1873_solutions

# Stronger assertion is similar to the one given in

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Unformatted text preview: . How many real solutions does the equation x 3  ax  b  0 have in the case 27b 2  4a 3  0? What if 27b 2  4a 3  0? 3. Use Scientific Notebook to find the exact form of the solutions of the equation 125x 3 300x 2  195x 28  4 7  0 and show graphically that only one of these solutions is real. Exercises on Inequalities 1. Prove that if x and y are positive real numbers then their product xy is positive.We assume that x  0 and y  0. Since 0  x and y  0 it follows from the order axiom for the real number system that 0y  xy and the fact that 0y  0 allows us to deduce that 0  xy. 2. Prove that if x and y are negative real numbers then their product xy is positive. If x and y are negative then x and y are positive and since xy  x y the fact that xy is positive follows from Exercise 1. 3. Given real numbers a and b, prove that |a | |b | |a b |. Since a  b a it follows from the triangle inequality that 74 b |a |  |b  a b| |b |  |a b| |a | |b | |a b |. ||a | |b || |a b |. |a | |b | |a b |. and therefore 4. Given real numbers a and b, prove that From Exercise 3 we know that and |b | |a | |b a |  |a b | Since ||a | |b || is one of the numbers |a | |b | and |b | |a | the inequality ||a | |b || |a b | follows at once. 5. In each of the following cases, find the numbers x for which the given inequality is true. Compare your answers with the answers given by Scientific Notebook a. |2x 3 | |6 x |. Method 1: We separate the problem into three cases as illustrated in the figure: Case 1: When x 3 2 3 2 the inequality |2x 6 |6 x | says that 3 which tells us that x Case 2: When 3  x 2 3| 2x 6 x 3. 6 the inequality |2x 3| |6 x | says that 3 6 x 2x which tells us that x 3. Case 3: When x  6 the inequality |2x 3 | |6 x | says that 2x 3 x 6 which tells us that x 3 (which is impossible). The set of numbers x for which the inequality |2x 3 | |6 x | holds is therefore 3, 3 Þ 3 , 3  3, 3 . 2 2 Method 2: We look first at the equation |2x 3 |  |6 x | which says that either 2x 3  6 x or 2x 3  x 6. The latter condition says that either x  3 or x  3 and so we separate the problem into the cases indicated by the next figure 3 3 This method leads us even more quickly to the solutions set 3, 3 . We omit the details. b. ||x | 5 |  |x 6 |. We provide one solution here. Another approach is suggested by the solution provided below for part c. We begin by looking at the equation ||x | 5 |  |x 6 | which says that either |x | 5  x 6 or |x | 5  6 x. When x 0 these equations say that either x 5  x 6 (which is impossible) or x 5  6 x, which tells us that x  11 . 2 75 When x  0 the equations say that either x 5  x 6 or x 5  6 x which are both impossible. Therefore the only value of x at which the inequality ||x | 5 |  |x 6 | can switch from true to false or from false to true is 11 and by looking at a specimen value of x less than 2 11 and a specimen value greater then 11 we see that the inequality holds if and only if x  11 . 2 2 2 c. |2|x | 5| |4 |x 1...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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