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Unformatted text preview: Math 25: Advanced Calculus Fall 2010 Homework # 5  Solutions 5.1: Show that max { x,y } =  x y  2 + x + y 2 . What expression would give min { x,y } ? Solution 1: We claim that min { x,y } = x + y 2  x y  2 . We know that either x ≥ y or x ≤ y . We will prove the claims by examining these two cases separately. Case 1: x ≥ y . In this case,  x y  = x y . We compute that x + y 2 +  x y  2 = x + y 2 + x y 2 = x = max { x,y } , and x + y 2  x y  2 = x + y 2 x y 2 = y = min { x,y } . Case 2: y ≥ x . In this case,  x y  = y x . We compute that x + y 2 +  x y  2 = x + y 2 + y x 2 = y = max { x,y } , and x + y 2  x y  2 = x + y 2 y x 2 = x = min { x,y } . In either case, we have established that max { x,y } = x + y 2 +  x y  2 min { x,y } = x + y 2  x y  2 . Solution 2: First, we observe that  x  = max { x, x } so that x ≤  x  and x ≤  x  for all x ∈ R . We consider two cases: Either a + b ≥ 0 or a + b < 0. Case 1: a + b ≥ 0. In this case,  a + b  = a + b ≤  a  +  b  by our observation. Case 2: a + b < 0. In this case,  a + b  = ( a + b ) = a + b ≤  a  +  b  by our observation. 1 5.2: In class, we saw that for all real numbers a and b ,  a + b  ≤  a  +  b  . Give a careful proof of this fact by examining all possible cases (...
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This note was uploaded on 03/18/2012 for the course MAT MAT 25 taught by Professor Stevenklee during the Fall '10 term at UC Davis.
 Fall '10
 StevenKlee
 Calculus

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