Calculus Solutions 8th Edition.pdf - CONTENTS Introduction 1 Chapter 1 Functions 2 Chapter 2 Limits and Continuity.

# Calculus Solutions 8th Edition.pdf - CONTENTS Introduction...

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This preview shows page 1 out of 676 pages. Unformatted text preview: CONTENTS Introduction .............................................................. 1 Chapter 1. Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Chapter 2. Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 Chapter 3. The Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 Chapter 4. Logarithmic and Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . 99 Chapter 5. Analysis of Functions and Their Graphs . . . . . . . . . . . . . . . . . . . . . . . . 139 Chapter 6. Applications of the Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 Chapter 7. Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 Chapter 8. Applications of the Definite Integral in Geometry, Science, and Engineering . . . . . . . . . . . . . . . . . . . . . . . . . 256 Chapter 9. Principles of Integral Evaluation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 Chapter 10. Mathematical Modeling with Differential Equations . . . . . . . . . . . . . . 343 Chapter 11. Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361 Chapter 12. Analytic Geometry in Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .408 Chapter 13. Three-Dimensional Space; Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448 Chapter 14. Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 490 Chapter 15. Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524 Chapter 16. Multiple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573 Chapter 17. Topics in Vector Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 608 Appendix A. Real Numbers, Intervals, and Inequalities . . . . . . . . . . . . . . . . . . . . . . . 640 Appendix B. Absolute Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 647 Appendix C. Coordinate Planes and Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 650 Appendix D. Distance, Circles, and Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . 658 Appendix E. Trigonometry Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 668 Appendix F. Solving Polynomial Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674 CALCULUS: A New Horizon from Ancient Roots EXERCISE SET FOR INTRODUCTION 1. (a) x = 0.123123123 . . .; 1000x = 123.123123123 . . . = 123 + x; 999x = 123; x = (b) x = 12.7777 . . .; 10x = 127.7777 . . ., so 9x = 10x − x = 115; x = (c) x = 38.07818181 . . .; 100x = 3807.818181 . . .; 99x = 3769.74; 376974 41886 20943 3769.74 = = = x= 99 9900 1100 550 537 4296 = 0.4296000 . . . = 0.4296 = 10000 1250 (d) 115 9 41 123 = 999 333 repeats 2. 3. 4. z }| { 22 = 3. 142857 . . . (a) π is irrational, and thus has a nonrepeating decimal expansion, whereas 7 22 (b) >π 7 Ã Ã √ ! √ ! 333 63 17 + 15 5 22 63 17 + 15 5 223 355 √ √ < < < (b) (a) < 71 106 25 7 + 15 5 113 7 25 7 + 15 5 Ã √ ! 63 17 + 15 5 333 √ (d) (c) 106 25 7 + 15 5 ¶2 µ ¶2 16 256 2 8 D = r = r . The area of a circle of radius r is 9 9 81 πr2 so 256/81 was the approximation used for π. µ (a) If r is the radius, then D = 2r so (b) 256/81 ≈ 3.16049, 22/7 ≈ 3.14268, and π ≈ 3.14159 so 256/81 is worse than 22/7. 5. The first series, taken to ten terms, adds to 3.0418; the second, as printed, adds to 3.1416. 6. 1 1 1 1 1 1 1 = 0.111111 . . . = + + + + + + ... 9 10 100 1000 10000 100000 1000000 1 8 5 1 8 5 2 = 0.185185 . . . = + + + + + + ... (b) 27 10 100 1000 10000 100000 1000000 3 1 1 1 1 1 14 = 0.311111 . . . = + + + + + + ... (c) 45 10 100 1000 10000 100000 1000000 (a) 6 3 6 3 6 3 7 = 0.636363 . . . = + + + + + + ... 11 10 100 1000 10000 100000 1000000 2 4 2 4 2 4 8 = 0.242424 . . . = + + + + + + ... (b) 33 10 100 1000 10000 100000 1000000 4 1 6 6 6 6 5 = 0.416666 . . . = + + + + + + ... (c) 12 10 100 1000 10000 100000 1000000 7. (a) 8. (a) 1, 2, 1.75, 1.7321 (b) 1, 3, 2.33, 2.238, 2.2361 9. (a) 1, 4, 2.875, 2.6549, 2.6458 (b) 1, 25.5, 13.7, 8.69, 7.22, 7.0726, 7.0711 10. (a) Let x1 = 12 (a + b), x2 = 21 (a + x1 ), x3 = 12 (a + x2 ), etc. Then b > x1 > x2 > · · · > xn−1 > xn > a so all the xi ’s are distinct, there are infinitely many of them and they all lie between a and b. (b) x = 0.99999 . . ., 10x = 9.99999 . . ., 9x = 9, x = 1 (c) (1.999999 . . .)/2 = 0.999999 . . . = 1; yes it is consistent, as all three are equal. (d) 10x = 9 + x, so x = 9/9 = 1. They are equal. 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 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 (b) 1983; \$32,000 (c) the first two years; the curve is steeper (downhill) (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 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) Breaks could be caused by war, pestilence, flood, earthquakes, for example. (b) C decreases for eight hours, takes a jump upwards, and then repeats. (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. (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. 8. 9. (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 12. (c) 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 = 12 (1320 − 2πr) = 21 (1320 − 2π(80 + x)) = 660 − 80π − πx. 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 6= 3 (b) √ √ x ≤ − 3 or x ≥ 3 Exercise Set 1.2 4. 4 (c) 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 6= 0 (a) x 6= − sin x 6= 1, so x 6= (2n + 21 )π, n = 0, ±1, ±2, . . . (e) 7 5 1 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 (b) (d) (e) −2 < x < 2 x 6= −1 cos x ≤ 1 < 2, 2 − cos x > 0, all x 5. (a) x ≤ 3 (b) −2 ≤ x ≤ 2 6. (a) x ≥ 7. (a) yes (c) no (vertical line test fails) 2 3 (b) − 3 3 ≤x≤ 2 2 (c) x ≥ 0 (d) all x (e) all x (c) x ≥ 0 (d) x 6= 0 (e) x ≥ 0 (b) yes (d) 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. T 11. h 12. w t t t 5 13. 14. 10 15 (a) 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; ½ 2x + 1, x < 0 f (x) = 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 (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; ½ 8 − 2x, x < 5/2 f (x) = 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 + 1 so g(x) = 3(x − 2) − (x + 1) = 2x − 7; x < −1 7 − 2x, 5 − 4x, −1 ≤ x < 2 g(x) = 2x − 7, x≥2 (a) V = (8 − 2x)(15 − 2x)x (b) −∞ < x < +∞, −∞ < V < +∞ (c) 0 < x < 4 (d) minimum value at x = 0 or at x = 4; maximum value somewhere in between (can be approximated by zooming with graphing calculator) 16. (a) x = 3000 tan θ (c) 0 ≤ θ < π/2, 0 ≤ x < +∞ (b) θ 6= nπ + π/2 for n an integer, −∞ < n < ∞ (d) 3000 ft 6000 6 0 0 17. (i) x = 1, −2 causes division by zero (ii) g(x) = x + 1, all x 18. (i) x = 0 causes division by zero (ii) g(x) = 19. (a) 25◦ F 20. If v = 48 then −60 = WCI = 1.6T − 55; thus T = (−60 + 55)/1.6 ≈ −3◦ F. 21. 22. √ x + 1 for x ≥ 0 (b) 2◦ F (c) −15◦ F √ If v = 8 then −10 = WCI = 91.4 + (91.4 − T )(0.0203(8) − 0.304 8 − 0.474); thus √ T = 91.4 + (10 + 91.4)/(0.0203(8) − 0.304 8 − 0.474) and T = 5◦ F The WCI is given by three formulae, but the first work with the data. Hence √ and third don’t √ −15 = WCI = 91.4 + (91.4 − 20)(0.0203v − 0.304 v − 0.474); set x = v so that v = x2 and obtain 0.0203x2 − 0.304x − 0.474 + (15 + 91.4)/(91.4 − 20) = 0. Use the quadratic formula to find the two roots. Square them to get v and discard the spurious solution, leaving v ≈ 25. 23. Let t denote time in minutes after 9:23 AM. Then D(t) = 1000 − 20t ft. EXERCISE SET 1.3 1. (e) seems best, though only (a) is bad. y 2. (e) seems best, though only (a) is bad and (b) is not good. y 0.5 0.5 -1 1 x -1 1 -0.5 -0.5 x Exercise Set 1.3 6 3. (b) and (c) are good; (a) is very bad. 4. (b) and (c) are good; (a) is very bad. y y 15 -11 14 -12 13 -13 12 -14 x -2 -1 0 1 2 5. [−3, 3] × [0, 5] -2 -1 0 1 2 x 6. [−4, 2] × [0, 3] y y 2 4 1 2 -3 7. -2 -3 -1 1 2 -2 -1 1 x x 3 (a) window too narrow, too short (c) good window, good spacing (b) window wide enough, but too short (d) window too narrow, too short y -5 -100 5 10 15 20 x -200 -300 -400 -500 (e) window too narrow, too short 8. (a) window too narrow (c) good window, good tick spacing (b) window too short (d) window too narrow, too short y 50 -16 -12 -8 -4 4 x -50 -100 -150 -200 -250 (e) shows one local minimum only, window too narrow, too short 7 Chapter 1 10. [6, 12] × [−100, 100] 9. [−5, 14] × [−60, 40] y y 100 40 20 50 -5 5 x 10 8 -20 10 x 12 -50 -40 -100 -60 12. [−1000, 1000] × [−13, 13] 11. [−0.1, 0.1] × [−3, 3] y y 3 10 2 5 1 x -0.1 x 1000 -1000 0.1 -1 -5 -2 -10 -3 13. [−250, 1050] × [−1500000, 600000] y -1000 14. [−3, 20] × [−3500, 3000] y 1000 x 1000 -500000 5 10 x 15 -1000 -2000 15. [−2, 2] × [−20, 20] 16. [1.6, 2] × [0, 2] y y 20 2 10 1.5 x -2 -1 1 1 2 0.5 -10 x -20 1.6 17. depends on graphing utility 1.7 1.8 y 6 6 4 4 2 -2 2 2 -2 2 18. depends on graphing utility y -4 1.9 4 x -4 -2 2 -2 -4 -4 -6 -6 4 x Exercise Set 1.3 19. 8 (a) f (x) = √ √ (b) f (x) = − 16 − x2 16 − x2 y y 4 3 2 1 -4 -4 -2 2 4 x 4 y (d) 4 4 3 2 -4 2 -1 -2 -3 -4 x y (c) -2 -2 2 4 2 x 1 -2 1 -4 2 3 4 (e) No; the vertical line test fails. 20. √ (b) y = ± x2 + 1 p (a) y = ±3 1 − x2 /4 y 4 2 -4 -2 2 4 x -2 -4 21. y (a) y (b) (c) y 1 1 x -1 x x -1 (d) 1 1 -1 2 y (e) y 1 (f ) y 1 1 π 2π x −π 1 π —1 x x -1 1 -1 y 22. 1 x -1 1 -1 23. The portions of the graph of y = f (x) which lie below the x-axis are reflected over the x-axis to give the graph of y = |f (x)|. 9 Chapter 1 24. Erase the portion of the graph of y = f (x) which lies in the left-half plane and replace it with the reflection over the y-axis of the portion in the right-half plane (symmetry over the y-axis) and you obtain the graph of y = f (|x|). 25. (b) (a) for example, let a = 1.1 y y 3 2 1 µ 1 26. They are identical. 27. 2 3 x y y 15 10 x 5 x -1 1 2 3 28. This graph is very complex. We show three views, small (near the origin), medium and large: (a) (b) y (c) y y 10 x 2 -1 x -50 -1 1000 x 40 29. (a) (b) y y 1 1.5 0.5 1 x -3 -2 -1 1 2 3 -0.5 0.5 -1 -3 (c) 1 -1 -2 2 3 x y (d) y 1.5 1.5 1 0.5 -1 1 1 2 3 x 0.5 x -2 -1 1 Exercise Set 1.3 10 30. y 1 x 2 31. (a) stretches or shrinks the graph in the y-direction; flips it if c changes sign (b) As c increases, the parabola moves down and to the left. If c increases, up and right. y y c=2 4 c=1 2 -1 c=2 6 c=1 4 x 1 8 2 c = -1.5 -2 c = -1.5 -2 -1 1 x 2 (c) The graph rises or falls in the y-direction with changes in c. y c = +2 8 c = .5 6 c = -1 4 2 -2 32. -1 1 2 x y (a) 2 x 1 2 —2 (b) x-intercepts at x = 0, a, b. Assume a < b and let a approach b. The two branches of the curve come together. If a moves past b then a and b switch roles. y y y 3 3 2 2 2 1 1 3 x -1 -2 -3 1 2 1 1 3 -1 a = 0.5 b = 1.5 -2 -3 2 3 x 1 -1 a=1 b = 1.5 -2 -3 a = 1.5 b = 1.6 2 3 x 11 Chapter 1 33. The curve oscillates between the lines y = x and y = −x with increasing rapidity as |x| increases. 34. The curve oscillates between the lines y = +1 and y = −1, infinitely many times in any neighborhood of x = 0. y y 30 20 10 x -30 -20 -10 -10 10 20 x 30 —2 4 -20 —1 —1 -30 35. Plot f (x) on [−10, 10]; then on [−1, 0], [−0.7, −0.6], [−0.65, −0.64], [−0.646, −0.645]; for the other root use [4, 5], [4.6, 4.7], [4.64, 4.65], [4.645, 4.646]; roots −0.6455, 4.6455. 36. Plot f (x) on [−10, 10]; then on [−4, −3], [−3.7, −3.6], [−3.61, −3.60], [−3.606, −3.605]; for the other root use [3, 4], [3.6, 3.7], [3.60, 3.61], [3.605, 3.606]; roots 3.6055, −3.6055. EXERCISE SET 1.4 1. (a) (b) y -1 y 2 1 1 0 1 x 2 -1 (c) 1 2 (d) y y 1 -1 2. 2 1 x 2 -4 y (a) -2 2 1 x -2 2 y 3 y (d) 1 1 x 2 -1 x 2 y (b) x -2 (c) x 3 3 x -1 1 Exercise Set 1.4 3. 12 (a) y (b) 1 y 1 -2 -1 1 2 x -0.5 0.5 1 1.5 1 2 3 x -1 -1 (c) (d) y y 1 1 -1 1 2 3 x -1 x -1 -1 5. Translate right 2 units, and up one unit. 4. y y 1 -2 x 2 10 -2 6. Translate left 1 unit, reflect over x-axis, and translate up 2 units. y 2 4 6 x 7. Translate left 1 unit, stretch vertically by a factor of 2, reflect over x-axis, translate down 3 units. y -8 -6 -4 -2 2 4 6 x -20 -40 x 1 -60 -80 –2 -100 8. Translate right 3 units, compress vertically by a factor of 12 , and translate up 2 units. 9. y = (x + 3)2 − 9; translate left 3 units and down 9 units. y y 15 10 5 -8 2 -6 -4 -2 2 -5 4 x x 13 Chapter 1 10. y = (x + 3)2 − 19; translate left 3 units and down 19 units. 11. y = −(x − 1)2 + 2; translate right 1 unit, reflect over x-axis, translate up 2 units. y y –4 x 2 –5 -2 -1 1 2 3 4 x -2 -4 -6 12. y = 21 [(x − 1)2 + 2]; translate left 1 unit and up 2 units, compress vertically by a factor of 21 . 13. Translate left 1 unit, reflect over x-axis, translate up 3 units. y y 2 1 2 1 2 4 6 8 10 12 x x 1 14. Translate right 4 units and up 1 unit. 15. Compress vertically by a factor of 12 , translate up 1 unit. y y 2 4 1 4 x 10 1 16. Stretch vertically by a factor of reflect over x-axis. √ 3 and 2 3 x 17. Translate right 3 units. y 10 y 2 x -1 2 -10 4 6 x Exercise Set 1.4 14 18. Translate right 1 unit and reflect over x-axis. 19. Translate left 1 unit, reflect over x-axis, translate up 2 units. y y 12 10 8 6 4 2 2 -2 x 2 -4 -3 -2 -1 -4 20. y = 1 − 1/x; reflect over x-axis, translate up 1 unit. 1 -2 -4 -6 -8 x 2 21. Translate left 2 units and down 2 units. y y 5 -4 -2 x 2 x -2 -5 22. Translate right 3 units, reflect over x-axis, translate up 1 unit. 23. Stretch vertically by a factor of 2, translate right 1 unit and up 1 unit. y y 1 4 x 5 -1 2 x 2 24. y = |x − 2|; translate right 2 units. 25. Stretch vertically by a factor of 2, reflect over x-axis, translate up 2 units. y y 2 4 1 3 2 4 2 x 1 -2 2 -1 x 15 Chapter 1 26. Translate right 2 units and down 3 units. 27. Translate left 1 unit and up 2 units. y y x 2 3 2 –2 1 x -3 -2 -1 1 28. Translate right 2 units, reflect over x-axis. y 1 x 4 –1 29. y (a) (b) y = 2 ½ 0 if x ≤ 0 2x if 0 < x x -1 30. 1 y x 2 –5 31. x2 + 2x + 1, all x; 2x − x2 − 1, all x; 2x3 + 2x, all x; 2x/(x2 + 1), all x 32. 3x − 2 + |x|, all x; 3x − 2 − |x|, all x; 3x|x| − 2|x|, all x; (3x − 2)/|x|, all x 6= 0 33. √ √ 3 x − 1, x ≥ 1; x − 1, x ≥ 1; 2x − 2, x ≥ 1; 2, x > 1 34. (2x2 + 1)/x(x2 + 1), all x 6= 0; −1/x(x2 + 1), all x 6= 0; 1/(x2 + 1), all x 6= 0; x2 /(x2 + 1), all x 6= 0 35. (a) 3 (b) 9 (c) 2 (d) 2 36. (a) π − 1 (b) 0 (c) −π 2 + 3π − 1 (d) 1 Exercise Set 1.4 37. 38. 16 (a) t4 + 1 (b) t2 + 4t + 5 (c) x2 + 4x + 5 (e) x2 + 2xh + h2 + 1 (f ) x2 + 1 (g) x + 1 (a) (e) √ √ 4 p√ 5s + 2 (b) x (f ) 0 √ (c) 3 5x √ (g) 1/ 4 x x+2 1 +1 x2 (h) 9x2 + 1 (d) √ (d) 1/ x (h) |x − 1| 39. 2x2 −2x+1, all x; 4x2 +2x, all x 41. 1 − x, x ≤ 1; 43. 1 1 1 1 , x 6= , 1; − − , x 6= 0, 1 1 − 2x 2 2x 2 44. 45. x−6 + 1 46. 47. (a) g(x) = 48. (a) g(x) = x + 1, h(x) = x2 (b) g(x) = 1/x, h(x) = x − 3 49. (a) g(x) = x2 , h(x) = sin x (b) g(x) = 3/x, h(x) = 5 + cos x 50. (a) g(x) = 3 sin x, h(x) = x2 (b) g(x) = 3x2 + 4x, h(x) = sin x 51. (a) f (x) = x3 , g(x) = 1 + sin x, h(x) = x2 (b) f (x) = 52. (a) f (x) = 1/x, g(x) = 1 − x, h(x) = x2 (b) f (x) = |x|, g(x) = 5 + x, h(x) = 2x 53. √ √ 1 − x2 , |x| ≤ 1 x, h(x) = x + 2 40. 2 − x6 , all x; −x6 + 6x4 − 12x2 + 8, all x 42. p√ x2 x2 + 3 − 3, |x| ≥ x 1 , x 6= 0; + x, x 6= 0 +1 x x x+1 (b) g(x) = |x|, h(x) = x2 − 3x + 5 1 -3 -2 -1 1 2 3 x -1 -2 -3 -4 55. Note that f (g(−x)) = f (−g(x)) = f (g(x)), so f (g(x)) is even. y f (g(x)) 1 –3 1 –1 –1 –3 √ 54. {−2, −1, 0, 1, 2, 3} y 2 x √ √ 6; x, x ≥ 3 x, g(x) = 1 − x, h(x) = √ 3 x 17 Chapter 1 56. Note that g(f (−x)) = g(f (x)), so g(f (x)) is even. y 3 g( f (x)) 1 x –3 1 –1 3 –1 –2 57. f (g(x)) = 0 when g(x) = ±2, so x = ±1.4; g(f (x)) = 0 when f (x) = 0, so x = ±2. 58. f (g(x)) = 0 at x = −1 and g(f (x)) = 0 at x = −1 59. 6xh + 3h2 3(x + h)2 − 5 − (3x2 − 5) = = 6x + 3h h h 60. 2xh + h2 + 6h (x + h)2 + 6(x + h) − (x2 + 6x) = = 2x + h + 6 h h 61. 1/(x + h) − 1/x x − (x + h) −1 = = h xh(x + h) x(x + h) 62. 1/(x + h)2 − 1/x2 x2 − (x + h)2 2x + h = 2 =− 2 h x h(x + h)2 x (x + h)2 63. (a) the origin 64. (a) (b) the x-axis (c) the y-axis (b) y (c) y x 65. (a) 66. (a) x f (x) −3 1 −2 −5 −1 −1 0 0 1 −1 2 −5 3 1 x f (x) (b) (b) y (a) even y x x −3 1 −2 5 −1 −1 0 0 1 1 2 −5 y x x 67. (d) none (b) odd (c) odd (d) neither 68. neither; odd; even 69. (a) (c) f (−x) = (−x)2 = x2 = f (x), even f (−x) = | − x| = |x| = f (x), even (e) f (−x) = (f ) f (−x) = 2 = f (x), even (b) f (−x) = (−x)3 = −x3 = −f (x), odd (d) f (−x) = −x + 1, neither x3 + x (−x)3 − (−x) = − = −f (x), odd 1 + (−x)2 1 + x2 3 -1 Exercise Set 1.4 70. (a) x-axis, because x = 5(−y)2 + 9 gives x = 5y 2 + 9 (b) x-axis, y-axis, and origin, because x2 − 2(−y)2 = 3, (−x)2 − 2y 2 = 3, and (−x)2 − 2(−y)2 = 3 all give ...
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