Solution Manual for Calculus , 8th Edition
- Mathematics
- 1118 pages
- Shared May 2025
Maximize your understanding with Solution Manual for Calculus, 8th Edition, a detailed solutions manual for all your textbook problems.
Preview 16 of 1118 pages
See the first 16 pages exactly as they appear in the document.
Page 1 Read page 1 as text
1 FUNCTIONS AND MODELS 1.1 Four Ways to Represent a Function 1. The functions ( ) = + √ 2 − and ( ) = + √ 2 − give exactly the same output values for every input value, so and are equal. 2. ( ) = 2 − − 1 = ( − 1) − 1 = for − 1 6 = 0 , so and [where ( ) = ] are not equal because (1) is unde fi ned and (1) = 1 . 3. (a) The point (1 3) is on the graph of , so (1) = 3 . (b) When = − 1 , is about − 0 2 , so ( − 1) ≈ − 0 2 . (c) ( ) = 1 is equivalent to = 1 When = 1 , we have = 0 and = 3 . (d) A reasonable estimate for when = 0 is = − 0 8 . (e) The domain of consists of all -values on the graph of . For this function, the domain is − 2 ≤ ≤ 4 , or [ − 2 4] . The range of consists of all -values on the graph of . For this function, the range is − 1 ≤ ≤ 3 , or [ − 1 3] . (f ) As increases from − 2 to 1 , increases from − 1 to 3 . Thus, is increasing on the interval [ − 2 1] . 4. (a) The point ( − 4 − 2) is on the graph of , so ( − 4) = − 2 . The point (3 4) is on the graph of , so (3) = 4 . (b) We are looking for the values of for which the -values are equal. The -values for and are equal at the points ( − 2 1) and (2 2) , so the desired values of are − 2 and 2 . (c) ( ) = − 1 is equivalent to = − 1 . When = − 1 , we have = − 3 and = 4 . (d) As increases from 0 to 4 , decreases from 3 to − 1 . Thus, is decreasing on the interval [0 4] . (e) The domain of consists of all -values on the graph of . For this function, the domain is − 4 ≤ ≤ 4 , or [ − 4 4] . The range of consists of all -values on the graph of . For this function, the range is − 2 ≤ ≤ 3 , or [ − 2 3] . (f ) The domain of is [ − 4 3] and the range is [0 5 4] . 5. From Figure 1 in the text, the lowest point occurs at about ( ) = (12 − 85) . The highest point occurs at about (17 115) . Thus, the range of the vertical ground acceleration is − 85 ≤ ≤ 115 . Written in interval notation, we get [ − 85 115] . 6. Example 1: A car is driven at 60 mi h for 2 hours. The distance traveled by the car is a function of the time . The domain of the function is { | 0 ≤ ≤ 2 } , where is measured in hours. The range of the function is { | 0 ≤ ≤ 120 } , where is measured in miles. uplic ible 9
Page 2 Page 3 Read page 3 as text
10 ¤ CHAPTER 1 FUNCTIONS AND MODELS Example 2: At a certain university, the number of students on campus at any time on a particular day is a function of the time after midnight. The domain of the function is { | 0 ≤ ≤ 24 } , where is measured in hours. The range of the function is { | 0 ≤ ≤ } , where is an integer and is the largest number of students on campus at once. Example 3: A certain employee is paid $8 00 per hour and works a maximum of 30 hours per week. The number of hours worked is rounded down to the nearest quarter of an hour. This employee’s gross weekly pay is a function of the number of hours worked . The domain of the function is [0 30] and the range of the function is { 0 2 00 4 00 238 00 240 00 } . 240 pay hours 0.25 0.50 0.75 0 29.50 29.75 30 2 4 238 236 7. No, the curve is not the graph of a function because a vertical line intersects the curve more than once. Hence, the curve fails the Vertical Line Test. 8. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [ − 2 2] and the range is [ − 1 2] . 9. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [ − 3 2] and the range is [ − 3 − 2) ∪ [ − 1 3] . 10. No, the curve is not the graph of a function since for = 0 , ± 1 , and ± 2 , there are in fi nitely many points on the curve. 11. (a) When = 1950 , ≈ 13 8 ◦ C , so the global average temperature in 1950 was about 13 8 ◦ C . (b) When = 14 2 ◦ C , ≈ 1990 . (c) The global average temperature was smallest in 1910 (the year corresponding to the lowest point on the graph) and largest in 2005 (the year corresponding to the highest point on the graph). (d) When = 1910 , ≈ 13 5 ◦ C , and when = 2005 , ≈ 14 5 ◦ C . Thus, the range of is about [13 5 , 14 5] . 12. (a) The ring width varies from near 0 mm to about 1 6 mm , so the range of the ring width function is approximately [0 1 6] . (b) According to the graph, the earth gradually cooled from 1550 to 1700, warmed into the late 1700s, cooled again into the late 1800s, and has been steadily warming since then. In the mid-19th century, there was variation that could have been associated with volcanic eruptions. 13. The water will cool down almost to freezing as the ice melts. Then, when the ice has melted, the water will slowly warm up to room temperature. post , in
Page 4 Read page 4 as text
SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 11 14. Runner A won the race, reaching the fi nish line at 100 meters in about 15 seconds, followed by runner B with a time of about 19 seconds, and then by runner C who fi nished in around 23 seconds. B initially led the race, followed by C, and then A. C then passed B to lead for a while. Then A passed fi rst B, and then passed C to take the lead and fi nish fi rst. Finally, B passed C to fi nish in second place. All three runners completed the race. 15. (a) The power consumption at 6 AM is 500 MW which is obtained by reading the value of power when = 6 from the graph. At 6 PM we read the value of when = 18 obtaining approximately 730 MW (b) The minimum power consumption is determined by fi nding the time for the lowest point on the graph, = 4 or 4 AM . The maximum power consumption corresponds to the highest point on the graph, which occurs just before = 12 or right before noon. These times are reasonable, considering the power consumption schedules of most individuals and businesses. 16. The summer solstice (the longest day of the year) is around June 21, and the winter solstice (the shortest day) is around December 22. (Exchange the dates for the southern hemisphere.) 17. Of course, this graph depends strongly on the geographical location! 18. The value of the car decreases fairly rapidly initially, then somewhat less rapidly. 19. As the price increases, the amount sold decreases. 20. The temperature of the pie would increase rapidly, level off to oven temperature, decrease rapidly, and then level off to room temperature. 21. or du cessi
Page 5 Read page 5 as text
12 ¤ CHAPTER 1 FUNCTIONS AND MODELS 22. (a) (b) (c) (d) 23. (a) (b) 9:00 AM corresponds to = 9 . When = 9 , the temperature is about 74 ◦ F . 24. (a) (b) The blood alcohol concentration rises rapidly, then slowly decreases to near zero. Note that the BAC in this exercise is measured in mg mL , not percent. 25. ( ) = 3 2 − + 2 (2) = 3(2) 2 − 2 + 2 = 12 − 2 + 2 = 12 ( − 2) = 3( − 2) 2 − ( − 2) + 2 = 12 + 2 + 2 = 16 ( ) = 3 2 − + 2 ( − ) = 3( − ) 2 − ( − ) + 2 = 3 2 + + 2 ( + 1) = 3( + 1) 2 − ( + 1) + 2 = 3( 2 + 2 + 1) − − 1 + 2 = 3 2 + 6 + 3 − + 1 = 3 2 + 5 + 4 2 ( ) = 2 · ( ) = 2(3 2 − + 2) = 6 2 − 2 + 4 post , in
Page 6 Read page 6 as text
SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 13 (2 ) = 3(2 ) 2 − (2 ) + 2 = 3(4 2 ) − 2 + 2 = 12 2 − 2 + 2 ( 2 ) = 3( 2 ) 2 − ( 2 ) + 2 = 3( 4 ) − 2 + 2 = 3 4 − 2 + 2 [ ( )] 2 = 3 2 − + 2 2 = 3 2 − + 2 3 2 − + 2 = 9 4 − 3 3 + 6 2 − 3 3 + 2 − 2 + 6 2 − 2 + 4 = 9 4 − 6 3 + 13 2 − 4 + 4 ( + ) = 3( + ) 2 − ( + ) + 2 = 3( 2 + 2 + 2 ) − − + 2 = 3 2 + 6 + 3 2 − − + 2 26. A spherical balloon with radius + 1 has volume ( + 1) = 4 3 ( + 1) 3 = 4 3 ( 3 + 3 2 + 3 + 1) . We wish to fi nd the amount of air needed to in fl ate the balloon from a radius of to + 1 . Hence, we need to fi nd the difference ( + 1) − ( ) = 4 3 ( 3 + 3 2 + 3 + 1) − 4 3 3 = 4 3 (3 2 + 3 + 1) . 27. ( ) = 4 + 3 − 2 , so (3 + ) = 4 + 3(3 + ) − (3 + ) 2 = 4 + 9 + 3 − (9 + 6 + 2 ) = 4 − 3 − 2 , and (3 + ) − (3) = (4 − 3 − 2 ) − 4 = ( − 3 − ) = − 3 − . 28. ( ) = 3 , so ( + ) = ( + ) 3 = 3 + 3 2 + 3 2 + 3 , and ( + ) − ( ) = ( 3 + 3 2 + 3 2 + 3 ) − 3 = (3 2 + 3 + 2 ) = 3 2 + 3 + 2 . 29. ( ) − ( ) − = 1 − 1 − = − − = − ( − ) = − 1( − ) ( − ) = − 1 30. ( ) − (1) − 1 = + 3 + 1 − 2 − 1 = + 3 − 2( + 1) + 1 − 1 = + 3 − 2 − 2 ( + 1)( − 1) = − + 1 ( + 1)( − 1) = − ( − 1) ( + 1)( − 1) = − 1 + 1 31. ( ) = ( + 4) ( 2 − 9) is de fi ned for all except when 0 = 2 − 9 ⇔ 0 = ( + 3)( − 3) ⇔ = − 3 or 3 , so the domain is { ∈ R | 6 = − 3 3 } = ( −∞ − 3) ∪ ( − 3 3) ∪ (3 ∞ ) . 32. ( ) = (2 3 − 5) ( 2 + − 6) is de fi ned for all except when 0 = 2 + − 6 ⇔ 0 = ( + 3)( − 2) ⇔ = − 3 or 2 , so the domain is { ∈ R | 6 = − 3 2 } = ( −∞ − 3) ∪ ( − 3 2) ∪ (2 ∞ ) . 33. ( ) = 3 √ 2 − 1 is de fi ned for all real numbers. In fact 3 ( ) , where ( ) is a polynomial, is de fi ned for all real numbers. Thus, the domain is R or ( −∞ ∞ ) . 34. ( ) = √ 3 − − √ 2 + is de fi ned when 3 − ≥ 0 ⇔ ≤ 3 and 2 + ≥ 0 ⇔ ≥ − 2 . Thus, the domain is − 2 ≤ ≤ 3 , or [ − 2 3] . 35. ( ) = 1 4 √ 2 − 5 is de fi ned when 2 − 5 0 ⇔ ( − 5) 0 . Note that 2 − 5 6 = 0 since that would result in division by zero. The expression ( − 5) is positive if 0 or 5 . (See Appendix A for methods for solving inequalities.) Thus, the domain is ( −∞ 0) ∪ (5 ∞ ) . or du cessi
Page 7 Read page 7 as text
14 ¤ CHAPTER 1 FUNCTIONS AND MODELS 36. ( ) = + 1 1 + 1 + 1 is de fi ned when + 1 6 = 0 [ 6 = − 1 ] and 1 + 1 + 1 6 = 0 . Since 1 + 1 + 1 = 0 ⇔ 1 + 1 = − 1 ⇔ 1 = − − 1 ⇔ = − 2 , the domain is { | 6 = − 2 , 6 = − 1 } = ( −∞ − 2) ∪ ( − 2 − 1) ∪ ( − 1 ∞ ) . 37. ( ) = 2 − √ is de fi ned when ≥ 0 and 2 − √ ≥ 0 . Since 2 − √ ≥ 0 ⇔ 2 ≥ √ ⇔ √ ≤ 2 ⇔ 0 ≤ ≤ 4 , the domain is [0 4] . 38. ( ) = √ 4 − 2 . Now = √ 4 − 2 ⇒ 2 = 4 − 2 ⇔ 2 + 2 = 4 , so the graph is the top half of a circle of radius 2 with center at the origin. The domain is | 4 − 2 ≥ 0 = | 4 ≥ 2 = { | 2 ≥ | |} = [ − 2 2] . From the graph, the range is 0 ≤ ≤ 2 , or [0 2] . 39. The domain of ( ) = 1 6 − 2 4 is the set of all real numbers, denoted by R or ( −∞ ∞ ) . The graph of f is a line with slope 1 6 and y -intercept − 2 4 . 40. Note that ( ) = 2 − 1 + 1 = ( + 1)( − 1) + 1 = − 1 for + 1 6 = 0 , i.e., 6 = − 1 . The domain of is the set of all real numbers except − 1 . In interval notation, we have ( −∞ − 1) ∪ ( − 1 ∞ ) . The graph of is a line with slope 1 , -intercept − 1 , and a hole at = − 1 . 41. ( ) = + 2 if 0 1 − if ≥ 0 ( − 3) = − 3 + 2 = − 1 , (0) = 1 − 0 = 1 , and (2) = 1 − 2 = − 1 . 42. ( ) = 3 − 1 2 if 2 2 − 5 if ≥ 2 ( − 3) = 3 − 1 2 ( − 3) = 9 2 , (0) = 3 − 1 2 (0) = 3 , and (2) = 2(2) − 5 = − 1 . 43. ( ) = + 1 if ≤ − 1 2 if − 1 ( − 3) = − 3 + 1 = − 2 , (0) = 0 2 = 0 , and (2) = 2 2 = 4 . post , in
Page 8 Read page 8 as text
SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 15 44. ( ) = − 1 if ≤ 1 7 − 2 if 1 ( − 3) = − 1 , (0) = − 1 , and (2) = 7 − 2(2) = 3 . 45. | | = if ≥ 0 − if 0 so ( ) = + | | = 2 if ≥ 0 0 if 0 Graph the line = 2 for ≥ 0 and graph = 0 (the -axis) for 0 46. ( ) = | + 2 | = + 2 if + 2 ≥ 0 − ( + 2) if + 2 0 = + 2 if ≥ − 2 − − 2 if − 2 47. ( ) = | 1 − 3 | = 1 − 3 if 1 − 3 ≥ 0 − (1 − 3 ) if 1 − 3 0 = 1 − 3 if ≤ 1 3 3 − 1 if 1 3 48. | | = if ≥ 0 − if 0 and | + 1 | = + 1 if + 1 ≥ 0 − ( + 1) if + 1 0 = + 1 if ≥ − 1 − − 1 if − 1 so ( ) = | | + | + 1 | = + ( + 1) if ≥ 0 − + ( + 1) if − 1 ≤ 0 − + ( − − 1) if − 1 = 2 + 1 if ≥ 0 1 if − 1 ≤ 0 − 2 − 1 if − 1 or du cessi
Page 9 Read page 9 as text
16 ¤ CHAPTER 1 FUNCTIONS AND MODELS 49. To graph ( ) = | | if | | ≤ 1 1 if | | 1 , graph = | | (Figure 16) for − 1 ≤ ≤ 1 and graph = 1 for 1 and for − 1 . We could rewrite f as ( ) = 1 if − 1 − if − 1 ≤ 0 if 0 ≤ ≤ 1 1 if 1 . 50. ( ) = | | − 1 = | | − 1 if | | − 1 ≥ 0 − ( | | − 1) if | | − 1 0 = | | − 1 if | | ≥ 1 − | | + 1 if | | 1 = − 1 if | | ≥ 1 and ≥ 0 − − 1 if | | ≥ 1 and 0 − + 1 if | | 1 and ≥ 0 − ( − ) + 1 if | | 1 and 0 = − 1 if ≥ 1 − − 1 if ≤ − 1 − + 1 if 0 ≤ 1 + 1 if − 1 0 51. Recall that the slope of a line between the two points ( 1 1 ) and ( 2 2 ) is = 2 − 1 2 − 1 and an equation of the line connecting those two points is − 1 = ( − 1 ) . The slope of the line segment joining the points (1 − 3) and (5 7) is 7 − ( − 3) 5 − 1 = 5 2 , so an equation is − ( − 3) = 5 2 ( − 1) . The function is ( ) = 5 2 − 11 2 , 1 ≤ ≤ 5 . 52. The slope of the line segment joining the points ( − 5 10) and (7 − 10) is − 10 − 10 7 − ( − 5) = − 5 3 , so an equation is − 10 = − 5 3 [ − ( − 5)] . The function is ( ) = − 5 3 + 5 3 , − 5 ≤ ≤ 7 . 53. We need to solve the given equation for . + ( − 1) 2 = 0 ⇔ ( − 1) 2 = − ⇔ − 1 = ± √ − ⇔ = 1 ± √ − . The expression with the positive radical represents the top half of the parabola, and the one with the negative radical represents the bottom half. Hence, we want ( ) = 1 − √ − . Note that the domain is ≤ 0 . 54. 2 + ( − 2) 2 = 4 ⇔ ( − 2) 2 = 4 − 2 ⇔ − 2 = ± √ 4 − 2 ⇔ = 2 ± √ 4 − 2 . The top half is given by the function ( ) = 2 + √ 4 − 2 , − 2 ≤ ≤ 2 . 55. For 0 ≤ ≤ 3 , the graph is the line with slope − 1 and -intercept 3 , that is, = − + 3 . For 3 ≤ 5 , the graph is the line with slope 2 passing through (3 0) ; that is, − 0 = 2( − 3) , or = 2 − 6 . So the function is ( ) = − + 3 if 0 ≤ ≤ 3 2 − 6 if 3 ≤ 5 56. For − 4 ≤ ≤ − 2 , the graph is the line with slope − 3 2 passing through ( − 2 0) ; that is, − 0 = − 3 2 [ − ( − 2)] , or = − 3 2 − 3 . For − 2 2 , the graph is the top half of the circle with center (0 0) and radius 2 . An equation of the circle post , in
Page 10 Read page 10 as text
SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 17 is 2 + 2 = 4 , so an equation of the top half is = √ 4 − 2 . For 2 ≤ ≤ 4 , the graph is the line with slope 3 2 passing through (2 0) ; that is, − 0 = 3 2 ( − 2) , or = 3 2 − 3 . So the function is ( ) = − 3 2 − 3 if − 4 ≤ ≤ − 2 √ 4 − 2 if − 2 2 3 2 − 3 if 2 ≤ ≤ 4 57. Let the length and width of the rectangle be and . Then the perimeter is 2 + 2 = 20 and the area is = . Solving the fi rst equation for in terms of gives = 20 − 2 2 = 10 − . Thus, ( ) = (10 − ) = 10 − 2 . Since lengths are positive, the domain of is 0 10 . If we further restrict to be larger than , then 5 10 would be the domain. 58. Let the length and width of the rectangle be and . Then the area is = 16 , so that = 16 . The perimeter is = 2 + 2 , so ( ) = 2 + 2(16 ) = 2 + 32 , and the domain of is 0 , since lengths must be positive quantities. If we further restrict to be larger than , then 4 would be the domain. 59. Let the length of a side of the equilateral triangle be . Then by the Pythagorean Theorem, the height of the triangle satis fi es 2 + 1 2 2 = 2 , so that 2 = 2 − 1 4 2 = 3 4 2 and = √ 3 2 . Using the formula for the area of a triangle, = 1 2 ( base )( height ) , we obtain ( ) = 1 2 ( ) √ 3 2 = √ 3 4 2 , with domain 0 . 60. Let the length, width, and height of the closed rectangular box be denoted by , , and , respectively. The length is twice the width, so = 2 . The volume of the box is given by = . Since = 8 , we have 8 = (2 ) ⇒ 8 = 2 2 ⇒ = 8 2 2 = 4 2 , and so = ( ) = 4 2 . 61. Let each side of the base of the box have length , and let the height of the box be . Since the volume is 2 , we know that 2 = 2 , so that = 2 2 , and the surface area is = 2 + 4 . Thus, ( ) = 2 + 4 (2 2 ) = 2 + (8 ) , with domain 0 . 62. The area of the window is = + 1 2 1 2 2 = + 2 8 , where is the height of the rectangular portion of the window. The perimeter is = 2 + + 1 2 = 30 ⇔ 2 = 30 − − 1 2 ⇔ = 1 4 (60 − 2 − ) . Thus, ( ) = 60 − 2 − 4 + 2 8 = 15 − 1 2 2 − 4 2 + 8 2 = 15 − 4 8 2 − 8 2 = 15 − 2 + 4 8 . Since the lengths and must be positive quantities, we have 0 and 0 . For 0 , we have 2 0 ⇔ 30 − − 1 2 0 ⇔ 60 2 + ⇔ 60 2 + . Hence, the domain of is 0 60 2 + . 63. The height of the box is and the length and width are = 20 − 2 , = 12 − 2 . Then = and so ( ) = (20 − 2 )(12 − 2 )( ) = 4(10 − )(6 − )( ) = 4 (60 − 16 + 2 ) = 4 3 − 64 2 + 240 . The sides , , and must be positive. Thus, 0 ⇔ 20 − 2 0 ⇔ 10 ; 0 ⇔ 12 − 2 0 ⇔ 6 ; and 0 . Combining these restrictions gives us the domain 0 6 . or du cessi
Page 11 Read page 11 as text
18 ¤ CHAPTER 1 FUNCTIONS AND MODELS 64. We can summarize the monthly cost with a piecewise de fi ned function. ( ) = 35 if 0 ≤ ≤ 400 35 + 0 10( − 400) if 400 65. We can summarize the amount of the fi ne with a piecewise de fi ned function. ( ) = 15(40 − ) if 0 ≤ 40 0 if 40 ≤ ≤ 65 15( − 65) if 65 66. For the fi rst 1200 kWh , ( ) = 10 + 0 06 . For usage over 1200 kWh , the cost is ( ) = 10 + 0 06(1200) + 0 07( − 1200) = 82 + 0 07( − 1200) . Thus, ( ) = 10 + 0 06 if 0 ≤ ≤ 1200 82 + 0 07( − 1200) if 1200 67. (a) (b) On $14 , 000 , tax is assessed on $4000 , and 10%($4000) = $400 . On $26 , 000 , tax is assessed on $16 , 000 , and 10%($10 , 000) + 15%($6000) = $1000 + $900 = $1900 . (c) As in part (b), there is $1000 tax assessed on $20,000 of income, so the graph of is a line segment from (10 , 000 0) to (20 , 000 1000) . The tax on $30,000 is $2500, so the graph of for 20 , 000 is the ray with initial point (20 , 000 1000) that passes through (30 , 000 2500) . 68. One example is the amount paid for cable or telephone system repair in the home, usually measured to the nearest quarter hour. Another example is the amount paid by a student in tuition fees, if the fees vary according to the number of credits for which the student has registered. 69. is an odd function because its graph is symmetric about the origin. is an even function because its graph is symmetric with respect to the -axis. post , in
Page 12 Read page 12 as text
SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 19 70. is not an even function since it is not symmetric with respect to the -axis. is not an odd function since it is not symmetric about the origin. Hence, is neither even nor odd. is an even function because its graph is symmetric with respect to the -axis. 71. (a) Because an even function is symmetric with respect to the -axis, and the point (5 3) is on the graph of this even function, the point ( − 5 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 ( − 5 − 3) must also be on its graph. 72. (a) If is even, we get the rest of the graph by re fl ecting about the -axis. (b) If is odd, we get the rest of the graph by rotating 180 ◦ about the origin. 73. ( ) = 2 + 1 . ( − ) = − ( − ) 2 + 1 = − 2 + 1 = − 2 + 1 = − ( ) . Since ( − ) = − ( ) , is an odd function. 74. ( ) = 2 4 + 1 . ( − ) = ( − ) 2 ( − ) 4 + 1 = 2 4 + 1 = ( ) . Since ( − ) = ( ) , is an even function. 75. ( ) = + 1 , so ( − ) = − − + 1 = − 1 . Since this is neither ( ) nor − ( ) , the function is neither even nor odd. 76. ( ) = | | . ( − ) = ( − ) | − | = ( − ) | | = − ( | | ) = − ( ) Since ( − ) = − ( ) , is an odd function. or du cessi
Page 13 Read page 13 as text
20 ¤ CHAPTER 1 FUNCTIONS AND MODELS 77. ( ) = 1 + 3 2 − 4 . ( − ) = 1+3( − ) 2 − ( − ) 4 = 1+3 2 − 4 = ( ) . Since ( − ) = ( ) , is an even function. 78. ( ) = 1 + 3 3 − 5 , so ( − ) = 1 + 3( − ) 3 − ( − ) 5 = 1 + 3( − 3 ) − ( − 5 ) = 1 − 3 3 + 5 Since this is neither ( ) nor − ( ) , the function is neither even nor odd. 79. (i) If and are both even functions, then ( − ) = ( ) and ( − ) = ( ) . Now ( + )( − ) = ( − ) + ( − ) = ( ) + ( ) = ( + )( ) , so + is an even function. (ii) If and are both odd functions, then ( − ) = − ( ) and ( − ) = − ( ) . Now ( + )( − ) = ( − ) + ( − ) = − ( ) + [ − ( )] = − [ ( ) + ( )] = − ( + )( ) , so + is an odd function. (iii) If is an even function and is an odd function, then ( + )( − ) = ( − ) + ( − ) = ( ) + [ − ( )] = ( ) − ( ) , which is not ( + )( ) nor − ( + )( ) , s o + is neither even nor odd. (Exception: if is the zero function, then + will be odd . If is the zero function, then + will be even .) 80. (i) If and are both even functions, then ( − ) = ( ) and ( − ) = ( ) . Now ( )( − ) = ( − ) ( − ) = ( ) ( ) = ( )( ) , so is an even function. (ii) If and are both odd functions, then ( − ) = − ( ) and ( − ) = − ( ) . Now ( )( − ) = ( − ) ( − ) = [ − ( )][ − ( )] = ( ) ( ) = ( )( ) , so is an even function. (iii) If is an even function and is an odd function, then ( )( − ) = ( − ) ( − ) = ( )[ − ( )] = − [ ( ) ( )] = − ( )( ) , so is an odd function. 1.2 Mathematical Models: A Catalog of Essential Functions 1. (a) ( ) = log 2 is a logarithmic function. (b) ( ) = 4 √ is a root function with = 4 . (c) ( ) = 2 3 1 − 2 is a rational function because it is a ratio of polynomials. (d) ( ) = 1 − 1 1 + 2 54 2 is a polynomial of degree 2 (also called a quadratic function ). (e) ( ) = 5 is an exponential function. (f ) ( ) = sin cos 2 is a trigonometric function. post , in
Page 14 Read page 14 as text
SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS ¤ 21 2. (a) = is an exponential function (notice that is the exponent ). (b) = is a power function (notice that is the base ). (c) = 2 (2 − 3 ) = 2 2 − 5 is a polynomial of degree 5 . (d) = tan − cos is a trigonometric function. (e) = (1 + ) is a rational function because it is a ratio of polynomials. (f ) = √ 3 − 1 (1 + 3 √ ) is an algebraic function because it involves polynomials and roots of polynomials. 3. We notice from the fi gure that and are even functions (symmetric with respect to the -axis) and that is an odd function (symmetric with respect to the origin). So (b) = 5 must be . Since is fl atter than near the origin, we must have (c) = 8 matched with and (a) = 2 matched with . 4. (a) The graph of = 3 is a line (choice ). (b) = 3 is an exponential function (choice ). (c) = 3 is an odd polynomial function or power function (choice ). (d) = 3 √ = 1 3 is a root function (choice ). 5. The denominator cannot equal 0, so 1 − sin 6 = 0 ⇔ sin 6 = 1 ⇔ 6 = 2 + 2 . Thus, the domain of ( ) = cos 1 − sin is | 6 = 2 + 2 , an integer . 6. The denominator cannot equal 0, so 1 − tan 6 = 0 ⇔ tan 6 = 1 ⇔ 6 = 4 + . The tangent function is not de fi ned if 6 = 2 + . Thus, the domain of ( ) = 1 1 − tan is | 6 = 4 + , 6 = 2 + , an integer . 7. (a) An equation for the family of linear functions with slope 2 is = ( ) = 2 + , where is the -intercept. (b) (2) = 1 means that the point (2 1) is on the graph of . We can use the point-slope form of a line to obtain an equation for the family of linear functions through the point (2 1) . − 1 = ( − 2) , which is equivalent to = + (1 − 2 ) in slope-intercept form. (c) To belong to both families, an equation must have slope = 2 , so the equation in part (b), = + (1 − 2 ) , becomes = 2 − 3 . It is the only function that belongs to both families. or du cessi
Page 15 Read page 15 as text
22 ¤ CHAPTER 1 FUNCTIONS AND MODELS 8. All members of the family of linear functions ( ) = 1 + ( + 3) have graphs that are lines passing through the point ( − 3 1) . 9. All members of the family of linear functions ( ) = − have graphs that are lines with slope − 1 . The -intercept is . 10. The vertex of the parabola on the left is (3 0) , so an equation is = ( − 3) 2 + 0 . Since the point (4 2) is on the parabola, we’ll substitute 4 for and 2 for to fi nd . 2 = (4 − 3) 2 ⇒ = 2 , so an equation is ( ) = 2( − 3) 2 . The -intercept of the parabola on the right is (0 1) , so an equation is = 2 + + 1 . Since the points ( − 2 2) and (1 − 2 5) are on the parabola, we’ll substitute − 2 for and 2 for as well as 1 for and − 2 5 for to obtain two equations with the unknowns and . ( − 2 2) : 2 = 4 − 2 + 1 ⇒ 4 − 2 = 1 (1) (1 − 2 5) : − 2 5 = + + 1 ⇒ + = − 3 5 (2) 2 · (2) + (1) gives us 6 = − 6 ⇒ = − 1 . F rom (2) , − 1 + = − 3 5 ⇒ = − 2 5 , so an equation is ( ) = − 2 − 2 5 + 1 . 11. Since ( − 1) = (0) = (2) = 0 , has zeros of − 1 , 0 , and 2 , so an equation for is ( ) = [ − ( − 1)]( − 0)( − 2) , or ( ) = ( + 1)( − 2) . Because (1) = 6 , we’ll substitute 1 for and 6 for ( ) . 6 = (1)(2)( − 1) ⇒ − 2 = 6 ⇒ = − 3 , so an equation for is ( ) = − 3 ( + 1)( − 2) . 12. (a) For = 0 02 + 8 50 , the slope is 0 02 , which means that the average surface temperature of the world is increasing at a rate of 0 02 ◦ C per year. The -intercept is 8 50 , which represents the average surface temperature in ◦ C in the year 1900. (b) = 2100 − 1900 = 200 ⇒ = 0 02(200) + 8 50 = 12 50 ◦ C 13. (a) = 200 , so = 0 0417 ( + 1) = 0 0417(200)( + 1) = 8 34 + 8 34 . The slope is 8 34 , which represents the change in mg of the dosage for a child for each change of 1 year in age. (b) For a newborn, = 0 , so = 8 34 mg. post , in
Page 16 Read page 16 as text
SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS ¤ 23 14. (a) (b) The slope of − 4 means that for each increase of 1 dollar for a rental space, the number of spaces rented decreases by 4 . The -intercept of 200 is the number of spaces that would be occupied if there were no charge for each space. The -intercept of 50 is the smallest rental fee that results in no spaces rented. 15. (a) (b) The slope of 9 5 means that increases 9 5 degrees for each increase of 1 ◦ C . (Equivalently, increases by 9 when increases by 5 and decreases by 9 when decreases by 5 .) The -intercept of 32 is the Fahrenheit temperature corresponding to a Celsius temperature of 0 . 16. (a) Let = distance traveled (in miles) and = time elapsed (in hours). At = 0 , = 0 and at = 50 minutes = 50 · 1 60 = 5 6 h, = 40 . Thus we have two points: (0 0) and 5 6 40 , so = 40 − 0 5 6 − 0 = 48 and so = 48 . (b) (c) The slope is 48 and represents the car’s speed in mi h. 17. (a) Using in place of and in place of , we fi nd the slope to be 2 − 1 2 − 1 = 80 − 70 173 − 113 = 10 60 = 1 6 . So a linear equation is − 80 = 1 6 ( − 173) ⇔ − 80 = 1 6 − 173 6 ⇔ = 1 6 + 307 6 307 6 = 51 16 . (b) The slope of 1 6 means that the temperature in Fahrenheit degrees increases one-sixth as rapidly as the number of cricket chirps 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 = 1 6 (150) + 307 6 = 76 16 ◦ F ≈ 76 ◦ F . 18. (a) Let denote the number of chairs produced in one day and the associated cost. Using the points (100 2200) and (300 4800) , we get the slope 4800 − 2200 300 − 100 = 2600 200 = 13 . So − 2200 = 13( − 100) ⇔ = 13 + 900 . (b) The slope of the line in part (a) is 13 and it represents the cost (in dollars) of producing each additional chair. (c) The -intercept is 900 and it represents the fi xed daily costs of operating the factory. 19. (a) We are given change in pressure 10 feet change in depth = 4 34 10 = 0 434 . Using for pressure and for depth with the point ( ) = (0 15) , we have the slope-intercept form of the line, = 0 434 + 15 . or du cessi
1102 more pages in the full document
Unlock it once and it stays in your library, ready to chat with or turn into flashcards.
Unlock the full documentMore Mathematics study materials
See all
Exam GuidesTest Bank for Precalculus, 6th Edition
Mathematics
1051 pages175downloadsFeb 2025
Textbook GuidesSolution Manual For Introduction To Management Science, 11th Edition
Mathematics
437 pages167downloadsFeb 2025
Textbook GuidesSolution Manual For Derivatives Markets, 3rd Edition
Mathematics
358 pages143downloadsFeb 2025
Textbook GuidesSolution Manual for Geometry, 1st Edition
Mathematics
446 pages141downloadsFeb 2025
Textbook GuidesSolution Manual for Beginning Algebra, 7th Edition
Mathematics
479 pages139downloadsFeb 2025
Textbook GuidesSolution Manual for Developmental Mathematics, 1st Edition
Mathematics
1623 pages137downloadsFeb 2025
Textbook GuidesSolution Manual for Elementary and Intermediate Algebra: Graphs and Models, 5th Edition
Mathematics
1278 pages126downloadsFeb 2025
Exam GuidesTest Bank for Calculus: Concepts and Contexts, Enhanced 4th Edition
Mathematics
687 pages122downloadsFeb 2025
Study this document with CramX
Once it is in your library, every tool can work from it.