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Unformatted text preview: 1.7. Let x = 1 ,y = 1. Then 1 /x = 1 6 < 1 = 1 /y . The statement is true if we add the condition that y > 0. For if x > y > 0, then x/y > y/y = 1, so 1 /y > 1 /x . (Note that this depends crucially on the fact that x and y are positive.) But this yields 0 > 1 /x 1 /y , and this implies that 1 /x > 1 /y , which is what we want. 1.21. To get from a ( x + y )( x y ) + b ( x y ) = 0 to a ( x + y ) + b = 0, we must divide by x y , but this is only valid if x 6 = y , which is contrary to the later supposition that x = y . 1.45. (a) We must check that f ( x ) is welldefined for 2 < x < 4. If x > 2, then we have x 1 > 1 0, so  x 1  = x 1. If x > 2, then  x  = x , so  x  1 = x 1. Thus f is welldefined on (2 , 4). Further, we see that the domain of f is R since any real number is either less than 4 or greater than 2. Hence the rules determine a function from R to R . (b) f is not welldefined at 0 since  1  =  1  = 1 6 = 1 =    1. Hence the rules do not determine a function from...
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This document was uploaded on 03/25/2011.
 Spring '10
 Math

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