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Chapter 4 9

# Chapter 4 9 - SECTION 4.1 MAXIMUM AND MINIMUM VALUES 271...

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Unformatted text preview: SECTION 4.1 MAXIMUM AND MINIMUM VALUES 271 67. (a) 044 From the graph. it appears that the absolute maximum value is about f(0.75) 2 0.32. and the absolute minimum value is f(0) 2 f(1) 2 0: that is. at both endpoints. , 1—293 (z—2m2)+(2x—2m2) 330—4732 (b)f(\$)=m‘\$2 :’ fItIH'wT—w” 2m sm- Sof’(:1c)20 => 3002433220 => a:(3—4a:)20 :> 1:200rg.f(0)2f(1)20[minimum]. and 10(2) : g 2— (gr : % [maximum]. 68. (a) From the graph. it appears that the absolute maximum value is about ' f(5.76) 2 0.58. and the absolute minimum value is about 0 k ‘ 2,, f(3.67) : 20.58. (b) f(x) 2 col : f’(:c) _ (2 "I'SinmU—Sin 36) — (cos w)(cosx) w 24—3in (2+sinac)2 (2+sinar)2' S l l 77r 11w 7_7r a—ﬁ/2_‘; ~> of (at) 0 smar 2 a: 6 or 6 .Now f( 6 ) — 3/2 — ¢§ [minimum]. and f(111r) 2 £2 : % [maximum]. mass 1000 69. The density is deﬁned as p 2 volume 2 V(T) (in g/cm3). But a critical point of p will also be a critical point . d d . . . . . of V [Since ﬁ 2 —1000V’2% and V IS never 0]. and V IS easrer to differentlate than p. V(T) : 999.87 2 0.06426T + 0.0085043T2 — 0.0000679T3 => V/(T) 2 —0.06426 + 0.0170086T — 0.0002037T2. Setting this equal to 0 and using the quadratic formula to ~0.0170086 :: \/0.01700862 ~ 4 - 0.0002037 - 0.06426 ﬁnd T. we get T 2 2(—0.0002037) a: 3.9665OC or 79.5318OC. Since we are only interested in the region 0°C 3 T g 300C. we check the density p at the endpoints and at 3.966500 1000 1000 1000 0 R m1.00013; 30 a: ‘ m _ ; _ R5 % . . ( I 99987 p( ) 10037628 0 99625 p(3 9665) 9997447 1 000255 So water has its maximum density at about 3.966500 ...
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