 = θ5 = 2.991563136 rad so θ ≈ 171◦ . Let f (θ) = θ − 3 sin(θ/2), then f (θ) = 1 − 1.5 cos(θ/2) so θn+1 = θn − r2 (θ − sin θ)/2 = πr2 /4 so θ − sin θ − π/2 = 0. Let f (θ) = θ − sin θ − π/2, then f (θ) = 1 − cos θ so θn − sin θn − π/2 . θn+1 = 1 − cos θn θ1 = 2, θ2 = 2.339014106, θ3 = 2.310063197, · · · , θ5 = θ6 = 2.309881460 rad; θ ≈ 132◦ . 201 Chapter 6 27. If x = 1, then y 4 + y = 1, y 4 + y − 1 = 0. Graph z = y 4 and z = 1 − y to see that they intersect near y 4 + yn − 1 . y = −1 and y = 1. Let f (y ) = y 4 + y − 1, then f (y ) = 4y 3 + 1 so yn+1 = yn − n 3 4yn + 1 If y1 = −1, then y2 = −1.333333333, y3 = −1.235807860, · · · , y6 = y7 = −1.220744085; if y1 = 1, then y2 = 0.8, y3 = 0.731233596, · · · , y6 = y7 = 0.724491959. 28. If x = 1, then 2y − cos y = 0. Graph z = 2y and z = cos y to see that they intersect near y = 0.5. Let 2yn − cos yn f (y ) = 2y − cos y , then f (y ) = 2 + sin y so yn+1 = yn − . 2 + sin yn y1 = 0.5, y2 = 0.450626693, y3 = 0.450183648, y4 = y5 = 0.450183611. 29. 30. 5000 (1 + i)25 − 1 ; set f (i) = 50i − (1 + i)25 + 1, f (i) = 50 − 25(1 + i)24 ; solve i f (i) = 0. Set i0 = .06 and ik+1 = ik − 50i − (1 + i)25 + 1 / 50 − 25(1 + i)24 . Then i1 = 0.05430, i2 = 0.05338, i3 = 0.05336, · · ·, i = 0.053362. S (25) = 250000 = 0.5 (a) x1 = 2, x2 = 5.3333, x3 = 11.055, x4 = 22.293, x5 = 44.676 0 15 0 (b) 31. (a) (b) x1 = 0.5, x2 = −0.3333, x3 = 0.0833, x4 = −0.0012, x5 = 0.0000 (and xn = 0 for n ≥ 6) x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 0.5000 −0.7500 0.2917 −1.5685 −0.4654 0.8415 −0.1734 2.7970 1.2197 0.1999 The sequence xn must diverge, since if it did converge then f (x) = x2 + 1 = 0 would have a solution. It seems the xn are oscillating back and forth in a quasi-cyclical fashion. EXERCISE SET 6.5 1. 2. f (−3) = f (3...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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