Anton Bivens Davis CALCULUS early transcendentals 7th edition ch01

# Calculus - Early Transcendentals

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1 CHAPTER 1 Functions EXERCISE SET 1.1 1. (a) around 1943 (b) 1960; 4200 (c) no; you need the year’s population (d) war; marketing techniques (e) news of health risk; social pressure, antismoking campaigns, increased taxation 2. (a) 1989; \$35,600 (b) 1975, 1983; \$32,000 (c) the Frst two years; the curve is steeper (downhill) 3. (a) 2 . 9 , 2 . 0 , 2 . 35 , 2 . 9 (b) none (c) y =0 (d) 1 . 75 x 2 . 15 (e) y max =2 . 8at x = 2 . 6; y min = 2 . 2at x =1 . 2 4. (a) x = 1 , 4 (b) none (c) y = 1 (d) x , 3 , 5 (e) y max =9at x =6; y min = x 5. (a) x , 4 (b) none (c) x 2; 4 x (d) y min = 1; no maximum value 6. (a) x =9 (b) none (c) x 25 (d) y min = 1; no maximum value 7. (a) Breaks could be caused by war, pestilence, ﬂood, earthquakes, for example. (b) C decreases for eight hours, takes a jump upwards, and then repeats. 8. (a) Yes, if the thermometer is not near a window or door or other source of sudden temperature change. (b) No; the number is always an integer, so the changes are in movements (jumps) of at least one unit. 9. (a) The side adjacent to the building has length x ,so L = x +2 y . Since A = xy = 1000 , L = x + 2000 /x . (b) x> 0 and x must be smaller than the width of the building, which was not given. (c) 120 80 20 80 (d) L min 89 . 44 ft 10. (a) V = lwh =(6 2 x )(6 2 x ) x (b) ±rom the Fgure it is clear that 0 <x< 3. (c) 20 0 03 (d) V max 16 in 3

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2 Chapter 1 11. (a) V = 500 = πr 2 h so h = 500 2 . Then C =(0 . 02)(2) 2 +(0 . 01)2 πrh =0 . 04 2 +0 . 02 500 2 . 04 2 + 10 r ; C min 4 . 39 at r 3 . 4 ,h 13 . 8 . 7 4 1.5 6 (b) C . 02)(2)(2 r ) 2 . 01)2 . 16 r 2 + 10 r . Since 0 . 04 π< 0 . 16, the top and bottom now get more weight. Since they cost more, we diminish their sizes in the solution, and the cans become taller. 7 4 1.5 5.5 (c) r 3 . 1 cm, h 16 . 0 cm, C 4 . 76 cents 12. (a) The length of a track with straightaways of length L and semicircles of radius r is P = (2) L + (2)( ) ft. Let L = 360 and r =80toget P = 720 + 160 π = 1222 . 65 ft. Since this is less than 1320 ft (a quarter-mile), a solution is possible. (b) P =2 L +2 = 1320 and 2 r x + 160, so L = 1 2 (1320 2 )= 1 2 (1320 2 π (80 + x )) = 660 80 π πx. 450 0 0 100 (c) The shortest straightaway is L = 360, so x =15 . 49 ft. (d) The longest straightaway occurs when x =0,so L = 660 80 π = 408 . 67 ft. EXERCISE SET 1.2 1. (a) f (0) = 3(0) 2 2= 2; f (2) = 3(2) 2 2 = 10; f ( 2)=3( 2) 2 2 = 10; f (3) = 3(3) 2 2 = 25; f ( 2) 2 2=4; f (3 t ) = 3(3 t ) 2 2=27 t 2 2 (b) f (0) = 2(0) = 0; f (2) = 2(2) = 4; f ( 2 )=2 ( 2) = 4; f (3) = 2(3) = 6; f ( 2 2; f (3 t )=1 / 3 t for t> 1 and f (3 t )=6 t for t 1. 2. (a) g (3) = 3+1 3 1 ; g ( 1) = 1+1 1 1 ; g ( π π +1 π 1 ; g ( 1 . 1) = 1 . 1 . 1 1 = 0 . 1 2 . 1 = 1 21 ; g ( t 2 1) = t 2 t 2 1 1 = t 2 t 2 2 (b) g (3) = 3+1 = 2; g ( 1 )=3 ; g ( π π +1; g ( 1 . 1 ; g ( t 2 1 )=3i f t 2 < 2 and g ( t 2 1) = t 2 1+1= | t | if t 2 2. 3. (a) x 6 =3 (b) x ≤− 3or x 3 (c) x 2 2 x + 5 = 0 has no real solutions so x 2 2 x + 5 is always positive or always negative. If x = 0, then x 2 2 x +5=5 > 0; domain: ( −∞ , + ). (d) x 6 (e) sin x 6 =1,so x 6 =(2 n + 1 2 ) π , n , ± 1 , ± 2 ,...
Exercise Set 1.2 3 4. (a) x 6 = 7 5 (b) x 3 x 2 must be nonnegative; y = x 3 x 2 is a parabola that crosses the x -axis at x =0 , 1 3 and opens downward, thus 0 x 1 3 (c) x 2 4 x 4 > 0, so x 2 4 > 0 and x 4 > 0, thus x> 4; or x 2 4 < 0 and x 4 < 0, thus 2 <x< 2

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Anton Bivens Davis CALCULUS early transcendentals 7th edition ch01

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