Ans-13 - Answers, Assignment #13 Exercises 3.3 1. Set f (x)...

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Answers, Assignment #13 Exercises 3.3 1. Set f ( x )= x 3 - 4 x + 2. Then f ( - 3) = - 13 ,f ( - 2) = 2; f has a zero in ( - 3 , - 2); f ( - 1)=5 (0) = 2 (1) = - 1; f has a zero in (0 , 1); f (2) = 2; f has a zero in (1 , 2). 2. Set f ( x ) = sin x + 2 cos x - x 2 . Then f (0) = 1 ( π/ 2)=1 - π 2 4 < 0. Therefore, f has a zero in (0 ,π/ 2) = sin x + 2 cos x = x 2 for some x in (0 2). 3. Set f ( x x 2 - 2. Then f (0) = - 2 (2) = 2. Therefore, f has a zero in (0 , 2) = x 2 =2 for some x (0 , 2). 4. (a) False. The statement says that a continuous function on a domain D has a maximum value. There are many counterexamples: (1) f ( x 1 x on D =(0 , ). (2) f ( x x on (0 , 1). A continuous function on a FINITE CLOSED INTERVAL (i.e., compact) DOES have a maximum value (and a minimum value). (b) False. Let f ( x 1 x on D , 1]. D is bounded, f ( D )=[1 , ) is not. (c) True. If k = f ( a ) or if k = f ( b ), then we are done. If f ( a ) <k<f ( b ), then apply Thoeorem 13. (d) False. f ( x x on (0 , 1). (e) False. f ( x x on (0 , 1). f has neither a maximum nor a minimum value. 5. (a) False. f ( x ) 1 on (0 , 1); (0 , 1) is open, f [(0 , 1)] = { 1 } is closed. (b) False. Take D = R and f ( x e x . D is closed and f ( D )=(0 , ) is open. (c) True. f - 1 [ f ( D )] = D . By Theorem 10 and its Corollary, the inverse image of an open set is open. (d) False. Take D =( - π,π ) and f ( x ) = sin x . D is not closed and f ( D )=[ - 1 , 1] is closed. The example in (a) will also work. (e) False. The example in (a) works here; (0 , 1) is not compact, f [(0 , 1)] = { 1 } is compact. (f) False. f ( x ) = sin x on ( -∞ , ); ( -∞ , ) is not bounded, f [( -∞ , )] = [ - 1 , 1] is bounded.
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This note was uploaded on 01/06/2011 for the course MATH 3333 taught by Professor Staff during the Spring '08 term at University of Houston.

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Ans-13 - Answers, Assignment #13 Exercises 3.3 1. Set f (x)...

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