Mathematical Thinking: Problem-Solving and Proofs (2nd Edition)

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Unformatted text preview: 2.5. (a) We claim x = ( y- b ) /m is a solution. We have m y- b m + b = ( y- b ) + b = y, so this is indeed a solution to y=mx+b as claimed. To show uniqueness, suppose x 1 and x 2 are solutions. Then mx 1 + b = y = mx 2 + b mx 1 = mx 2 x 1 = x 2 . In other words, all solutions to y = mx + b are the same (i.e. the solution is unique). (b) Given y and m , set x = 0 and b = y . Then mx + b = m 0 + y = 0 + y = y. 2.17. Since g ( x ) = g (- x ) for all such x 6 = 0, we have x 2 + x f ( x )- 1 =- x 2 +- x f (- x )- 1 . This yields x + x f ( x )- 1 +- x f (- x )- 1 = 0 x ( f ( x )- 1)( f (- x )- 1) + x ( f (- x )- 1) + x ( f ( x )- 1) = 0 x [( f ( x ) f (- x )- f ( x )- f (- x ) + 1) + ( f (- x )- 1) + ( f ( x )- 1)] = 0 ( f ( x ) f (- x )- f ( x )- f (- x ) + 1) + ( f (- x )- 1) + ( f ( x )- 1) = 0 f ( x ) f (- x )- 1 = 0 f ( x ) f (- x ) = 1 . 2.22. There are x,y ( R ) such that x < y but f ( x ) f ( y ). An example of a function that is not an increasing function nor a decreasing function would be any constant...
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HW5solns - 2.5. (a) We claim x = ( y- b ) /m is a solution....

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