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Unformatted text preview: Math 3110 Homework 10 Solutions Exercise 11.1.1 We can compute  f ( x ) f (1)  = x 1 + x 1 1 + 1 = 2 x 2(1 + x ) 1 + x 2(1 + x ) = x 1 2( x + 1) . Given > 0, let δ = min { 1 , 2 } . If  x 1  < δ ≤ 1, then we have  x 1  < 1, from which < x < 2, and so 1 < x + 1 < 3. Hence,  f ( x ) f (1)  = x 1 2( x + 1) <  x 1  2 < δ 2 < . Exercise 11.4.4 By exercise 2.4.1, max( a, b ) = a + b +  a b  2 . If we let a = f ( x ) and b = g ( x ), this holds for any particular value of x , so max( f, g )( x ) = f ( x ) + g ( x ) +  f ( x ) g ( x )  2 . We know that f ( x ) g ( x ) is continuous by linearity. By question 11.4.3,  f ( x ) g ( x )  is continuous. By linearity again (twice if you like), max( f, g )( x ) = f ( x ) + g ( x ) +  f ( x ) g ( x )  2 is continous. Similarly, min( f, g )( x ) = f ( x ) + g ( x )  f ( x ) g ( x )  2 is also continuous by linearity. Alternatively, we could do this one directly from the definition of continuity. We show that the functions continuous at an arbitrary point x = x . Let > 0 be given. Since f and g are continuous at x , there are δ 1 , δ 2 > 0 such that if  x x  < δ 1 , then  f ( x ) f ( x )  < and if  x x  < δ 2 , then  g ( x ) g ( x )  < . Let δ = min { δ 1 , δ 2 } . If  x x  < δ , then  x x  < δ 1 and  x x  < δ 2 , so  f ( x ) f ( x )  < and  g ( x ) g ( x )  < . Assume without loss of generality that f ( x ) ≥ g ( x ). If  x x  < δ and f ( x ) ≥ g ( x ), then  max( f, g )( x ) max( f, g )( x )  =  f ( x ) f ( x )  < , which is what we needed to show. 1 Otherwise, g ( x ) > f ( x ), and we have  max( f, g )( x ) max( f, g )( x )  =  g ( x ) f ( x )  . We compute g ( x ) f ( x ) ≥ g ( x ) g ( x ) > . Furthermore, g ( x ) f ( x ) ≤ f ( x ) f ( x ) < . Therefore, < g ( x ) f ( x ) < , which gives  g ( x ) f ( x )  < , and we are done....
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This note was uploaded on 09/29/2011 for the course MATH 3110 at Cornell.
 '08
 RAMAKRISHNA
 Math

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