solutionshw5-555-2011

# solutionshw5-555-2011 - Solutions homework 5. Page 128...

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

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.

Unformatted text preview: Solutions homework 5. Page 128 Problem 3: Using the substitution y =- x we get g ( y )- g (- c ) y- (- c ) = f (- y )- f ( c ) y- (- c ) =- f ( x )- f ( c ) x- c . Now letting y → - c- is the same as letting x → c + , from which the problem follows. Page 128 Problem 4. As g ( x ) = 0 for x ≥ 0 we clearly have g r (0) = 0. On the other hand ( g ◦ f )( x ) = x sin( 1 x ) for x > 0 and ( g ◦ f ))(0) = 0. Hence ( g ◦ f )( x )- ( g ◦ f )(0) x- = sin( 1 x ) , for x > 0. It is not difficult to se that lim x → + sin( 1 x ) doesn’t exist (consider e.g. what happens if x n = 1 πn and x n = 1 π 2 +2 πn ). Page 129 Problem 2. True, if f has a maximum at an interior pt c , then for any neigh- borhood V ⊂ S of c we have f ( x ) ≤ f ( c ) on V . Page 132 Problem 2. a. If there exist a < b in I such that f ( a ) = f ( b ), then by Rolle’s theorem there exists a < c < b such that f ( c ) = 0, which contradicts the assumption that f ( x ) 6 = 0 for all x ....
View Full Document

## This note was uploaded on 02/05/2012 for the course MATH 555 taught by Professor Staff during the Spring '07 term at South Carolina.

### Page1 / 2

solutionshw5-555-2011 - Solutions homework 5. Page 128...

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

View Full Document
Ask a homework question - tutors are online