Chapter 5: Introduction to Limits -...
Transcript of Chapter 5: Introduction to Limits -...
CPM Educational Program © 2012 Chapter 5: Page 1 Pre-Calculus with Trigonometry
Chapter 5: Introduction to Limits Lesson 5.1.1 5-1. 3. Decreases 4. Decreases 5. y = 1
x 5-3. a. y = kx2 b. 3 = k !22
34 = k
y = 34 x
2
c. y = 34 ! ("3)
2 # y = 34 !9 =
274
Review and Preview 5.1.1 5-4. f (x) = 1
x f (x) = 2x
5-5. f (x) = k(x2 ! 4)
2 = k(32 ! 4)2 = 5k25 = k
f (4) = 25 (4
2 ! 4) = 25 (16 ! 4) =
25 "12 =
245 = 4.8
5-6. y = k
x3
0.35 = k!0.753
0.35 " !0.9086 = !0.318 = k
y = !0.318x3
y = !0.31817.113 = !0.318
2.5768 = !0.123
CPM Educational Program © 2012 Chapter 5: Page 2 Pre-Calculus with Trigonometry
5-7. a. sin ! "
3( ) = ! sin "3 = ! 3
2
cos ! "3( ) = cos "3 = 1
2
tan ! "3( ) = ! tan "
3 = ! 31 = ! 3
csc ! "3( ) = 1
sin(! " 3)= ! 1
32
= ! 23= ! 2 3
3
sec ! "3( ) = 1
cos ! "3( ) =
112= 2
cot ! "3( ) = 1
tan(! " 3)= ! 1
3= ! 3
3
b. sin !3"( ) = ! sin 3" = !0 = 0cos !3"( ) = cos 3" = !1tan !3"( ) = ! tan 3" = !0 = 0
csc !3"( ) = 1sin(!3" )
= 10# und.
sec !3"( ) = 1cos(!3" )
= 1!1
= !1
cot !3"( ) = 1tan(!3" )
= 10# und.
5-8. Length of interval = 3! (!1) = 4
h(x)!1
3
" dx = f (x)!1
3
" dx + 5 # 4 = 17 + 20 = 37
5-9. Volume of Florida grapefruit = 4
3 ! " r3
Diameter of Texas grapefruit = 2r where r = diameter of Florida grapefruit.
Volume of Texas grapefruit = 43 ! 2r( )3 = 4
3 !8r3 = 8 " 43 ! " r3( )
# Texas grapefruit is 8 times as big as Florida grapefruit, but only costs 7 times as much, thus making it the better deal.
5-10. a. 7!
6 ,11!6 b. 5!
6 ,11!6
5-11. a. (x+2)(x+2)(x+2)
(x+2) = (x + 2)(x + 2) = (x + 2)2 x ! "2
b. (x!2)(x!3)(x!2)(x!2) =
x!3x!2 x " 2
c. 2(x+y)(x!y)(x+y) = 2 x ! y( ) x " !y
5-12. y = k
x + a!!!!!x = length of tube!!!!!a = cross-section area of tube
CPM Educational Program © 2012 Chapter 5: Page 3 Pre-Calculus with Trigonometry
Lesson 5.1.2 5-13. Numerator and denominate must be polynomials. x and x are not polynomials. 5-14. a. 20
17 = 1317 b. 19
7 = 14+57 = 14
7 + 57 = 2
57
5-15. a. f (x) = 1
x b. f (x) = 1x+2 + 3
This is the graph of f (x) shifted 2 units to the left and 3 units up. 5-16. g(x) = 1+ 3
x!1 a. b. Graph b. h(x) = 2x!5
x!3 = 2(x!3)+1x!3 = 2(x!3)
x!3 + 1x!3 = 2 +
1x!3
5-17. It is the graph of y = 1
x shifted up 2 units and right 3 units. y = 1x!3 + 2
5-18. Multiply the numerator and denominator by xy2 . 5-19.
a. x2
x2! x"1y2 +yx"2y"1+xy2
= yy !
xy2 +x2yy"1+x3y2
= xy3+x2y2
1+x3y3 b. y3
y3! x2y3+y"3
y"3+x= x2y6 +1
1+xy3
CPM Educational Program © 2012 Chapter 5: Page 4 Pre-Calculus with Trigonometry
5-20.
a. u = 1x +
xy2
= y2
xy2+ x2
xy2= x2 +y2
xy2 b. xy2
x2 +y2
c. x2 +y2
xy2!"#
$%&'2
= xy2
x2 +y2!"#
$%&2= (xy2 )2
(x2 +y2 )2= x2y4
(x2 +y2 )2
5-21.
a. u = x + y!1 = x + 1y =
xy+1y
u!1 = 1u =
1xy+1 y =
yxy+1
b. u = ab!1 + a!2b = ab +
ba2
= a3+b2
a2b
u!1 = 1u =
1a3+b2 a2b
= a2ba3+b2
c. u = ab!1 + a!2b = ab +
ba2
= a3+b2
a2b
u!1 = 1u2
= 1
a3+b2
a2b"#$
%&'2 = a2b
a3+b2( )2 = a4b2
(a3+b2 )2
Review and Preview 5.1.2 5-22. Example:
Vertical asymptote at x = !7 " y = 1x+7
Horizontal asymptote at x = 5" y = 1x+7 + 5
5-23.
2x(3x!1)(3x!1) ! 1(3x!1)(3x!1) +
3(3x!1) =
6x2 !2x!3x+1+3(3x!1) = 6x2 !5x+4
(3x!1)
5-24. a. Any angle in the 4th quadrant will satisfy this. Examples: 11!
6 , " !3
b. sin x = ! 12 " x = 7#
6 ,11#6
c. Any angle in the 3rd quadrant will satisfy this. Examples: 7!6 ,5!4
d. Approximately 6!7 e. Not possible, it does not satisfy the Fundamental Pythagorean Identity.
cos2 ! + sin2 ! = 1
(0.9)2 + (0.8)2 = 0.81+ 0.64 = 1.45 " 1
CPM Educational Program © 2012 Chapter 5: Page 5 Pre-Calculus with Trigonometry
5-25. a. y = k
x2 !3
1 = k!2
!2 = k
y = !2x2 !3
b. f (x) = !2x2 !3
f (!3) = !2(!3)2 !3
= !26 = ! 1
3
f (2) = !222 !3
= !21 = !2
f 12( ) = !2
(1 2)2 !3= !21 4!12 4 =
!2!11 4 = !2 " ! 4
11( ) = 811
f ! 1a( ) = !2
(!1 a)2 !3= !21 a2 !3a2 a2
= !2 " ! a2
1!3a2( ) = !2a2
1!3a2
5-26. a. A(d) = k
d2 b. A(d) = k
(3d )2= k9d2
= 19 !
kd2
It decreases 19 as much.
5-27. a. sec! = "2
1cos! = "2
1 = "2 cos!
" 12 = cos!
! = 2#3 ,
4#3
b. cot! = 31tan! = 3
1 = 3 tan!13= tan!
! = "6 ,
7"6
5-28.
a. 3 x3 + 4x2x !1 x2
= 3+ 4x4 x3
2x3 !1 x2= 3+ 4x
4
x3" x2
2x3 !1= 3+ 4x
4
2x4 ! x
b. 1 x2 + y2
x2 !1 y2= 1+ x
2y2 x2
x2y2 !1 y2= 1+ x
2y2
x2" y2
x2y2 !1= y2 + x2y4
x4y2 ! x2
c. 1 x4y2 +1 x3y1 x !1 x3
= 1+ xy x4y2
x2 !1 x3= 1+ xyx4y2
" x3
x2 !1= 1+ xyx3y2 ! xy2
CPM Educational Program © 2012 Chapter 5: Page 6 Pre-Calculus with Trigonometry
Lesson 5.1.3 5-29. See graph at right. The x-intercept of the first function is the x-value of the
vertical asymptote of the second (x = 3) . 5-30. See graph at right. a. x = 2, x = !2 b. Yes, y = 0 c. x < !2 or x > 2 d. !2 < x < 2 5-31. See graph at right. f (x) = sec(x)! a. (2!n,1) and (! + 2!n, "1), n an integer .
b. x ! "2 + "n, n an integer
c. Range: cos x! ["1,1]; sec x! y # 1 or y $ "1 5-32. See graph at right. 1
f (x)
5-33. See graph below. Cosecant: Domain: x ! "n , n an integer Range: y ! "1 or y # 1
CPM Educational Program © 2012 Chapter 5: Page 7 Pre-Calculus with Trigonometry
Review and Preview 5.1.3 5-34. a. T = 0.0725P b. k = 0.0725 5-35. a. f (100) = 2(100)+3
100+1 = 203101 = 2.0099
f (1000) = 2(1000)+31000+1 = 2003
1001 = 2.0010
f (10000) = 2(10000)+310000+1 = 20003
10001 = 2.0001
b. y = 2
c. Yes, since x ! "1, x = "1 is the vertical asymptote. d. f (x) = 2x+3
x+1 = 2 + 1x+1 This is the graph of 1
x , shifted one unit left and two units up . 5-36. Possible answer: g(x) = (x +1)(x ! 2) 5-37. 3
4 x2 5-38. c +1 = 25 !10c + c2
0 = c2 !11c + 240 = (c ! 8)(c ! 3)c = 3, 8
Check 3+1 = 5 ! 3 = 2
8 +1 = 3 " 5 ! 8 = !3Solution : c = 3
5-39. a(x) = 1
h(x) , b(x) =1f (x)
5-40. a. d = (5 ! (!2))2 + (!2 ! 5)2
d = 72 + 72 = 98 = 7 2
b. slope = m = !2!55!(!2) =
!77 = !1
Point-Slope Form: y ! 5 = !(x ! (!2))y ! 5 = !(x + 2)
OR y ! (!2) = !(x ! 5)y + 2 = !(x ! 5)
Slope-Intercept Form: y + 2 = !(x ! 5)
y + 2 = !x + 5y = !x + 3
CPM Educational Program © 2012 Chapter 5: Page 8 Pre-Calculus with Trigonometry
Lesson 5.2.1 5-41. See graph at right. f (x) = 1
x!2 + 3 lim
x!"f (x) = 3
5-42. a. He thinks that he will get to 3. b. She thinks that she will get to 1. 5-43. a. She thinks she is getting infinitely far up. b. + means approaching from the right. – means approaching from the left. c. lim
x!3"f (x) = 1 , lim
x!3+f (x) = 3
5-44. See graph at right. f (x) = 2
x!4 !1 a. lim
x!"f (x) = #1
b. limx!4"
f (x) = "#
c. limx!4+
f (x) = +"
5-45.
a. limx!"
3x3( ) = 0 b. lim
x!" 5x4( ) = " c. lim
x!" # 9
x( ) = 0
d. limx!"
sin x No limit.
sin x oscillates between –1 and 1, but does not approach any specific number. e. An example is f (x) = cos x . 5-46. a. g(x) = 2x+4
x+3 = 2(x+3)!2(x+3) = !2
x+3 + 2 b. lim g(x)
x!"= 2,!lim g(x)
x!#3#= ",!lim g(x)
x!#3+= #"
5-47. x must approach ! from the left and must approach !" from the right, therefore we do
not need extra notation. x cannot approach ! from the right.
CPM Educational Program © 2012 Chapter 5: Page 9 Pre-Calculus with Trigonometry
Review and Preview 5.2.1 5-48. See graph at right. lim
x!2f (x) = 22 + 3 = 7
5-49. In the first, the limit is approaching –2, and in the second, the limit is approaching 2 from
the left side. 5-50. y = k
x2
limx!"
1x2
= 0
As x!" , 1x2
gets smaller and smaller.
5-51. a. If u = sin! " u3#4u
u2= 3 b. u3 ! 4u = 3u2
u3 ! 3u2 ! 4u = 0u(u2 ! 3u ! 4) = 0u(u ! 4)(u +1) = 0
u = !1, 0, 4But u = sin" # 0So u = !1, 4
c. sin! = "1 or sin! = 4
! = 3#2 , !!!!!!!! sin! $ 4
5-52. a. 16
3 b. 3n2
10n2 !5
5-53. a. S(r) = kr2 b. 16! = k " (2)2
16! = 4kk = 4!
S(r) = 4! " r2
SA = S(3) = 4! " 33 = 36! 5-54. a. 2 b. 5.152
CPM Educational Program © 2012 Chapter 5: Page 10 Pre-Calculus with Trigonometry
Lesson 5.2.1 5-55. f. If the procedures are followed accurately, all the last acute angles should be very close to 60º. g. They should say 60º. Once this angle is reached, the obtuse angle will be 120º. This is, of
course, interesting since theoretically this angle is never reached. When it is bisected the resulting angle will equal 60º. Do not simply accept as justification that the pattern continues.
5-56. a. They are alternate interior angles. b. 180! ! 20! = 160!; a2 = 160!
2 = 80! c. d. 60.000 is obtained by n = 18.
Students can repeat this process by using the ANS key on their calculator. Enter 20 and push Í. At the next line use (180 – Z)/2. Repeatedly hitting Í will generate the sequence.
e. lim
x!"an = 60 f. The measure approaches 60º.
5-57.
n Angle an Measure of angle a1 Error: 60 – angle measure
1 a1 20 40
2 a2 80 –20
3 a3 50 10
4 a4 65 –5
5 a5 57.5 2.5
6 a6 61.25 –1.25
7 a7 59.375 0.625
8 a8 60.3125 –0.3125
n Measure of an
1 20 2 80 3 50 4 65 5 57.5 6 61.25 7 59.375 8 60.3125
CPM Educational Program © 2012 Chapter 5: Page 11 Pre-Calculus with Trigonometry
Review and Preview 5.2.2 5-58.
g(x) = ax 10 << a 1>a
Example Sketch
limx! "
g(x) 0 ! limx! "#
g(x) ! 0 5-59. See graph at right. f (x) = 2x ! 3
limx!"
(2x # 3) = " # 3 = "
limx! #"
(2x # 3) = 0 # 3 = #3
5-60. The limit is 60; it will never actually equal 60 unless the original acute angle was 60º. 5-61. a. S = kp
m b. 60 c. The score would be infinite.
5-62.
sin 60! = 0.866
sin !3 = 0.866
They are the same because 60! = !
3 .
5-63. a. 4 and 5 should be multiplied, not added. log 4 + log 5 = log 20 b. Base should not change. log3 7 + log3 7 = log3 49 c. Log of a product is the SUM of the logs. log(4 !5) = log 4 + log 5 d. No rule for log(4 + 5) . 5-64. f (x) = 1
x!2 ! 3 , asymptotes at x = 2 and y = !3
CPM Educational Program © 2012 Chapter 5: Page 12 Pre-Calculus with Trigonometry
5-65. a. Closest that Jade can be is when sin! = "1 . ! d = 25("1) + 35 = 10 feet. Farthest away
that Jade can be is when sin! = 1 . ! d = 25(1) + 35 = 60 feet b. 35 = 25 sin ! "t
15( ) + 350 = 25 sin ! "t
15( )0 = sin ! "t
15( )sin#1 0 = sin#1 sin ! "t
15( )( )2! = ! "t
15
30! = ! " tt = 30 seconds
c. 60 = 25 sin ! "t15( ) + 35
25 = 25 sin ! "t15( )
sin#1 1 = sin#1 sin ! "t15( )( )
15! " !2 = ! "t
15 " 15!
t = 152 = 7.5 seconds
Second time = 30 sec + 7.5 sec= 37.5 sec
d. 5-66.
a. x2y3 +1 x2
1 y3 + x= x4y3 +1 x2
1+ xy3 y3= x4y3 +1
x2! y3
1+ xy3= x4y6 + y3
x2 + x3y3
b. 1 x + x2
1 x2 + x= 1+ x3 x1+ x3 x2
= 1+ x3
x! x2
1+ x3= x
CPM Educational Program © 2012 Chapter 5: Page 13 Pre-Calculus with Trigonometry
Lesson 5.2.3 5-67. a. He thinks the height will be y = 2 . b. She thinks she will approach y = 2 . Benny: lim
x!3"f (x) = 2 ; Bertha: lim
x!3+f (x) = 2
c. f (x) =!x2 + 4 x " 32x ! 3 x > 3
#$%&
limx!1+
f (x) = 3
5-68. Right- and left-hand limits must be equal for a limit to exist. 5-69.
a. f (x) = x2 ! 2 for x " 2
! 12 (x ! 2) + 4 for x > 2
#$%
&% b. i. !" , ii. ! , iii. 4, iv. 2
c. It does not exist since the left hand limit and right hand limit are not the same. 5-70.
Example: f (x) =!x2 + 4 x " 32x ! 3 x > 3
#$%&
5-71. a. lim
x! 3"f (x) = 6.2, lim
x! 3+f (x) = 6.2
b. The limit exists since the limits as x approaches 3 from both the left and right sides are equal.
Review and Preview 5.2.3 5-72. lim
x!3"f (x) = "2, lim
x!3+f (x) = "1
limx!3
f (x) does not exist as the limits from the left and right are not equal.
5-73. See graph at right. f (x) = ! 1
x!3 + 2
limx!"#
f (x) = limx!"#
" 1x"3 + 2 = "0 + 2 = 2
x
y
f(x)
CPM Educational Program © 2012 Chapter 5: Page 14 Pre-Calculus with Trigonometry
5-74. See graph at right. 5-75.
limx!"
1x2 # 4
= 0
5-76.
x + x!3
x!2 + x!1"#$
%&'
1
= x +1 x3
1 x2 +1 x= x4 +1 x3
1+ x x2= x4 +1
x3( x2
1+ x= x4 +1x + x2
5-77.
42b6 = a, 15 = 42
b6 !b9
15 = 42b3, 514 = b3
b = 514
3 " 0.709, a = 425
143( )6
" 329.28
5-78. a. Possible estimate: lim
x!1g(x) " 0.42062
b. See graph at right. limit ≈ 0.42061984 5-79. See graph at right. a. lim
x!3x2 " 9 = 0
b. limx!3+
1x2 "9
= # c. limx!3"
1x2 "9
= "# d. limx!3
1x2 "9
Limit does not exist as left and right limits are not equal.
CPM Educational Program © 2012 Chapter 5: Page 15 Pre-Calculus with Trigonometry
Lesson 5.2.4 5-80. a. Yes, the limit exists, since the limit from the left and right are equal. lim
x!3 g(x) = 2
b. limx!3
h(x) = 2
c. h(3) = 3 d. No, the limit is the number f(x) approaches as x gets closer and closer to a. 5-81. a. No, f(–3) and the limit do not exist. b. Continuous c. No, limit ≠ f(2). d. No, f(4) does not exist. 5-82. a. b. c. 5-83. a. x ! 0 b. See graph at right. c. lim
x!0sin x = 0, lim
x!0x = 0
The numerator and denominator of f(x) are both approaching 0. d. lim
x!0sin xx = 1
5-84. a. n! , n " 0 b. f (x) = sin x
x oscillates as x!" but gets closer to y = 0 . c. The line y = 0 IS a horizontal asymptote. d. lim
x!0sin xx = 1
x
y
a x
y
a x
y
a
CPM Educational Program © 2012 Chapter 5: Page 16 Pre-Calculus with Trigonometry
Review and Preview 5.2.4 5-85. See graph at right. 1
f (x) =1
x3!3x!2 x " !1, 2
Vertical Asymptotes of f (x) # x = !1, x = 2
5-86. lim
x!3x
x2 +1= 332 +1
= 310
5-87. lim
x!" 2sin xx = sin" 2
" 2 = 1" 2 =
2"
5-88.
k10 = x
10!h
k(10 ! h) = 10xk(10!h)
10 = x (radius of cross-section)
Area of cross-section = ! k(10"h)10( )2
= !100 k
2(10 " h)2
5-89. a. cos 60
! = 12 b. sin 3!4 = 2
2
c. sin 3!2 = "1 d. cos 90! = 0
CPM Educational Program © 2012 Chapter 5: Page 17 Pre-Calculus with Trigonometry
5-90. a. csc2 x + sec2 x ! tan2 x ! cot2 x = 2
(csc2 x ! cot2 x) + (sec2 x ! tan2 x) = 21+1 = 22 = 2
b. sec x!11!cos x = sec x
sec x!11!cos x
1+cos x1+cos x( ) = sec x
sec x+1!1!cos x1!cos2 x
= sec x
sec x!cos xsin2 x
= sec x
1!cos2 xcos xsin2 x
= sec x
sin2 xcos xsin2 x
= sec x
1cos x = sec x
5-91. log2 A + log2 B = 4 ! log2(AB) = 4
log3 A " log3 B = 2 ! log3AB( ) = 2
24 = AB 32 = AB( ) = 16
B2( )AB = 16 9 = 16
B2
A = 16B B = 4
3 A = 164/3 = 12
5-92. a. A = kw
5!GPA
14.40 = 10k5!3.5 =
10k1.5 = 20
3 k" k = 2.16
A = 2.16w5!GPA
b. 20 = 2.16(15)5!GPA
5 !GPA = 2.16(15)20 = 1.62
GPA = 5 !1.62 ! 3.38
5-93. For x = –2: 2(!2) + a = (!2)2 ! 2
!4 + a = 2a = 6
For x = 3: 32 ! 2 = !3+ b7 = !3+ b10 = b
CPM Educational Program © 2012 Chapter 5: Page 18 Pre-Calculus with Trigonometry
Lesson 5.2.5 5-94. i. lim
x!"#f (x) = 0 ii. lim
x!"2"f (x) = 2 iii. lim
x!"2+f (x) = 3
iv. limx!"2
f (x) = no limit v. limx!2
f (x) = 3 vi. limx!"
f (x) = #"
5-95. lim
x!"#f (x) = 0 , lim
x!"4"f (x) = # , lim
x!"4+f (x) = # , lim
x!"4f (x) does not exist
limx!"1
f (x) = 1 , limx!1"
f (x) = "1 , limx!1+
f (x) = 2 , limx!1
f (x) does not exist
5-96. When x is very small, AC and AB! are almost the same length and parallel. Therefore as
x changes, the lengths of AC and AB! change at similar rates. 5-97. a. cos(x) approaches 1 while x approaches 0, so this is 1
little = big . b. x approaches 0 while 2x approaches 1, so this is
1big =little .
5-98. a. x ≠ 0 b. Both approach 0. c. See graph at right. lim
x!0f (x) = 0
d. Yes, y = 0 . Review and Preview 5.2.5 5-99. See graph at right.
limx!0
1" cos xx2
= 0.5
5-100. See graph at right. a. lim
x!2+f (x) = 3
b. limx!2"
f (x) = 1
c. limx!2
f (x)
The limit does not exist because the limits from the left and the right sides are different.
CPM Educational Program © 2012 Chapter 5: Page 19 Pre-Calculus with Trigonometry
5-101. See graph at right. Example (one of many): 5-102. a. 22 = 4
limx!2
f (x) = 4Therefore the graph is continuous at x = 2.
b. limx!5
f (x) = 4
5-103. See graph at right. Find the area under the curve. Distance traveled in the first five seconds: feet50510 =! Distance traveled in the t – 5 seconds: 1
2 (10 + 2t)(t ! 5) = 200 feet(10 + 2t)(t ! 5) = 40010t ! 50 + 2t2 !10t = 4002t2 = 450t2 = 225t = 15
5-104. a. y = 2 + 1
2x!3 b. y = 2, x = 32 c. lim
x!"4x /x#5/x2x /x#3/x = limx!"
42 = 2
d. Vertical asymptote because x ! 1.5 . We know the function will be approaching positive or negative infinity. f (1.51) = 4(1.51)!5
2(1.51)!3 = 52!!"!! limx#1.5+
= +$
5-105. a. See graph at right. b. x ! 1 c. x ! "1, 1, 3 d. lim
x!"#h(x) = 0
5-106. g(x) is the graph of f (x) with a vertical shift of 3 units. Therefore Bertha thinks that she
is getting closer to Benny’s height + 3, or 2 + 3 = 5 units. 5-107.
CPM Educational Program © 2012 Chapter 5: Page 20 Pre-Calculus with Trigonometry
a. h = kdg b. 2.4 = k(70!50)
0.30.72 = 20k0.036 = k
c. h = 0.036(70!30)1.5
= 0.036(40)1.5
= 1.441.5 = 0.96 BTU
CPM Educational Program © 2012 Chapter 5: Page 21 Pre-Calculus with Trigonometry
Closure Problems CL 5-108. a. lim
x!1"f (x) # 0.999+2
0.999"1 = "$!!!!! limx!1+
f (x) # 61.00012 "1
= $ Therefore limx!1
f (x) does not exist.
b. limx!2
f (x) = 622 "1
= 63 = 2
c. limx!"#
f (x) = x /x+2/xx /x"1/x = 1
1 = 1
CL 5-109. Yes, for all. CL 5-110. y = x2 ! 2x ! 8 = (x ! 4)(x + 2) This is a parabola opening up with x-intercepts at 4 and –2. Therefore f (x) will have vertical asymptotes
at x = 4 and x = !2 CL 5-111.
limx!0
f (x)x =
11"sin x "
11+sin xx =
(1+sin x)"(1"sin x)(1"sin x)(1+sin x)
x =2 sin x1"sin2 x
x = 2 sin xx cos2 x
= sin xx # 2
cos2 x= 1 # 2
cos2 (0)= 2
CL 5-112.
a. 2 b. 2 c. –2 d. 3 e. ∞ f. –1
CL 5-113.
m = kpqt!!!!!7 = k(5)(10)
4!!!!!k = 14
50 =725 21 =
725 p(2 p)
9
63 = 1425 p
2
63!2514 = p2
9!252 = p
3!52= 15 2
2 = p
CL 5-114. a. f (0.0001) = f (!0.0001) = !0.1111
b. f (x) =13+x !
13
x =3!(3+x)3(3+x)x =
!x3(3+x)x = ! 1
9+3x
c. limx!0
" 19+3(0) = " 1
9
CPM Educational Program © 2012 Chapter 5: Page 22 Pre-Calculus with Trigonometry
CL 5-115. They will have to agree that they think they are headed to the same place. CL 5-116. a. 3(!1)y!1 ! 2(!1) + 2y!1 = 1
!3y!1 + 2 + 2y!1 = 1!y!1 = !1
1y = 1
y = 1
b. 3xy !
2x +
2y = 1
3! 2y + 2x = xy3+ 2x = xy + 2y3+ 2x = y(x + 2)2x + 3x + 2
= y
CL 5-117. See graph at right. a. x ≠ 3 b. 3x =
c. limx!"
#5x + 2x # 3
= #5
y = #5
CL 5-118. a. See graph at right below. b. Find the area under the curve from 0 to 2t t= = . c. He is approximating the area under the curve by using right-endpoint rectangles with
width = 0.1.
d. 6(0.1k)k=1
20
!
e. 0.5[6(0)2 + 12(0.1)2 + 12(0.2)2 + .... + 12(1.9)2 + 6(2)2] f. Right-endpoint rectangles give 17.22 mi, left=14.82, trap=16.02
x
y
ft/sec
seconds 2 1
25