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Discrete Mathematics with Graph Theory (3rd Edition) 63

# Discrete Mathematics with Graph Theory (3rd Edition) 63 -...

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Section 3.2 and Xl = X2. Thus I is one-to-one. Next y E rngl +-+ y = I(x) for some X E A +-+ there is an X E A such that y = 5 - -1 1 +x +-+ there is an X E A such that y - 5 = --1 1 +x +-+ there is an x E A such that (y - 5)(1 + x) = -1 +-+ y # 5. 61 Thus rng 1= B = {y E R I y # 5} and I has an inverse B - A. To find a formula for 1-1 (x), 1 1 let y = 1-1 (x), x E B. Then x = I(y) = 5 - -1-' so x - 5 = --1-' (x - 5)(1 + y) = -1 +y +y and, since x # 5, 1 + y = -~5 and 1-1 (x) = y = -1- ~5' x- x- 3Xl 3 X 2 (c) Suppose I(xt} = I(X2). Then 2 1 2 l' so 6XlX2 + 3Xl = 6X lX2 + 3X2 and Xl + X2 + Xl = X2. Thus I is one-to-one. Now y E rng I +-+ y = I (x) for some x E A +-+ there is an x E A such that y = 2X 3 : 1 +-+ there is an x E A such that 2xy + y = 3x +-+ there is an x E A such that x(2y - 3) = -y 3 +-+ y # 2' Thus rng I = B = {y E R I
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Unformatted text preview: !} and I has an inverse B -A: To find a formula for 1-1 (x), let y = I-l(x), x E B. Then x = I(y) = 2y3! l' so 2xy +x = 3y and y(2x - 3) = -x. Since -x x E B, we know 2x - 3 # 0; thus, y = 2x _ 3 = I-l(x). Xl -3 X2 - 3 (d) Suppose I(Xl) = I(X2). Then --3 = --3' SOXlX2+3x l-3x 2-9 = XlX2-3xl +3X 2-9, Xl + X2 + 6Xl = 6X2 and Xl = X2. Thus I is one-to-one. Next y E rngl +-+ y = I(x) for some x E A x-3 +-+ there is an x E A such that y = --3 . x+ +-+ there is an x E A such that y(x + 3) = x - 3 +-+ there is an x E A such that x -yx = 3y + 3 +-+ there is an x E A such that x(1 -y) = 3(y + 1) +-+y#1. Thus rngl = B = {y E R I y # 1} and I has an inverse B -A. Let y = I-l(x) with x E B. Then x = I(y) = y - 3, so x(y + 3) = y -3, xy + 3x = y -3, y(1 -x) = 3(x + 1). Since y+3 ...J. 1 I-l( ) _ _ 3(1 + x) x,, x -y-1 . -x...
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