Business Calc Homework w answers_Part_91

Business Calc Homework w answers_Part_91 - Appendix A2 dx 0...

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Appendix A2 dx 0 at d 2 4 dx dy and at , and vertical tangents 0, 0 3 3 d d 5 at , at , and at . 3 3 dy 0 For 0 (or 2 ), becomes and l'Hpital's Rule dx 0 dy d 451 There are horizontal tangents 0, s Appendix A2 (pp. 581584) 1. Step 1: The formula holds for n ...x x Step 2: Suppose x 1 2 1, because x1 x1 1 Then, x1 x2 x1 ... x2 k x2 xk ... x1 . xk . xk ... xk 1 by the leads to dy dx 0 triangle inequality. sin (0) 4 sin (0) cos (0) cos (0) 2 cos 2 (0) 2 sin 2 (0) 0, so this is So, by the transitivity of , x1 x1 x2 ... xk 1 original formula for all n. x2 ... xk 1 . another horizontal tangent line. Horizontal tangents: At 0 or 2 , we have r 0 and the Cartesian 0. The mathematical induction principle now guarantees the 2. Step 1: The formula holds for n 1 r (1 r)(1 1 r r) 1 1 1, because r2 . r coordinates are (0, 0), so the tangent is y At are 2 , we have r 3 3 2 3 2 cos , sin 2 3 2 3 3 and the Cartesian coordinates 2 3 3 3 , , so the tangent is 4 4 Step 2: Suppose 1 1 r 1 r ... 1 r2 rk r k 1(1 r ... 1 rk 1 1 rk 1 . Then 1 r r2 rk 3 3 . y 4 4 , we again have r At 3 1 3 and the Cartesian 2 3 , 4 3 4 3 rk 1 rk 1 1 r r) 1 rk 2 . 1 r The mathematical induction principle now guarantees the original formula for every positive integer n. , so 3. Step 1: The formula holds for n Step 2: Suppose d k (x ) dx d k 1 d (x ) (x dx dx coordinates are the tangent is y 3 4 3 4 cos , sin 2 3 2 3 3 4 3 1, because d (x) dx 1. . kx k 1. Then x k) xk x kx k d k (x ) dx xk (k Vertical tangents: At are 1 1 cos , sin 2 3 2 3 1 3 , , so the tangent is x 4 4 1 . 4 3 d (x) dx x kx k 1 and the Cartesian coordinates 2 1 xk 1)x k. , we have r The mathematical induction principle now guarantees the original formula for any positive integer n. 4. Step 1: The formula holds for n 1, because f (x1) f (x1). Step 2: Suppose f (x1x2 ... xk) f (x1) f (x2) ... f (xk). Then by the given property, f (x1x2 ... xk 1) f (x1x2 ... xk) f (xk 1) f (x1) f (x2) ... f (xk 1). The mathematical induction principle now guarantees the original formula for every positive integer n. 5. Step 1: The formula holds for n Step 2: Suppose 2 31 2 32 2 31 2 32 At , we have r 2 and the Cartesian coordinates are ( 2, 0), so the tangent is x 2. (2 cos , 2 sin ) At are x y x 5 , we have r 3 1 5 1 5 cos , sin 2 3 2 3 1 and the Cartesian coordinates 2 1 , 4 3 4 , so the tangent is 0, 1 . In summary, the horizontal tangents are y 4 3 4 3 1, because 2 3k 2 3 1 1 . 3 , and y 1 . 4 3 4 3 , and the vertical tangents are ... 2 3k 1 1 1 3k ... 3 2 3k 1 1 1 3k 1. 1 . Then 3k 2 3k 1 2 and x 1 1 The mathematical induction principle now guarantees the original formula for all positive integers n. 452 Appendix A2 9. Step 1: The formula holds for n 3 6 27 4 5 6 7 343 6, because 12 1(1 1/2)(1 3 1) 6. Experiment: n n! n3 1 1 1 2 2 8 1, because 1. k2 1)2 1) 1) 3 k(k 1/2)(k 3 1) 24 120 720 5040 64 125 216 Step 2: Suppose 12 Then 12 22 k(k k(k 22 ... ... (k . Step 1: The inequality holds for n 720 216. k 3. Then (k 1 3 3 1/2)(k 3 1/2)(k (k 3(k 1)2 1)2 Step 2: Suppose k! For k 1 k2 1)! (k 1)k 3. 2 and (k 1) . So (k 1)(k 2 4, k 2 k 2 1 (since k k2 k3 (9/2)k 2 3 (13/2)k 3 1), and so k 1)k , (k 3 k 2 2k 1 (k of (k 1)! 1) , and thus by the transitivity (k 1)3. 3/2)(k 3 1) 2) (k 1)[(k The mathematical induction principle now guarantees the original inequality for all n 7. Experiment: n 2n n2 1 2 1 2 4 4 3 8 9 4 16 16 5 32 25 6 64 36 6. 1/2][(k 3 1) 1] . The mathematical induction principle now guarantees the original formula for all positive integers n. 10. Step 1: The formula holds for n 13 1(1 2 1) 2 1, because 1. ... (k (k (k 1 (k 2) 2 Step 1: The inequality holds for n Step 2: Suppose 2k since 1 k 5, because 32 3, k 2k 2 1 k 25. Step 2: Suppose 13 Then 13 23 k(k 2 23 ... 1) 2 1)2 4 k3 1)3 1)3 1)(k 1)2 . k(k 2 1) 2 . k 2. For k 1 , and so k2 ,2 k 1. Then by the k 2(k k2 4 (k transitivity of 2k 1 2k + 1. And thus 2k 1 (k 1)2. 2k 2k k2 1)2 (k 4 2)2 The mathematical induction principle now guarantees the original inequality for all n 5. 3, because 2 2 2k 1 4 2 8 3 k 1)(k 2 (k 1)2 8. Step 1: The inequality holds for n Step 2: Suppose 2k by the transitivity of 1 . Then 2k 1 8 1 . 8 The mathematical induction principle now guarantees the original formula for all positive integers n. and the fact that 1 , and 4 1 k 1 1 ,2 . 8 8 11. (a) Step 1: The formula holds for n 1, because 1 The mathematical induction principle now guarantees the original inequality for n 3. k 1 (ak (ak i k 1 i 1 k 1 1 ak) i k 1 ak bk k 1 i 1 a1 i b1 . Step 2: Suppose (ak k 1 i 1 bk ) k 1 ak bk. Then k 1 k 1 bk) ak i 1 k 1 k 1 bk k 1 i (ak i bk) ai 1 (ai bi 1 1 bi 1) ak bk. The mathematical induction principle now guarantees the original formula for every positive integer n. Appendix A3 (b) Step 1: The formula holds for n 1, because 1 453 s Appendix A3 (pp. 584592) 1. a 4 9 k 1 (ak (ak 1 bk) i k 1 ak bk k 1 i 1 a1 i b1 . c 1 2 b x 4 7 Step 2: Suppose (ak k 1 i 1 bk ) k 1 ak bk. Then k 1 k 1 bk) k 1 i 1 k 1 ak i k 1 i (ak k 1 i bk) ai 1 (ai bi 1 bi 1) 1 Step 1: x Step 2: bk 1 2 1 2 4 9 c 3 1 x 2 4 9 x 1 2 1 2 1 2 1 2 1 . 18 4 7 1 . 14 ak bk. k 1 i 1 1 , or 18 The value of 2. a 2.7591 which assures x 4 is the smaller value, 7 b x The mathematical induction principle now guarantees the original formula for every positive integer n. x (c) Step 1: The formula holds for n 1, because 3.2391 k 1 cak i 1 c i k 1 ak c 1 1 ca1. Step 1: x i 1 3 3 3 x 2.7591 x 3 3 Step 2: Suppose cak k 1 k 1 ak. Then cak k 1 i 1 k 1 cak k 1 cak ak c c c ak i k 1 i 1 k 1 ak 1 i Step 2: or 0.2409, 0.2391. 3 cak . 1 3 3.2391 1 ak The value of 2.7591 0.2391. x which assures x 3.2391 is the smaller value, The mathematical induction principle how guarantees the original formula for every positive integer n. (d) Step 1: The formula c k 1 n 3. Step 1: x n c holds for n 1, 3 3 x x 3 3 3 because k 1 c c 1 Step 2: From the graph, 1 c i k 1 2.61 0.39, 0.39. c. or i c. Then c (i 1) c. 4. Step 1: x ( 1) 1 3 Step 2: Suppose c i 1 3.41 x 0.41; thus x 1 1 i c k 1 The mathematical induction principle now guarantees the original formula for every positive integer n. 12. Step 1: The formula holds for n number x), because x1 xk 1 Step 2: From the graph, 0.77, or 9 25 1 (and every real 1 0.36. 16 7 9 9 16 9 0.36; thus 25 25 1 x1 k x. x k Step 2: Suppose x x . Then xk x xk x x k 1. k xk x The mathematical induction principle now guarantees the original formula for every positive integer n (and every real number x). 5. Step 1: (2x 2) ( 6) 0.02 2x 4 0.02 0.02 2x 4 0.02 4.02 2x 3.98 2.01 x 1.99 Step 2: x ( 2) x 2 2 x 2 0.01. 6. Step 1: x 1 1 0.1 0.1 0.9 x 1 1.1 0.81 0.19 x 0.21 Step 2: x 0 x x 1 1 0.1 x 1 1.21 0.19. 454 Appendix A3 11. (a) lim (sin x) x1 7. Step 1: 19 x 3 1 1 19 x 3 1 19 x 4 4 19 x 16 2 4 x 19 16 15 x 3 or 3 x 15 Step 2: x 10 x 10 10 x 10. Then 10 3 7, or 10 15 5; thus 5. sin 1 sin 1 0.01 0.841 0.01 sin 1 sin x 0.01) 1 (b) Step 1: sin x sin x 0.01 0.01 sin 1 x 0.01 sin 1 sin 1 (sin 1 sin 8. Step 1: x 2 4 0.5 0.5 x 2 4 0.5 2 3.5 x 4.5 3.5 x 4.5 4.5 x 3.5, for x near 2. Step 2: x ( 2) x 2 2 x 2. Then 2 4.5 4.5 2 0.1213, or 2 3.5 2 3.5 0.1292; thus 4.5 2 0.121. 9. (a) lim x 5 (sin 1 x 1 . 1 0.01) 1 Step 2: x 1 1 x Then 1 1 Choose Alternately, graph y1 1 sin sin 1 sin sin 1 1 (sin 1 0.01) 0.01) 0.0182, or (sin 1 (sin 1 0.01) 0.01) 1 0.0188. (sin 1 x 2 6x 5 x 5 lim x 5 (x 5)(x 1) x 5 lim (x x 5 1) 0.018. sin x, y2 sin 1 1, and 4, x (b) Step 1: x2 5. 6x 5 x 5 ( 4) (x 0.05 4 0.05 5 5. 5 y3 x x sin 1 0.98175 1.01878 0.018. 1. The curve intersects the lines at 1 1 0.01825 and at 0.01878. We may choose Step 2: x Then 0.05 4.05 5.05 ( 5) 5 5)(x 1) x 5 x x 1 3.95, x 4.95, x x 5 4.95 2x) 4) 2x) 2 2 x 5. 5.05 0.05, or 0.05. [0.9, 1.1] by [0.78, 0.88] 5 10. (a) lim f (x) x1 0.05; thus 2 and 2, so lim f (x) x1 lim (4 x1 lim f (x) x1 lim (6x x1 2. 2x 0.5 (b) Step 1: x 1: (4 x 2 0.5 0.5 2 3 ; 4 x 1: (6x x 4) 6 6 2 0.5 0.5 6x 13 . 12 6 0.5 Step 2: x 1 Then 1 1 x 3 4 x 1 . 1 1 . Choose 12 1 , or 1 4 1 . 12 13 12 Appendix A3 x 4 x x2 4 1 ( 1)2 4 1 3 x x2 x 4 1 3 455 12. (a) lim 2 x 1 x 13. Step 1: For x 1 1, x 2 x2 1 1 x x 1 1 1 1 x 1 1 x2 1 (b) Step 1: 0.1 1 3 0.1 Step 2: x 52 for x near 1. 30 13 2 52 x x , 30 30 1 1 x 1. 1 1 1 , 1 1 1 1. 1}, , 1 near x 1. 0.1 13 7 30 30 x2 4 28 13 2 7 2 x x x 30 30 30 28 7 Then x 2 x 0 30 30 1 Then or which using the quadratic formula implies x x 15 13 15 7 15 7 901 421 421 0.7883 or 5.0740, and also 1.1551 15 13 Choose min {1 that is, the smaller of the two distances. 901 13 15 7 421 x 901 15 14. Step 1: . 1 x2 1 3 1 3 1 3 3 1 3 1 3 3 1 x2 1 x2 1 3 1 3 1 x2 1 3 3.4628. Thus Step 2: x Then or ( 1) 1 1 x 1 or Step 2: x x 15 13 x 1. 901 3 3 3 3 3 1 3 x2 3 1 3 x x x 1 1 3 3 , for x near 3 3. 901 28 0.1551, 13 15 421 1 7 22 421 0.2117. 7 3 3 3 3 x 3 3 1 3 3 3 . 1 Then or 3 3 3 3 1 3 , Choose Alternately, graph y1 y3 x x 1 3 0.155. x x2 4 1 3. , y2 1 3 0.1, and Choose min 15. (a) (5 I x5 0.1. The curve intersects the lines at 1 1 0.15513 and at 0.21167. We may choose (b) lim x 3 2 3 1 3 , 3 1 3 2 3 . 1.15513 0.78833 0.155. ) 5 (5, 5 5 ) 2 ) 0 , 4) 0 L, show that x c 2 16. (a) 4 I x4 (4 (4 4 x (b) lim [ 1.5, 0] by [ 0.15, 0.55] 17. If L, c, and k are real numbers and lim f (x) xc for any k f (x) 0, there is a k L . 0, let x x c 0 such that 0 Proof: For any is a k k . Since lim f (x) xc L, there 0 such that 0 . Therefore, 0 c f (x) k f (x) L k L . ...
View Full Document

Ask a homework question - tutors are online