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Unformatted text preview: .y E rngj ~ y == f(:1;) for some x EA ~ there is an X E.A such that y = 1 + ~4 x.' 1 ++ there is an X E A such that y 1 = X _ 4 ++ there is an X E A such that (y ~ l)(x  4) = 1 ++y#L Thus rng f = B = {y E R I y # I} .and f has an inverse B + A, To find a formula f~r fl(x), 1 1 let y = fl(x), x E B. Then x = f(y) = 1 + 4' so xI = 4' (x l)(y 4) = 1 . y yand, since x # 1, y  4 = ~1 and.J:l(x) = y = 4 + ~1. x xl 1 1 1 (b) Supposef(xl) = f(X2). Then51= 51,so 1= 1' l+Xl = 1+x2 . + Xi + X2 + Xl + X2...
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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