• date post

06-May-2015
• Category

## Documents

• view

1.997

19

Embed Size (px)

description

capitulo 3

### Transcript of solucionario de purcell 3

• 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.