1 FUNCTIONS AND MODELS1.1 Four Ways to Represent a FunctionIn exercises requiring estimations or approximations, your answers may vary slightly from the answers given here.1. (a) The point (31. 32) is on the graph of 1, so 1(31) = 32.(b) When r = 2, j is about 2.8, so 1(2) E 2.8.(c) 1(r) = 2 is equivalent to j = 2. When j = 2, we have r = 33 and r = 1.(d) Reasonable estimates for r when j = 0 are r = 32.5 and r = 0.3.(e) The domain of 1 consists of all r-values on the graph of 1. For this function, the domain is 33 $ r $ 3, or [33. 3].The range of 1 consists of all j-values on the graph of 1. For this function, the range is 32 $ j $ 3, or [32. 3].(f ) As r increases from 31 to 3, j increases from 32 to 3. Thus, 1 is increasing on the interval [31. 3].3. From Figure 1 in the text, the lowest point occurs at about (t. o) = (12. 385). The highest point occurs at about (17. 115).Thus, the range of the vertical ground acceleration is 385 $ o $ 115. Written in interval notation, we get [385. 115].5. No, the curve is not the graph of a function because a vertical line intersects the curve more than once. Hence, the curve failsthe Vertical Line Test.7. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [33. 2] and the rangeis [33. 32)[31. 3].9. The persons weight increased to about 160 pounds at age 20 and stayed fairly steady for 10 years. The persons weightdropped to about 120 pounds for the next 5 years, then increased rapidly to about 170 pounds. The next 30 years saw a gradualincrease to 190 pounds. Possible reasons for the drop in weight at 30 years of age: diet, exercise, health problems.11. The water will cool down almost to freezing as the icemelts. Then, when the ice has melted, the water willslowly warm up to room temperature.13. Of course, this graph depends strongly on thegeographical location!15. As the price increases, the amount sold decreases. 17.910 CHAPTER 1 FUNCTIONS AND MODELS19. (a) (b) From the graph, we estimate the number ofcell-phone subscribers worldwide to be about92 million in 1995 and 485 million in 1999.21. 1(r) = 3r23r + 2.1(2) = 3(2)232 + 2 = 12 32 + 2 = 12.1(32) = 3(32)23(32) + 2 = 12 + 2 + 2 = 16.1(o) = 3o23o + 2.1(3o) = 3(3o)23(3o) + 2 = 3o2+o + 2.1(o + 1) = 3(o + 1)23(o + 1) + 2 = 3(o2+ 2o + 1) 3o 31 + 2 = 3o2+ 6o + 3 3o + 1 = 3o2+ 5o + 4.21(o) = 2 1(o) = 2(3o23o + 2) = 6o232o + 4.1(2o) = 3(2o)23(2o) + 2 = 3(4o2) 32o + 2 = 12o232o + 2.1(o2) = 3(o2)23(o2) + 2 = 3(o4) 3o2+ 2 = 3o43o2+ 2.[1(o)]2= 3o23o + 22= 3o23o + 2

3o23o + 2

= 9o433o3+ 6o233o3+o232o + 6o232o + 4 = 9o436o3+ 13o234o + 4.1(o +/) = 3(o +/)23(o +/) + 2 = 3(o2+ 2o/ +/2) 3o 3/ + 2 = 3o2+ 6o/ + 3/23o 3/ + 2.23. 1(r) = 4 + 3r 3r2, so 1(3 +/) = 4 + 3(3 +/) 3(3 +/)2= 4 + 9 + 3/ 3(9 + 6/ +/2) = 4 33/ 3/2,and 1(3 +/) 31(3)/= (4 33/ 3/2) 34/= /(33 3/)/= 33 3/.25. 1(r) 31(o)r 3o=1r 3 1or 3o=o 3rror 3o =o 3rro(r 3o) = 31(r 3o)ro(r 3o) = 3 1or27. 1(r) = r(3r 31) is dened for all r except when 0 = 3r 31Cr =13, so the domainis r M R | r 6=13= 3". 13

13. "

.29. 1(t) =It +3It is dened when t D 0. These values of t give real number results forIt, whereas any value of t gives a realnumber result for3It. The domain is [0. ").31. /(r) = 1

4Ir235r is dened when r235r0 C r(r 35)0. Note that r235r 6= 0 since that would result indivision by zero. The expression r(r 35) is positive if r < 0 or r5. (See Appendix A for methods for solvinginequalities.) Thus, the domain is (3". 0)(5. ").SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION 1133. 1(r) = 5 is dened for all real numbers, so the domain is R, or (3". ").The graph of 1 is a horizontal line with j-intercept 5.35. 1(t) = t236t is dened for all real numbers, so the domain is R, or(3". "). The graph of 1 is a parabola opening upward since the coefcientof t2is positive. To nd the t-intercepts, let j = 0 and solve for t.0 = t236t = t(t 36) i t = 0 and t = 6. The t-coordinate of thevertex is halfway between the t-intercepts, that is, at t = 3. Since1(3) = 3236 3 = 39, the vertex is (3. 39).37. o(r) =Ir 35 is dened when r 35 D 0 or r D 5, so the domain is [5. ").Since j =Ir 35i j2= r 35 i r = j2+5, we see that o is thetop half of a parabola.39. G(r) = 3r + |r|r. Since |r| =

r if r D 03r if r < 0 , we haveG(r) =

3r +rrif r03r 3rrif r < 0 =

4rrif r02rrif r < 0 =

4 if r02 if r < 0Note that G is not dened for r = 0. The domain is (3". 0)(0. ").41. 1(r) =

r + 2 if r < 01 3r if r D 0The domain is R.43. 1(r) =

r + 2 if r $ 31r2if r31Note that for r = 31, both r + 2 and r2are equal to 1.The domain is R.45. Recall that the slope i of a line between the two points (r1. j1) and (r2. j2) is i = j23j1r23r1 and an equation of the lineconnecting those two points is j 3j1 = i(r 3r1). The slope of this line segment is 7 3(33)5 31= 52, so an equation isj 3(33) =52(r 31). The function is 1(r) =52r 3 112 , 1 $ r $ 5.12 CHAPTER 1 FUNCTIONS AND MODELS47. We need to solve the given equation for j. r + (j 31)2= 0C(j 31)2= 3r Cj 31 = I3r Cj = 1 I3r. The expression with the positive radical represents the top half of the parabola, and the one with the negativeradical represents the bottom half. Hence, we want 1(r) = 1 3I3r. Note that the domain is r $ 0.49. For 0 $ r $ 3, the graph is the line with slope 31 and j-intercept 3, that is, j = 3r + 3. For 3 < r $ 5, the graph is the linewith slope 2 passing through (3. 0); that is, j 30 = 2(r 33), or j = 2r 36. So the function is1(r) =

3r + 3 if 0 $ r $ 32r 36 if 3 < r $ 551. Let the length and width of the rectangle be 1 and W. Then the perimeter is 21 + 2W = 20 and the area is = 1W.Solving the rst equation for W in terms of 1 gives W = 20 3212= 10 31. Thus, (1) = 1(10 31) = 101312. Sincelengths are positive, the domain of is 0 < 1 < 10. If we further restrict 1 to be larger than W, then 5 < 1 < 10 would bethe domain.53. Let the length of a side of the equilateral triangle be r. Then by the Pythagorean Theorem, the height j of the triangle satisesj2+

12r

2= r2, so that j2= r23 14r2=34r2and j = I32 r. Using the formula for the area of a triangle, =12(base)(height), we obtain (r) =12(r)

I32 r

= I34 r2, with domain r0.55. Let each side of the base of the box have length r, and let the height of the box be /. Since the volume is 2, we know that2 = /r2, so that / = 2r2, and the surface area is o = r2+ 4r/. Thus, o(r) = r2+ 4r(2r2) = r2+ (8r), withdomain r0.57. The height of the box is r and the length and width are 1 = 20 32r, W = 12 32r. Then \ = 1Wr and so\ (r) = (20 32r)(12 32r)(r) = 4(10 3r)(6 3r)(r) = 4r(60 316r +r2) = 4r3364r2+ 240r.The sides 1, W, and r must be positive. Thus, 10C 20 32r0Cr < 10;W0C12 32r0Cr < 6; and r0. Combining these restrictions gives us the domain 0 < r < 6.59. (a) (b) On $14,000, tax is assessed on $4000, and 10%($4000) = $400.On $26,000, tax is assessed on $16,000, and10%($10,000) + 15%($6000) = $1000 + $900 = $1900.(c) As in part (b), there is $1000 tax assessed on $20,000 of income, sothe graph of T is a line segment from (10,000. 0) to (20,000. 1000).The tax on $30,000 is $2500, so the graph of T for r20,000 isthe ray with initial point (20,000. 1000) that passes through(30,000. 2500).SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS1361. 1 is an odd function because its graph is symmetric about the origin. o is an even function because its graph is symmetric withrespect to the j-axis.63. (a) Because an even function is symmetric with respect to the j-axis, and the point (5. 3) is on the graph of this even function,the point (35. 3) must also be on its graph.(b) Because an odd function is symmetric with respect to the origin, and the point (5. 3) is on the graph of this odd function,the point (35. 33) must also be on its graph.65. 1(r) =rr2 + 1.1(3r) =3r(3r)2 + 1 =3rr2 + 1 = 3rr2 + 1 = 31(r).So 1 is an odd function.67. 1(r) =rr + 1, so 1(3r) =3r3r + 1 =rr 31.Since this is neither 1(r) nor 31(r), the function 1 isneither even nor odd.69. 1(r) = 1 + 3r23r4.1(3r) = 1 + 3(3r)23(3r)4= 1 + 3r23r4= 1(r).So 1 is an even function.1.2 Mathematical Models: A Catalog of Essential Functions1. (a) 1(r) =5Ir is a root function with n = 5.(b) o(r) =I1 3r2 is an algebraic function because it is a root of a polynomial.(c) /(r) = r9+r4is a polynomial of degree 9.(d) v(r) = r2+ 1r3 +r is a rational function because it is a ratio of polynomials.(e) c(r) = tan2r is a trigonometric function.(f ) t(r) = log10r is a logarithmic function.3. We notice from the gure that o and / are even functions (symmetric with respect to the j-axis) and that 1 is an odd function(symmetric with respect to the origin). So (b)

j = r5must be 1. Since o is atter than / near the origin, we must have(c)

j = r8matched with o and (a)

j = r2matched with /.14 CHAPTER 1 FUNCTIONS AND MODELS5. (a) An equation for the family of linear functions with slope 2is j = 1(r) = 2r +/, where / is the j-intercept.(b) 1(2) = 1 means that the point (2. 1) is on the graph of 1. We can use thepoint-slope form of a line to obtain an equation for the family of linearfunctions through the point (2. 1). j 31 = i(r 32), which is equivalentto j = ir + (1 32i) in slope-intercept form.(c) To belong to both families, an equation must have slope i = 2, so the equation in part (b), j = ir + (1 32i),becomes j = 2r 33. It is the only function that belongs to both families.7. All members of the family of linear functions 1(r) = c 3r have graphsthat are lines with slope 31. The j-intercept is c.9. Since 1(31) = 1(0) = 1(2) = 0, 1 has zeros of 31, 0, and 2, so an equation for 1 is 1(r) = o[r 3(31)](r 30)(r 32),or 1(r) = or(r + 1)(r 32). Because 1(1) = 6, well substitute 1 for rand 6 for 1(r).6 = o(1)(2)(31)i 32o = 6io = 33, so an equation for 1 is 1(r) = 33r(r + 1)(r 32).11. (a) 1 = 200, so c = 0.04171(o + 1) = 0.0417(200)(o + 1) = 8.34o + 8.34. The slope is 8.34, which represents thechange in mg of the dosage for a child for each change of 1 year in age.(b) For a newborn, o = 0, so c = 8.34 mg.13. (a) (b) The slope of 95 means that 1 increases 95 degrees for each increaseof 1

C. (Equivalently, 1 increases by 9 when C increases by 5and 1 decreases by 9 when C decreases by 5.) The 1-intercept of32 is the Fahrenheit temperature corresponding to a Celsiustemperature of 0.SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS 1515. (a) Using ` in place of r and T in place of j, we nd the slope to beT23T1`23`1 =80 370173 3113 = 1060 = 16. So a linearequation is T 380 =16(` 3173)CT 380 =16` 3 1736CT =16` + 3076

3076= 51.16.(b) The slope of 16 means that the temperature in Fahrenheit degrees increases one-sixth as rapidly as the number of cricketchirps per minute. Said differently, each increase of 6 cricket chirps per minute corresponds to an increase of 1

F.(c) When ` = 150, the temperature is given approximately by T =16(150) + 3076= 76.16

F E 76

F.17. (a) We are givenchange in pressure10 feet change in depth = 4.3410= 0.434. Using 1 for pressure and d for depth with the point(d. 1) = (0. 15), we have the slope-intercept form of the line, 1 = 0.434d + 15.(b) When 1 = 100, then 100 = 0.434d + 15C0.434d = 85Cd =850.434 E 195.85 feet. Thus, the pressure is100 lbin2at a depth of approximately 196 feet.19. (a) The data appear to be periodic and a sine or cosine function would make the best model. A model of the form1(r) = o cos(/r) +c seems appropriate.(b) The data appear to be decreasing in a linear fashion. A model of the form1(r) = ir +/ seems appropriate.Some values are given to many decimal places. These are the results given by several computer algebra systems rounding is left to the reader.21. (a)A linear model does seem appropriate.(b) Using the points (4000. 14.1) and (60,000. 8.2), we obtainj 314.1 =8.2 314.160,000 34000 (r 34000) or, equivalently,j E 30.000105357r + 14.521429.(c) Using a computing device, we obtain the least squares regression line j = 30.0000997855r + 13.950764.The following commands and screens illustrate how to nd the least squares regression line on a TI-83 Plus.Enter the data into list one (L1) and list two (L2). Press to enter the editor.Find the regession line and store it in Y1. Press .16 CHAPTER 1 FUNCTIONS AND MODELSNote from the last gure that the regression line has been stored in Y1 and that Plot1 has been turned on (Plot1 ishighlighted). You can turn on Plot1 from the Y= menu by placing the cursor on Plot1 and pressing or bypressing .Now press to produce a graph of the data and the regressionline. Note that choice 9 of the ZOOM menu automatically selects a windowthat displays all of the data.(d) When r = 25,000, j E 11.456; or about 11.5 per 100 population.(e) When r = 80,000, j E 5.968; or about a 6% chance.(f ) When r = 200,000, j is negative, so the model does not apply.23. (a)A linear model does seem appropriate.(b)Using a computing device, we obtain the least squaresregression line j = 0.089119747r 3158.2403249,where r is the year and j is the height in feet.(c) When r = 2000, the model gives j E 20.00 ft. Note that the actual winning height for the 2000 Olympics is less than thewinning height for 1996so much for that prediction.(d) When r = 2100, j E 28.91 ft. This would be an increase of 9.49 ft from 1996 to 2100. Even though there was an increaseof 8.59 ft from 1900 to 1996, it is unlikely that a similar increase will occur over the next 100 years.25. Using a computing device, we obtain the cubicfunction j = or3+/r2+cr +d witho = 0.0012937, / = 37.06142, c = 12,823,and d = 37,743,770. When r = 1925,j E 1914 (million).SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS171.3 New Functions from Old Functions1. (a) If the graph of 1 is shifted 3 units upward, its equation becomes j = 1(r) + 3.(b) If the graph of 1 is shifted 3 units downward, its equation becomes j = 1(r) 33.(c) If the graph of 1 is shifted 3 units to the right, its equation becomes j = 1(r 33).(d) If the graph of 1 is shifted 3 units to the left, its equation becomes j = 1(r + 3).(e) If the graph of 1 is reected about the r-axis, its equation becomes j = 31(r).(f ) If the graph of 1 is reected about the j-axis, its equation becomes j = 1(3r).(g) If the graph of 1 is stretched vertically by a factor of 3, its equation becomes j = 31(r).(h) If the graph of 1 is shrunk vertically by a factor of 3, its equation becomes j =131(r).3. (a) (graph 3) The graph of 1 is shifted 4 units to the right and has equation j = 1(r 34).(b) (graph 1) The graph of 1 is shifted 3 units upward and has equation j = 1(r) + 3.(c) (graph 4) The graph of 1 is shrunk vertically by a factor of 3 and has equation j =131(r).(d) (graph 5) The graph of 1 is shifted 4 units to the left and reected about the r-axis. Its equation is j = 31(r + 4).(e) (graph 2) The graph of 1 is shifted 6 units to the left and stretched vertically by a factor of 2. Its equation isj = 21(r + 6).5. (a) To graph j = 1(2r) we shrink the graph of 1horizontally by a factor of 2.The point (4. 31) on the graph of 1 corresponds to thepoint

12 4. 31

= (2. 31).(b) To graph j = 1

12r

we stretch the graph of 1horizontally by a factor of 2.The point (4. 31) on the graph of 1 corresponds to thepoint (2 4. 31) = (8. 31).(c) To graph j = 1(3r) we reect the graph of 1 aboutthe j-axis.The point (4. 31) on the graph of 1 corresponds to thepoint (31 4. 31) = (34. 31).(d) To graph j = 31(3r) we reect the graph of 1 aboutthe j-axis, then about the r-axis.The point (4. 31) on the graph of 1 corresponds to thepoint (31 4. 31 31) = (34. 1).18CHAPTER 1 FUNCTIONS AND MODELS7. The graph of j = 1(r) =I3r 3r2 has been shifted 4 units to the left, reected about the r-axis, and shifted downward1 unit. Thus, a function describing the graph isj = 31 . .. .reectabout r-axis1(r + 4). .. .shift4 units left31. .. .shift1 unit leftThis function can be written asj = 31(r + 4) 31 = 3

3(r + 4) 3(r + 4)231 = 3

3r + 12 3(r2 + 8r + 16) 31 = 3I3r235r 34 319. j = 3r3: Start with the graph of j = r3and reectabout the r-axis. Note: Reecting about the j-axisgives the same result since substituting 3r for r givesus j = (3r)3= 3r3.11. j = (r + 1)2: Start with the graph of j = r2and shift 1 unit to the left.13. j = 1 + 2 cos r: Start with the graph of j = cos r, stretch vertically by a factor of 2, and then shift 1 unit upward.15. j = sin(r2): Start with the graph of j = sinr and stretch horizontally by a factor of 2.SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS 1917. j =Ir + 3 : Start with the graph ofj =Ir and shift 3 units to the left.19. j =12(r2+ 8r) =12(r2+ 8r + 16) 38 =12(r + 4)238: Start with the graph of j = r2, compress vertically by afactor of 2, shift 4 units to the left, and then shift 8 units downward.0 0 0 021. j = 2(r + 1): Start with the graph of j = 1r, shift 1 unit to the left, and then stretch vertically by a factor of 2.23. j = |sinr|: Start with the graph of j = sinr and reect all the parts of the graph below the r-axis about the r-axis.25. This is just like the solution to Example 4 except the amplitude of the curve (the 30

N curve in Figure 9 on June 21) is14 312 = 2. So the function is 1(t) = 12 + 2 sin

2r365(t 380). March 31 is the 90th day of the year, so the model gives1(90) E 12.34 h. The daylight time (5:51 AM to 6:18 PM) is 12 hours and 27 minutes, or 12.45 h. The model value differsfrom the actual value by 12.45312.3412.45E 0.009, less than 1%.27. (a) To obtain j = 1(|r|), the portion of the graph of j = 1(r) to the right of the j-axis is reected about the j-axis.(b) j = sin|r| (c) j = |r|20 CHAPTER 1 FUNCTIONS AND MODELS29. 1(r) = r3+ 2r2; o(r) = 3r231. 1 = R for both 1 and o.(1 +o)(r) = (r3+ 2r2) + (3r231) = r3+ 5r231, 1 = R.(1 3o)(r) = (r3+ 2r2) 3(3r231) = r33r2+ 1, 1 = R.(1o)(r) = (r3+ 2r2)(3r231) = 3r5+ 6r43r332r2, 1 = R.

1o

(r) = r3+ 2r23r231 , 1 = r | r 6= 1I3since 3r231 6= 0.31. 1(r) = r231, 1 = R; o(r) = 2r + 1, 1 = R.(a) (1o)(r) = 1(o(r)) = 1(2r + 1) = (2r + 1)231 = (4r2+ 4r + 1) 31 = 4r2+ 4r, 1 = R.(b) (o1)(r) = o(1(r)) = o(r231) = 2(r231) + 1 = (2r232) + 1 = 2r231, 1 = R.(c) (11)(r) = 1(1(r)) = 1(r231) = (r231)231 = (r432r2+ 1) 31 = r432r2, 1 = R.(d) (oo)(r) = o(o(r)) = o(2r + 1) = 2(2r + 1) + 1 = (4r + 2) + 1 = 4r + 3, 1 = R.33. 1(r) = 1 33r; o(r) = cos r. 1 = R for both 1 and o, and hence for their composites.(a) (1o)(r) = 1(o(r)) = 1(cos r) = 1 33 cos r.(b) (o1)(r) = o(1(r)) = o(1 33r) = cos(1 33r).(c) (11)(r) = 1(1(r)) = 1(1 33r) = 1 33(1 33r) = 1 33 + 9r = 9r 32.(d) (oo)(r) = o(o(r)) = o(cos r) = cos(cos r) [Note that this is not cos r cos r.]35. 1(r) = r + 1r, 1 = {r | r 6= 0}; o(r) = r + 1r + 2, 1 = {r | r 6= 32}(a) (1o)(r) = 1(o(r)) = 1

r + 1r + 2

= r + 1r + 2 +1r + 1r + 2= r + 1r + 2 + r + 2r + 1= (r + 1)(r + 1) + (r + 2)(r + 2)(r + 2)(r + 1)=

r2+ 2r + 1

+

r2+ 4r + 4

(r + 2)(r + 1)=2r2+ 6r + 5(r + 2)(r + 1)Since o(r) is not dened for r = 32 and 1(o(r)) is not dened for r = 32 and r = 31,the domain of (1o)(r) is 1 = {r | r 6= 32. 31}.(b) (o1)(r) = o(1(r)) = o

r + 1r

=

r + 1r

+ 1

r + 1r

+ 2 =r2+ 1 +rrr2+ 1 + 2rr=r2+r + 1r2 + 2r + 1 = r2+r + 1(r + 1)2Since 1(r) is not dened for r = 0 and o(1(r)) is not dened for r = 31,the domain of (o1)(r) is 1 = {r | r 6= 31. 0}.(c) (11)(r) = 1(1(r)) = 1

r + 1r

= r + 1r

+1r + 1i= r + 1r +1i2+1i= r + 1r +rr2 + 1= r(r)

r2+ 1

+ 1

r2+ 1

+r(r)r(r2 + 1)= r4+r2+r2+ 1 +r2r(r2 + 1)= r4+ 3r2+ 1r(r2 + 1). 1 = {r | r 6= 0}SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS 21(d) (oo)(r) = o(o(r)) = o

r + 1r + 2

=r + 1r + 2 + 1r + 1r + 2 + 2 =r + 1 + 1(r + 2)r + 2r + 1 + 2(r + 2)r + 2=r + 1 +r + 2r + 1 + 2r + 4 = 2r + 33r + 5Since o(r) is not dened for r = 32 and o(o(r)) is not dened for r = 353,the domain of (oo)(r) is 1 = r | r 6= 32. 353.37. (1o/)(r) = 1(o(/(r))) = 1(o(r 31)) = 1(2(r 31)) = 2(r 31) + 1 = 2r 3139. (1o/)(r) = 1(o(/(r))) = 1(o(r3+ 2)) = 1[(r3+ 2)2]= 1(r6+ 4r3+ 4) = (r6 + 4r3 + 4) 33 =Ir6 + 4r3 + 141. Let o(r) = r2+ 1 and 1(r) = r10. Then (1o)(r) = 1(o(r)) = 1(r2+ 1) = (r2+ 1)10= 1(r).43. Let o(r) =3Ir and 1(r) =r1 +r. Then (1o)(r) = 1(o(r)) = 1( 3Ir) =3Ir1 +3Ir = 1(r).45. Let o(t) = cos t and 1(t) =It. Then (1o)(t) = 1(o(t)) = 1(cos t) =Icos t = &(t).47. Let /(r) = r2, o(r) = 3i, and 1(r) = 1 3 r. Then(1o/)(r) = 1(o(/(r))) = 1(o(r2)) = 1

3i2

= 1 33i2= 1(r).49. Let /(r) = Ir, o(r) = sec r, and 1(r) = r4. Then(1o/)(r) = 1(o(/(r))) = 1(o(Ir)) = 1(secIr) = (secIr)4= sec4(Ir) = 1(r).51. (a) o(2) = 5, because the point (2. 5) is on the graph of o. Thus, 1(o(2)) = 1(5) = 4, because the point (5. 4) is on thegraph of 1.(b) o(1(0)) = o(0) = 3(c) (1o)(0) = 1(o(0)) = 1(3) = 0(d) (o1)(6) = o(1(6)) = o(6). This value is not dened, because there is no point on the graph of o that hasr-coordinate 6.(e) (oo)(32) = o(o(32)) = o(1) = 4(f ) (11)(4) = 1(1(4)) = 1(2) = 3253. (a) Using the relationship distance = rate time with the radius v as the distance, we have v(t) = 60t.(b) = v2i( v)(t) = (v(t)) = (60t)2= 3600t2. This formula gives us the extent of the rippled area(in cm2) at any time t.55. (a) From the gure, we have a right triangle with legs 6 and d, and hypotenuse c.By the Pythagorean Theorem, d2+ 62= c2ic = 1(d) =Id2 + 36.(b) Using d = vt, we get d = (30 kmhr)(t hr) = 30t (in km). Thus,d = o(t) = 30t.(c) (1o)(t) = 1(o(t)) = 1(30t) = (30t)2 + 36 =I900t2 + 36. This function represents the distance between thelighthouse and the ship as a function of the time elapsed since noon.22 CHAPTER 1 FUNCTIONS AND MODELS57. (a)1(t) =

0 if t < 01 if t D 0(b)\ (t) =

0 if t < 0120 if t D 0so \ (t) = 1201(t).(c) Starting with the formula in part (b), we replace 120 with 240 to reect thedifferent voltage. Also, because we are starting 5 units to the right of t = 0,we replace t with t 35. Thus, the formula is \ (t) = 2401(t 35).59. If 1(r) = i1r + /1 and o(r) = i2r + /2, then(1o)(r) = 1(o(r)) = 1(i2r +/2) = i1(i2r +/2) +/1 = i1i2r +i1/2 +/1.So 1o is a linear function with slope i1i2.61. (a) By examining the variable terms in o and /, we deduce that we must square o to get the terms 4r2and 4r in /. If we let1(r) = r2+c, then (1o)(r) = 1(o(r)) = 1(2r + 1) = (2r + 1)2+c = 4r2+ 4r + (1 +c). Since/(r) = 4r2+ 4r + 7, we must have 1 +c = 7. So c = 6 and 1(r) = r2+ 6.(b) We need a function o so that 1(o(r)) = 3(o(r)) + 5 = /(r). But/(r) = 3r2+ 3r + 2 = 3(r2+r) + 2 = 3(r2+r 31) + 5, so we see that o(r) = r2+r 31.63. (a) If 1 and o are even functions, then 1(3r) = 1(r) and o(3r) = o(r).(i) (1 +o)(3r) = 1(3r) +o(3r) = 1(r) +o(r) = (1 +o)(r), so 1 +o is an even function.(ii) (1o)(3r) = 1(3r) o(3r) = 1(r) o(r) = (1o)(r), so 1o is an even function.(b) If 1 and o are odd functions, then 1(3r) = 31(r) and o(3r) = 3o(r).(i) (1 +o)(3r) = 1(3r) +o(3r) = 31(r) + [3o(r)] = 3[1(r) +o(r)] = 3(1 +o)(r),so 1 +o is an odd function.(ii) (1o)(3r) = 1(3r) o(3r) = 31(r) [3o(r)] = 1(r) o(r) = (1o)(r), so 1o is an even function.65. We need to examine /(3r)./(3r) = (1o)(3r) = 1(o(3r)) = 1(o(r)) [because o is even] = /(r)Because /(3r) = /(r), / is an even function.SECTION 1.4 GRAPHING CALCULATORS AND COMPUTERS231.4 Graphing Calculators and Computers1. 1(r) =Ir335r2(a) [35. 5] by [35. 5](There is no graph shown.)(b) [0. 10] by [0. 2] (c) [0. 10] by [0. 10]The most appropriate graph is produced in viewing rectangle (c).3. Since the graph of 1(r) = 5 + 20r 3r2is aparabola opening downward, an appropriate viewingrectangle should include the maximum point.5. 1(r) =4I81 3r4 is dened when 81 3r4D 0Cr4$ 81C |r| $ 3, so the domain of 1 is [33. 3]. Also0 $4I81 3r4 $4I81 = 3, so the range is [0. 3].7. The graph of 1(r) = r33225r is symmetric with respect to the origin.Since 1(r) = r33225r = r(r23225) = r(r + 15)(r 315), thereare r-intercepts at 0, 315, and 15. 1(20) = 3500.9. The period of o(r) = sin(1000r) is2r1000 E 0.0063 and its range is[31. 1]. Since 1(r) = sin2(1000r) is the square of o, its range is[0. 1] and a viewing rectangle of [30.01. 0.01] by [0. 1.1] seemsappropriate.11. The domain of j =Ir is r D 0, so the domain of 1(r) = sinIr is [0. ")and the range is [31. 1]. With a little trial-and-error experimentation, we ndthat an Xmax of 100 illustrates the general shape of 1, so an appropriateviewing rectangle is [0. 100] by [31.5. 1.5].24 CHAPTER 1 FUNCTIONS AND MODELS13. The rst term, 10 sinr, has period 2 and range [310. 10]. It will be the dominant term in any large graph ofj = 10 sinr + sin100r, as shown in the rst gure. The second term, sin100r, has period2r100 =r50 and range [31. 1].It causes the bumps in the rst gure and will be the dominant term in any small graph, as shown in the view near theorigin in the second gure.15. We must solve the given equation for j to obtain equations for the upper andlower halves of the ellipse.4r2+ 2j2= 1C2j2= 1 34r2Cj2= 1 34r22Cj =

1 34r2217. From the graph of j = 3r236r +1and j = 0.23r 32.25 in the viewingrectangle [31. 3] by [32.5. 1.5], it isdifcult to see if the graphs intersect.If we zoom in on the fourth quadrant,we see the graphs do not intersect.19. From the graph of 1(r) = r339r234, we see that there is one solutionof the equation 1(r) = 0 and it is slightly larger than 9. By zooming in orusing a root or zero feature, we obtain r E 9.05.21. We see that the graphs of 1(r) = r2and o(r) = sinr intersect twice. Onesolution is r = 0. The other solution of 1 = o is the r-coordinate of thepoint of intersection in the rst quadrant. Using an intersect feature orzooming in, we nd this value to be approximately 0.88. Alternatively, wecould nd that value by nding the positive zero of /(r) = r23sinr.Note: After producing the graph on a TI-83 Plus, we can nd the approximate value 0.88 by using the following keystrokes:. The 1 is just a guess for 0.88.SECTION 1.4 GRAPHING CALCULATORS AND COMPUTERS 2523. o(r) = r310 is larger than 1(r) = 10r2whenever r100.25. We see from the graphs of j = |sinr 3r| and j = 0.1 that there aretwo solutions to the equation |sinr 3r| = 0.1: r E 30.85 andr E 0.85. The condition |sinr 3r| < 0.1 holds for any r lyingbetween these two values, that is, 30.85 < r < 0.85.27. (a) The root functions j =Ir,j =4Ir and j =6Ir(b) The root functions j = r,j =3Ir and j =5Ir(c) The root functions j =Ir, j =3Ir,j =4Ir and j =5Ir(d) For any n, the nth root of 0 is 0 and the nth root of 1 is 1; that is, all nth root functions pass through the points (0. 0)and (1. 1). For odd n, the domain of the nth root function is R, while for even n, it is {r M R | r D 0}. Graphs of even root functions look similar to that ofIr, while those of odd root functions resemble that of3Ir. As n increases, the graph ofqIr becomes steeper near 0 and atter for r1.29. 1(r) = r4+cr2+r. If c < 31.5, there are three humps: two minimum pointsand a maximum point. These humps get atter as c increases, until at c = 31.5two of the humps disappear and there is only one minimum point. This singlehump then moves to the right and approaches the origin as c increases.31. j = rn23i. As n increases, the maximum of thefunction moves further from the origin, and getslarger. Note, however, that regardless of n, thefunction approaches 0 as r