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
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.
- Spring '07