A 1989 35600 b 1983 32000 c the rst two years

Info iconThis preview shows page 1. Sign up to view the full content.

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: ased taxation 2. (a) 1989; $35,600 (b) 1983; $32,000 (c) the first two years; the curve is steeper (downhill) 3. (a) −2.9, −2.0, 2.35, 2.9 (d) −1.75 ≤ x ≤ 2.15 4. (a) x = −1, 4 (d) x = 0, 3, 5 5. (a) x = 2, 4 (b) none (c) x ≤ 2; 4 ≤ x (d) ymin = −1; no maximum value 6. (a) x = 9 (b) none (c) x ≥ 25 (d) ymin = 1; no maximum value 7. (a) (b) none (c) y = 0 (e) ymax = 2.8 at x = −2.6; ymin = −2.2 at x = 1.2 (b) none (c) y = −1 (e) ymax = 9 at x = 6; ymin = −2 at x = 0 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) The number is always an integer, so the changes are in movements (jumps) of at least one unit. 9. (a) If the side adjacent to the building has length x then L = x + 2y . 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) (d) Lmin ≈ 89.44 120 20 80 80 10. (a) V = lwh = (6 − 2x)(6 − 2x)x (b) From the figure it is clear that 0 < x < 3. (c) (d) Vmax ≈ 16 20 0 3 0 2 3 Chapter 1 11. 500 . Then πr2 500 C = (0.02)(2)πr2 + (0.01)2πrh = 0.04πr2 + 0.02πr 2 πr 10 = 0.04πr2 + ; Cmin ≈ 4.39 at r ≈ 3.4, h ≈ 13.8. r (a) V = 500 = πr2 h so h = 7 1.5 6 4 10 (b) C = (0.02)(2)(2r)2 + (0.01)2πrh = 0.16r2 + . Since r 0.04π < 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 1.5 5.5 4 (c) 12. r ≈ 3.1, h ≈ 16.0, C ≈ 4.76 (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 + 160π = 1222.65 ft. Since this is less than 1320 ft (a quarter-mile), a solution is possible. (b) P = 2L + 2πr = 1320 and 2r = 2x + 160, so L = 1 (1320 − 2πr) = 1 (1320 − 2π (80 + x)) = 660 − 80π − πx. 2 2 450 0 100 0 (c) The shortest straightaway is L = 360, so x = 15.49 ft. (d) The longest straightaway occurs when x = 0, so L = 660 − 80π = 408.67 ft. EXERCISE SET 1.2 1. 2. 3. 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 (3t) = 3(3t)2 − 2 = 27t2 − 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 (3t) = 1/3t for t > 1 and f (3t) = 6t for t ≤ 1. (a) −1 + 1 π+1 −1.1 + 1 −0.1 1 3+1 = 2; g (−1) = = 0; g (π ) = ; g (−1.1) = = = ; 3−1 −1 − 1 π−1 −1.1 − 1 −2.1 21 t2 t2 − 1 + 1 =2 g (t2 − 1) = 2 t −1−1 t −2 √ √ (b) g (3) = 3 + 1 = 2; g (−1) = 3; g (π ) = π + 1; g (−1.1) = 3; g (t2 − 1) = 3 if t2 < 2 and √ g (t2 − 1) = t2 − 1 + 1 = |t| if t2 ≥ 2. (a) g (3) = (a) x=3 (b) √ √ x ≤ − 3 or x ≥ 3 Exercise Set 1.2 (c) 4 x2 − 2x + 5 = 0 has no real solutions so x2 − 2x + 5 is always positive or always negative. If x = 0, then x2 − 2x + 5 = 5 > 0; domain: (−∞, +∞). (d) x = 0 4. (a) x=− sin x = 1, so x = (2n + 1 )π , 2 n = 0, ±1, ±2, . . . (e) 7 5 1 (b) x − 3x2 must be nonnegative; y = x − 3x2 is a parabola that crosses the x-axis at x = 0, and 3 1 opens downward, thus 0 ≤ x ≤ 3 2 x −4 > 0, so x2 − 4 > 0 and x − 4 > 0, thus x > 4; or x2 − 4 < 0 and x − 4 < 0, thus (c) x−4 (d) −2 < x < 2 x = −1 (e) cos x ≤ 1 < 2, 2 − cos x > 0, all x 5. (a) x ≤ 3 6. (a) x ≥ 7. (a) yes (c) no (vertical line test fails) 8. The sine of θ/2 is (L/2)/10 (side opposite over hypotenuse), so that L = 20 sin(θ/2). 9. The cosine of θ is (L − h)/L (side adjacent over hypotenuse), so h = L(1 − cos θ). 10. (b) −2 ≤ x ≤ 2 2 3 (b) − (c) x ≥ 0 T (e) all x (c) x ≥ 0 3 3 ≤x≤ 2 2 (d) all x (d) x = 0 (e) x ≥ 0 (b) yes (d) no (vertical line test fails) 11. h 12. w t t t 5 13. (a) 10 15 If x < 0, then |x| = −x so f (x) = −x + 3x + 1 = 2x + 1. If x ≥ 0, then |x| = x so f (x) = x + 3x + 1 = 4x + 1; f (x) = 2x + 1, x < 0 4x + 1, x ≥ 0 (b) If x < 0, then |x| = −x and |x − 1| = 1 − x so g (x) = −x + 1 − x = 1 − 2x. If 0 ≤ x < 1, then |x| = x and |x − 1| = 1 − x so g (x) = x + 1 − x = 1. If x ≥ 1, then |x| = x and |x − 1| = x − 1 so g (x) = x + x − 1 = 2x − 1; x<0 1 − 2x, 1, 0≤x<1 g (x) = 2x − 1, x≥1 14. (a) If x < 5/2, then |2x − 5| = 5 − 2x so f (x) = 3+(5 − 2x) = 8 − 2x. If x ≥ 5/2, then |2x − 5| = 2x − 5 so f (x) = 3 + (2x − 5) = 2x − 2; f (x) = 8 − 2x, x < 5/2 2x − 2, x ≥ 5/2 5 Chapter 1 (b) 15. If x < −1, then |x − 2| = 2 − x and |x + 1| = −x − 1 so g (x) = 3(2 − x) − (−x − 1) = 7 − 2x. If −1 ≤ x < 2, then |x − 2| = 2 − x and |x + 1| = x + 1 so g (x) = 3(2 − x) − (x + 1) = 5 − 4x. If x ≥ 2, then |x − 2| = x − 2 and |x + 1| = x +...
View Full Document

This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

Ask a homework question - tutors are online