ma1012_hw1_201012_updated

ma1012_hw1_201012_updated - y 5 < t < or + 5 < y c...

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Unformatted text preview: y 5 < t < or + 5 < y c 5 + < 1 + 5 Ê 4 1 3. + 5 a1y c15cc1< c6. 5< Ê<4œ 1!or 6. c œœ c1Ê 5 œ10<or!tor ct < yc26.Ê t œ 0 or t œ 2 t /2<)Ê NT.c)t5œ<1 1cÊc 1t6Êyc2/2 < tx/2 < 6/2 Ê t œx <œ 2 œ < 6œ or1 < + < 3. 1 +2 1 xc œÊ< 1ct c2 Êt c x NT. b) ) = 5 is a counter 1/2  y > 6) ple. ÊNNT. y never true givenexample. (Actually, never true given that 4  y  6) 1/6 (Actually, NT. 1/2. x < 6 Êthat 4> 1/x  1/6 Ê 1/6 < 1/x < 1/2. 1/6d< 1/x < 2 < c) 4 < y <From17. c91x<3s cÊ < 1,8>c 3s > 1/6 9Ê 1/6 >3s œ < 1/2or c6(1/6)c 25 Ê < œ 7 or Êœ < 6/x < 3. < ec1/x < 1/2 Ê 6(1/6)9 < 6(1/x)or c3s œ Ê 1 <<6/xs<œ 7 Ê œ 25 < 6(1/x) s 6(1/2) s 1 25 NT. 2 4. 1/x 6(1/2) Ê y 1/x 3.7 9 NT. )6 Ê a), < c <y œ 5 1/2 Ê 4 < yc 6 Ê25c 4. 7 , yc 3sÊÊ or 8< 3s œ > 8# Ê 6c3s œ c # < œ # c # Ê c # s 3s œ 1/6œ 2 1/6 2 or # 6 6 6 6 d)4 < y <) 6 Ê a), c1x< y6cÊ <x1, 6 Ê (x y <4) < 2 yMathematics I: > 2 Ê cx < c1 Ê cx + 4 < 2 Ê c(x c 4) < 2. NT.f From y < < < 5 < Ê 4 < c 6 Ê and6. < x < 6 Ê x Homework 2 ,c 4) < 2 and 2 NT. < 6 Ê x > 2 Ê cx < c2 Ê cx + < < 2 Ê c(x c 4) < 2. Ê < x 2 6. 42 NT. c < y 1s < inequalities 2. c1 < + 1 1 Section +1 <e) y or sc(x1 < 4) c 2Ê 10 Ê1 or |+<12. !œ 4) < 2 < orœ 1 c s0 < y c 4 < 2. | s œ < c c 1 T 1c1 5 c s < c or s c(x y c 5 s Ê4 s + œ Ê 1.1 Ê Numbers 3 c ! Real c 4)1œ21and5 + 1c he+pairÊ 1 |œ y2c 4#s<œ1 cÊ 1 œ and c(x or 4) œ 2 Ê |sxœ 44or< 2. 0 and the Real Line < of x œ18. # Ê œ 4 # # s2 0 # S cx f) + 5) # gc1 NT.c 5 + 5) 6Ê (1/2)(> 1 + > c6(1/2)(y63.5cx5) < (1/2)(1 c2 < Ê So< y/2ection c2 <Realc6 < cx < 2. NT. (1/2)(y c < x <# < (1/2)(1 ccx Ê < < Ê c c + < c2. But + 5) 2. 2 c6 << 3. < 1.1 2 or Numbers and the Real Line < y 2 5 < 1 Ê c2 + 5) 5) 2 y/2 < < ) )(c1 Ê c< < cx < c2. But c2 < 2. So c6 < cx < c2 < 2 or c6 < cx < 2. c6 6 Thomas’ Calculus 11th Edition x /44>g) cNT.hFrom 1/6 <xy<<cŸÊc1(2) > c1(x) < Ê1(6) < 1/y 6 < cx < c2 1/y > Êa), 4c2 < 66 Ê 1/4 > 1/y > 1/6 c 1/6 Ê < 1/4. (x)or c2s 1/64 Ê 19.<<29.or<s1/4.4 2; c2s interval ( 2ß or s c c2; < 1(6) ) ÊNT.s2 < 1/y 2s 2 c or c6 c2  x 2; solution 4 Ê s c 2 2) Ÿ  xh) 2; solution y c 5 < 1cÊ 2) c 5 > c1 Ê y > 4 Ê cy < c4 Ê cy + 5 < 1 Ê c(y c 5) < 1.  NT. c1 < interval ( 2ß 5 > c1 Ê y (> 4 Ê Chapter y Section < 1 5, 9, 13,c 5), < 9, 27, 33, 35, 37, and 41. c r [2ß 4 ) Ê tion intervals 1c_ß c2] y < c_1, cy + 5 1: Ê c(y 17 11. Chapter c= necessarily1olutionc 5 < 1 Preliminaries ß _) c(y 1 <5) < 15and (y c 5) < 1 Ê | y c 5 | < 1. Also, 1of inequalities Chapter 1. andpairc 5)rtrue. Given: 5 | y < y c 5 < s true,y intervals (necessarily<[2 Ê | y 4. NT Preliminariesy c 5c 5) < 1The _ß c2]inequalities cc c 1.c < 1. 2 c1 <Ê (y < 1. Not c (y of 1 5 < 1. The true. c NNT = t necessarily pair Given:2 Ÿ x Ÿ 2; solution interval [c2ß 2] qqqqñïïïïïïïïñqqqqp x Ÿ x Ÿ 2; solution 20. c< yc252] 1 Ê c1 + 5 < y c 5 + 5 < 1 +qqqqñïïïïïïïïñqqqqpc2 interval [ c ß < x a 1 5 Êection 1.1 Real Numbers and the Real Line 4< 2 2 5 < y2)  + )(x<NT. cTIP: < y < 6. 5(x c 6) 1 + Ê 12(x c 2) 5 S c2 y < 6. 3 " c 5 " 5 c 6) 5" Ê 4 7 (x c 3 # or c(s)b NNT. #y = 54is c 2)  c(x # Ê 12(x cnever 5(x c 6) that 4  y  6) 3) 11. s(x counter "example. (Actually, 2)  true given 7 Ê a c#" or s c 6) Section 1.1 Real Numbers and the 3 5 b never c2 5s b that or c(s 6) 5. c2x € 4 Ê x true given 3 4  y  b 3) " Ê s c # or cs # 30. 3 ple. (Actually, 7 s 512x 56)24s  5x21. 30 3 a),7x 11 Ÿ 6 or < c2Ê 6t7Ÿ #4; 6 Ê y > 4. c cor Ÿ c c; c Ê t c c #3 55 x1, c 7 y 3 29. 2s 6 Ê cc Ÿ 4 Ê 12x y 2 or  Ÿ c2; 4 4] 7x  c6 or x [ Real Ê cc Êr s [c2 s 5x Ÿ t(x4 1 y 3 4 or #2 2s t Ÿ Ê solution interval 2s ß ; or c2sSection 1.1 2 or s Numbers and the Real Line ,ŸÊcc#Ÿ <c)ÊNT.>From Ÿ4;c5 ss< c24 o29.Ÿc 30<Ê<solution4intervalc2ß64] Ÿ c2; < y 4. #4 ïïïïïïñqqqqqqñïïïïïïî s Ê c7 7 ˆc_ßc3x‘ a),c #Ê y c#5 < 1, Ê 4 < y < 6 Ê y < 6. s ïïïïïïïïïñqqqqqqqqp x < 3 1ß < x Ÿ 1 ‰ tion c 3x c 5 NT. c # d 6.Ÿ8intervals)6 Ê y <Fromr c c _ïïïïïïïïïñqqqqqqqqp x c7/2 6 Ê 6. 7‘ 5 or4 < y < 7 ,6Ê1 x olution intervals (c_ß c2] r [2ß _) c_ß c r <c ß _‰ c5/2 solution intervals ˆ ïïïïïïñqqqqqqñïïïïïïî s 1 s # xb 12 c7/2 c5/2 c 4x s 0 2 or 4 1 Ê Ÿ e < t 1 #+ 2st 5 y4 b3x1 <NT. ( t< c1; cŸ 12b2; c 1c(4xc 520) < 1 (c_ß 0 < r [2ß < 2.  2 Ÿ or 4+ 2s Ê122.Ê Ê cxb5<24 <c6x 1Ê3s y  c1; 1Ÿ 24+b 6x c2] y c 4_) 1 <b 2 5 c )Ê c3 cc120) y 2 1 b2.c + olution intervals 1 Ê c 1 4 1 12. y < b s  Ê 3x + 1 b ct 5 Ÿ 4 b 2 22otal: 10 exercises.1) + 5) < (1/2)(y c 5 + 5) < (1/2)(1 + 5) Ê 2 < y/2 ðïïïïïïïïðqqqqp t f NT. 1olution < 0 + 5) c y < qqqqt 4 c( qqqqqqqqqñïïïïïïïïî x (tion + 45Ÿ 10x cß 3x1)Ê <8xÊ(1/2)(1or xc3Ê c1 y/2 Ÿ 3. 7 c cinterval) Ê c s+ 5) c 5 € 1 Ê( (1/2)( x 15x 5) $ Ÿ c3 "(ccT Ê xŸxinterval 10x ß5Ê < 22 < x qqqqðïïïïïïïïðqqqqpc3 3. 5 b 6x1 or intervals c_ß c <r c ßhapter€ 2 2 c 5 solution < (1/2)(y € 12 Ÿ 1 [2 4 1 x c(" c 5 € 1 Ê qqqqqqqqqñïïïïïïïïî x c 7. € Ÿ c s b"(( x)cor5c(s 2]cc104€ Ÿ or Ÿ Preliminariescx € 0 or c€ 2 22/5 1 31. b 4 C _)6 s1 4 c x) > 1/6 Ê xÊ 3# # or x 30.4 g) NT. From a),3) < y"< Ê Ê 1/4 ># 1/y cs " # 1/6 < 1/yx<31/4. cSection 1.1 7Real Numbers 1 " 5c c22/5 and the Line 3 S s c or Real 30. s _) 4 > 1/yor x € Ê solution intervals (c_ß !) r (2ßb 3 # or c(s b 3) # Ê ection#1.1cs #Numbers and the Real Real Line 3 x  0 > 1/6 2; 1/65< 1/y < 1/4. qqqqqqqqqñïïïïïïïïî x Ê )> c # cr<<Ÿc 57x; cyor x € 122/5 c(y c55) <Ê (cy7 < c4r Êß _y + 5 < 1 Ê c(y c 5) < 1. c Ê Ê y c 5 2; c solution intervals ! h s 4 Ê co ys y4 Ê #< 1 0 + 5 <c> Ê1 Ê y > 4 1. c_ß 11) (2 c ) NT. 1 > y c4 7  œ c1 Ê or  3yor c#œ €32(3b 3)c% 6c cor 1cÊc3 s6) ; 3y s pair r2; Ÿ c # c 3 3 13. "Ê 3x   Êc 3Ê 7 12(x o 3y 8. b3(2 7"x) ccÊlso,2sÊ54(x 3sÊ7€ #34or2x311 1.sThe11#c of1 5(x 6);c(y c 5) < 1 and (y c 5) < 1 Ê | y c 5 | < 1. b 11. 3y 2)  4 y s 33c y  4 (s 23. x) c11cc y‘œ5"b Ê _ #2  c Ê s  y ; 1 € 6 b 2xsolution intervals<ˆy2sc5(y2sÊ6 s (xc ß5s ‰Ê or 2;Ÿ c2) inequalities1. ïïïïïïñqqqqqqñïïïïïïî s 29. of inequalities c < € r <c 5 or 7 c 5) 2s 4 5 y œ c 4 2s 5 < 1.5Thec(2 c 3x) €71 #Ê _ß3x c 5) or #‰€olution intervals ˆc_ß5c <‘ r <c 5 ß _c7/2 pair A 29. cor c or c # 3ˆ <y1 and<(yŸ c < 1 Ê | y c | 7 ‰ð 11 3x €s or€5 o s Ÿ csolution interval 1ß 3x s ïïïïïïñqqqqqqñïïïïï c5/2 6 0 # Ê interval ˆ1 11 Answers: ïïïïïïïïïð € 57x c3x or 3 or 3x € 7 # ; tionÊ c #5xrß Ê‰ 0 € x 2 c 12x€ 5 or35x c 3x) qqqqqqqqp x ïïïïïïïïï0qqqqqqqqp x 32. Ê intervals c(2 30 Ê Ê c6 € x  c 7 3x c 24 c7/2 c5/2 73 s#olution œ 105 or y(c_ß4c2]_)[2ß _) r 0 c œ 1 or x € 3olution intervals (c_ßyœ c œ7c7 Ê y œ 10 or y œ c4 7 y x 3 c 7 or y c 3 ;œ c14.Ê c < s c2] r [2ß solution .intervals ˆ4 _ß cy ‘ r32œ 7c1 ‰ c 3€ 3 ; c Ê x 7 Ê xc # ßor or x ïïïïïïñqqqqqqñïïïïïïî s _ # 5 c2x € € c c c5 or x c7/2 tion 31. 1 c x c_ß7c5.(r 7 x) b 5 12 " cx €10c x2y€ 2c c(" ïïïïïïðqqqqqqor c5/2 intervals ( € 1 or 1) " ˆ ßx_‰ 1 Ê 31. ðïïïïïïî x € o 2x c 12. cc 6 intervals c_ß b 20) 7 _ b 3 7 Ê 9.5xy œ c"4 "7x b c6Ê12ys3olutionÊ bcÊcc(4xc1)1r ˆ4 ß5Êc x)  y9 1 cx € 07/3 x € 2 ïïïïïïðqqqqqqðïïïïïïî x r2y b 5# Ê 24. c c "2y2c7Ÿ  4 3x3 Ê(y6 c2;  orŸ 24 ‰c6x€ 1 c c2;  # b" 4 5x 6 c4" Ê 2t œ c1 or " œ c9 Ê 5 œ c " or 7t œ c " 3ß _) cs t b 5 œ 4 Ê30. b & œ b 3 intervalorc33)Ê (œ 4 !s t (2 œ or 1 or 2t #œ c9 Ê t œ c "qqqqðc 9 c1 or 2t  bor 15. 2t b # or 4 (s 2t b c_ß ) r 2t s# 30. s 2; c(s œ 3) ( b2t c2) s Ê c x 0 3 x € or solution intervals # 22 7/3 b # qqqqp y œ ß _) ) t (2yïïïïïïïïð2 tionʀ".1 c(10 ‰sß 2) solution 5c5xc3 0Ê# ßx 7& Ê c c Êor c#€c # solution intervals (cðqqqqpc3 # qqqqðïïïïïïïï _ß#!orrqqqqqqqqqñïïïïïïïïî x 8 c( r 1 c c4 Ÿ 10x € 2 5 Ÿ 0 Ê" Ÿ1 ïïïïïïïïïñqqqqqqqqp x 1 c interval c 3" c xx)o€#c53 3x 4 € 7 or Ê c x  #x or x 2; x c3 Ê c Ÿ 1 x 6 ˆror1c3x c 5 Ê " sÊ cx ïïïïïïïïïñqqqqqqqqp 5 b6 c2 9s Ÿ c ; c22/5 1 Ê Êc b 1 r # or# 1 or 2t œ # œ c or" s 2cc b Ÿ 1 or c ˆ c9‰ Ê 1t s 9. r # #bot œŸorr # ; 1rb1#c2 1 Ê 32. 2 orc € 5 solution#t3x)! or (c_ßc‰c€(25ßtor ) r b 17œ 2 or x 1 0 1 x œ c1 c(2 r intervalsÊœc!) r 1 Ê 3x € t ‰ 1 5 t 1c € or 1 c t 3x<Ê œ ‰ c t œ orc 3xt€ 2; ors33. 1cintervals cˆˆ 7_ß2 71 3r_c05or_ or ct r bc2ŸÊ 2t œ 0 or t œ 2 Ê c tœ ˆ1or cc #‘œ c ‘ Ê ct ßœ ! < 5#or c(2 c 3x) € 5 Ê c3x € 3 or 3x € 7 olution ïïïïïïñqqqqqqñïïïïïïî s z z œ ïïïïïïñqqqqqqñïïïïïïî s #_ 25. 16.intervals 0 c_ß # r c # z r z c r 3x 4 solution ŸŸ œ 3œ c_ߟ 32. r [1 5 ß _ 0 Ÿ z 10; 0. 61 xor .ŸŸ2 c3;$solution3x  ;12x (y 16ÊŸ0c2 Ÿ 3x € 12 5c intervals 1 or Ê c 27 Ÿ Ÿ16 c51Ê1cxÊÊŸ (c1 2x Êc 1 Ÿ 110 c3] Ÿ c ߟ 2) ÊintervalsŸc_ß c3] r [1ß _) c7/2 c7/2 c5/2 c5/2  0 Ÿ 13. y 5Ê orc z r 3 3; 5 5 4 7 5x c c1 or x € 3Êr 8xŸ œ c10; x solution (; 7 4 orx Ÿ œ c2 Ê t œ 0 13. œ 2 ct 4 or t Ê x  c1 or x € 3 7 2 3sinterval [0c 3s œ 2olution Ê 3sqqqqqqqqq 9 25 c c3x ˆ _ €  14x cc 3x 298 oror3r 10] c sc (xÊ interval [077 corc3s c7c Ê Ê3ss œ c x qqqqqqqqq2ïïïïïïïïî x œ s 5 8 ß intervals9 5 c œ 9r c 810] ‰ œ œ œ 6 7 s 25 tionÊ € olution(2Ê 3x)2#€c_ß c1)œor ß#ß3or3s3x €ðïïïïïïïïî oror œ 3s 7œ c 25 Ê s œ 7 orðs œ 25 ïïïïïïð qqqqqq ïïïïïïî x ð #c 2 6 7/3 6 ( s ïïïïïïðqqqqqqðïïïïï " €x5 2c1ˆ 5 x € ‰ € c 83c 3sœ 7x)"€3 1x)Ê olution c2y€ 0102c_ßœ c4r ˆ 3 ß _‰ # c1 or c31. 1 c x 17.1 x € 3"or c2 y c 3€ 1cÊ # 0 x œ €oror € #c1)c 6 c31.7 ;€ 5 orc 1 c 2or c 3r cx72 Ê or x ( x y 2 " 1 y r c( € intervals œ Ê 7 8. 3(2 c x) 25 14. b x) 7" €6 c 3x € 6 c "‰ € or c1 7/3 3 2(3 34. 3z Ê œ c ˆ 5 b 3z 67 œ c # orr c3s œ c €c2 Ÿ 5€c67or Ÿor25 solution2x Ÿ 5 Êc_ß !) r Ÿ ß2; ) 3r b 2x 3r 3 Ê s œ 7 or s 5 2 c c # c r #s 0 2;x € 6 3! x _ (2 _ ðqqqqqqð solution ror" Êc0126.11Ê0 x0 ˆ 3rß2 r c2o2;czc1 2; œ 4Ê_ßœ cor r ïïïïïïïïïïïïïïïðqqqqqqqqp x c_ß or €€ 5 € c o b s € (c x ŸbÊ r 3 3 1_ 7 s 1Ê! 3rÊ Ÿ intervals ( 5 Ÿ 3z œ 1 intervals € 5xˆ1rÊ1)Ÿ#3cœ€ solution1intervalsÊors sœ )0or3 sŸxz )1 Ê s œ 4 or s œ 0 ïïïïïïî x c 1 33. Ÿ 25 ÊÊ c 18..3z‰ Ê 1xÊ1 ‰ #2 sœ Ÿ orsrð#3 1c r €27r (2ß œ ! œ 5 7 1 s c œ Êorr brŸ 2€c1b ( Ÿ c2 c# # # or3 1 œc c1 5 ïïïïïïïïï qqqqqqqqp  " # 12 or # ˆ 2t 15. 7# b‰5 œ 45 < #2t2b5 ‘ œ 51 or #2t rb1 ‰ 1 Ê c9 Ê t œ r 7/3 t œ c 9 or tion intervals<(c_ß ") solution intervalorc33.2&rb" c4 Ê c ˆ œ3c1#or 2t œ ðqqqqqqðcb 1 Ÿ0c2 r qqqqñïïïïïïïïñqqqqp z b 3ß tion interval s œ ß42or s œ 30s_ c 3 ‘ r ß olution intervals (c#_ß0 ) r ˆ 7 ß _# ïïïïïïr1 1 2 or ïïïïïïî # s ‰ qqqqñïïïïïïïïñqqqqp 2/3 c ïïïïïïðqqqqqqðïïïïïïî r 7/3 # 2 cz or # œ ! Ê r 1 or r2Ÿ c3;€ 5 or c(2 c 3x) € 5" Ê 3] 3x[1ß _or 3x2/3 7 Ê 2 _ß c 3) 7/3 32. rb1c 3x € 5 or solution intervals (cc c or3ror€c3; solution intervals (c_ß c3] r [1ß _) c1 2 32. c 3x c(2 c 3x) € 5 Ê r 3x1€ 3 r Ÿ3x € 7c € r b" 1 7b 2# .or 2x ˆ #" ‰ 19. Ê 2c2ß12) " 2 or r 5x Ÿ c2 (c2ß 2) x 1 2; c c interval (rÊ x 7 2; solution interval solution 7x b c  c c 7 7b 1Ê 9 5x # Ê 6 or € 3 ; Ê x 16.c1 c tc1 1#or 16c t; œ c1 Ê ct œ ! or c7 œ c2 Ê t œ 0 or t œ 2  x  xœ or x € t " 5 # r " 3k " È" 27. 10 9È" c‰" 5È2 Ê c  5 "  5 k 2 2 1 .ˆ 2 3 x   " 3 "7 Ê c" Ê Ê x Ÿ c3;7solution  c€" 2 ;c7 cc3] rc c_ 2‰ # x  # 2 ‰ c " x ÊÊ  334. x 1r or# r"Êolution # 3rx#cintervals k(xc_ßÈ#2€Ê[1x ßÈ)c ˆÊ Ȁ; x € # c 5 c €ˆcorc intervals#(c3_ß (x #_ß 3r 1) _‰ 7 2_  x r c# 5# k r ˆ # ß Ê 5 5 #6s35.c5x ointervals c1)€ x 7 ß r ˆ 3 olution ïïïïïïðqqqqqqðïïïïïïî x 3r s ïïïïïïðqqqqqqðïïïïïïî x ‰€ 2 34. or 2 Ÿ x Ÿ 2; ŠcÈ2 È2‹3r 2 2c2 3 Ÿ22; solution 3c " € 2 c 5 c "7 [ qqqqñïïïïïïïïñqqqqp c 7/3 7/3 c7 1 5 tion intervalsolutionß interval € Ÿ ßx2] Ê r € 7 orÈ  È 5[c22߉ qqqqqq2ïïïïïïðqqqqqqpxx 1 25 ð 20. c c c  9 ;or 8 c 5 œ c19 ˆ Ê 3r € 7 or17. 58 olution‰interval Š3s interval Ê ß c3s œ c # 3or c3s œ 7 25 Ê s œqqqqñïïïïïïïïñqqqqp x x c 7 3s 5œ 5 solution interval‹ 7 5 2] r 23r # 2 7 cÈ c #È# 2 Ê s ˆ2 2 2 x 2 25 qqqqqqðïïïïïïðqqqqqqp 3r 5 3c c2 2 6 or s œ 6 c# 37  x  2 ; solution 27‰ r 2 7ß Êß 56cxc ˆ 3r c4interval qqqqñïïïïïïïïñqqqqp 5 € c 5 Ê r € 3 or r  1 . €12 c")5 r 7 ß _c 16 5 € 5 or c x È# €s 5 or intervals _ß 2x ˆ 12x ‰ 3x cÈ # 5 c " 10.olutionrb" c r" " (c5 rb1 5 b rb3 1 16 4 2 1Ê 1 or c ˆc1 ‰ 1 b 1 b 1 b or Ÿb 1 Ÿ c2 7ïïïïïïðqqqqqqðïïïïïïî r 33. # 3 c ˆ ‰ #2 Ê r 2 or r 2 1 r _ß 7 33. c1 7/3 ïïïïïïðqqqqqqðïïïïï s s Ê 3r r 3r c or sÊ €x7 or sÊ 1 sœ  r olution intervals (cc2 ") r ˆ 3 ß _‰ 3 t5 € 5 ox c # c14x tÊ3 x1#solutionr#interval2cð2ßÊ # œ 2 or # x Ÿk 3 Ê € Ÿ c 2 2r qqqqqqqqq 1 t4] 4; solution œ c1 7/3 21. or c tœ or ŸÊ cœ 4] x#ŸÊc21Ÿ Êk 285  218.5 4; Ÿ c32;xŸÊc11c[cïïïïïïïïîc3]!rÊ ß[s qqqqqqqqq2ïïïïïïïïî x Ê x c3 rŸcŸ orc13 c3 "Ê 2 intervals € x € " 2 # 71 4 r1Ÿ 3; solution ( Ÿ 1 (c_ß interval _2ß 4 or s œ 0 ð 28.r 1 or xr Ÿ # ‰ solutionkintervals (2 _ß x Ÿ c2; ß _)[1 ) 2 2 Ê) r ˆ ßx_€ 2 Ÿ kx Ê 2 c or c3] r ïïïïïïðqqqqqqðïïïïïïî r [1 xx # 7 solution intervals 1c_ß " 4 Ê c Ê (  x c 4  3 Ê Ê  336. ( Ÿ 1 3; # € 7  3. tion 35. x# [c2Êck2]k r [2ß _) cÈ2 2 x  È2 ; ïïïïïïñqqqqqqñïïïïïïî r tioninterval (c_ß4] xx  È2 3 Ê 2 interval # ß c1 7/3 c È s€ x € (# ïïïïïïñqqqqqqñïïïïïïî r Ê 2 c1; b7  735.2x_ß2;#solution_) È2 ˆ 2 c‰ 2 2  x  È2 qqqqðïïïïïïïïðqqqqp x 1  t b 22 1 2Ê c319. tcolution2Ê solution‰  tÊ (c2ßinterval Êðßïïïïïïïï2qqqqp2/7 72 22. c1 2tx interval Ê c3c2] r [2ßk 2) 1 c  2 € x € 7 Ê 7 3 x  2solution3rinterval ˆ 7 ßx2 interval c1; qqqq 2; 2  È 2; 3r 2 ‰ 2  kx ð; x c2 2 2 r 34. (c3cc 3€ Š " € ß 5 c ˆ 5 c " solution5r34. "1) ccÈ25ˆ or 2‹"‰ € 5 € 5 interval 5 or c qqqqqqðïïïïïï 2 2/7 ðïïïïïïïïðqqqqp ðtqqqqqqp x 5 lution interval È ß qqqq ‹ # s3r È27interval 2c3olution interval ŠcÈ2ß È23 olution x 3r ( ; s3 c1) ß qqqqðïïïïïïïïðqqqqp t È È# qqqqqqðïïïïïïðqqqqq x #  # Ê kxk  xk 3ÊÊ 2  3r  3 or 2  cx  3 7 cÈ # c c3 c 1 cÈ x  9 Ê 2  k 2 3 . 7c r c # or cc 3 €Ê r € 7 or r  or r  1 5 Ê Ê€ È# Ê 3r 37. 5 5o  x x  9 5 5 2c 5 kxk 3 r3 Ê 1  x  3 or 2  cx  3 c1 € 4€ 5 € 5 20. c2 Ÿ5x Ÿ 2; Ê   solution interval [3 22 2] cß qqqqñïïïïïïïïñqqqqp#x qqqqðïïïïïïïïðqqqqp t solution intervalsŠ ÈintervalsPreliminaries_‰cß1 ‰ qqqqqqðïïïïïïðqqqqqqp x ðïïïïïïî r 2  x  3 or c3 c Chapter intervals (cr ˆ "ß r ˆ 7 _ x s2ß È2;‹ 1 (c_ß ") _ß 7 ) c2 4 olution 2 c3 olution ïïïïïïc2 ðqqqqqqðïïïïïïî r ïïïïïïðqqqqqq È c %  intervals x#4 Ê 2 23. x 3) 11 c  4 Ê 332;  c2; Ê 1  y  11 È# 3y 4 Ÿ ( ß c 3 r 3y Ê 2 Ê 1 x  11 x  c  3y x 3 or c ; 3 y3 Ê tion 36. c 7  c3Ê 2)Ÿ k(2kß %  x 27 or36. Ÿ c3 x# 3y  11 kxqqqqð 2 3or;ðqqqqðïïïïðqqqp x 7/3 7/3 ïïïï x Ÿ c2; # c1 c1 4 3ß c Ê 2 Ÿ k Ê c3 Ÿ x c2 2 3 qqqqðïïïïðqqqqðïïïïðqqqp x olution # 11 21. ssolution intervals) (cc2 2) r (2ß13) 2] _ 11 1 1 lution interval ;ˆ1interval.(0 olutioncrx[2ß 3 ˆÊ111 Ê Ÿ x Ÿ 4; ‰2 < 1 Ê ¹ x cïïïïïïñqqqqqqñïïïïïïî r< x <21. 3 1  ys 41. xß c‰37 c_ß ct# interval < ßsolution tinterval (c_ß c2] r [2ßc2ß< 1 c2 c 1 < x c c3 1 Ê 2 ïïïïïïñqqqqqqñïïïïï 3Ÿ +1 ¹ 3 x  cÊ x c1 Ÿ 4 1 43 ‰ ˆ c 2 solution interval [ 2_) 4]2 Ê 2 2 < 2 c0 3 # 4 2 # k Ê #x 2 or È Ÿ c2; È È; È2 Ê2 cÈ2  x2 È2 ; 4 Ÿ x Ê 35. xx  # Ê #xkÊ kx k2Ê c 2 Ÿ k 35. x  k  x  x " c2 2 " " " # 1" xÊ c 1 Ê x "c S ck_ß solutionx# x)." # or 3" (0 1)  #  " < 3  1 x  c2]0 3 ß _ Ê "  x 1È È  k 42 # solution intervals2ko<thes38.Ê <[2is<the interval‹ßcx  " Ê "  x  " or "  cx  " qqqqqqðïïïïïïðqqqqqqp x ( 2 olution interval 2ßÊÈ2ß Èxk r Šc Šc 2 ïïïïïïñqqqqqqñïïïïïïî r ðqqqqqqp x 46   3 1" 37. b 5  #" 9 c1.2 c1"kxk bbÊ2"Êcc32;  c4 Ê c3  3  2 2;qqqqqqðïïïïïï 2y 4 " x olution interval 2y245 1 33  y 3 ‹ ct # cc3  3 # Ê Ê c1  tc4  c Êx 2 6#or 2 1;x  2y  # 22. 9 2È " 24. 2y y cc " " 37. 4  x  9 Ê" 2  kxk  3 Ê 2  x  3 or 2  # cÈ# È# È# c cx  3 3  x  # or c #  x  c 3 ; " Ê lution interval ("3ß "3 orsc3 " 3 x xc2;# (or3ßß c2)9 x  c 3 ; 1 qqqqðïïïïïïïïðqqqqp y ðïïïïïïïïðqqqqp y olution  # qqqq1 Êy ˆ c#x c2) solution interval (c 3 c1) 2 c2; ‰ ˆ "  interval c c #  qqqq ðïïïïïïïïðqqqqp t 3 Ê c3  4  c tion intervals 2.cx# ßc x cr ß0# ‰Ê x c x + 1Ê 2  x x c qqqqxïïïïðc 1 ðc2 cˆqqqp x 3 Ê x c2 2 or x Ÿ c1. ‰ c3 c1 23 3 Ê 2  x  3 or 4  cxÊ 3ˆ " or‰c3  ð3 c3c2; qqqq ïïïïðx cc3 4 "‰ ¹ 3 " 2 ¹ 2  x Ê or # 3 2ðïïïïðqqqqðïïïï2 qqqp ðïïïïðqqqp x 2 2 4  x olution2  kxk (c3ß c2) r (2ß 3) ˆc " ß c 3  3 ß # 9 Ê intervals solution intervals 2 # r qqqqðïïïïðqqqq c1/2 qqqq 1/3 c1/3 1/2 s ð x # # Ÿ kxk Êïïïïïïïïðqqqqp (c3ß c2) r (2ß 3) c3 qqqqðkxx 2 solution intervals2; y c c 1/3 1/2 c2 2 1/2 3 1/3 qqqqðïïïïðqqqqðïïïïð or ‰c3 2Ê 2 2; is Ê o cˆ36.3 4136.x  zx  intervalor x Ÿ c1] x Ÿ ß x Ÿ 2  T Ÿ solution Ê 2 c k Ê xor 2 [2 c2; 3 z r x x c orhe 43 Ÿ x xc z 2 Ÿ 3(c_ß c1. rc2 _) c Ê 11 c3 c2 2 3 2 1 Ÿ 5 c 1 Ÿ 1 s2 0 25.intervalÊ c 1 7c_ß10;ß0_5 3y  11 ÊŸ1z y10; 3 ; Ê Ÿ 5 c12Ÿ (5c_ß ( z2] 4Ê 2] 3 )[2ߟ 2 Ê 0 Ÿ interval cŸ Ê Ÿ 23. c%  3y 0 ŸŸ 1 rc r z _) c  Ÿ [2 ïïïïïïñqqqqqqñïïïïïïî r ïïïïïïñqqqqqqñïïïïïïî r 1)#  4 "Ê kx c(1"k 3ßsc "Ê c2  x solution intervalsolution22) r (2ß 3)# " c 1  2 c  olution qqqqð " kx c 1k " 2 "Ê " c 1  ïïïïðqqqqðïïïïðqqqp x c2 3 2 2 # c 2 39. s kx  # Ê 2# 2 38. 9 x  4 Ê 3s(x c k1)  4 Ê  11   3 or xx 3 c 2" olution $if a  [0 x " ðïïïïïïïï 0lutionŸ   3. [0ß 10]if a olution interval 0.1ß10]‰ x#  "  c"qqqqqqcÊ "  2  "ðqqqqp x  " cŸ  x 10;4 solution interval (c"ß interval 3ˆ ß " 3 1 z interval True 3; 0; False ) 38. x #3 or 3  c " " 4 Ê 3  kxk  c1 # 3 # Ê" c1x xc "3; solution interval (c"ß $) ; 9 qqqqqqðïïïïïïïïðqqqqp x Ê 3  x  # or c #   3 # c1 3  kxk x#  2 "Ê k  kx ÊÊ  cxx x o  cx x 3  3 2 " x   or 2 c "   3 " "37. 4"37.x 4 9 "Ê 9  kx2  3" k  "3   2  " " r3 or 2  cx c " ; Ê # Ê 3  3z Ê  " " or 3 # 3 9  x  1 Ÿ 2 Ê 26.Ÿc1 Ÿ c "c r ˆß" Ê Ÿ 2; Ÿ 2yŸ 3 Ê Ê c3 z y  c ð s 4 44. intervals ˆc #x 3 ‰k 1xŸ22#1)k œ 1  qqqq 2 Ÿ 3z c olutionkx c ck1œxc#2 "or 33ÊccxcŸ‰z 2;c6c x3z Í# #1 c x c 2 Ÿ  Ÿ Ÿ 1 2; ïïïïðqqqqðïïïïðqqqp x 24. 12c3ßx3zcb or  3# x3c c2; #  c4 3 " x 2; " 2y 3  c  c  5(x33 Ê 1 Í # "0Í "  1/3 1/2 qqqqðïïïïðqqqqðïïïïð "Ê2È # ‰ 3)#  # Êkx br3c  2 c # ; Ê "  x # o k " Ê xolution interval < solution intervals ˆc # ß c 3 ‰ r ˆ 3 ß #c1/2 c1/3qqqqðïïïïïïïïðqqqqp c1/3 " 3 <solution s(x b 3)cß #(2) rc2)3)‘  È ‘olution intervals c ( k(2b2 2) c1/2 1/3 1/ lution2 interval 2;c 2 ß 2# 40. solution #interval 3ßcx2ßßc3k(2ß 3) 2 qqqqñïïïïïïïïñqqqqp ðqqqqðïïïïðqqqp x z x qqqqñïïïïïïïïñqqqqp y intervals"3 " 3 c Ê 3 ß r ( qqqqðïïïï zc3 ðqqqqðïïïïðqqqp 3 ÊÈ 3 Ÿ z Ÿ c s" ïïïï 3È c 2 qqqqðc2 c2/3 c 2  x b 3  È # or 3 3 c 3È  solution intervals ˆc2" ß c c‰ r ˆÈ2 ‰ x  c3 bÈ 2 ; ß# qqqqðïïïïðqqqq 3 c3c2/3c22 2 322 3 # È2  x  1/2 b È1/3 cðïïïïðqqqp x 45.  k b œ (akb b)  x kb b k x c 1orb b);c c 2 c 2 Ê a 3b 39. (x c 1) (1) 4 a Ê bkkxÊ qqqqñïïïïïïïïñqqqqp z c 1  2 or c2  œ 2 (a c3 c c3 c 2 ; 1/3 1/2 È2ß 3 b Èz c (x c # kx Ê c2  x c qqqqp x tion interval Šc3 xbothcsolution ‹ 2/3 39. b)# $) 1)2  4 Ê È c 1k  2ðïïïïïïïïð1  2 z Ê38. " " c x#" 73;squared"2  k" "(a 1("7 c30c 752Ÿ" 23"Ê qqqqqq " qqqqqqðïïïïïïïïðqqqqp x c1    "s#1 Ÿ "5equal k bŠ "ß "" È x c " b 02 z Ÿ710;È " interval" c Ê " 5 ß  or " 5‹ c" 25. c " " 3 interval Ÿ c3 c " #€ 5 c3 b È#qqqqqqðïïïïïïïïðqqqqp x " " " Ê 27. colutionc 5x  Ê Ÿ  Êx 38. c  c x 3 c 1 Ÿ k Ê #cÊ c c1 3 3  x c Êx interval ïïïïïïïïðqqq #  3 c x  # 9Ê ab1 kx2k c2 # 3Ê# #xk€ c€  c    # or 3 € x  # (x c 1)#  4 Ê(2) c 9 #k ab# Ê4kak kb" x c#1 Êx23 #1 # 3x x # 3; solution # # x (c"ß $) kx " Ÿ 4 " " œ k c3 c È# qqqqqqð1 c3 b È # c 3 " s"  2or2 c2 #oc [0ß# x3  c " ; olution interval " 10] ;   52 7 " 5 Ê interval (ßc"ß $) ; solution interval ˆ 2 ß 2 ‰ # qqqqqqðïïïïïïïïðqqqqp x 2 € Ê 3  x 3 # x  #‰ r 2 x # c 3 Ê c # € 5 #( # solution or Ê# 7Ê1 x  ; 3;3) kak œ a Ê 7ˆœ ca,5 so kak œ a# ; likewise, kbk œ b#   x x solution intervalkak7 x  c1 qqqqðïïïïðqqqqðïïïïðqqqp x 3 40. (x b 3) solution intervals ˆcÈß2ˆc ‰ ßr ˆ" ‰ " ‰ˆ " ß " ‰ 7 5  # solution intervals c " " c " ß r Ê kx b 3k  5" qqqqðïïïïðqqqqðïïïïðqqqp x 3# 3# È È # 40. (x # 3)# 33z# for kx # 2‰ c x and c1/3 x 1/3 1/3 and (4) x# Ÿ y# 2 Ÿ 3z c#1xŸ Ÿ Èy3c1 x Ÿ yŸÊ all nonnegative real numbers 1/2 c1/2 Letc1/3 ka b b1/2 1/2 y. œ k 26. c implies 2Ê   3 Ê ;c 2 Ÿ 2 È2  x b 3  È2 or c3 c È2borxŸ#c3 b È2b 3k  z Ÿ 2; 5 Êc # 3 # # x œ  b k and a (x 2 c   3 Let 1 28.2kolution interval€È k 1Ÿ È  xkb 3  Èb or c a bÈ k x  c3 b È ; 3)# k b k k s b k2bk2so1 € a <Ê 2 ß ‘ k 3 ( bÊ x2" 3 b x s4 ( c qqqqñïïïïïïïïñqqqqp z mbers x and#y.ÊÊxxyœ#k aabÈÊ cthat #kbb"bÊ2ca2ak2b kbÊ 1 € k# € bk Ÿ k3 kc kb2 .ïïïïïïïï2qqqqp x olution interval c3#Ê È2ß4 13x3 21kÊ  2x  c2  c 1  2 7 qqqqqqð ð 39. (x39. 1) cc3 c x4x Ê k  7 2 2 Ê x c 1 x 2 c [email protected] c 3‹ c 4 2 c2/3 (x xŠ 1)  k cc k È2  x b 2  È2 or c3 c È22  x 2solution È2 ; Šc3 c È2ß 2 3 b È2‹ È# c 2‰ c3 c qqqqqq 3  x  2; c ‰ b interval c3 ïïïïïïïïðqqqqp x kÊ x € € 2 Ê Ê  x  2; 2 € x € Ê ˆ 2 ß 23c"ß $solution interval ˆïïïïïïïïðqqqqp x ðb È# ðqqqqp x ðïïïïïïïïðqqq .€ c x " Ê 3; qqqqðïïïïïïïïðqqqqp x € 77 solution 3; 7 1 2# c1  c1  xsolution interval ( kabk œ ab $) kaqqqqð 7 ß x b 0, interval 707 and )c"ß œ k kbk . qqqqqqðïïïïïïïï Ê c3 b È # 46. If7 a 0 and then solution interval ( ab 2 c3 c È # 2/7 2 qqqqqq 1 2/7 c1 c 3 3 solution interval Šc3 c Ècß " b c 2‹ " Ê c 7  c "  c 5 qqqqqq€ 5 2 c3 3 È "  7 " ðïïïïïïïïðqqqqp x n interval ˆ 2 ß 2‰ If a 27. b qqqqðxïïïïïïïïabk œ xab œx(ca)(cb)€ xkakÈ#k . 0 and #  0, then ab#€ 0 and #k ðqqqqp # Ê # c3 c kb# œ 7 c3 b È # 2 2/7 ...
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This note was uploaded on 12/26/2010 for the course IE 1012 taught by Professor Franciscoperez during the Fall '10 term at ITESM.

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