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Unformatted text preview: Homework 3 Solutions Kyle Chapman October 20, 2011 31.4 Suppose 2 x + 1 /x < 3. Then there are two cases, either x is positive, or negative. If x is positive, then multiplying both sides of the equation by x leaves the orientation of the inequality, giving us 2 x 2 + 1 < 3 x so 2 x 2 3 x + 1 < 0 meaning (2 x 1)( x 1) < 0. This means exactly one of the two must be negative. Since in this case x > 0 we know 2 x > x so 2 x 1 > x 1 so if the first term is negative, then the second is also. Thus, the first case is exactly (2 x 1 > > x 1 so 2 x > 1 > x so 1 / 2 < x < 1. This fully solves the first case. For the second case, x being negative means multiplying through by x reverses the inequality. This gives us the equation (2 x 1)( x 1) > 0. Now, both terms must have the same sign, and since x < 0, x 1 < 1 < 0 so one term must be negative making both negative. Thus, 2 x 1 < 0 forms the solution space for the second case, which is x < 1 / 2. We have to restrict this to negative entries, and so we get that all negative x values are solutions. Thus, the full solution is (∞ , 0) ∪ (1 / 2 , 1). 31.6 We can observe that  z  = √ z 2 . Then it is simply a matter of observing  xy  = p ( xy ) 2 = p x 2 y 2 = √ x 2 p y 2 =  x  y  . This method is a bit over powered, however, so it may be more satisfying to use a....
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This note was uploaded on 12/26/2011 for the course MATH 8 taught by Professor Bigelow during the Fall '08 term at UCSB.
 Fall '08
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