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# hw10 - Math 3110 Homework 10 Solutions Exercise 11.1.1 We...

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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 0 < 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 0 . Let > 0 be given. Since f and g are continuous at x 0 , there are δ 1 , δ 2 > 0 such that if | x - x 0 | < δ 1 , then | f ( x ) - f ( x 0 ) | < and if | x - x 0 | < δ 2 , then | g ( x ) - g ( x 0 ) | < . Let δ = min { δ 1 , δ 2 } . If | x - x 0 | < δ , then | x - x 0 | < δ 1 and | x - x 0 | < δ 2 , so | f ( x ) - f ( x 0 ) | < and | g ( x ) - g ( x 0 ) | < . Assume without loss of generality that f ( x 0 ) g ( x 0 ). If | x - x 0 | < δ and f ( x ) g ( x ), then | max( f, g )( x ) - max( f, g )( x 0 ) | = | f ( x ) - f ( x 0 ) | < , which is what we needed to show. 1

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Otherwise, g ( x ) > f ( x ), and we have | max( f, g )( x ) - max( f, g )( x 0 ) | = | g ( x ) - f ( x 0 ) | . We compute g ( x ) - f ( x 0 ) g ( x ) - g ( x 0 ) > - . Furthermore, g ( x ) - f ( x 0 ) f ( x ) - f ( x 0 ) < . Therefore, - < g ( x ) - f ( x 0 ) < , which gives | g ( x ) - f ( x 0 ) | < , and we are done. For any particular value of x , max( f ( x ) , g ( x )) and min( f ( x ) , g ( x )) are just f ( x ) and g ( x ) in some order. Hence, max( f ( x ) , g ( x )) + min( f ( x ) , g ( x )) = f ( x ) + g ( x ), from which we have min( f ( x ) , g ( x )) = f ( x ) + g ( x ) - max( f ( x ) , g ( x )). Then by Theorem 11.4C, min( f ( x ) , g ( x )) is continuous because it is a linear combination of continuous functions. (We could have repeated the reasoning to show that max( f ( x ) , g ( x )) is continuous, but citing a theorem is faster.) Exercise 11.5.1 (a) Since x is a real number, there is a sequence { a n }
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