chap03-sols.pdf - 3.1 SOLUTIONS CHAPTER THREE Solutions for...

This preview shows page 1 out of 124 pages.

Unformatted text preview: 3.1 SOLUTIONS CHAPTER THREE Solutions for Section 3.1 Exercises 1. The derivative, ′ (), is defined as ′ () = lim ℎ→0 If () = 7, then ′ () = lim ℎ→0 ( + ℎ) − () . ℎ 7−7 0 = lim = 0. ℎ→0 ℎ ℎ 2. The definition of the derivative says that ′ () = lim ℎ→0 Therefore, ′ () = lim ℎ→0 ( + ℎ) − () . ℎ [17( + ℎ) + 11] − [17 + 11] 17ℎ = lim = 17. ℎ→0 ℎ ℎ 3. The is in the exponent and we haven’t learned how to handle that yet. 4. ′ = 32 . (power rule) 5. ′ = −1 . ′ (power rule) 11 6. = 12 . 7. ′ = 1110 . 8. ′ = 11−12 . 9. ′ = −12−13 . 10. ′ = 3.22.2 . 11. ′ = − 34 −7∕4 . 12. ′ = 34 1∕3 . 13. ′ = 34 −1∕4 . = 2 + 5. 15. ′ () = 32 − 6 + 8. 14. 16. ′ () = −4−5 . 1 17. Since () = 5 = −5 , we have ′ () = −5−6 . 1 18. Since () = − 6.1 = −−6.1 , we have ′ () = −(−6.1)−7.1 = 6.1−7.1 . 1 7 19. Since = 7∕2 = −7∕2 , we have = − −9∕2 . 2 √ 1 −1∕2 1∕2 = . 20. Since = = , we have 2 21. ′ () = 14 −3∕4 . 1 1 22. Since ℎ() = √ = −1∕3 , we have ℎ′ () = − −4∕3 . 3 3 √ 1 1 3 23. Since () = = 3∕2 = −3∕2 , we have ′ () = − −5∕2 . 2 3 223 224 Chapter Three /SOLUTIONS 24. ℎ() = ⋅ ln = , so ℎ′ () = . 25. ′ = 61∕2 − 25 −1∕2 . 26. ′ () = 6 − 4. 27. ′ = 17 + 12−1∕2 . 28. ′ = 2 − 1 . 22 29. The power rule gives ′ () = 203 − 2 . 3 3 30. ℎ′ () = 6−4 + −1∕2 2 31. ′ = −123 − 122 − 6. 32. ′ = 154 − 25 −1∕2 − 33. ′ = 6 − 6 3∕2 + 7 . 2 2 . 3 √ 3 1 = 1∕2 + −1∕2 . ( + 1) = 1∕2 + 1∕2 ⋅ 1 = 3∕2 + 1∕2 , we have 2 2 √ 35. Since = 3∕2 (2 + ) = 23∕2 + 3∕2 1∕2 = 23∕2 + 2 , we have = 31∕2 + 2. 34. Since = 36. Since ℎ() = 3 4 + = 3−1 + 4−2 , we have ℎ′ () = −3−2 − 8−3 . 2 37. Since ℎ() = ( −1∕2 − −2 ) = −1∕2 − −2 = 1∕2 − −1 , we have ℎ′ () = 38. = + 1 , so ′ = 1 − 1 . 2 ) ) 1 1( 1 −1 1 ( + −1 , so ′ () = 1 − −2 = + = 3 3 3 3 3 ) ( 1 1 5 3 1 3 − − − ′ −1 −2 40. () = 2 + + 2 = 2 − − 2 . 2 2 √ 41. = √ − √1 = − √1 39. () = 43. 44. 45. 46. ) 2 − 1 . 2 + 213∕2 . √ (1 + ) 1∕2 ⋅ 1 + 1∕2 1∕2 3∕2 3 1 Since () = = = 2 + 2 = −3∕2 + −1∕2 , we have ′ () = − −5∕2 − −3∕2 . 2 2 2 2 32 2 + − ′ () = + Since () = = + = + −1 , we have ′ () = −−2 . + Since ℎ() = = + , we have ℎ′ () = . Since 4∕3, , and are all constants, we have ′ = 42. 1 √ 2 ( 1 −1∕2 + −2 . 2 4 8 = (2) = . 3 3 47. Since is a constant times , we have ∕ = 32 . 48. Since , , and are all constants, we have = (2) + (1) + 0 = 2 + . 49. Since and are constants, we have 1 = 0 + −1∕2 = √ . 2 2 50. We have Δ = 4.02 − 4 = 0.02. From the tangent line approximation we get: Δ ≈ ′ (4)Δ = 7(0.02) = 0.14. 3.1 SOLUTIONS 225 Thus, (4.02) = (4) + Δ ≈ 5 + 0.14 = 5.14. 51. We have Δ = 3.92 − 4 = −0.08. From the tangent line approximation we get: Δ ≈ ′ (4)Δ = 7(−0.08) = −0.56. Thus, (3.92) = (4) + Δ ≈ 5 − 0.56 = 4.44. 52. We have Δ = 5.03 − 5 = 0.03. From the tangent line approximation we get: Δ ≈ ′ (5)Δ = −2(0.03) = −0.06. Thus, (5.03) = (5) + Δ ≈ 3 − 0.06 = 2.94. 53. We have Δ = 1.95 − 2 = −0.05. From the tangent line approximation we get: Δ ≈ ′ (2)Δ = −3(−0.05) = 0.15. Thus, (1.95) = (2) + Δ ≈ −4 + 0.15 = −3.85. 54. We have Δ = −2.99 − (−3) = 0.01. From the tangent line approximation we get: Δ ≈ ′ (−3)Δ = 2(0.01) = 0.02. Thus, (−2.99) = (−3) + Δ ≈ −4 + 0.02 = −3.98. 55. We have Δ = 2.99 − 3 = −0.01. From the tangent line approximation we get: Δ ≈ ′ (3)Δ = −2(−0.01) = 0.02. Thus, (2.99) = (3) + Δ ≈ −4 + 0.02 = −3.98. 56. We have Δ = 1.04 − 1 = 0.04. Differentiating,we get ′ () = 43 − 2, so ′ (1) = 2. Δ ≈ ′ (1)Δ = 2(0.04) = 0.08. Thus, (1.04) = (1) + Δ ≈ 3 + 0.08 = 3.08. 57. We have Δ = 0.97 − 1 = −0.03. Differentiating,we get ′ () = 32 + 2, so ′ (1) = 5. Δ ≈ ′ (1)Δ = 5(−0.03) = −0.15. Thus, (0.97) = (1) + Δ ≈ −4 + (−0.15) = −4.15. Problems 58. We use the power rule to differentiate the polynomial term by term to get ′ () = 243 − 2. Similarly, ′′ () = 722 and ′′′ () = 144. 59. (a) We use the power rule to get ′ () = 76 . Similarly, ′′ () = 425 and ′′′ () = 2104 . (b) By computing the first three derivatives, we observe that with each higher derivative the exponent decreases by 1. Eventually the exponent will be 0, which means that this derivative is a constant function. The next derivative will be zero. Therefore, we have (8) () = 0. 226 Chapter Three /SOLUTIONS 60. Differentiating, we have ∕ = 32 + 4 − 1, so ( ) 2 = = (32 + 4 − 1) = 6 + 4, 2 and 3 = 3 ( 2 2 ) = (6 + 4) = 6. 61. First, we rewrite the given function as () = 1∕2 + −1∕2 . Differentiating, we have ∕ = (1∕2)−1∕2 − (1∕2)−3∕2 , so ( ) ( ) 1 −1∕2 1 −3∕2 1 3 2 = − = − −3∕2 + −5∕2 , = 2 2 2 4 4 and ) ( ( ) 3 1 3 3 15 2 = = − −3∕2 + −5∕2 = −5∕2 − −7∕2 . 3 2 4 4 8 8 62. So far,√we can only take the derivative of powers of and the sums of constant multiples of powers of . Since we cannot write + 3 in this form, we cannot yet take its derivative. 63. The is in the exponent and we have not learned how to handle that yet. 64. ′ () = (−1) + −(+1) , differentiating term-by-term and using the power rule. 65. ′ = 6, differentiating term-by-term and using the power rule. 66. We cannot write 1 32 +4 as the sum of powers of multiplied by constants. 67. ′ = −2∕33 , differentiating term-by-term and using the power rule 68. We multiply out () = (3 + 8)(2 − 5) = 62 + − 40, and differentiate term by term to get ′ () = 12 + 1 and ′′ () = 12. 69. Differentiating gives ′ () = 62 − 4 so ′ (1) = 6 − 4 = 2. Thus the equation of the tangent line is ( − 1) = 2( − 1) or = 2 − 1. 70. To calculate the equation of the tangent line to = () = 2 + 3 − 5 at = 2, we need to find the -coordinate and the slope at = 2. The -coordinate is = (2) = 22 + 3(2) − 5 = 5, so a point on the line is (2, 5). The slope is found using the derivative: ′ () = 2 + 3. At the point = 2, we have Slope = ′ (2) = 2(2) + 3 = 7. The equation of the line is − 5 = 7( − 2) = 7 − 9. When we graph the function and the line together, the line = 7 − 9 appears to lie tangent to the curve = 2 + 3 − 5 at the point = 2 as we expect. 71. The slope of the tangent line is the value of the first derivative at = 2. Differentiating gives ( 3 ) ( ) 1 3 4 −1 4 = − − 2 3 2 3 4 1 = ⋅ 32 − (−1)−2 2 3 3 4 = 2 + 2 . 2 3 3.1 SOLUTIONS For = 2, ′ (2) = 227 3 2 1 4 = 6 + = 6.333 (2) + 2 3 3(2)2 and 23 4 2 − = 4 − = 3.333. 2 3(2) 3 To find the -intercept for the tangent line equation at the point (2, 3.333), we substitute in the general equation, = +, and solve for . (2) = 3.333 = + 6.333(2) −9.333 = . The tangent line has the equation = −9.333 + 6.333. 72. (a) We have (2) = 8, so a point on the tangent line is (2, 8). Since ′ () = 32 , the slope of the tangent is given by = ′ (2) = 3(2)2 = 12. Thus, the equation is − 8 = 12( − 2) or = 12 − 16. (b) See Figure 3.1. The tangent line lies below the function () = 3 , so estimates made using the tangent line are underestimates. 30 = 3 = 12 − 16 20 10 −2 2 4 −10 Figure 3.1 73. (a) The slope of at = 1 is negative, so ′ (1) < 0. On the other hand, the slopes of at = −1, = 0, and = 4 are all positive, with the least steep slope occurring at = 0 and the steepest slope occurring at = 4. Therefore, we must have ′ (1) < ′ (0) < ′ (−1) < ′ (4). (b) The derivative function of is given by ′ () = 32 − 6 + 2. ′ Evaluating at = −1, 0, 1, 4, we have (1) = −1, ′ (0) = 2, ′ (−1) = 11 and ′ (4) = 26 which confirms the ordering of the four quantities. 74. ′ = 32 − 18 − 16 5 = 32 − 18 − 16 0 = 32 − 18 − 21 0 = 2 − 6 − 7 0 = ( + 1)( − 7) = −1 or = 7. When = −1, = 7; when = 7, = −209. Thus, the two points are (−1, 7) and (7, −209). 228 Chapter Three /SOLUTIONS 75. Decreasing means ′ () < 0: ′ () = 43 − 122 = 42 ( − 3), so ′ () < 0 when < 3 and ≠ 0. Concave up means ′′ () > 0: ′′ () = 122 − 24 = 12( − 2) so ′′ () > 0 when 12( − 2) > 0 < 0 or > 2. So, both conditions hold for < 0 or 2 < < 3. 76. The graph increases when ∕ > 0: = 54 − 5 > 0 5(4 − 1) > 0 so 4 > 1 so > 1 or < −1. The graph is concave up when 2 ∕2 > 0: 2 = 203 > 0 so > 0. 2 We need values of where { > 1 or < −1} AND { > 0}, which implies > 1. Thus, both conditions hold for all values of larger than 1. 77. (a) Since the power of will go down by one every time you take a derivative (until the exponent is zero after which the derivative will be zero), we can see immediately that (8) () = 0. (b) (7) () = 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 ⋅ 0 = 5040. 78. (a) Differentiating, we have ′ () = 32 − 1, so ′ (−1) = 3(−1)2 − 1 = 2. (b) Taking the second derivative, we have ′′ () = 6, so ′′ (−1) = 6(−1) = −6. (c) Since ′ (−1) = 2, we see that () is increasing at = −1. The only function that is increasing at = −1 is (III), so (III) must be the graph of (). 79. (a) Differentiating, we have ′ () = 62 + 6, so ′ (−1) = 6(−1)2 + 6(−1) = 0. (b) Taking the second derivative, we have ′′ () = 12 + 6, so ′′ (−1) = 12(−1) + 6 = −6. (c) Since ′ (−1) = 0, we see that () has a horizontal tangent line at = −1. This means it could match (I) or (IV). Since ′′ (−1) = −6, we see that () is concave down at = −1, so () matches (IV). 80. (a) Differentiating, we have ′ () = 43 − 2 − 2, so ′ (−1) = 4(−1)3 − 2(−1) − 2 = −4. (b) Taking the second derivative, we have ′′ () = 122 − 2, so ′′ (−1) = 12(−1)2 − 2 = 10. (c) Since ′ (−1) = −4, () is decreasing at = −1. The only function which is decreasing at = −1 is (II), so (II) must be the graph of (). 3.1 SOLUTIONS 229 81. (a) Differentiating, we have ℎ′ () = 83 + 242 + 30 + 14, so ℎ′ (−1) = 8(−1)3 + 24(−1)2 + 30(−1) + 14 = 0. (b) Taking the second derivative, we have ℎ′′ () = 242 + 48 + 30, so ℎ′′ (−1) = 24(−1)2 + 48(−1) + 30 = 6. (c) Since ℎ′ (−1) = 0, we see that ℎ() has a horizontal tangent line at = −1. This means it could match (I) or (IV). Since ℎ′′ (−1) = 6, we see that ℎ() is concave up at = −1, so ℎ() matches (I). 82. Since () = 700 − 32 , we have (5) = 700 − 3(25) = 625 cm. Since ′ () = −6, we have ′ (5) = −30 cm/year. In the year 2021, the sand dune will be 625 cm high and eroding at a rate of 30 centimeters per year. 83. (a) We have (2) = 25∕2 = 12.5 liters, which means that the volume of gas in the balloon is 12.5 liters when the outside pressure is 2 atmospheres. (b) First, observe that ( ) 25 25 ′ ( ) = = (25 −1 ) = −25 −2 = − 2 , so when the pressure is 2 atmospheres 25 ′ (2) = − 2 = −6.25. 2 Since is measured in atmospheres and is measured in liters, the units of ′ (2) are liters per atmosphere. Therefore, our calculation indicates that, at an outside pressure of 2 atmospheres, the volume of the gas in the balloon decreases at a rate of about 6.25 liters for each atmosphere by which the outside pressure is increased. (c) Since ′ (1) = −25∕12 = −25 liters per atmosphere and ′ (2) = −25∕22 = −6.25 liters per atmosphere, we see that the volume of the balloon decreases faster when the outside pressure is 1 atmosphere. 84. (a) Velocity () = = (1250 − 162 ) = −32. Since ≥ 0, the ball’s velocity is negative. This is reasonable, since its height is decreasing. (b) Acceleration () = = (−32) = −32. So its acceleration is the negative constant −32. (c) The ball hits the ground when its height = 0. This gives 1250 − 162 = 0 = ±8.84 seconds We discard = −8.84 because time is nonnegative. So the ball hits the ground 8.84 seconds after its release, at which time its velocity is (8.84) = −32(8.84) = −282.88 feet/sec = −192.84 mph. 85. (a) The average velocity between = 0 and = 2 is given by Average velocity = (2) − (0) −4.9(22 ) + 25(2) + 3 − 3 33.4 − 3 = = = 15.2 m/sec. 2−0 2−0 2 (b) Since ′ () = −9.8 + 25, we have Instantaneous velocity = ′ (2) = −9.8(2) + 25 = 5.4 m/sec. (c) Acceleration is given ′′ () = −9.8. The acceleration at = 2 (and all other times) is the acceleration due to gravity, which is −9.8 m/sec2 . (d) We can use a graph of height against time to estimate the maximum height of the tomato. See Figure 3.2. Alternately, we can find the answer analytically. The maximum height occurs when the velocity is zero and () = −9.8 + 25 = 0 when = 2.6 sec. At this time the tomato is at a height of (2.6) = 34.9. The maximum height is 34.9 meters. 230 Chapter Three /SOLUTIONS height (m) 34.9 2.6 5.2 (sec) Figure 3.2 (e) We see in Figure 3.2 that the tomato hits ground at about = 5.2 seconds. Alternately, we can find the answer analytically. The tomato hits the ground when () = −4.92 + 25 + 3 = 0. We solve for using the quadratic formula: = −25 ± √ (25)2 − 4(−4.9)(3) 2(−4.9) √ −25 ± 683.8 = −9.8 = −0.12 and = 5.2. We use the positive values, so the tomato hits the ground at = 5.2 seconds. 86. (a) Given ℎ() = () − (), this means ℎ′ () = ′ () − ′ (), so ℎ(0) = (0) − (0) = 2000 − 1500 = 500 ℎ′ (0) = ′ (0) − ′ (0) = 11 − 13.5 = −2.5. This tells us that initially there are 500 more acre-feet of water in the first reservoir than the second, but that this difference drops at an initial rate of 2.5 acre-feet per day. (b) Assume ℎ′ is constant, this means ℎ′ () = −2.5. It also means that ℎ is a linear function: ℎ() = Starting value + Rate of change × Time ⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟ ⏟⏟⏟ ℎ(0) ℎ′ (0) = 500 − 2.5. To find zeros of ℎ, we write: ℎ() = 0 500 − 2.5 = 0 = 200. This tells us the difference in water level will be zero—that is, the water levels will be equal—after 200 days. 87. (a) Using the power rule, we have = (7700)(0.67)0.67−1 = 5159−0.33 . (b) We have || = (5159)(0.1)−0.33 = 11030 m3 ∕sec per km3 . ||=0.1 For glacial lakes with volumes near 0.1 km3 , a small increase of Δ km3 in volume corresponds to a change in the maximum outflow rate of about 11, 030Δ m3 ∕sec. Larger lakes have a larger maximum outflow during a jökulhlaup. 3.1 SOLUTIONS 231 88. (a) Let = () = . Then we have ′ () = −1 . Thus, Δ ≈ −1 Δ. Dividing through by = , we get: Δ −1 Δ Δ ≈ = . (b) The percent change in from = is Δ∕. The percent change in is Δ∕. Thus, we have near = : Percent change in ≈ ⋅ Percent change in . 89. We have = 2 . Thus, the multiplier for percent error is the power 2, so Δ Δ ≈2 = 2(5%) = 10%. The percent error in is approximately 2(5%) = 10%. 90. We have = 43 ∕3, so the multiplier for percent error is the power 3: Δ Δ ≈3 . To get Δ ∕ = 3%, we take Δ∕ = 1%. Thus, for 3% in accuracy in , we need approximately 1% accuracy in . 91. We have = 2 ∕20, so the multiplier for the percent change is the power 2: Δ Δ ≈2 = 2(10%) = 20%. The percent change in is approximately 2(10%) = 20%. 92. Since = 3 , the multiplier for the percent change is the power 3. An increase from 9 to 10 is 11%. Therefore, we have: Δ Δ ≈3 = 3(11%) = 33%. The percent change in power is approximately 3(11%) = 33%. Thus, small differences in average wind speed are important in deciding where to locate wind turbines. 93. (a) Using the power rule we have ′ () = 2 (0.07)−1∕3 = 0.0467−1∕3 . 3 (b) We have ′ (30) = (0.0467)30−1∕3 = 0.015 mm per meter. (c) We can approximate this difference using the derivative ′ (30) as follows Δℎ ≈ ′ (30)Δ = 0.015 ⋅ 6 = 0.09 mm. The runoff is about 0.09 mm deeper at a distance of 6 meters farther down the slope. To check, we can calculate this difference exactly; it is Δℎ = (36) − (30) = 0.07(36)2∕3 − 0.07(30)2∕3 = 0.0873 mm. 94. (a) We have ( ) ( −2 ) 2 1 = = (−2)−3 = − 3 . = 2 (b) The derivative ∕ is the rate of change of acceleration due to the pull of gravity with respect to distance from the center of the earth. The further away from the center of the earth, the weaker the pull of gravity is. So is decreasing and therefore its derivative, ∕, is negative. (c) By part (a), 2 || 2(6.67 ⋅ 10−20 )(6 ⋅ 1024 ) 2 || −6 km∕sec ≈ −3.05 ⋅ 10 . =− =− 3 | | ||=6400 km ||=6400 (6400)3 (d) Since the magnitude of ∕ is small, the value of is not changing much near = 6400 km. It is reasonable to assume that is a constant near the surface of the earth. 232 Chapter Three /SOLUTIONS √ 95. (a) = 2 (b) Since ( 1) ) ( 2 2 1 − 12 = √ . = √ 2 , so = √ 2 is positive, the period increases as the length increases. 96. (a) = 2 = 2. (b) This is the formula for the circumference of a circle. (c) ′ () ≈ (+ℎ)−() for small ℎ. When ℎ > 0, the numerator of the difference quotient denotes the area of the region ℎ contained between the inner circle (radius ) and the outer circle (radius + ℎ). See figure below. As ℎ approaches 0, this area can be approximated by the product of the circumference of the inner circle and the “width” of the region, i.e., ℎ. Dividing this by the denominator, ℎ, we get ′ = the circumference of the circle with radius . ✛ ℎ We can also think about the derivative of as the rate of change of area for a small change in radius. If the radius increases by a tiny amount, the area will increase by a thin ring whose area is simply the circumference at that radius times the small amount. To get the rate of change, we divide by the small amount and obtain the circumference. 97. If () = , then ′ () = −1 . This means ′ (1) = ⋅ 1−1 = ⋅ 1 = , because any power of 1 equals 1. 98. Since is proportional to 3 , we have = 3 for some constant . Thus, ∕ = (32 ) = 32 . Thus, ∕ is proportional to 2 . 99. Since () = , ′ () = −1 . We know that ′ (2) = ()2−1 = 3, and ′ (4) = ()4−1 = 24. Therefore, ()4−1 ()2−1 ′ (4) 24 = ′ (2) 3 ( )−1 4 = =8 2 2−1 = 8, and thus = 4. Substituting = 4 into the expression for ′ (2), we get 3 = (4)(8), or = 3∕32. 100. (a) We see that (1) = ⋅ 1 = , and the graph of 2 + 3 goes through the point (1, 4), so () is continuous when = 4. (b) No, () does not have a derivative at (1, 4). There is a corner there. We can see this without a graph by noticing that 2 ( + 3) = 2 has a value of 2 at = 1, but (4) = 4 has a value of 4 at = 1. 101. (a) Differentiating () and (), we have ′ () = 32 + 6 − 2 and ′ () = 32 + 6 − 2. (b) The graph of () is a vertical shift of the graph of () down by 5 units, so it has the same slope as () for every . Since derivatives measure slope, it follows that ′ () = ′ (). (c) Any function whose graph is a vertical shift of () or () has the same derivative as () or (). 3.1 SOLUTIONS 233 102. (a) To show that () is even, we need to show (−) = () for all . We have (−) = (−)4 − 3(−)2 + 1 = 4 − 32 + 1 = (). (b) The derivative of () is ′ () = 43 − 6. To show that ′ () is odd, we need to show (−) = − () for all . We have ′ (−) = 4(−)3 − 6(−) = −43 + 6 = −(43 − 6) = − ′ (). (c) If () is a polynomial that is an even function, then all terms are even powers of . This means all terms of ′ () will be odd powers (since the power rule subtracts one from each power), so ′ (−) = − ′ () meaning ′ () is an odd function. Strengthen Your Understanding 2 ( + ) = 2 for any constant a. 104. The function can be written as () = −2 so the power rule gives ′ () = −2−3 = −2∕3 . 103. Since the derivative of a constant is zero, 105. One possible example is () = 2 and () = 3. More generally, () = 2 + and () = 3 + work for any and . 106. Any function of the form () = 2 + , where is a positive constant works. One possibility is () = 2 + 1. 107. If () = 32 , we have ′ () = 6 and ′′ () = 6. Other answers are possible. 108. True. Since ( )∕ = −1 , the derivative of a power function is a power function, so the derivative of a polynomial is a polynomial. 109. False, since ( ) ( −2 ) −2 = = −2−3 = 3 . 2 110. True. The slope of () + () at = 2 is the sum of the derivatives, ′ (2) + ′ (2) = 3.1 + 7.3 = 10.4. 111. True. Since ′′ () > 0 and ′′ () > 0 for all , we have ′′ () + ′′ () > 0 for all , which means that () + () is concave up. 112. False. Let () = 22 and () = 2 . Then () − () = 2 , which is concave up for all . Additional Problems (online only) 113. The slopes of the tangent lines to = 2 − 2 + 4 are given by ′ = 2 − 2. A line through the origin has equation = . So, at the tangent point, 2 − 2 + 4 = where = ′ = 2 − 2. 2 − 2 + 4 = (2 − 2) 2 − 2 + 4 = 22 − 2 −2 + 4 = 0 −( + 2)( − 2) = 0 = 2, −2. Thus, the points of tangency are (2, 4) and (−2, 12). The lines through these points and the origin are = 2 and = −6, respectively. Graphically, this can be seen in Figure 3.3. = 2 − 2 + 4 (−2, 12) = 2 = −6 (2, 4) Figure 3.3 234 Chapter Three /SOLUTIONS 114. ′ () = 122 + 12 − 23 ≥ 1 122 + 12 − 24 ≥ 0 12(2 + − 2) ≥ 0 12( + 2)( − 1) ≥ 0. Hence ≥ 1 or ≤ −2. 115. (a) We have () = 2 − . We see that ′ () = 2 − 1 < 0 when < 12 . So is decreasing when < 12 . (b) We have () = 1∕2 − , so ′ () = 1 −1∕2 −1 < 0 2 1 −1∕2 <1 2 −1∕2 < 2 1 1∕2 > 2 1 > . 4 Thus () is d...
View Full Document

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture