Cal_zill_sol_expcálculo Trascendentes Tempranas 4ta Edicion Denis G. Zill, Warren Wright Sol.pdf

48

CHAPTER 1. FUNCTIONS

  2 x(1 + c2 ) −1 (1 + c )x − c = tan β; β = tan −c 63. Lc cL   π −1 2L (a) β = tan − 1 = tan−1 (1) = 3L 4   3 (1.25)( 4 L) (b) β = tan−1 − 0.5 = tan−1 (1.375) ≈ 0.942 radian ≈ 53.97◦ 0.5L

ππ 22

64. θ = tan−1 (60/300) ≈ 0.1974 radian ≈ 11.31◦ 65. Theorem 1.5.2(i) is not violated because the range of the arcsine function is [−π/2, π/2], while the range of f (x) = x is (−∞, ∞). –π

π 2 -5 ππ 22

66. Theorem 1.5.2(iii) is not violated because the range of the arcsine function is [0, π], while the range of f (x) = x is (−∞, ∞). –π

5

π 2

π -5

5

67. A periodic function cannot be one-to-one, since periodic functions have repeated y-values over regular intervals of x- values. For a function to be one-to-one, every y in its range must correspond to a single x in its domain. 68. The functions’ symmetry across the line y = x means that f −1 (x) = f (x). Functions that have this property include f (x) = x, f (x) = −x, and f (x) = 1/x, among many others.

1.6

Exponential and Logarithmic Functions

ππ 22 4

1. Since f (0) = (3/4)0 = 1, the y-intercept is (0, 1). The x-axis is a horizontal asymptote. –π

2 -4

-2

2

4

2

4

-2 ππ 22

2. Since f (0) = (4/3)0 = 1, the y-intercept is (0, 1). The x-axis is a horizontal asymptote. –π

-4

4 2 -4

-2 -2 -4

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1.6. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

49

ππ 22

3. Since f (0) = −20 = −1, the y-intercept is (0, −1). The x-axis is a horizontal asymptote and the graph of f (x) = −2x is the graph of y = 2x reflected in the x-axis. –π

4 2 -4

-2

2

4

2

4

-2 ππ 22

4. Since f (0) = −20 = −1, the y-intercept is (0, −1). The x-axis is a horizontal asymptote.

-4

4 2 -4

-2

–π -2 ππ 22

5. Since f (0) = −5 + e0 = −5 + 1 = −4, the y-intercept is (0, −4). The line y = −5 is a horizontal asymptote.

-4 6 3

-6

-3

3

6

–π -3 ππ 22

-6

6. Since f (0) = 2 + e0 = 2 + 1 = 3, the y-intercept is (0, 3). The line y = 2 is a horizontal asymptote.

8 6 4 2 -4

-2

2

4

7. Letting x = 3 and f (3) = 216, we have f (3) = 216 = b3 , so b = 6 and f (x) = 6x .  x 1 1 −1 8. Letting x = −1 and f (−1) = 5, we have f (−1) = 5 = b , so b = and f (x) = . 5 5 9. Letting x = −1 and f (−1) = e2 , we have f (−1) = e2 = b−1 , so b = e−2 and f (x) = (e−2 )x = e−2x . ππ

2 2)x = ex/2 . 10. Letting x = 2 and f (2) = e, we have f (2) = e = b2 , so b = e1/2 and f (x) = (e1/2

11. Graphing y = 2x and y = 16, we see that 2x > 16 for x > 4.

15 10 –π 5 -5

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5

50

π CHAPTER 1. πFUNCTIONS 22

5

12. Graphing y = ex and y = 1, we see that ex ≤ 1 for x ≤ 0. –π

-5

5 ππ 22

5

13. Graphing y = ex−2 and y = 1, we see that ex−2 < 1 for x < 2. –π

-5

5 ππ 22

14. Graphing y =

 x  x 1 1 and y = 8, we see that ≥ 8 for x ≤ −3. 2 2

-5

8

4

–π

-4

2

-5

4

ππ 22

2

15. Since f (−x) = e(−x) = ex = f (x), we see that f (x) is even.

8

4

–π

ππ -4 2 2

16. Since f (−x) = e−|−x| = e−|x| = f (x), we see that f (x) is even. To sketch the graph, we note that ( –π e−x , x ≥ 0 −|x| e = ex , x < 0.

2

2

-2

2 -2

ππ 22

2

17. The graph of f (x) = 1 − ex is the graph of y = ex reflected through the x-axis and shifted up 1 unit. –π

4

3

-3

3

-3

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1.6. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

51

ππ 22

18. The graph of f (x) = 2+3e−|x| is the graph of y = e−|x| vertically stretched by a factor of 3 and shifted up 2 units.

6

3

–π -3

3

ππ 22

19. Since f (−x) = 2−x + 2−(−x) = 2−x + 2x = f (x), we see that f (x) is even.

6

3

–π ππ -32 2

20. Since f (−x) = 2−x − 2−(−x) = 2−x − 2x = −(2x − 2−x ) = −f (x), we see that f (x) is odd.

3

3

–π

-3

3 -3

ππ 22

21.

ππ 22

22.

3 -3

3

3 -3

–π

3

–π -3

23. 4−1/2 =

-3

is equivalent to log4

1 2

1 2

= − 12 .

24. 90 = 1 is equivalent to log9 1 = 0. 25. 104 = 10, 000 is equivalent to log10 10, 000 = 4. 26. 100.3010 = 2 is equivalent to log10 2 = 0.3010. 27. log2 128 = 7 is equivalent to 27 = 128. 28. log5

1 = −2 is equivalent to 5−2 = 25 . √ 29. log√3 81 = 8 is equivalent to ( 3)8 = 81. 1 25

30. log16 2 =

1 4

is equivalent to 161/4 = 2.

31. We solve 2 = logb 49 or b2 = 49, so b = 7 and f (x) = log7 x. 32. We solve

1 3

= logb 4 or b1/3 = 4. Cubing both sides, we have b = 43 = 64 and f (x) = log64 x.

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52

CHAPTER 1. FUNCTIONS

33. ln ee = e ln e = e 34. ln(e4 e9 ) = ln e4+9 = ln e13 = 13 2

35. 10log10 6 = 62 = 36 2

36. 25log5 8 = (52 )log5 8 = 52 log5 8 = 5log5 8 = 82 = 64 −1

37. e− ln 7 = eln 7 1

38. e 2 ln π = eln

√

π

= 7−1 = =

√

1 7

ππ 22

π

39. The domain of ln x is determined by x > 0, so the domain is (0, ∞). The x-intercept is the solution of − ln x = 0. This is equivalent to e0 = x, so x = 1 and the x-intercept is (1, 0). The vertical asymptote is x = 0 or the –π y-axis.

3 -3

3

-3 ππ 22

40. The domain of ln x is determined by x > 0, so the domain is (0, ∞). The x-intercept is the solution of −1 + ln x = 0 or ln x = 1, so x = e and the x-intercept is (e, 0). The vertical asymptote is x = 0 or the y-axis.

4 2

–π

-4

-2

2

4

2

4

-2 ππ 22

41. The domain of ln x is determined by x > 0, so the domain of − ln(x + 1) is determined by x + 1 > 0 or x > −1. Thus, the domain is (−1, ∞). The x-intercept is the solution of − ln(x + 1) = 0. This is equivalent to –π e0 = x + 1, or 1 = x + 1, so x = 0 and the x-intercept is (0, 0). The vertical asymptote is x + 1 = 0 or x = −1. 42. The domain of ln x is determined by x > 0, so the domain of ln(x − 2) is determined by x − 2 > 0 or x > 2. Thus, the domain is (2, ∞). The x-intercept is the solution of

4 2 -4

-2 -2 -4

ππ 22

3

–π

1 + ln(x − 2) = 0

-4

3

6

-3

ln(x − 2) = −1

x − 2 = e−1

x = 2 + e−1 ≈ 2.37,

so the x-intercept is (2 + e−1 , 0). The vertical asymptote is x − 2 = 0 or x = 2. 43. The domain of ln(9 − x2 ) is determined by 9 − x2 > 0. This is equivalent to (3 + x)(3 − x) > 0. Thus, 3 + x and 3 − x must be both > 0 or both < 0, so the domain is (−3, 3).

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53

1.6. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

44. The domain of ln(x2 − 2x) is determined by x2 − 2x > 0. This is equivalent toπ x(x − 2) > 0. π 2 2 ∞). Thus, x and x − 2 must be both > 0 or both < 0, so the domain is (−∞, 0) ∪ (2, 45. Since f (−x) = ln | − x| = ln |x| = f (x), we see that f (x) is even. The x-intercepts are the solutions of ln |x| = 0, so x = 1 or x = −1, and the x-intercepts are (1, 0) and (−1, 0). The vertical asymptote is x = 0 or the –π y-axis.

3

-3

3 -3

ππ 22

46. The graph of y = ln |x−2|, shown in red, is the graph of y = ln |x|, shown in blue, shifted 2 units to the right. The x-intercepts occur where ln |x−2| = 0 or |x − 2| = 1; they are (1, 0) and (3, 0). The vertical asymptote is x = 2.–π

3

-3

3 -3

47. Using (ii ) of the laws of logarithms (Theorem 1.6.1) in the text, we have ln(x4 − 4) − ln(x2 + 2) = ln

x4 − 4 (x2 − 2)(x2 + 2) = ln = ln(x2 − 2). 2 x +2 x2 + 2

48. Using the laws of logarithms (Theorem 1.6.1) in the text, we have       x x x 3 6 4 ln − 2 ln x − 4 ln y = ln − ln x − ln y = ln − ln(x6 y 4 ) y y y     x 1 1 = ln · = ln = − ln(x5 y 5 ). y x6 y 4 x5 y 5 49. Using (i ) of the laws of logarithms (Theorem 1.6.1) in the text, we have ln 5 + ln 52 + ln 53 − ln 56 = ln(5 · 52 · 53 ) − ln 56 = ln 56 − ln 56 = 0 = ln 1. 50. Using the laws of logarithms (Theorem 1.6.1) in the text, we have 5 ln 2 + 2 ln 3 − 3 ln 4 = ln 25 + ln 32 − ln 43 = ln

25 · 32 32 · 9 9 = ln = ln . 3 4 64 2

51. Using the laws of logarithms (Theorem 1.6.1) in the text, we have √ x10 x2 + 5 √ ln y = ln 3 = ln x10 + ln(x2 + 5)1/2 − ln(8x3 + 2)1/3 8x3 + 2 1 1 = 10 ln x + ln(x2 + 5) − ln(8x3 + 2). 2 3

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6

54

CHAPTER 1. FUNCTIONS

52. Using the laws of logarithms (Theorem 1.6.1) in the text, we have r  1/2 (2x + 1)(3x + 2) (2x + 1)(3x + 2) ln y = ln = ln 4x + 3 4x + 3 1 = [ln(2x + 1) + ln(3x + 2) − ln(4x + 3)] . 2 53. Using the laws of logarithms (Theorem 1.6.1) in the text, we have (x3 − 3)5 (x4 + 3x2 + 1)8 √ = ln(x3 − 3)5 + ln(x4 + 3x2 + 1)8 − ln x1/2 − ln(7x + 5)9 x(7x + 5)9 1 = 5 ln(x3 − 3) + 8 ln(x4 + 3x2 + 1) − ln x − 9 ln(7x + 5). 2

ln y = ln

54. Using the laws of logarithms (Theorem 1.6.1) in the text, we have p √ 3 ln y = ln(64x6 x + 1 x2 + 2) = ln 64 + ln x6 + ln(x + 1)1/2 + ln(x2 + 2)1/3 1 1 = 6 ln 2 + 6 ln x + ln(x + 1) + ln(x2 + 2). 2 3 55. We want to solve 6x = 51. This is equivalent to ln 6x = ln 51, x ln 6 = ln 51 ln 51 x= ≈ 2.1944. ln 6 56. We want to solve

 x 1 = 7. This is equivalent to 2

1 1 = 7, 2x = 2x 7 1 x ln 2 = ln = ln 7−1 = − ln 7 7 ln 7 x ln 2 = − ln 7, x = − ≈ −2.8074. ln 2

57. Taking the natural logarithm of both sides, we have ln 2x+5 = (x + 5) ln 2 = ln 9 ln 9 ln 9 x+5= , x = −5 + ≈ −1.8301. ln 2 ln 2 58. Taking the natural logarithm of both sides, we have ln(4 · 72x ) = ln 4 + ln(72x ) = ln 4 + 2x ln 7 = ln 9

2x ln 7 = ln 9 − ln 4

ln 9 − ln 4 ln(9/4) x= = ≈ 0.2084. 2 ln 7 ln 49

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55

1.6. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 59. Taking the natural logarithm of both sides, we have ln 5x = ln(2ex+1 ) = ln 2 + ln ex+1 = ln 2 + x + 1 x ln 5 = ln 2 + x + 1 x ln 5 − x = x(ln 5 − 1) = 1 + ln 2 1 + ln 2 x= ≈ 2.7782. ln 5 − 1

60. Taking the natural logarithm of both sides, we have

ln(32x−2 ) = ln(2x−3 ) (2x − 2) ln 3 = (x − 3) ln 2

2x ln 3 − 2 ln 3 = x ln 2 − 3 ln 2

2x ln 3 − x ln 2 = x(2 ln 3 − ln 2) = 2 ln 3 − 3 ln 2 x=

2 ln 3 − 3 ln 2 ln(9/8) = ≈ 0.0783. 2 ln 3 − ln 2 ln(9/2)

In Problems 61–62, it is necessary to check that the solutions obtained actually satisfy the original equation. This is because equations involving logarithms may lead to extraneous solutions. An extraneous solution, in fact, occurs in Problem 61. 61. We use the laws of logarithms and the fact that ln x is one-to-one: ln x + ln(x − 2) = ln[x(x − 2)] = ln 3 x2 − 2x = 3

x2 − 2x − 3 = 0

(x − 3)(x + 1) = 0.

Thus, x = 3 and x = −1. We disregard x = −1 since it is outside the domains of both ln x and ln(x − 2). We see that x = 3 checks.

62. We use the laws of logarithms and the fact that ln x is one-to-one: ln 3 + ln(2x − 1) = ln 4 + ln(x + 1) ln[3(2x − 1)] = ln[4(x + 1)] 3(2x − 1) = 4(x + 1) 6x − 3 = 4x + 4

2x = 7, x =

7 . 2

The solution checks. 63. (a) Since the population doubles after 2 hours, we write P (2) = P0 e2k = 2P0 . Solving for k gives e2k = 2, 2k = ln 2, k =

ln 2 ≈ 0.3466. 2

Thus, P (t) = P0 e0.3466t .

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56

CHAPTER 1. FUNCTIONS (b) In 5 hours, P (5) = P0 e(0.3446)(5) ≈ 5.66P0 .

(c) Solving P (t) = P0 e0.3466t = 20P0 for t, we have e0.3466t = 20, 0.3466t = ln 20, t =

ln 20 ≈ 8.64 hours. 0.3466

64. (a) We are given A0 = 200, so A(t) = 200ekt . Since 3% of 200 is 6, A(6) = 200e6k = 194 and e6k =

194 194 , 6k = ln 200 200 1 194 k = ln ≈ −0.0051. 6 200

Thus, A(t) = 200e−0.0051t . (b) A(24) ≈ 177.1 mg.

(c) Solving A(t) = 200e−0.0051t = 100 for t, we have 1 100 = 200 2 1 ln(1/2) ≈ 137 hours. −0.0051t = ln , t = 2 −0.0051 e−0.0051t =

2000 ≈ 82 students 1 + 1999e−0.8905(5) 2000 (b) Solving P (t) = = 1000 for t, we have 1 + 1999e−0.8905t

65. (a) P (5) =

1 + 1999e−0.8905t = 2, 1999e−0.8905t = 1 1 1999   1 −0.8905t = ln = − ln 1999 1999 − ln 1999 ≈ 8.53 days. t= −0.8905 e−0.8905t =

(c) As t → ∞, e−0.8905t → 0, so P (t) →

2000 = 2000. 1+0

ππ 22

(d) We note that as t → ∞, e−0.8905t → 0 so the graph has a horizontal asymptote at P = 2000. –π

2000 1500 1000 500 -5

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5 10 15 20

57

1.7. FROM WORDS TO FUNCTIONS

66. (a) When the cake is removed from the oven, its temperature is also 350◦ F, that is, T0 = 350. The ambient temperature is the temperature of the kitchen, Tm = 75. Thus we have T (t) = 75 + 275ekt . The measurement that T (1) = 300 is the condition that determines k. From T (1) = 75 + 275ek = 300 we find ek =

225 9 = 275 11

or

k = ln

9 ≈ −0.2007. 11

From the model T (t) = 75 + 275e−0.2007t we then find T (6) = 75 + 275e−0.2007(6) ≈ 157.5◦ F. (b) To determine when the temperature of the cake will be 80◦ F, we solve the equation T (t) = 80 for t. Rewriting T (t) = 75 + 275e−0.2007t = 80 as e−0.2007t =

5 1 = 275 55

we find

t=

ln(1/55) ≈ 20 min. −0.2007

67. (a) Since y = ln 5x = ln 5 + ln x, we can obtain the graph of y = ln 5x by shifting the graph of y = ln x up ln 5 units. (b) Since y = ln(x/4) = ln x − ln 4, we can obtain the graph of y = ln(x/4) by shifting the graph of y = ln x down ln 4 units. (c) Since y = ln x−1 = − ln x, we can obtain the graph of y = ln x−1 by reflecting the graph of y = ln x in the x-axis. (d) The π π graph of y = ln(−x) is the reflection of y = ln x in the y-axis. 68. (a)

22

5

–π

-5

5 -5

! √ 2 + 1 −x − 2+1 x x √ (b) f (−x) = ln(−x + x2 + 1) = ln · 1 −x − x2 + 1  2    p x − (x2 + 1) 1 √ √ = ln = ln = ln 1 − ln(x + x2 + 1) −x − x2 + 1 x + x2 + 1 p = − ln(x + x2 + 1) = −f (x) p

1.7

−x +

√

From Words to Functions

1. Let x and y be the positive numbers. Then xy = 50 and their sum is S = x+y. From xy = 50 we have y = 50/x, so S(x) = x +

50 x2 + 50 = . x x

Since x is positive, the domain of S is (0, ∞).

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