Anton Bivens Davis CALCULUS early transcendentals 7th edition ch01

# Calculus - Early Transcendentals

• Homework Help
• PresidentHackerCaribou10582
• 43
• 100% (3) 3 out of 3 people found this document helpful

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

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 . 0 , 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 0 0 3 (d) V max 16 in 3

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

View Full Document
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 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)( 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 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) = 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. 2. (a) g (3) = 3 + 1 3 1 = 2; g ( 1) = 1 + 1 1 1 = 0; g ( π ) = π + 1 π 1 ; g ( 1 . 1) = 1 . 1 + 1 1 . 1 1 = 0 . 1 2 . 1 = 1 21 ; g ( t 2 1) = t 2 1 + 1 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) = 3; g ( t 2 1) = 3 if t 2 < 2 and g ( t 2 1) = t 2 1 + 1 = | t | if t 2 2. 3. (a) x = 3 (b) x ≤ − 3 or 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: ( −∞ , + ).
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern