57328-0136147054_App

# 57328-0136147054_App - Appendix A A.1: |3 17| = | 14| =...

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Appendix A A.1: | 3 17 | = |− 14 | = ( 14) = 14. A.2: 3 | + | 17 | = ( 3) + 17 = 3 + 17 = 20. A.3: ± ± ± ± 0 . 25 1 4 ± ± ± ± = ± ± ± ± 1 2 ± ± ± ± = ² 1 2 ³ = 1 2 . A.4: | 5 |−|− 7 | =5 7= 2. A.5: | ( 5)(4 9) | = | ( 5)( 5) | = | 25 | = 25. A.6: 6 | | 4 | + 2 | = 6 4+2 =1. A.7: | ( 3) 3 | = 27 | = 27. A.8: ± ± ± 3 3 ± ± ± =3 3 because 3 > 3. A.9: According to Mathematica , π 3 . 14159265358979323846264338327950288419716939937510582097494 , whereas 22 7 3 . 14285714285714285714285714285714285714285714285714285714286 . Therefore π 22 7 < 0 , and thus ± ± ± ± π 22 7 ± ± ± ± = 22 7 π. For a more elegant proof of the inequality, see Miscellaneous Problems 110 and 111 of Chapter 7. A.10: −| 7 4 | = (7 4) = 3. A.11: If x< 3then x 3 < 0, so in this case | x 3 | = ( x 3) = 3 x . A.12: Given | x 7 | < 1, it follows from the fourth property of absolute value in (3) that 1 <x 7 < 1, and thus 6 <x< 8. Therefore | x 5 | + | x 10 | =( x 5) + (10 x )=5 . A.13: 2 x 7 < 3: 2 4; 2. Solution set: ( −∞ , 2). A.14: 1 4 x> 2: 4 1 2; 1 4 .S o l u t i o n s e t : ( , 1 4 ) . A.15: 3 x 4 = 17: 3 x = 21; x = 7. Solution set: [7 , + ). 1920

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A.16: 2 x +5 5 9: 2 x 5 4; x 5 2. Solution set: ( −∞ , 2]. A.17: 2 3 x< 7: 3 5; 3 x> 5; 5 3 .S o l u t i o n s e t : ( 5 3 , + ) . A.18: 6 5 9: 5 15; 5 15; 3. Solution set: ( , 3). A.19: 3 < 2 x < 7: 8 < 2 2; 4 <x< 1. Solution set: ( 4 , 1). A.20: 4 5 3 x 5 5 10: 9 5 3 x 5 15; 3 5 x 5 5. Solution set: [ 3 , 5]. A.21: 6 5 5 2 2: 11 5 2 3; 3 < 2 x 5 11; 3 2 <x 5 11 2 o l u t i o n s e t : ( 3 2 , 11 2 ± . A.22: 3 < 1 5 7: 2 < 5 6; 6 < 5 2; 6 5 2 5 o l u t i o n s e t : ( 6 5 , 2 5 ) . A.23: If | 3 2 x | < 5, then by the fourth property of absolute value in (3) we have 5 < 3 2 5. Thus 8 < 2 2, and hence 2 < 2 8. Therefore 1 4. So the solution set of the original inequality is ( 1 , 4). A.24: If | 5 x +3 | 5 4, then by an extension of the fourth property of absolute value in (3) we have 4 5 5 x 5 4. Thus 7 5 5 x 5 1, and so 7 5 5 x 5 1 5 . Thus the solution set of the original inequality is the interval ² 7 5 , 1 5 ± . A.25: We will Fnd the complement of the solution set—those real numbers not in the solution set—by solving instead the inequality | 1 3 x | 5 2. By an extension of the fourth property of absolute value in (3), we then have 2 5 1 3 x 5 2. It follows that 3 5 3 x 5 1, and hence 1 5 3 x 5 3. So the complement of the solution set of the original equality is the closed interval ² 1 3 , 1 ± . Therefore the solution set of the inequality | 1 3 x | > 2is( , 1 3 ) (1 , + ). A.26: Given: 1 < | 7 x 1 | < 3. Case 1: 7 x 1 = 0. This is equivalent to the assertion that x = 1 7 ; moreover, the given inequality takes the form 1 < 7 x 1 < 3; that is, 2 < 7 4, so that 2 7 4 7 . Hence the solutions we obtain in Case 1 are the real numbers in the interval ( 2 7 , 4 7 ) .C a s e2 :7 x 1 < 0. This is equivalent to the assertion that 1 7 ; moreover, the given inequality takes the form 1 < 1 7 3. Hence 0 < 7 2, so that 2 < 7 0, and therefore 2 7 0. So the solutions we obtain in Case 2 are the real numbers in the interval ( 2 7 , 0 ) . These two cases are exclusive and exhaustive, and so the solution set
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57328-0136147054_App - Appendix A A.1: |3 17| = | 14| =...

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