1. January 27, 2005 11:42 L24-ch01 Sheet number 1 Page number 1 blackCHAPTER 1FunctionsEXERCISE SET 1.11. (a) 2.9,2.0, 2.35, 2.9 (b) none (c) y = 0(d) 1.75 x 2.15 (e) ymax = 2.8at x = 2.6; ymin = 2.2 at x = 1.22. (a) x = 1, 4 (b) none (c) y = 1(d) x = 0, 3, 5 (e) ymax = 9 at x = 6; ymin = 2 at x = 03. (a) yes (b) yes(c) no (vertical line test fails) (d) no (vertical line test fails)4. (a) The natural domain of f is x= 1, and for g it is the set of all x. Whenever x= 1,f(x) = g(x), but they have different domains.(b) The domain of f is the set of all x 0; the domain of g is the same.5. (a) around 1943 (b) 1960; 4200(c) no; you need the years population (d) war; marketing techniques(e) news of health risk; social pressure, antismoking campaigns, increased taxation6. (a) around 1983 (b) 1966(c) the former (d) no, it appears to be levelling out7. (a) 1999, $34,400 (b) 1985, $37,000(c) second year; graph has a larger (negative) slope18. (a) In thousands, approximately43.2 37.86=5.46per yr, or $900/yr(b) The median income during 1993 increased from $37.8K to $38K (K for kilodollars; all figuresapproximate). During 1996 it increased from $40K to $42K, and during 1999 it decreasedslightly from $43.2K to $43.1K. Thus the average rate of change measured on January 1 was(40 - 37.8)/3 for the first three-yr period and (43.2 - 40)/3 for the second-year period, andhence the median income as measured on January 1 increased more rapidly in the secondthree-year period. Measured on December 31, however, the numbers are (42 - 38)/3 and (43.1- 42)/3, and the former is the greater number. Thus the answer to the question depends onwhere in the year the median income is measured.(c) 19939. (a) f(0) = 3(0)22 = 2; f(2) = 3(2)22 = 10; f(2) = 3(2)22 = 10; f(3) = 3(3)22 = 25;2) = 3(f(2)2 2 = 4; f(3t) = 3(3t)2 2 = 27t2 22) = 2(b) f(0) = 2(0) = 0; f(2) = 2(2) = 4; f(2) = 2(2) = 4; f(3) = 2(3) = 6; f(2;f(3t) = 1/3t for t1 and f(3t) = 6t for t 1.10. (a) g(3) =3 + 13 1= 2; g(1) =1 + 11 1= 0; g() = + 1 1; g(1.1) =1.1 + 11.1 1=0.12.1=121;g(t2 1) = t2 1 + 1t2 1 1= t2t2 2(b) g(3) =3 + 1 = 2; g(1) = 3;g() = + 1; g(1.1) = 3; g(t2 1) = 3 if t22 andg(t2 1) =t2 1 + 1 = |t| if t2 2.

2. January 27, 2005 11:42 L24-ch01 Sheet number 2 Page number 2 black2 Chapter 13 or x 11. (a) x= 3 (b) x 3(c) x2 2x + 5 = 0 has no real solutions so x2 2x + 5 is always positive or always negative. Ifx = 0, then x2 2x+5 = 50; domain: (,+).(d) x= 0 (e) sin x= 1, so x= (2n + 12), n = 0,1,2, . . .12. (a) x= 75(b) x 3x2 must be nonnegative; y = x 3x2 is a parabola that crosses the x-axis at x = 0, 13and opens downward, thus 0 x 13(c) x2 4x 40, so x2 40 and x 40, thus x4; or x2 40 and x 40, thus2x2(d) x= 1 (e) cos x 12, 2 cosx0, all x13. (a) x 3 (b) 2 x 2 (c) x 0 (d) all x (e) all x14. (a) x 23(b) 32 x 32 (c) x 0 (d) x= 0 (e) x 015. (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.16. (a) Yes, if the thermometer is not near a window or door or other source of sudden temperaturechange.(b) No; the number is always an integer, so the changes are in movements (jumps) of at leastone unit.17.h 18. Ttt19. (a) x = 2, 4 (b) none (c) x 2; 4 x (d) ymin = 1; no maximum value20. (a) x = 9 (b) none (c) x 25 (d) ymin = 1; no maximum value21. The cosine of is (L h)/L (side adjacent over hypotenuse), so h = L(1 cos ).22. The sine of /2 is (L/2)/10 (side opposite over hypotenuse), so that L = 20 sin(/2).23. (a) If x0, then |x| = x so f(x) = x + 3x+1 = 2x + 1. If x 0, then |x| = x sof(x) = x + 3x+1 = 4x + 1;f(x) =2x + 1, x04x + 1, x 0(b) If x0, then |x| = x and |x1| = 1x so g(x) = x+1x = 12x. If 0 x1, then|x| = x and |x1| = 1x so g(x) = x+1x = 1. If x 1, then |x| = x and |x1| = x1so g(x) = x + x 1 = 2x 1;g(x) =1 2x, x01, 0 x12x 1, x 1 3. January 27, 2005 11:42 L24-ch01 Sheet number 3 Page number 3 blackExercise Set 1.1 324. (a) If x5/2, 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, x5/22x 2, x 5/2(b) If x1, then |x2| = 2x and |x+1| = x1 so g(x) = 3(2x)(x1) = 72x.If 1 x2, then |x2| = 2x and |x+1| = x+1 so g(x) = 3(2x)(x+1) = 54x.If x 2, then |x 2| = x 2 and |x + 1| = x + 1 so g(x) = 3(x 2) (x+1) = 2x 7;g(x) =7 2x, x15 4x, 1 x22x 7, x 225. (a) V = (8 2x)(15 2x)x(b) 0 x 4(c) 0 V 91(d) As x increases, V increases and then decreases; themaximum value could be approximated by zooming inon the graph.1000 4026. (a) V = (6 2x)2x(b) 0x3(c) 0V16(d) A x increases, V increases and then decreases; themaximum value occurs somewhere on 0x3,and can be approximated by zooming with a graphingcalculator.200 3027. (a) The side adjacent to the buildinghas length x, so L = x + 2y.(b) A = xy = 1000, so L = x + 2000/x.(c) all x= 012020 8080(d) L 89.44 ft28. (a) x = 3000 tan (b) = n + /2 for any integer n,n(c) 3000 ft60000 6029. (a) V = 500 = r2h so h =500r2 . ThenC = (0.02)(2)r2 + (0.01)2rh = 0.04r2 + 0.02r500r2= 0.04r2 +10r; Cmin 4.39 cents at r 3.4 cm,h 13.8cm71.5 64 4. January 27, 2005 11:42 L24-ch01 Sheet number 4 Page number 4 black4 Chapter 1(b) C = (0.02)(2)(2r)2 + (0.01)2rh = 0.16r2 +10r. Since0.040.16, the top and bottom now get more weight.Since they cost more, we diminish their sizes in thesolution, and the cans become taller.71.5 5.54(c) r 3.1 cm, h 16.0 cm, C 4.76 cents30. (a) The length of a track with straightaways of length L and semicircles of radius r isP = (2)L + (2)(r) ft. Let L = 360 and r =8 0 to get P = 720 + 160 = 1222.65 ft.Since this is less than 1320 ft (a quarter-mile), a solution is possible.(b) P = 2L + 2r = 1320 and 2r = 2x + 160, soL= 12(1320 2r) = 12(1320 2(80 + x))= 660 80 x.4500 1000(c) The shortest straightaway is L = 360, so x = 15.49 ft.(d) The longest straightaway occurs when x = 0,so L = 660 80 = 408.67 ft.31. (i) x = 1,2 causes division by zero (ii) g(x) = x + 1, all x32. (i) x = 0 causes division by zero (ii) g(x) =x + 1 for x 033. (a) 25F (b) 13F (c) 5F34. If v = 48then 60 = WCT 1.4157T 30.6763; thus T 21F when WCT = 60.35. As in the previous exercise, WCT 1.4157T 30.6763; thus T 15F when WCT = 10.36. The WCT is given by two formulae, but the first doesnt work with the data. Hence5 = WCT = 27.2v0.16 + 48.17 and v 66F.EXERCISE SET 1.21. (e) seems best,though only (a) is bad.0.50.5y1 1x2. (e) seems best, thoughonly (a) is bad and (b)is not good.0.50.5y1 1x3. (b) and (c) are good;(a) is very bad.141312y1 1x 5. January 27, 2005 11:42 L24-ch01 Sheet number 5 Page number 5 blackExercise Set 1.2 54. (b) and (c) are good;(a) is very bad.11121314y2 1 0 1 2x5. [3, 3] [0, 5]2yx6. [4, 2] [0, 3]2yx7. (a) window too narrow, too short(b) window wide enough, but too short(c) good window, good spacing200400y5 5 10 20x(d) window too narrow, too short(e) window too narrow, too short8. (a) window too narrow(b) window too short(c) good window, good tick spacing50100250y16 8 4 4x(d) window too narrow, too short(e) shows one local minimum only,window too narrow, too short9. [5, 14] [60, 40]402060y5 5 10x10. [6, 12] [100, 100]10050y10 12x11. [0.1, 0.1] [3, 3]322y0.1x0.112. [1000, 1000] [13, 13]1055y1000x100013. [250, 1050] [1500000, 600000]500000y1000 1000x 6. January 27, 2005 11:42 L24-ch01 Sheet number 6 Page number 6 black6 Chapter 114. [3, 20] [3500, 3000]1000100020005 10 15xy15. [2, 2] [20, 20]20101020yx2 1 1 216. [1.6, 2] [0, 2]210.5y1.6 1.7 1.8 2x17. depends on graphing utility6422y4 2 4x18. depends on graphing utility6422y4 2 2 4x19. (a) f(x) =16 x2312 2 4xy16 x2(b) f(x) = 13xy4 2 2 4(c)22xy2 2(d)411 4yx(e) No; the vertical linetest fails.1 x2/420. (a) y = 332 23xyx2 + 1(b) y = 424 2 2 424xy 7. January 27, 2005 11:42 L24-ch01 Sheet number 7 Page number 7 blackExercise Set 1.2 721. (a)1 1xy1(b)1x12y(c)1 1x1y (d)c 2c1yx(e)c1yxC1(f)1x11y122.1xy1 123. The portions of the graph of y = f(x) which lie below the x-axis are reflected over the x-axis togive the graph of y = |f(x)|.24. Erase the portion of the graph of y = f(x) which lies in the left-half plane and replace it with thereflection over the y-axis of the portion in the right-half plane (symmetry over the y-axis) and youobtain the graph of y = f(|x|).25. (a) for example, let a = 1.1ayx(b)321y1 2 3x26. They are identical.xy27.15105y1 1 2 3x 8. January 27, 2005 11:42 L24-ch01 Sheet number 8P age number 8 black8 Chapter 128. This graph is very complex. We show three views, small (near the origin), medium and large:(a) yx1 21(b)50y x10(c)1000y40x29. (a)1y3 1 1 3x(b)10.53 1 1 30.5xy(c)1.50.5y1 1 2 3x(d)10.52 1 1xy30. y12x31. (a) stretches or shrinks thegraph in the y-direction;reflects it over the x-axisif c changes sign422c = 2c = 1c = 1.51 1xy(b) As c increases, the lowerpart of the parabolamoves down and to theleft; as c decreases, themotion is down and tothe right.82c = 2c = 1c = 1.51 2xy(c) The graph rises or fallsin the y-direction withchanges in c.8642c = 2c = 0.5c = 12 1 2xyAs c increases, the lower part of the parabola moves down and to the left; as c decreases, themotion is down and to the right 9. January 27, 2005 11:42 L24-ch01 Sheet number 9 Page number 9 blackExercise Set 1.3 932. (a) y221 2x(b) x-intercepts at x = 0, a, b. Assume ab and let a approach b. The two branches of thecurve come together. If a moves past b then a and b switch roles.31131 2 3xa = 0.5b = 1.5y31131 2 3xa = 1b = 1.5y31132 3xa = 1.5b = 1.6y33. The curve oscillates betweenthe lines y = x and y = x withincreasing rapidity as |x| increases.30101030y30 10 10 30x34. The curve oscillates between the linesy = +1 and y = 1, infinitely manytimes in any neighborhood of x = 0.y2 41xEXERCISE SET 1.31. (a)101y1 1 2x(b)21y1 2 3x(c) 1y1 1 2x(d)2y4 2 2x 10. January 27, 2005 11:42 L24-ch01 Sheet number 10 Page number 10 black10 Chapter 12. (a) yx2 1(b)x1y1(c)x1 3y1(d)x1y113. (a)11yx2 1 1 2(b)11y1x(c)1 y1 1 2 31x(d)11y1 1 2 3x4.1 y1 1 2 31x5. Translate left 1 unit, stretch vertically bya factor of 2, reflect over x-axis, translatedown 3 units.6 2 2 62060y x6. Translate right 3 units,12compress vertically bya factor of , andtranslate up 2 units.yx247. y = (x + 3)2 9;translate left 3 unitsand down 9 units.15108 4 2 25xy8. y = (x + 3)2 19;translate left 3 unitsand down 19 units.yx54 11. January 27, 2005 11:42 L24-ch01 Sheet number 11 Page number 11 blackExercise Set 1.3 119. y = (x 1)2 + 2;translate right 1 unit,reflect over x-axis,translate up 2 units.22 1 1 326xy10. y = 12[(x 1)2 + 2];12translate right 1 unitand up 2 units,compress verticallyby a factor of .yx21111. Translate left 1 unit,reflect over x-axis,translate up 3 units.212 8xy12. Translate right 4 unitsand up 1 unit.yx414 1013. Compress verticallyby a factor of 12,translate up 1 unit.2y1 2 3x14. Stretch vertically bya factor of3 andreflect over x-axis.y x1215. Translate right 3 units.1010y4 6x16. Translate right 1 unitand reflect over x-axis.yx22 4417. Translate left 1 unit,reflect over x-axis,translate up 2 units.626y3 1 2x18. y = 1 1/x;reflect over x-axis,translate up 1 unit.yx55219. Translate left 2 unitsand down 2 units.24 2xy 12. January 27, 2005 11:42 L24-ch01 Sheet number 12 Page number 12 black12 Chapter 120. Translate right 3 units, reflectover x-axis, translate up 1 unit.yx11521. Stretch vertically by a factor of 2,translate right 1/2 unit and up 1 unit.yx42222. y = |x 2|; translate right 2 units.yx212 423. Stretch vertically by a factor of 2,reflect over x-axis, translate up 1 unit.312 21xy24. Translate right 2 unitsand down 3 units.y x2225. Translate left 1 unitand up 2 units.31y3 1 1x26. Translate right 2 units,reflect over x-axis.yx11427. (a) 2y1 1x(b) y =0 if x 02x if 0x28. yx52x 1, x 1; (f g)(x) =29. (f + g)(x) = 3x 1, x 1; (fg)(x) = 2x 2, x 1;(f/g)(x) = 2, x130. (f + g)(x) = (2x2 + 1)/[x(x2 + 1)], all x= 0; (f g)(x) = 1/[x(x2 + 1)], all x= 0; (fg)(x) =1/(x2 + 1), all x= 0; (f/g)(x) = x2/(x2 + 1), all x= 031. (a) 3 (b) 9 (c) 2 (d) 232. (a) 1 (b) 0 (c) 2 + 3 1 (d) 1 13. January 27, 2005 11:42 L24-ch01 Sheet number 13 Page number 13 blackExercise Set 1.3 1333. (a) t4 + 1 (b) t2 + 4t + 5 (c) x2 + 4x + 5 (d)1x2 + 1(e) x2 + 2xh + h2 + 1 (f) x2 + 1 (g) x + 1 (h) 9x2 + 134. (a)5s + 2 (b)5x (d) 1/x + 2 (c) 3x(e) 4 x (f) 0 (g) 1/ 4 x (h) |x 1|35. (f g)(x) = 1 x, x 1; (g f)(x) =1 x2, |x| 136. (f g)(x) =x2 + 3 3, |x| 6; (g f)(x) =x, x 337. (f g)(x) =11 2x, x=12, 1; (g f)(x) = 12x 12, x= 0, 138. (f g)(x) = xx2 + 1, x= 0; (g f)(x) =1x+ x, x= 039. x6 + 1 40. xx + 141. (a) g(x) =x, h(x) = x + 2 (b) g(x) = |x|, h(x) = x2 3x + 542. (a) g(x) = x + 1, h(x) = x2 (b) g(x) = 1/x, h(x) = x 343. (a) g(x) = x2, h(x) = sin x (b) g(x) = 3/x, h(x) = 5 + cos x44. (a) g(x) = 3sinx, h(x) = x2 (b) g(x) = 3x2 + 4x, h(x) = sin x45. (a) f(x) = x3, g(x) = 1 + sin x, h(x) = x2 (b) f(x) =x, g(x) = 1 x, h(x) = 3 x46. (a) f(x) = 1/x, g(x) = 1 x, h(x) = x2 (b) f(x) = |x|, g(x) = 5+x, h(x) = 2x47. yx22 22448. {2,1, 0, 1, 2, 3}49. Note thatf(g(x)) = f(g(x)) = f(g(x)),so f(g(x)) is even.f (g(x))xy13 1 11350. Note that g(f(x)) = g(f(x)),so g(f(x)) is even.xy11231133g( f (x)) 14. January 27, 2005 11:42 L24-ch01 Sheet number 14 Page number 14 black14 Chapter 151. f(g(x)) = 0 when g(x) = 2, so x = 1.4; g(f(x)) = 0 when f(x) = 0, so x = 2.52. f(g(x)) = 0 at x = 1 and g(f(x)) = 0 at x = 153.3w2 5 (3x2 5)w x=3(w x)(w + x)w x= 3w + 3x3(x + h)2 5 (3x2 5)h=6xh + 3h2h= 6x + 3h;54. w2 + 6w (x2 + 6x)w x= w + x + 6(x + h)2 + 6(x + h) (x2 + 6x)h=2xh + h2 + 6hh= 2x + h + 6;55.1/w 1/xw x= x wwx(w x)= 1xw;1/(x + h) 1/xh= x (x + h)xh(x + h)=1x(x + h)56.1/w2 1/x2w x= x2 w2x2w2(w x)= x + wx2w2 ;1/(x + h)2 1/x2h= x2 (x + h)2x2h(x + h)2 = 2x + hx2(x + h)257. neither; odd; even58. (a) x 3 2 1 0 1 2 3f(x) 1 5 1 0 1 5 1(b) x 3 2 1 0 1 2 3f(x) 1 5 1 0 1 5 159. (a)y (b)xxy(c)xy60. (a)y (b)xxy61. (a) even (b) odd (c) odd (d) neither62. (a) the origin (b) the x-axis (c) the y-axis (d) none 15. January 27, 2005 11:42 L24-ch01 Sheet number 15 Page number 15 blackExercise Set 1.3 1563. (a) f(x) = (x)2 = x2 = f(x), even (b) f(x) = (x)3 = x3 = f(x), odd(c) f(x) = | x| = |x| = f(x), even (d) f(x) = x + 1, neither(e) f(x) =(x)5 (x)1 + (x)2 = x5 x1 + x2 = f(x), odd(f) f(x) = 2 = f(x), even64. (a) x-axis, because x = 5(y)2 + 9 gives x = 5y2 + 9(b) x-axis, y-axis, and origin, because x2 2(y)2 = 3, (x)2 2y2 = 3, and(x)2 2(y)2 = 3 all give x2 2y2 = 3(c) origin, because (x)(y) = 5 gives xy = 565. (a) y-axis, because (x)4 = 2y3 + y gives x4 = 2y3 + y(b) origin, because (y) =(x)3 + (x)2 gives y = x3 + x2(c) x-axis, y-axis, and origin because (y)2 = |x| 5, y2 = | x| 5, and (y)2 = | x| 5 allgive y2 = |x| 566. (a) x-axis(b) origin, both axes(c) origin67. (a) y-axis(b) none(c) origin, both axes68. 34 4369. 23 3270. (a) Whether we replace x with x, y with y, or both, we obtain the same equation, so byTheorem 1.3.3 the graph is symmetric about the x-axis, the y-axis and the origin.(b) y = (1 x2/3)3/2(c) For quadrant II, the same; for III and IV use y = (1 x2/3)3/2. (For graphing it may behelpful to use the tricks that precede Exercise 29 in Section 1.2.)71.52y1 2x72. yx2273. (a)1yCxO c o(b)2yOxC c o 16. January 27, 2005 11:42 L24-ch01 Sheet number 16 Page number 16 black16 Chapter 174. (a)x1y11(b)x1y12 1 23(c)x1y13(d)xy1C c/275. Yes, e.g. f(x) = xk and g(x) = xn where k and n are integers.76. If x 0 then |x| = x and f(x) = g(x). If x0 then f(x) = |x|p/q if p is even and f(x) = |x|p/qif p is odd; in both cases f(x) agrees with g(x).EXERCISE SET 1.41. (a) y = 3x + b (b) y = 3x + 6 (c)1010y2 2xy = 3x + 6y = 3x + 2y = 3x 42. Since the slopes are negative reciprocals, y = 13x + b.3. (a) y = mx + 2(b) m = tan = tan 135 = 1, so y = x + 2 (c)5431m = 1.52 1 2xym = 1 m = 14. (a) y = mx (b) y = m(x 1)(c) y = 2 + m(x 1) (d) 2x + 4y = C5. Let the line be tangent to the circle at the point (x0, y0) where x20+ y20 = 9. The slope of thetangent line is the negative reciprocal of y0/x0 (why?), so m = x0/y0 and y = (x0/y0)x + b.Substituting the point (x0, y0) as well as y0 = 9 x20we get y = 9 x0x 9 x20. 17. January 27, 2005 11:42 L24-ch01 Sheet number 17 Page number 17 blackExercise Set 1.4 176. Solve the simultaneous equations to get the point (2, 1/3) of intersection. Then y = 13 +m(x+2).7. The x-intercept is x = 10 so that with depreciation at 10% per year the final value is always zero,and hence y = m(x 10). The y-intercept is the original value.y2 6 10x8. A line through (6,1) has the form y + 1 = m(x 6). The intercepts are x = 6 + 1/m andy = 6m 1. Set (6 + 1/m)(6m+1) = 3, or 36m2 + 15m + 1 = (12m + 1)(3m + 1) = 0 withroots m = 1/12,1/3; thus y + 1 = (1/3)(x 6) and y + 1 = (1/12)(x 6).9. (a) The slope is 1.51 1 223xy(b) The y-intercept is y = 1.421 14xy(c) They pass through the point (4, 2).yx6226 4(d) The x-intercept is x = 1.31131 2xy10. (a) horizontal linesxy(b) The y-intercept is y = 1/2.xy 222 18. January 27, 2005 11:42 L24-ch01 Sheet number 18P age number 18 black18 Chapter 1(c) The x-intercept is x = 1/2.xy11(d) They pass through (1, 1).xy211(1, 1)11. (a) VI (b) IV (c) III (d) V (e) I (f) II12. In all cases k must be positive, or negative values would appear in the chart. Only kx3 decreases,so that must be f(x). Next, kx2 grows faster than kx3/2, so that would be g(x), which grows fasterthan h(x) (to see this, consider ratios of successive values of the functions). Finally, experimenta-tion(a spreadsheet is handy) for values of k yields (approximately) f(x) = 10x3, g(x) = x2/2,h(x) = 2x1.5.13. (a)30102 1 21030xy2 1 21040y x(b)422y42 4x610y4 2 4x(c)211 2 32xy1y1 2 3x 19. January 27, 2005 11:42 L24-ch01 Sheet number 19 Page number 19 blackExercise Set 1.4 1914. (a)xy24040xy240280(b)xy2424xy2224(c)xy14232 2 42y x15. (a)105510y2 2x(b)642y2 1 2x(c)105510y2 2x 20. January 27, 2005 11:42 L24-ch01 Sheet number 20 Page number 20 black20 Chapter 116. (a)xy23(b)xy22(c)xy2217. (a)62y3 1 1x(b)80202080y1 2 3 4 5x(c)y10403x(d)40202040y1 2 3 4 5x18. (a)xy11(b)216 2 212xy(c)xy2424(d)yx24619. (a)1.50.51y3 2 1x(b)10.50.51y1 3x 21. January 27, 2005 11:42 L24-ch01 Sheet number 21 Page number 21 blackExercise Set 1.4 21(c)301020y1 1 3x(d)301020y2 1 2x20. (a)1010y1 3 4x(b)y845 3x(c)106106y1 1x(d)105y2 2x21. y = x2 + 2x = (x + 1)2 1105y2 2x22. (a)1y2 2x(b)1y2 2x23. (a) Nm (b) k = 20 Nm(c) V (L) 0.25 0.5 1.0 1.5 2.0P (N/m2) 80 103 40 103 20 103 13.3 103 10 103(d)302010P(N/m2)10 20V(m3) 22. January 27, 2005 11:42 L24-ch01 Sheet number 22 Page number 22 black22 Chapter 124. If the side of the square base is x and the height of the container is y then V = x2y = 100; minimizeA = 2x2 + 4xy = 2x2 + 400/x. A graphing utility with a zoom feature suggests that the solutionis a cube of side 10013cm.25. (a) F = k/x2 so 0.0005 = k/(0.3)2 and k = 0.000045 Nm2.(b) F = 0.000005 N(c)t5 10F105(d) When they approach one another, the force becomes infinite; when they get far apart it tendsto zero.26. (a) 2000 = C/(4000)2, so C = 3.2 1010 lbmi2(b) W = C/50002 = (3.2 1010)/(25 106) = 1280 lb.(c)150005000W2000 8000x(d) No, but W is very small when x is large.27. (a) II; y = 1, x = 1, 2 (b) I; y = 0, x = 2, 3(c) IV; y = 2 (d) III; y = 0, x = 228. The denominator has roots x = 1, so x2 1 is the denominator. To determine k use the point(0,1) to get k = 1, y = 1/(x2 1).29. (a) y = 3 sin(x/2)(b) y = 4 cos 2x(c) y = 5 sin 4x30. (a) y = 1 + cos x(b) y = 1 + 2sinx(c) y = 5 cos 4x31. (a) y = sin(x + /2)(b) y = 3 + 3 sin(2x/9)(c) y = 1 + 2 sin(2(x /4))2 sin(120t)32. V = 12033. (a) 3, /2yx3126(b) 2, 2222 4xy(c) 1, 4yx3212c 4c 6c 23. January 27, 2005 11:42 L24-ch01 Sheet number 23 Page number 23 blackExercise Set 1.5 2334. (a) 4, yx2243 9 f l(b) 1/2, 2/3yx0.40.28 c(c) 4, 6yx4224i 15c221c235. (a) Asin(t + ) = Asin(t) cos + Acos(t) sin = A1 sin(t) + A2 cos(t)(b) A1 = Acos ,A2 = Asin , so A =A21+ A22and = tan1(A2/A1).13/2, = tan1 1(c) A = 532;x =1325sin2t + tan1 13210^ 61036. three; x = 0, x = 1.895533 33EXERCISE SET 1.51. (a) f(g(x)) = 4(x/4) = x, g(f(x)) = (4x)/4 = x, f and g are inverse functions(b) f(g(x)) = 3(3x 1)+1 = 9x 2= x so f and g are not inverse functions(c) f(g(x)) = 3 (x3 + 2) 2 = x, g(f(x)) = (x 2) + 2 = x, f and g are inverse functions(d) f(g(x)) = (x1/4)4 = x, g(f(x)) = (x4)1/4 = |x|= x, f and g are not inverse functions2. (a) They are inverse functions.22 22(b) The graphs are not reflections ofeach other about the line y = x.22 22 24. January 27, 2005 11:42 L24-ch01 Sheet number 24 Page number 24 black24 Chapter 1(c) They are inverse functions providedthe domain of g is restricted to [0,+)50 50(d) They are inverse functions providedthe domain of f(x) is restrictedto [0,+)20 203. (a) yes (b) yes (c) no (d) yes (e) no (f) no4. (a) no, the horizontal line test fails63 32(b) yes, by the horizontal line test101 3105. (a) yes; all outputs (the elements of row two) are distinct(b) no; f(1) = f(6)6. (a) Since the point (0, 0) lies on the graph, no other point on the line x = 0 can lie on the graph,by the vertical line test. Thus the hour hand cannot point straight up or straight down. Thusnoon, midnight, 6AM and 6PM are impossible. To show other times are possible, suppose(a, b) lies on the graph with a= 0. Then the function y = Ax1/3 passes through (0, 0) and(a, b) provided A = b/a1/3.(b) same as (a)(c) Then the minute hand cannot point to 6 or 12, so in addition to (a), times of the form 1:00,1:30, 2:00, 2:30, , 12:30 are also impossible.7. (a) f has an inverse because the graph passes the horizontal line test. To compute f1(2) startat 2 on the y-axis and go to the curve and then down, so f1(2) = 8; similarly, f1(1) = 1and f1(0) = 0.(b) domain of f1 is [2, 2], range is [8, 8] (c)842 1 248yx8. (a) the horizontal line test fails(b) 3x 1; 1 x 2; and 2 x4.9. y = f1(x), x = f(y) = 7y 6, y =17(x + 6) = f1(x) 25. January 27, 2005 11:42 L24-ch01 Sheet number 25 Page number 25 blackExercise Set 1.5 2510. y = f1(x), x = f(y) = y + 1y 1, xy x = y + 1, (x 1)y = x + 1, y = x + 1x 1= f1(x)11. y = f1(x), x = f(y) = 3y3 5, y = 3 (x + 5)/3 = f1(x)12. y = f1(x), x = f(y) = 5 4y + 2, y =14(x5 2) = f1(x)13. y = f1(x), x = f(y) = 3 2y 1, y = (x3 + 1)/2 = f1(x)14. y = f1(x), x = f(y) =5y2 + 1, y =5 xx= f1(x)3/x = f1(x)15. y = f1(x), x = f(y) = 3/y2, y = 16. y = f1(x), x = f(y) =2y, y 0y2, y0, y = f1(x) =x/2, x 0x, x017. y = f1(x), x = f(y) =5/2 y, y21/y, y 2, y = f1(x) =5/2 x, x1/21/x, 0x 1/218. y = p1(x), x = p(y) = y3 3y2 + 3y 1 = (y 1)3, y = x1/3 + 1 = p1(x)19. y = f1(x), x = f(y) = (y + 2)4 for y 0, y = f1(x) = x1/4 2 for x 1620. y = f1(x), x = f(y) =y + 3 for y 3, y = f1(x) = x2 3 for x 03 2y for y 3/2, y = f1(x) = (3 x2)/2 for x 021. y = f1(x), x = f(y) = 22. y = f1(x), x = f(y) = 3y2 + 5y 2 for y 0, 3y2 + 5y 2 x = 0 for y 0,y = f1(x) = (5 +12x + 49)/6 for x 223. y = f1(x), x = f(y) = y 5y2 for y 1, 5y2 y + x = 0 for y 1,y = f1(x) = (1+1 20x)/10 for x 424. (a) C =59(F 32)(b) how many degrees Celsius given the Fahrenheit temperature(c) C = 273.15 C is equivalent to F = 459.67 F, so the domain is F 459.67, the rangeis C 273.1525. (a) y = f(x) = (6.214 104)x (b) x = f1(y) =1046.214y(c) how many meters in y milesx)26. (a) f(g(x))= f(x)2 = x, x1;= (g(f(x))= g(x2)=x2 = x, x1(b)xyy = f (x)y = g(x)(c) no, because f(g(x)) = x for every x in the domain of g is not satisfied(the domain of g is x 0) 26. January 27, 2005 11:42 L24-ch01 Sheet number 26 Page number 26 black26 Chapter 127. (a) f(f(x)) =3 3 x1 x1 3 x1 x=3 3x 3 + x1 x 3 + x= x so f = f1(b) symmetric about the line y = x28. y = f1(x), x = f(y) = ay2 + by + c, ay2 + by + c x = 0, use the quadratic formula to gety =b b2 4a(c x)2a;(a) f1(x) =b +b2 4a(c x)2a(b) f1(x) =b b2 4a(c x)2a29. if f1(x) = 1, then x = f(1) = 2(1)3 + 5(1) + 3 = 1030. if f1(x) = 2, then x = f(2) = (2)3/[(2)2 +1] = 8/531. f(f(x)) = x thus f = f1 so the graph is symmetric about y = x.32. (a) Suppose x1= x2 where x1 and x2 are in the domain of g and g(x1), g(x2) are in the domain off then g(x1)= g(x2) because g is one-to-one so f(g(x1))= f(g(x2)) because f is one-to-onethus f g is one-to-one because (f g)(x1)= (f g)(x2) if x1= x2.(b) f, g, and f g all have inverses because they are all one-to-one. Let h = (f g)1 then(f g)(h(x)) = f[g(h(x))] = x, apply f1 to both sides to get g(h(x)) = f1(x), then applyg1 to get h(x) = g1(f1(x)) = (g1 f1)(x), so h = g1 f1.33.xy34. Suppose that g and h are both inverses of f then f(g(x)) = x, h[f(g(x))] = h(x), buth[f(g(x))] = g(x) because h is an inverse of f so g(x) = h(x).5 4335. tan = 4/3, 0/2; use the triangle shown toget sin = 4/5, cos = 3/5, cot = 3/4, sec = 5/3,csc = 5/42.612.436. sec = 2.6, 0/2; use the triangle shown to getsin = 2.4/2.6 = 12/13, cos = 1/2.6 = 5/13,tan = 2.4 = 12/5, cot = 5/12, csc = 13/1237. (a) 0 x (b) 1 x 1(c) /2x/2 (d) x+ 27. January 27, 2005 11:42 L24-ch01 Sheet number 27 Page number 27 blackExercise Set 1.5 2738. Let = sin1(3/4) thensin = 3/4, /20and (see figure) sec = 4/734739. Let = cos1(3/5),sin 2 = 2 sin cos = 2(4/5)(3/5) = 24/255 4340. (a) sin(cos1 x) =1 x21x1 x2cos1 x(b) tan(cos1 x) =1 x2x1cos1 xx1 x2(c) csc(tan1 x) =1 + x2x1x1 + x2tan1 x(d) sin(tan1 x) = x1 + x21x1 + x2tan1 x41. (a) cos(tan1 x) =1 1 + x21x1 + x2tan1 x(b) tan(cos1 x) =1 x2xx1cos1 x1 x2(c) sin(sec1 x) =x2 1x1xsec1 xx2 1(d) cot(sec1 x) =1 x2 11xx2 1sec1 x 28. January 27, 2005 11:42 L24-ch01 Sheet number 28P age number 28 black28 Chapter 142. (a) x 1.00 0.80 0.6 0.40 0.20 0.00 0.20 0.40 0.60 0.80 1.00sin1 x 1.57 0.93 0.64 0.41 0.20 0.00 0.20 0.41 0.64 0.93 1.57cos1 x 3.14 2.50 2.21 1.98 1.77 1.57 1.37 1.16 0.93 0.64 0.00(b) yx11(c) yx32110.5 143. (a)c/20.5 0.5xyc/2(b)xyc/2c/244. (a) x = sin1(0.37) 2.7626 rad (b) = 18 0 + sin1(0.61) 217.645. (a) x = + cos1(0.85) 3.6964 rad (b) = cos1(0.23) 76.746. (a) x = tan1(3.16) 1.8773 (b) = 18 0 tan1(0.45) 155.847. (a) sin1 0.91, so it is not in the domain of sin1 x(b) 1 sin1 x 1 is necessary, or 0.841471 x 0.84147148. sin 2 = gR/v2 = (9.8)(18)/(14)2 = 0.9, 2 = sin1(0.9) or 2 = 18 0 sin1(0.9) so = 12sin1(0.9) 32 or = 90 12sin1(0.9) 58. The ball will have a lowerparabolic trajectory for = 32 and hence will result in the shorter time of flight.c y49. (a) yxc/210 10xc/25(b) The domain of cot1 x is (,+), the range is (0, ); the domain of csc1 x is (,1][1,+), the range is [/2, 0) (0, /2].50. (a) y = cot1 x; if x0 then 0y/2 and x = cot y, tan y = 1/x, y = tan1(1/x);if x0 then /2y and x = cot y = cot(y ), tan(y ) = 1/x, y = + tan1 1x(b) y = sec1 x, x = sec y, cos y = 1/x, y = cos1(1/x)(c) y = csc1 x, x = csc y, sin y = 1/x, y = sin1(1/x)51. (a) 55.0 (b) 33.6 (c) 25.852. (b) = sin1 RR + h= sin1 637816, 378 23 29. January 27, 2005 11:42 L24-ch01 Sheet number 29 Page number 29 blackExercise Set 1.5 2953. (a) If = 90, then sin = 1,1 sin2 sin2 =1 sin2 = cos ,D = tan tan = (tan 23.45)(tan 65) 0.93023374 so h 21.1 hours.(b) If = 270, then sin = 1, D = tan tan 0.93023374 so h 2.9 hours.54. 42 = 22 + 32 2(2)(3) cos , cos = 1/4, = cos1(1/4) 10455. y = 0 when x2 = 6000v2/g, x = 10v60/g = 100030 for v = 400 and g = 32;30, = tan1(3/tan = 3000/x = 3/30) 29.abx56. = , cot = xa + band cot = xbso = cot1 xa + b cot1 xb57. (a) Let = sin1(x) then sin = x, /2 /2. But sin() = sin and/2 /2 so sin() = (x) = x, = sin1 x, = sin1 x.(b) proof is similar to that in Part (a)58. (a) Let = cos1(x) then cos = x, 0 . But cos( ) = cos and0 so cos( ) = x, = cos1 x, = cos1 x(b) Let = sec1(x) for x 1; then sec = x and /2 . So 0 /2 and = sec1 sec( ) = sec1(sec ) = sec1 x, or sec1(x) = sec1 x.1 xsin1 x1 x259. (a) sin1 x = tan1 x1 x2(see figure)(b) sin1 x + cos1 x = /2; cos1 x = /2 sin1 x = /2 tan1 x1 x260. tan( + ) =tan + tan 1 tan tan ,tan(tan1 x + tan1 y)=tan(tan1 x) + tan(tan1 y)1 tan(tan1 x) tan(tan1 y)= x + y1 xyso tan1 x + tan1 y = tan1 x + y1 xy61. (a) tan1 12+ tan1 13= tan1 1/2 + 1/31 (1/2) (1/3)= tan11 = /4(b) 2 tan1 13= tan1 13+ tan1 13= tan1 1/3 + 1/31 (1/3) (1/3)= tan1 34,2 tan1 13+ tan1 17= tan1 34+ tan1 17= tan1 3/4 + 1/71 (3/4) (1/7)= tan11 = /462. sin(sec1 x) = sin(cos1(1/x)) = 30. 1 1x2=x2 1|x| 31. January 27, 2005 11:42 L24-ch01 Sheet number 30 Page number 30 black30 Chapter 1EXERCISE SET 1.61. (a) 4 (b) 4 (c) 1/42. (a) 1/16 (b) 8 (c) 1/33. (a) 2.9690 (b) 0.03414. (a) 1.8882 (b) 0.93815. (a) log2 16 = log2(24) = 4 (b) log2132= log2(25) = 5(c) log44 = 1 (d) log9 3 = log9(91/2) = 1/26. (a) log10(0.001) = log10(103) = 3 (b) log10(104) = 4e) = ln(e1/2) = 1/2(c) ln(e3) = 3 (d) ln(7. (a) 1.3655 (b) 0.30118. (a) 0.5229 (b) 1.14479. (a) 2 lna +12ln b +12ln c = 2r + s/2 + t/2 (b) ln b 3 lna ln c = s 3r t10. (a)13ln c ln a ln b = t/3 r s (b)12(ln a + 3lnb 2 lnc) = r/2 + 3s/2 t11. (a) 1 + log x +12log(x 3) (b) 2 ln|x| + 3 ln sin x 12ln(x2 + 1)12. (a)13log(x + 2) log cos 5x (b)12ln(x2 + 1) 12ln(x3 + 5)13. log24(16)3= log(256/3) 14. logx log(sin3 2x) + log 100 = logx100sin3 2x15. ln3 x(x + 1)2cos x16. 1 + x = 103 = 1000, x = 99917.x = 101 = 0.1, x = 0.01 18. x2 = e4, x = e219. 1/x = e2, x = e2 20. x = 7 21. 2x = 8 ,x = 422. log10 x3 = 30, x3 = 1030, x = 1010 23. log10 x = 5, x = 10524. ln 4x ln x6 = ln2, ln4x5 = ln2,4x5 = 2, x5 = 2, x = 5 225. ln 2x2 = ln3, 2x2 = 3, x2 = 3/2, x =3/2 (we discard 3/2 because it does not satisfy theoriginal equation)26. ln 3x = ln2, xln3 = ln2, x =ln 2ln 327. ln 52x = ln3, 2x ln5 = ln3, x = ln 32 ln52x = 5/3, 2x = ln(5/3), x = 1228. eln(5/3) 32. January 27, 2005 11:42 L24-ch01 Sheet number 31 Page number 31 blackExercise Set 1.6 3129. e3x = 7/2, 3x = ln(7/2), x =13ln(7/2)30. ex(1 2x) = 0 so ex = 0 (impossible) or 1 2x = 0, x = 1/231. ex(x+2) = 0 so ex = 0 (impossible) or x+2 = 0, x = 232. e2x ex 6 = (ex 3)(ex +2) = 0 so ex = 2 (impossible) or ex = 3, x = ln333. e2x 3ex +2 = (ex 2)(ex 1) = 0 so ex = 2, x = ln 2 or ex = 1, x = 034. (a) yx222 6(b) yx22 235. (a) yx6422 4(b) yx24 2436. (a) y x110(b) yx1337. log2 7.35 = (log 7.35)/(log 2) = (ln 7.35)/(ln 2) 2.8777;log5 0.6 = (log 0.6)/(log 5) = (ln 0.6)/(ln 5) 0.317438. 100 2539. 20 3340. (a) Let X = logb x and Y = loga x. Then bX = x and aY = x so aY = bX, or aY/X = b, whichmeans loga b = Y/X. Substituting for Y and X yieldsloga xlogb x= loga b, logb x =loga xloga b.(b) Let x = a to get logb a = (loga a)/(loga b) = 1/(loga b) so (loga b)(logb a) = 1.(log2 81)(log3 32) = (log2[34])(log3[25]) = (4 log2 3)(5 log3 2) = 20(log2 3)(log3 2) = 20 33. January 27, 2005 11:42 L24-ch01 Sheet number 32 Page number 32 black32 Chapter 141. x = y 1.4710, x = y 7.857181 8042. Since the units are billions, one trillion is 1,000 units. Solve 1000 = 0.051517(1.1306727)x for x bytaking common logarithms, resulting in 3 = log 0.051517 + x log 1.1306727, which yields x 80.4,so the debt first reached one trillion dollars around 1980.43. (a) no, the curve passes through the origin (b) y = 2x/45)x(c) y = 2x (d) y = (51 2044. (a) As x + the function grows very slowly, but it is always increasing and tends to +. Asx 1+ the function tends to .(b)1 255xy45. log(1/2)0 so 3 log(1/2)2 log(1/2)46. Let x = logb a and y = logb c, so a = bx and c = by.First, ac = bxby = bx+y or equivalently, logb(ac) = x + y = logb a + logb c.Secondly, a/c = bx/by = bxy or equivalently, logb(a/c) = x y = logb a logb c.Next, ar = (bx)r = brx or equivalently, logb ar = rx = r logb a.Finally, 1/c = 1/by = by or equivalently, logb(1/c) = y = logb c.47. 75et/125 = 15, t = 125 ln(1/5) = 125 ln 5 201 days.48. (a) If t = 0, then Q = 12 grams(b) Q = 12e0.055(4) = 12e0.22 9.63 grams(c) 12e0.055t = 6, e0.055t = 0.5, t = (ln 0.5)/(0.055) 12.6 hours 34. January 27, 2005 11:42 L24-ch01 Sheet number 33 Page number 33 blackExercise Set 1.7 3349. (a) 7.4; basic (b) 4.2; acidic (c) 6.4; acidic (d) 5.9; acidic50. (a) log[H+] = 2.44, [H+] = 102.44 3.6 103 mol/L(b) log[H+] = 8.06, [H+] = 108.06 8.7 109 mol/L51. (a) 140 dB; damage (b) 120 dB; damage(c) 80 dB; no damage (d) 75 dB; no damage52. Suppose that I1 = 3I2 and 1 = 10 log10 I1/I0, 2 = 10 log10 I2/I0. ThenI1/I0 = 3I2/I0, log10 I1/I0 = log10 3I2/I0 = log10 3 + log10 I2/I0, 1 = 10 log10 3 + 2,1 2 = 10 log10 3 4.8decib els.53. Let IA and IB be the intensities of the automobile and blender, respectively. Thenlog10 IA/I0 = 7 and log10 IB/I0 = 9.3, IA = 107I0 and IB = 109.3I0, so IB/IA = 102.3 200.54. The decibel level of the nth echo is 120(2/3)n;120(2/3)n10 if (2/3)n1/12, n log(1/12)log(2/3)=log 12log 1.5 6.13 so 6 echoes can be heard.55. (a) logE = 4.4 + 1.5(8.2) = 16.7,E = 1016.7 5 1016 J(b) Let M1 and M2 be the magnitudes of earthquakes with energies of E and 10E,respectively. Then 1.5(M2 M1) = log(10E) logE = log 10 = 1,M2 M1 = 1/1.5 = 2/3 0.67.56. Let E1 and E2 be the energies of earthquakes with magnitudes M and M + 1, respectively. ThenlogE2 logE1 = log(E2/E1) = 1.5,E2/E1 = 101.5 31.6.EXERCISE SET 1.71. The sum of the squares for the residuals for line I is approximately12 + 12 + 12 + 02 + 22 + 12 + 12 + 12 = 10, and the same for line II is approximately02 + (0.4)2 + (1.2)2 + 02 + (2.2)2 + (0.6)2 + (0.2)2 + 02 = 6.84; line II is the regression line.2. (a) The data appear to be periodic, so a trigonometric model may be appropriate.(b) The data appear to lie near a parabola, so a quadratic model may be appropriate.(c) The data appear to lie near a line, so a linear model may be appropriate.(d) The data appear to be randomly distributed, so no model seems to be appropriate.3. Least squares line S = 0.6414t 1054.517, correlation coefficient 0.57541989 1992 1995 1998 2001 2004240230220tS4. (a) p = 0.0154T + 4.19 (b) 3.42 atm5. (a) Least squares line p = 0.0146T + 3.98, correlation coefficient 0.9999(b) p = 3.25 atm (c) T = 272C 35. January 27, 2005 11:42 L24-ch01 Sheet number 34 Page number 34 black34 Chapter 16. (a) p = 0.0203 T + 5.54, r = 0.9999 (b) T = 273(c) 1.05 p = 0.0203(T)(1.1) + 5.54 and p = 0.0203T + 5.54, subtract to get.05p = 0.1(0.0203T), p = 2(0.0203T). But p = 0.0203T + 5.54,equate the right hand sides, get T = 5.54/0.0203 273C7. (a) R = 0.00723T + 1.55 (b) T = 214C8. (a) y = 0.0236x + 4.79 (b) T = 203C9. (a) S = 0.50179w 0.00643 (b) S = 8 ,w = 16 lb10. (a) S = 0.756w 0.0133(b) 2S = 0.756(w+5)0.0133, subtract one equation from the other to get S = 5(0.756) = 3.7811. (a) Let h denote the height in inches and y the number of rebounds per minute. Theny = 0.0174h 1.0549, r = 0.7095(b)y (c) The least squares line is a fair model for these data,65 750.30.20.1hsince the correlation coefficient is 0.7095.12. (a) Let h denote the height in inches and y the weight in pounds.Then y = 5.45h 218; r = 0.8074(b)y (c) 223 lb70 80250225200175h13. (a) (0 + b 0)2 + (1 + b 0)2 + (1 + b 1)2 = 3b2 + 2b + 1(b) The function of Part (a) is a parabola in the variable b, and has its minimum when b = 13,hence the desired line is y = x 13.14. Let the points be (A,B) and (A,C). Let the regression line pass through the point (A, b) and haveslope m.The line y = m(x A) + b is such a line.The sum of the squares of the residues is (b B)2 + (b C)2 which is easily seen to be minimizedat b = (B +C)/2, which means the line passes through the point midway between the two originalgiven data points.Since the slope of the line is arbitary, and since the residual errors are independent of the slope,it follows that there are infinitely many such lines of regression, and they all pass through themidpoint. 36. January 27, 2005 11:42 L24-ch01 Sheet number 35 Page number 35 blackExercise Set 1.7 3515. Let the points be (A,B), (A,C) and (A,B) and let the line of regression have the formy = m(x A) + b. Then the sum of the squares of the residues is(B b)2 + (C b)2 + (m(x A) + b B)2.Choose m and b so that the line passes through the point (A,B), which makes the third termzero, and passes through the midpoint between (A,B) and (A,C), which minimizes the sum(B b)2 + (C b)2 (see Exercise 15). These two conditions together minimize the sum of thesquares of the residuals, and they determine the slope and one point, and thus they determine theline of regression.16. Let the four vertices of the trapezoid be (A,B), (A,B), (A, C) and (A, C). From Exercise 14 itfollows that the sum of the squares of residuals at the two points (A,B) and (A,B) is minimized ifthe line of regression passes through the midpoint between (A,B) and (A,B). Similarly the sumof the sum of the squares of the residuals at the two points (A,B) and (A, C) is minimized byhaving the line of regression pass through the midpoint between these two points. Altogether wehave determined two points through which the line of regression must pass, and hence determinedthe line.17. (a) H 20000/110 182 km/s/Mly(b) One light year is 9.408 1012 km andt = dv=1H=120km/s/Mly=9.408 1018km20km/s= 4.704 1017 s = 1.492 1010 years.(c) The Universe would be even older.18. The population is modeled by P = 0.0045833t2 16.378t + 14635; in the year 2000 the popula-tionwould be P(2000) = 212, 200, 000. This is far short; in 1990 the population of the US wasapproximately 250,000,000.19. As in Example 4, a possible model is of the form T = D + Asin Bt CB. Since the longestday has 993 minutes and the shortest has 706, take 2A = 993 706 = 287 or A = 143.5. Themidpoint between the longest and shortest days is 849.5 minutes, so there is a vertical shift ofD = 8 49.5. The period is about 365.25 days, so 2/B = 365.25 or B = /183. Note that the sinefunction takes the value 1 when t CB= 91.8125, and T is a minimum at about t = 0. Thusthe phase shift CB 91.5. Hence T = 8 49.5+143.5 sin 183t 2is a model for the temperature.20. As in Example 4, a possible model for the fraction f of illumination is of the formf = D + Asin Bt CB. Since the greatest fraction of illumination is 1 and the least is 0,2A = 1, A = 1/2. The midpoint of the fraction of illumination is 1/2, so there is a vertical shiftof D = 1/2. The period is approximately 30 days, so 2/B = 30 or B = /15. The phase shift isapproximately 49/2, so C/B = 49/2, and f = 1/2 + 1/2 sin 15t 492d21. t = 0.4452.30 25022. (a) t = 0.373 r1.5(b) 238,000 km(c) 1.89 days 37. January 27, 2005 11:42 L24-ch01 Sheet number 36 Page number 36 black36 Chapter 1EXERCISE SET 1.81. (a) x + 1 = t = y 1, y = x + 2642yt = 2t = 42 4t = 0t = 1t = 3t = 5x(c) t 0 1 2 3 4 5x 1 0 1 2 3 4y 1 2 3 4 5 62. (a) x2 + y2 = 11yt = 0.5t = 0.75 t = 0.25t = 1 x1 1t = 0(c) t 0 0.2500 0.50 0.7500 1x 1 0.7071 0.00 0.7071 1y 0 0.7071 1.00 0.7071 03. t = (x + 4)/3; y = 2x + 10128 6xy4. t = x + 3; y = 3x + 2,3 x 0xy(0, 2)(3, 7)5. cos t = x/2, sin t = y/5;x2/4 + y2/25 = 155 55xy6. t = x2; y = 2x2 + 4,x 0184xy7. cos t = (x 3)/2,sin t = (y 2)/4;(x 3)2/4 + (y 2)2/16 = 1762xy8. sec2 t tan2 t = 1;x2 y2 = 1, x 1and y 01xy9. cos 2t = 1 2 sin2 t;x = 1 2y2,1 y 111 11xy10. t = (x 3)/4;y = (x 3)2 9xy(3, 9) 38. January 27, 2005 11:42 L24-ch01 Sheet number 37 Page number 37 blackExercise Set 1.8 3711. x/2 + y/3 = 1, 0 x 2, 0 y 3xy2312. y = x 1, x 1, y 011xy13. x = 5 cos t, y = 5 sin t, 0 t 257.5 7.5514. x = cos t, y = sin t, t 3/211 1115. x = 2, y = t20 3116. x = 2 cos t, y = 3 sin t, 0 t 232 2317. x = t2, y = t, 1 t 110 1118. x = 1 + 4cos t, y = 3 + 4sint, 0 t 227 8819. (a) 1435 80(b) t 0 1 2 3 4 5x 0 5.5 8 4.5 8 32.5y 1 1.5 3 5.5 9 13.53.(c) x = 0 when t = 0, 22(d) for 0t2(e) at t = 2 39. January 27, 2005 11:42 L24-ch01 Sheet number 38P age number 38 black38 Chapter 120. (a) 52 140(b) y is always 1 since cos t 1(c) greater than 5, since cos t 121. (a) 30 205(b) o1 1O22. (a) 1.72.3 2.31.7(b)610 10^23. (a) Eliminate t to get x x0x1 x0= y y0y1 y0(b) Set t = 0 to get (x0, y0); t = 1 for (x1, y1).(c) x = 1+t, y = 2 + 6t (d) x = 2 t, y = 4 6t24. (a) x = 3 2t, y = 4 + 5t, 0 t 1 (b) x = at, y = b(1 t), 0 t 125. (a) |RP|2 = (xx0)2+(yy0)2 = t2[(x1x0)2+(y1y0)2] and |QP|2 = (x1x0)2+(y1y0)2,so r = |R P| = |Q P|t = qt.(b) t = 1/2 (c) t = 3/426. x = 2+t, y = 1 + 2t(a) (5/2, 0) (b) (9/4,1/2) (c) (11/4, 1/2)27. (a) IV, because x always increases whereas y oscillates.(b) II, because (x/2)2 + (y/3)2 = 1, an ellipse.(c) V, because x2 + y2 = t2 increases in magnitude while x and y keep changing sign.(d) VI; examine the cases t1 and t1 and you see the curve lies in the first, second andfourth quadrants only.(e) III because y0.(f) I; since x and y are bounded, the answer must be I or II; but as t runs, say, from 0 to , xgoes directly from 2 to 2, but y goes from 0 to 1 to 0 to 1 and back to 0, which describesI but not II. 40. January 27, 2005 11:42 L24-ch01 Sheet number 39 Page number 39 blackExercise Set 1.8 3928. (a) from left to right (b) counterclockwise (c) counterclockwise(d) As t travels from to 1, the curve goes from (near) the origin in the third quadrantand travels up and left. As t travels from 1 to + the curve comes from way down inthe second quadrant, hits the origin at t = 0, and then makes the loop clockwise and finallyapproaches the origin again as t +.(e) from left to right(f) Starting, say, at (1, 0), the curve goes up into the first quadrant, loops back through theorigin and into the third quadrant, and then continues the figure-eight.29. The two branches corresponding to 1 t 0 and 0 t 1 coincide.30. (a) Eliminate t t0t1 t0to obtain y y0x x0= y1 y0x1 x0.(b) from (x0, y0) to (x1, y1)(c) x = 3 2(t 1), y = 1 + 5(t 1)50 5231. (a) x ba= y dc(b)321y1 2 3x32. (a) If a = 0 the line segment is vertical; if c = 0 it is horizontal.(b) The curve degenerates to the point (b, d).33. 62 6234. 105 10535. 30 3036. 60 6037.21y0.5 1x38. x = 1/2 4t, y = 1/2 for 0 t 1/4x = 1/2, y= 1/2 4(t 1/4) for 1/4 t 1/2x = 1/2 + 4(t 1/2), y= 1/2 for 1/2 t 3/4x = 1/2, y= 1/2 + 4(t 3/4) for 3/4 t 1 41. January 27, 2005 11:42 L24-ch01 Sheet number 40 Page number 40 black40 Chapter 139. (a) x = 4 cos t, y = 3 sin t (b) x = 1 + 4cos t, y = 2 + 3sint(c) 34 4355 3140. (a) t = x/(v0 cos ),so y = x tan gx2/(2v20 cos2 ).(b)1000060002000x40000 80000y2t,41. (a) From Exercise 40, x = 4002t 4.9t2.y = 400(b) 16,326.53 m(c) 65,306.12 m42. (a) 1525 2515(b) 1525 2515(c) 1525 2515a = 3, b = 21525 2515a = 2, b = 31525 2515a = 2, b = 7 42. January 27, 2005 11:42 L24-ch01 Sheet number 41 Page number 41 blackExercise Set 1.8 4143. Assume that a= 0 and b= 0; eliminate the parameter to get (x h)2/a2 + (y k)2/b2 = 1. If|a| = |b| the curve is a circle with center (h, k) and radius |a|; if |a|= |b| the curve is an ellipsewith center (h, k) and major axis parallel to the x-axis when |a||b|, or major axis parallel to they-axis when |a||b|.(a) ellipses with a fixed center and varying axes of symmetry(b) (assume a= 0 and b= 0) ellipses with varying center and fixed axes of symmetry(c) circles of radius 1 with centers on the line y = x 144. Refer to the diagram to get b = a, = a/b but = + /2 so = /2 = (a/b 1) /2x= (a b) cos b sin = (a b) cos + b cosa bb,y= (a b) sin b cos = (a b) sin b sina bb.b a bxy a45. (a)aa aayx(b) Use b = a/4 in the equations of Exercise 44 to getx = 34a cos + 14a cos 3, y = 34a sin 14a sin 3;but trigonometric identities yield cos 3 = 4 cos3 3 cos , sin 3 = 3 sin 4 sin3 ,so x = a cos3 , y = a sin3 .(c) x2/3 + y2/3 = a2/3(cos2 + sin2 ) = a2/346. (a)111xya = 1/233113xya = 355115xya = 5(b) (x a)2 + y2 = (2a cos2 t a)2 + (2a cos t sin t)2= 4a2 cos4 t 4a2 cos2 t + a2 + 4a2 cos2 t sin2 t= 4a2 cos4 t 4a2 cos2 t + a2 + 4a2 cos2 t(1 cos2 t) = a2,a circle about (a, 0) of radius a 43. January 27, 2005 11:42 L24-ch01 Sheet number 42 Page number 42 black42 Chapter 1REVIEW EXERCISES, CHAPTER 11. yx55 512. (a) f(2) = 2, g(3) = 2 (b) x = 3, 3 (c) x2, x3(d) the domain is 5 x 5 and the range is 5 y 4(e) the domain is 4 x 4.1, the range is 3 y 5(f) f(x) = 0 at x = 3, 5; g(x) = 0 at x = 3, 23.7050400 2 4 6tT4. Assume that the paint is applied in a thin veneer of uniform thickness, so that the quantity of paintto be used is proportional to the area covered. If P is the amount of paint to be used, P = kr2.The constant k depends on physical factors, such as the thickness of the paint, absorption of thewood, etc.5. (a) If the side has length x and height h, then V = 8 = x2h, so h = 8/x2. Then the costC = 5x2 + 2(4)(xh) = 5x2 + 64/x.(b) The domain of C is (0,+) because x can be very large (just take h very small).6. (a) Suppose the radius of the uncoated ball is r and that of the coated ball is r + h. Then the4343plastic has volume 43equal to the difference of the volumes, i.e.V = (r+h)3r3 = h[3r2+3rh+h2] in3. But r = 3 and hence V = 43h[27+9h+h2].(b) 0h7. (a) The base has sides (10 2x)/2 and 6 2x, and the height is x, so V = (6 2x)(5 x)x ft3.(b) From the picture we see that x5 and 2x6, so 0x3.(c) 3.57 ft 3.79 ft 1.21 ft8. (a) d =(x 1)2 + 1/x2;(b) x0, 0x+(c) d 0.8 2 at x 1.3821.61.210.8y0.5 1 2 3x9.12y2 1 2x 44. January 27, 2005 11:42 L24-ch01 Sheet number 43 Page number 43 blackReview Exercises, Chapter 1 4310. (a) On the interval [20, 30]the curve seems tame,30000020000010 20 30xy(b) but seen close up on the interval[1/1000, +1/1000] we see that thereis some wiggling near the origin.0.0005 0.001xy0.000010.0000211. x 4 3 2 1 0 1 2 3 4f(x) 0 1 2 1 3 2 3 4 4g(x) 3 2 1 3 1 4 4 2 0(f g)(x) 4 3 2 1 1 0 4 2 3(g f)(x) 1 3 4 4 2 1 2 0 312. f g(x) = 1/|x| with domain x= 0, and g f(x) is nowhere defined, with domain .13. f(g(x)) = (3x+2)2+1, g(f(x)) = 3(x2+1)+2, so 9x2+12x+5 = 3x2+5, 6x2+12x = 0, x = 0,214. (a) (3 x)/x(b) no; the definition of f(g(x)) requires g(x) to be defined, so x= 1, and f(g(x)) requiresg(x)= 1, so we must have g(x)= 1, i.e. x= 0; whereas h(x) only requires x= 015. When f(g(h(x))) is defined, we require g(h(x))= 1 and h(x)= 0. The first requirement is(2). For all other x, f g h = 1/(2x2).equivalent to x= 1, the second is equivalent to x= 16. g(x) = x2 + 2x17. (a) even odd = odd (b) a square is even(c) even + odd is neither (d) odd odd = even18. (a) y = |x 1|, y = |(x) 1| = |x + 1|,y = 2|x + 1|, y = 2|x + 1| 3,y = 2|x + 1| + 3(b) yx33 1 2119. (a) The circle of radius 1 centered at (a, a2); therefore, the family of all circles of radius 1 withcenters on the parabola y = x2.(b) All parabolas which open up, have latus rectum equal to 1 and vertex on the line y = x/2. 45. January 27, 2005 11:42 L24-ch01 Sheet number 44 Page number 44 black44 Chapter 120. Let y = ax2 +bx+c. Then 4a+2b+c = 0, 64a+8b+c = 18, 64a8b+c = 18, from which b = 010x2 65and 60a = 18, or finally y = 3.21. (a)602020y100 300t(b) when2365(t 101) =32, or t = 374.75, which is the same date as t = 9.75, so during thenight of January 10th-11th(c) from t = 0 to t = 70.58and from t = 313.92 to t = 365 (the same date as t = 0) , for a totalof about 122 days22. Let y = A+B sin(at+b). Since the maximum and minimum values of y are 35 and 5, A+B = 35and AB = 5, A = 20, B = 15. The period is 12 hours, so 12a = 2 and a = /6. The maximumoccurs at t = 2, so 1 = sin(2a + b) = sin(/3 + b), /3 + b = /2, b = /2 /3 = /6 andy = 20 + 15 sin(t/6 + /6).23. When x = 0 the value of the green curve is higher than that of the blue curve, therefore the bluecurve is given by y = 1 + 2sinx.The points A,B,C,D are the points of intersection of the two curves, i.e. where1+2 sin x = 2 sin(x/2)+2 cos(x/2). Let sin(x/2) = p, cos(x/2) = q. Then 2 sin x = 4 sin(x/2) cos(x/2)(basic trigonometric identity, so the equation which yields the points of intersection becomes1+4pq = 2p+2q, 4pq2p2q+1 = 0, (2p1)(2q1) = 0; thus whenever either sin(x/2) = 1/2 orcos(x/2) = 1/2, i.e. when x/2 = /6, 5/6,/3. Thus A has coordinates (2/3, 13), B hascoordinates (/3,1+3), C has coordinates (2/3,1+3), and D has coordinates (5/3, 13).24. (a) R = R0 is the R-intercept, R0k is the slope, and T = 1/k is the T-intercept(b) 1/k = 273, or k = 1/273(c) 1.1 = R0(1 + 20/273), or R0 = 1.025(d) T = 126.55C25. (a) f(g(x)) = x for all x in the domain of g, and g(f(x)) = x for all x in the domain of f.(b) They are reflections of each other through the line y = x.(c) The domain of one is the range of the other and vice versa.(d) The equation y = f(x) can always be solved for x as a function of y. Functions with noinverses include y = x2, y = sin x.26. (a) For sin x, /2 x /2; for cos x, 0 x ; for tan x, /2x/2; for sec x,0 x/2 or /2x . 46. January 27, 2005 11:42 L24-ch01 Sheet number 45 Page number 45 blackReview Exercises, Chapter 1 45(b) yy = sin1 xy = sin xx11c/2yx1cy = cos1 xy = cos xyx2y = tan xc/2 c/22y = tan1 xyxy = sec1 x y = sec x21c/2y = sec1 xy = sec x27. (a) x = f(y) = 8y3 1; y = f1(x) =x + 181/3=12(x + 1)1/3(b) f(x) = (x 1)2; f does not have an inverse because f is not one-to-one, for examplef(0) = f(2) = 1.(c) x = f(y) = (ey)2 + 1; y = f1(x) = lnx 1 = 12ln(x 1)(d) x = f(y) = y + 2y 1; y = f1(x) = x + 2x 1(e) x = f(y) = sin1 2yy; y =12 + sin1 x(f) x =11 + 3tan1 y; y = tan1 x3x28. It is necessary and sufficient that the graph of f pass the horizontal line test. Suppose to the con-trarythat ah + bch + d= ak + bck + dfor h= k. Then achk+bck+adh+bd = achk+adk+bch+bd, bc(hk) =ad(h k). It follows from h= k that ad bc = 0. These steps are reversible, hence f1 exists ifand only if ad bc= 0, and if so, thenx = ay + bcy + d, xcy + xd = ay + b, y(cx a) = b xd, y = b xdcx a= f1(x)29. Draw equilateral triangles of sides 5, 12, 13, and 3, 4, 5. Then sin[cos1(4/5)] = 3/5,sin[cos1(5/13)] = 12/13, cos[sin1(4/5)] = 3/5, cos[sin1(5/13)] = 12/13(a) cos[cos1(4/5) + sin1(5/13)]= cos(cos1(4/5)) cos(sin1(5/13))sin(cos1(4/5)) sin(sin1(5/13))=451213 35513=3365.(b) sin[sin1(4/5) + cos1(5/13)]= sin(sin1(4/5)) cos(cos1(5/13))+cos(sin1(4/5)) sin(cos1(5/13))=45513+351213=5665. 47. January 27, 2005 11:42 L24-ch01 Sheet number 46 Page number 46 black46 Chapter 130. (a) yxc/21(b) yxc/21(c) yxc/25(d) yxc/2131. y = 5 ft = 60 in, so 60 = log x, x = 1060 in 1.58 1055 mi.32. y = 100 mi = 12 5280 100 in, so x = log y = log 12 + log 5280 + log 100 6.8018 in33. 3 lne2x(ex)3+ 2 exp(ln 1) = 3 ln e2x + 3 ln(ex)3 + 2 1 = 3(2x) + (3 3)x+2 = 15x + 234. Y = ln(Cekt) = lnC + lnekt = lnC + kt, a line with slope k and Y -intercept lnC35. (a) yx224(b) The curve y = ex/2 sin 2x has xintercepts at x = /2, 0, /2, , 3/2. It intersects thecurve y = ex/2 at x = /4, 5/4 and it intersects the curve y = ex/2 at x = /4, 3/4.36. (a)205v1 2 3 4 5t(b) limt(1 e1.3t) = 1,and thus as t 0, v 24.61 ft/s.(c) For large t the velocity approaches c = 24.61.(d) No; but it comes very close (arbitrarily close).(e) 3.009 s37. (a)200100N10 30 50t(b) N = 80 when t = 9.35 yrs(c) 220 sheep 48. January 27, 2005 11:42 L24-ch01 Sheet number 47 Page number 47 blackReview Exercises, Chapter 1 4738. (a) The potato is done in the interval 27.65t32.71.(b) 91.54 min. The oven temperature is always 400 F, so the difference between the oventemperature and the potato temperature is D = 400 T. Initially D = 325, so solveD = 75 + 325/2 = 237.5 for t, t 22.76.39. (a) The function ln x x0.2 is negative at x = 1 and positive at x = 4, so it is reasonable toexpect it to be zero somewhere in between. (This will be established later in this book.)(b) x = 3.654 and 3.32105 10511 1 3 5135xy40. (a) If xk = ex then k ln x = x, orln xx=1k. The steps are reversible.(b) By zooming it is seen that the maximum value of y isapproximately 0.368(actually , 1/e), so there are two distinctsolutions of xk = ex whenever k1/0.368 2.717.(c) x 1.15541. (a) 1.90 1.92 1.94 1.96 1.98 2.00 2.02 2.04 2.06 2.08 2.103.4161 3.4639 3.5100 3.5543 3.5967 3.6372 3.6756 3.7119 3.7459 3.7775 3.8068(b) y = 1.9589x 0.2910(c) y 3.6372 = 1.9589(x 2), or y = 1.9589x 0.2806(d) As one zooms in on the point (2, f(2))the two curves seem to converge to one line.3.81.9 2.23.442. (a) 0.10 0.08 0.06 0.04 0.02 0.00 0.02 0.04 0.06 0.08 0.100.9950 0.9968 0.9982 0.9992 0.9998 1.0000 0.9998 0.9992 0.9982 0.9968 0.995012(b) y = 13.669x2 + 2.865x + 0.733(c) y = x2 + 1(d) As one zooms in on the point (0, f(0)) thetwo curves seem to converge to one curve. 1.21.7 1.70.8 49. January 27, 2005 11:42 L24-ch01 Sheet number 48P age number 48 black48 Chapter 143. The data are periodic, so it is reasonable that a trigonometric function might approximate them.A possible model is of the form T = D + Asin Bt CB. Since the highest level is 1.032meters and the lowest is 0.045, take 2A = 1.032 0.042 = 0.990 or A = 0.495. The midpointbetween the lowest and highest levels is 0.537 meters, so there is a vertical shift of D = 0.537.The period is about 12 hours, so 2/B = 12 or B = /6. The phase shift CB 6.5. HenceT = 0.537 + 0.495 sin6is a model for the temperature.(t 6.5)10 20 30 40 50 60 7010.80.60.40.2tT44. (a) y = (415/458)x + 9508/1603 0.906x + 5.931(b) 85.6745. x(t) =2 cos t, y(t) = 2 sin t, 0 t 3/246. x = f(1 t), y = g(1 t) 47.2112y1 1 2x48. (a) The three functions x2, tan(x) and ln(x) are all increasing on the indicated interval, and thusso is f(x) and it has an inverse.(b) The asymptotes for f(x) are x = 0, x = /2. The asymptotes for f1(x) are y = 0, y = /2.3#x = 63xyy = xy = f (x)x = f (y) 50. January 27, 2005 11:43 L24-ch02 Sheet number 1 Page number 49 blackCHAPTER 2Limits and ContinuityEXERCISE SET 2.11. (a) 0 (b) 0 (c) 0 (d) 32. (a) + (b) + (c) + (d) undef3. (a) (b) (c) (d) 14. (a) 1 (b) (c) does not exist (d) 25. for all x0= 4 6. for all x0= 6, 313. (a) 2 1.5 1.1 1.01 1.001 0 0.5 0.9 0.99 0.9990.1429 0.2105 0.3021 0.3300 0.3330 1.0000 0.5714 0.3690 0.3367 0.333710 1The limit is .490 20The limit is 1/3.(b) 2 1.5 1.1 1.01 1.001 1.00010.4286 1.0526 6.344 66.33 666.3 6666.3501 20The limit is +.(c) 0 0.5 0.9 0.99 0.999 0.99991 1.7143 7.0111 67.001 667.0 6667.0050 51. January 27, 2005 11:43 L24-ch02 Sheet number 2 Page number 50 black50 Chapter 214. (a) 0.25 0.1 0.001 0.0001 0.0001 0.001 0.1 0.250.5359 0.5132 0.5001 0.5000 0.5000 0.4999 0.4881 0.47210.60.25 0.250The limit is 1/2.(b) 0.25 0.1 0.001 0.00018.4721 20.488 2000.5 200011000 0.250The limit is +.(c) 0.25 0.1 0.001 0.00017.4641 19.487 1999.5 2000000.25 0 The limit is .10015. (a) 0.25 0.1 0.001 0.0001 0.0001 0.001 0.1 0.252.7266 2.9552 3.0000 3.0000 3.0000 3.0000 2.9552 2.726630.25 0.252The limit is 3. 52. January 27, 2005 11:43 L24-ch02 Sheet number 3 Page number 51 blackExercise Set 2.1 51(b) 0 0.5 0.9 0.99 0.999 1.5 1.1 1.01 1.0011 1.7552 6.2161 54.87 541.1 0.1415 4.536 53.19 539.5601.5 060The limit does not exist.16. (a) 0 0.5 0.9 0.99 0.999 1.5 1.1 1.01 1.0011.5574 1.0926 1.0033 1.0000 1.0000 1.0926 1.0033 1.0000 1.00001.51.5 01The limit is 1.(b) 0.25 0.1 0.001 0.0001 0.0001 0.001 0.1 0.251.9794 2.4132 2.5000 2.5000 2.5000 2.5000 2.4132 1.97942.50.25 0.252The limit is 5/2.17. msec = x2 1x + 1= x 1 which gets close to 2 as x gets close to 1, thus y 1 = 2(x + 1) ory = 2x 118. msec = x2x= x which gets close to 0 as x gets close to 0 (doh!), thus y = 019. msec = x4 1x 1= x3 + x2 + x + 1 which gets close to 4 as x gets close to 1, thus y 1 = 4(x 1)or y = 4x 320. msec = x4 1x + 1= x3x2+x1 which gets close to 4 as x gets close to 1, thus y1 = 4(x+1)or y = 4x 3 53. January 27, 2005 11:43 L24-ch02 Sheet number 4 Page number 52 black52 Chapter 221. (a) The limit appears to be 3.3.51 12.5(b) The limit appears to be 3.3.50.001 0.0012.5(c) The limit does not exist.3.50.000001 0.0000012.522. (a) 0.01 0.001 0.0001 0.000011.666 0.1667 0.1667 0.17(b) 0.000001 0.0000001 0.00000001 0.000000001 0.00000000010 0 0 0 0(c) it can be misleading23. (a) The plot over the interval [a, a] becomes subject to catastrophic subtraction if a is smallenough (the size depending on the machine).(c) It does not.24. (a) the mass ofthe object while at rest(b) the limiting mass as the velocity approaches the speed ofligh t; the mass is unbounded25. (a) The length ofthe rod while at rest(b) The limit is zero. The length ofthe rod approaches zeroEXERCISE SET 2.21. (a) 6 (b) 13 (c) 8 (d) 16 (e) 2 (f) 1/2(g) The limit doesnt exist because the denominator tends to zero but the numerator doesnt.(h) The limit doesnt exist because the denominator tends to zero but the numerator doesnt.2. (a) 0(b) The limit doesnt exist because lim f doesnt exist and lim g does.(c) 0 (d) 3 (e) 0(f) The limit doesnt exist because the denominator tends to zero but the numerator doesnt.(g) The limit doesnt exist becausef(x) is not defined for 0 x2.(h) 1 54. January 27, 2005 11:43 L24-ch02 Sheet number 5 Page number 53 blackExercise Set 2.3 533. 6 4. 27 5. 3/4 6. -3 7. 4 8. 129. -4/5 10. 0 11. 3 12. 1 13. 3/2 14. 4/315. + 16. 17. does not exist 18. +19. 20. does not exist 21. + 22. 23. does not exist 24. 25. + 26. does not exist27. + 28. + 29. 6 30. 431. (a) 2(b) 2(c) 232. (a) -2(b) 0(c) does not exist33. (a) 3(b) yx4134. (a) 6(b) F(x) = x 335. (a) Theorem 2.2.2(a) doesnt apply; moreover one cannot add/subtract infinities.(b) limx0+1x 1x2= limx0+x 1x2= 36. limx01x+1x2= limx0x + 1x2 = + 37. limx0xx =x + 4 + 21438. limx0x2x = 0x + 4 + 239. The left and/or right limits could be plus or minus infinity; or the limit could exist, or equal anypreassigned real number. For example, let q(x) = x x0 and let p(x) = a(x x0)n where n takeson the values 0, 1, 2.40. (a) If x0 then f(x) = ax + bx ax + bx2x= b, so the limit is b.(b) Similarly if x0 then f(x) = a, so the limit is a.(c) Since the left limit is a and the right limit is b, the limit can only exist if a = b, in which casef(x) = a for all x= 0 and the limit is a.EXERCISE SET 2.31. (a) (b) +2. (a) 2(b) 03. (a) 0(b) 14. (a) does not exist(b) 0 55. January 27, 2005 11:43 L24-ch02 Sheet number 6 Page number 54 black54 Chapter 25. (a) 12 (b) 21 (c) 15 (d) 25(e) 2 (f) 3/5 (g) 0(h) The limit doesnt exist because the denominator tends to zero but the numerator doesnt.6. (a) 20 (b) 0 (c) + (d) (e) (42)1/3 (f) 6/7 (g) 7 (h) 7/123 7. 8. + 9. + 10. + 11. 3/212. 5/2 13. 0 14. 0 15. 0 16. 5/317. 51/3/2 18. 3/2 19. 5 20.5 21. 1/66 23.22. 1/3 24.3 25. 26. +27. 1/7 28. 4/729. It appears that limt+n(t) = +, and limt+e(t) = c.30. (a) It is the initial temperature ofthe potato (400 F).(b) It is the ambient temperature, i.e. the temperature ofthe room.31. (a) + (b) 5 32. (a) 0 (b) 633. limx2 + 3 x)x+(x2 + 3 + xx2 +3+x= limx+3 x2 +3+x= 034. limx+x2 3x x)(x2 3x + xx2 3x + x= limx+3x x2 3x + x= 3/235. limx+ x2 + ax + xx2 + ax xx2 + ax + x= limx+ axx2 + ax + x= a/236. limx+x2 + ax x2 + bxx2 + ax +x2 + bxx2 + ax +x2 + bx= limx+ (a b)xx2 + ax +x2 + bx= a b237. limx+p(x) = (1)n and limxp(x) = +38. If mn the limits are both zero. If m = n the limits are both 1. If nm the limits are(1)n+m and +, respectively.39. If mn the limits are both zero. If m = n the limits are both equal to am, the leading coefficientof p. If nm the limits are where the sign depends on the sign of am and whether n is evenor odd.40. (a) p(x) = q(x) = x (b) p(x) = x, q(x) = x2(c) p(x) = x2, q(x) = x (d) p(x) = x + 3, q(x) = x41. If mn the limit is 0. If m = n the limit is 3. If mn and n m is odd, then the limit is+; if mn and n m is even, then the limit is .42. If mn the limit is zero. If m = n the limit is cm/dm. If nm the limit is + if cmdm0and if cmdm0. 56. January 27, 2005 11:43 L24-ch02 Sheet number 7 Page number 55 blackExercise Set 2.3 5543. + 44. 0 45. + 46. 047. 1 48. 1 49. 1 50. 151. 52. 53. 54. +55. 1 56. 1 57. + 58. 59. (a)4 8 12 16 202001601208040tv(b) limtv = 1901 limt0.168te= 190, so the asymptote is v = c = 190 ft/sec.(c) Due to air resistance (and other factors) this is the maximum speed that a sky diver canattain.60. (a) 50371.7/(151.3 + 181.626) 151.3 million(b)2000 2050 2100 2150350300250200tP(c) limtp(t) =50371.7151.33 + 181.626 limt e0.031636(t1950) =50371.7151.33 333 million(d) The population becomes stable at this number.61. limx+f(x) = limxf(x) = L62. (a) Make the substitution t = 1/x to see that they are equal.(b) Make the substitution t = 1/x to see that they are equal.63. x + 1x= 1+1x, so limx+(x + 1)xxx = e from Figure 1.6.6.64. If x is large enough, then 1 +1x 0, and so|x + 1|x|x|x =1 +1xx, hencelimx|x + 1|x|x|x = limx1 +1xx= e by Figure 1.6.6.65. Set t = x, then get limt1 +1tt= e by Figure 1.6.6. 57. January 27, 2005 11:43 L24-ch02 Sheet number 8 Page number 56 black56 Chapter 266. Set t = x then limt+1 +1tt= e67. Same as limx+1 1xx=1eby Exercise 65.68. Same as limt+|t 1|t|t|t =1eby Exercise 64.69. limx+x +2x3x limx+x3x which is clearly +.70. If x 1 then2x 2, consequently |x|+2x |x|2. But |x|2 gets arbitrarily large, and hence|x| 2x3xgets arbitrarily small, since the exponent is negative. Thus limx|x| +2x3x= 0.71. Set t = 1/x, then limt1 +1tt= e 72. Set t = 1/x then limt+1 +1tt= e73. Set t = 1/x, then limt+1 +1tt=1e74. Set x = 1/t, then limt1 +1tt=1e.75. Set t = 1/(2x), then limt+1 1t6t= e676. Set t = 1/(2x), then limt+1 +1t6t= e677. Set t = 1/(2x), then limt1 1t6t=1e678. Set t = 1/(2x), then limt+1 1t6t=1e679. f(x) = x+2+2x 2,so lim(f(x) (x+2)) = 0xand f(x) is asymptotic to y = x + 2.159312 6 3 9 153915xx = 2yy = x + 280. f(x) = x2 1 + 3/x,so limx[f(x) (x2 1) = 0]and f(x) is asymptotic to y = x2 1.5314 2 2 42xyy = x2 1 58. January 27, 2005 11:43 L24-ch02 Sheet number 9 Page number 57 blackExercise Set 2.4 5781. f(x) = x2 + 1 + 2/(x 3)so limx[f(x) (x2 +1)] = 0and f(x) is asymptotic to y = x2 + 1.1264 2 2 4612xx = 3yy = x2 + 182. f(x) = x3 +32(x 1) 32(x + 1)so limx[f(x) x3] = 0and f(x) is asymptotic to y = x3.1552 215xyy = x3x = 1x = 183. f(x) sin x = 0 and f(x) is asymptotic to y = sin x.534 2 84xyy = sin xx = 184. Note that the function is not defined for 1x 1. Forx outside this interval we have f(x) =x2 +2x 1whichsuggests that lim[f(x)|x| ] = 0 (this can be checked withx42a CAS) and hence f(x) is asymptotic to y = |x| . 4 2 2 424xyy = |x|x = 1EXERCISE SET 2.41. (a) |f(x) f(0)| = |x + 2 2| = |x|0.1 ifand only if |x|0.1(b) |f(x) f(3)| = |(4x 5) 7| = 4|x 3|0.1 ifand only if |x 3|(0.1)/4 = 0.0025(c) |f(x)f(4)| = |x216| if |x4|. We get f(x) = 16+ = 16.001 at x = 4.000124998,which corresponds to = 0.000124998; and f(x) = 16 = 15.999 at x = 3.999874998, forwhich = 0.000125002. Use the smaller : thus |f(x) 16| provided |x 4|0.000125(to six decimals).2. (a) |f(x) f(0)| = |2x + 3 3| = 2|x|0.1 ifand only if |x|0.05(b) |f(x) f(0)| = |2x + 3 3| = 2|x|0.01 ifand only if |x|0.005(c) |f(x) f(0)| = |2x + 3 3| = 2|x|0.0012 ifand only if |x|0.00063. (a) x1 = (1.95)2 = 3.8025, x2 = (2.05)2 = 4.2025(b) = min ( |4 3.8025|, |4 4.2025|) = 0.1975 59. January 27, 2005 11:43 L24-ch02 Sheet number 10 Page number 58 black58 Chapter 24. (a) x1 = 1/(1.1) = 0.909090 . . . , x2 = 1/(0.9) = 1.111111 . . .(b) = min( |1 0.909090|, |1 1.111111|) = 0.0909090 . . .5. |(x3 4x+5)2|0.05, 0.05(x3 4x+5)20.05,1.95x3 4x + 52.05; x3 4x + 5 = 1.95 atx = 1.0616, x3 4x + 5 = 2.05 at x = 0.9558; =min (1.0616 1, 1 0.9558) = 0.04422.20.9 1.11.96.5x+1 = 3.5 at x = 2.25,5x+1 = 4.5 at x = 3.85, so = min(3 2.25, 3.85 3) = 0.7552 407. With the TRACE feature of a calculator we discover that (to five decimal places) (0.87000, 1.80274)and (1.13000, 2.19301) belong to the graph. Set x0 = 0.87 and x1 = 1.13. Since the graph of f(x)rises from left to right, we see that if x0xx1 then 1.80274f(x)2.19301, and therefore1.8f(x)2.2. So we can take = 0.13.8. From a calculator plot we conjecture that limx0f(x) = 2. Using the TRACE feature we see thatthe points (0.2, 1.94709) belong to the graph. Thus if 0.2x0.2 then 1.95f(x) 2 andhence |f(x) L|0.050.1 = .9. |2x 8| = 2|x 4|0.1 if |x 4|0.05, = 0.0510. |5x 2 13| = 5|x 3|0.01 if |x 3|0.002, = 0.00211.x2 9x 3 6= |x + 3 6| = |x 3|0.05 if |x 3|0.05, = 0.0512.4x2 12x + 1+ 2= |2x 1 + 2| = |2x + 1|0.05 ifx +12 0.025, = 0.02513. On the interval [1, 3] we have |x2 +x+2| 14, so |x3 8| = |x2||x2 +x+2| 14|x2|0.001provided |x 2|0.001 114; but 0.00005 0.00114, so for convenience we take = 0.00005 (thereis no need to choose an optimal ).14. Sincex 2| =x0, ||x 4||x + 2| |x 4|20.001 if |x 4|0.002, = 0.00215. if 1 then |x|3, so1x 15=|x 5|5|x||x 5|150.05 if |x 5|0.75, = 3/416. |x 0| = |x|0.05 if |x|0.05, = 0.05 60. January 27, 2005 11:43 L24-ch02 Sheet number 11 Page number 59 blackExercise Set 2.4 5917. (a) limx4f(x) = 3(b) |10f(x) 30| = 10|f(x) 3|0.005 provided |f(x) 3|0.0005, which is true for|x 3|0.0001, = 0.000118. (a) limx3f(x) = 7; limx3g(x) = 5(b) |7f(x) 21|0.03 is equivalent to |f(x) 7|0.01, so let= 0.01 in condition (i): thenwhen |x 3| = 0.012 = 0.0001, it follows that |f(x) 7|0.01, or |3f(x) 21|0.03.19. It suffices to have |10f(x)+2x38| |10f(x)30|+2|x4|0.01, by the triangle inequality. Toensure |10f(x)30|0.005 use Exercise 17 (with= 0.0005) to get = 0.0001. Then |x4|yields |10f(x)+2x38| 10|f(x)3|+2|x4| (10)0.0005+(2)0.0001 0.005+0.00020.0120. Let = 0.0009. By the triangle inequality |3f(x) + g(x) 26| 3|f(x) 7| + |g(x) 5| 3 0.0009 + 0.0072 = 0.03 + 0.00720.06.21. |3x 15| = 3|x 5| if |x 5|13, = 1322. |7x + 5 + 2| = 7|x + 1| if |x + 1|17, = 1723.2x2 + xx 1= |2x| if |x|12, = 1224.x2 9x + 3 (6)= |x + 3| if |x + 3|, = 1225. |12f(x) 3| = |x + 2 3| = |x 1| if0|x 1|, = 26. |9 2x 5| = 2|x 2| if0|x 2|, = 27. (a) |(3x2 + 2x 20 300| = |3x2 + 2x 320| = |(3x + 32)(x 10)| = |3x + 32| |x 10|(b) If |x 10|1 then |3x + 32|65, since clearly x11(c) = min(1, /65); |3x + 32| |x 10|65 |x 10|65 /65 = 28. (a)283x + 1 4=28 12x 43x + 1=12x + 243x + 1=123x + 1 |x 2|(b) If |x2|4 then 2x6, so x can be very close to 1/3, hence123x + 1is not bounded.(c) If |x 2|1 then 1x3 and 3x + 14, so123x + 1124= 3(d) = min(1, /3);123x + 1 |x 2|3 |x 2|3 /3 = 29. if 1 then |2x2 2| = 2|x 1||x + 1|6|x 1| if |x 1|16 , = min(1, 16)30. If 1 then |x2 + x 12| = |x + 4| |x 3|5|x 3| if |x 3|/5, = min(1, 15)31. If 12and |x (2)| then 52 x32, x + 112, |x + 1|12; then1x + 1 (1)=|x + 2||x + 1|2|x + 2| if |x + 2| 12, = min12,1232. If 14then2x + 3x 8=|6x 3||x| 6|x 12|14= 24|x 12| if |x 12|/24, = min(14, 24 ) 61. January 27, 2005 11:43 L24-ch02 Sheet number 12 Page number 60 black60 Chapter 2x 2| =33. |x 2)(x + 2 x + 2=x 4 x + 212|x 4| if |x 4|2, = 234. If x2 then |f(x) 5| = |9 2x 5| = 2|x 2| if |x 2|12, 1 = 12. If x2 then|f(x) 5| = |3x 1 5| = 3|x 2| if |x 2|13, 2 = 13Now let = min(1, 2) then forany x with |x 2|, |f(x) 5|35. (a) |f(x) L| =1x20.1 if x 10, N =10(b) |f(x) L| =xx + 1 1=1x + 1 0.01 if x + 1100, N = 99(c) |f(x) L| =1x311000if |x|10, x10, N = 10(d) |f(x) L| =xx + 1 1=1x + 1 0.01 if |x + 1|100, x 1100, x101,N = 10136. (a)1x3 0.1, x101/3, N = 101/3 (b)1x3 0.01, x1001/3, N = 1001/3(c)1x3 0.001, x10, N = 1037. (a) x211 + x211 = 1 , x1 = ; x221 + x22= 1 , x2 =1 (b) N =1 1 (c) N = 38. (a) x1 = 1/3; x2 = 1/3 (b) N = 1/3 (c) N = 1/339.1x20.01 if |x|10, N = 1040.1x + 20.005 if |x + 2|200, x198, N = 19841.xx + 1 1=1x + 1 0.001 if |x + 1|1000, x999, N = 99942.4x 12x + 5 2=112x + 5 0.1 if |2x + 5|110, 2x105, N = 52.543.1x + 2 0 0.005 if |x + 2|200, x 2200, x202, N = 20244.1x2 0.01 if |x|10, x10, x10, N = 1045.4x 12x + 5 2=112x + 5 0.1 if |2x+5|110, 2x5110, 2x115, x57.5, N = 57.546.xx + 1 1=1x + 1 0.001 if |x + 1|1000, x 11000, x1001, N = 1001 62. January 27, 2005 11:43 L24-ch02 Sheet number 13 Page number 61 blackExercise Set 2.4 6147.1x2if |x| 1 , N =1 48.1x + 2if |x + 2| 1, x + 2 1, x 1 2, N =1 249.4x 12x + 5 2=112x + 5if |2x + 5| 11, 2x 5 11, 2x11 5, x112 52,N = 52 11250.xx + 1 1=1x + 1if |x + 1| 1, x 1 1, x1 1, N = 1 151.2 xx 1 2=2 x 1ifx 1 2,x1 +2, x1 +22, N1 +2252. 2x if xlog2 , Nlog2 53. (a)1x2100 if |x| 110(b)1|x 1|1000 if |x 1| 11000(c)1(x 3)21000 if |x 3| 11010(d) 1x410000 if x4 110000, |x| 11054. (a)1(x 1)210 ifand only if |x 1| 1 10(b)1(x 1)21000 ifand only if |x 1| 11010(c)1(x 1)2100000 ifand only if |x 1| 11010055. if M 0 then1(x 3)2 M, 0(x 3)2 1M, 0|x 3| 1 M, =1 M56. if M 0 then1(x 3)2 M, 0(x 3)2 1M, 0|x 3| 1 M, =1 M57. if M 0 then1|x| M, 0|x| 1M, =1M58. if M 0 then1|x 1| M, 0|x 1| 1M, =1Mx4 M, 0x4 1M59. if M 0 then 1, |x| 1(M)1/4 , =1(M)1/460. if M 0 then1x4 M, 0x4 1M, x 1M1/4 , =1M1/461. if x2 then |x + 1 3| = |x 2| = x 2 if2x2 + , = 62. if x1 then |3x+25| = |3x3| = 3|x1| = 3(1x) if1 x13, 1 13x1, = 13 63. January 27, 2005 11:43 L24-ch02 Sheet number 14 Page number 62 black62 Chapter 263. if x4 thenx 4 if x 42, 4x4 + 2, = 264. if x0 thenx if x2, 2x0, = 265. if x2 then |f(x) 2| = |x 2| = x 2 if2x2 + , = 66. if x2 then |f(x) 6| = |3x 6| = 3|x 2| = 3(2 x) if2 x13, 2 13x2, = 13 67. (a) if M 0 and x1 then11 xM, x 1 1M, 1x1 1M, = 1M(b) if M 0 and x1 then11 xM, 1 x 1M, 1 1M x1, =1M68. (a) if M 0 and x0 then1xM, x 1M, 0x 1M, =1M(b) if M 0 and x0 then1xM, x 1M,1M x0, = 1M69. (a) Given any M 0 there corresponds N 0 such that if xN then f(x)M, x + 1M,xM 1, N = M 1.(b) Given any M 0 there corresponds N 0 such that if xN then f(x)M, x + 1M,xM 1, N = M 1.70. (a) Given any M 0 there corresponds N 0 such that if xN then f(x)M, x2 3M,x M + 3, N =M + 3.(b) Given any M 0 there corresponds N 0 such that if xN then f(x)M, x3 +5M,x(M 5)1/3, N = (M 5)1/3.71. if 2 then |x 3|2, 2x 32, 1x5, and |x2 9| = |x+3||x 3|8|x 3| if|x 3|18, = min2, 1872. (a) We dont care about the value of f at x = a, because the limit is only concerned with valuesof x near a. The condition that f be defined for all x (except possibly x = a) is necessary,because ifsome points were excluded then the limit may not exist; for example, let f(x) = xif1 /x is not an integer and f(1/n) = 6. Then limx0f(x) does not exist but it would ifthepoints 1/n were excluded.(b) if x0 thenx is not defined (c) yes; if 0.01 then x0, sox is defined73. (a) 0.4 amperes (b) [0.3947, 0.4054] (c) 37.5 + ,37.5 (d) 0.0187 (e) It becomes infinite.EXERCISE SET 2.51. (a) no, x = 2 (b) no, x = 2 (c) no, x = 2 (d) yes(e) yes (f) yes2. (a) no, x = 2 (b) no, x = 2 (c) no, x = 2 (d) yes(e) no, x = 2 (f) yes3. (a) no, x = 1, 3 (b) yes (c) no, x = 1 (d) yes(e) no, x = 3 (f) yes 64. January 27, 2005 11:43 L24-ch02 Sheet number 15 Page number 63 blackExercise Set 2.5 634. (a) no, x = 3 (b) yes (c) yes (d) yes(e) no, x = 3 (f) yes5. (a) 3 (b) 3 6. 2/57. (a) yx3(b) yx11 3(c) yx111(d) yx2 38. f(x) = 1/x, g(x) = 65. 0 if x = 0sin1xif x= 09. (a) Ct1$42(b) One second could cost you one dollar.10. (a) no; disasters (war, flood, famine, pestilence, for example) can cause discontinuities(b) continuous(c) not usually continuous; see Exercise 9(d) continuous11. none 12. none 13. none 14. x = 2, 2 15. x = 0,1/216. none 17. x = 1, 0, 1 18. x = 4, 0 19. none 20. x = 1, 021. none; f(x) = 2x + 3 is continuous on x4 and f(x) = 7+16xis continuous on 4x;limx4f(x) = limx4+f(x) = f(4) = 11 so f is continuous at x = 422. limx1f(x) does not exist so f is discontinuous at x = 123. (a) f is continuous for x1, and for x1; limx1f(x) = 5, limx1+f(x) = k, so if k = 5 then f iscontinuous for all x(b) f is continuous for x2, and for x2; limx2f(x) = 4k, limx2+f(x) = 4+k, so if 4k = 4+k,k = 4/3 then f is continuous for all x 66. January 27, 2005 11:43 L24-ch02 Sheet number 16 Page number 64 black64 Chapter 224. (a) f is continuous for x3, and for x3; limx3f(x) = k/9, limx3+f(x) = 0, so if k = 0 then fis continuous for all x(b) f is continuous for x0, and for x0; limx0f(x) doesnt exist unless k = 0, and ifso thenlimx0 f(x) = +; limx0+f(x) = 9, so no value of k25. f is continuous for x1, 1x2 and x2; limx1f(x) = 4, limx1+f(x) = k, so k = 4 isrequired. Next, limx2f(x) = 3m+ k = 3m+4, limx2+f(x) = 9, so 3m+4 = 9,m = 5/3 and f iscontinuous everywhere if k = 4,m = 5/326. (a) no, f is not defined at x = 2 (b) no, f is not defined for x 2(c) yes (d) no, f is not defined for x 227. (a) yxc(b) yxc28. (a) f(c) = limxcf(x)(b) limx1f(x) = 2 limx1g(x) = 1yx11 1yx11(c) Define f(1) = 2 and redefine g(1) = 1.x0 f(x) = 1= +1 = limx0+29. (a) x = 0, limf(x) so the discontinuity is not removable(b) x = 3; define f(3) = 3 = limx3f(x), then the discontinuity is removable(c) f is undefined at x = 2; at x = 2, limx2f(x) = 1, so define f(2) = 1 and f becomescontinuous there; at x = 2, limx2does not exist, so the discontinuity is not removable30. (a) f is not defined at x = 2; limx2f(x) = limx2x + 2x2 + 2x + 4=13, so define f(2) =13and fbecomes continuous there(b) limx2f(x) = 1= 4 = limx2+f(x), so f has a nonremovable discontinuity at x = 2(c) limx1f(x) = 8= f(1), so f has a removable discontinuity at x = 1 67. January 27, 2005 11:43 L24-ch02 Sheet number 17 Page number 65 blackExercise Set 2.5 6531. (a) discontinuity at x = 1/2, not removable;at x = 3, removableyx555(b) 2x2 + 5x 3 = (2x 1)(x + 3)32. (a) there appears to be one discontinuitynear x = 1.5243 34(b) one discontinuity at x 1.5233. For x0, f(x) = x3/5 = (x3)1/5 is the composition (Theorem 2.5.6) ofthe two continuousfunctions g(x) = x3 and h(x) = x1/5 and is thus continuous. For x0, f(x) = f(x) which is thecomposition ofthe continuous functions f(x) (for positive x) and the continuous function y = x.Hence f(x) is continuous for all x0. At x = 0, f(0) = limx0f(x) = 0.34. x4 +7x2 +1 10, thus f(x) is the composition ofthe polynomial x4 +7x2 +1, the square rootx, and the function 1/x and is therefore continuous by Theorem 2.5.6.35. (a) Let f(x) = k for x= c and f(c) = 0; g(x) = l for x= c and g(c) = 0. If k = l then f +g iscontinuous; otherwise its not.(b) f(x) = k for x= c, f(c) = 1; g(x) = l= 0 for x= c, g(c) = 1. If kl = 1, then fg iscontinuous; otherwise its not.36. A rational function is the quotient f(x)/g(x) oft wo polynomials f(x) and g(x). By Theorem 2.5.2f and g are continuous everywhere; by Theorem 2.5.3 f/g is continuous except when g(x) = 0.37. Since f and g are continuous at x = c we know that limxcf(x) = f(c) and limxcg(x) = g(c). In thefollowing we use Theorem 2.2.2.(a) f(c) + g(c) = limxcf(x) + limxcg(x) = lim(f(x) + g(x)) so f + g is continuous at x = c.xc(b) same as (a) except the + sign becomes a sign(c) f(c) g(c) =limxcf(x)limxc= limg(x)xcf(x) g(x) so f g is continuous at x = c38. h(x) = f(x) g(x) satisfies h(a)0, h(b)0. Use the Intermediate Value Theorem or Theorem2.5.8.39. Ofcourse such a function must be discontinuous. Let f(x) = 1 on 0 x1, and f(x) = 1 on1 x 2. 68. January 27, 2005 11:43 L24-ch02 Sheet number 18 Page number 66 black66 Chapter 240. (a) (i) no (ii) yes (b) (i) no (ii) no (c) (i) no (ii) no41. The cone has volume r2h/3. The function V (r) = r2h (for variable r and fixed h) gives thevolume ofa right circular cylinder ofheigh t h and radius r, and satisfies V (0)r2h/3V(r).By the Intermediate Value Theorem there is a value c between 0 and r such that V (c) = r2h/3,so the cylinder ofradius c (and height h) has volume equal to that ofthe cone.42. A square whose diagonal has length r has area f(r) = r2/2. Note thatf(r) = r2/2r2/22r2 = f(2r). By the Intermediate Value Theorem there must be a valuec between r and 2r such that f(c) = r2/2, i.e. a square ofdiagonal c whose area is r2/2.43. If f(x) = x3 + x2 2x then f(1) = 2, f(1) = 0. Use the Intermediate Value Theorem.44. Since limxp(x) = and limx+p(x) = + (or vice versa, ifthe leading coefficient of p isnegative), it follows that for M = 1 there corresponds N10, and for M = 1 there is N20,such that p(x)1 f or xN1 and p(x)1 f or xN2. Choose x1N1 and x2N2 and useTheorem 2.5.8 on the interval [x1, x2] to find a solution of p(x) = 0.45. For the negative root, use intervals on the x-axis as follows: [2,1]; since f(1.3)0 andf(1.2)0, the midpoint x = 1.25 of[ 1.3,1.2] is the required approximation ofthe root.For the positive root use the interval [0, 1]; since f(0.7)0 and f(0.8)0, the midpoint x = 0.75of[0 .7, 0.8] is the required approximation.46. x = 1.22 and x = 0.72.102 1510.7 0.8147. For the negative root, use intervals on the x-axis as follows: [2,1]; since f(1.7)0 andf(1.6)0, use the interval [1.7,1.6]. Since f(1.61)0 and f(1.60)0 the midpointx = 1.605 of[ 1.61,1.60] is the required approximation ofthe root. For the positive root usethe interval [1, 2]; since f(1.3)0 and f(1.4)0, use the interval [1.3, 1.4]. Since f(1.37)0and f(1.38)0, the midpoint x = 1.375 of[1 .37, 1.38] is the required approximation.48. x = 1.603 and x = 1.3799.11.7 1.6211.3 1.40.549. x = 2.24 69. January 27, 2005 11:43 L24-ch02 Sheet number 19 Page number 67 blackExercise Set 2.6 6750. Set f(x) = ax 1+ bx 3. Since limx1+f(x) = + and limx3f(x) = there exist x11 andx23 (with x2x1) such that f(x)1 for 1xx1 and f(x)1 f or x2x3. Choosex3 in (1, x1) and x4 in (x2, 3) and apply Theorem 2.5.8 on [x3, x4].51. The uncoated sphere has volume 4(x 1)3/3 and the coated sphere has volume 4x3/3. Ifthevolume ofthe uncoated sphere and ofthe coating itselfare the same, then the coated sphere hastwice the volume ofthe uncoated sphere. Thus 2(4(x1)3/3) = 4x3/3, or x36x2+6x2 = 0,with the solution x = 4.847 cm.52. Let g(t) denote the altitude ofthe monk at time t measured in hours from noon of day one, andlet f(t) denote the altitude ofthe monk at time t measured in hours from noon of day two. Theng(0)f(0) and g(12)f(12). Use Exercise 38.53. We must show limxcf(x) = f(c). Let 0; then there exists 0 such that if |x c| then|f(x) f(c)|. But this certainly satisfies Definition 2.4.1.54. (a) yx10.40.2 0.8(b) Let g(x) = xf(x). Then g(1) 0 and g(0) 0; by the Intermediate Value Theorem thereis a solution c in [0, 1] of g(c) = 0.EXERCISE SET 2.61. none 2. x = 3. n, n = 0,1,2, . . .4. x = n + /2, n = 0,1,2, . . . 5. x = n, n = 0,1,2, . . .6. none 7. 2n + /6, 2n + 5/6, n = 0,1,2, . . .8. x = n + /2, n = 0,1,2, . . . 9. [1, 1]10. (,1] [1,) 11. (0, 3) (3,+)12. (, 0) (0,+), and if f is defined to be e at x = 0, then continuous for all x13. (,1] [1,) 14. (3, 0) (0,)15. (a) sin x, x3 + 7x + 1 (b) |x|, sin x (c) x3, cos x, x + 1(d)x,3 + x, sin x, 2x (e) sin x, sin x (f) x5 2x3 + 1, cos x16. (a) Use Theorem 2.5.6. (b) g(x) = cos x, g(x) =1x2 + 1, g(x) = x2 + 117. coslimx+1x= cos0 = 1 18. sinlimx+x2 3x= sin3= 3219. sin1limx+x1 2x= sin112= 6 70. January 27, 2005 11:43 L24-ch02 Sheet number 20 Page number 68 black68 Chapter 220. lnlimx+x + 1x= ln(1) = 021. 3 lim0sin 33= 3 22.12limh0sin hh=1223.lim0+1lim0+sin = + 24.lim0sin lim0sin = 025.tan 7xsin 3x=73 cos 7xsin 7x7x3xsin 3xso limx0tan 7xsin 3x=73(1)(1)(1) =7326.sin 6xsin 8x=68sin 6x6x8xsin 8x, so limx0sin 6xsin 8x=68=3427.15limx0+x limx0+sin xx= 0 28.13limx0sin xx2=1329.limx0xlimx0sin x2x2= 030.sin h1 cos h=sin h1 cos h1 + cos h1 + cos h=sin h(1 + cos h)1 cos2 h=1 + cos hsin h; no limit31. t21 cos2 t=tsin t2, so limt0t21 cos2 t= 132. cos( 12 x) = sin(12) sin x = sin x, so limx0xcos12 = 1 x33. 21 cos 1 + cos 1 + cos = 2(1 + cos )1 cos2 =sin 2(1 + cos ) so lim021 cos = (1)22 = 234.1 cos 3hcos2 5h 11 + cos 3h1 + cos 3h=sin2 3hsin2 5h11 + cos 3h, solimx01 cos 3hcos2 5h 1= limx0sin2 3hsin2 5h11 + cos 3h= 352 12= 95035. limx0+sin1x= limt+sin t; limit does not exist36. limx0x 3 limx0sin xx= 337.2 cos 3x cos 4xx=1 cos 3xx+1 cos 4xx. Note that1 cos 3xx=1 cos 3xx1 + cos 3x1 + cos 3x=sin2 3xx(1 + cos 3x)=sin 3xxsin 3x1 + cos 3x. Thuslimx02 cos 3x cos 4xx= limx0sin 3xxsin 3x1 + cos 3x+ limx0sin 4xxsin 4x1 + cos 4x= 0+0 = 0 71. January 27, 2005 11:43 L24-ch02 Sheet number 21 Page number 69 blackExercise Set 2.6 6938.tan 3x2 + sin2 5xx2 =3cos 3x2sin 3x23x2 + 52 sin2 5x(5x)2 , solimit = limx03cos 3x2 limx0sin 3x23x2 + 25 limx0sin 5x5x2= 3+25 = 2839. a/b 40. k2s41. 5.1 5.01 5.001 5.0001 5.00001 4.9 4.99 4.999 4.9999 4.999990.098845 0.099898 0.99990 0.099999 0.100000 0.10084 0.10010 0.10001 0.10000 0.10000The limit is 0.1.42. 2.1 2.01 2.001 2.0001 2.00001 1.9 1.99 1.999 1.9999 1.999990.484559 0.498720 0.499875 0.499987 0.499999 0.509409 0.501220 0.500125 0.500012 0.500001The limit is 0.5.43. 1.9 1.99 1.999 1.9999 1.99999 2.1 2.01 2.001 2.0001 2.000010.898785 0.989984 0.999000 0.999900 0.999990 1.097783 1.009983 1.001000 1.000100 1.000010The limit is 1.44. 0.9 0.99 0.999 0.9999 0.99999 1.1 1.01 1.001 1.0001 1.000010.405086 0.340050 0.334001 0.333400 0.333340 0.271536 0.326717 0.332667 0.333267 0.333327The limit is 1/3.45. Since limx0sin(1/x) does not exist, no conclusions can be drawn.46. k = f(0) = limx0sin 3xx= 3 limx0sin 3x3x= 3, so k = 347. limx0 f(x) = k limx0sin kxkx cos kx= k, limx0+f(x) = 2k2, so k = 2k2, k =1248. No; sin x/|x| has unequal one-sided limits.49. (a) limt0+sin tt= 1 (b) limt01 cos tt= 0 (Theorem 2.6.4)(c) sin( t) = sin t, so limx xsin x= limt0tsin t= 150. cos2 t= sin t, so limx2cos(/x)x 2= limt0( 2t) sin t4t= limt0 2t4limt0sin tt= 451. t = x 1; sin(x) = sin(t + ) = sin t; and limx1sin(x)x 1= limt0sin tt= 52. t = x /4; tan x 1 =2 sin tcos t sin t; limx/4tan x 1x /4= limt02 sin tt(cos t sin t)= 253. t = x /4,cos x sin xx /4= 2 sin tt; limx/4cos x sin xx /42 lim= t0sin tt2= 72. January 27, 2005 11:43 L24-ch02 Sheet number 22 Page number 70 black70 Chapter 254. limx0h(x) = L = h(0) so h is continuous at x = 0.Apply the Theorem to h g to obtain on the one hand h(g(0)) = L, and on the otherh(g(x)) =f(g(x))g(x) , x= 0L, x = 0Since f(g(x)) = x and g = f1 this shows that limt0xf1(x)= L55. limx0xsin1 x= limx0sin xx= 156. tan(tan1 x) = x, so limx0tan1 xx= limx0xtan x= ( limx0cos x) limx0xsin x= 157. 5 limx0sin1 x5x= 5 limx05xsin 5x= 558. limx11x + 1limx1sin1(x 1)x 1=12limx1x 1sin(x 1)=1259. 3 limx0e3x 13x= 360. With y = ln(1 + x), eln(1+x) = 1+x, x = ey 1, so limx0ln(1 + x)x= limx0xex 1= 161. ( limx0x) limx0ex2 1x2 = 0 1 = 0 62. 5 limx0ln(1 + 5x)5x= 5 1 = 5 (Exercise 60)63. |x| x cos50x |x| 64. x2 x2 sin503 x x265. limx0f(x) = 1 by the Squeezing Theorem11 101xyy = cos xy = f (x)y = 1 x266. limx+f(x) = 0 by the Squeezing Theoremyx1467. Let g(x) = 1xand h(x) =1x; thus limx+sin xx= 0 by the Squeezing Theorem.68. yxyx 73. January 27, 2005 11:43 L24-ch02 Sheet number 23 Page number 71 blackExercise Set 2.6 7169. (a) sin x = sin t where x is measured in degrees, t is measured in radians and t = x180. Thuslimx0sin xx= limt0sin t(180t/)= 180.70. cos x = cos t where x is measured in degrees, t in radians, and t = x180. Thuslimx01 cos xx= limt01 cos t(180t/)= 0.71. (a) sin 10 = 0.17365 (b) sin 10 = sin 18 18= 0.1745372. (a) cos = cos 2 = 1 2 sin2(/2) 1 2(/2)2 = 1 12 2(b) cos 10 = 0.98481(c) cos 10 = 1 12 182 0.9847773. (a) 0.08749 (b) tan 5 36= 0.0872774. (a) h = 52.55 ft(b) Since is small, tan 180is a good approximation.(c) h 52.36 ft75. (a) Let f(x) = x cos x; f(0) = 1, f (/2) = /2. By the IVT there must be a solution off(x) = 0.(b) yx1.510.50y = xy = cos xc/2(c) 0.73976. (a) f(x) = x + sin x 1; f(0) = 1, f (/6) = /6 1/20. By the IVT there must be asolution of f(x) = 0 in the interval.(b) yy = 1 sin x (c) x = 0.511x0.50c/6y = x 74. January 27, 2005 11:43 L24-ch02 Sheet number 24 Page number 72 black72 Chapter 277. (a) Gravity is stronger at the poles than at the equator.30 60 909.849.829.80fg(b) Let g() be the given function. Then g(38)9.8 and g(39)9.8, so by the IntermediateValue Theorem there is a value c between 38 and 39 for which g(c) = 9.8 exactly.78. (a) does not exist (b) the limit is zero(c) For part (a) consider the fact that given any 0 there are infinitely many rational numbersx satisfying |x| and there are infinitely many irrational numbers satisfying the samecondition. Thus ifthe limit were to exist, it could not be zero because ofthe rational numbers,and it could not be 1 because ofthe irrational numbers, and it could not be anything elsebecause of all the numbers. Hence the limit cannot exist. For part (b) use the SqueezingTheorem with +x and x as the squeezers.REVIEW EXERCISES, CHAPTER 21. (a) 1 (b) no limit (c) no limit(d) 1 (e) 3 (f) 0(g) 0 (h) 2 (i) 1/22. (a) 0.222 . . . , 0.24390, 0.24938, 0.24994, 0.24999, 0.25000; for x= 2, f(x) =1x + 2,so the limit is 1/4.(b) 1.15782, 4.22793, 4.00213, 4.00002, 4.00000, 4.00000; to prove,usetan 4xx=sin 4xx cos 4x=4cos 4xsin 4x4x, the limit is 4.3. (a) x 1 0.1 0.01 0.001 0.0001 0.00001 0.000001f(x) 1.000 0.443 0.409 0.406 0.406 0.405 0.405(b) yx0.51 14. x 3.1 3.01 3.001 3.0001 3.00001 3.000001f(x) 5.74 5.56 5.547 5.545 5.5452 5.54518A CAS yields 5.545177445 75. January 27, 2005 11:43 L24-ch02 Sheet number 25 Page number 73 blackReview Exercises, Chapter 2 735. 1 6. For x= 1,x3 x2x 1= x2, so limx1x3 x2x 1= 17. If x= 3 then3x + 9x2 + 4x + 3=3x + 1with limit 328. 9.253=32310.x2 + 4 2x2x2 + 4 + 2 x2 + 4 + 2= x2x2(x2 + 4 + 2)=1 x2 + 4 + 2, solimx0x2 + 4 2x2 = limx01 x2 + 4 + 2=1411. (a) y = 0 (b) none (c) y = 212. (a)5, no limit,10,10, no limit, +, no limit(b) 1, +1,1,1, no limit, 1, +113. 1 14. 2 15. 3 k16. lim0tan1 cos = tanlim01 cos = tanlim01 cos2 (1 + cos )= tan0 = 017. + 18. ln(2 sin cos ) ln tan = ln 2 + 2 ln cos so the limit is ln 2.19.1 +3xx=1 +3xx/3(3)so the limit is e320.1 + axbx= 1 + axx/a(ab)so the limit is eab21. $2,001.60, $2,009.66, $2,013.62, $2013.7523. (a) f(x) = 2x/(x 1) (b) yx101024. Given any window ofheigh t 2 centered at the point x = a, y = L there exists a width 2 suchthat the window ofwidth 2 and height 2 contains all points ofthe graph ofthe function for x inthat interval.25. (a) limx2f(x) = 5 (b) 0.004526. 0.07747 (use a graphing utility) 76. January 27, 2005 11:43 L24-ch02 Sheet number 26 Page number 74 black74 Chapter 227. (a) |4x 7 1|0.01, 4|x 2|0.01, |x 2|0.0025, let = 0.0025(b)4x2 92x 3 6 0.05, |2x + 3 6|0.05, |x 1.5|0.025, take = 0.025(c) |x2 16|0.001; if 1 then |x + 4|9 if |x 4|1; then |x2 16| = |x 4||x + 4| 9|x 4|0.001 provided |x 4|0.001/9, take = 0.0001, then |x2 16|9|x 4| 9(0.0001) = 0.00090.00128. (a) Given 0 then |4x 7 1| provided |x 2|/4, take = /4(b) Given 0 the inequality4x2 92x 3 6holds if |2x + 3 6|, |x 1.5|/2, take = /229. Let= f(x0)/20; then there corresponds 0 such that if |xx0| then |f(x)f(x0)|,f(x) f(x0), f(x)f(x0) = f(x0)/20 f or x0 xx0 + .30. (a) x 1.1 1.01 1.001 1.0001 1.00001 1.000001f(x) 0.49 0.54 0.540 0.5403 0.54030 0.54030(b) cos 131. (a) f is not defined at x = 1, continuous elsewhere(b) none(c) f is not defined at x = 0,332. (a) continuous everywhere except x = 3(b) defined and continuous for x 1, x 1(c) continuous for x033. For x2 f is a polynomial and is continuous; for x2 f is a polynomial and is continuous. Atx = 2, f(2) = 13= 13 = limx2+f(x) so f is not continuous there.35. f(x) = 1 f or a x a + b2and f(x) = 1 f or a + b2 x b36. If, on the contrary, f(x0)0 for some x0 in [0, 1], then by the Intermediate Value Theorem wewould have a solution of f(x) = 0 in [0, x0], contrary to the hypothesis.37. f(6) = 185, f(0) = 1, f(2) = 65; apply Theorem 2.4.8 twice, once on [6, 0] and once on [0, 2] 77. January 27, 2005 11:43 L24-ch03 Sheet number 1 Page number 75 blackCHAPTER 3The Derivative75EXERCISE SET 3.11. (a) mtan = (50 10)/(15 5)= 40/10= 4 m/s(b)t (s)410 20v (m/s)2. (a) mtan (90 0)/(10 2)= 90/8= 11.25 m/s(b) mtan (140 0)/(10 4)= 140/6 23.33 m/s3. (a) mtan = (600 0)/(20 2.2)= 600/17.8 33.71 m/s(b) mtan (820 600)/(20 16)= 220/4= 55 m/ sThe speed is increasing with time.4. (a) (10 10)/(3 0) = 0 cm/s(b) t = 0, t = 2, and t = 4.2 (horizontal tangent line)(c) maximum: t = 1 (slope0) minimum: t = 3 (slope0)(d) (3 18)/(4 2) = 7.5 cm/s (slope of estimated tangent line to curve at t = 3)5. From the figure:tst0 t1 t2(a) The particle is moving faster at time t0 because the slope of the tangent to the curve at t0 isgreater than that at t2.(b) The initial velocity is 0 because the slope of a horizontal line is 0.(c) The particle is speeding up because the slope increases as t increases from t0 to t1.(d) The particle is slowing down because the slope decreases as t increases from t1 to t2.6.tst0 t17. It is a straight line with slope equal to the velocity.8. (a) decreasing (slope of tangent line decreases with increasing time)(b) increasing (slope of tangent line increases with increasing time)(c) increasing (slope of tangent line increases with increasing time)(d) decreasing (slope of tangent line decreases with increasing time) 78. January 27, 2005 11:43 L24-ch03 Sheet number 2 Page number 76 black76 Chapter 39. (a) msec = f(1) f(0)1 0=21= 2(b) mtan = limx10f(x1) f(0)x1 0= limx102x21 0x1 0= limx102x1 = 0(c) mtan = limx1x0f(x1) f(x0)x1 x0= limx1x02x21 2x20x1 x0= limx1x0(2x1 + 2x0)= 4x0(d)3212 1 1 21xySecantTangent10. (a) msec = f(2) f(1)2 1=23 131= 7(b) mtan = limx11f(x1) f(1)x1 1= limx11x31 1x1 1= limx11(x1 1)(x21+ x1 + 1)x1 1= limx11(x21+ x1 +1) = 3(c) mtan = limx1x0f(x1) f(x0)x1 x0= limx1x0x31 x30x1 x0= limx1x0(x21+ x1x0 + x20)= 3x20(d)TangentxySecant5911. (a) msec = f(3) f(2)3 2=1/3 1/21= 16(b) mtan = limx12f(x1) f(2)x1 2= limx121/x1 1/2x1 2= limx122 x12x1(x1 2)= limx1212x1= 14(c) mtan = limx1x0f(x1) f(x0)x1 x0= limx1x01/x1 1/x0x1 x0= limx1x0x0 x1x0x1(x1 x0)= limx1x01x0x1= 1x20(d)xySecant1 Tangent4 79. January 27, 2005 11:43 L24-ch03 Sheet number 3 Page number 77 blackExercise Set 3.1 7712. (a) msec = f(2) f(1)2 1=1/4 11= 34(b) mtan = limx11f(x1) f(1)x1 1= limx111/x21 1x1 1= limx111 x21x21(x1 1)= limx11(x1 + 1)x21= 2(c) mtan = limx1x0f(x1) f(x0)x1 x0= limx1x01/x21 1/x20x1 x0= limx1x0x20 x21x20x21(x1 x0)= limx1x0(x1 + x0)x20x21= 2x30(d)xyTangent Secant1213. (a) mtan = limx1x0f(x1) f(x0)x1 x0= limx1x0(x21 1) (x20 1)x1 x0= limx1x0(x21 x20)x1 x0= limx1x0(x1 + x0) = 2x0(b) mtan = 2(1) = 214. (a) mtan = limx1x0f(x1) f(x0)x1 x0= limx1x0(x21+ 3x1 + 2) (x20+ 3x0 + 2)x1 x0= limx1x0(x21 x20) + 3(x1 x0)x1 x0= limx1x0(x1 + x0 +3) = 2x0 + 3(b) mtan = 2(2)+3 = 715. (a) mtan = limx1x0f(x1) f(x0)x1 x0= limx1x0x1 x0x1 x0= limx1x01 x1 +x0=12x0(b) mtan =121=1216. (a) mtan = limx1x0f(x1) f(x0)x1 x0= limx1x0x1 1/1/x0x1 x0= limx1x0x0 x1x0x1 (x1 x0)= limx1x01 x0x1 +x1 (x0 )= 13/202x(b) mtan = 12(4)3/2 = 11617. (a) 72F at about 4:30 P.M. (b) about (67 43)/6 = 4F/h(c) decreasing most rapidly at about 9 P.M.; rate of change of temperature is about 7F/h(slope of estimated tangent line to curve at 9 P.M.) 80. January 27, 2005 11:43 L24-ch03 Sheet number 4 Page number 78 black78 Chapter 318. For V = 10 the slope of the tangent line is about 0.25 atm/L, for V = 25 the slope isabout 0.04 atm/L.19. (a) during the first year after birth(b) about 6 cm/year (slope of estimated tangent line at age 5)(c) the growth rate is greatest at about age 14; about 10 cm/