hw7[1] - HOMEWORK 7: SOLUTIONS/HINTS (17.12a) Let x 2 ( a;b...

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HOMEWORK 7: SOLUTIONS/HINTS (17.12a) Let x 2 ( a;b ), then as Q \ ( a;b ) ± ( a;b ) is dense there is a sequence of rational numbers r n converging to x . Since f is continuous and f ( r n ) = 0 for all n , we now get f ( x ) = f (lim r n ) = lim f ( r n ) = 0 : (17.13a) Let x 2 R , ± 2 (0 ; 1) and let ² > 0. If x is rational (resp. irrational) then there is an irrational (resp. rational) number y 2 R such that j x ² y j < ² and consequently we get in either case that j f ( x ) ² f ( y ) j = 1 > ±: So f cannot be continuous at x: (18.4) De±ne f ( x ) := 1 x ² x 0 ; for x 2 S: Then f is continuous and as the sequence x n 2 S converging to x 0 , we get lim j f ( x n ) j = lim 1 j x n ² x 0 j = + 1 ; which shows that f is unbounded on S: (18.5) De±ne h ( x ) := g ( x ) ² f ( x ). Then h is a continuous function on [ a;b ] and h ( a ) = g ( a ) ² f ( a ) ³ 0 and h ( b ) = g ( b ) ² f ( b ) ´ 0. By the Intermediate Value Theorem there exists an x 0 2 [ a;b ] such that h ( x 0 ) = 0 and hence g ( x 0 ) = f ( x 0 ) : (18.6) Take f ( x ) = cos( x ), g ( x ) = x , a = 0 and b = ± 2 in problem (18.5), then there is an x 0 2 [0 ; ± 2 ] such that x 0 = cos( x 0 ) : Since the equation does not hold for x = 0 or x = ± 2 , it follows that x 0
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hw7[1] - HOMEWORK 7: SOLUTIONS/HINTS (17.12a) Let x 2 ( a;b...

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