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Chapter 4 137

# Chapter 4 137 - CHAPTER 4 REVIEW 399 0 17 22 I(b To ﬁnd...

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Unformatted text preview: CHAPTER 4 REVIEW 399 0 17 22 I (b) To ﬁnd the maximum height, set 12(t) on 3 < t g 17 equal to 0. —32(t — 3) + 270 = 0 => t1 = 11.4375 5 and the maximum height is 5(t1) = —16(t1 — 3)2 + 270(t1 — 3) + 270 : 14090625 ft. (c) To find the time to land, set 801) = —18(t 1 22) + 424 = 0. Then it 4 22 : 4—1284 : 23.5. so 1‘ m 45.6 s. 79. (a) First note that 90 mi/h : 90 x % ft/s 2 132 ft/s. Then a(t) = 4 ft/s2 :> v(t) 2 415+ C. but 13(0) : 0 => C : 0. Now 4t : 132 when t : 1% : 33 s. so it takes 33 s to reach 132 ft/s. Therefore. taking 3(0) : 0. we have 3(t) : 2752, 0 S t S 33. So 5(33) 2 2178 ft. 15 minutes : 15(60) : 900 S. so for 33 < t g 933 we have v(t) : 132 ft/s => 3(933) : 132(900) + 2178 : 120,978 ft = 22.9125 mi. (b) As in part (a). the train accelerates for 33 s and travels 2178 ft while doing so. Similarly, it decelerates for 33 s and travels 2178 ft at the end of its trip. During the remaining 900 — 66 = 834 s it travels at 132 ft/s. so the distance traveled is 132 ~ 834 = 110.088 ft. Thus. the total distance is 2178 +110.088 + 2178 = 114.444 ft: 21.675 mi. (C) 45 mi 2 45(5280) : 237.600 ft. Subtract 2(2178) to take care of the speeding up and slowing down. and we have 233.244 ft at 132 ft/s for a trip of 233.244/132 : 1767 s at 90 mi/h. The total time is 1767 + 2(33) : 1833 s = 30 min 33 s : 30.55 min. ((1) 37.5(60) : 2250 s. 2250 1 2(33) 2 2184 s at maximum speed. 2184(132) + 2(2178) : 292.644 total feet or 292.644/5280 : 55.425 mi. 4 Review CONCEPT CHECK 1. A function f has an absolute maximum at a: 2 c if f (c) is the largest function value on the entire domain of f. whereas f has a local maximum at c if f (c) is the largest function value when a: is near c. See Figure 4 in Section 4.1. 2. (a) See Theorem 4.1.3. (b) See the Closed Interval Method before Example 8 in Section 4.1. 3. (a) See Theorem 4.1.4. (b) See Deﬁnition 4.1.6. 4. (a) See Rolle’s Theorem at the beginning of Section 4.2. (b) See the Mean Value Theorem in Section 4.2. Geometric interpretation—there is some point P on the graph of a function f [on the interval (a. b)] where the tangent line is parallel to the secant line that connects (a. f ((1)) and (b. f(b)). 5. (a) See the ND Test before Example I in Section 4.3. (b) See the Concavity Test before Example 4 in Section 4.3. ...
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