lab6_solns_f11

# lab6_solns_f11 - Math 116 - Lab 6 - Solutions - Fall 2011....

This preview shows pages 1–2. Sign up to view the full content.

Math 116 - Lab 6 - Solutions - Fall 2011. 1. a) f 00 ( π 4 ) = - 4 < 0 concave down. 1. b) f 00 (1) = 25 e - 5 > 0 concave up. 1. c) f 00 (12) = - 1 144 < 0 concave down. 2. a) f 00 ( x ) = 1 x . Since the domain of f ( x ) = x ln( x ) is x (0 , ), we have f 00 ( x ) > 0 for every x in the domain. Therefore, there are no critical points. 2. b) f 00 ( x ) = cosh( x ) > 0 for every x . So there are no second order critical points. 2. c) f 00 ( x ) = sinh( x ). The second order critical point is the solution of f 00 ( x ) = 0, which is x = 0; it is an inﬂection point as f 00 ( x ) changes sign at x = 0. 2. d) f 0 ( x ) = e - x (1+ e - x ) 2 = 1 ( e x/ 2 + e - x/ 2 ) 2 = 1 4 [cosh( x/ 2)] - 2 . f 00 ( x ) = - 1 4 1 cosh 3 ( x 2 ) sinh( x 2 ). Hence the solution of f 00 ( x ) = 0 is x = 0; it’s easy to verify that it is an inﬂection point. 3. f 00 ( x ) = b 2 e - bx > 0 for every x . So the answer is for any a and b . 4. a) f 0 ( x ) = x 3 - 4 x = x ( x - 2)( x + 2). So the solutions of f 0 ( x ) = 0 are x = 0, x = 2 and x = - 2. Since f 00 ( x ) = 3 x 2 - 4, it is easy to verify that f 00 (0) < 0, whereas f 00 (2) > 0 and f 00 ( - 2) > 0. Hence the second derivative test tells us that x = 0 is a local maximum, while x = 2 and x = - 2 are local minima. 4. b)

This preview has intentionally blurred sections. Sign up to view the full version.

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

## This note was uploaded on 12/28/2011 for the course MATH 116 taught by Professor Robertandre during the Spring '11 term at Waterloo.

### Page1 / 4

lab6_solns_f11 - Math 116 - Lab 6 - Solutions - Fall 2011....

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online