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

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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.

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

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