# Mathematics for Economists

• No School
• NONE 0
• Notes
• 50

This preview shows page 11 - 14 out of 50 pages.

13.17 Suppose not; that is, suppose that for every n there exists an x n [ B 1 6 n ( x p ) with f ( x n ) # 0. Since || x n 2 x p || , 1 6 n , x n x p . Since f is continuous, f ( x n ) f ( x p ). Since each f ( x n ) # 0, f ( x p ) 5 lim f ( x n ) # 0, but f ( x p ) . 0. (See Theorem 12.4). 13.18 From Theorem 12.5 conclude that a function f : R k R m is continuous if and only if its coordinate functions are continuous. From Theorems 12.2 and 12.3 and this observation, conclude that if f and g are continuous, then the coordinate functions of f 1 g and f ? g are continuous.
62 MATHEMATICS FOR ECONOMISTS 13.19 = : Suppose f 5 ( f 1 , . . . , f m ) is continuous at x p . Let h x n j be a sequence converging to x p . Then, f ( x n ) f ( x p ). By Theorem 12.5, each f i ( x n ) f i ( x p ). So each f i is continuous. = : Reverse the above argument. 13.20 Suppose x n 5 ( x n 1 , x n 2 , . . . , x nk ) x 0 5 ( x 01 , x 02 , . . . , x 0 k ). By Theorem 12.5, x ni x 0 i for each i . This says h ( x 1 , . . . , x k ) 5 x i is continuous. Any monomial g ( x 1 , . . . , x k ) 5 Cx n 1 1 ? ? ? x n k k is a product of such h s (for positive integer n i s) and is continuous by Theorem 13.4. Any polynomial is a sum of monomials and is continuous by Theorem 13.4. 13.21 Let h t n j n 5 1 denote a sequence with limit t p . Then g ( t n ) 5 f ( t n , a 2 , . . . , a k ) f ( t p , a 2 , . . . , a k ) 5 g ( t p ), so g is continuous. For f ( x, y ) 5 xy 2 6 ( x 4 1 y 4 ), f ( t, a ) 5 ta 2 6 ( t 4 1 a 4 ). If a 0 this sequence clearly converges to 0 6 a 4 5 0 as t converges to 0. If a 5 0, f ( t, 0) 5 0 for all t . Similar arguments apply to continuity in the second coordinate. But f ( t, t ) 5 t 3 6 ( t 4 1 t 4 ), which grows as 1 6 t as t 0. 13.22 a ) If y [ f ( U 1 ), then there is an x [ U 1 such that f ( x ) 5 y . Since U 1 , U 2 , x [ U 2 , so y [ f ( U 2 ). b ) If x [ f 2 1 ( V 1 ), then there is a y [ V 1 such that f ( x ) 5 y . Now y [ V 2 , so x [ f 2 1 ( V 2 ). c ) If x [ U , then there is a y [ f ( U ) such that f ( x ) 5 y . Thus x [ f 2 1 ( f ( U )). d ) If y [ f ( f 2 1 ( V )), then there is an x [ f 2 1 ( V ) such that f ( x ) 5 y , so f ( x ) [ V . e ) x [ f 2 1 ( V c ) if and only if there is a y [ V c such that f ( x ) 5 y . Since f ( x ) 5 y [ V c , f ( x ) V . Thus x f 2 1 ( V ), and so x [ ( f 2 1 ( V )) c . 13.23 f ( x ) Domain Range One-to-one f 2 1 ( y ) Onto a ) 3 x 2 7 R R Yes 1 3 ( y 1 7) Yes b ) x 2 2 1 R [ 2 1 , ) No No c ) e x R (0 , ) Yes ln( y ) No d ) x 3 2 x R R No Yes e ) x 6 ( x 2 1 1) R [ 2 1 2 , 1 2 ] No No f ) x 3 R R Yes y 1 6 3 Yes g ) 1 6 x R 2 h 0 j R 2 h 0 j Yes 1 6 y No h ) p x 2 1 [1 , ) [0 , ) Yes y 2 1 1 No i ) xe 2 x R ( 2‘ , 1 6 e ] No No
ANSWERS PAMPHLET 63 13.24 a ) f ( x ) 5 log x , g ( x ) 5 x 2 1 1; b ) f ( x ) 5 x 2 , g ( x ) 5 sin x ; c ) f ( x ) 5 (cos x, sin x ), g ( x ) 5 x 3 ; d ) f ( x ) 5 x 3 1 x , g ( x ) 5 x 2 y . Chapter 14 14.1 a ) ] f ] x 5 8 xy 2 3 y 3 1 6 , ] f ] y 5 4 x 2 2 9 xy 2 . b ) ] f ] x 5 y, ] f ] y 5 x. c ) ] f ] x 5 y 2 , ] f ] y 5 2 xy. d ) ] f ] x 5 2 e 2 x 1 3 y , ] f ] y 5 3 e 2 x 1 3 y . e ) ] f ] x 5 2 2 y ( x 1 y ) 2 , ] f ] y 5 2 x ( x 1 y ) 2 . f ) ] f ] x 5 6 xy 2 7 p y, ] f ] y 5 3 x 2 2 7 x 2 p y . 14.2 For the Cobb-Douglas function, ] f ] x i 5 k a i x a i 2 1 i x a j j 5 a i q x . For the CES function, ] f ] x i 5 c i hk ( c 1 x 2 a 1 1 c 2 x 2 a 2 ) 2 ( h 6 a ) 2 1 x 2 a 2 1 i .