{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

[Solutions Manual] Anton Bivens Davis CALCULUS early transcendentals 7th edition

[Solutions Manual] Anton Bivens Davis CALCULUS early transcendentals 7th edition

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 first two years; the curve is steeper (downhill) 3. (a) − 2 . 9 , − 2 . , 2 . 35 , 2 . 9 (b) none (c) y = 0 (d) − 1 . 75 ≤ x ≤ 2 . 15 (e) y max = 2 . 8 at x = − 2 . 6; y min = − 2 . 2 at x = 1 . 2 4. (a) x = − 1 , 4 (b) none (c) y = − 1 (d) x = 0 , 3 , 5 (e) y max = 9 at x = 6; y min = − 2 at x = 0 5. (a) x = 2 , 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) From the figure it is clear that 0 < x < 3. (c) 20 3 (d) V max ≈ 16 in 3 2 Chapter 1 11. (a) V = 500 = πr 2 h so h = 500 πr 2 . Then C = (0 . 02)(2) πr 2 + (0 . 01)2 πrh = 0 . 04 πr 2 + 0 . 02 πr 500 πr 2 = 0 . 04 πr 2 + 10 r ; C min ≈ 4 . 39 at r ≈ 3 . 4 , h ≈ 13 . 8 . 7 4 1.5 6 (b) C = (0 . 02)(2)(2 r ) 2 + (0 . 01)2 πrh = 0 . 16 r 2 + 10 r . Since . 04 π < . 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)( πr ) ft. Let L = 360 and r = 80 to get 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 πr = 1320 and 2 r = 2 x + 160, so L = 1 2 (1320 − 2 πr ) = 1 2 (1320 − 2 π (80 + x )) = 660 − 80 π − πx. 450 100 (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) = 3( √ 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 √ 2; f (3 t ) = 1 / 3 t for t > 1 and f (3 t ) = 6 t for t ≤ 1....
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern