solucionario de purcell 3

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  • 1.Applications of the CHAPTER3Derivative3 3.1 Concepts Review 7. ( x) = 2 x + 3; 2x + 3 = 0 when x = .21. continuous; closed and bounded 3Critical points: 2, , 122. extreme 39 (2) = 2, = , (1) = 42 43. endpoints; stationary points; singular points9Maximum value = 4, minimum value = 4. f (c) = 0; f (c) does not exist41 6 8. G ( x) = (6 x 2 + 6 x 12) = ( x 2 + x 2); Problem Set 3.1 5 51. Endpoints: 2 , 4x 2 + x 2 = 0 when x = 2, 1 Singular points: noneCritical points: 3, 2, 1, 3 Stationary points: 0, 29 7G (3) = , G (2) = 4, G (1) = , G (3) = 9 Critical points: 2, 0, 2, 4 5 5Maximum value = 9,2. Endpoints: 2 , 47 Singular points: 2 minimum value = Stationary points: 0 5 Critical points: 2, 0, 2, 4 9.f ( x) = 3 x 2 3; 3x 2 3 = 0 when x = 1, 1.3. Endpoints: 2 , 4Critical points: 1, 1 Singular points: nonef(1) = 3, f(1) = 1 Stationary points: 1, 0,1, 2,3No maximum value, minimum value = 1 Critical points: 2, 1, 0,1, 2,3, 4(See graph.)4. Endpoints: 2 , 4 Singular points: none Stationary points: none Critical points: 2, 4 5.f ( x) = 2 x + 4; 2 x + 4 = 0 when x = 2. Critical points: 4, 2, 0 f(4) = 4, f(2) = 0, f(0) = 4 Maximum value = 4, minimum value = 010.f ( x) = 3 x 2 3; 3x 2 3 = 0 when x = 1, 1. 136. h( x) = 2 x + 1; 2 x + 1 = 0 when x = . Critical points: , 1, 1, 3 221 3 17 Critical points: 2, , 2f = , f (1) = 3, f (1) = 1, f (3) = 192 2 8 11 Maximum value = 19, minimum value = 1 h(2) = 2, h = , h(2) = 6 24 1 Maximum value = 6, minimum value = 4 154 Section 3.1 Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2. 12x2x 11. h(r ) = ; h(r ) is never 0; h(r ) is not defined 15. g ( x) = ; = 0 when x = 0. 2 2r2 (1 + x )(1 + x 2 ) 2 when r = 0, but r = 0 is not in the domain onCritical point: 0 [1, 3] since h(0) is not defined. g(0) = 1 Critical points: 1, 3As x , g ( x) 0+ ; as x , g ( x) 0+. Note that lim h(r ) = and lim h( x) = .x 0 x 0+ Maximum value = 1, no minimum value No maximum value, no minimum value.(See graph.) 2x2x 12. g ( x) = ; = 0 when x = 02 2(1 + x ) (1 + x 2 ) 2 Critical points: 3, 0, 1 1 1 g(3) =, g(0) = 1, g(1) =10 2 1 Maximum value = 1, minimum value =101 x216.f ( x) = ; 13.f '( x) = 4x 4x 3 (1 + x 2 )2( = 4 x x2 1 ) 1 x2= 0 when x = 1, 1 = 4 x ( x 1)( x + 1) (1 + x 2 )24 x ( x 1)( x + 1) = 0 when x = 0,1, 1 . Critical points: 1, 1, 4 Critical points: 2, 1, 0,1, 2 114 f (1) = , f (1) = , f (4) = 22 17f ( 2 ) = 10 ; f ( 1) = 1 ; f ( 0 ) = 2 ; f (1) = 1 ; 1f ( 2 ) = 10Maximum value = , 2 Maximum value: 101 Minimum value: 1 minimum value = 2 14.f ' ( x ) = 5 x 4 25 x 2 + 20 17. r ( ) = cos ; cos = 0 when =+ k (4 = 5 x 5x + 4)22 = 5 ( x 2 4 )( x 2 1)Critical points: , 4 6 = 5 ( x 2 )( x + 2 )( x 1)( x + 1) 1 1r = , r = 5 ( x 2 )( x + 2 )( x 1)( x + 1) = 0 when 426 2 x = 2, 1,1, 21 1Maximum value = , minimum value = Critical points: 3, 2, 1,1, 2 221941f ( 3) = 79 ; f ( 2 ) = ; f ( 1) = ;18. s (t ) = cos t + sin t ; cos t + sin t = 0 when 333513f (1) =; f ( 2) = tan t = 1 or t = + k . 3 34353 Maximum value: Critical points: 0,, 3 4 Minimum value: 79 3 s(0) = 1, s = 2, s ( ) = 1 . 4 Maximum value = 2,minimum value = 1 Instructors Resource Manual Section 3.1155 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 3. x 125. g ' ( ) = 2 ( sec tan ) + 2 sec 19. a ( x) = ; a ( x) does not exist when x = 1.x 1= sec ( tan + 2 ) Critical points: 0, 1, 3 sec ( tan + 2 ) = 0 when = 0 . a(0) = 1, a(1) = 0, a(3) = 2 Maximum value = 2, minimum value = 0Consider the graph: y3(3s 2) 20. f ( s ) = ; f ( s ) does not exist when s = 2 . 3s 23Critical points: 1, 2 , 4 13 f(1) = 5, f 3( ) 2 = 0, f(4) = 10 x Maximum value = 10, minimum value = 04411 21. g ( x) =; f ( x) does not exist when x = 0.3x2 / 3 Critical points: 1, 0, 27 Critical points: , 0, g(1) = 1, g(0) = 0, g(27) = 3 44 Maximum value = 3, minimum value = 1 2 222g =; g ( 0) = 0 ; g = 2 4 164 16 22. s (t ) = ; s (t ) does not exist when t = 0. 3/ 5 5t2 2 Critical points: 1, 0, 32Maximum value: ; Minimum value: 0 16 s(1) = 1, s(0) = 0, s(32) = 4 Maximum value = 4, minimum value = 0( 2 + t ) t 2 / 3 t 5/ 3 (1)5 23. H ' ( t ) = sin t26. h ' ( t ) = 3 sin t = 0 when( 2 + t )2 t = 0, , 2 ,3 , 4 ,5 , 6 , 7 ,8 5 10 2 t2/3 (2 + t ) t t2/3 + t Critical points: 0, , 2 ,3 , 4 ,5 , 6 , 7 ,83 = 3 3 = H ( 0 ) = 1 ; H ( ) = 1 ; H ( 2 ) = 1 ;( 2 + t )2( 2 + t )2 H ( 3 ) = 1 ; H ( 4 ) = 1 ; H ( 5 ) = 1 ; 2t 2 / 3 ( t + 5 ) = H ( 6 ) = 1 ; H ( 7 ) = 1 ; H ( 8 ) = 1 3( 2 + t ) 2Maximum value: 1 h ' ( t ) is undefined when t = 2 and h ' ( t ) = 0 Minimum value: 1 when t = 0 or t = 5 . Since 5 is not in the 24. g ' ( x ) = 1 2 cos x interval of interest, it is not a critical point. Critical points: 1, 0,81 1 2 cos x = 0 cos x =whenh ( 1) = 1 ; h ( 0 ) = 0 ; h ( 8 ) = 162 55 5Maximum value: 16 ; Minimum value: 1 x= , , ,5 3 3 3 3 5 5f ( x) = 3 x 2 12 x + 1;3x 2 12 x + 1 = 0 Critical points: 2 , , , , , 2 27. a.3 3 3 3 33 33 5 5when x = 2 and x = 2 +. g ( 2 ) = 2 ; g = 3;33 3 3 33 33 Critical points: 1, 2 ,2+,5 g = + 3 ; g = 3 ;33 33 3 3 33 5 5 f(1) = 6, f 2 2.04, g =+ 3 ; g ( 2 ) = 2 3 3 3 5 33 Maximum value: + 3 f 2+ 26.04, f(5) = 183 3 5 Maximum value 2.04; Minimum value: 33minimum value 26.04 156 Section 3.1 Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 4. ( x3 6 x 2 + x + 2)(3x 2 12 x + 1) 29. Answers will vary. One possibility: b.g ( x) =;y32x 6x + x + 2 533 g '( x) = 0 when x = 2 and3 33 x = 2+ . g ( x) does not exist when 3 5 5xf(x) = 0; on [1, 5], f(x) = 0 when x 0.4836 and x 0.7172 33 5Critical points: 1, 0.4836, 2 , 330. Answers will vary. One possibility:33 y0.7172, 2 +, 53g(1) = 6, g(0.4836) = 0, 533 g2 2.04, g(0.7172) = 0,3 33 g2+ 26.04, g(5) = 185x3 Maximum value 26.04,minimum value = 0528. a.f ( x) = x cos x; on [1, 5], x cos x = 0 when31. Answers will vary. One possibility:3yx = 0, x = , x =2 2 5 3Critical points: 1, 0, , , 52 2f(1) 3.38, f(0) = 3, f 3.57,2x5 3 f 2.71, f(5) 2.51 2 Maximum value 3.57,5minimum value 2.7132. Answers will vary. One possibility: (cos x + x sin x + 2)( x cos x) y b.g ( x) = ;cos x + x sin x + 2 5 3 g ( x ) = 0 when x = 0, x = , x=22g ( x) does not exist when f(x) = 0;on [1, 5], f(x) = 0 when x 3.455x3Critical points: 1, 0, , 3.45, , 52 25g(1) 3.38, g(0) = 3, g 3.57,2 3 g(3.45) = 0, g 2.71, g(5) 2.51 2 Maximum value 3.57;minimum value = 0 Instructors Resource Manual Section 3.1157 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5. 33. Answers will vary. One possibility: 3.2 Concepts Reviewy1. Increasing; concave up 5 2.f ( x) > 0; f ( x) < 03. An inflection point5x 4.f (c) = 0; f (c) does not exist.5 Problem Set 3.234. Answers will vary. One possibility: 1.f ( x) = 3; 3 > 0 for all x. f(x) is increasingyfor all x. 51 2. g ( x) = 2 x 1; 2x 1 > 0 when x > . g(x) is21 increasing on , and decreasing on2 5x 1 , . 25 3. h(t ) = 2t + 2; 2t + 2 > 0 when t > 1. h(t) is 35. Answers will vary. One possibility:increasing on [1, ) and decreasing ony ( , 1].5 4.f ( x) = 3x 2 ; 3 x 2 > 0 for x 0 .f(x) is increasing for all x.5. G ( x ) = 6 x 2 18 x + 12 = 6( x 2)( x 1)5xSplit the x-axis into the intervals ( , 1), (1, 2),(2, ).33 3Test points: x = 0, , 3; G (0) = 12, G = ,522 2G (3) = 12 36. Answers will vary. One possibility:G(x) is increasing on ( , 1] [2, ) andydecreasing on [1, 2]. 5 6.f (t ) = 3t 2 + 6t = 3t (t + 2)Split the x-axis into the intervals ( , 2),(2, 0), (0, ).Test points: t = 3, 1, 1; f (3) = 9, 5 x f (1) = 3, f (1) = 9f(t) is increasing on ( , 2] [0, ) and5decreasing on [2, 0].7. h( z ) = z 3 2 z 2 = z 2 ( z 2)Split the x-axis into the intervals ( , 0), (0, 2),(2, ).Test points: z = 1, 1, 3; h(1) = 3, h(1) = 1, h(3) = 9h(z) is increasing on [2, ) and decreasing on( , 2]. 158 Section 3.2Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 6. 8.f ( x) = 2 x 16.f ( x) = 12 x 2 + 48 x = 12 x( x + 4); f ( x) > 0 when 3 xx < 4 and x > 0.Split the x-axis into the intervals ( , 0), (0, 2),f(x) is concave up on ( , 4) (0, ) and(2, ).concave down on (4, 0); inflection points areTest points: 1, 1, 3; f (1) = 3, f (1) = 1,(4, 258) and (0, 2).1 f (3) = 17. F ( x) = 2sin 2 x 2 cos 2 x + 4 = 6 4 cos 2 x;27f(x) is increasing on (0, 2] and decreasing on6 4 cos 2 x > 0 for all x since 0 cos 2 x 1.( , 0) [2, ).F(x) is concave up for all x; no inflection points.18. G ( x) = 48 + 24 cos 2 x 24sin 2 x 9. H (t ) = cos t ; H (t ) > 0 when 0 t < and 23= 24 + 48cos 2 x; 24 + 48cos 2 x > 0 for all x. < t 2.G(x) is concave up for all x; no inflection points. 2 3H(t) is increasing on 0, , 2 and 19.f ( x) = 3 x 2 12; 3 x 2 12 > 0 when 2 2x &lt