6
INVERSE FUNCTIONS:
Exponential, Logarithmic, and Inverse Trigonometric Functions
6.1 Inverse Functions and Their Derivatives
1. (a) See Definition 1.
(b) It must pass the Horizontal Line Test.
2. (a) −1 () =
⇔ () = for any in . The domain of −1 is and the range of −1 is .
(b) See the steps in Box 5.
(c) Reflect the graph of about the line = .
3. is not one­to­one because 2 6= 6, but (2) = 20 = (6).
4. is one­to­one because it never takes on the same value twice.
5. We could draw a horizontal line that intersects the graph in more than one point. Thus, by the Horizontal Line Test, the
function is not one­to­one.
6. No horizontal line intersects the graph more than once. Thus, by the Horizontal Line Test, the function is one­to­one.
7. No horizontal line intersects the graph more than once. Thus, by the Horizontal Line Test, the function is one­to­one.
8. We could draw a horizontal line that intersects the graph in more than one point. Thus, by the Horizontal Line Test, the
function is not one­to­one.
9. The graph of () = 2 − 3 is a line with slope 2. It passes the Horizontal Line Test, so is one­to­one.
Algebraic solution: If 1 6= 2 , then 21 6= 22
⇒ 21 − 3 6= 22 − 3 ⇒ (1 ) 6= (2 ), so is one­to­one.
10. The graph of () = 4 − 16 is symmetric with respect to the ­axis. Pick any ­values equidistant from 0 to find two equal
function values. For example, (−1) = −15 and (1) = −15, so is not one­to­one.
11. No horizontal line intersects the graph of () = 3 + 4 more than once. Thus, by the Horizontal Line Test, the function is
one­to­one.
Algebraic solution: If 1 6= 2 , then 31 6= 32
12. The graph of () =
13. () = 1 − sin .
⇒ 31 + 4 6= 32 + 4 ⇒ (1 ) 6= (2 ), so is one­to­one.
√
3
passes the Horizontal Line Test, so is one­to­one.
(0) = 1 and () = 1, so is not one­to­one.
14. The graph of () = 4 − 1 passes the Horizontal Line Test when is restricted to the interval [0,10], so is one­to­one.
15. A football will attain every height up to its maximum height twice: once on the way up, and again on the way down.
Thus, even if 1 does not equal 2 , (1 ) may equal (2 ), so is not 1­1.
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CHAPTER 6
INVERSE FUNCTIONS
16. is not 1­1 because eventually we all stop growing and therefore, there are two times at which we have the same height.
17. (a) Since is 1­1, (6) = 17
⇔ −1 (17) = 6.
(b) Since is 1­1, −1 (3) = 2 ⇔ (2) = 3.
18. First, we must determine such that () = 3. By inspection, we see that if = 1, then (1) = 3. Since is 1­1 ( is an
increasing function), it has an inverse, and −1 (3) = 1. If is a 1­1 function, then ( −1 ()) = , so ( −1 (2)) = 2.
√
√
⇒ 0 () = 1 + 1(2 ) 0 on (0 ∞). So is increasing and hence, 1­1. By inspection,
√
(4) = 4 + 4 = 6, so −1 (6) = 4.
19. () = +
20. (a) is 1­1 because it passes the Horizontal Line Test.
(b) Domain of = [−3 3] = Range of −1 . Range of = [−1 3] = Domain of −1 .
(c) Since (0) = 2, −1 (2) = 0.
(d) Since (−17) ≈ 0, −1 (0) ≈ −17.
21. We solve = 59 ( − 32) for : 95 = − 32
⇒ = 95 + 32. This gives us a formula for the inverse function, that
is, the Fahrenheit temperature as a function of the Celsius temperature . ≥ −45967 ⇒
9
≥ −49167
5
9
+ 32 ≥ −45967
5
⇒ ≥ −27315, the domain of the inverse function.
0
22. =
1 − 2 2
2
2
⇒ 1 − 2 = 02
⇒
2
20
=
1
−
2
2
20
2
⇒ = 1 − 02 .
⇒ = 1− 2
2
2
This formula gives us the speed of the particle in terms of its mass , that is, = −1 ().
23. = () = 5 − 4
⇒
4 = 5 −
⇒
= 14 (5 − ). Interchange and : = 14 (5 − ). So
−1 () = 14 (5 − ) = 54 − 14 .
24. We write = () =
5 + 3 = 6 − 7
so −1 () =
6 − 3
and solve for : (5 + 7) = 6 − 3 ⇒ 5 + 7 = 6 − 3 ⇒
5 + 7
⇒ (5 + 3) = 6 − 7
⇒ =
6 − 7
6 − 7
. Interchanging and gives =
,
5 + 3
5 + 3
6 − 7
.
5 + 3
25. First note that () = 1 − 2 , ≥ 0, is one­to­one. We first write = 1 − 2 , ≥ 0, and solve for :
2 = 1 − ⇒ =
√
−1 () = 1 − .
√
√
1 − (since ≥ 0). Interchanging and gives = 1 − , so the inverse function is
26. Completing the square, we have () = 2 − 2 = (2 − 2 + 1) − 1 = ( − 1)2 − 1 and, with the restriction ≥ 1,
√
is one­to­one. We write = ( − 1)2 − 1, ≥ 1, and solve for : − 1 = + 1 (since ≥ 1 ⇔ − 1 ≥ 0),
√
√
√
so = 1 + + 1. Interchanging and gives = 1 + + 1, so −1 () = 1 + + 1.
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°
⇒
SECTION 6.1
27. First write = () = 2 +
INVERSE FUNCTIONS AND THEIR DERIVATIVES
¤
563
√
√
+ 1 and note that ≥ 2. Solve for : − 2 = + 1 ⇒ ( − 2)2 = + 1 ⇒
= ( − 2)2 − 1 ( ≥ 2). Interchanging and gives = ( − 2)2 − 1 so −1 () = ( − 2)2 − 1 with domain ≥ 2.
28. = () = 1 +
√
2 + 3
( ≥ 1)
⇒ −1=
√
2 + 3
⇒ ( − 1)2 = 2 + 3 ⇒ ( − 1)2 − 2 = 3 ⇒
= 13 ( − 1)2 − 23 . Interchange and : = 13 ( − 1)2 − 23 . So −1 () = 13 ( − 1)2 − 23 . Note that the domain of −1
is ≥ 1.
29. We solve = 2 +
5
√
√
3
3
for : 5 = 2 + ⇒
gives the inverse function =
√
3
5
−2 .
3
√
3
= 5 − 2 ⇒ = 5 − 2 . Interchanging and
√
1−
√ , the domain is ≥ 0. (0) = 1 and as increases, decreases. As → ∞,
1+
√
√
√
1 − 1
−1
1 − 1
√ · √ = √
→
= −1, so the range of is −1 ≤ 1. Thus, the domain of −1 is −1 ≤ 1.
1
1 + 1
1 + 1
√
√
√
√
√
√
√
1−
√
=
⇒ (1 + ) = 1 − ⇒ + = 1 − ⇒
+ = 1− ⇒
1+
2
2
√
√
1−
1−
1−
⇒ =
(1 + ) = 1 − ⇒
=
. Interchange and : =
. So
1+
1+
1+
30. For () =
−1 () =
31. = () =
1−
1+
2
with −1 ≤ 1.
√
2 − 3
.
4 + 3 ( ≥ 0) ⇒ 2 = 4 + 3 ⇒ =
4
Interchange and : =
2 − 3
2 − 3
. So −1 () =
( ≥ 0). From
4
4
the graph, we see that and −1 are reflections about the line = .
32. = () = 2 − 4 ( ≥ 0)
⇒ 4 = 2 −
⇒
√
√
= 4 2 − [since ≥ 0]. Interchange and : = 4 2 − .
√
So −1 () = 4 2 − ( ≤ 2). From the graph, we see that and −1 are
reflections about the line = .
33. Reflect the graph of about the line = . The points (−1 −2), (1 −1),
(2 2), and (3 3) on are reflected to (−2 −1), (−1 1), (2 2), and (3 3)
on −1 .
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CHAPTER 6
INVERSE FUNCTIONS
34. Reflect the graph of about the line = .
35. (a) = () =
√
1 − 2
(0 ≤ ≤ 1 and note that ≥ 0) ⇒
⇒ 2 = 1 − 2 ⇒ = 1 − 2 . So
2 = 1 − 2
√
−1 () = 1 − 2 , 0 ≤ ≤ 1. We see that −1 and are the same
function.
(b) The graph of is the portion of the circle 2 + 2 = 1 with 0 ≤ ≤ 1 and
0 ≤ ≤ 1 (quarter­circle in the first quadrant). The graph of is symmetric
with respect to the line = , so its reflection about = is itself, that is,
−1 = .
√
36. (a) = () = 3 1 − 3
3 = 1 − 3 ⇒ 3 = 1 − 3 ⇒
√
= 3 1 − 3 . So −1 () = 3 1 − 3 . We see that and −1 are the
⇒
same function.
(b) The graph of is symmetric with respect to the line = , so its reflection
about = is itself, that is, −1 = .
37. (a) 1 6= 2
⇒ 31 6= 32
⇒ (1 ) 6= (2 ), so is one­to­one.
1
.
(b) 0 () = 32 and (2) = 8 ⇒ −1 (8) = 2, so ( −1 )0 (8) = 1 0 ( −1 (8)) = 1 0 (2) = 12
⇒ = 13 . Interchanging and gives = 13 ,
so −1 () = 13 . Domain −1 = range() = .
(c) = 3
(e)
Range( −1 ) = domain( ) = .
(d) −1 () = 13 ⇒ ( −1 )0 () = 13 −23
1
as in part (b).
( −1 )0 (8) = 13 14 = 12
38. (a) 1 6= 2
⇒ 1 − 2 6= 2 − 2 ⇒
(b) (6) = 2, so −1 (2) = 6. Also 0 () =
(c) =
⇒
√
√
1 − 2 6= 2 − 2 ⇒ (1 ) 6= (2 ), so is 1­1.
2
1
1
1
1
√
, so ( −1 )0 (2) = 0 −1
= 0
=
= 4.
( (2))
(6)
14
−2
√
− 2 ⇒ 2 = − 2 ⇒ = 2 + 2.
(e)
Interchange and : = 2 + 2. So −1 () = 2 + 2.
Domain = [0 ∞), range = [2 ∞).
(d) −1 () = 2 + 2 ⇒ ( −1 )0 () = 2 ⇒ ( −1 )0 (2) = 4.
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°
SECTION 6.1
39. (a) Since ≥ 0, 1 6= 2
⇒ 21 6= 22
⇒ 9 − 21 6= 9 − 22
(b) 0 () = −2 and (1) = 8 ⇒ −1 (8) = 1, so ( −1 )0 (8) =
√
⇒ 2 = 9 − ⇒ = 9 − .
√
√
Interchange and : = 9 − , so −1 () = 9 − .
INVERSE FUNCTIONS AND THEIR DERIVATIVES
¤
⇒ (1 ) 6= (2 ), so is 1­1.
1
1
1
1
= 0
=
=− .
0 ( −1 (8))
(1)
−2
2
(c) = 9 − 2
(e)
Domain( −1 ) = range ( ) = [0 9].
Range( −1 ) = domain ( ) = [0 3].
√
⇒ ( −1 )0 (8) = − 12 as in part (b).
(d) ( −1 )0 () = −1 2 9 −
40. (a) 1 6= 2
⇒ 1 − 1 6= 2 − 1 ⇒
1
1
6=
1 − 1
2 − 1
⇒ (1 ) 6= (2 ), so is 1­1.
1
(b) −1 (2) = 32 since 32 = 2. Also 0 () = −1( − 1)2 , so ( −1 )0 (2) = 1 0 32 = −4
= − 14 .
(c) = 1( − 1) ⇒ − 1 = 1
and : = 1 + 1. So
−1
⇒ = 1 + 1. Interchange
(e)
() = 1 + 1, 0 (since 1).
Domain = (0 ∞), range = (1 ∞)
(d) ( −1 )0 () = −12 , so ( −1 )0 (2) = − 14 .
41. () = 3 + 3 sin + 2 cos
⇒ 0 () = 32 + 3 cos − 2 sin . Observe that (0) = 2, so that −1 (2) = 0.
By Exercise 83, we have ( −1 )0 (2) =
42. (0) = 2
1
1
1
1
1
= 0
=
=
= .
0 ( −1 (2))
(0)
3(0)2 + 3 cos 0 − 2 sin 0
3(1)
3
⇒ −1 (2) = 0, and () = 3 + 3 sin + 2 cos ⇒ 0 () = 32 + 3 cos − 2 sin and 0 (0) = 3.
Thus, ( −1 )0 (2) =
1
1
1
= 0
= .
0 ( −1 (2))
(0)
3
⇒ −1 (3) = 0, and () = 3 + 2 + tan(2) ⇒ 0 () = 2 + 2 sec2 (2) and
0
0 (0) = 2 · 1 = 2 . Thus, −1 (3) = 1 0 −1 (3) = 1 0 (0) = 2.
43. (0) = 3
44. (1) = 3
⇒ −1 (3) = 1, and () =
( −1 )0 (3) =
45. (4) = 5
√
32 + 4
7
and 0 (1) = . Thus,
3 + 4 + 4 ⇒ 0 () = √
6
2 3 + 4 + 4
1
1
6
1
= 0
=
= .
0 ( −1 (3))
(1)
76
7
⇒ −1 (5) = 4. Thus, ( −1 )0 (5) =
46. is an increasing function, so it has an inverse.
(−1 )0 (8) =
1
1
3
1
= 0
=
= .
0 ( −1 (5))
(4)
23
2
(2) = 8 ⇔ −1 (8) = 2. Thus,
1
1
1
= 0
= .
0 ( −1 (8))
(2)
5
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CHAPTER 6
47. () =
(3) =
INVERSE FUNCTIONS
√
√
1 + 3 ⇒ 0 () = 1 + 3 0, so is an increasing function and it has an inverse. Since
3
3√
1 + 3 = 0, −1 (0) = 3. Thus, ( −1 )0 (0) =
3
⇒ −1 (2) = 3. Thus, ( −1 )0 (2) =
48. (3) = 2
0 () = −
( −1 )0 ()
[ −1 ()]2
⇒ 0 (2) = −
49. We see that the graph of = () =
Enter =
1
1
1
1
= 0
= √
= √ .
0 ( −1 (0))
(3)
1 + 33
28
1
1
1
= 0
= 9. Hence, () = −1
0 ( −1 (2))
(3)
()
⇒
( −1 )0 (2)
9
= − 2 = −1.
(3)
[ −1 (2)]2
√
3 + 2 + + 1 is increasing, so is 1­1.
3 + 2 + + 1 and use your CAS to solve the equation for . You
will likely get two (irrelevant) solutions involving imaginary expressions, as well
as one which can be simplified to
√
√
√
3 √
= −1 () = − 64 3 − 272 + 20 − 3 + 272 − 20 + 3 2
√ √
1 23 − 8 − 2 13
3 274 − 402 + 16 or, equivalently,
,
6
2 13
√
where = 1082 + 12 48 − 1202 + 814 − 80.
where = 3
50. Since sin(2) = 0, () = sin is not one­to­one. 0 () = cos 0 on − 2 2 , so is increasing and hence 1­1 on
− 2 2 . Let = −1 () = sin−1 so that sin = . Differentiating sin = implicitly with respect to gives us
cos
=1 ⇒
we have
1
=
. Now cos2 + sin2 = 1 ⇒ cos = ± 1 − sin2 , but since cos 0 on − 2 2 ,
cos
1
1
=
.
= √
1 − 2
1 − sin2
51. (a) If the point ( ) is on the graph of = (), then the point ( − ) is that point shifted units to the left. Since
is 1­1, the point ( ) is on the graph of = −1 () and the point corresponding to ( − ) on the graph of is
( − ) on the graph of −1 . Thus, the curve’s reflection is shifted down the same number of units as the curve itself is
shifted to the left. So an expression for the inverse function is −1 () = −1 () − .
(b) If we compress (or stretch) a curve horizontally, the curve’s reflection in the line = is compressed (or stretched)
vertically by the same factor. Using this geometric principle, we see that the inverse of () = () can be expressed as
−1 () = (1) −1 ().
52. (a) We know that 0 () =
00 () = −
1
. Thus,
0 (())
00 (()) · 0 ()
00 (()) · [1 0 ( 0 ())]
00 (())
00 (())
=−
=−
=−
.
2
0
2
0
2
[ (())]
[ (())]
0 (())[ 0 (())]
[ 0 (())]3
3
(b) is increasing ⇒ 0 (()) 0 ⇒ [ 0 (())] 0. is concave upward ⇒ 00 (()) 0.
So 00 () = −
00 (())
0, which implies that [ ’s inverse] is concave downward.
[ 0 (())]3
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°
SECTION 6.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES
6.2 Exponential Functions and Their Derivatives
1. (a) () = , 0
(b)
(c) (0 ∞)
(d) See Figures 4(c), 4(b), and 4(a), respectively.
2. (a) The number is the value of such that the slope of the tangent line at = 0 on the graph of = is exactly 1.
(b) ≈ 271828
(c) () =
3. All of these graphs approach 0 as → −∞, all of them pass through the point
(0 1), and all of them are increasing and approach ∞ as → ∞. The larger the
base, the faster the function increases for 0, and the faster it approaches 0 as
→ −∞.
4. The graph of − is the reflection of the graph of about the ­axis, and the
graph of 8− is the reflection of that of 8 about the ­axis. The graph of 8
increases more quickly than that of for 0, and approaches 0 faster
as → −∞.
5. The functions with base greater than 1 (3 and 10 ) are increasing, while those
with base less than 1
1
3
and
1
10
are decreasing. The graph of
reflection of that of 3 about the ­axis, and the graph of
1
10
1
3
is the
is the reflection of
that of 10 about the ­axis. The graph of 10 increases more quickly than that of
3 for 0, and approaches 0 faster as → −∞.
6. Each of the graphs approaches ∞ as → −∞, and each approaches 0 as
→ ∞. The smaller the base, the faster the function grows as → −∞, and
the faster it approaches 0 as → ∞.
7. We start with the graph of = 3 (Figure 12) and shift
1 unit upward to get the graph of () = 3 + 1.
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CHAPTER 6
INVERSE FUNCTIONS
(Figure 3) and stretch vertically by a factor of 2 to obtain the graph of = 2 12 . Then
we shift the graph 3 units downward to get the graph of () = 2 12 − 3.
8. We start with the graph of =
1
2
9. We start with the graph of = (Figure 14) and reflect about the y­axis to get the graph of = − . Then we reflect the
graph about the x­axis to get the graph of = −− .
10. We start with the graph of = 4 (Figure 3) and shift
2 units to the left to get the graph of = 4+2 .
11. We start with the graph of = (Figure 14) and reflect about the ­axis to get the graph of = − . Then we compress
the graph vertically by a factor of 2 to obtain the graph of = 12 − and then reflect about the ­axis to get the graph
of = − 12 − . Finally, we shift the graph one unit upward to get the graph of = 1 − 12 − .
12. We start with the graph of = (Figure 14) and
reflect the portion of the graph in the first quadrant
about the ­axis to obtain the graph of = || .
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°
SECTION 6.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES
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569
13. (a) To find the equation of the graph that results from shifting the graph of = two units downward, we subtract 2 from the
original function to get = − 2.
(b) To find the equation of the graph that results from shifting the graph of = two units to the right, we replace with
− 2 in the original function to get = −2 .
(c) To find the equation of the graph that results from reflecting the graph of = about the x­axis, we multiply the original
function by −1 to get = − .
(d) To find the equation of the graph that results from reflecting the graph of = about the y­axis, we replace with − in
the original function to get = − .
(e) To find the equation of the graph that results from reflecting the graph of = about the x­axis and then about the y­axis,
we first multiply the original function by −1 (to get = − ) and then replace with − in this equation to
get = −− .
14. (a) This reflection consists of first reflecting the graph about the ­axis (giving the graph with equation = − )
and then shifting this graph 2 · 4 = 8 units upward. So the equation is = − + 8.
(b) This reflection consists of first reflecting the graph about the ­axis (giving the graph with equation = − )
and then shifting this graph 2 · 2 = 4 units to the right. So the equation is = −(−4) .
2
15. (a) The denominator is zero when 1 − 1− = 0
⇔
2
1− = 1
1 − 2 = 0
⇔
⇔
= ±1. Thus,
2
1 −
has domain { | 6= ±1} = (−∞ −1) ∪ (−1 1) ∪ (1 ∞).
1 − 1−2
1+
(b) The denominator is never equal to zero, so the function () = cos has domain , or (−∞ ∞).
the function () =
16. (a) The function () =
√
10 − 100 has domain | 10 − 100 ≥ 0 = | 10 ≥ 102 = { | ≥ 2} = [2 ∞).
(b) The sine and exponential functions have domain , so () = sin( − 1) also has domain .
17. Use = with the points (1 6) and (3 24).
4 = 2
6 = 1
= 6
and 24 = 3
⇒ 24 =
⇒
⇒ = 2 [since 0] and = 62 = 3. The function is () = 3 · 2 .
18. Given the ­intercept (0 2), we have = = 2 . Using the point 2 29 gives us 29 = 22
= 13
6 3
[since 0]. The function is () = 2 13 or () = 2(3)− .
⇒
1
= 2
9
19. In this question, we know that x = 1, So for the function f :
and for the function g:
x = 1 ⇒ f (1) = 1 2 = 1 m
(1)
x = 1 ⇒ g(1) = 2 1 = 2 m
(2)
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⇒
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CHAPTER 6
INVERSE FUNCTIONS
20. We see from the graphs that for less than about 18, () = 5 () = 5 , and then near the point (18 171) the curves
intersect. Then () () from ≈ 18 until = 5. At (5 3125) there is another point of intersection, and for 5 we
see that () (). In fact, increases much more rapidly than beyond that point.
21. The graph of finally surpasses that of at ≈ 358.
22. We graph = and = 1,000,000,000 and determine where
= 1 × 109 . This seems to be true at ≈ 20723, so 1 × 109
for 20723.
23. lim (1001) = ∞ by (3), since 1001 1.
→∞
24. By (3), if 1, lim = 0, so lim (1001) = 0.
→−∞
→−∞
1−0
3 − −3
1 − −6
=1
= lim
=
3
−3
→∞
→∞ 1 + −6
+
1+0
25. Divide numerator and denominator by 3 : lim
2
26. Let = −2 . As → ∞, → −∞. So lim − = lim = 0 by (11).
→∞
→−∞
27. Let = 3(2 − ). As → 2+ , → −∞. So lim 3(2−) = lim = 0 by (11).
→2+
→−∞
28. Let = 3(2 − ). As → 2− , → ∞. So lim 3(2−) = lim = ∞ by (11).
→2−
→∞
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°
SECTION 6.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES
29. Since −1 ≤ cos ≤ 1 and −2 0, we have −−2 ≤ −2 cos ≤ −2 . We know that lim (−−2 ) = 0 and
→∞
lim −2 = 0, so by the Squeeze Theorem, lim (−2 cos ) = 0.
→∞
30.
→∞
lim
→(2)+
sec = 0 since sec → −∞ as → (2)+ .
31. () = −2
⇒ 0 () = −2( ) = −2
32. () = +
⇒ 0 () = + −1
33. () = (32 − 5)
PR
⇒
0 () = (32 − 5)( )0 + (32 − 5)0 = (32 − 5) + (6 − 5)
= [(32 − 5) + (6 − 5)] = (32 + − 5)
34. By the Quotient Rule, =
3
3
⇒ 0 =
35. By (9), =
2
36. () = −
37. = tan
1 −
CR
2
⇒ 0 =
3
(3 ) = 32 .
⇒ 0 () = − ·
⇒ 0 = tan
(1 − ) − (− )
− 2 + 2
=
=
.
2
2
(1 − )
(1 − )
(1 − )2
2
2
( − ) = − (2 − 1)
(tan ) = (sec2 )tan
√
3
= 13 . Then
1 −23
1
( ) =
( ) =
=
=
.
3
3
3
3 ( + 1)2
3 ( + 1)2
38. Let = () = + 1 and = () =
39. () =
2
2 +
(2 + ) 2 + (2) − 2 (2 + )
4 + 23 + 2 2 + 22 − 23 − 2 2
=
2
2
( + )
(2 + )2
0 () =
(3 + 2 )
4 + 22
=
(2 + )2
(2 + )2
=
40. () =
QR
⇒
√
2
· +1 ⇒
√
2
2
2
2
2
√ √
1
+ 1 + +1 ·
· +1 ·
= · +1 · 2 + +1 · √
2
2
√
2
2
1
4 + 1
√
= +1 2 + √
or +1
2
2
0 () =
41. Using the Product Rule and the Chain Rule, = 2 −3
⇒
0 = 2 −3 (−3) + −3 (2) = −3 (−32 + 2) = −3 (2 − 3).
42. () = tan(1 + 2 )
⇒ 0 () = sec2 (1 + 2 ) · (1 + 2 )0 = 22 sec2 (1 + 2 )
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°
¤
571
572
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CHAPTER 6
INVERSE FUNCTIONS
43. () = sin
⇒ 0 () = (cos ) · + (sin ) · = ( cos + sin )
44. () = (−1)
⇒ 0 () = (−1)
45. By (9), () = sin 2
(−1)
( − 1)(1) − (1)
= (−1)
=−
− 1
( − 1)2
( − 1)2
⇒
0 () = sin 2 ( sin 2)0 = sin 2 ( · 2 cos 2 + sin 2 · 1) = sin 2 (2 cos 2 + sin 2)
46. = sin 2 + sin(2 )
0 = sin 2
⇒
sin 2 + cos(2 ) 2 = sin 2 (cos 2) · 2 + cos(2 ) 2 · 2 = 2 cos 2 sin 2 + 22 cos(2 )
√
2
47. () = sec
⇒
√
√
2 1
sec 2
sec 2 = sec (sec 2 )−12
2
√
√
√
1
2
2
= sec √
· sec 2 tan 2 · 2 = sec 2 tan 2 sec
2 sec 2
√
sec 2
0 () =
√
48. () = 1 2 − 1
⇒
√
1
1
−1
1
1
0 () = 1 · √
= −1 ;
= −−2 = − 2
· 2 + 2 − 1 · 1 · − 2
2 2 − 1
√
3
2
2
−1
− +1
√
−
= 1 √
or 1
2
2 − 1
2 2 − 1
⇒
1 +
(1 + ) − ( )
(1 + − )
·
·
0 () = cos
=
cos
=
cos
1 +
(1 + )2
1 +
(1 + )2
(1 + )2
1 +
49. () = sin
2
2
50. () = sin ( )
2
2
2
2
⇒ 0 () = sin ( ) · 2 sin(2 ) · cos(2 ) · 2 = 4 sin(2 ) cos(2 )sin ( )
51. = cos + sin
⇒ 0 = (− sin ) + (cos )( ) + cos = (cos − sin ) + cos , so
0 (0) = 0 (cos 0 − sin 0) + cos 0 = 1(1 − 0) + 1 = 2. An equation of the tangent line to the curve = cos + sin at
the point (0 1) is − 1 = 2( − 0) or = 2 + 1.
52. =
1+
1 +
⇒
At 0 12 , 0 =
53.
0 =
(1 + )(1) − (1 + )
1 + − −
1 −
=
=
2
2
(1 + )
(1 + )
(1 + )2
1
1
= , and an equation of the tangent line is − 12 = 14 ( − 0) or = 14 + 12 .
(1 + 1)2
4
( ) =
( − ) ⇒ ·
·
0
1
−
· = 1 − 0
2
= 1 − 0
⇒ 0 −
· 1 − · 0
= 1 − 0 ⇒
2
−
=
⇒ 0 1 −
2
⇒ ·
0
· = 1−
2
⇒
−
( − )
=
0 = 2
2 −
−
2
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°
¤
SECTION 6.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES
⇒ 0 + · 1 + + 0 = 0 ⇒ 0 ( + ) = − −
54. + = 1
(0 1), 0 = −
⇒ 0 = −
+
. At
+
+1·1
= −( + 1), so an equation for the tangent line is − 1 = −( + 1)( − 0), or = −( + 1) + 1.
0+1
55. = + −2
⇒ 0 = − 12 −2 ⇒ 00 = + 14 −2 , so
2 00 − 0 − = 2 + 14 −2 − − 12 −2 − + −2 = 0.
56. = − + −
⇒ 0 = −− + − − − = ( − )− − −
⇒
00 = ( − )− − − + − = ( − 2)− + − ,
so 00 + 2 0 + = ( − 2)− + − + 2 ( − )− − − + − + − = 0.
57. =
⇒ 0 =
00 = 2 , so if = satisfies the differential equation 00 + 6 0 + 8 = 0,
⇒
then 2 + 6 + 8 = 0; that is, (2 + 6 + 8) = 0. Since 0 for all , we must have 2 + 6 + 8 = 0,
or ( + 2)( + 4) = 0, so = −2 or −4.
58. =
⇒ 0 =
⇒ 00 = 2 . Thus, + 0 = 00
√
⇔ + = 2
⇔
(2 − − 1) = 0 ⇔ = 1 ±2 5 , since 6= 0.
59. () = 2
⇒ 0 () = 22
000 () = 22 · 22 = 23 2
60. () = −
⇒ 00 () = 2 · 22 = 22 2
⇒ ···
⇒
⇒ () () = 2 2
⇒ 0 () = (−− ) + − = (1 − )−
⇒
00 () = (1 − )(−− ) + − (−1) = ( − 2)−
⇒ 000 () = ( − 2)(−− ) + − = (3 − )−
(4) () = (3 − )(−− ) + − (−1) = ( − 4)−
⇒
···
⇒
⇒ () () = (−1) ( − )− .
So 1000 − = ( − 1000)− .
61. (a) () = + is continuous on and (−1) = −1 − 1 0 1 = (0), so by the Intermediate Value Theorem,
+ = 0 has a solution in (−1 0).
(b) () = + ⇒ 0 () = + 1, so +1 = −
+
. Using 1 = −05, we get 2 ≈ −0566311,
+ 1
3 ≈ −0567143 ≈ 4 , so the solution is −0567143 to six decimal places.
62. () =
573
2
1 +
⇒ 0 () = −
2
1
, so (0) = 1 and 0 (0) = − .
2
(1 + )2
Thus, () (0) + 0 (0)( − 0) = 1 − 12 . We need
2
2
1
− 01 1 −
+ 01, which is true when
1 +
2
1 +
−1423 1423. Note that to ensure the accuracy, we have rounded the
smaller value up and the larger value down.
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°
574
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CHAPTER 6
INVERSE FUNCTIONS
63. Half of 760 RNA copies per mL, corresponding to = 1, is 380 RNA copies per mL. Using the graph of in Figure 11, we
estimate that it takes about 35 additional days for the patient’s viral load to decrease to 38 RNA copies per mL.
64. (a)
(b) Using a graphing calculator, we obtain the exponential
curve () = 3689301(106614) .
(c) Using the TRACE and zooming in, we find that the bacteria count
doubles from 37 to 74 in about 1087 hours.
65. (a) lim () = lim
→∞
1
→∞ 1 + −
=
1
= 1, since 0 ⇒ − → −∞ ⇒ − → 0. As time increases, the
1+·0
proportion of the population that has heard the rumor approaches 1; that is, everyone in the population has heard the rumor.
(b) () = (1 + − )−1
⇒
−
= −(1 + − )−2 (−− ) =
(1 + − )2
(c)
From the graph of () = (1 + 10−05 )−1 , it seems that () = 08
(indicating that 80% of the population has heard the rumor) when
≈ 74 hours.
66. (a)
The displacement function is squeezed between the other two functions. This
is because −1 ≤ sin 4 ≤ 1 ⇒ −8−2 ≤ 8−2 sin 4 ≤ 8−2 .
(b) The maximum value of the displacement is about 66 cm, occurring at ≈ 036 s. It occurs just before the graph of the
displacement function touches the graph of 8−2 (when = 8 ≈ 039).
(c) The velocity of the object is the derivative of its displacement function, that is,
−2
8
sin 4 = 8 −2 cos 4(4) + sin 4 − 12 −2
If the displacement is zero, then we must have sin 4 = 0 (since the exponential term in the displacement function is
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°
SECTION 6.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES
(d)
¤
575
always positive). The first time that sin 4 = 0 after = 0 occurs at = 4 . Substituting this into our expression for the
velocity, and noting that the second term vanishes, we get 4 = 8−8 cos 4 · 4 · 4 = −32−8 ≈ −216 cms.
The graph indicates that the displacement is less than 2 cm from equilibrium
whenever is larger than about 28.
67. () =
(1 + 2 ) − (2)
(2 − 2 + 1)
( − 1)2
, [0 3]. 0 () =
=
=
. 0 () = 0 ⇒
2
2
2
2
2
1+
(1 + )
(1 + )
(1 + 2 )2
( − 1)2 = 0 ⇔ = 1. 0 () exists for all real numbers since 1 + 2 is never equal to 0. (0) = 1,
(1) = 2 1359, and (3) = 3 10 2009. So (3) = 3 10 is the absolute maximum value and (0) = 1 is the
absolute minimum value.
68. () = 2 , [−3 1].
0 () = 2
1
2
+ 2 (1) = 2
1
2
+ 1 . 0 () = 0 ⇔
1
+1 =0
2
⇔ = −2.
(−3) = −3−32 ≈ −0669, (−2) = −2−1 ≈ −0736, and (1) = 12 ≈ 1649. So (1) = 12 is the absolute
maximum value and (−2) = −2 is the absolute minimum value.
69. () = −
⇒ 0 () = 1 − = 0 ⇔ = 1 ⇔ = 0. Now 0 () 0 for all 0 and 0 () 0 for all
0, so the absolute maximum value is (0) = 0 − 1 = −1.
70. () =
⇒ 0 () =
−
= 0 ⇔ ( − 1) = 0 ⇒ = 1. Now 0 () 0 ⇔
2
− 1 0 ⇔ 1 and 0 () 0 ⇔
−
0 ⇔
2
−
0 ⇔ − 1 0 ⇔ 1. Thus there is an absolute
2
minimum value of (1) = at = 1.
⇒ 0 () = (22 ) + 2 (1) = 2 (2 + 1). Thus, 0 () 0 if − 12 and 0 () 0 if − 12 .
So is increasing on − 12 ∞ and is decreasing on −∞ − 12 .
71. (a) () = 2
(b) 00 () = 2 (2) + (2 + 1) · 22 = 22 [1 + (2 + 1)] = 22 (2 + 2) = 42 ( + 1). 00 () 0 ⇔ −1
and 00 () 0 ⇔ −1. Thus, is concave upward on (−1 ∞) and is concave downward on (−∞ −1).
(c) There is an inflection point at −1 −−2 , or −1 −12 .
72. (a) () =
2
⇒ 0 () =
2 − (2)
( − 2)
( − 2)
=
=
. 0 () 0 ⇔ 0 or 2, so is
(2 )2
4
3
increasing on (−∞ 0) and (2 ∞). 0 () 0 ⇔ 0 2, so is decreasing on (0 2).
(b) 00 () =
3 [ · 1 + ( − 2) ] − ( − 2) · 32
2 [( − 1) − 3( − 2)]
(2 − 4 + 6)
=
=
.
3
2
6
( )
4
2 − 4 + 6 = (2 − 4 + 4) + 2 = ( − 2)2 + 2 0, so 00 () 0 and is CU on (−∞ 0) and (0 ∞).
(c) There are no changes in concavity and, hence, there are no points of inflection.
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°
576
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CHAPTER 6
INVERSE FUNCTIONS
73. = () = −1(+1)
C. No symmetry D.
−1 ( + 1) → −∞,
A. = { | 6= −1} = (−∞ −1) ∪ (−1, ∞) B. No ­intercept; ­intercept = (0) = −1
lim −1(+1) = 1 since −1( + 1) → 0, so = 1 is a HA.
→±∞
lim −1(+1) = ∞ since −1( + 1) → ∞, so = −1 is a VA.
→−1−
E. 0 () = −1(+1) ( + 1)2
⇒ 0 () 0 for all except 1, so
is increasing on (−∞ −1) and (−1 ∞). F. No extreme values
G. 00 () =
lim −1(+1) = 0 since
→−1+
H.
−1(+1)
−1(+1) (−2)
−1(+1) (2 + 1)
+
=
−
( + 1)4
( + 1)3
( + 1)4
⇒
00 () 0 ⇔ 2 + 1 0 ⇔ − 12 , so is CU on (−∞ −1)
and −1 − 12 , and CD on − 12 , ∞ . has an IP at − 12 , −2 .
74. = () = − sin , 0 ≤ ≤ 2
A. = B. ­intercept: (0) = 0; ­intercepts: () = 0 ⇔ sin = 0 ⇔
= 0, , and 2. C. No symmetry D. No asymptote E. 0 () = − cos + sin (−− ) = − (cos − sin ).
0 () = 0 ⇔ cos = sin ⇔ = 4 , 5
. 0 () 0 if is in 0 4 or 5
2 [ is increasing] and
4
4
0 () 0 if is in 4 5
[ is decreasing]. F. Local maximum value 4 and local minimum value 5
4
4
G. 00 () = − (− sin − cos ) + (cos − sin )(−− ) = − (−2 cos ). 00 () 0 ⇔ −2 cos 0 ⇔
[ is CU] and 00 () 0 ⇔
H.
cos 0 ⇒ is in 2 3
2
cos 0 ⇒ is in 0 2 or 3
2 2 [ is CD].
IP at 2 + 2 +
75. = 1(1 + − )
A. = B. No ­intercept; ­intercept = (0) = 12 C. No symmetry
−
1
) = 0 since lim − = ∞, so has horizontal asymptotes
D. lim 1(1 + − ) = 1 +
0 = 1 and lim 1(1 +
→∞
→−∞
0
− −2
= 0 and = 1. E. () = −(1 +
F. No extreme values G. 00 () =
)
→−∞
−
(−
−
)=
− 2
(1 +
) . This is positive for all , so is increasing on .
(1 + − )2 (−− ) − − (2)(1 + − )(−− )
− (− − 1)
=
−
4
(1 + )
(1 + − )3
The second factor in the numerator is negative for 0 and positive for 0,
H.
and the other factors are always positive, so is CU on (−∞, 0) and CD
on (0 ∞). IP at 0, 12
76. The function () = cos is periodic with
period 2, so we consider it only on the interval
[0 2]. We see that it has local maxima of about
(0) ≈ 272 and (2) ≈ 272, and a local
minimum of about (314) ≈ 037. To find the
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°
SECTION 6.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES
¤
577
exact values, we calculate 0 () = − sin cos . This is 0 when − sin = 0 ⇔ = 0, or 2 (since we are only
considering ∈ [0 2]). Also 0 () 0 ⇔ sin 0 ⇔ 0 . So (0) = (2) =
(both maxima) and () = cos = 1 (minimum). To find the inflection points, we calculate and graph
00 () =
(− sin cos ) = − cos cos − sin (cos )(− sin ) = cos sin2 − cos . From the graph of 00 (),
we see that has inflection points at ≈ 090 and at ≈ 538. These ­coordinates correspond to inflection points
(090 186) and (538 186).
3
77. () = − → 0 as → −∞, and
() → ∞ as → ∞. From the graph,
it appears that has a local minimum of
about (058) = 068, and a local
maximum of about (−058) = 147.
To find the exact values, we calculate
3
0 () = 32 − 1 − , which is 0 when 32 − 1 = 0 ⇔ = ± √13 . The negative solution corresponds to the local
√ 3
√
√
maximum − √13 = (−1 3) − (−1 3) = 2 39 , and the positive solution corresponds to the local minimum
1
√
3
√
3
√
√
= (1 3) − (1 3) = −2 39 . To estimate the inflection points, we calculate and graph
00 () =
3
3
3
3
2
3 − 1 − = 32 − 1 − 32 − 1 + − (6) = − 94 − 62 + 6 + 1 .
From the graph, it appears that 00 () changes sign (and thus has inflection points) at ≈ −015 and ≈ −109. From the
graph of , we see that these ­values correspond to inflection points at about (−015 115) and (−109 082).
78. () = − with = 001, = 4, and = 007. We will find the
zeros of 00 for () = − .
0 () = (−− ) + − (−1 ) = − (− + −1 )
00 () = − (−−1 + ( − 1)−2 ) + (− + −1 )(−− )
= −2 − [− + ( − 1) + 2 2 − ]
= −2 − (2 2 − 2 + 2 − )
Using the given values of and gives us 00 () = 2 −007 (000492 − 056 + 12). So 00 () = 001 00 () and its zeros
600
are = 0 and the solutions of 000492 − 056 + 12 = 0, which are 1 = 200
7 2857 and 2 = 7 8571.
At 1 minutes, the rate of increase of the level of medication in the bloodstream is at its greatest and at 2 minutes, the rate of
decrease is the greatest.
79. Let = 0135 and = −2802. Then () =
+ 1 = 0 ⇔ = −
⇒ 0 () = ( · · + · 1) = ( + 1). 0 () = 0 ⇔
1
≈ 036 h. (0) = 0, (−1) = − −1 = − ≈ 00177, and (3) = 33 ≈ 000009.
[continued]
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°
¤
578
CHAPTER 6
INVERSE FUNCTIONS
The maximum average BAC during the first three hours is about 00177 gdL and it occurs at approximately 036 h
(214 min).
80. (a) As || → ∞, = −2 (22 ) → −∞, and → 0. The HA is = 0. Since takes on its maximum value at = 0, so
2
2
does . Showing this result using derivatives, we have () = − (2 )
2
2
⇒ 0 () = − (2 ) (−2 ).
0 () = 0 ⇔ = 0. Because 0 changes from positive to negative at = 0, (0) = 1 is a local maximum. For
inflection points, we find 00 () = −
00 () = 0 ⇔ 2 = 2
−1
2
2
2
2
1 −2(22 )
· 1 + − (2 ) (−2 ) = 2 − (2 ) (1 − 22 ).
2
⇔ = ±. 00 () 0 ⇔ 2 2
⇔ − .
So is CD on (− ) and CU on (−∞ −) and ( ∞). IP at (± −12 ).
(b) Since we have IP at = ±, the inflection points move away from the ­axis as increases.
From the graph, we see that as increases, the graph tends to spread out and
(c)
there is more area between the curve and the ­axis.
81.
( + ) =
5
5
=
= 5 − (−5) = 10
0
82.
83.
1
1
+1
1
+ =
+ − (0 + 1) =
+−1
+1
+
1
+
1
0
1
−5
2
0
−5
=
2
1
1
1
1
− = − − = − −2 + 0 = (1 − −2 )
0
0
2
84. Let = −4 . Then = −43 and 3 = − 14 , so
85. Let = 1 + . Then = , so
86.
87.
88.
4
3 − =
√
√
1 + =
= 23 32 + = 23 (1 + )32 + .
(1 + )2
=
( + − )2 = (2 + 2 + −2 ) = 12 2 + 2 − 12 −2 +
(4 + )5
1 + 2 + 2
=
= 4 +
=
=
4
− 14 = − 14 + = − 14 − + .
(− + 2 + ) = −− + 2 + +
5 = 16 6 + = 16 (4 + )6 +
89. Let = 1 − . Then = − and = −, so
=
(1 − )2
1
(−) = −
2
90. Let = sin . Then = cos , so
−2 = −(−−1 ) + =
sin cos =
1
1
+ =
+ .
1 −
= + = sin + .
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 6.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES
91. Let = 1, so = −12 . When = 1, = 1; when = 2, = 12 . Thus,
2
1
92.
1
=
2
12
1
√
12
(−) = − 1 = −(12 − ) = − .
1√
1+1
1√
1 + −
= 1 + −
12
− − =
=
1
+
(−)
−
= −
0
0
2
1+1
√
= − 23 [(1 + 1)32 − 232 ] = 43 2 − 23 (1 + 1)32
= − 23 32
2
2
2
2−
2 2
= 12 (−−4 + 1)
= 12 −−
1
93. avg = 2 −
0
0
0
1
1
[ − ( 2 − 2)] = −1 ( − 2 + 2) = − 13 3 + 2 −1
= − 13 + 2 − −1 + 13 − 2 = − −1 − 23 + 4 = − −1 + 10
3
94. Area =
95. Area =
1
−1
1
1 3
− = 13 3 − 0 = 13 3 − − 13 − 1 = 13 3 − + 23 ≈ 4644
0
96. 00 () = 3 + 5 sin
⇒ 0 () = 3 − 5 cos +
⇒ 2 = 0 (0) = 3 − 5 +
0 () = 3 − 5 cos + 4 ⇒ () = 3 − 5 sin + 4 +
⇒ = 4, so
⇒ 1 = (0) = 3 +
⇒ = −2,
so () = 3 − 5 sin + 4 − 2.
97. =
1
0
1
1
( )2 = 0 2 = 12 2 0 = 2 2 − 1
98. The shell has radius circumference 2, and
2
height − , so =
1
2
2− .
0
Let = . Thus, = 2 , so
2
=
1
0
1
− = −− 0 = (1 − 1).
99. First Figure
Second Figure
Third Figure
√
Let = , so = 2 and = 2 . When = 0, = 0; when = 1, = 1. Thus,
1 √
1
1
1 = 0 = 0 (2 ) = 2 0 .
2 =
1
0
1
2 = 2 0 .
Let = sin , so = cos . When = 0, = 0; when = 2 , = 1. Thus,
2
2
1
1
3 = 0 sin sin 2 = 0 sin (2 sin cos ) = 0 (2 ) = 2 0 .
Since 1 = 2 = 3 , all three areas are equal.
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°
¤
579
580
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CHAPTER 6
INVERSE FUNCTIONS
100. Let () = with = 450268 and = 112567, and () = population after hours. Since () = 0 (),
3
0
() = (3) − (0) is the total change in the population after three hours. Since we start with 400 bacteria, the
population will be
3
3
3
(3) = 400 + 0 () = 400 + 0 = 400 + 0 = 400 + 3 − 1
≈ 400 + 11,313 = 11,713 bacteria
101. The rate is measured in liters per minute. Integrating from = 0 minutes to = 60 minutes will give us the total amount of oil
that leaks out (in liters) during the first hour.
60
60
() = 0 100−001
[ = −001, = −001]
0
−06
−06
(−100 ) = −10,000 0
= −10,000(−06 − 1) 45119 4512 L
= 100 0
102. The rate is measured in kilograms per year. Integrating from = 0 years (2000) to = 20 years (2020) will give us the net
change in biomass from 2000 to 2020.
20
0
60,000−06
=
(1 + 5−06 )2
=
1+5−12
6
60,000 1
− 3
2
= 1 + 5−06
= −3−06
1+5−12
20,000
20,000
20,000
16,666
=
−
−12
1
+
5
6
6
Thus, the predicted biomass for the year 2020 is approximately 25,000 + 16,666 = 41,666 kg.
103.
30
30
−30
0 − = 0
(−)
1
0
−30
= 0 −
= 0 (−−30 + 1)
() =
0
= −
= − −
1
The integral
30
0
() represents the total amount of urea removed from the blood in the first 30 minutes of dialysis.
2
104. (a) erf() = √
0
−2
=
2
2
−
0
− =
−2
0
0
⇒
2
− =
0
√
erf() By Property 5 of definite integrals in Section 4.2,
2
−2
+
, so
√
√
2
2
√
erf() −
erf() = 12 [erf() − erf()].
− −
− =
2
2
0
2
2
2
2
2
2
(b) = erf() ⇒ 0 = 2 erf() + erf 0 () = 2 + · √ −
2
[by FTC1] = 2 + √ .
105. We use Theorem 6.1.7. Note that (0) = 3 + 0 + 0 = 4, so −1 (4) = 0. Also 0 () = 1 + . Therefore,
−1 0
(4) =
1
1
1
1
= 0
=
= .
0 ( −1 (4))
(0)
1 + 0
2
106. We recognize this limit as the definition of the derivative of the function () = sin at = , since it is of the form
lim
→
() − ()
. Therefore, the limit is equal to 0 () = (cos )sin = −1 · 0 = −1.
−
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°
SECTION 6.3
LOGARITHMIC FUNCTIONS
¤
From the graph, it appears that is an odd function ( is undefined for = 0).
107.
To prove this, we must show that (−) = −().
1
1 − 1 1
1 − (−1)
1 − 1
1 − 1(−)
=
=
· 1 = 1
(−) =
1(−)
(−1)
1
1+
1+
+1
1 + 1
1 − 1
=−
= −()
1 + 1
so is an odd function.
108. We’ll start with = −1 and graph () =
1
for = 01, 1, and 5.
1 +
From the graph, we see that there is a horizontal asymptote = 0 as → −∞
and a horizontal asymptote = 1 as → ∞. If = 1, the y­intercept is 0 12 .
As gets smaller (close to 0), the graph of moves left. As gets larger, the graph
of moves right.
As changes from −1 to 0, the graph of is stretched horizontally. As
changes through large negative values, the graph of is compressed horizontally.
(This takes care of negatives values of .)
If is positive, the graph of is reflected through the y­axis.
Last, if = 0, the graph of is the horizontal line = 1(1 + ).
6.3 Logarithmic Functions
1. (a) It is defined as the inverse of the exponential function with base , that is, log =
(b) (0 ∞)
(c)
(d) See Figure 1.
⇔ = .
2. (a) The natural logarithm is the logarithm with base , denoted ln .
(b) The common logarithm is the logarithm with base 10, denoted log .
(c) See Figure 3.
3. (a) log3 81 = log3 34 = 4
4. (a) ln
1
= ln −2 = −2
2
1
= log3 3−4 = −4
(b) log3 81
(b) ln
√
= ln 12 = 12
(c) log9 3 = log9 912 = 12
50
= ln(50 ) = 50
(c) ln ln
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°
581
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CHAPTER 6
INVERSE FUNCTIONS
5. (a) log2 30 − log2 15 = log2
30
15
= log2 2 = 1
(b) log3 10 − log3 5 − log3 18 = log3
10
5
2
18
104
4
5 · 104
= log5 5−4 = −4
− log3 18 = log3 2 − log3 18 = log3
= log3
1
9
= log3 3−2 = −2
(c) 2 log5 100 − 4 log5 50 = log5 1002 − log5 504 = log5
1002
504
−2
3
(b) −2 ln 5 = ln 5
6. (a) 3 ln 2 = ln 2 = 23 = 8
= log5
3
1
= 5−2 = 25
7. (a) log10 2 3 = log10 2 + log10 3 + log10
[Law 1]
= 2 log10 + 3 log10 + log10
[Law 3]
4
(b) ln √
= ln 4 − ln(2 − 4)12
2 − 4
(c) ln(ln ) = ln(3) = 3
[Law 2]
= 4 ln − 12 ln[( + 2)( − 2)]
[Law 3]
= 4 ln − 12 [ln( + 2) + ln( − 2)]
[Law 1]
= 4 ln − 12 ln( + 2) − 12 ln( − 2)
8. (a) ln
12
1
3
3
3
= ln
= ln
−3
−3
2
−3
[Law 3]
= 12 [ln 3 + ln − ln( − 3)]
[Laws 1 and 2]
= 12 ln 3 + 12 ln − 12 ln( − 3)
(b) log2 (3 + 1) 3 ( − 3)2 = log2 (3 + 1) + log2 3 ( − 3)2
= log2 (3 + 1) + log2 ( − 3)23
= log2 (3 + 1) + 23 log2 ( − 3)
[Law 1]
[Law 3]
√
9. (a) log10 20 − 13 log10 1000 = log10 20 − log10 100013 = log10 20 − log10 3 1000
= log10 20 − log10 10 = log10 20
= log10 2
10
(b) ln − 2 ln + 3 ln = ln − ln 2 + ln 3 = ln
10. (a) ln 10 + 2 ln 5 = ln 10 + ln 52
[by Law 3]
= ln [(10)(25)]
[by Law 1]
3
+ ln 3 = ln 2
2
= ln 250
4
(b) log10 4 + log10 − 13 log10 ( + 1) = log10 (4) − log10 ( + 1)13 = log10 √
3
+1
11. (a) 3 ln( − 2) − ln(2 − 5 + 6) + 2 ln( − 3) = ln( − 2)3 − ln [( − 2)( − 3)] + ln( − 3)2
( − 2)3 ( − 3)2
= ln[( − 2)2 ( − 3)]
= ln
( − 2)( − 3)
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°
SECTION 6.3
LOGARITHMIC FUNCTIONS
(b) log − log + log = log − log + log = log
12. (a) log5 10 =
ln 10
≈ 1430677
ln 5
(b) log15 12 =
13. (a) log3 12 =
ln 12
≈ 2261860
ln 3
(b) log12 6 =
14. To graph the functions, we use log2 =
ln 12
≈ 0917600
ln 15
ln 6
≈ 0721057
ln 12
ln
ln
, log4 =
, etc. These
ln 2
ln 4
graphs all approach −∞ as → 0+ , and they all pass through the point
(1 0). Also, they are all increasing, and all approach ∞ as → ∞. The
smaller the base, the larger the rate of increase of the function (for 1)
and the closer the approach to the ­axis (as → 0+ ).
15. To graph these functions, we use log15 =
ln
ln
and log50 =
.
ln 15
ln 50
These graphs all approach −∞ as → 0+ , and they all pass through the
point (1 0). Also, they are all increasing, and all approach ∞ as → ∞.
The functions with larger bases increase extremely slowly, and the ones with
smaller bases do so somewhat more quickly. The functions with large bases
approach the ­axis more closely as → 0+ .
16. We see that the graph of ln is the reflection of the graph of about the
line = , and that the graph of log8 is the reflection of the graph of 8
about the same line. The graph of 8 increases more quickly than that of .
Also note that log8 → ∞ as → ∞ more slowly than ln .
17. We need x such that log2 x = 25 cm ⇐⇒ x = 2 25 = 33 554 432 cm = 335.5443 km
18.
From the graphs, we see that () = 01 () = ln for approximately 0 306, and then () () for
306 343 × 1015 (approximately). At that point, the graph of finally surpasses the graph of for good.
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°
¤
583
584
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CHAPTER 6
INVERSE FUNCTIONS
19. (a) Shift the graph of = log10 five units to the left to
(b) Reflect the graph of = ln about the ­axis to obtain
the graph of = − ln .
obtain the graph of = log10 ( + 5). Note the vertical
asymptote of = −5.
= log10
= log10 ( + 5)
20. (a) Reflect the graph of = ln about the ­axis to obtain
= ln
= − ln
(b) Reflect the portion of the graph of = ln to the right
the graph of = ln (−).
of the ­axis about the ­axis. The graph of = ln ||
is that reflection in addition to the original portion.
= ln
= ln (−)
= ln
= ln ||
21. (a) The domain of () = ln + 2 is 0 and the range is .
(b) = 0 ⇒ ln + 2 = 0 ⇒ ln = −2 ⇒ = −2
(c) We shift the graph of = ln two units upward.
22. (a) The domain of () = ln( − 1) − 1 is 1 and the range is .
(b) = 0 ⇒ ln( − 1) − 1 = 0 ⇒ ln( − 1) = 1 ⇒
− 1 = 1
⇒ =+1
(c) We shift the graph of = ln one unit to the right and one unit downward.
23. (a) ln(4 + 2) = 3
⇒ ln(4+2) = 3
⇒ 4 + 2 = 3
⇒ 4 = 3 − 2 ⇒ = 14 (3 − 2) ≈ 4521
(b) 2−3 = 12 ⇒ ln 2−3 = ln 12 ⇒ 2 − 3 = ln 12 ⇒ 2 = 3 + ln 12 ⇒ = 12 (3 + ln 12) ≈ 2742
24. (a) log2 (2 − − 1) = 2
⇒ 2 − − 1 = 22 = 4 ⇒ 2 − − 5 = 0 ⇒
√
1 ± (−1)2 − 4(1)(−5)
1 ± 21
=
=
.
2(1)
2
√
√
1 − 21
1 + 21
≈ −1791 and 2 =
≈ 2791.
Solutions are 1 =
2
2
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°
SECTION 6.3
LOGARITHMIC FUNCTIONS
¤
585
(b) 1 + 4+1 = 20 ⇒ 4+1 = 19 ⇒ ln 4+1 = ln 19 ⇒ 4 + 1 = ln 19 ⇒ 4 = −1 + ln 19 ⇒
= 14 (−1 + ln 19) ≈ 0486
2
25. (a) ln + ln( − 1) = 0
⇒ ln[( − 1)] = 0 ⇒ ln[ −] = 0 ⇒ 2 − = 1 ⇒ 2 − − 1 = 0. The
√
√
1 ± (−1)2 − 4(1)(−1)
1± 5
1− 5
quadratic formula gives =
=
, but we note that ln
is undefined because
2(1)
2
2
√
√
1+ 5
1− 5
0. Thus, =
≈ 1618.
2
2
(b) 51−2 = 9 ⇒ ln 51−2 = ln 9 ⇒ (1 − 2) ln 5 = ln 9 ⇒ 1 − 2 =
26. (a) ln(ln ) = 0
(b)
⇒ ln(ln ) = 0
ln 9
ln 5
⇒ =
1
ln 9
−
≈ −0183
2
2 ln 5
⇒ ln = 1 ⇒ = ≈ 2718
60
= 4 ⇒ 60 = 4(1 + − ) ⇒ 15 = 1 + −
1 + −
⇒ 14 = −
⇒ ln 14 = ln −
⇒
ln 14 = − ⇒ = − ln 14 ≈ −2639
27. (a) 2 − 3 + 2 = 0
⇔ ( − 1)( − 2) = 0 ⇔ = 1 or = 2 ⇔ = ln 1 or = ln 2, so = 0 or ln 2.
(b) = 10 ⇔ ln = ln 10 ⇔ ln = = ln 10 ⇔ ln = ln(ln 10) ⇔ = ln ln 10
28. (a) 3+1 =
⇔ = 13 (ln − 1)
⇔ 3 + 1 = ln
(b) log2 () = ⇔ = 2
29. (a) ln(1 + 3 ) − 4 = 0
⇔ = 2
⇔ ln(1 + 3 ) = 4 ⇔ 1 + 3 = 4
(b) 21 = 42 ⇔ 1 = 21 ⇔
30. (a) 21−3 = 99
1
1
= ln 21 ⇔ =
≈ 03285.
ln 21
⇔ (1 − 3) ln 2 = ln 99 ⇔ 1 − 3 =
ln 99
1
1−
≈ −18765
=
3
ln 2
+1
(b) ln
=2 ⇔
31. (a) ln 0
⇒ 0
is 0 1.
+1
= 2
√
⇔ 3 = 4 − 1 ⇔ = 3 4 − 1 ≈ 37704.
ln 99
ln 2
⇔ 3 = 1 −
ln 99
ln 2
⇔ + 1 = 2 ⇔ (2 − 1) = 1 ⇔ =
⇔
1
≈ 01565
2 − 1
⇒ 1. Since the domain of () = ln is 0, the solution of the original inequality
(b) 5 ⇒ ln ln 5 ⇒ ln 5
32. (a) 1 3−1 2
⇒ ln 1 3 − 1 ln 2 ⇒ 0 3 − 1 ln 2 ⇒ 1 3 1 + ln 2 ⇒
1
1
3 3 (1 + ln 2)
(b) 1 − 2 ln 3 ⇒ −2 ln 2 ⇒ ln −1 ⇒ −1
33. If is the intensity of the 1989 San Francisco earthquake, then log10 () = 71
⇒
log10 (16) = log10 16 + log10 () = log10 16 + 71 ≈ 83.
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°
586
¤
CHAPTER 6
INVERSE FUNCTIONS
34. Let 1 and 2 be the intensities of the music and the mower. Then 10 log10
log10
1
2
= log10
1 0
2 0
35. (a) = () = 100 · 23
= log10
⇒
1
0
− log10
= 23
100
2
0
⇒ log2
Formula, we can write this as = −1 () = 3 ·
1
0
2
= 120 and 10 log10
= 106, so
0
1
= 1014 ≈ 25.
2
= 12 − 106 = 14 ⇒
100
=
3
⇒ = 3 log2
. Using the Change of Base
100
ln(100)
. This function tells us how long it will take to obtain
ln 2
bacteria (given the number ).
(b) = 50,000 ⇒ = −1 (50,000) = 3 ·
ln 50,000
ln 500
100
=3
≈ 269 hours
ln 2
ln 2
= 1 − − ⇒ − = 1 −
⇒
0
0
⇒ = − ln 1 −
. This formula gives the time (in seconds) needed after a discharge to
− = ln 1 −
0
0
36. (a) We write = 0 (1 − − ) and solve for :
obtain a given charge .
090
= −50 ln(01) ≈ 1151 seconds. It will take
(b) We set = 090 and = 50 to get = −50 ln 1 −
0
approximately 115 seconds—just shy of two minutes—to recharge the capacitors to 90% of capacity.
√
37. lim ln( − 1) = −∞ since
→1+
√
− 1 → 0+ as → 1+ .
38. As → 2− , 8 − 4 = (8 − 3 ) → 0+ since is positive and 8 − 3 → 0+ . Thus, lim log5 (8 − 4 ) = −∞.
→2−
39. lim ln(cos ) = ln 1 = 0. [ln(cos ) is continuous at = 0 since it is the composite of two continuous functions.]
→0
40. lim ln(sin ) = −∞ since sin → 0+ as → 0+ .
→0+
41. lim [ln(1 + 2 ) − ln(1 + )] = lim ln
→∞
→∞
1 + 2
= ln
1+
1 + 2
→∞ 1 +
lim
= ln
1
+
= ∞, since the limit in
→∞ 1 + 1
lim
parentheses is ∞.
42. lim [ln(2 + ) − ln(1 + )] = lim ln
→∞
→∞
2+
1+
= lim ln
→∞
2 + 1
1 + 1
= ln
1
= ln 1 = 0
1
43. () = ln(4 − 2 ).
= | 4 − 2 0 = | 2 4 = { | || 2} = (−2 2)
44. () = log2 (2 + 3).
= | 2 + 3 0 = { | ( + 3) 0} = { | 0 and + 3 0} ∪ { | 0 and + 3 0}
= { | 0 and −3} ∪ { | 0 and −3} = (−∞ −3) ∪ (0 ∞)
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°
SECTION 6.3
45. (a) For () =
√
3 − 2 , we must have 3 − 2 ≥ 0 ⇒ 2 ≤ 3
⇒
LOGARITHMIC FUNCTIONS
¤
587
2 ≤ ln 3 ⇒ ≤ 12 ln 3.
Thus, the domain of is (−∞ 12 ln 3].
(b) = () =
√
3 − 2
[note that ≥ 0] ⇒ 2 = 3 − 2
⇒ 2 = 3 − 2
⇒ 2 = ln(3 − 2 ) ⇒
= 12 ln(3 − 2 ). Interchange and : = 12 ln(3 − 2 ). So −1 () = 12 ln(3 − 2 ). For the domain of −1 ,
√
√
√
√
we must have 3 − 2 0 ⇒ 2 3 ⇒ || 3 ⇒ − 3 3 ⇒ 0 ≤ 3 since ≥ 0. Note
√
that the domain of −1 , [0 3 ), equals the range of .
46. (a) For () = ln(2 + ln ), we must have 2 + ln 0
⇒ ln −2 ⇒ −2 . Thus, the domain of
is (−2 ∞).
(b) = () = ln(2 + ln ) ⇒ = 2 + ln ⇒ ln = − 2 ⇒ = −2 . Interchange and : = −2 .
So −1 () = −2 . The domain of −1 , as well as the range of , is .
47. (a) We must have − 3 0
⇔ 3 ⇔ ln 3. Thus, the domain of () = ln( − 3) is (ln 3 ∞).
(b) = ln( − 3) ⇒ = − 3 ⇒ = + 3 ⇒ = ln( + 3), so −1 () = ln( + 3).
Now + 3 0 ⇒ −3, which is true for any real , so the domain of −1 is .
48. (a) By (6), ln 300 = 300 and ln(300 ) = 300.
(b) A calculator gives ln 300 = 300 and an error message for ln(300 ) because 300 is larger than most calculators can
evaluate.
49. We solve = 3 ln( − 2) for : 3 = ln( − 2)
⇒ 3 = − 2 ⇒ = 2 + 3 . Interchanging and gives the
inverse function = 2 + 3 .
50. = () = log4 (3 + 2)
√
So −1 () = 3 4 − 2.
√
√
⇒ 4 = 3 + 2 ⇒ 3 = 4 − 2 ⇒ = 3 4 − 2. Interchange and : = 3 4 − 2.
51. We solve = 1− for : ln = ln 1−
function = 1 − ln .
52. = (ln )2 , ≥ 1, ln =
53. = () = 32−4
⇒ ln = 1 − ⇒ = 1 − ln . Interchanging and gives the inverse
√
√
⇒ = . Interchange and : = is the inverse function.
⇒ log3 = 2 − 4 ⇒ 2 = log3 + 4 ⇒ = 12 log3 + 2. Interchange and :
= 12 log3 + 2. So −1 () = 12 log3 + 2.
54. We solve =
1 − −
for : (1 + − ) = 1 − −
1 + −
− (1 + ) = 1 −
⇒ − =
1−
1+
⇒ + − = 1 − −
⇒ − = ln
1−
1+
⇒ = − ln
⇒ − + − = 1 −
1−
or, equivalently,
1+
−1
1−
1+
1+
= ln
. Interchanging and gives the inverse function = ln
.
= ln
1+
1−
1−
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°
⇒
588
¤
CHAPTER 6
INVERSE FUNCTIONS
33
1
1
1
⇔ 2 ln 3 = − ln 3 ⇔ − 2 ln 3, so is increasing on − 2 ln 3 ∞ .
⇒ 0 () = 33 − . Thus, 0 () 0
55. () = 3 −
2 13
56. = 2 − −3
4 92
⇔
33
⇔
⇒ 00 = 2 − 9−3 . Thus, 00 0 ⇔ 2 9−3
⇔ 14 ln 92 , so is concave downward on −∞ 14 ln 92 .
⇒ 0 = 2 + 3−3
⇔ 4 ln 92
⇔ 32 1 ⇔
⇔
57. (a) We have to show that − () = (−).
√
√
−1
= ln
− () = − ln + 2 + 1 = ln + 2 + 1
= ln
+
1
√
2 + 1
√
√
√
− 2 + 1
− 2 + 1
1
√
√
= ln 2 + 1 − = (−)
·
= ln 2
2 −1
2
2
−
+ +1 − +1
Thus, is an odd function.
√
√
(b) Let = ln + 2 + 1 . Then = + 2 + 1 ⇔ ( − )2 = 2 + 1 ⇔ 2 − 2 + 2 = 2 + 1 ⇔
2 = 2 − 1 ⇔ =
2 − 1
= 12 ( − − ). Thus, the inverse function is −1 () = 12 ( − − ).
2
58. Let − be the point where the tangent meets the curve. The tangent has slope −− and is perpendicular to the line
2 − = 8, which has slope 2. So −− = − 12 ⇒ − = 12 ⇒ = 2 ⇒ = ln( ) = ln 2. Thus, the point on
the curve is ln 2 12 and the equation of the tangent is − 12 = − 12 ( − ln 2) or + 2 = 1 + ln 2.
59. 1 ln = 2
1
· ln = ln 2 ⇒ 1 = ln 2, a contradiction, so the given equation has no
ln
⇒ ln(1 ln ) = ln(2) ⇒
solution. The function () = 1 ln = (ln )1 ln = 1 = for all 0, so the function () = 1 ln is the constant
function () = .
60. (a) lim ln = lim
→∞
→∞
ln ln
2
= lim (ln ) = ∞ since (ln )2 → ∞ as → ∞.
→∞
− ln
2
(b) lim − ln = lim ln
= lim −(ln ) = 0 since −(ln )2 → −∞ as → 0+ .
→0+
→0+
(c) lim 1 = lim
→0+
→0+
→0+
ln 1
ln
→ −∞ as → 0+ . Note that as → 0+ , ln is a large
= lim (ln ) = 0 since
→0+
negative number and is a small positive number, so (ln ) → −∞.
− ln
(d) lim (ln 2)− ln = lim ln(ln 2)
= lim − ln ln(ln 2) = 0 since − ln ln(ln 2) → −∞ as → ∞.
→∞
→∞
→∞
61. (a) Let 0 be given. We need such that | − 0| when . But
Then
⇒ log ⇒ | − 0| = , so lim = 0.
→−∞
(b) Let 0 be given. We need such that when . But
Then
⇔ log . Let = log .
⇒ log
⇔ log . Let = log .
⇒ , so lim = ∞.
→∞
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°
SECTION 6.4 DERIVATIVES OF LOGARITHMIC FUNCTIONS
62. (a) () = −
¤
⇒ () = 0 () = −− = −()
(b) (0) = 0 = , so is the initial velocity.
(c) () = − = 2 ⇒ − = 12
⇒ − = ln 12 = − ln 2 ⇒ = (ln 2)
⇒ 0 2 − 2 − 2 ≤ 1. Now 2 − 2 − 2 ≤ 1 gives 2 − 2 − 3 ≤ 0 and hence
√
√
( − 3)( + 1) ≤ 0. So −1 ≤ ≤ 3. Now 0 2 − 2 − 2 ⇒ 1 − 3 or 1 + 3. Therefore,
√
√
ln(2 − 2 − 2) ≤ 0 ⇔ −1 ≤ 1 − 3 or 1 + 3 ≤ 3.
63. ln(2 − 2 − 2) ≤ 0
64. (a) The primes less than 25 are 2, 3, 5, 7, 11, 13, 17, 19, and 23. There
are 9 of them, so (25) = 9. We use the sieve of Eratosthenes, and
arrive at the figure at right. There are 25 numbers left over, so
(100) = 25.
25
()
. We compute (100) =
≈ 115,
ln
100 ln 100
(1000) ≈ 116, 104 ≈ 113, 105 ≈ 110, 106 ≈ 108,
and 107 ≈ 107.
(b) Let () =
(c) By the Prime Number Theorem, the number of primes less than a billion, that is, (109 ), should be close to
109 ln 109 ≈ 48,254,942. In fact, (109 ) = 50,847,543, so our estimate is off by about 51%. Do not attempt this
calculation at home.
6.4 Derivatives of Logarithmic Functions
1. The differentiation formula for logarithmic functions,
2. () = ln(3 + 2 )
⇒ 0 () =
1
1
2
(3 + 2 ) =
·
· 2 =
3 + 2
3 + 2
3 + 2
3. () = ln(2 + 3 + 5)
⇒ 0 () =
4. () = ln −
0 () = ·
5. () = sin(ln )
⇒
2 + 3
1
1
·
(2 + 3 + 5) = 2
· (2 + 3) = 2
2 + 3 + 5
+ 3 + 5
+ 3 + 5
1
+ (ln ) · 1 − 1 = 1 + ln − 1 = ln
⇒ 0 () = cos(ln ) ·
1
cos(ln )
ln = cos(ln ) · =
6. () = ln(sin2 ) = ln(sin )2 = 2 ln |sin |
7. () = ln
1
⇒ 0 () =
Another solution: () = ln
1
(log ) =
, is simplest when = because ln = 1.
ln
1
1
⇒ 0 () = 2 ·
1
· cos = 2 cot
sin
1
1
1
= − 2 =− .
1
1
= ln 1 − ln = − ln ⇒ 0 () = − .
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
589
590
¤
8. =
CHAPTER 6
INVERSE FUNCTIONS
1
= (ln )−1
ln
⇒ 0 = −1(ln )−2 ·
−1
1
=
(ln )2
9. () = ln(−2 ) = ln + ln −2 = ln − 2
10. () =
1
−2
√
1
1
1
· = √
(1 + ln ) = √
1 + ln ⇒ 0 () = 12 (1 + ln )−12
2 1 + ln
2 1 + ln
⇒ 0 () = (ln )2 cos + sin · 2 ln ·
11. () = (ln )2 sin
12. () = ln
√
1
1
· 2 = 2
2 + 1 = ln(2 + 1)12 = 12 ln(2 + 1) ⇒ 0 () = · 2
2 +1
+1
⇒ 0 =
13. = log8 (2 + 3)
14. = log10 sec
15. () =
0 () =
2 sin
1
= ln ln cos +
√
√ 2
2
1
1
·
· √
= 2
2 + 1 ⇒ 0 () = √
+1 = √
2
2
2
+
1
+1
+1 2 +1
Or: () = ln
⇒ 0 =
ln
1 + ln(2)
2 + 3
1
1
·
(2 + 3) = 2
· (2 + 3) = 2
(2 + 3) ln 8
( + 3) ln 8
( + 3) ln 8
1
tan
1
·
(sec ) =
· sec tan =
sec (ln 10)
sec (ln 10)
ln 10
⇒
1
1
[1 + ln(2)] · 1 − ln · 2
·2
[1 + ln(2) − ln ]
1 + (ln 2 + ln ) − ln
1 + ln 2
=
=
=
2
[1 + ln(2)]
[1 + ln(2)]2
[1 + ln(2)]2
[1 + ln(2)]2
16. = ln(csc − cot )
0 =
⇒ 0 () =
⇒
1
1
csc (csc − cot )
(csc − cot ) =
(− csc cot + csc2 ) =
= csc
csc − cot
csc − cot
csc − cot
17. () = 5 + 5
⇒ 0 () = 54 + 5 ln 5
18. () = sin(2 )
⇒ 0 () = cos(2 ) · 2 ln 2 + sin(2 ) · 1 = 2 ln 2 cos(2 ) + sin(2 )
19. () = 2 log2
1
+ log2 · 2 ln 2 = 2
⇒ () = 2
ln 2
0
Note that log2 (ln 2) =
2
20. () = ln
2
0 () = ln ·
1
+ log2 (ln 2) .
ln 2
ln
(ln 2) = ln by the change of base formula. Thus, 0 () = 2
ln 2
1
+ ln .
ln 2
⇒
2
2
2
1
(2 ln ) = ln 2 · + (ln ) · 2 = ln ( + 2 ln ) = ln (1 + 2 ln )
√
(2 + 1)4
= ln + ln(2 + 1)4 − ln 3 2 − 1 = ln + 4 ln(2 + 1) − 13 ln(2 − 1) ⇒
2 − 1
21. () = ln √
3
0 () =
1
1
1
1
8
2
1
+4· 2
· 2 − ·
·2= + 2
−
+1
3 2 − 1
+1
3(2 − 1)
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 6.4 DERIVATIVES OF LOGARITHMIC FUNCTIONS
22. = ln
0 =
¤
√
√
1 + 2
= ln 1 + 2 − ln 1 − 2 = 12 ln(1 + 2) − 12 ln(1 − 2) ⇒
1 − 2
1
1
1
1
1
1
·
·2− ·
· (−2) =
+
2 1 + 2
2 1 − 2
1 + 2
1 − 2
23. = ln 3 − 25
⇒ 0 =
1
−104
· (−104 ) =
5
3 − 2
3 − 25
24. = ln(− + − ) = ln(− (1 + )) = ln(− ) + ln(1 + ) = − + ln(1 + )
0 = −1 +
⇒
−1 − + 1
1
=
=−
1+
1+
1+
2
2
2
2
25. () = +ln = · ln = · = ⇒
0 () = ·
2
2
2
2
2
2
+ ·
() = · ·
(2 ) + · 1 = · · 2 +
2
2
2
= 22 + = (22 + 1)
26. = log2 ( log5 )
⇒
1
1
( log5 ) =
=
( log5 )(ln 2)
( log5 )(ln 2)
0
Note that log5 (ln 5) =
28. () = 3cos 2
1
1
1
+ log5 =
+
.
ln 5
( log5 )(ln 5)(ln 2) (ln 2)
1
1 + ln
ln
1
(ln 5) = ln by the change of base formula. Thus, 0 =
+
=
.
ln 5
ln ln 2
ln 2
ln ln 2
27. Using Formula 7 and the Chain Rule, () = 4
0 () = 4 (ln 4)
·
= −1
⇒ 0 () = 3cos 2 ln 3
⇒
4
= 4 (ln 4) −−2 = − (ln 4) 2 .
(cos 2) = −2(sin 2) 3cos 2 ln 3
√
√
2
1
1
2
2
√
√
√
ln + + 1 =
+ +1 =
·
· 1+
29.
+ 2 + 1
+ 2 + 1
2 2 + 1
√
√ 2
1
+ 2 + 1
1
1
+1
√
√
√
· √
+√
=
·
= √
=
+ 2 + 1
2 + 1
2 + 1
+ 2 + 1
2 + 1
2 + 1
30.
ln
√
√
1
1
1 − cos
=
ln 1 − cos − ln 1 + cos =
ln(1 − cos ) − ln(1 + cos )
1 + cos
2
2
1
1
1
1
·
· sin − ·
· (− sin )
2 1 − cos
2 1 + cos
sin
1
sin
1 sin (1 + cos ) + sin (1 − cos )
+
=
=
2 1 − cos
1 + cos
2
(1 − cos )(1 + cos )
1
1 2 sin
1 sin + sin cos + sin − sin cos
=
=
= csc
=
2
2
1 − cos2
2 sin
sin
=
√
√
2 + ln
1
1
√
⇒
ln ⇒ 0 = · + (ln ) √ =
2
2
√
√
√
√
ln
2 − (2 + ln )
2 (1) − (2 + ln )(1 )
2 − (2 + ln )(1 )
√
√ 2
00 =
=− √
=
=
4
(2 )
(4)
4
31. =
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°
591
592
¤
CHAPTER 6
32. =
ln
1 + ln
INVERSE FUNCTIONS
⇒ 0 =
(1 + ln )(1) − (ln )(1)
1
=
(1 + ln )2
(1 + ln )2
[(1 + ln )2 ]
=−
[(1 + ln )2 ]2
[Reciprocal Rule]
00
=−
⇒
· 2(1 + ln ) · (1) + (1 + ln )2
2 (1 + ln )4
(1 + ln )[2 + (1 + ln )]
3 + ln
=− 2
2 (1 + ln )4
(1 + ln )3
=−
33. = ln |sec |
⇒ 0 =
34. = ln(1 + ln )
⇒ 0 =
[(1 + ln )]
=−
[(1 + ln )]2
00
1 − ln( − 1)
35. () =
0
() =
=
1
1
sec =
sec tan = tan ⇒ 00 = sec2
sec
sec
1
1
1
· =
1 + ln
(1 + ln )
[Reciprocal Rule] = −
⇒
(1) + (1 + ln )(1)
1 + 1 + ln
2 + ln
=− 2
=− 2
2 (1 + ln )2
(1 + ln )2
(1 + ln )2
⇒
−1
( − 1)[1 − ln( − 1)] +
− 1 − ( − 1) ln( − 1) +
−1 =
−1
=
2
2
[1
−
ln(
−
1)]
( − 1)[1 − ln( − 1)]2
[1 − ln( − 1)]
[1 − ln( − 1)] · 1 − ·
2 − 1 − ( − 1) ln( − 1)
( − 1)[1 − ln( − 1)]2
Dom() = { | − 1 0 and 1 − ln( − 1) 6= 0} = { | 1 and ln( − 1) 6= 1}
= | 1 and − 1 6= 1 = { | 1 and 6= 1 + } = (1 1 + ) ∪ (1 + ∞)
36. () =
√
2 + ln = (2 + ln )12
⇒ 0 () =
1
1
1
(2 + ln )−12 · = √
2
2 2 + ln
Dom() = { | 2 + ln ≥ 0} = { | ln ≥ −2} = { | ≥ −2 } = [−2 ∞).
37. () = ln(2 − 2)
⇒ 0 () =
2( − 1)
1
(2 − 2) =
.
2 − 2
( − 2)
Dom( ) = { | ( − 2) 0} = (−∞ 0) ∪ (2 ∞).
38. () = ln ln ln
⇒ 0 () =
1
1
1
·
· .
ln ln ln
Dom( ) = { | ln ln 0} = { | ln 1} = { | } = ( ∞).
1
1
1
( + ln ) =
1+
.
+ ln
+ ln
1
1
1
1+
=
(1 + 1) = 1 · 2 = 2.
Substitute 1 for to get 0 (1) =
1 + ln 1
1
1+0
39. () = ln( + ln )
40. () = cos(ln 2 )
⇒ 0 () =
⇒ 0 () = − sin(ln 2 )
Substitute 1 for to get 0 (1) = −
1
2 sin(ln 2 )
ln 2 = − sin(ln 2 ) 2 (2) = −
.
2 sin(ln 12 )
= −2 sin 0 = 0.
1
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°
SECTION 6.4 DERIVATIVES OF LOGARITHMIC FUNCTIONS
41. = ln(2 − 3 + 1)
0 =
⇒
1
· (2 − 3)
2 − 3 + 1
⇒
¤
593
0 (3) = 11 · 3 = 3, so an equation of a tangent line at
(3 0) is − 0 = 3( − 3), or = 3 − 9.
42. = 2 ln
⇒ 0 = 2 ·
1
+ (ln )(2) ⇒ 0 (1) = 1 + 0 = 1 , so an equation of a tangent line at (1 0) is
− 0 = 1( − 1), or = − 1.
43. () = sin + ln
⇒ 0 () = cos + 1.
This is reasonable, because the graph shows that increases when 0 is
positive, and 0 () = 0 when has a horizontal tangent.
44. =
ln
0 (1) =
⇒ 0 =
(1) − ln
1 − ln
=
.
2
2
1−0
1−1
= 1 and 0 () =
= 0 ⇒ equations of tangent
12
2
lines are − 0 = 1( − 1) or = − 1 and − 1 = 0( − )
or = 1.
45. () = + ln(cos )
0 ( 4 ) = 6
⇒
⇒
0 () = +
− tan 4 = 6
46. () = log (32 − 2)
⇒
⇒
0 () =
1
· (− sin ) = − tan .
cos
−1 =6
⇒
= 7.
1
· 6. 0 (1) = 3
(32 − 2) ln
⇒
1
·6 = 3
ln
⇒
2 = ln
⇒
= 2 .
47. = (2 + 2)2 (4 + 4)4
⇒ ln = ln[(2 + 2)2 (4 + 4)4 ] ⇒ ln = 2 ln(2 + 2) + 4 ln(4 + 4) ⇒
1
163
1
1 0
4
=2· 2
· 2 + 4 · 4
· 43 ⇒ 0 = 2
+ 4
⇒
+2
+4
+2
+4
163
4
0 = (2 + 2)2 (4 + 4)4
+
2 + 2
4 + 4
48. =
− cos2
2 + + 1
⇒ ln = ln
− cos2
2 + + 1
⇒
ln = ln − + ln | cos |2 − ln(2 + + 1) = − + 2 ln | cos | − ln(2 + + 1) ⇒
1
1 0
1
2 + 1
= −1 + 2 ·
(− sin ) − 2
(2 + 1) ⇒ 0 = −1 − 2 tan − 2
⇒
cos
++1
++1
2 + 1
− cos2
0 = − 2
1 + 2 tan + 2
++1
++1
49. =
−1
4 + 1
12
−1
⇒ ln = ln 4
+1
1 1
1 0
1 1
=
−
· 43
2−1
2 4 + 1
1
1
ln( − 1) − ln(4 + 1) ⇒
2
2
23
1
23
1
−1
− 4
−
⇒ 0 =
⇒ 0 =
2( − 1) + 1
4 + 1 2 − 2
4 + 1
⇒ ln =
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
594
¤
50. =
CHAPTER 6
INVERSE FUNCTIONS
√ 2 −
( + 1)23
2
⇒
⇒ ln = ln 12 − ( + 1)23
2
1
1 0
1 1
= · + 2 − 1 + ·
⇒
2
3 +1
√
2
2
2
1
1
0 =
+ 2 − 1 +
+ 2 − 1 +
⇒ 0 = − ( + 1)23
2
3 + 3
2
3 + 3
ln = 12 ln + (2 − ) + 23 ln( + 1) ⇒
51. =
⇒ ln = ln
⇒ ln = ln ⇒ 0 = (1) + (ln ) · 1 ⇒ 0 = (1 + ln ) ⇒
0 = (1 + ln )
⇒ ln =
53. = sin
⇒ ln = ln sin
sin
=
+ ln cos
0
0
1
1
= − 2 ln +
1
ln ⇒
52. = 1
1
0
= (sin ) · + (ln )(cos ) ⇒
0
sin sin
=
+ ln cos
⇒
⇒
1 − ln
1
⇒ 0 = 1
2
ln = sin ln ⇒
√
√
⇒ ln = ln
⇒ ln = ln 12
55. = (cos )
⇒ ln = ln(cos )
⇒ ln = ln cos ⇒
54. =
⇒ ln = 12 ln ⇒
√
0 = 12 + 12 ln ⇒ 0 = 12
(1 + ln )
1
1
1 0
1
= · + ln ·
2
2
⇒
1 0
1
=·
· (− sin ) + ln cos · 1 ⇒
cos
sin
0 = ln cos −
⇒ 0 = (cos ) (ln cos − tan )
cos
1
1 0
1
= ln ·
· cos + ln sin ·
sin
ln sin
cos
ln sin
0 = ln ·
+
⇒ 0 = (sin )ln ln cot +
sin
56. = (sin )ln
57. = ln
⇒ ln = ln(sin )ln
⇒ ln = ln · ln sin ⇒
⇒ ln = ln ln = (ln )2
58. = (ln )cos
0 = (ln )cos
⇒
⇒ ln = cos ln(ln ) ⇒
cos
59. = ln(2 + 2 )
ln
0
= 2 ln
2 ln
1
⇒ 0 = ln
1
1
0
= cos ·
· + (ln ln )(− sin ) ⇒
ln
− sin ln ln
⇒ 0 =
1
2 + 2
2 + 2 0
(2 + 2 ) ⇒ 0 = 2
+ 2
2 0 + 2 0 − 2 0 = 2 ⇒ (2 + 2 − 2) 0 = 2 ⇒ 0 =
60. =
0 =
⇒ ln = ln
⇒ ·
⇒ 2 0 + 2 0 = 2 + 2 0
⇒
2
2 + 2 − 2
1
1
+ (ln ) · 0 = · · 0 + ln
⇒ 0 ln −
0
= ln −
ln −
ln −
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
⇒
⇒
¤
SECTION 6.4 DERIVATIVES OF LOGARITHMIC FUNCTIONS
61. () = ln( − 1)
⇒ 0 () =
(4) () = −2 · 3( − 1)−4
1
= ( − 1)−1
( − 1)
⇒ ···
⇒ 00 () = −( − 1)−2
⇒ 000 () = 2( − 1)−3
⇒ () () = (−1)−1 · 2 · 3 · 4 · · · · · ( − 1)( − 1)− = (−1)−1
595
⇒
( − 1)!
( − 1)
62. = 8 ln , so 9 = 8 0 = 8(87 ln + 7 ). But the eighth derivative of 7 is 0, so we now have
8(87 ln ) = 7(8 · 76 ln + 86 ) = 7(8 · 76 ln ) = 6(8 · 7 · 65 ln ) = · · · = (8! 0 ln ) = 8!
ln
63. () = √
⇒ 0 () =
√
√
2 − ln
(1) − (ln )[1(2 )]
=
232
⇒
232 (−1) − (2 − ln )(312 )
3 ln − 8
=
0 ⇔ ln 83
43
452
and CD on (0 83 ). The inflection point is 83 83 −43 .
00 () =
64. () = ln , 0 () = ln + 1 = 0 when ln = −1
⇔ 83 , so is CU on (83 ∞)
⇔ = −1 . 0 () 0 ⇔ ln + 1 0 ⇔
ln −1 ⇔ 1. 0 () 0 ⇔ ln + 1 0 ⇔ 1. Therefore, there is an absolute minimum value
of (1) = (1) ln(1) = −1.
65. = () = ln(sin )
A. = { in | sin 0} =
∞
=−∞
(2 (2 + 1) ) = · · · ∪ (−4 −3) ∪ (−2 −) ∪ (0 ) ∪ (2 3) ∪ · · ·
B. No ­intercept; ­intercepts: () = 0 ⇔ ln(sin ) = 0 ⇔ sin = 0 = 1 ⇔
integer .
C. is periodic with period 2. D.
lim
→(2)+
() = −∞ and
lim
→[(2+1)]−
= 2 + 2 for each
() = −∞, so the lines
cos
= cot , so 0 () 0 when 2 2 + 2 for each
sin
integer , and 0 () 0 when 2 + 2 (2 + 1). Thus, is increasing on 2 2 + 2 and
= are VAs for all integers . E. 0 () =
decreasing on 2 + 2 (2 + 1) for each integer .
F. Local maximum values 2 + 2 = 0, no local minimum.
H.
G. 00 () = − csc2 0, so is CD on (2 (2 + 1)) for
each integer No IP
66. = ln(tan2 )
A. = { | 6= 2} B. ­intercepts + 4 , no ­intercept. C. (−) = (), so the curve is
symmetric about the ­axis. Also ( + ) = (), so is periodic with period , and we consider parts D–G only for
− 2 2 D. lim ln(tan2 ) = −∞ and
→0
= ± 2 are VA. E. 0 () =
lim
→(2)−
ln(tan2 ) = ∞,
lim
→−(−2)+
ln(tan2 ) = ∞, so = 0,
sec2
2 tan sec2
=2
0 ⇔
2
tan
tan
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°
596
¤
CHAPTER 6
INVERSE FUNCTIONS
tan 0 ⇔ 0 2 , so is increasing on 0 2 and
decreasing on − 2 0 F. No maximum or minimum
4
2
=
sin cos
sin 2
H.
−8 cos 2
0
sin2 2
⇔ cos 2 0 ⇔ − 4 4 , so is CD on − 4 0 and
0 4 and CU on − 2 − 4 and 4 2 . IP are ± 4 0 .
G. 0 () =
67. = () = ln(1 + 2 )
­axis. D.
⇒ 00 () =
A. = B. Both intercepts are 0 C. (−) = (), so the curve is symmetric about the
lim ln(1 + 2 ) = ∞, no asymptotes. E. 0 () =
→±∞
2
0 ⇔
1 + 2
H.
0, so is increasing on (0 ∞) and decreasing on (−∞ 0)
F. (0) = 0 is a local and absolute minimum.
G. 00 () =
2(1 + 2 ) − 2(2)
2(1 − 2 )
=
0 ⇔
2
2
(1 + )
(1 + 2 )2
|| 1, so is CU on (−1 1), CD on (−∞ −1) and (1 ∞).
IP (1 ln 2) and (−1 ln 2).
68. = () = ln(1 + 3 )
A. 1 + 3 0 ⇔ 3 −1 ⇔ −1, so = (−1 ∞). B. ­intercept:
(0) = ln 1 = 0; ­intercept: () = 0 ⇔ ln(1 + 3 ) = 0 ⇔ 1 + 3 = 0
symmetry D.
lim () = −∞, so = −1 is a VA E. 0 () =
→−1+
⇔ 3 = 0 ⇔ = 0 C. No
32
. 0 () 0 on (−1 0) and (0 ∞)
1 + 3
[ 0 () = 0 at = 0], so by Exercise 3.3.79, is increasing on (−1 ∞). F. No extreme values
G. 00 () =
(1 + 3 )(6) − 32 (32 )
(1 + 3 )2
H.
3[2(1 + 3 ) − 33 ]
3(2 − 3 )
=
3
2
(1 + )
(1 + 3 )2
√
√
00 () 0 ⇔ 0 3 2, so is CU on 0 3 2 and is CD on (−1 0)
√
√
and 3 2 ∞ . IP at (0 0) and 3 2 ln 3
=
69. We use the CAS to calculate 0 () =
00 () =
2 + sin + cos
and
2 + sin
22 sin + 4 sin − cos2 + 2 + 5
. From the graphs, it
2 (cos2 − 4 sin − 5)
seems that 0 0 (and so is increasing) on approximately the intervals
(0 27), (45 82) and (109 143). It seems that 00 changes sign
(indicating inflection points) at ≈ 38, 57, 100 and 120.
Looking back at the graph of () = ln(2 + sin ), this implies that the inflection points have approximate coordinates
(38 17), (57 21), (100 27), and (120 29).
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 6.4 DERIVATIVES OF LOGARITHMIC FUNCTIONS
70. We see that if ≤ 0, () = ln(2 + ) is only defined for 2 −
⇒ ||
¤
597
√
−, and
() =
lim
() = −∞, since ln → −∞ as → 0. Thus, for 0, there are vertical asymptotes at
lim
√
√
→ −+
→− −−
√
= ± , and as decreases (that is, || increases), the asymptotes get further apart. For = 0,
lim () = −∞, so there is a vertical asymptote at = 0. If 0, there is no asymptote. To find the maxima, minima, and
→0
inflection points, we differentiate: () = ln(2 + ) ⇒ 0 () =
1
(2), so by the First Derivative Test there is a
2 +
local and absolute minimum at = 0. Differentiating again, we get
00 () =
2( − 2 )
−2
1
2
(2)
+
2
−
+
(2)
= 2
. Now if
2 +
( + )2
≤ 0, this is always negative, so is concave down on both of the intervals
on which it is defined. If 0, then 00 changes sign when = 2 ⇔
√
√
= ± . So for 0 there are inflection points at ± , and as
increases, the inflection points get further apart.
From the graph, it appears that the curves = ( − 4)2 and = ln intersect
71.
just to the left of = 3 and to the right of = 5, at about = 53. Let
() = ln − ( − 4)2 . Then 0 () = 1 − 2( − 4), so Newton’s method
says that +1 = − ( ) 0 ( ) = −
ln − ( − 4)2
. Taking
1 − 2( − 4)
0 = 3, we get 1 ≈ 2957738, 2 ≈ 2958516 ≈ 3 , so the first solution is 2958516, to six decimal places. Taking 0 = 5,
we get 1 ≈ 5290755, 2 ≈ 5290718 ≈ 3 , so the second (and final) solution is 5290718, to six decimal places.
We use Newton’s method with () = ln(4 − 2 ) − and
72.
0 () =
1
2
(−2) − 1 = −1 −
. The formula is
4 − 2
4 − 2
+1 = − ( ) 0 ( ). From the graphs it seems that the solutions
occur at approximately = −19 and = 11. However, if we use 1 = −19
as an initial approximation to the first solution, we get 2 ≈ −2009611, and
() = ln( − 2)2 − is undefined at this point, making it impossible to calculate 3 . We must use a more accurate first
estimate, such as 1 = −195. With this approximation, we get 1 = −195, 2 ≈ −11967495, 3 ≈ −1964760,
4 ≈ 5 ≈ −1964636. Calculating the second solution gives 1 = 11, 2 ≈ 1058649, 3 ≈ 1058007,
4 ≈ 5 ≈ 1058006. So, correct to six decimal places, the two solutions of the equation ln(4 − 2 ) = are = −1964636
and = 1058006.
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
598
¤
CHAPTER 6
INVERSE FUNCTIONS
⇒ 0 () = 1 ⇒ 00 () = −12 . The linear approximation to ln near 1 is
73. (a) Let () = ln
ln ≈ (1) + 0 (1)( − 1) = ln 1 + 11 ( − 1) = − 1.
(c)
(b)
From the graph, it appears that the linear approximation is
accurate to within 01 for between about 062 and 151.
74. (a) = with = 4502714 × 10−20 and = 1029953851,
where is measured in thousands of people. The fit appears to be very good.
(b) For 1800: 1 =
5308 − 3929
7240 − 5308
= 1379, 2 =
= 1932.
1800 − 1790
1810 − 1800
So 0 (1800) ≈ (1 + 2 )2 = 16555 thousand peopleyear.
For 1850: 1 =
23,192 − 17,063
31,443 − 23,192
= 6129, 2 =
= 8251.
1850 − 1840
1860 − 1850
So 0 (1850) ≈ (1 + 2 )2 = 719 thousand peopleyear.
(c) Using 0 () = ln (from Formula 7) with the values of and from part (a), we get 0 (1800) ≈ 15685 thousand
peopleyear and 0 (1850) ≈ 68607. These estimates are somewhat less than the ones in part (b).
(d) (1870) ≈ 41,94656. The difference of 34 million people is most likely due to the Civil War (1861–1865).
75.
4
2
3
= 3
4
2
4
4
1
= 3 ln || = 3(ln 4 − ln 2) = 3 ln = 3 ln 2
2
2
76. Let = 5 + 1, so = 5 . When = 0, = 1; when = 3, = 16. Thus,
16
1
1
1
1
=
ln ||
= (ln 16 − ln 1) = ln 16.
5
5
5
5
1
=
5 + 1
2
2
1
1
1
1 5
1
= − ln |8 − 3| = − ln 2 − − ln 5 = (ln 5 − ln 2) = ln
8 − 3
3
3
3
3
3 2
1
0
77.
16
3
1
1
1
Or: Let = 8 − 3. Then = −3 , so
2
2 1
2
− 3
1
1
1
1 5
1
=
= − ln || = − ln 2 − − ln 5 = (ln 5 − ln 2) = ln .
8
−
3
3
3
3
3
3
2
1
5
5
78.
2
9
9
9
√
1
1
81
+2+
= 12 2 + 2 + ln 4 =
+ 18 + ln 9 − (8 + 8 + ln 4)
+ √
=
2
4
4
= 85
+ ln 94
2
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 6.4 DERIVATIVES OF LOGARITHMIC FUNCTIONS
79.
3
1
32 + 4 + 1
= 20 + ln 3
80. Let = ln . Then =
1
, so
81. Let = ln . Then =
cos(ln )
=
⇒
(ln )2
=
cos = sin + = sin(ln ) + .
2 =
1 3
1
+ = (ln )3 + .
3
3
82. Let = 3 + 1. Then = 3 2 and 13 = 2 , so
83.
2
=
3 + 1
1
1
1
1
= ln || + = ln | 3 + 1| + .
3
3
3
sin cos
sin 2
=
2
= 2. Let = cos . Then = − sin , so
1 + cos2
1 + cos2
2 = −2
= −2 · 12 ln(1 + 2 ) + = − ln(1 + 2 ) + = − ln(1 + cos2 ) + .
1 + 2
Or: Let = 1 + cos2 .
84. Let = + 1. Then = , so
85.
4
0
2 =
+ 1
=
= ln || + = ln( + 1) + .
4
1
15
1
16
2
−
=
=
ln 2
ln
2
ln
2
ln
2
0
86. Let = 2 . Then = 2 , so
2
2 = 12
2 =
1 2
1 2
+ =
2 + .
2 ln 2
2 ln 2
1
(ln |sin | + ) =
cos = cot
sin
cos
(b) Let = sin . Then = cos , so cot =
=
= ln || + = ln |sin | + .
sin
87. (a)
88.
(ln )2
ln
=
⇔ ln = (ln )2
⇔ 0 = (ln )2 − ln ⇔
0 = ln (ln − 1) ⇔ ln = 0 or 1 ⇔ = 0 or 1 [1 or ]
(ln )2
ln
−
= 12 (ln )2 − 13 (ln )3
=
1
1
1
1
1
= 2 − 3 − (0 − 0) = 6
√
89. The cross­sectional area is 1 + 1
1
0
¤
3
3
1
3 2
3
27
3 + 4 +
=
=
+ 4 + ln || =
+ 12 + ln 3 −
+ 4 + ln 1
2
2
2
1
1
2
= ( + 1). Therefore, the volume is
= [ln( + 1)]10 = (ln 2 − ln 1) = ln 2.
+1
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°
599
600
¤
CHAPTER 6
INVERSE FUNCTIONS
90. Using cylindrical shells, we get =
3
0
91. =
2
=
1
1000
600
=
3
2
= ln(1 + 2 ) 0 = ln 10.
2
+1
1000
600
1000
1
= ln | |
= (ln 1000 − ln 600) = ln 1000
= ln 53 .
600
600
Initially, = , where = 150 kPa and = 600 cm3 , so = (150)(600) = 90,000 kPa · cm3 . Thus,
= 90,000 ln 53 kPa · cm3 = 90,000 ln 53
3
N
1
1000 2
m
m
100
= 90 ln 53 N·m ≈ 45.974 J
92. 00 () = −2 , 0
⇒ 0 () = −1 +
⇒ () = − ln + + . 0 = (1) = + and
0 = (2) = − ln 2 + 2 + = − ln 2 + 2 − = − ln 2 +
⇒ = ln 2 and = − ln 2. So
() = − ln + (ln 2) − ln 2.
⇒ 0 () = 2 + 1. If = −1 , then (1) = 2 ⇒ (2) = 1, so
93. () = 2 + ln
0 (2) = 1 0 ((2)) = 1 0 (1) = 13 .
⇒ 0 () = + 1. = −1 and (1) = ⇒ () = 1, so 0 () = 1 0 (1) = 1( + 1).
94. () = + ln
95. The curve and the line will determine a region when they intersect at two or
more points. So we solve the equation (2 + 1) = ⇒ = 0 or
± −4()( − 1)
1
2 + − 1 = 0 ⇒ = 0 or =
=±
− 1.
2
Note that if = 1, this has only the solution = 0, and no region is
determined. But if 1 − 1 0 ⇔ 1 1 ⇔ 0 1, then there are two solutions. [Another way of seeing this
is to observe that the slope of the tangent to = (2 + 1) at the origin is 0 = 1 and therefore we must have 0 1.]
Note that we cannot just integrate between the positive and negative roots, since the curve and the line cross at the origin.
Since and (2 + 1) are both odd functions, the total area is twice the area between the curves on the interval
0 1 − 1 . So the total area enclosed is
2
√1−1
0
1
√1−1
2
2
1
−
=
2
ln(
+
1)
−
2
2
0
2 + 1
1
1
−1+1 −
−1
− (ln 1 − 0)
= ln
1
= ln
+ − 1 = − ln − 1
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 6.4 DERIVATIVES OF LOGARITHMIC FUNCTIONS
¤
601
96. Let = . Then = , and as → ∞, → ∞.
1
1
1+
= lim 1 +
= lim 1 +
= by Equation 9.
→∞
→∞
→∞
Therefore, lim
97. If () = ln (1 + ), then 0 () =
Thus, lim
→0
1
, so 0 (0) = 1.
1+
ln(1 + )
()
() − (0)
= lim
= lim
= 0 (0) = 1.
→0
→0
−0
98. (a) = log2
2
⇒
[ constant] =
1
1
2
=
·
ln 2
2
ln 2
As increases, the rate of change of difficulty decreases.
(b) = log2
2
⇒
[ constant] =
1
−2
1
·
=−
· −2 −2 =
2 ln 2 2
ln 2
2
ln 2
The negative sign indicates that difficulty decreases with increasing width. While the magnitude of the rate of change
1
1
=
decreases
as
increases
, the rate of change itself
decreases with increasing width that is, −
ln 2 ln 2
increases (gets closer to zero from the negative side) with increasing values of .
(c) The answers to (a) and (b) agree with intuition. For fixed width, the difficulty of acquiring a target increases, but less and
less so, as the distance to the target increases. Similarly, for a fixed distance to a target, the difficulty of acquiring the target
decreases, but less and less so, as the width of the target increases.
99. =
⇒ 0 = ln , so the slope of the tangent line to the curve = at the point ( ) is ln . An equation of
this tangent line is then − = ln ( − ). If is the ­intercept of this tangent line, then 0 − = ln ( − ) ⇒
−1
−1
= 1 . The distance between ( 0) and ( 0) is | − |, and
−1 = ln ( − ) ⇒
= − ⇒ | − | =
ln
ln |ln |
this distance is the constant
1
for any . [Note: The absolute value is needed for the case 0 1 because ln is
|ln |
negative there. If 1, we can write − = 1(ln ) as the constant distance between ( 0) and ( 0).]
100. =
⇒ 0 = ln , so the slope of the tangent line to the curve = at the point (0 0 ) is 0 ln . An equation
of this tangent line is then − 0 = 0 ln ( − 0 ). Since this tangent line must pass through (0 0), we have
0 − 0 = 0 ln (0 − 0 ), or 0 = 0 (ln ) 0 . Since (0 0 ) is a point on the exponential curve = , we also have
0 = 0 . Equating the expressions for 0 gives 0 = 0 (ln ) 0
⇒ 1 = (ln ) 0
⇒ 0 = 1(ln ).
So 0 = 0 = 0 ln [by Formula 6.3.7] = (1(ln )) ln = 1 = .
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°
602
¤
CHAPTER 6
INVERSE FUNCTIONS
6.2* The Natural Logarithmic Function
1. (a) ln
√
= ln()12 = 12 ln() = 12 (ln + ln ) = 12 ln + 12 ln [assuming that the variables are positive]
4
(b) ln √
= ln 4 − ln(2 − 4)12
2 − 4
[Law 2]
= 4 ln − 12 ln[( + 2)( − 2)]
[Law 3]
= 4 ln − 12 [ln( + 2) + ln( − 2)]
[Law 1]
= 4 ln − 12 ln( + 2) − 12 ln( − 2)
2. (a) ln
(b) ln
2
= ln 2 − ln( 3 4 ) = 2 ln − (ln 3 + ln 4 ) = 2 ln − 3 ln − 4 ln
3 4
12
3
3
1
3
= ln
= ln
−3
−3
2
−3
[Law 3]
= 12 [ln 3 + ln − ln( − 3)]
[Laws 1 and 2]
= 12 ln 3 + 12 ln − 12 ln( − 3)
3. (a) ln − 2 ln + 3 ln = ln − ln 2 + ln 3 = ln
3
+ ln 3 = ln 2
2
√
4
(b) ln 4 + ln − 13 ln( + 1) = ln(4 · ) − ln( + 1)13 = ln(4) − ln 3 + 1 = ln √
3
+1
4. (a) ln 10 + 2 ln 5 = ln 10 + ln 52
[by Law 3]
= ln [(10)(25)]
[by Law 1]
= ln 250
(b) 3 ln( − 2) − ln(2 − 5 + 6) + 2 ln( − 3) = ln( − 2)3 − ln [( − 2)( − 3)] + ln( − 3)2
( − 2)3 ( − 3)2
= ln[( − 2)2 ( − 3)]
= ln
( − 2)( − 3)
5. (a) ln 3 + 13 ln 8 = ln 3 + ln 813 = ln 3 + ln 2 = ln(3 · 2) = ln 6
(b) 13 ln( + 2)3 + 12 ln − ln(2 + 3 + 2)2 = ln[( + 2)3 ]13 + 12 ln
= ln( + 2) + ln
(2 + 3 + 2)2
√
2 + 3 + 2
√
( + 2)
= ln
( + 1)( + 2)
√
= ln
+1
[by Laws 3, 2]
[by Law 3]
[by Law 1]
Note that since ln is defined for 0, we have + 1, + 2, and 2 + 3 + 2 all positive, and hence their logarithms
are defined.
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 6.2* THE NATURAL LOGARITHMIC FUNCTION
6. Reflect the graph of = ln about the ­axis to obtain
¤
7. Reflect the graph of = ln about the ­axis to obtain
the graph of = ln (−).
the graph of = − ln .
= ln
= ln (−)
8. Reflect the portion of the graph of = ln to the right
= ln
= − ln
= ln
= ln( + 3)
9.
of the ­axis about the ­axis. The graph of = ln ||
is that reflection in addition to the original portion.
= ln
= ln ||
√
10. lim ln( − 1) = −∞ since
→1+
√
− 1 → 0+ as → 1+ .
11. lim ln(cos ) = ln 1 = 0. [ln(cos ) is continuous at = 0 since it is the composite of two continuous functions.]
→0
12. lim ln(sin ) = −∞ since sin → 0+ as → 0+ .
→0+
13. lim [ln(1 + 2 ) − ln(1 + )] = lim ln
→∞
→∞
parentheses is ∞.
14. lim [ln(2 + ) − ln(1 + )] = lim ln
→∞
→∞
15. () = 3 ln
⇒ 0 () = 3 ·
16. () = ln −
⇒
17. () = ln(2 + 3 + 5)
1 + 2
= ln
1+
2+
1+
1 + 2
→∞ 1 +
lim
= lim ln
→∞
2 + 1
1 + 1
= ln
1
+
= ∞, since the limit in
→∞ 1 + 1
lim
= ln
1
= ln 1 = 0
1
1
+ (ln ) · 32 = 2 + 32 ln = 2 (1 + 3 ln )
0 () = ·
1
+ (ln ) · 1 − 1 = 1 + ln − 1 = ln
⇒ 0 () =
2 + 3
1
1
·
(2 + 3 + 5) = 2
· (2 + 3) = 2
2 + 3 + 5
+ 3 + 5
+ 3 + 5
1
1
2
(3 + 2 ) =
·
· 2 =
3 + 2
3 + 2
3 + 2
18. () = ln(3 + 2 )
⇒ 0 () =
19. () = sin(ln )
⇒ 0 () = cos(ln ) ·
20. () = ln(cos )
⇒ 0 () =
603
1
cos(ln )
ln = cos(ln ) · =
1
1
·
cos =
(− sin ) = −tan
cos
cos
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
604
¤
CHAPTER 6
21. () = ln
1
INVERSE FUNCTIONS
Another solution: () = ln
22. =
1
= (ln )−1
ln
23. () = sin ln(5)
24. () =
1
1
⇒ 0 () =
1
1
1
= − 2 =− .
1
1
= ln 1 − ln = − ln ⇒ 0 () = − .
⇒ 0 = −1(ln )−2 ·
⇒ 0 () = sin ·
−1
1
=
(ln )2
sin · 5
sin
1
·
(5) + ln(5) · cos =
+ cos ln(5) =
+ cos ln(5)
5
5
√
1
1
1
· = √
(1 + ln ) = √
1 + ln ⇒ 0 () = 12 (1 + ln )−12
2 1 + ln
2 1 + ln
2 sin
1
25. () = (ln ) sin ⇒ () = (ln ) cos + sin · 2 ln · = ln ln cos +
26. () = ln
2
0
2
√
√ 2
2
1
1
·
· √
= 2
2 + 1 ⇒ 0 () = √
+1 = √
2
+1
+ 1
2 + 1 2 2 + 1
Or: () = ln
√
1
1
· 2 = 2
2 + 1 = ln(2 + 1)12 = 12 ln(2 + 1) ⇒ 0 () = · 2
2 +1
+1
⇒ 0 = 2(ln tan ) ·
27. = (ln tan )2
1
2(ln tan ) sec2
2 ln tan
· sec2 =
or 0 =
tan
tan
sin cos
28. = ln(tan2 ) = ln(tan )2 = 2 ln tan
29. () =
0 () =
ln
1 + ln(2)
2
1
cos 1
sec2 = 2
=
[or 2 csc sec ]
tan
sin cos2
sin cos
⇒
1
1
[1 + ln(2)] · 1 − ln · 2
·2
[1 + ln(2) − ln ]
1 + (ln 2 + ln ) − ln
1 + ln 2
=
=
=
[1 + ln(2)]2
[1 + ln(2)]2
[1 + ln(2)]2
[1 + ln(2)]2
30. () = ln
0 () =
⇒ 0 = 2
−
= ln( − ) − ln( + ) ⇒
+
−( + ) − ( − )
1
−2
1
(−1) −
=
= 2
−
+
( − )( + )
− 2
√
(2 + 1)4
= ln + ln(2 + 1)4 − ln 3 2 − 1 = ln + 4 ln(2 + 1) − 13 ln(2 − 1) ⇒
2 − 1
31. () = ln √
3
0 () =
32. = ln
0 =
1
1
1
1
8
2
1
+4· 2
· 2 − ·
·2= + 2
−
+1
3 2 − 1
+1
3(2 − 1)
√
√
1 + 2
= ln 1 + 2 − ln 1 − 2 = 12 ln(1 + 2) − 12 ln(1 − 2) ⇒
1 − 2
1
1
1
1
1
1
·
·2− ·
· (−2) =
+
2 1 + 2
2 1 − 2
1 + 2
1 − 2
33. = ln 3 − 25
⇒ 0 =
1
−104
· (−104 ) =
5
3 − 2
3 − 25
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 6.2* THE NATURAL LOGARITHMIC FUNCTION
34. = ln(csc − cot )
0 =
35.
36.
¤
605
⇒
1
1
csc (csc − cot )
(csc − cot ) =
(− csc cot + csc2 ) =
= csc
csc − cot
csc − cot
csc − cot
√
√
1
1
2
√
√
ln + 2 + 1 =
+ 2 + 1 =
·
· 1+ √
+ 2 + 1
+ 2 + 1
2 2 + 1
√
√ 2
1
1
1
+1
+ 2 + 1
√
√
√
√
√
·
+
=
·
= √
=
+ 2 + 1
2 + 1
2 + 1
+ 2 + 1
2 + 1
2 + 1
ln
√
√
1
1
1 − cos
=
ln 1 − cos − ln 1 + cos =
ln(1 − cos ) − ln(1 + cos )
1 + cos
2
2
1
1
1
1
·
· sin − ·
· (− sin )
2 1 − cos
2 1 + cos
sin
1
sin
1 sin (1 + cos ) + sin (1 − cos )
+
=
=
2 1 − cos
1 + cos
2
(1 − cos )(1 + cos )
1
1 2 sin
1 sin + sin cos + sin − sin cos
=
=
= csc
=
2
1 − cos2
2 sin2
sin
=
√
√
2 + ln
1
1
√
⇒
ln ⇒ 0 = · + (ln ) √ =
2
2
√
√
√
√
ln
2 − (2 + ln )
2 (1) − (2 + ln )(1 )
2 − (2 + ln )(1 )
√
√ 2
00 =
=− √
=
=
4
(2 )
(4)
4
37. =
38. =
ln
1 + ln
⇒ 0 =
(1 + ln )(1) − (ln )(1)
1
=
(1 + ln )2
(1 + ln )2
[(1 + ln )2 ]
=−
[(1 + ln )2 ]2
00
=−
⇒ 0 =
40. = ln(1 + ln )
[(1 + ln )]
=−
[(1 + ln )]2
() =
=
· 2(1 + ln ) · (1) + (1 + ln )2
2 (1 + ln )4
1
1
sec =
sec tan = tan ⇒ 00 = sec2
sec
sec
⇒ 0 =
00
0
=−
(1 + ln )[2 + (1 + ln )]
3 + ln
=− 2
2 (1 + ln )4
(1 + ln )3
39. = ln |sec |
41. () =
[Reciprocal Rule]
⇒
1 − ln( − 1)
1
1
1
· =
1 + ln
(1 + ln )
[Reciprocal Rule] = −
⇒
(1) + (1 + ln )(1)
1 + 1 + ln
2 + ln
=− 2
=− 2
2 (1 + ln )2
(1 + ln )2
(1 + ln )2
⇒
−1
( − 1)[1 − ln( − 1)] +
− 1 − ( − 1) ln( − 1) +
−1 =
−1
=
[1 − ln( − 1)]2
( − 1)[1 − ln( − 1)]2
[1 − ln( − 1)]2
[1 − ln( − 1)] · 1 − ·
2 − 1 − ( − 1) ln( − 1)
( − 1)[1 − ln( − 1)]2
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
[continued]
606
¤
CHAPTER 6
INVERSE FUNCTIONS
Dom() = { | − 1 0 and 1 − ln( − 1) 6= 0} = { | 1 and ln( − 1) 6= 1}
= | 1 and − 1 6= 1 = { | 1 and 6= 1 + } = (1 1 + ) ∪ (1 + ∞)
42. () =
√
2 + ln = (2 + ln )12
⇒ 0 () =
1
1
1
(2 + ln )−12 · = √
2
2 2 + ln
Dom() = { | 2 + ln ≥ 0} = { | ln ≥ −2} = { | ≥ −2 } = [−2 ∞).
43. () = ln(2 − 2)
⇒ 0 () =
2( − 1)
1
(2 − 2) =
.
2 − 2
( − 2)
Dom( ) = { | ( − 2) 0} = (−∞ 0) ∪ (2 ∞).
44. () = ln ln ln
⇒ 0 () =
1
1
1
·
· .
ln ln ln
Dom( ) = { | ln ln 0} = { | ln 1} = { | } = ( ∞).
1
1
1
( + ln ) =
1+
.
+ ln
+ ln
1
1
1
1+
=
(1 + 1) = 1 · 2 = 2.
Substitute 1 for to get 0 (1) =
1 + ln 1
1
1+0
45. () = ln( + ln )
46. () =
00 () =
ln
⇒ 0 () =
1
− (ln )(1)
1 − ln
0
⇒ () =
=
2
2
1
− (1 − ln )(2)
2 −
⇒ 00 () =
, so
(2 )2
− − 0
1
= − 3.
4
47. () = sin + ln
⇒ 0 () = cos + 1.
This is reasonable, because the graph shows that increases when 0 is
positive, and 0 () = 0 when has a horizontal tangent.
48. () = ln(2 + + 1)
⇒ 0 () =
1
(2 + 1). Notice from
2 + + 1
the graph that is increasing when 0 () is positive.
49. = sin(2 ln )
⇒ 0 = cos(2 ln ) ·
2
2
. At (1 0), 0 = cos 0 · = 2, so an equation of the tangent line is
1
− 0 = 2 · ( − 1), or = 2 − 2.
50. = ln(3 − 7)
⇒ 0 =
1
· 32
3 − 7
− 0 = 12( − 2) or = 12 − 24.
⇒ 0 (2) =
12
= 12, so an equation of a tangent line at (2 0) is
8−7
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°
¤
SECTION 6.2* THE NATURAL LOGARITHMIC FUNCTION
⇒ 0 =
51. = ln(2 + 2 )
1
2 + 2
2 + 2 0
(2 + 2 ) ⇒ 0 = 2
+ 2
2 0 + 2 0 − 2 0 = 2 ⇒ (2 + 2 − 2) 0 = 2 ⇒ 0 =
⇒
2
2 + 2 − 2
⇒ 1 + 0 = cos + 0 sin ⇒ 0 (1 − sin ) = cos − 1 ⇒
52. ln = ln + ln = sin
0 =
⇒ 2 0 + 2 0 = 2 + 2 0
607
cos − 1 cos − 1
=
1 − sin
1 − sin
53. () = ln( − 1)
⇒ 0 () =
(4) () = −2 · 3( − 1)−4
1
= ( − 1)−1
( − 1)
⇒ ···
⇒ 00 () = −( − 1)−2
⇒ 000 () = 2( − 1)−3
⇒ () () = (−1)−1 · 2 · 3 · 4 · · · · · ( − 1)( − 1)− = (−1)−1
⇒
( − 1)!
( − 1)
54. = 8 ln , so 9 = 8 0 = 8(87 ln + 7 ). But the eighth derivative of 7 is 0, so we now have
8(87 ln ) = 7(8 · 76 ln + 86 ) = 7(8 · 76 ln ) = 6(8 · 7 · 65 ln ) = · · · = (8! 0 ln ) = 8!
55. = () = ln(sin )
A. = { in | sin 0} =
∞
=−∞
(2 (2 + 1) ) = · · · ∪ (−4 −3) ∪ (−2 −) ∪ (0 ) ∪ (2 3) ∪ · · ·
B. No ­intercept; ­intercepts: () = 0 ⇔ ln(sin ) = 0 ⇔ sin = 0 = 1 ⇔
integer .
C. is periodic with period 2. D.
lim
→(2)+
() = −∞ and
lim
→[(2+1)]−
= 2 + 2 for each
() = −∞, so the lines
cos
= cot , so 0 () 0 when 2 2 + 2 for each
sin
integer , and 0 () 0 when 2 + 2 (2 + 1). Thus, is increasing on 2 2 + 2 and
decreasing on 2 + 2 (2 + 1) for each integer .
H.
F. Local maximum values 2 + 2 = 0, no local minimum.
= are VAs for all integers . E. 0 () =
G. 00 () = − csc2 0, so is CD on (2 (2 + 1)) for
each integer No IP
56. = ln(tan2 )
A. = { | 6= 2} B. ­intercepts + 4 , no ­intercept. C. (−) = (), so the curve is
symmetric about the ­axis. Also ( + ) = (), so is periodic with period , and we consider parts D–G only for
− 2 2 D. lim ln(tan2 ) = −∞ and
→0
lim
→(2)−
ln(tan2 ) = ∞,
sec2
2 tan sec2
=2
0 ⇔
2
tan
tan
tan 0 ⇔ 0 2 , so is increasing on 0 2 and
decreasing on − 2 0 F. No maximum or minimum
lim
→−(−2)+
ln(tan2 ) = ∞, so = 0,
= ± 2 are VA. E. 0 () =
4
2
=
sin cos
sin 2
H.
−8 cos 2
0
sin2 2
⇔ cos 2 0 ⇔ − 4 4 , so is CD on − 4 0 and
0 4 and CU on − 2 − 4 and 4 2 . IP are ± 4 0 .
G. 0 () =
⇒ 00 () =
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°
608
¤
CHAPTER 6
INVERSE FUNCTIONS
57. = () = ln(1 + 2 )
­axis. D.
A. = B. Both intercepts are 0 C. (−) = (), so the curve is symmetric about the
lim ln(1 + 2 ) = ∞, no asymptotes. E. 0 () =
→±∞
2
0 ⇔
1 + 2
H.
0, so is increasing on (0 ∞) and decreasing on (−∞ 0)
F. (0) = 0 is a local and absolute minimum.
G. 00 () =
2(1 + 2 ) − 2(2)
2(1 − 2 )
=
0 ⇔
2
2
(1 + )
(1 + 2 )2
|| 1, so is CU on (−1 1), CD on (−∞ −1) and (1 ∞).
IP (1 ln 2) and (−1 ln 2).
58. = () = ln(1 + 3 )
A. 1 + 3 0 ⇔ 3 −1 ⇔ −1, so = (−1 ∞). B. ­intercept:
(0) = ln 1 = 0; ­intercept: () = 0 ⇔ ln(1 + 3 ) = 0 ⇔ 1 + 3 = 0
symmetry D.
lim () = −∞, so = −1 is a VA E. 0 () =
→−1+
⇔ 3 = 0 ⇔ = 0 C. No
32
. 0 () 0 on (−1 0) and (0 ∞)
1 + 3
[ 0 () = 0 at = 0], so by Exercise 3.3.79, is increasing on (−1 ∞). F. No extreme values
G. 00 () =
(1 + 3 )(6) − 32 (32 )
(1 + 3 )2
H.
3[2(1 + 3 ) − 33 ]
3(2 − 3 )
=
3
2
(1 + )
(1 + 3 )2
√
√
00 () 0 ⇔ 0 3 2, so is CU on 0 3 2 and is CD on (−1 0)
√
√
and 3 2 ∞ . IP at (0 0) and 3 2 ln 3
=
59. We use the CAS to calculate 0 () =
00 () =
2 + sin + cos
and
2 + sin
22 sin + 4 sin − cos2 + 2 + 5
. From the graphs, it
2 (cos2 − 4 sin − 5)
seems that 0 0 (and so is increasing) on approximately the intervals
(0 27), (45 82) and (109 143). It seems that 00 changes sign
(indicating inflection points) at ≈ 38, 57, 100 and 120.
Looking back at the graph of () = ln(2 + sin ), this implies that the inflection points have approximate coordinates
(38 17), (57 21), (100 27), and (120 29).
60. We see that if ≤ 0, () = ln(2 + ) is only defined for 2 −
⇒ ||
√
−, and
() =
lim
() = −∞, since ln → −∞ as → 0. Thus, for 0, there are vertical asymptotes at
lim
√
√
→ −+
→− −−
√
= ± , and as decreases (that is, || increases), the asymptotes get further apart. For = 0,
lim () = −∞, so there is a vertical asymptote at = 0. If 0, there is no asymptote. To find the maxima, minima, and
→0
inflection points, we differentiate: () = ln(2 + ) ⇒ 0 () =
1
(2), so by the First Derivative Test there is a
2 +
local and absolute minimum at = 0. Differentiating again, we get
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°
SECTION 6.2* THE NATURAL LOGARITHMIC FUNCTION
00 () =
1
2 +
¤
609
2( − 2 )
−2
(2) + 2 − 2 +
(2) = 2
. Now if
( + )2
≤ 0, this is always negative, so is concave down on both of the intervals
on which it is defined. If 0, then 00 changes sign when = 2 ⇔
√
√
= ± . So for 0 there are inflection points at ± , and as
increases, the inflection points get further apart.
From the graph, it appears that the curves = ( − 4)2 and = ln intersect
61.
just to the left of = 3 and to the right of = 5, at about = 53. Let
() = ln − ( − 4)2 . Then 0 () = 1 − 2( − 4), so Newton’s method
says that +1 = − ( ) 0 ( ) = −
ln − ( − 4)2
. Taking
1 − 2( − 4)
0 = 3, we get 1 ≈ 2957738, 2 ≈ 2958516 ≈ 3 , so the first solution is 2958516, to six decimal places. Taking 0 = 5,
we get 1 ≈ 5290755, 2 ≈ 5290718 ≈ 3 , so the second (and final) solution is 5290718, to six decimal places.
We use Newton’s method with () = ln(4 − 2 ) − and
62.
0 () =
1
2
(−2) − 1 = −1 −
. The formula is
4 − 2
4 − 2
+1 = − ( ) 0 ( ). From the graphs it seems that the solutions
occur at approximately = −19 and = 11. However, if we use 1 = −19
as an initial approximation to the first solution, we get 2 ≈ −2009611, and
() = ln( − 2) − is undefined at this point, making it impossible to calculate 3 . We must use a more accurate first
2
estimate, such as 1 = −195. With this approximation, we get 1 = −195, 2 ≈ −11967495, 3 ≈ −1964760,
4 ≈ 5 ≈ −1964636. Calculating the second solution gives 1 = 11, 2 ≈ 1058649, 3 ≈ 1058007,
4 ≈ 5 ≈ 1058006. So, correct to six decimal places, the two solutions of the equation ln(4 − 2 ) = are = −1964636
and = 1058006.
63. = (2 + 2)2 (4 + 4)4
⇒ ln = ln[(2 + 2)2 (4 + 4)4 ] ⇒ ln = 2 ln(2 + 2) + 4 ln(4 + 4) ⇒
1
163
1 0
1
4
=2· 2
· 2 + 4 · 4
· 43 ⇒ 0 = 2
+ 4
⇒
+2
+4
+2
+4
163
4
+
0 = (2 + 2)2 (4 + 4)4
2 + 2
4 + 4
64. =
( + 1)4 ( − 5)3
( − 3)8
⇒ ln || = 4 ln | + 1| + 3 ln | − 5| − 8 ln | − 3| ⇒
4
3
8
0
=
+
−
+1
−5
−3
65. =
−1
4 + 1
⇒ 0 =
( + 1)4 ( − 5)3
( − 3)8
12
−1
⇒ ln = ln 4
+1
1 1
1 1
1 0
=
−
· 43
2−1
2 4 + 1
3
8
4
+
−
+1
−5
−3
1
1
ln( − 1) − ln(4 + 1) ⇒
2
2
3
2
1
23
1
−1
0
0
−
⇒ =
− 4
⇒ =
2( − 1) 4 + 1
4 + 1 2 − 2
+1
⇒ ln =
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
¤
610
CHAPTER 6
INVERSE FUNCTIONS
32
cos
1
0
=4 3
+2
−
⇒ ln || = 4 ln 3 + 1 + 2 ln |sin | − 13 ln ||. So
+1
sin
3
(3 + 1)4 sin2
1
122
0 =
+
2
cot
−
.
3 + 1
3
13
66. =
67.
4
2
(3 + 1)4 sin2
13
3
= 3
4
2
⇒
4
4
1
= 3 ln || = 3(ln 4 − ln 2) = 3 ln = 3 ln 2
2
2
68. Let = 5 + 1, so = 5 . When = 0, = 1; when = 3, = 16. Thus,
16
1
1
1
1
=
ln ||
= (ln 16 − ln 1) = ln 16.
5
5
5
5
1
=
5 + 1
2
2
1
1
1
1 5
1
= − ln |8 − 3| = − ln 2 − − ln 5 = (ln 5 − ln 2) = ln
8 − 3
3
3
3
3
3
2
1
0
69.
16
3
1
1
1
Or: Let = 8 − 3. Then = −3 , so
2
2 1
2
− 3
1
1
1
1 5
1
=
= − ln || = − ln 2 − − ln 5 = (ln 5 − ln 2) = ln .
8
−
3
3
3
3
3
3
2
1
5
5
2
9
9
9
√
1
81
1
+ 18 + ln 9 − (8 + 8 + ln 4)
70.
+ √
=
+2+
= 12 2 + 2 + ln 4 =
2
4
4
= 85
+ ln 94
2
71.
3
1
32 + 4 + 1
=
3
3
1
3 2
3
27
3 + 4 +
=
+ 4 + ln || =
+ 12 + ln 3 −
+ 4 + ln 1
2
2
2
1
1
= 20 + ln 3
72. Let = ln . Then =
73. Let = ln . Then =
1
, so
6
=
ln
(ln )2
=
⇒
ln 6
ln 6
1
= ln ||
= ln ln 6 − ln 1 = ln ln 6
1
1
2 =
1 3
1
+ = (ln )3 + .
3
3
74. Let = 3 + 1. Then = 3 2 and 13 = 2 , so
75.
2
=
3 + 1
1
1
1
1
= ln || + = ln | 3 + 1| + .
3
3
3
sin cos
sin 2
=
2
= 2. Let = cos . Then = − sin , so
1 + cos2
1 + cos2
2 = −2
= −2 · 12 ln(1 + 2 ) + = − ln(1 + 2 ) + = − ln(1 + cos2 ) + .
1 + 2
Or: Let = 1 + cos2 .
1
76. Let = ln . Then = , so
cos(ln )
=
cos = sin + = sin(ln ) + .
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°
SECTION 6.2* THE NATURAL LOGARITHMIC FUNCTION
¤
1
(ln |sin | + ) =
cos = cot
sin
cos
=
= ln || + = ln |sin | + .
(b) Let = sin . Then = cos , so cot =
sin
77. (a)
78.
(ln )2
ln
=
⇔ ln = (ln )2
⇔ 0 = (ln )2 − ln ⇔
0 = ln (ln − 1) ⇔ ln = 0 or 1 ⇔ = 0 or 1 [1 or ]
ln
(ln )2
−
= 12 (ln )2 − 13 (ln )3
=
1
1
1
= 2 − 13 − (0 − 0) = 16
√
79. The cross­sectional area is 1 + 1
1
0
2
= ( + 1). Therefore, the volume is
= [ln( + 1)]10 = (ln 2 − ln 1) = ln 2.
+1
80. Using cylindrical shells, we get =
3
0
81. =
2
1
=
1000
600
=
3
2
= ln(1 + 2 ) 0 = ln 10.
2 + 1
1000
600
1000
1
= ln | |
= (ln 1000 − ln 600) = ln 1000
= ln 53 .
600
600
Initially, = , where = 150 kPa and = 600 cm3 , so = (150)(600) = 90,000 kPa · cm3 . Thus,
3
N
1
m
= 90,000 ln 53 kPa · cm3 = 90,000 ln 53 1000 2
m
100
= 90 ln 53 N·m ≈ 45.974 J
82. 00 () = −2 , 0
⇒ 0 () = −1 +
⇒ () = − ln + + . 0 = (1) = + and
0 = (2) = − ln 2 + 2 + = − ln 2 + 2 − = − ln 2 +
⇒ = ln 2 and = − ln 2. So
() = − ln + (ln 2) − ln 2.
83. () = 2 + ln
⇒ 0 () = 2 + 1. If = −1 , then (1) = 2 ⇒ (2) = 1, so
0 (2) = 1 0 ((2)) = 1 0 (1) = 13 .
84. (a) Let () = ln
⇒ 0 () = 1 ⇒ 00 () = −12 . The linear approximation to ln near 1 is
ln ≈ (1) + 0 (1)( − 1) = ln 1 + 11 ( − 1) = − 1.
(b)
(c)
From the graph, it appears that the linear approximation is
accurate to within 01 for between about 062 and 151.
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°
611
612
¤
CHAPTER 6
85. (a)
INVERSE FUNCTIONS
We interpret ln 15 as the area under the curve = 1 from = 1 to
= 15. The area of the rectangle is 12 · 23 = 13 . The area of the
5
trapezoid is 12 · 12 1 + 23 = 12
. Thus, by comparing areas, we
5
.
observe that 13 ln 15 12
(b) With () = 1, = 10, and ∆ = 005, we have
15
ln 15 = 1 (1) ≈ (005)[ (1025) + (1075) + · · · + (1475)]
1
1
1
≈ 04054
+ 1075
+ · · · + 1475
= (005) 1025
1
1
1
1 0
1
12 − 1
, = − 2 . The slope of is
= − . Let be the ­coordinate of the point on = with slope − .
2−1
2
2
√
1
1
Then − 2 = −
⇒ 2 = 2 ⇒ = 2 since 0. Therefore the tangent line is given by
2
√
√
⇒ = − 12 + 2.
− √12 = − 12 − 2
86. (a) =
(b)
Since the graph of = 1 is concave upward, the graph lies above the
tangent line, that is, above the line segment . Now || = − 12 +
and || = −1 +
1
2
√
2
√
2. So the area of the trapezoid is
√
√
1 √
− 2 + 2 + −1 + 2 1 = − 34 + 2 ≈ 06642. So ln 2 area
of trapezoid 066.
87.
The area of is
The area of is
1
1
1
1
and so + + · · · +
+1
2
3
1
= ln .
1
1
1
and so 1 + + · · · +
2
−1
1
= ln .
1
1
88. If () = ln( ), then 0 () = (1 )(−1 ) = . But if () = ln , then 0 () = . So and must differ by a
constant: ln( ) = ln + . Put = 1: ln(1 ) = ln 1 +
⇒ = 0, so ln( ) = ln .
89. The curve and the line will determine a region when they intersect at two or
more points. So we solve the equation (2 + 1) = ⇒ = 0 or
± −4()( − 1)
1
2 + − 1 = 0 ⇒ = 0 or =
=±
− 1.
2
Note that if = 1, this has only the solution = 0, and no region is
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°
SECTION 6.3* THE NATURAL EXPONENTIAL FUNCTION
¤
613
determined. But if 1 − 1 0 ⇔ 1 1 ⇔ 0 1, then there are two solutions. [Another way of seeing this
is to observe that the slope of the tangent to = (2 + 1) at the origin is 0 = 1 and therefore we must have 0 1.]
Note that we cannot just integrate between the positive and negative roots, since the curve and the line cross at the origin.
Since and (2 + 1) are both odd functions, the total area is twice the area between the curves on the interval
0 1 − 1 . So the total area enclosed is
2
√1−1
0
√1−1
−
= 2 12 ln(2 + 1) − 12 2 0
2
+1
1
1
−1+1 −
−1
− (ln 1 − 0)
= ln
1
+ − 1 = − ln − 1
= ln
90.
From the graphs, we see that () = 01 () = ln for approximately 0 306, and then () () for
306 343 × 1015 (approximately). At that point, the graph of finally surpasses the graph of for good.
91. If () = ln (1 + ), then 0 () =
Thus, lim
→0
1
, so 0 (0) = 1.
1+
ln(1 + )
()
() − (0)
= lim
= lim
= 0 (0) = 1.
→0
→0
−0
6.3* The Natural Exponential Function
The function value at = 0 is 1 and the slope at = 0 is 1;
1.
that is, if () = , then 0 (0) = 1.
3
−2
2. (a) ln 15 = 15 by (4).
(b) 3 ln 2 = ln 2 = 23 = 8
(c) −2 ln 5 = ln 5
1
= ln −2 = −2
2
50
= ln(50 ) = 50
4. (a) ln ln
(b) ln
√
= ln 12 = 12
(c) ln sin = sin
3. (a) ln
3
(b) ln(ln ) = ln(3) = 3
1
= 5−2 = 25
(c) +ln = ln =
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
614
¤
CHAPTER 6
INVERSE FUNCTIONS
5. (a) ln(4 + 2) = 3
⇒ ln(4+2) = 3
⇒ 4 + 2 = 3
⇒ 4 = 3 − 2 ⇒ = 14 (3 − 2) ≈ 4521
(b) 2−3 = 12 ⇒ ln 2−3 = ln 12 ⇒ 2 − 3 = ln 12 ⇒ 2 = 3 + ln 12 ⇒ = 12 (3 + ln 12) ≈ 2742
6. (a) ln(2 − 1) = 3
2
⇒ ln( −1) = 3
√
⇒ 2 = 3 + 1 ⇒ = ± 3 + 1 ≈ ±4592
⇒ 2 − 1 = 3
(b) 1 + 4+1 = 20 ⇒ 4+1 = 19 ⇒ ln 4+1 = ln 19 ⇒ 4 + 1 = ln 19 ⇒ 4 = −1 + ln 19 ⇒
= 14 (−1 + ln 19) ≈ 0486
2
7. (a) ln + ln( − 1) = 0
⇒ ln[( − 1)] = 0 ⇒ ln[ −] = 0 ⇒ 2 − = 1 ⇒ 2 − − 1 = 0. The
√
√
1 ± (−1)2 − 4(1)(−1)
1± 5
1− 5
=
, but we note that ln
is undefined because
quadratic formula gives =
2(1)
2
2
√
√
1+ 5
1− 5
0. Thus, =
≈ 1618.
2
2
(b) − −2 = 1 ⇒ − 1 = −2
⇒ ln( − 1) = ln −2
⇒ ln( − 1) = −2 ⇒
= − 12 ln( − 1) ≈ −0271
8. (a) ln(ln ) = 0
(b)
⇒ ln(ln ) = 0
⇒ ln = 1 ⇒ = ≈ 2718
60
= 4 ⇒ 60 = 4(1 + − ) ⇒ 15 = 1 + −
1 + −
⇒ 14 = −
⇒ ln 14 = ln −
⇒
ln 14 = − ⇒ = − ln 14 ≈ −2639
9. (a) 2 − 3 + 2 = 0
⇔ ( − 1)( − 2) = 0 ⇔ = 1 or = 2 ⇔ = ln 1 or = ln 2, so = 0 or ln 2.
(b) = 10 ⇔ ln = ln 10 ⇔ ln = = ln 10 ⇔ ln = ln(ln 10) ⇔ = ln ln 10
10. (a) 3+1 =
⇔ 3 + 1 = ln
⇔ = 13 (ln − 1)
(b) ln(2 + 1) = 2 − ln ⇒ ln + ln(2 + 1) = ln 2 ⇒ ln [(2 + 1)] = ln 2
√
−1 + 1 + 82
[since 0].
22 + − 2 = 0 ⇒ =
4
11. (a) ln(1 + 3 ) − 4 = 0
⇔ ln(1 + 3 ) = 4 ⇔ 1 + 3 = 4
(b) 21 = 42 ⇔ 1 = 21 ⇔
12. (a) 31(−4) = 7
13. (a) ln 0
⇒ 0
+1
= 2
⇒
√
⇔ 3 = 4 − 1 ⇔ = 3 4 − 1 ≈ 37704.
1
1
= ln 21 ⇔ =
≈ 03285.
ln 21
⇒ ln 31(−4) = ln 7 ⇒
+1
(b) ln
=2 ⇔
⇒ 22 + = 2
ln 3
1
ln 3 = ln 7 ⇒ − 4 =
−4
ln 7
⇒ =4+
⇔ + 1 = 2 ⇔ (2 − 1) = 1 ⇔ =
ln 3
≈ 45646
ln 7
1
≈ 01565
2 − 1
⇒ 1. Since the domain of () = ln is 0, the solution of the original inequality
is 0 1.
(b) 5 ⇒ ln ln 5 ⇒ ln 5
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°
SECTION 6.3* THE NATURAL EXPONENTIAL FUNCTION
14. (a) 1 3−1 2
¤
615
⇒ ln 1 3 − 1 ln 2 ⇒ 0 3 − 1 ln 2 ⇒ 1 3 1 + ln 2 ⇒
1
1
3 3 (1 + ln 2)
(b) 1 − 2 ln 3 ⇒ −2 ln 2 ⇒ ln −1 ⇒ −1
15. We start with the graph of = (Figure 2) and reflect about the y­axis to get the graph of = − . Then we reflect the
graph about the x­axis to get the graph of = −− .
16. We start with the graph of = (Figure 2) and reflect
the portion of the graph in the first quadrant about the
­axis to obtain the graph of = || .
17. We start with the graph of = (Figure 2) and reflect about the ­axis to get the graph of = − . Then we compress
the graph vertically by a factor of 2 to obtain the graph of = 12 − and then reflect about the ­axis to get the graph
of = − 12 − . Finally, we shift the graph one unit upward to get the graph of = 1 − 12 − .
18. We start with the graph of = (Figure 2) and reflect about the ­axis to get the graph of = − . Then shift the graph
upward one unit to get the graph of = 1 − . Finally, we stretch the graph vertically by a factor of 2 to obtain the graph of
= 2(1 − ).
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°
616
¤
CHAPTER 6
19. (a) For () =
INVERSE FUNCTIONS
√
3 − 2 , we must have 3 − 2 ≥ 0 ⇒ 2 ≤ 3
⇒
2 ≤ ln 3 ⇒ ≤ 12 ln 3.
Thus, the domain of is (−∞ 12 ln 3].
(b) = () =
√
3 − 2
[note that ≥ 0] ⇒ 2 = 3 − 2
⇒ 2 = 3 − 2
⇒ 2 = ln(3 − 2 ) ⇒
= 12 ln(3 − 2 ). Interchange and : = 12 ln(3 − 2 ). So −1 () = 12 ln(3 − 2 ). For the domain of −1 ,
√
√
√
√
we must have 3 − 2 0 ⇒ 2 3 ⇒ || 3 ⇒ − 3 3 ⇒ 0 ≤ 3 since ≥ 0. Note
√
that the domain of −1 , [0 3 ), equals the range of .
20. (a) For () = ln(2 + ln ), we must have 2 + ln 0
⇒ ln −2 ⇒ −2 . Thus, the domain of
is (−2 ∞).
(b) = () = ln(2 + ln ) ⇒ = 2 + ln ⇒ ln = − 2 ⇒ = −2 . Interchange and : = −2 .
So −1 () = −2 . The domain of −1 , as well as the range of , is .
21. We solve = 3 ln( − 2) for : 3 = ln( − 2)
⇒ 3 = − 2 ⇒ = 2 + 3 . Interchanging and gives the
inverse function = 2 + 3 .
22. = (ln )2 , ≥ 1, ln =
√
√
⇒ = . Interchange and : = is the inverse function.
23. We solve = 1− for : ln = ln 1−
⇒ ln = 1 − ⇒ = 1 − ln . Interchanging and gives the inverse
function = 1 − ln .
24. We solve =
1 − −
for : (1 + − ) = 1 − −
1 + −
− (1 + ) = 1 −
⇒ − =
1−
1+
⇒ + − = 1 − −
⇒ − = ln
1−
1+
⇒ = − ln
⇒ − + − = 1 −
⇒
1−
or, equivalently,
1+
−1
1+
1+
1−
. Interchanging and gives the inverse function = ln
.
= ln
= ln
1+
1−
1−
25. Divide numerator and denominator by 3 : lim
3 − −3
→∞ 3 + −3
= lim
1 − −6
→∞ 1 + −6
=
1−0
=1
1+0
2
26. Let = −2 . As → ∞, → −∞. So lim − = lim = 0 by (6).
→∞
→−∞
27. Let = 3(2 − ). As → 2+ , → −∞. So lim 3(2−) = lim = 0 by (6).
→2+
→−∞
28. Let = 3(2 − ). As → 2− , → ∞. So lim 3(2−) = lim = ∞ by (6).
→2−
→∞
29. Since −1 ≤ cos ≤ 1 and −2 0, we have −−2 ≤ −2 cos ≤ −2 . We know that lim (−−2 ) = 0 and
→∞
lim −2 = 0, so by the Squeeze Theorem, lim (−2 cos ) = 0.
→∞
→∞
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°
SECTION 6.3* THE NATURAL EXPONENTIAL FUNCTION
30.
lim
→(2)+
sec = 0 since sec → −∞ as → (2)+ .
31. () = −2
⇒ 0 () = −2( ) = −2
32. () = +
⇒ 0 () = + −1
33. () = (32 − 5)
PR
⇒
0 () = (32 − 5)( )0 + (32 − 5)0 = (32 − 5) + (6 − 5)
= [(32 − 5) + (6 − 5)] = (32 + − 5)
34. By the Quotient Rule, =
3
3
⇒ 0 =
35. By (9), =
2
36. () = −
37. = tan
1 −
CR
2
⇒ 0 =
3
(3 ) = 32 .
⇒ 0 () = − ·
⇒ 0 = tan
(1 − ) − (− )
− 2 + 2
=
=
.
(1 − )2
(1 − )2
(1 − )2
2
2
( − ) = − (2 − 1)
(tan ) = (sec2 )tan
√
3
= 13 . Then
1 −23
1
(
( ) =
=
=
)
=
.
3
3 3 ( + 1)2
3 3 ( + 1)2
38. Let = () = + 1 and = () =
39. () =
2
2 +
(2 + ) 2 + (2) − 2 (2 + )
4 + 23 + 2 2 + 22 − 23 − 2 2
=
2
2
( + )
(2 + )2
0 () =
4 + 22
(3 + 2 )
=
(2 + )2
(2 + )2
=
40. () =
QR
⇒
√
2
· +1 ⇒
√
2
2
2
2
2
√ √
1
+ 1 + +1 ·
· +1 ·
= · +1 · 2 + +1 · √
2
2
√
2
2
1
+
1
4
√
or +1
= +1 2 + √
2
2
0 () =
41. Using the Product Rule and the Chain Rule, = 2 −3
⇒
0 = 2 −3 (−3) + −3 (2) = −3 (−32 + 2) = −3 (2 − 3).
42. () = tan(1 + 2 )
43. () = sin
⇒ 0 () = sec2 (1 + 2 ) · (1 + 2 )0 = 22 sec2 (1 + 2 )
⇒ 0 () = (cos ) · + (sin ) · = ( cos + sin )
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°
¤
617
618
¤
CHAPTER 6
44. () = (−1)
INVERSE FUNCTIONS
⇒ 0 () = (−1)
45. By (9), () = sin 2
0
⇒
0
sin 2
( sin 2) = sin 2 ( · 2 cos 2 + sin 2 · 1) = sin 2 (2 cos 2 + sin 2)
() =
46. = sin 2 + sin(2 )
0 = sin 2
(−1)
( − 1)(1) − (1)
= (−1)
=−
2
− 1
( − 1)
( − 1)2
⇒
sin 2 + cos(2 ) 2 = sin 2 (cos 2) · 2 + cos(2 ) 2 · 2 = 2 cos 2 sin 2 + 22 cos(2 )
√
2
47. () = sec
⇒
√
√
2 1
sec 2
sec 2 = sec (sec 2 )−12
2
√
√
√
1
2
2
= sec √
· sec 2 tan 2 · 2 = sec 2 tan 2 sec
2 sec 2
√
2
0 () = sec
√
48. () = 1 2 − 1
⇒
√
1
1
1
−1
1
= −1 ;
= −−2 = − 2
· 2 + 2 − 1 · 1 · − 2
0 () = 1 · √
2 2 − 1
√
3
2
2 − 1
− +1
√
−
= 1 √
or 1
2
2 − 1
2 2 − 1
⇒
1 +
(1 + ) − ( )
(1 + − )
·
·
=
cos
=
cos
0 () = cos
1 +
(1 + )2
1 +
(1 + )2
(1 + )2
1 +
49. () = sin
2
2
50. () = sin ( )
2
2
2
2
⇒ 0 () = sin ( ) · 2 sin(2 ) · cos(2 ) · 2 = 4 sin(2 ) cos(2 )sin ( )
51. = cos + sin
⇒ 0 = (− sin ) + (cos )( ) + cos = (cos − sin ) + cos , so
0 (0) = 0 (cos 0 − sin 0) + cos 0 = 1(1 − 0) + 1 = 2. An equation of the tangent line to the curve = cos + sin at
the point (0 1) is − 1 = 2( − 0) or = 2 + 1.
52. =
1+
1 +
⇒
At 0 12 , 0 =
53.
0 =
(1 + )(1) − (1 + )
1 + − −
1 −
=
=
2
2
(1 + )
(1 + )
(1 + )2
1
1
= , and an equation of the tangent line is − 12 = 14 ( − 0) or = 14 + 12 .
(1 + 1)2
4
( ) =
( − ) ⇒ ·
·
0
1
−
· = 1 − 0
2
= 1 − 0
⇒ 0 −
· 1 − · 0
= 1 − 0 ⇒
2
−
=
⇒ 0 1 −
2
⇒ ·
0
· = 1−
2
⇒
−
( − )
0 = 2
=
2 −
−
2
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°
¤
SECTION 6.3* THE NATURAL EXPONENTIAL FUNCTION
⇒ 0 + · 1 + + 0 = 0 ⇒ 0 ( + ) = − −
54. + = 1
(0 1), 0 = −
⇒ 0 = −
+
. At
+
+1·1
= −( + 1), so an equation for the tangent line is − 1 = −( + 1)( − 0), or = −( + 1) + 1.
0+1
55. = + −2
⇒ 0 = − 12 −2 ⇒ 00 = + 14 −2 , so
2 00 − 0 − = 2 + 14 −2 − − 12 −2 − + −2 = 0.
56. = − + −
⇒ 0 = −− + − − − = ( − )− − −
⇒
00 = ( − )− − − + − = ( − 2)− + − ,
so 00 + 2 0 + = ( − 2)− + − + 2 ( − )− − − + − + − = 0.
57. =
⇒ 0 =
00 = 2 , so if = satisfies the differential equation 00 + 6 0 + 8 = 0,
⇒
then 2 + 6 + 8 = 0; that is, (2 + 6 + 8) = 0. Since 0 for all , we must have 2 + 6 + 8 = 0,
or ( + 2)( + 4) = 0, so = −2 or −4.
58. =
⇒ 0 =
⇒ 00 = 2 . Thus, + 0 = 00
√
⇔ + = 2
⇔
(2 − − 1) = 0 ⇔ = 1 ±2 5 , since 6= 0.
59. () = 2
⇒ 0 () = 22
000 () = 22 · 22 = 23 2
60. () = −
⇒ 00 () = 2 · 22 = 22 2
⇒ ···
⇒
⇒ () () = 2 2
⇒ 0 () = (−− ) + − = (1 − )−
⇒
00 () = (1 − )(−− ) + − (−1) = ( − 2)−
⇒ 000 () = ( − 2)(−− ) + − = (3 − )−
(4) () = (3 − )(−− ) + − (−1) = ( − 4)−
⇒
···
⇒
⇒ () () = (−1) ( − )− .
So 1000 − = ( − 1000)− .
61. (a) () = + is continuous on and (−1) = −1 − 1 0 1 = (0), so by the Intermediate Value Theorem,
+ = 0 has a solution in (−1 0).
(b) () = + ⇒ 0 () = + 1, so +1 = −
+
. Using 1 = −05, we get 2 ≈ −0566311,
+ 1
3 ≈ −0567143 ≈ 4 , so the solution is −0567143 to six decimal places.
62. () =
619
2
1 +
⇒ 0 () = −
2
1
, so (0) = 1 and 0 (0) = − .
2
(1 + )2
Thus, () (0) + 0 (0)( − 0) = 1 − 12 . We need
2
2
1
− 01 1 −
+ 01, which is true when
1 +
2
1 +
−1423 1423. Note that to ensure the accuracy, we have rounded the
smaller value up and the larger value down.
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°
620
¤
CHAPTER 6
INVERSE FUNCTIONS
63. (a) lim () = lim
→∞
1
→∞ 1 + −
=
1
= 1, since 0 ⇒ − → −∞ ⇒ − → 0. As time increases, the
1+·0
proportion of the population that has heard the rumor approaches 1; that is, everyone in the population has heard the rumor.
(b) () = (1 + − )−1
⇒
−
= −(1 + − )−2 (−− ) =
(1 + − )2
(c)
From the graph of () = (1 + 10−05 )−1 , it seems that () = 08
(indicating that 80% of the population has heard the rumor) when
≈ 74 hours.
64. (a)
The displacement function is squeezed between the other two functions. This
is because −1 ≤ sin 4 ≤ 1 ⇒ −8−2 ≤ 8−2 sin 4 ≤ 8−2 .
(b) The maximum value of the displacement is about 66 cm, occurring at ≈ 036 s. It occurs just before the graph of the
displacement function touches the graph of 8−2 (when = 8 ≈ 039).
(c) The velocity of the object is the derivative of its displacement function, that is,
−2
8
sin 4 = 8 −2 cos 4(4) + sin 4 − 12 −2
If the displacement is zero, then we must have sin 4 = 0 (since the exponential term in the displacement function is
always positive). The first time that sin 4 = 0 after = 0 occurs at = 4 . Substituting this into our expression for the
(d)
velocity, and noting that the second term vanishes, we get 4 = 8−8 cos 4 · 4 · 4 = −32−8 ≈ −216 cms.
The graph indicates that the displacement is less than 2 cm from equilibrium
whenever is larger than about 28.
65. () =
(1 + 2 ) − (2)
(2 − 2 + 1)
( − 1)2
0
,
[0
3].
()
=
=
=
. 0 () = 0 ⇒
1 + 2
(1 + 2 )2
(1 + 2 )2
(1 + 2 )2
( − 1)2 = 0 ⇔ = 1. 0 () exists for all real numbers since 1 + 2 is never equal to 0. (0) = 1,
(1) = 2 1359, and (3) = 3 10 2009. So (3) = 3 10 is the absolute maximum value and (0) = 1 is the
absolute minimum value.
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°
SECTION 6.3* THE NATURAL EXPONENTIAL FUNCTION
66. () = 2 , [−3 1].
0 () = 2
1
2
+ 2 (1) = 2
1
2 + 1 .
0 () = 0 ⇔
1
2 + 1 = 0
¤
621
⇔ = −2.
(−3) = −3−32 ≈ −0669, (−2) = −2−1 ≈ −0736, and (1) = 12 ≈ 1649. So (1) = 12 is the absolute
maximum value and (−2) = −2 is the absolute minimum value.
67. () = −
⇒ 0 () = 1 − = 0 ⇔ = 1 ⇔ = 0. Now 0 () 0 for all 0 and 0 () 0 for all
0, so the absolute maximum value is (0) = 0 − 1 = −1.
68. () =
⇒ 0 () =
−
= 0 ⇔ ( − 1) = 0 ⇒ = 1. Now 0 () 0 ⇔
2
− 1 0 ⇔ 1 and 0 () 0 ⇔
−
0 ⇔
2
−
0 ⇔ − 1 0 ⇔ 1. Thus there is an absolute
2
minimum value of (1) = at = 1.
⇒ 0 () = (22 ) + 2 (1) = 2 (2 + 1). Thus, 0 () 0 if − 12 and 0 () 0 if − 12 .
So is increasing on − 12 ∞ and is decreasing on −∞ − 12 .
69. (a) () = 2
(b) 00 () = 2 (2) + (2 + 1) · 22 = 22 [1 + (2 + 1)] = 22 (2 + 2) = 42 ( + 1). 00 () 0 ⇔ −1
and 00 () 0 ⇔ −1. Thus, is concave upward on (−1 ∞) and is concave downward on (−∞ −1).
(c) There is an inflection point at −1 −−2 , or −1 −12 .
70. (a) () =
2
⇒ 0 () =
2 − (2)
( − 2)
( − 2)
=
=
. 0 () 0 ⇔ 0 or 2, so is
2
2
4
( )
3
increasing on (−∞ 0) and (2 ∞). 0 () 0 ⇔ 0 2, so is decreasing on (0 2).
(b) 00 () =
3 [ · 1 + ( − 2) ] − ( − 2) · 32
2 [( − 1) − 3( − 2)]
(2 − 4 + 6)
=
=
.
3
2
6
( )
4
2 − 4 + 6 = (2 − 4 + 4) + 2 = ( − 2)2 + 2 0, so 00 () 0 and is CU on (−∞ 0) and (0 ∞).
(c) There are no changes in concavity and, hence, there are no points of inflection.
71. = () = −1(+1)
C. No symmetry D.
−1 ( + 1) → −∞,
A. = { | 6= −1} = (−∞ −1) ∪ (−1, ∞) B. No ­intercept; ­intercept = (0) = −1
lim −1(+1) = 1 since −1( + 1) → 0, so = 1 is a HA.
→±∞
lim −1(+1) = ∞ since −1( + 1) → ∞, so = −1 is a VA.
→−1−
E. 0 () = −1(+1) ( + 1)2
⇒ 0 () 0 for all except 1, so
H.
is increasing on (−∞ −1) and (−1 ∞). F. No extreme values
G. 00 () =
lim −1(+1) = 0 since
→−1+
−1(+1) (−2)
−1(+1) (2 + 1)
−1(+1)
+
=−
4
3
( + 1)
( + 1)
( + 1)4
⇒
00 () 0 ⇔ 2 + 1 0 ⇔ − 12 , so is CU on (−∞ −1)
and −1 − 12 , and CD on − 12 , ∞ . has an IP at − 12 , −2 .
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°
622
¤
CHAPTER 6
INVERSE FUNCTIONS
72. = () = 2 −
A. = B. ­intercept: (0) = 0; ­intercepts: () = 0 ⇒ 2 =
= 0 C. No symmetry D.
⇒ = 1 ⇒
lim 2 − = 0, so = 0 is a HA. No VA. E. 0 () = 22 − = (2 − 1),
→−∞
so 0 () 0 ⇔ 12
⇔ ln 12 = − ln 2 and 0 () 0 ⇔
12 ⇔ ln 12 , so is decreasing on −∞ ln 12
and increasing on ln 12 ∞ .
H.
2
F. Local minimum value ln 12 = 2 ln(12) − ln(12) = 12 − 12 = − 14
G. 00 () = 42 − = (4 − 1), so 00 () 0 ⇔ 14 ⇔
ln 14 and 00 () 0 ⇔ ln 14 . Thus, is CD on −∞ ln 14 and
2
3
CU on ln 14 ∞ . IP at ln 14 14 − 14 = ln 14 − 16
73. = 1(1 + − )
A. = B. No ­intercept; ­intercept = (0) = 12 C. No symmetry
1
D. lim 1(1 + − ) = 1 +
= 1 and lim 1(1 + − ) = 0 since lim − = ∞, so has horizontal asymptotes
0
→∞
→−∞
→−∞
= 0 and = 1. E. 0 () = −(1 + − )−2 (−− ) = − (1 + − )2 . This is positive for all , so is increasing on .
F. No extreme values G. 00 () =
(1 + − )2 (−− ) − − (2)(1 + − )(−− )
− (− − 1)
=
(1 + − )4
(1 + − )3
The second factor in the numerator is negative for 0 and positive for 0,
H.
and the other factors are always positive, so is CU on (−∞, 0) and CD
on (0 ∞). IP at 0, 12
74. The function () = cos is periodic with
period 2, so we consider it only on the interval
[0 2]. We see that it has local maxima of about
(0) ≈ 272 and (2) ≈ 272, and a local
minimum of about (314) ≈ 037. To find the
exact values, we calculate 0 () = − sin cos . This is 0 when − sin = 0 ⇔ = 0, or 2 (since we are only
considering ∈ [0 2]). Also 0 () 0 ⇔ sin 0 ⇔ 0 . So (0) = (2) =
(both maxima) and () = cos = 1 (minimum). To find the inflection points, we calculate and graph
00 () =
(− sin cos ) = − cos cos − sin (cos )(− sin ) = cos sin2 − cos . From the graph of 00 (),
we see that has inflection points at ≈ 090 and at ≈ 538. These ­coordinates correspond to inflection points
(090 186) and (538 186).
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°
SECTION 6.3* THE NATURAL EXPONENTIAL FUNCTION
¤
623
3
75. () = − → 0 as → −∞, and
() → ∞ as → ∞. From the graph,
it appears that has a local minimum of
about (058) = 068, and a local
maximum of about (−058) = 147.
To find the exact values, we calculate
3
0 () = 32 − 1 − , which is 0 when 32 − 1 = 0 ⇔ = ± √13 . The negative solution corresponds to the local
√ 3
√
√
maximum − √13 = (−1 3) − (−1 3) = 2 39 , and the positive solution corresponds to the local minimum
1
√
3
√
3
√
√
= (1 3) − (1 3) = −2 39 . To estimate the inflection points, we calculate and graph
00 () =
3
3
3
3
2
3 − 1 − = 32 − 1 − 32 − 1 + − (6) = − 94 − 62 + 6 + 1 .
From the graph, it appears that 00 () changes sign (and thus has inflection points) at ≈ −015 and ≈ −109. From the
graph of , we see that these ­values correspond to inflection points at about (−015 115) and (−109 082).
76. () = − with = 001, = 4, and = 007. We will find the
zeros of 00 for () = − .
0 () = (−− ) + − (−1 ) = − (− + −1 )
00 () = − (−−1 + ( − 1)−2 ) + (− + −1 )(−− )
= −2 − [− + ( − 1) + 2 2 − ]
= −2 − (2 2 − 2 + 2 − )
Using the given values of and gives us 00 () = 2 −007 (000492 − 056 + 12). So 00 () = 001 00 () and its zeros
2857 and 2 = 600
8571.
are = 0 and the solutions of 000492 − 056 + 12 = 0, which are 1 = 200
7
7
At 1 minutes, the rate of increase of the level of medication in the bloodstream is at its greatest and at 2 minutes, the rate of
decrease is the greatest.
77. Let = 0135 and = −2802. Then () =
+ 1 = 0 ⇔ = −
⇒ 0 () = ( · · + · 1) = ( + 1). 0 () = 0 ⇔
1
≈ 036 h. (0) = 0, (−1) = − −1 = − ≈ 00177, and (3) = 33 ≈ 000009.
The maximum average BAC during the first three hours is about 00177 gdL and it occurs at approximately 036 h
(214 min).
78. (a) As || → ∞, = −2 (22 ) → −∞, and → 0. The HA is = 0. Since takes on its maximum value at = 0, so
2
2
does . Showing this result using derivatives, we have () = − (2 )
2
2
⇒ 0 () = − (2 ) (−2 ).
0 () = 0 ⇔ = 0. Because 0 changes from positive to negative at = 0, (0) = 1 is a local maximum. For
−1
2
2
2
2
2
2
1
inflection points, we find 00 () = − 2 − (2 ) · 1 + − (2 ) (−2 ) = 2 − (2 ) (1 − 22 ).
00 () = 0 ⇔ 2 = 2
⇔ = ±. 00 () 0 ⇔ 2 2
⇔ − .
So is CD on (− ) and CU on (−∞ −) and ( ∞). IP at (± −12 ).
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°
¤
624
CHAPTER 6
INVERSE FUNCTIONS
(b) Since we have IP at = ±, the inflection points move away from the ­axis as increases.
From the graph, we see that as increases, the graph tends to spread out and
(c)
there is more area between the curve and the ­axis.
79.
1
( + ) =
0
80.
81.
1
1
+1
1
+ =
+ − (0 + 1) =
+−1
+1
+1
+1
0
5
=
= 5 − (−5) = 10
−5
5
−5
2
0
=
2
1
1
1
1
− = − − = − −2 + 0 = (1 − −2 )
0
0
2
82. Let = −4 . Then = −43 and 3 = − 14 , so
83. Let = 1 + . Then = , so
84.
85.
86.
4
3 − =
√
√
1 + =
= 23 32 + = 23 (1 + )32 + .
(1 + )2
=
( + − )2 = (2 + 2 + −2 ) = 12 2 + 2 − 12 −2 +
(4 + )5
1 + 2 + 2
=
= 4 +
=
=
4
− 14 = − 14 + = − 14 − + .
(− + 2 + ) = −− + 2 + +
5 = 16 6 + = 16 (4 + )6 +
87. Let = 1 − . Then = − and = −, so
=
(1 − )2
1
(−) = −
2
88. Let = sin . Then = cos , so
−2 = −(−−1 ) + =
sin cos =
1
1
+ =
+ .
1 −
= + = sin + .
89. Let = 1, so = −12 . When = 1, = 1; when = 2, = 12 . Thus,
2
1
90.
1
=
2
12
1
√
12
(−) = − 1 = −(12 − ) = − .
1√
1+1
1√
1 + −
= 1 + −
12
− − =
=
1
+
(−)
−
= −
0
0
2
1+1
√
= − 23 [(1 + 1)32 − 232 ] = 43 2 − 23 (1 + 1)32
= − 23 32
2
2
2
2−
2 2
= 12 −−
= 12 (−−4 + 1)
1
91. avg = 2 −
0
0
0
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°
SECTION 6.3* THE NATURAL EXPONENTIAL FUNCTION
¤
625
1
1
[ − ( 2 − 2)] = −1 ( − 2 + 2) = − 13 3 + 2 −1
= − 13 + 2 − −1 + 13 − 2 = − −1 − 23 + 4 = − −1 + 10
3
92. Area =
93. Area =
1
−1
1
1 3
− = 13 3 − 0 = 13 3 − − 13 − 1 = 13 3 − + 23 ≈ 4644
0
94. 00 () = 3 + 5 sin
⇒ 0 () = 3 − 5 cos +
⇒ 2 = 0 (0) = 3 − 5 +
0 () = 3 − 5 cos + 4 ⇒ () = 3 − 5 sin + 4 +
⇒ = 4, so
⇒ 1 = (0) = 3 +
⇒ = −2,
so () = 3 − 5 sin + 4 − 2.
95. =
1
0
1
1
( )2 = 0 2 = 12 2 0 = 2 2 − 1
96. The shell has radius circumference 2, and
2
height − , so =
1
2
2− .
0
Let = . Thus, = 2 , so
2
=
1
0
1
− = −− 0 = (1 − 1).
√
Let = , so = 2 and = 2 . When = 0, = 0; when = 1, = 1. Thus,
1 √
1
1
1 = 0 = 0 (2 ) = 2 0 .
97. First Figure
Second Figure
2 =
Third Figure
1
0
1
2 = 2 0 .
Let = sin , so = cos . When = 0, = 0; when = 2 , = 1. Thus,
2
2
1
1
3 = 0 sin sin 2 = 0 sin (2 sin cos ) = 0 (2 ) = 2 0 .
Since 1 = 2 = 3 , all three areas are equal.
98. Let () = with = 450268 and = 112567, and () = population after hours. Since () = 0 (),
3
0
() = (3) − (0) is the total change in the population after three hours. Since we start with 400 bacteria, the
population will be
3
3
3
(3) = 400 + 0 () = 400 + 0 = 400 + 0 = 400 + 3 − 1
≈ 400 + 11,313 = 11,713 bacteria
99. The rate is measured in liters per minute. Integrating from = 0 minutes to = 60 minutes will give us the total amount of oil
that leaks out (in liters) during the first hour.
60
0
60
100−001
[ = −001, = −001]
−06
−06
(−100 ) = −10,000 0
= −10,000(−06 − 1) 45119 4512 L
= 100 0
() =
0
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°
626
¤
CHAPTER 6
INVERSE FUNCTIONS
100. The rate is measured in kilograms per year. Integrating from = 0 years (2000) to = 20 years (2020) will give us the net
change in biomass from 2000 to 2020.
20
0
60,000−06
=
(1 + 5−06 )2
=
1+5−12
6
60,000 1
− 3
2
= 1 + 5−06
= −3−06
1+5−12
20,000
20,000
20,000
16,666
=
−
−12
1
+
5
6
6
Thus, the predicted biomass for the year 2020 is approximately 25,000 + 16,666 = 41,666 kg.
101.
30
0
30
−30
0 − = 0
() =
(−)
0
1
−30
= 0 −
= 0 (−−30 + 1)
= −
= − −
1
The integral
30
0
() represents the total amount of urea removed from the blood in the first 30 minutes of dialysis.
2
102. (a) erf() = √
0
2
− =
2
2
−
2
− =
0
⇒
2
2
− + − , so
− =
0
0
2
− −
0
2
− =
0
√
erf() By Property 5 of definite integrals in Section 4.2,
2
√
√
√
erf() −
erf() = 12 [erf() − erf()].
2
2
2
2
2
2
2
2
(b) = erf() ⇒ 0 = 2 erf() + erf 0 () = 2 + · √ −
2
[by FTC1] = 2 + √ .
From the graph, it appears that is an odd function ( is undefined for = 0).
103.
To prove this, we must show that (−) = −().
1
1 − 1 1
1 − (−1)
1 − 1
1 − 1(−)
=
=
· 1 = 1
(−) =
1(−)
(−1)
1
1+
1+
+1
1 + 1
1 − 1
=−
= −()
1 + 1
so is an odd function.
104. We’ll start with = −1 and graph () =
1
for = 01, 1, and 5.
1 +
From the graph, we see that there is a horizontal asymptote = 0 as → −∞
and a horizontal asymptote = 1 as → ∞. If = 1, the y­intercept is 0 12 .
As gets smaller (close to 0), the graph of moves left. As gets larger, the graph
of moves right.
[continued]
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°
SECTION 6.4* GENERAL LOGARITHMIC AND EXPONENTIAL FUNCTIONS
¤
627
As changes from −1 to 0, the graph of is stretched horizontally. As
changes through large negative values, the graph of is compressed horizontally.
(This takes care of negatives values of .)
If is positive, the graph of is reflected through the y­axis.
Last, if = 0, the graph of is the horizontal line = 1(1 + ).
105. Using the second law of logarithms and Equation 5, we have ln( ) = ln − ln = − = ln( − ). Since ln is a
one­to­one function, it follows that = − .
106. Using the third law of logarithms and Equation 5, we have ln = = ln = ln ( ) . Since ln is a one­to­one
function, it follows that = ( ) .
6.4* General Logarithmic and Exponential Functions
1. (a) = ln
(b) The domain of () = is .
(c) The range of () = [ 6= 1] is (0 ∞).
(d) (i) See Figure 1.
(ii) See Figure 3.
(iii) See Figure 2.
2. (a) log is the number such that = .
(b) The domain of () = log is (0 ∞).
(c) The range of () = log is .
(d) See Figure 11.
√
3. Since = ln , 4− = − ln 4 .
2
√
4. Since = ln , 5 = 5 ln .
2
5. Since = ln , 10 = ln 10 .
6. Since = ln , (tan )sec = sec ln tan .
1
= log3 3−4 = −4
(b) log3 81
7. (a) log3 81 = log3 34 = 4
(c) log9 3 = log9 912 = 12
8. The Laws of Logarithms, used in several solutions for this section, are listed on Reference Page 4.
(a) log10
√
10 = log10 1012 = 12 by the cancellation equation after (5).
(b) log10 40 + log10 25 = log10 [(40)(25)]
[by Law 1]
= log10 100
= log10 102 = 2
(c) log2 30 − log2 15 = log2
30
15
[by the cancellation equation after (5)]
= log2 2 = 1
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°
628
¤
CHAPTER 6 INVERSE FUNCTIONS
9. (a) log3 10 − log3 5 − log3 18 = log3
10
5
2
18
104
54 · 104
= log5 5−4 = −4
− log3 18 = log3 2 − log3 18 = log3
= log3
1
9
= log3 3−2 = −2
(b) 2 log5 100 − 4 log5 50 = log5 1002 − log5 504 = log5
10. (a) log
1002
504
= log5
1
1
1
= −1 since −1 = . [Or: log = log −1 = −1]
(b) 10(log10 4 + log10 7) = 10log10 4 · 10log10 7 = 4 · 7 = 28
[Or: 10(log10 4 + log10 7) = 10log10 (4·7) = 10log10 28 = 28]
11. All of these graphs approach 0 as → −∞, all of them pass through the point
(0 1), and all of them are increasing and approach ∞ as → ∞. The larger the
base, the faster the function increases for 0, and the faster it approaches 0 as
→ −∞.
12. The functions with base greater than 1 (3 and 10 ) are increasing, while those
are decreasing. The graph of 13 is the
1
reflection of that of 3 about the ­axis, and the graph of 10
is the reflection of
with base less than 1
1
3
and
1
10
that of 10 about the ­axis. The graph of 10 increases more quickly than that of
3 for 0, and approaches 0 faster as → −∞.
13. (a) log5 10 =
ln 10
≈ 1430677
ln 5
14. To graph the functions, we use log2 =
(b) log3 12 =
ln 12
≈ 2261860
ln 3
(c) log12 6 =
ln 6
≈ 0721057
ln 12
ln
ln
, log4 =
, etc. These
ln 2
ln 4
graphs all approach −∞ as → 0+ , and they all pass through the point
(1 0). Also, they are all increasing, and all approach ∞ as → ∞. The
smaller the base, the larger the rate of increase of the function (for 1)
and the closer the approach to the ­axis (as → 0+ ).
15. To graph these functions, we use log15 =
ln
ln
and log50 =
.
ln 15
ln 50
These graphs all approach −∞ as → 0+ , and they all pass through the
point (1 0). Also, they are all increasing, and all approach ∞ as → ∞.
The functions with larger bases increase extremely slowly, and the ones with
smaller bases do so somewhat more quickly. The functions with large bases
approach the ­axis more closely as → 0+ .
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°
SECTION 6.4* GENERAL LOGARITHMIC AND EXPONENTIAL FUNCTIONS
¤
629
16. We see that the graph of ln is the reflection of the graph of about the
line = , and that the graph of log8 is the reflection of the graph of 8
about the same line. The graph of 8 increases more quickly than that of .
Also note that log8 → ∞ as → ∞ more slowly than ln .
17. Use = with the points (1 6) and (3 24).
4 = 2
6 = 1
= 6
and 24 = 3
⇒ 24 =
⇒
⇒ = 2 [since 0] and = 62 = 3. The function is () = 3 · 2 .
18. Given the ­intercept (0 2), we have = = 2 . Using the point 2 29 gives us 29 = 22
= 13
6 3
[since 0]. The function is () = 2 13 or () = 2(3)− .
19. (a) 2 ft = 24 in, (24) = 242 in = 576 in = 48 ft.
⇒
1
= 2
9
⇒
(24) = 224 in = 224 (12 · 5280) mi ≈ 265 mi
(b) 3 ft = 36 in, so we need such that log2 = 36 ⇔ = 236 = 68,719,476,736. In miles, this is
68,719,476,736 in ·
1 mi
1 ft
·
≈ 1,084,5877 mi.
12 in 5280 ft
20. We see from the graphs that for less than about 18, () = 5 () = 5 , and then near the point (18 171) the curves
intersect. Then () () from ≈ 18 until = 5. At (5 3125) there is another point of intersection, and for 5 we
see that () (). In fact, increases much more rapidly than beyond that point.
21. lim (1001) = ∞ by Figure 1, since 1001 1.
→∞
22. By Figure 1, if 1, lim = 0, so lim (1001) = 0.
→−∞
2
23. lim 2− = lim 2
→∞
→−∞
→−∞
[where = −2 ] = 0
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°
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¤
CHAPTER 6 INVERSE FUNCTIONS
24. Let = 2 − 5 + 6. As → 3+ , = ( − 2)( − 3) → 0+ . lim log10 2 − 5 + 6 = lim log10 = −∞
→3+
[analogous to (4) in Section 6.2*].
→0+
25. () = 5 + 5
⇒ 0 () = 54 + 5 ln 5
26. () = sin(2 )
⇒ 0 () = cos(2 ) · 2 ln 2 + sin(2 ) · 1 = 2 ln 2 cos(2 ) + sin(2 )
27. Using Formula 4 and the Chain Rule, () = 4
0 () = 4 (ln 4)
⇒
4
= 4 (ln 4) −−2 = − (ln 4) 2 .
(cos 2) = −2(sin 2) 3cos 2 ln 3
2 2
2
2
2
2
2
4 = sec2 4 · 4 ln 4
( ) = 2 ln 4 sec2 4 · 4
⇒ 0 () = sec 2 4
2
29. () = tan(4 )
30. () = (1 + 10ln )6
⇒
(1 + 10ln ) = 6(1 + 10ln )5 10ln ln 10
(ln ) = 6 ln 10(1 + 10ln )5 · 10ln
0 () = 6(1 + 10ln )5
31. = log8 (2 + 3)
⇒ 0 =
2 + 3
1
1
·
(2 + 3) = 2
· (2 + 3) = 2
(2 + 3) ln 8
( + 3) ln 8
( + 3) ln 8
√
1
1
1
√
1
√ =
⇒ 0 () = √
= √
2(ln 10)
ln 10
ln 10 2
Or: () = log10
33. = log4 sin
√
1
1 1
=
= log10 12 = 12 log10 ⇒ 0 () =
2 ln 10
2 (ln 10)
⇒ 0 = ·
34. = log2 ( log5 )
0 =
= −1
⇒ 0 () = 3cos 2 ln 3
28. () = 3cos 2
32. () = log10
1
cot
· cos + log4 sin · 1 =
+ log4 sin
sin ln 4
ln 4
⇒
1
1
( log5 ) =
( log5 )(ln 2)
( log5 )(ln 2)
Note that log5 (ln 5) =
35. =
·
1
1
1
+ log5 =
+
.
ln 5
( log5 )(ln 5)(ln 2) (ln 2)
1
1 + ln
ln
1
(ln 5) = ln by the change of base formula. Thus, 0 =
+
=
.
ln 5
ln ln 2
ln 2
ln ln 2
⇒ ln = ln
⇒ ln = ln ⇒ 0 = (1) + (ln ) · 1 ⇒ 0 = (1 + ln ) ⇒
0 = (1 + ln )
1
ln ⇒
36. = 1
⇒ ln =
37. = sin
⇒ ln = ln sin
0 =
sin
+ ln cos
⇒
0
1
1
= − 2 ln +
1 − ln
1
⇒ 0 = 1
2
1
0
= (sin ) · + (ln )(cos ) ⇒
sin
+ ln cos
0 = sin
⇒
ln = sin ln ⇒
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°
SECTION 6.4* GENERAL LOGARITHMIC AND EXPONENTIAL FUNCTIONS
√
38. =
√
⇒ ln = ln
⇒ ln = ln 12
39. = (cos )
⇒ ln = ln(cos )
⇒ ln = ln cos ⇒
⇒ ln = 12 ln ⇒
√
0 = 12 + 12 ln ⇒ 0 = 12
(1 + ln )
¤
1
1
1 0
1
= · + ln ·
2
2
631
⇒
1 0
1
=·
· (− sin ) + ln cos · 1 ⇒
cos
sin
⇒ 0 = (cos ) (ln cos − tan )
0 = ln cos −
cos
1
1 0
1
= ln ·
· cos + ln sin ·
sin
ln sin
ln sin
cos
+
⇒ 0 = (sin )ln ln cot +
0 = ln ·
sin
40. = (sin )ln
41. = ln
⇒ ln = ln(sin )ln
⇒ ln = ln ln = (ln )2
42. = (ln )cos
0 = (ln )cos
43. = 10
⇒ ln = ln · ln sin ⇒
⇒
⇒ ln = cos ln(ln ) ⇒
cos
ln
0
= 2 ln
⇒
2 ln
1
⇒ 0 = ln
1
1
0
= cos ·
· + (ln ln )(− sin ) ⇒
ln
− sin ln ln
⇒ 0 = 10 ln 10, so at (1 10), the slope of the tangent line is 101 ln 10 = 10 ln 10, and its equation is
− 10 = 10 ln 10( − 1), or = (10 ln 10) + 10(1 − ln 10).
44. () = cos = (ln )cos
⇒
1
0 () = ln cos ln (− sin ) + cos
cos
− sin ln
= cos
This is reasonable, because the graph shows that increases when
0 () is positive.
45.
4
0
46.
47.
2 =
4
1
15
16
1
2
−
=
=
ln 2
ln 2
ln 2
ln 2
0
5
+ 5 = 16 6 + ln15 5 +
(ln )(ln 10)
1
ln
1
log10
=
=
. Now put = ln , so = , and the expression becomes
ln 10
1 1 2
1
1
=
(ln )2 + .
+ 1 =
ln 10
ln 10 2
2 ln 10
log10
and we get
= 12 ln 10(log10 )2 + .
Or: The substitution = log10 gives =
ln 10
48. Let = 2 . Then = 2 , so
2
2 = 12
2 =
1 2
1 2
+ =
2 + .
2 ln 2
2 ln 2
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°
632
¤
CHAPTER 6 INVERSE FUNCTIONS
49. Let = sin . Then = cos and
50. Let = 2 + 1. Then = 2 ln 2 , so
3sin cos =
3 =
1
1
1
=
ln || + =
ln(2 + 1) + .
ln 2
ln 2
ln 2
2
=
2 +1
1 sin
3
+ =
3
+ .
ln 3
ln 3
0
1
5
2
2
5
−
−
51. =
(2 − 5 ) +
(5 − 2 ) =
+
ln 2
ln 5 −1
ln 5
ln 2 0
−1
0
1
15
2
1
12
5
1
1
−
−
−
+
−
−
−
=
ln 2
ln 5
ln 2
ln 5
ln 5
ln 2
ln 5
ln 2
0
=
1
1
16
−
5 ln 5
2 ln 2
52. Using disks, the volume is =
1
0
1
[10− ]2 = 0 10−2 . To evaluate the integral, we let = −2 ⇒
= −2 , = 0 ⇒ = 0, and = 1 ⇒ = −2, so we have
=−
2
−2
0
10 = −
−2
99
1
10
(10−2 − 1) =
=−
2 ln 10
2
ln
10
200
ln 10
0
53. We see that the graphs of = 2 and = 1 + 3− intersect at ≈ 06. We
let () = 2 − 1 − 3− and calculate 0 () = 2 ln 2 + 3− ln 3, and
using the formula +1 = − ( ) 0 ( ) (Newton’s method), we get
1 = 06, 2 ≈ 3 ≈ 0600967. So, correct to six decimal places, the
solution occurs at = 0600967.
54. =
0 =
⇒ ln = ln
1
1
+ (ln ) · 0 = · · 0 + ln
⇒ 0 ln −
0
= ln −
⇒
ln −
ln −
55. = () = log4 (3 + 2)
√
So −1 () = 3 4 − 2.
56. lim − ln = lim
→0+
⇒ ·
→0+
√
√
⇒ 4 = 3 + 2 ⇒ 3 = 4 − 2 ⇒ = 3 4 − 2. Interchange and : = 3 4 − 2.
ln − ln
2
= lim −(ln ) = 0 since − (ln )2 → −∞ as → 0+ .
→0+
57. If is the intensity of the 1989 San Francisco earthquake, then log10 () = 71
⇒
log10 (16) = log10 16 + log10 () = log10 16 + 71 ≈ 83.
58. Let 1 and 2 be the intensities of the music and the mower. Then 10 log10
1
0
2
= 120 and 10 log10
= 106, so
0
1
1 0
1
2
log10
= log10
= log10
− log10
= 12 − 106 = 14 ⇒
2
2 0
0
0
1
= 1014 ≈ 25.
2
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°
SECTION 6.4* GENERAL LOGARITHMIC AND EXPONENTIAL FUNCTIONS
¤
633
59. We find with the loudness formula from Exercise 58, substituting 0 = 10−12 and = 50:
50 = 10 log10
10−12
respect to : = 10 log10
0
=
0 (50) =
10
ln 10
60. (a) () = 0
⇔ = 10−7 wattm2 . Now we differentiate with
10−12
1
10
1
1
= 10
=
. Substituting = 10−7 , we get
(0 ) ln 10 0
ln 10
⇔ 5 = log10
1
10−7
⇒
10−12
⇔ 105 =
108
dB
≈ 434 × 107
.
ln 10
wattm2
⇒ 0 () = 0 (ln ) = (0 ) ln = () ln
(b) We substitute 0 = 8, = 038 and = 20 into the first expression for 0 () above:
0 (20) = 8(ln 038)(038)20 ≈ −305 × 10−8 .
(c) The average value of the function () between = 0 and = 20 is
20
0
()
1
=
20 − 0
20
20
8(038) =
⇒
61. (a) = log2
2
0
20
2(03820 − 1)
2 (038)
≈ 041.
=
5 ln 038 0
5 ln 038
[ constant] =
1
1
2
=
·
ln 2
2
ln 2
As increases, the rate of change of difficulty decreases.
(b) = log2
2
⇒
[ constant] =
1
−2
1
·
=−
· −2 −2 =
2 ln 2 2
ln 2
2
ln 2
The negative sign indicates that difficulty decreases with increasing width. While the magnitude of the rate of change
1
1
=
decreases with increasing width that is, −
decreases
as
increases
, the rate of change itself
ln 2 ln 2
increases (gets closer to zero from the negative side) with increasing values of .
(c) The answers to (a) and (b) agree with intuition. For fixed width, the difficulty of acquiring a target increases, but less and
less so, as the distance to the target increases. Similarly, for a fixed distance to a target, the difficulty of acquiring the target
decreases, but less and less so, as the width of the target increases.
62. () = (143653 × 109 ) · (101395)
⇒ 0 () = (143653 × 109 ) · (101395) (ln 101395). The units for 0 () are
millions of people per year. The rates of increase for 1920, 1950, and 2000 are 0 (20) ≈ 2625, 0 (50) ≈ 3978, and
0 (100) ≈ 7953, respectively.
63. Half of 760 RNA copies per mL, corresponding to = 1, is 380 RNA copies per mL. Using the graph of in Figure 9, we
estimate that it takes about 35 additional days for the patient’s viral load to decrease to 38 RNA copies per mL.
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°
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¤
CHAPTER 6 INVERSE FUNCTIONS
64. (a)
(b) Using a graphing calculator, we obtain the exponential
curve () = 3689301(106614) .
(c) Using the TRACE and zooming in, we find that the bacteria count
doubles from 37 to 74 in about 1087 hours.
65. (a) = with = 4502714 × 10−20 and = 1029953851,
where is measured in thousands of people. The fit appears to be very good.
(b) For 1800: 1 =
5308 − 3929
7240 − 5308
= 1379, 2 =
= 1932.
1800 − 1790
1810 − 1800
So 0 (1800) ≈ (1 + 2 )2 = 16555 thousand peopleyear.
For 1850: 1 =
23,192 − 17,063
31,443 − 23,192
= 6129, 2 =
= 8251.
1850 − 1840
1860 − 1850
So 0 (1850) ≈ (1 + 2 )2 = 719 thousand peopleyear.
(c) Using 0 () = ln (from Formula 4) with the values of and from part (a), we get 0 (1800) ≈ 15685 thousand
peopleyear and 0 (1850) ≈ 68607. These estimates are somewhat less than the ones in part (b).
(d) (1870) ≈ 41,94656. The difference of 34 million people is most likely due to the Civil War (1861–1865).
66. Let = . Then = , and as → ∞, → ∞.
1
1
= lim 1 +
= lim 1 +
= by Equation 9.
Therefore, lim 1 +
→∞
→∞
→∞
67. =
⇒ 0 = ln , so the slope of the tangent line to the curve = at the point ( ) is ln . An equation of
this tangent line is then − = ln ( − ). If is the ­intercept of this tangent line, then 0 − = ln ( − ) ⇒
−1
−1
= 1 . The distance between ( 0) and ( 0) is | − |, and
= − ⇒ | − | =
−1 = ln ( − ) ⇒
ln
ln |ln |
this distance is the constant
1
for any . [Note: The absolute value is needed for the case 0 1 because ln is
|ln |
negative there. If 1, we can write − = 1(ln ) as the constant distance between ( 0) and ( 0).]
68. =
⇒ 0 = ln , so the slope of the tangent line to the curve = at the point (0 0 ) is 0 ln . An equation
of this tangent line is then − 0 = 0 ln ( − 0 ). Since this tangent line must pass through (0 0), we have
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 6.5
EXPONENTIAL GROWTH AND DECAY
¤
0 − 0 = 0 ln (0 − 0 ), or 0 = 0 (ln ) 0 . Since (0 0 ) is a point on the exponential curve = , we also have
0 = 0 . Equating the expressions for 0 gives 0 = 0 (ln ) 0
⇒ 1 = (ln ) 0
⇒ 0 = 1(ln ).
So 0 = 0 = 0 ln [by combining Property 6.3*.4 with Law 3 of Theorem 6.2*.3] = (1(ln )) ln = 1 = .
69. Using Definition 1 and the second law of exponents for , we have − = ( − ) ln = ln − ln =
ln
= .
ln
70. Using Definition 1, the first law of logarithms, and the first law of exponents for , we have
() = ln() = (ln + ln ) = ln + ln = ln ln = .
71. Let log = and log = . Then = and = .
(a) = = +
(b)
= = −
⇒ log () = + = log + log
⇒ log
(c) = ( ) =
= − = log − log
⇒ log ( ) = = log
6.5 Exponential Growth and Decay
1. The relative growth rate is
1
= 04159, so
= 04159 and by Theorem 2,
() = (0)04159 = 3804159 million cells. Thus, (2) = 3804159(2) 87 million cells.
2. (a) By Theorem 2, () = (0) = 50 . In 20 minutes
1
3
hour , there are 100 cells, so 13 = 503 = 100 ⇒
3 = 2 ⇒ 3 = ln 2 ⇒ = 3 ln 2 = ln(23 ) = ln 8.
(b) () = 50(ln 8) = 50 · 8
(c) (6) = 50 · 86 = 50 · 218 = 13,107,200 cells
(d)
=
(e) () = 106
⇒ 0 (6) = (6) = (ln 8) (6) ≈ 27,255,656 cellsh
⇔ 50 · 8 = 1,000,000 ⇔ 8 = 20,000 ⇔ ln 8 = ln 20,000 ⇔ =
3. (a) By Theorem 2, () = (0) = 50 . Now (15) = 50(15) = 975
15 = ln 195 ⇒
⇒ 15 = 975
50
ln 20,000
≈ 476 h
ln 8
⇒
1
= 15
ln 195 ≈ 19803. So () ≈ 5019803 cells.
(b) Using 19803 for , we get (3) = 5019803(3) = 19,01385 19,014 cells.
(c)
=
⇒ 0 (3) = · (3) = 19803 · 19,014 [from parts (a) and (b)] = 37,6534 37,653 cellsh
(d) () = 5019803 = 250,000 ⇒ 19803 =
=
250,000
50
⇒ 19803 = 5000 ⇒ 19803 = ln 5000 ⇒
ln 5000
430 h
19803
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°
635
636
¤
CHAPTER 6
INVERSE FUNCTIONS
4. (a) () = (0)
⇒ (2) = (0)2 = 400 and (6) = (0)6 = 25,600. Dividing these equations, we get
6 2 = 25,600400 ⇒ 4 = 64 ⇒ 4 = ln 26 = 6 ln 2 ⇒ = 32 ln 2 ≈ 10397, about 104% per hour.
(b) 400 = (0)2
⇒ (0) = 4002
3
⇒ (0) = 4003 ln 2 = 400 ln 2 = 40023 = 50.
(c) () = (0) = 50(32)(ln 2) = 50(ln 2 )(32)
⇒ () = 50(2)15
(d) (45) = 50(2)15(45) = 50(2)675 ≈ 5382 bacteria
(e)
= =
3
ln 2 (50(2)675 ) [from parts (a) and (b)] ≈ 5596 bacteriah
2
(f ) () = 50,000 ⇒ 50,000 = 50(2)15
=
⇒ 1000 = (2)1.5
⇒ ln 1000 = 1.5 ln 2 ⇒
ln 1000
≈ 6.64 h
1.5 ln 2
5. (a) Let the population (in millions) in the year be (). Since the initial time is the year 1750, we substitute − 1750 for in
Theorem 2, so the exponential model gives () = (1750)(−1750) . Then (1800) = 980 = 790(1800−1750)
(50)
980
790 =
⇒ ln 980
790 = 50
⇒
1
⇒ = 50
ln 980
790 ≈ 00043104. So with this model, we have
(1900) = 790(1900−1750) ≈ 1508 million, and (1950) = 790(1950−1750) ≈ 1871 million. Both of these
estimates are much too low.
(b) In this case, the exponential model gives () = (1850)(−1850)
⇒ (1900) = 1650 = 1260(1900−1850)
⇒
1
= (50) ⇒ = 50
ln 1650
≈ 0005393. So with this model, we estimate
ln 1650
1260
1260
(1950) = 1260(1950−1850) ≈ 2161 million. This is still too low, but closer than the estimate of (1950) in part (a).
(c) The exponential model gives () = (1900)(−1900)
⇒ (1950) = 2560 = 1650(1950−1900)
⇒
1
= (50) ⇒ = 50
ln 2560
≈ 0008785. With this model, we estimate
ln 2560
1650
1650
(2000) = 1650(2000−1900) ≈ 3972 million. This is much too low. The discrepancy is explained by the fact that the
world birth rate (average yearly number of births per person) is about the same as always, whereas the mortality rate
(especially the infant mortality rate) is much lower, owing mostly to advances in medical science and to the wars in the first
part of the twentieth century. The exponential model assumes, among other things, that the birth and mortality rates will
remain constant.
6. (a) Let () be the population (in millions) in the year . Since the initial time is the year 1950, we substitute − 1950 for in
Theorem 2, and find that the exponential model gives () = (1950)(−1950)
(1960) = 100 = 83(1960−1950)
⇒
100
= 10
83
⇒
1
⇒ = 10
ln 100
≈ 00186. With this model, we estimate
83
(1980) = 83(1980−1950) = 8330 ≈ 145 million, which is an underestimate of the actual population of 150 million.
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°
SECTION 6.5
(b) As in part (a), () = (1960)(−1960)
¤
EXPONENTIAL GROWTH AND DECAY
⇒ (1980) = 150 = 10020
⇒ 20 = ln 150
100
637
⇒
1
= 20
ln 32 ≈ 00203. Thus, (2000) = 10040 = 225 million, which is an overestimate of the actual population
of 214 million.
(c) As in part (a), () = (1980)(−1980)
⇒ (2000) = 214 = 15020
⇒ 20 = ln 214
150
⇒
1
= 20
ln 214
≈ 00178. Thus, (2010) = 15030 ≈ 256, which is an overestimate of the actual population of
150
243 million.
(d) Using the model in part (c), (2025) = 150(2025−1980) = 15045 334 million. This prediction is likely too high.
The model gave an overestimate for 2010, and the amount of overestimation is likely to compound as time increases.
7. (a) If = [N2 O5 ] then by Theorem 2,
(b) () = −00005 = 09
= −00005
⇒ () = (0)−00005 = −00005 .
⇒ −00005 = 09 ⇒ −00005 = ln 09 ⇒ = −2000 ln 09 ≈ 211 s
8. (a) The mass remaining after days is () = (0) = 50 . Since the half­life is 28 days, (28) = 5028 = 25
28 = 12
⇒ 28 = ln 12
⇒ = −(ln 2)28, so () = 50−(ln 2)28 = 50 · 2−28 .
(b) (40) = 50 · 2−4028 ≈ 186 mg
(c) () = 2 ⇒ 2 = 50 · 2−28
1
(−28) ln 2 = ln 25
⇒
(d)
⇒
−28
2
50 = 2
⇒
1
ln 2 ≈ 130 days
⇒ = −28 ln 25
9. (a) If () is the mass (in mg) remaining after years, then () = (0) = 100 .
(30) = 10030 = 12 (100) ⇒ 30 = 12
⇒ = −(ln 2)30 ⇒ () = 100−(ln 2)30 = 100 · 2−30
(b) (100) = 100 · 2−10030 ≈ 992 mg
1
(c) 100−(ln 2)30 = 1 ⇒ −(ln 2)30 = ln 100
⇒ = −30 lnln001
2 ≈ 1993 years
10. (a) If () is the mass after days and (0) = , then () = .
1
ln 0643. To find the half­life, we set the mass after
(300) = 300 = 0643 ⇒ 300 = 0643 ⇒ = 300
days equal to one­half of the original mass. Hence, (1300)(ln 0643) = 12 ⇔
=
1
(ln 0643) = ln 12
300
300 ln 12
471 days.
ln 0643
(b) (1300)(ln 0643) = 13 ⇔
1
1
300 (ln 0643) = ln 3
⇔ =
300 ln 13
746 days
ln 0643
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°
⇔
638
¤
CHAPTER 6
INVERSE FUNCTIONS
11. Let () be the level of radioactivity. Thus, () = (0)− and is determined by using the half­life:
(5730) = 12 (0) ⇒ (0)−(5730) = 12 (0) ⇒ −5730 = 12
⇒ −5730 = ln 12
If 74% of the 14 C remains, then we know that () = 074(0) ⇒ 074 = −(ln 2)5730
=−
⇒ =−
ln 12
ln 2
=
.
5730
5730
⇒ ln 074 = −
ln 2
5730
⇒
5730(ln 074)
≈ 2489 ≈ 2500 years.
ln 2
12. From Exercise 11, we have the model () = (0)− with = (ln 2)5730. Thus,
(68,000,000) = (0)−68000000 ≈ (0) · 0 = 0. There would be an undetectable amount of 14 C remaining for a
68­million­year­old dinosaur.
Now let () = 01% (0), so 0001(0) = (0)−
=
⇒ 0001 = −
⇒ ln 0001 = − ⇒
ln 0001
ln 0001
=
≈ 57,104, which is the maximum age of a fossil that we could date using 14 C.
−
−(ln 2)5730
13. Let measure time since a dinosaur died in millions of years, and let () be the amount of 40 K in the dinosaur’s bones at
time . Then () = (0)− and is determined by the half­life: (1250) = 12 (0) ⇒ (0)−(1250) = 12 (0) ⇒
−1250 = 12
⇒ −1250 = ln 12
⇒ =−
ln 12
ln 2
=
. To determine if a dinosaur dating of 68 million years is
1250
1250
possible, we find that (68) = (0)−(68) ≈ 0963(0), indicating that about 96% of the 40 K is remaining, which is
clearly detectable. To determine the maximum age of a fossil by using 40 K, we solve () = 01%(0) for .
(0)− = 0001(0) ⇔ − = 0001 ⇔ − = ln 0001 ⇔ =
ln 0001
≈ 12,457 million, or
−(ln 2)1250
12457 billion years.
14. From the information given, we know that
5 = 2(0)
= 2
⇒ = 2 by Theorem 2. To calculate we use the point (0 5):
⇒ = 5. Thus, the equation of the curve is = 52 .
15. (a) Using Newton’s Law of Cooling,
= ( − ), we have
= ( − 22). Now let = − 22 , so
(0) = (0) − 22 = 85 − 22 = 63 , so is a solution of the initial­value problem = with (0) = 63 and by
Theorem 2 we have () = (0) = 63 .
1
43
45
43
1
y(30) = 63e30k = 65 − 22 =⇒ e30k = 43
=⇒ k = 30
ln 43
, so y(t) = 63e 30 t ln 63 and y(45) = 63e 30 ln 63 ≈
63
63
35.5◦ C. Thus, T (45) ≈ 35.5 + 22 = 57.5◦ C.
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 6.5
1
43
1
EXPONENTIAL GROWTH AND DECAY
43
(b) T (t) = 40 =⇒ y(t) = 18. y(t) = 63e 30 t ln 63 = 18 =⇒ e 30 t ln 63 = 18
=⇒
63
30 ln 2
7
ln 43
63
1
t ln 43
30
63
¤
639
= ln 27 =⇒ t =
≈ 98 min.
16. Let () be the temperature of the body hours after 1:30 PM . Then (0) = 325 and (1) = 303. Using Newton’s Law of
Cooling,
= ( − ), we have
= ( − 20). Now let = − 20, so (0) = (0) − 20 = 325 − 20 = 125,
so is a solution to the initial value problem = with (0) = 125 and by Theorem 2 we have
() = (0) = 125 .
(1) = 303 − 20 ⇒ 103 = 125(1)
⇒ = ln 103
12.5 . The murder occurred when
⇒ = 103
12.5
17
() = 37 − 20 ⇒ 12.5 = 17 ⇒ = 12.5
17
⇒ = ln 12.5
17
ln 103
⇒ = ln 12.5
≈ −1588 h
12.5
≈ −95 minutes. Thus, the murder took place about 95 minutes before 1:30 PM, or 11:55 AM.
17.
= ( − 20). Letting = − 20, we get
= , so () = (0) . (0) = (0) − 20 = 5 − 20 = −15, so
(25) = (0)25 = −1525 , and (25) = (25) − 20 = 10 − 20 = −10, so −1525 = −10 ⇒ 25 = 23 . Thus,
1
ln 23 , so () = (0) = −15(125) ln(23) . More simply, 25 = 23
25 = ln 23 and = 25
=
2 25
3
⇒ () = −15 ·
2 25
3
(a) (50) = 20 + (50) = 20 − 15 ·
.
2 5025
3
(b) 15 = () = 20 + () = 20 − 15 ·
= 20 − 15 ·
2 25
3
⇒ 15 ·
2 2
3
2 25
3
2 125
3
◦
= 20 − 20
3 = 133̄ C
=5 ⇒
⇒ = 25 ln 13 ln 23 ≈ 6774 min.
(25) ln 23 = ln 13
18.
⇒ =
2 25
3
= 13
⇒
= ( − 20). Let = − 20. Then
= , so () = (0) (0) = (0) − 20 = 95 − 20 = 75,
so () = 75 . When () = 70,
= −1◦ Cmin. Equivalently,
= −1 when () = 50. Thus,
= () = 50 and 50 = () = 75 . The first relation implies = −150, so the second relation says
⇒ = −50 ln 23 ≈ 2027 min.
50 = 75−50 . Thus, −50 = 23 ⇒ −50 = ln 23
−1 =
19. (a) Let () be the pressure at altitude . Then =
⇒ () = (0) = 1013 .
1
⇒ = 1000
⇒
ln 8714
(1000) = 10131000 = 8714 ⇒ 1000 = ln 8714
1013
1013
1
8714
8714
() = 1013 1000 ln( 1013 ) , so (3000) = 10133 ln( 1013 ) ≈ 645 kPa.
6187
8714
(b) (6187) = 1013 1000 ln( 1013 ) ≈ 399 kPa
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
⇒
640
¤
CHAPTER 6
INVERSE FUNCTIONS
20. (a) Using = 0 1 +
with 0 = 2500, = 0045, and = 3, we have:
(i) Annually: = 1
(ii) Quarterly: = 4
(iii) Monthly: = 12
(iv) Weekly: = 52
(v) Daily: = 365
(vi) Hourly: = 365 · 24
(vii) Continuously:
1·3
0045
= 2500 1 +
= $285292
1
4·3
0045
= $285919
= 2500 1 +
4
12·3
0045
= $286062
= 2500 1 +
12
52·3
0045
= $286117
= 2500 1 +
52
365·3
0045
= $286132
= 2500 1 +
365
365·24·3
0045
= $286134
= 2500 1 +
365 · 24
3
= 2500(0045) = $286134
(b)
21. (a) Using = 0 1 +
with 0 = 4000, = 00175, and = 5, we have:
(i) Annually: = 1
(ii) Semiannually: = 2
(iii) Monthly: = 12
(iv) Weekly: = 52
(v) Daily: = 365
(vi) Continuously:
1·5
00175
= $436247
= 4000 1 +
1
2·5
00175
= $436411
= 4000 1 +
2
12·5
00175
= $436549
= 4000 1 +
12
52·5
00175
= $436570
= 4000 1 +
52
365·5
00175
= $436576
= 4000 1 +
365
= 4000(00175)5 = $436577
(b) = 00175 and (0) = 4000.
22. (a) 0 003 = 20
⇔ 003 = 2 ⇔ 003 = ln 2 ⇔ = 100
ln 2 ≈ 2310, so the investment will double in
3
about 2310 years.
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 6.6
INVERSE TRIGONOMETRIC FUNCTIONS
¤
641
(b) The annual interest rate in = 0 (1 + ) is . From part (a), we have = 0 003 . These amounts must be equal,
so (1 + ) = 003
⇒ 1 + = 003
⇒ = 003 − 1 ≈ 00305 = 305%, which is the equivalent annual
interest rate.
APPLIED PROJECT Controlling Red Blood Cell Loss During Surgery
1. Let () be the volume of RBCs (in liters) at time (in hours). Since the total volume of blood is 5 L, the concentration of
RBCs is 5. The patient bleeds 2 L of blood in 4 hours, so
2
1
=−
·
=−
4 5
10
From Section 6.5, we know that = has solution () = (0) . In this case, (0) = 45% of 5 = 94 and
1
= − 10
, so () = 94 −10 . At the end of the operation, the volume of RBCs is (4) = 94 −04 ≈ 151 L.
2. Let be the volume of blood that is extracted and replaced with saline solution. Let () be the volume of RBCs with the
9
9
ANH procedure. Then (0) is 45% of (5 − ), or 20
(5 − ), and hence () = 20
(5 − )−10 . We want
(4) ≥ 25% of 5 ⇔
9
(5 − )−04 ≥ 54
20
⇔ ≤ 5 − 25
04 ≈ 086 L. To maximize the
9
⇔ 5 − ≥ 25
04
9
effect of the ANH procedure, the surgeon should remove 086 L of blood and replace it with saline solution.
3. The RBC loss without the ANH procedure is (0) − (4) = 94 − 94 −04 ≈ 074 L. The RBC loss with the ANH procedure is
9
9
9
(5 − ) − 20
(5 − )−04 = 20
(5 − )(1 − −04 ). Now let = 5 − 25
04 [from Problem 2] to
(0) − (4) = 20
9
9
9
5 − 5 − 25
04 (1 − 04 ) = 20
· 25
04 (1 − 04 ) = 54 (04 − 1) ≈ 061 L. Thus, the ANH
get (0) − (4) = 20
9
9
procedure reduces the RBC loss by about 074 − 061 = 013 L (about 44 fluid ounces).
6.6 Inverse Trigonometric Functions
1. (a) sin−1 (05) = 6 because sin 6 = 05 and 6 is in the interval − 2 2 (the range of sin−1 ).
(b) cos −1 (−1) = because cos = −1 and is in the interval [0 ] (the range of cos−1 ).
√
2. (a) tan−1 3 = 3 because tan 3 =
√
3 and 3 is in the interval − 2 2 (the range of tan−1 ).
(b) sec−1 2 = 3 because sec 3 = 2 and 3 is in 0 2 ∪ 3
(the range of sec−1 ).
2
√
3. (a) csc −1 2 = 4 because csc 4 =
(b) cos−1
√
(the range of csc−1 ).
2 and 4 is in 0 2 ∪ 3
2
√
√
32 = 6 because cos 6 = 32 and 6 is in [0 ].
√
√
4. (a) cot−1 − 3 = 5
because cot 5
= − 3 and 5
is in (0 ) (the range of cot−1 ).
6
6
6
(b) arcsin 1 = 2 because sin 2 = 1 and 2 is in − 2 2 (the range of arcsin).
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
642
¤
CHAPTER 6
INVERSE FUNCTIONS
5. (a) In general, tan(arctan ) = for any real number . Thus, tan(arctan 10) = 10.
√
√
(b) arcsin(sin(54)) = arcsin −1 2 = − 4 because sin − 4 = −1 2 and − 4 is in − 2 2 .
6. (a) tan−1 tan
3
4
7. Let = sin−1 23
= tan−1 (−1) = −
4
√3
1
(b) cos arcsin
= cos
=
2
6
2
[see the figure].
2
Then tan sin−1 23 = tan = √ .
5
8. Let = arccos 35
[see the figure].
Then csc arccos 35 = csc = 54 .
5
9. Let = sin−1 13
[see the figure].
5
cos 2 sin−1 13
= cos 2 = cos2 − sin2
2 5 2
25
119
− 13 = 144
= 12
13
169 − 169 = 169
10. Let = tan−1 2 and = tan−1 3. Then
cos tan−1 2 + tan−1 3 = cos( + ) = cos cos − sin sin
1 1
2 3
= √ √ −√ √
5 10
5 10
−5
−1
−5
= √ = √ = √
50
5 2
2
11. Let = sin−1 . Then − 2 ≤ ≤ 2
⇒ cos ≥ 0, so cos(sin−1 ) = cos =
√
1 − sin2 = 1 − 2 .
12. Let = sin−1 . Then sin = , so from the triangle (which
illustrates the case 0), we see that
.
tan(sin−1 ) = tan = √
1 − 2
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 6.6
INVERSE TRIGONOMETRIC FUNCTIONS
¤
643
13. Let = tan−1 . Then tan = , so from the triangle (which
illustrates the case 0), we see that
.
sin(tan−1 ) = sin = √
1 + 2
14. Let = arccos . Then cos = , so from the triangle (which
illustrates the case 0), we see that
sin(2 arccos ) = sin 2 = 2 sin cos
√
√
= 2( 1 − 2 )() = 2 1 − 2
The graph of sin−1 is the reflection of the graph of
15.
sin about the line = .
The graph of tan−1 is the reflection of the graph of
16.
tan about the line = .
17. Let = cos−1 . Then cos = and 0 ≤ ≤
⇒ − sin
=1 ⇒
1
1
1
= −√
=−
= −
. [Note that sin ≥ 0 for 0 ≤ ≤ .]
sin
1 − 2
1 − cos2
18. (a) Let = sin−1 and = cos−1 . Then cos =
sin =
√
1 − 2 . So
√
1 − sin2 = 1 − 2 since cos ≥ 0 for − 2 ≤ ≤ 2 . Similarly,
sin(sin−1 + cos−1 ) = sin( + ) = sin cos + cos sin = · +
√
√
1 − 2 1 − 2
= 2 + (1 − 2 ) = 1
, and so sin−1 + cos−1 = 2 .
But − 2 ≤ sin−1 + cos−1 ≤ 3
2
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°
644
¤
CHAPTER 6
INVERSE FUNCTIONS
(b) We differentiate sin−1 + cos−1 = 2 with respect to , and get
1
√
(cos−1 ) = 0 ⇒
+
1 − 2
19. Let = cot−1 . Then cot =
1
(cos−1 ) = − √
.
1 − 2
⇒ − csc2
=1 ⇒
1
1
1
=− 2 =−
=−
.
csc
1 + cot2
1 + 2
. Differentiate with respect to :
20. Let = sec−1 . Then sec = and ∈ 0 2 ∪ 3
2
sec tan
tan =
1
1
1
= √
=
=
. Note that tan2 = sec2 − 1 ⇒
sec tan
2 − 1
sec sec2 − 1
=1 ⇒
sec2 − 1 since tan 0 when 0 2 or 3
2 .
21. Let = csc−1 . Then csc =
⇒ − csc cot
=1 ⇒
1
1
1
=− √
=−
=−
. Note that cot ≥ 0 on the domain of csc−1 .
csc cot
2 − 1
csc csc2 − 1
22. = tan−1 (2 )
⇒
0 =
1
2
1
2
( ) =
·
· 2 =
1 + (2 )2
1 + 4
1 + 4
5
1
(5) = √
·
1 − 252
1 − (5)2
23. () = sin−1 (5) ⇒ 0 () =
√
1
1
√
24. () = arccos ⇒ 0 () = −
√ 2 = − √1 −
1−
25. = (tan−1 )2
⇒ 0 = 2(tan−1 )1 ·
26. () = sec−1 ( ) ⇒ 0 () =
27. = tan−1
0 =
√
−1 ⇒
1 −12
2
1
=− √ √
2 1−
1
2 tan−1
(tan−1 ) = 2 tan−1 ·
=
2
1+
1 + 2
1
1
1
( ) = √
· = √
·
2
2
2
−
1
−1
( ) − 1
1
1
1
1
1
√
1
= · √
= √
· √
−1 =
√
2 ·
1 + ( − 1) 2 − 1
2 −1
2 − 1
1+
−1
28. () = ln(arctan(4 ))
0 () =
=
⇒
1
1
1
4
· (arctan(4 )) =
·
( )
·
arctan(4 )
arctan(4 ) 1 + (4 )2
1
1
43
·
· 43 =
4
8
8
arctan( ) 1 +
(1 + ) arctan(4 )
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 6.6
29. = arctan(cos )
⇒ 0 =
INVERSE TRIGONOMETRIC FUNCTIONS
1
sin
(− sin ) = −
1 + (cos )2
1 + cos2
√
⇒
1 + 2
30. = tan−1 −
¤
√ 2
1
+1−
1
√
√
√
1
−
=
√
2
2 + 1
1 + 2 − 2 2 + 1 + 2 + 1
2 + 1
1 + − 2 + 1
√
√
√
2 + 1 −
2 + 1 −
2 + 1 −
√
√
=
√
= √
=
2
2
2
2
2
2
2
2 1+ − +1
+1
2 + 1 (1 + ) − ( + 1)
2 (1 + ) 2 + 1 −
0 =
=
1
2(1 + 2 )
2
2
31. () = arcsin( )
32. () = arcsin(1)
2
⇒ 0 () = arcsin( ) ·
⇒
1
1
1
() =
=
1 − (1)2
1 − 12
0
1
1
√
= −
=−
| | 2 − 1
2 (2 − 1)
33. () = cot−1 () + cot−1 (1)
0 () = −
2
1
2arcsin( )
[arcsin( 2 )] = arcsin( ) ·
· 2 = √
2
2
1 − 4
1 − ( )
1
1
1
1
√ = −√
− 2 = −
4 − 2
1 − 12 4
⇒
1
1
1
1
1
2
1
1
=
−
·
−
= 0.
−
·
−
+ 2
=−
1 + 2
1 + (1)2
1 + 2
2 + 1
2
1 + 2
+1
Note that this makes sense because () =
34. = cos−1 (sin−1 )
35. = sin−1 +
⇒
√
1 − 2
3
for 0 and () =
for 0.
2
2
1
1
1
sin−1 = −
0 = −
·
·√
−1 2
−1 2
1 − 2
1 − (sin )
1 − (sin )
⇒
1
1
0 = · √
+ (sin−1 )(1) + (1 − 2 )−12 (−2) = √
+ sin−1 − √
= sin−1
2
1 − 2
1 − 2
1 − 2
12
1−
1−
= arctan
⇒
1+
1+
12
−12
1 1−
1
1−
1
(1 + )(−1) − (1 − )(1)
0 =
·
=
·
2 · 1 +
1− 2 1+
(1 + )2
1−
1+
1+
1
+
1+
12
1 1+
1
−2
1 + 1 (1 + )12
−2
· ·
·
=
·
=
·
1− 2 1−
1+
(1 + )2
2
2 (1 − )12 (1 + )2
+
1+
1+
−1
−1
=
= √
2(1 − )12 (1 + )12
2 1 − 2
36. = arctan
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
645
646
¤
CHAPTER 6
37. = tan−1
0 =
=
INVERSE FUNCTIONS
+ ln
1 −
−
= tan−1
+ ln
⇒
+
2
+
( + ) · 1 − ( − ) · 1
1
1
2
1
1
1
1 +
·
·
=
2 · + · − ·
2 +
2
( + )2
2 − ( + )2
1+
+
+
1
2
+
·
+
= 2
+ 2
( − )( + )
+ 2
− 2
38. () = arcsin( )
1
⇒ 0 () =
· = √
.
1 − 2
1 − ( )2
Domain() = { | −1 ≤ ≤ 1} = { | 0 ≤ 1} = (−∞ 0].
Domain( 0 ) = | 1 − 2 0 = | 2 1 = { | 2 0} = (−∞ 0).
1
2
⇒ 0 () = −
(−2) =
.
2
1
−
(3
− 2)2
1 − (3 − 2)
39. () = cos−1 (3 − 2)
Domain() = { | −1 ≤ 3 − 2 ≤ 1} = { | −4 ≤ −2 ≤ −2} = { | 2 ≥ ≥ 1} = [1 2].
Domain(0 ) = | 1 − (3 − 2)2 0 = | (3 − 2)2 1 = { | |3 − 2| 1}
= { | −1 3 − 2 1} = { | −4 −2 −2} = { | 2 1} = (1 2)
40.
tan−1(2 ) =
( + 2 ) ⇒
1
(2 0 + · 2) = 1 + · 2 0 + 2 · 1 ⇒
1 + (2 )2
2
2
2
2
0
0
2
0
−
2
=
1
+
−
⇒
−
2
= 1 + 2 −
1 + 4 2
1 + 4 2
1 + 4 2
1 + 4 2
0
=
1 + 2 −
2
1 + 4 2
2
− 2
1 + 4 2
41. () = sin−1
+
4
0 (2) = sin−1 12 = 6
42. = 3 arccos
2
or 0 =
√
16 − 2
⇒
1 + 4 2 + 2 + 4 4 − 2
2 − 2 − 25 3
⇒ 0 () = sin−1
4
+
⇒
−√
= sin−1
4
16 − 2
4 1 − (4)2
√
1
3
1
⇒ 0 = 3 −
= − 3. An equation of the tangent
, so at (1 ), 0 = −
2
1 − (2)2
2 1− 1
4
√
√
√
line is − = − 3 ( − 1), or = − 3 + + 3.
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 6.6
43. () =
INVERSE TRIGONOMETRIC FUNCTIONS
√
√
−12
1
1
arcsin
1 − 2
1 − 2 arcsin ⇒ 0 () = 1 − 2 · √
+ arcsin ·
(−2) = 1 − √
2
1 − 2
1 − 2
Note that 0 = 0 where the graph of has a horizontal tangent. Also note
that 0 is negative when is decreasing and 0 is positive when is
increasing.
44. () = arctan(2 − )
⇒ 0 () =
1
2
2 − 1
( − ) =
·
1 + (2 − )2
1 + (2 − )2
Note that 0 = 0 where the graph of has a horizontal tangent. Also note
that 0 is negative when is decreasing and 0 is positive when is
increasing.
45.
¤
lim sin−1 = sin−1 (−1) = − 2
→−1+
1
1 + 2
1 + 2
12 + 1
→ .
As → ∞, =
=
2
2
1 + 2
1 + 2
12 + 2
2
1 + 2
= lim arccos = arccos 12 = 3 .
lim arccos
→∞
→12
1 + 22
46. Let =
47. Let = . As → ∞, → ∞. lim arctan( ) = lim arctan = 2 by (8).
→∞
→∞
48. Let = ln . As → 0+ , → −∞. lim tan−1 (ln ) = lim tan−1 = − 2 by (8).
→0+
→−∞
49. If = −1 (), then () = . Differentiating implicitly with respect to and remembering that is a function of ,
we get 0 ()
50. (4) = 5
1
= 1, so
= 0
()
⇒
−1 0
() =
0
⇒ −1 (5) = 4. By Exercise 83, −1 (5) =
1
.
0 ( −1 ())
1
1
3
1
= 0
=
= .
0 ( −1 (5))
(4)
23
2
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°
647
648
¤
CHAPTER 6
INVERSE FUNCTIONS
110
,
=
,
10
1 − (10)2
110
210
=
=
(2) rads,
=
rads = 14 rads
= 6
1 − (10)2
1 − (610)2
= 2 fts, sin =
10
51.
⇒ = sin−1
= 4 revmin = 8 · 60 radh. From the diagram, we see that tan =
⇒ = tan−1
.
3
3
2
13
Thus, 8 · 60 =
kmh, and
=
=
.
So
=
8
·
60
·
3
1
+
1 + (3)2
3
52.
at = 1,
= 8 · 60 · 3 1 + 19 kmh = 1600 kmh.
53. = () = sin−1 (( + 1))
A. = { | −1 ≤ ( + 1) ≤ 1}. For −1 we have − − 1 ≤ ≤ + 1 ⇔
2 ≥ −1 ⇔ ≥ − 12 , so = − 12 ∞ B. Intercepts are 0 C. No symmetry
D. lim sin−1
→∞
+1
= lim sin−1
→∞
1
1 + 1
= sin−1 1 = 2 , so = 2 is a HA.
1
1
( + 1) −
√
0,
E. 0 () =
=
( + 1) 2 + 1
1 − [( + 1)]2 ( + 1)2
so is increasing on − 12 ∞ F. No local maximum or minimum,
− 12 = sin−1 (−1) = − 2 is an absolute minimum
H.
√
√
2 + 1 + ( + 1)/ 2 + 1
( + 1)2 (2 + 1)
3 + 2
0 on , so is CD on − 12 ∞ .
=−
2
32
( + 1) (2 + 1)
G. 00 () = −
54. = () = tan−1
−1
+1
A. = { | 6= −1} B. ­intercept = 1, ­intercept = (0) = tan−1 (−1) = − 4
E. 0 () =
lim tan−1
−1
+1
1 − 1
→±∞
→±∞
1 + 1
−1
−1
= − and lim tan−1
= .
Also lim tan−1
+
−
+
1
2
+
1
2
→−1
→−1
C. No symmetry D.
= lim tan−1
= tan−1 1 = 4 , so = 4 is a HA.
1
2
1
( + 1) − ( − 1)
0,
=
= 2
1 + [( − 1)( + 1)]2
( + 1)2
( + 1)2 + ( − 1)2
+1
so is increasing on (−∞ −1) and (−1 ∞) F. No extreme values
H.
2
G. 00 () = −2 2 + 1 0 ⇔ 0, so is CU on (−∞ −1)
and (−1 0), and CD on (0 ∞). IP at 0 − 4
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°
SECTION 6.6
55. = () = − tan−1
INVERSE TRIGONOMETRIC FUNCTIONS
¤
649
A. = B. Intercepts are 0 C. (−) = −(), so the curve is symmetric about the
origin. D. lim ( − tan−1 ) = ∞ and lim ( − tan−1 ) = −∞, no HA.
→∞
→−∞
But () − − 2 = − tan−1 + 2 → 0 as → ∞, and
() − + 2 = − tan−1 − 2 → 0 as → −∞, so = ± 2 are
slant asymptotes. E. 0 () = 1 −
1
2 + 1
=
2
2 + 1
H.
0, so is
increasing on . F. No extrema
G. 00 () =
(1 + 2 )(2) − 2 (2)
2
=
0 ⇔ 0, so
(1 + 2 )2
(1 + 2 )2
is CU on (0, ∞), CD on (−∞, 0). IP at (0, 0).
56. = () = arctan
C. No symmetry D.
A. = B. ­intercept: (0) = 0 = 1; no ­intercept since arctan is positive for all .
lim () = −2 [≈ 021], so = −2 is a HA.
→−∞
lim () = 2 [≈ 481], so = 2 is a
→∞
1
. 0 () 0 for all , so is increasing on . F. No extreme values
1 + 2
1
(1 + 2 )arctan
− arctan (2)
1 + 2
00
H.
G. () =
(1 + 2 )2
HA. E. 0 () = arctan
=
arctan (1 − 2)
(1 + 2 )2
00 () 0 for 12 , so is CU on −∞ 12 and is CD on 12 ∞ .
IP at 12 arctan 12 ≈ (05 159)
57. () = arctan(cos(3 arcsin )). We use a CAS to compute 0 and 00 , and to graph , 0 , and 00 :
From the graph of 0 , it appears that the only maximum occurs at = 0 and there are minima at = ±087. From the graph
of 00 , it appears that there are inflection points at = ±052.
58. First note that the function () = − sin−1 is only defined on the interval [−1 1], since sin−1 is only defined on that
√
interval. We differentiate to get 0 () = 1 − 1 − 2 . Now if ≤ 0, then 0 () ≥ 1, so there is no extremum and is
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
¤
650
CHAPTER 6
INVERSE FUNCTIONS
increasing on its domain. If 1, then 0 () 0, so there is no local extremum and is decreasing on its domain, and if
= 1, then there is still no extremum, since 0 () does not change sign at = 0. So we can only have local extrema if
√
√
0 1. In this case, is increasing where 0 () 0 ⇔
1 − 2 ⇔ || 1 − 2 , and decreasing where
√
√
√
1 − 2 || ≤ 1. has a maximum at = 1 − 2 and a minimum at = − 1 − 2 .
59. () =
2(2 + 1) + 3
3
22 + 5
=
= 2+ 2
2
+1
2 + 1
+1
⇒ () = 2 + 3 tan−1 +
2
implies that 0 () is defined for in (−1 1) and () = 2 sin−1 + for −1 1. By continuity, we
1 − 2
can extend the domain of to [−1 1]. Now (1) = 5 ⇒ 2 sin−1 1 + = 5 ⇒ = 5 − 2 2 = 5 − , so
60. 0 () = √
() = 2 sin−1 + 5 − .
61.
√3
4
√3
8
−
=
8
=
=
8
arctan
=
8
√
√
2
3
6
6
3
1 3
1 3 1 +
1√2
1
6
−1
−1 1
−1
√ − sin
=6
−√
− −
=6
= 3
62.
= 6 sin
√ = 6 sin
√
4
4
2
−1 2
2
2
1 − 2
−1 2
1√2
. When = 0, = 0; when = 12 , = 6 . Thus,
1 − 2
63. Let = sin−1 , so = √
12
0
sin−1
√
=
1 − 2
6
=
0
2
2
6
0
=
2
.
72
64. Let = 4. Then = 4 , so
√34
0
1
=
1 + 162
4
√3
0
√
√3
1
= 14 tan−1 0 = 14 tan−1 3 − tan−1 0 = 14 3 − 0 = 12
.
2
1+
65. Let = 1 + 2 . Then = 2 , so
1
1
2
= tan−1 + 12 ln|| +
+
= tan−1 +
2
2
1+
1+
= tan−1 + 12 ln1 + 2 + = tan−1 + 12 ln(1 + 2 ) + [since 1 + 2 0].
1+
=
1 + 2
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 6.6
INVERSE TRIGONOMETRIC FUNCTIONS
¤
651
66. Let = − cos . Then = sin , so
2
0
sin
=
1 + cos2
0
0
1
= tan−1 −1 = tan−1 0 − tan−1 (−1) = 0 − − 4 = 4 .
2
1
+
−1
67. Let = arctan . Then =
1
, so
2 + 1
68. Let = arctan . Then =
1
=
(2 + 1) arctan
1
√
=
2 − 4
2 = 13 3 + = 13 (arctan )3 + .
1
= ln || + = ln |arctan | + .
1
, so
1 − 2
1
1
, so
= 2
1 + 2
+1
69. Let = arcsin . Then = √
70. Let = 12 . Then = 12
(arctan )2
=
2 + 1
arcsin
√
=
1 − 2
1
2
√
=
2
4 2 − 1
⇒
1
=
2 (2)2 − 1
71. Let = 3 . Then = 32 and
2
√
=
1 − 6
= + = arcsin + .
√
= 12 sec−1 + = 12 sec−1 12 + .
2
−1
1
√3
= 13 sin−1 + = 13 sin−1 (3 ) + .
1 − 2
72. Let = , so = . When = 0, = 1; when = 1, = . Thus,
1
0
=
1 + 2
73. Let =
1
1
= tan−1 = tan−1 − tan−1 1 = tan−1 − 4 .
2
1+
1
√
√
2
√
. Then = √ and
= 2 tan−1 + = 2 tan−1 + .
=
2
1
+
2
(1 + )
74. Let = 2 . Then = 2 , so
=
1 + 4
1
2
1 + 2
= 12 tan−1 + = 12 tan−1 (2 ) + .
75. Let = . Then = , so
1
√
=
2
− 2
1
=
1 − ()2
√
+ .
= sin−1 + = sin−1
1 − 2
2
2
2
1
1
. By Formula 14, this is equal to
√
=
2
2 + 4
0
0 +4
2
2
12 tan−1 (2) 0 = 2 4 − 0 = 8 .
76. We use the disk method: =
77.
The integral represents the area below the curve = sin−1 on the interval
∈ [0 1]. The bounding curves are = sin−1 ⇔ = sin , = 0 and
= 1. We see that ranges between sin−1 0 = 0 and sin−1 1 = 2 . So we have to
integrate the function = 1 − sin between = 0 and = 2 :
1
0
sin−1 =
2
0
(1 − sin ) =
2
+ cos 2 − (0 + cos 0) = 2 − 1.
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
652
¤
CHAPTER 6
INVERSE FUNCTIONS
78. Let = arctan and = arctan . Then by the addition formula for the tangent (see Reference Page 2 in the textbook),
tan(arctan ) + tan(arctan )
+
tan + tan
=
⇒ tan( + ) =
1 − (tan )(tan )
1 − tan(arctan ) tan(arctan )
1 −
+
, since − 2 arctan + arctan 2 .
arctan + arctan = + = arctan
1 −
tan( + ) =
79. (a) arctan 12 + arctan 13 = arctan
1
+ 13
2
= arctan 1 =
4
1 − 12 · 13
(b) 2 arctan 13 + arctan 17 = arctan 13 + arctan 13 + arctan 17 = arctan
= arctan 34 + arctan 17 = arctan
80. (a) () = sin(sin−1 )
⇒
1
+ 13
3
+ arctan 17
1 − 13 · 13
3
+ 17
4
= arctan 1 =
4
1 − 34 · 17
(b) () = sin−1 (sin )
cos
1
cos
sin−1 (sin ) =
cos = √
=
2
2
|cos
|
cos
1 − sin
(c) 0 () =
(d) () = cos−1 (sin ), so
cos
cos
.
=−
0 () = −
|cos |
1 − sin2
Notice that () = 2 − () because
sin−1 + cos−1 = 2 for all .
2
4
2
4
√
−
= √
−
=0
2
2
1−
1−
2 1 − 2
1 − (1 − 22 )2
81. Let () = 2 sin−1 − cos−1 (1 − 22 ). Then 0 () = √
[since ≥ 0]. Thus 0 () = 0 for all ∈ [0 1). Thus () = . To find let = 0. Thus
2 sin−1 (0) − cos−1 (1) = 0 = . Therefore we see that () = 2 sin−1 − cos−1 (1 − 22 ) = 0 ⇒
2 sin−1 = cos−1 (1 − 22 ).
82. Let () = arcsin
−1
+1
√
− 2 arctan + 2 . Note that the domain of is [0 ∞). Thus
1
1
2
( + 1) − ( − 1)
1
1
· √ = √
−
−√
= 0. Then () = .
2
2
(
+
1)
1
+
2
(
+
1)
(
+ 1)
−1
1−
+1
0 () =
To find , we let = 0 ⇒ arcsin(−1) − 2 arctan 0 + 2 =
√
−1
= 2 arctan − 2 .
arcsin
+1
⇒ − 2 − 0 + 2 = 0 = . Thus, () = 0 ⇒
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
APPLIED PROJECT
83. = sec−1
⇒ sec = ⇒ sec tan
=1 ⇒
¤
653
1
=
. Now tan2 = sec2 − 1 = 2 − 1, so
sec tan
√
tan = ± 2 − 1. For ∈ 0 2 , ≥ 1, so sec = = || and tan ≥ 0 ⇒
For ∈
WHERE TO SIT AT THE MOVIES
√
2 − 1 ⇒
1
1
√
= √
=
.
2
−1
|| 2 − 1
2 , ≤ −1, so || = − and tan = −
1
1
1
1
√
√
=
=
= √
=
sec tan
− 2 − 1
(−) 2 − 1
|| 2 − 1
84. (a) Since |arctan(1)| 2 , we have 0 ≤ | arctan(1)| ≤ 2 || → 0 as → 0. So, by the Squeeze Theorem,
lim () = 0 = (0), so is continuous at 0.
→0
() − (0)
arctan(1) − 0
1
(b) Here
=
= arctan
. So (see Exercise 62 in Section 2.2 for a discussion of left­ and
−0
1
() − (0)
0
= lim arctan
= lim arctan = − , while
right­hand derivatives) −
(0) = lim
→−∞
−0
2
→0−
→0−
1
() − (0)
0
+
= lim arctan
(0) = lim
= lim arctan = . So 0 (0) does not exist.
→∞
−0
2
→0+
→0+
APPLIED PROJECT
Where to Sit at the Movies
1. |V P | = 2.7 + x cos α, |P T | = 10.5 − (1.2 + x sin α) = 9.3 − x sin α, and |P B| =
(1.2 + x sin α) − 3 = x sin α − 1.8. So, using the Pythagorean Theorem, we have
p
p
|V T | = |V P |2 + |P T |2 = (2.7 + x cos α)2 + (9.3 − x sin α)2 = a, and
|V B| =
p
p
|V P |2 + |P B|2 = (2.7 + x cos α)2 + (x sin α − 1.8)2 = b. Using the Law
of Cosines on 4V BT , we get
7.52 = a2 + b2 − 2ab cos θ ⇔ θ = arccos
2
a + b2 − 56.25
, as required.
2ab
2. From the graph of θ, it appears that the value of x which maximizes θ is x ≈ 2.3 m. Assuming
that the first row is at x = 0, the row closest to this value of x is the fourth row, at x = 2.7 m,
and from the graph, the viewing angle in this row seems to be about 0.84 radians, or about
48.2 ◦.
3. With a CAS,we type in the definition of θ, substitute in the proper values of a and b in terms of x and α = 20◦ = π9 radians,
and then use the differentiation command to find the derivative. We use a numerical root finder and find that the root of the
equation dθ/dx = 0 is x ≈ 2.3, as approximated in Problem 2.
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
654
¤
CHAPTER 6
INVERSE FUNCTIONS
4. From the graph in Problem 2, it seems that the average value of the function on the interval [0,18] is about 0.6. We can use a
1
CAS to approximate 18
´ 18
0
θ(x) dx ≈ 0.606 ≈ 34.7 ◦. (The calculation is much faster if we reduce the number of digits of
accuracy required.) The minimum value is θ(18) ≈ 0.36
. and, from Problem 2, the maximum value is about 0.84.
6.7 Hyperbolic Functions
1. (a) sinh 0 = 12 (0 − −0 ) = 0
2. (a) tanh 0 =
(b) cosh 0 = 12 (0 + −0 ) = 12 (1 + 1) = 1
(0 − −0 )2
=0
(0 + −0 )2
(b) tanh 1 =
1 − −1
2 − 1
≈ 076159
=
1 + −1
2 + 1
3. (a) cosh(ln 5) = 12 (ln 5 + −ln 5 ) = 12 5 + (ln 5 )−1 = 12 (5 + 5−1 ) = 12 5 + 15 = 13
5
(b) cosh 5 = 12 (5 + −5 ) ≈ 7420995
4. (a) sinh 4 = 12 (4 − −4 ) ≈ 2728992
(b) sinh(ln 4) = 12 (ln 4 − −ln 4 ) = 12 4 − (ln 4 )−1 = 12 (4 − 4−1 ) = 12 4 − 14 = 15
8
5. (a) sech 0 =
1
1
= =1
cosh 0
1
(b) cosh−1 1 = 0 because cosh 0 = 1.
6. (a) sinh 1 = 12 (1 − −1 ) ≈ 117520
√
√
(b) Using Equation 3, we have sinh−1 1 = ln 1 + 12 + 1 = ln 1 + 2 ≈ 088137.
8
8
5
5
13 3 −
+ −
− −
+5
= − − + + − =
−
7. 8 sinh + 5 cosh = 8
2
2
2
2
2
2
2
2
8. 22 + 3−2 = sinh 2 + cosh 2
22 + 3−2 =
2
2
+ −2
− −2
+
⇒
⇒ 22 + 3−2 =
2
2
2 −2 2 −2
−
+ +
2
2
2
2
⇒ 22 + 3−2 =
+ 2 − + −2
+
2
2
⇒
− +
+
= 2 and
= 3 ⇒ + = 4 and − + = 6 ⇒ 2 = 10 ⇒ = 5 and = −1.
2
2
Thus, 22 + 3−2 = − sinh 2 + 5 cosh 2.
9. sinh(ln ) =
1
−1
1
1 2 − 1
2 − 1
1 ln
1
1
= ( − −1 ) =
(
− ln
−
=
=
− − ln ) =
2
2
2
2
2
2
1
1
4
−4
1 ln 4
= (4 + −4 )
4 + ln
+ − ln =
2
2
2
8
8
1 +1
+1
1 4
1
=
=
+ 4 =
2
2
4
24
10. cosh(4 ln ) = cosh(ln 4 ) =
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 6.7
HYPERBOLIC FUNCTIONS
¤
11. sinh(−) = 12 [− − −(−) ] = 12 (− − ) = − 12 (− − ) = − sinh
12. cosh(−) = 12 [− + −(−) ] = 12 (− + ) = 12 ( + − ) = cosh
13. cosh + sinh = 12 ( + − ) + 12 ( − − ) = 12 (2 ) =
14. cosh − sinh = 12 ( + − ) − 12 ( − − ) = 12 (2− ) = −
15. sinh cosh + cosh sinh =
1
1
−
)
2 ( −
−
)
2 ( +
+
1
1
−
)
2 ( +
−
)
2 ( −
= 14 [(+ + − − −+ − −− ) + (+ − − + −+ − −− )]
= 14 (2+ − 2−− ) = 12 [+ − −(+) ] = sinh( + )
16. cosh cosh + sinh sinh =
1
2
( + − ) 12 ( + − ) + 12 ( − − ) 12 ( − − )
= 14 (+ + − + −+ + −− ) + (+ − − − −+ + −− )
= 14 (2+ + 2−− ) = 12 + + −(+) = cosh( + )
17. Divide both sides of the identity cosh2 − sinh2 = 1 by sinh2 :
cosh2
sinh2
1
−
=
2
2
sinh
sinh
sinh2
⇔ coth2 − 1 = csch2 .
cosh sinh
sinh cosh
+
sinh cosh + cosh sinh
sinh( + )
cosh cosh
cosh cosh
=
=
18. tanh( + ) =
sinh sinh
cosh cosh
cosh( + )
cosh cosh + sinh sinh
+
cosh cosh
cosh cosh
tanh + tanh
=
1 + tanh tanh
19. Putting = in the result from Exercise 15, we have
sinh 2 = sinh( + ) = sinh cosh + cosh sinh = 2 sinh cosh .
20. Putting = in the result from Exercise 16, we have
cosh 2 = cosh( + ) = cosh cosh + sinh sinh = cosh2 + sinh2 .
21. tanh(ln ) =
22.
(ln − − ln )2
− (ln )−1
(2 − 1)
sinh(ln )
− 1
2 − 1
− −1
= ln
=
= 2
= 2
=
=
−
ln
ln
−1
−1
cosh(ln )
(
+
)2
+ ( )
+
+ 1
( + 1)
+1
1
( + − ) + 12 ( − − )
1 + (sinh ) cosh
cosh + sinh
1 + tanh
=
=
= 21
= − = 2
1
−
−
1 − tanh
1 − (sinh ) cosh
cosh − sinh
( + ) − 2 ( − )
2
Or: Using the results of Exercises 13 and 14,
cosh + sinh
= − = 2
cosh − sinh
23. By Exercise 13, (cosh + sinh ) = ( ) = = cosh + sinh .
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
655
656
¤
CHAPTER 6
24. coth =
1
tanh
INVERSE FUNCTIONS
1
1
13
=
=
.
tanh
1213
12
⇒ coth =
sech2 = 1 − tanh2 = 1 −
12 2
13
25
= 169
5
⇒ sech = 13
[sech, like cosh, is positive].
1
13
=
.
513
5
cosh =
1
sech
⇒ cosh =
tanh =
sinh
cosh
⇒ sinh = tanh cosh ⇒ sinh =
csch =
1
sinh
⇒ csch =
1
5
=
.
125
12
25. sech =
1
cosh
⇒ sech =
1
3
= .
53
5
cosh2 − sinh2 = 1 ⇒ sinh2 = cosh2 − 1 =
1
3
= .
43
4
csch =
1
sinh
⇒ csch =
tanh =
sinh
cosh
⇒ tanh =
43
4
= .
53
5
coth =
1
tanh
⇒ coth =
1
5
= .
45
4
5 2
3
12 13
12
·
=
.
13 5
5
− 1 = 16
9
⇒ sinh = 43
[because 0].
26. (a)
27. (a) lim tanh = lim
− −
→∞ + −
→∞
·
−
1−0
1 − −2
=1
= lim
=
−
→∞
1 + −2
1+0
0−1
− −
2 − 1
=
= −1
· = lim 2
−
→−∞ +
→−∞
+1
0+1
(b) lim tanh = lim
→−∞
(c) lim sinh = lim
→∞
→∞
− −
=∞
2
− −
= −∞
→−∞
2
(d) lim sinh = lim
→−∞
(e) lim sech = lim
→∞
(f ) lim coth = lim
→∞
2
→∞ + −
+ −
→∞ − −
=0
·
−
1+0
1 + −2
= 1 [Or: Use part (a).]
= lim
=
→∞ 1 − −2
−
1−0
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 6.7
(g) lim coth = lim
cosh
(h) lim coth = lim
cosh
(i) lim csch = lim
2
→0+ sinh
→0+
→0− sinh
→0−
¤
= ∞, since sinh → 0 through positive values and cosh → 1.
= −∞, since sinh → 0 through negative values and cosh → 1.
→−∞ − −
→−∞
HYPERBOLIC FUNCTIONS
=0
1−0
1
sinh
− −
1 − −2
=
=
=
lim
=
lim
→∞
→∞
→∞
2
2
2
2
(j) lim
28. (a)
1
(cosh ) =
( + − ) = 12 ( − − ) = sinh
2
(b)
(tanh ) =
sinh
cosh
(c)
(csch ) =
1
sinh
=−
cosh
1
cosh
·
= − csch coth
=−
sinh sinh
sinh2
(d)
(sech ) =
1
cosh
=−
sinh
1
sinh
·
= − sech tanh
=−
2
cosh
cosh
cosh
(e)
(coth ) =
cosh
sinh
=
=
cosh cosh − sinh sinh
cosh2 − sinh2
1
=
=
= sech2
2
cosh
cosh2
cosh2
sinh sinh − cosh cosh
sinh2 − cosh2
1
=
=−
= − csch2
sinh2
sinh2
sinh2
29. Let = sinh−1 . Then sinh = and, by Example 1(a), cosh2 − sinh2 = 1
⇒ [with cosh 0]
√
√
cosh = 1 + sinh2 = 1 + 2 . So by Exercise 13, = sinh + cosh = + 1 + 2 ⇒
√
= ln + 1 + 2 .
√
cosh2 − 1 = 2 − 1. So, by Exercise 13,
√
√
= cosh + sinh = + 2 − 1 ⇒ = ln + 2 − 1 .
30. Let = cosh−1 . Then cosh = and ≥ 0, so sinh =
Another method: Write = cosh = 12 + − and solve a quadratic, as in Example 3.
31. (a) Let = tanh−1 . Then = tanh =
2
1+ =
2
−
1
1+
= ln
.
2
1−
( − − )2
sinh
2 − 1
=
· = 2
−
cosh
( + )2
+1
2
2
⇒ 1 + = (1 − ) ⇒
1+
=
1−
⇒ 2 + = 2 − 1 ⇒
1+
⇒ 2 = ln
⇒
1−
(b) Let = tanh−1 . Then = tanh , so from Exercise 22 we have
1+
1+
1+
1 + tanh
1
=
⇒ 2 = ln
2 =
⇒ = ln
.
1 − tanh
1−
1−
2
1−
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
657
658
¤
CHAPTER 6
INVERSE FUNCTIONS
32. (a) (i) = csch−1
⇔ csch = ( 6= 0)
(ii) We sketch the graph of csch−1 by reflecting the graph of csch (see Exercise 26)
about the line = .
2
⇒ − − = 2 ⇒
− −
√
1 ± 2 + 1
2
. But 0, so for 0,
( ) − 2 − = 0 ⇒ =
√
√
√
1 + 2 + 1
1 − 2 + 1
2 + 1
1
−1
and for 0, =
. Thus, csch = ln
+
.
=
||
(iii) Let = csch−1 . Then = csch =
(b) (i) = sech−1 ⇔ sech = and 0
(ii) We sketch the graph of sech−1 by reflecting the graph of sech (see Exercise 26)
about the line = .
2
⇒ + − = 2 ⇒
+ −
√
1 ± 1 − 2
. But 0 ⇒ 1.
( )2 − 2 + = 0 ⇔ =
√
√
√
1 − 1 − 2
1 ⇔ 1 − 1 − 2 ⇔ 1 − 1 − 2
This rules out the minus sign because
(iii) Let = sech−1 , so = sech =
1 − 2 + 2 1 − 2 ⇔ 2 ⇔ 1, but = sech ≤ 1.
√
√
1 + 1 − 2
1 + 1 − 2
⇒ sech−1 = ln
Thus, =
.
(c) (i) = coth−1 ⇔ coth =
(ii) We sketch the graph of coth−1 by reflecting the graph of coth (see Exercise 26)
about the line = .
(iii) Let = coth−1 . Then = coth =
− − = + −
2 = ln
+1
−1
+ −
− −
⇒
⇒ ( − 1) = ( + 1)−
⇒ coth−1 =
⇒ 2 =
1 +1
ln
2 −1
33. (a) Let = cosh−1 . Then cosh = and ≥ 0
1
1
1
= √
=
=
sinh
2 − 1
cosh2 − 1
⇒ sinh
⇒
=1 ⇒
[since sinh ≥ 0 for ≥ 0]. Or: Use Formula 4.
(b) Let = tanh−1 . Then tanh = ⇒ sech2
=1 ⇒
Or: Use Formula 5.
(c) Let = coth−1 . Then coth = ⇒ − csch2
by Exercise 17.
+1
−1
=1 ⇒
1
1
1
=
=
=
.
1 − 2
sech2
1 − tanh2
1
1
1
=−
=
=
1 − 2
csch2
1 − coth2
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
⇔
SECTION 6.7
34. (a) Let = sech−1 . Then sech =
⇒ − sech tanh
HYPERBOLIC FUNCTIONS
¤
659
=1 ⇒
1
1
1
=−
=−
=− √
. [Note that 0 and so tanh 0.]
2
sech tanh
1 − 2
sech 1 − sech
(b) Let = csch−1 . Then csch = ⇒ − csch coth
=1 ⇒
1
=−
. By Exercise 17,
csch coth
√
√
coth = ± csch2 + 1 = ± 2 + 1. If 0, then coth 0, so coth = 2 + 1. If 0, then coth 0,
√
1
1
=−
=− √
.
so coth = − 2 + 1. In either case we have
csch coth
|| 2 + 1
0 () = sinh(3) ·
(3) = sinh(3) · 3 = 3 sinh 3
35. () = cosh 3
⇒
36. () = cosh
⇒ 0 () = sinh + (cosh ) = (sinh + cosh ), or, using Exercise 13, ( ) = 2 .
37. () = sinh(2 )
⇒ 0 () = cosh(2 )
PR
2
38. () = sinh2 = (sinh )
39. () = sinh(ln )
⇒ 0 () = 2(sinh )1
⇒ 0 () = cosh(ln )
Or: () = sinh(ln ) =
40. () = ln(sinh )
(2 ) = 2 cosh(2 )
(sinh ) = 2 sinh cosh , or, using Exercise 19, sinh 2.
1
1 ln
1
1
1 2 + 1
2 + 1
ln =
+ − ln
=
+
=
=
2
2
2
22
2 + 1
1
1
1 ln
1
1
( − − ln ) =
−
⇒ 0 () =
1+ 2 =
2
2
2
22
⇒ 0 () =
1
1
sinh =
cosh = coth
sinh
sinh
√ √
√
√
41. () = tanh ⇒ 0 () = sech2
= sech2
42. () = tanh 2
⇒ 0 () = tanh 2 ·
1
√
2
√
sech2
√
=
2
tanh 2 = tanh 2 · sech2 (2) · 2 = 2tanh 2 sech2 (2)
PR
43. = sech tanh
⇒ 0 = sech · sech2 + tanh · (− sech tanh ) = sech3 − sech tanh2
44. = sech(tanh )
⇒ 0 = − sech(tanh ) tanh(tanh ) ·
45. () = coth
(tanh ) = − sech(tanh ) tanh(tanh ) · sech2
√
PR
2 + 1 ⇒
√
√
√
√
2
0 () = − csch2 2 + 1 12 (2 + 1)−12 · 2 + coth 2 + 1 (1) = coth 2 + 1 − √
csch2 2 + 1
2
+1
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
660
¤
CHAPTER 6
INVERSE FUNCTIONS
1 + sinh
1 − sinh
46. () =
0 () =
=
QR
⇒
(1 − sinh ) cosh − (1 + sinh )(− cosh )
cosh − sinh cosh + cosh + sinh cosh
=
(1 − sinh )2
(1 − sinh )2
2 cosh
(1 − sinh )2
2
1
(−2) = − √
⇒ 0 () =
·
1 + 42
1 + (−2)2
47. () = sinh−1 (−2)
48. () = tanh−1 (3 )
⇒ 0 () =
49. = cosh−1 (sec )
⇒
1
32
(3 ) =
·
1 − (3 )2
1 − 6
1
1
1
0 = √
(sec ) = √
· sec tan
·
· sec tan =
tan
sec2 − 1
tan2
50. = sech −1 (sin )
[since 0 ≤ 2] = sec
⇒
1
1
√
· cos
(sin ) = −
·
2
sin
cos2
sin 1 − sin
1
1
=−
· cos [since 0 2] = −
= − csc
sin · cos
sin
0 = −
51. () = cosh−1
√
1 + 2
⇒
√
1
1
2
0 () = √
·
1 + 2 =
= √
· √
√
2
2
2) − 1
2 ·
2
1
+
(1
+
1 + 2
2
1+
−1
= √
1 + 2
[since 0]
1
= √
1 + 2
√
1 − 2 = tanh−1 + 12 ln(1 − 2 ) ⇒
1
1
(−2) = tanh−1
+
0 = tanh−1 +
1 − 2
2 1 − 2
52. = tanh−1 + ln
53. = sinh−1 (3) −
0 = sinh−1
54.
55.
√
9 + 2
⇒
13
2
+
+√
− √
= sinh−1
−√
= sinh−1
2
2
2
2
3
3
3
2 9+
9+
9+
1 + (3)
1 + (sinh ) cosh
cosh + sinh
1 + tanh
=
=
= − [by Exercises 13 and 14] = 2 , so
1 − tanh
1 − (sinh ) cosh
cosh − sinh
√
2
4 1 + tanh
1
4
4 1 + tanh
= 2 = 2 . Thus,
=
( ) = 2 .
1 − tanh
1 − tanh
2
1
sech2
1 cosh2
arctan(tanh ) =
(tanh
)
=
=
1 + (tanh )2
1 + tanh2
1 + (sinh2 ) cosh2
=
1
1
[by Exercise 20] = sech 2
=
cosh 2
cosh2 + sinh2
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 6.7
HYPERBOLIC FUNCTIONS
¤
661
56. (a) Let = 003291765. A graph of the central curve,
= () = 21149 − 2096 cosh , is shown.
(b) (0) = 21149 − 2096 cosh 0 = 21149 − 2096(1) = 19053 m
(c) = 100 ⇒ 100 = 21149 − 2096 cosh ⇒
2096 cosh = 11149 ⇒ cosh =
= ± cosh−1
11149
2096
11149
2096
⇒
1
11149
⇒ = ± cosh−1
≈ ±7156 m. The points are approximately (±7156 100).
2096
(d) () = 21149 − 2096 cosh ⇒ 0 () = −2096 sinh · .
1
1
11149
11149
11149
= −2096 sinh ± cosh−1
= −2096 sinh ± cosh−1
≈ ∓36.
0 ± cosh−1
2096
2096
2096
So the slope at (7156 100) is about −36 and the slope at (−7156 100) is about 36.
2
2
gets large, and from Figure 5 or Exercise 27(a), tanh
2
tanh
(1) =
.
approaches 1. Thus, =
≈
2
2
2
57. As the depth of the water gets large, the fraction
For = cosh() with 0, we have the ­intercept equal to .
58.
As increases, the graph flattens.
59. (a) = 20 cosh(20) − 15
1
⇒ 0 = 20 sinh(20) · 20
= sinh(20). Since the right pole is positioned at = 7,
7
≈ 03572.
we have 0 (7) = sinh 20
7
, so
(b) If is the angle between the tangent line and the ­axis, then tan = slope of the line = sinh 20
7
= tan−1 sinh 20
≈ 0343 rad ≈ 1966◦ . Thus, the angle between the line and the pole is = 90◦ − ≈ 7034◦ .
60. We differentiate the function twice, then substitute into the differential equation: =
=
sinh
= sinh
cosh
⇒
2
=
cosh
. We evaluate the two sides
= cosh
2
2
2
cosh
and
RHS
=
=
cosh
,
=
1
+
=
1 + sinh2
separately: LHS =
2
⇒
by the identity proved in Example 1(a).
61. (a) From Exercise 60, the shape of the cable is given by = () =
­axis, so the lowest point is (0 (0)) =
cosh
. The shape is symmetric about the
0
and the poles are at = ±100. We want to find when the lowest
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°
662
¤
CHAPTER 6
INVERSE FUNCTIONS
kg­m
= 60 ⇒ = 60 = (60 m)(2 kgm)(98 ms2 ) = 1176
, or 1176 N (newtons).
s2
· 100
100
cosh
The height of each pole is (100) =
= 60 cosh
≈ 16450 m.
60
point is 60 m, so
(b) If the tension is doubled from to 2 , then the low point is doubled since
2
= 120. The height of the
= 60 ⇒
2
· 100
100
cosh
= 120 cosh
≈ 16413 m, just a slight decrease.
poles is now (100) =
2
120
62. (a) lim () = lim
→∞
→∞
tanh
=
lim tanh
=
·1
→∞
(b) Belly­to­earth:
= 98, = 0515, = 60, so the terminal velocity is
Feet­first:
= 98, = 0067, = 60, so the terminal velocity is
63. (a) = sinh + cosh
as → ∞,
→ ∞
=
60(98)
≈ 3379 ms.
0515
60(98)
≈ 9368 ms.
0067
⇒ 0 = cosh + sinh ⇒
00 = 2 sinh + 2 cosh = 2 ( sinh + cosh ) = 2
(b) From part (a), a solution of 00 = 9 is () = sinh 3 + cosh 3. Now −4 = (0) = sinh 0 + cosh 0 = , so
= −4. Also, 0 () = 3 cosh 3 − 12 sinh 3, so 6 = 0 (0) = 3 ⇒ = 2. Thus, = 2 sinh 3 − 4 cosh 3.
1
1
1 ln(sec +tan )
sec + tan +
+ − ln(sec +tan ) =
2
2
sec + tan
sec − tan
1
sec − tan
1
sec + tan +
=
sec + tan +
=
2
(sec + tan )(sec − tan )
2
sec2 − tan2
64. cosh = cosh[ln(sec + tan )] =
= 12 (sec + tan + sec − tan ) = sec
√
⇒ = sinh−1 1 = ln 1 + 2 , by Equation 3.
65. The tangent to = cosh has slope 1 when 0 = sinh = 1
Since sinh = 1 and = cosh =
√
√ √
1 + sinh2 , we have cosh = 2. The point is ln 1 + 2 , 2 .
66. () = tanh( sin ), where is a positive integer. Note that ( + 2) = (); that is, is periodic with period 2.
Also, from Figure 3, −1 tanh 1, so we can choose a viewing rectangle of [0 2] × [−1 1]. From the graph, we see
that () becomes more rectangular looking as increases. As becomes
large, the graph of approaches the graph of = 1 on the intervals
(2 (2 + 1)) and = −1 on the intervals ((2 − 1) 2).
67. Let = cosh . Then = sinh , so
68. Let = 1 + 4. Then = 4 , so
sinh cosh2 =
sinh(1 + 4) = 14
2 = 13 3 + = 13 cosh3 + .
sinh = 14 cosh + = 14 cosh(1 + 4) + .
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°
SECTION 6.7
69. Let =
√
. Then = √ and
2
70. Let = cosh . Then = sinh , and
71.
cosh
=
cosh2 − 1
cosh
=
sinh2
HYPERBOLIC FUNCTIONS
√
√
sinh
√
= sinh · 2 = 2 cosh + = 2 cosh + .
tanh =
sinh
=
cosh
1
cosh
·
=
sinh sinh
= ln || + = ln(cosh ) + .
coth csch = − csch +
72. Let = 2 + tanh . Then = sech2 , so
sech2
=
2 + tanh
1
= ln || + = ln |2 + tanh | + = ln (2 + tanh ) + [since 2 + tanh 1].
73. Let = 3. Then = 3 and
6
2
2
2
√
or
= cosh−1
= cosh−1 2 − cosh−1 43
2 −1
43
43
√
√
2
√
√
2
6+3 3
4+ 7
−1
2
√
= ln
= ln + − 1
= ln 2 + 3 − ln
= cosh
3
43
43
4+ 7
1
√
=
2 − 9
4
1
√
3 =
2 −9
9
43
74. Let = 4. Then = 4 and
1
0
4
1
4
4
√
1
1
√4
sinh−1 =
ln + 2 + 1
=
= 14 ln 4 + 17 − ln 1
2
4
4
0
0
+1
0
√
= 14 ln 4 + 17
1
√
=
162 + 1
75. Let = . Then = and
1 +
1
+
.
or ln
2
1 −
76. We want
1
0
=
1 − 2
= tanh−1 + = tanh−1 +
1 − 2
sinh = 1. To calculate the integral, we put = ,
so = , the upper limit becomes , and the equation becomes
1
1
[cosh − 1] = 1 ⇔ cosh − 1 = .
sinh = 1 ⇔
0
We plot the function () = cosh − − 1, and see that its positive root
1
lies at approximately = 162. So the equation 0 sinh = 1 holds
for ≈ 162.
77. (a) From the graphs, we estimate
that the two curves = cosh 2
and = 1 + sinh intersect at
= 0 and at = ≈ 0481.
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°
¤
663
664
¤
CHAPTER 6
INVERSE FUNCTIONS
(b) We have found the two solutions of the equation cosh 2 = 1 + sinh to be = 0 and = ≈ 0481. Note from the
first graph that 1 + sinh cosh 2 on the interval (0 ), so the area between the two curves is
=
0
(1 + sinh − cosh 2) = + cosh − 12 sinh 2 0
= [ + cosh − 12 sinh 2] − [0 + cosh 0 − 12 sinh 0] ≈ 00402
78. The area of the triangle with vertices , , and (cosh 0) is 12 sinh cosh , and the area under the curve 2 − 2 = 1, from
= 1 to = cosh , is
cosh √
2 − 1 . Therefore, the area of the shaded region is
1
cosh √
2 − 1 . So, by FTC1,
() = 12 sinh cosh − 1
√
0 () = 12 cosh2 + sinh2 − cosh2 − 1 sinh = 12 cosh2 + sinh2 − sinh2 sinh
= 12 cosh2 + sinh2 − sinh2 = 12 cosh2 − sinh2 = 12 (1) = 12
Thus () = 12 + , since 0 () = 12 . To calculate , we let = 0. Thus,
cosh 0 √
(0) = 12 sinh 0 cosh 0 − 1
2 − 1 = 12 (0) +
⇒
= 0. Thus () = 12 .
79. If + − = cosh( + )
[or sinh( + )], then
+ − = 2 + ± −− = 2 ± − − = 2 ± 2 − − . Comparing coefficients of
and − , we have = 2 (1) and = ± 2 − (2). We need to find and . Dividing equation (1) by equation (2)
gives us = ±2
=
⇒ = 12 ln ± . Solving equations (1) and (2) for gives us
⇒ () 2 = ln ±
2
2
and = ± , so
=±
2
2
⇒ 2 = ±4 ⇒ = 2
√
±.
() If 0, we use the + sign and obtain a cosh function, whereas if 0, we use the − sign and obtain a sinh
function.
In summary, if and have the same sign, we have + − = 2
opposite sign, then + − = 2
√
− sinh + 12 ln − .
√
cosh + 12 ln , whereas, if and have the
6.8 Indeterminate Forms and l'Hospital's Rule
H
Note: The use of l’Hospital’s Rule is indicated by an H above the equal sign: =
1. (a) lim
()
(b) lim
()
(c) lim
()
→ ()
→ ()
→ ()
is an indeterminate form of type
0
.
0
= 0 because the numerator approaches 0 while the denominator becomes large.
= 0 because the numerator approaches a finite number while the denominator becomes large.
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°
SECTION 6.8
(d) If lim () = ∞ and () → 0 through positive values, then lim
INDETERMINATE FORMS AND L’HOSPITAL’S RULE
()
→ ()
→
and () = 2 .] If () → 0 through negative values, then lim
()
→ ()
¤
665
= ∞. [For example, take = 0, () = 12 ,
= −∞. [For example, take = 0, () = 12 ,
and () = −2 .] If () → 0 through both positive and negative values, then the limit might not exist. [For example,
take = 0, () = 12 , and () = .]
(e) lim
()
→ ()
is an indeterminate form of type
∞
.
∞
2. (a) lim [ ()()] is an indeterminate form of type 0 · ∞.
→
(b) When is near , () is large and () is near 1, so ()() is large. Thus, lim [()()] = ∞.
→
(c) When is near , () and () are both large, so ()() is large. Thus, lim [()()] = ∞.
→
3. (a) When is near , () is near 0 and () is large, so () − () is large negative. Thus, lim [ () − ()] = −∞.
→
(b) lim [ () − ()] is an indeterminate form of type ∞ − ∞.
→
(c) When is near , () and () are both large, so () + () is large. Thus, lim [ () + ()] = ∞.
→
()
4. (a) lim [ ()]
→
is an indeterminate form of type 00 .
(b) If = [ ()]() , then ln = () ln (). When is near , () → ∞ and ln () → −∞, so ln → −∞.
Therefore, lim [ ()]() = lim = lim ln = 0, provided is defined.
→
→
→
(c) lim [()]() is an indeterminate form of type 1∞ .
→
(d) lim [ ()] () is an indeterminate form of type ∞0 .
→
(e) If = [ ()]() , then ln = () ln (). When is near , () → ∞ and ln () → ∞, so ln → ∞. Therefore,
lim [ ()]() = lim = lim ln = ∞.
→
→
→
(f ) lim () () = lim [ ()]1() is an indeterminate form of type ∞0 .
→
→
5. From the graphs of and , we see that lim () = 0 and lim () = 0, so l’Hospital’s Rule applies.
→2
→2
0
lim ()
0 (2)
18
9
() H
0 ()
→2
=
=
= 4 =
= lim 0
→2 ()
→2 ()
lim 0 ()
0 (2)
4
5
lim
→2
6. From the graphs of and , we see that lim () = 0 and lim () = 0, so l’Hospital’s Rule applies.
→2
→2
0
lim ()
0 (2)
15
3
() H
0 ()
→2
=
= 0
=
=−
= lim 0
0
→2 ()
→2 ()
lim ()
(2)
−1
2
lim
→2
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°
666
¤
CHAPTER 6
INVERSE FUNCTIONS
7. and = − 1 are differentiable and 0 = 6= 0 on an open interval that contains 0.
lim () = 0 and lim () = 0,
→0
→0
so we have the indeterminate form 00 and can apply l’Hospital’s Rule.
1
() H
0 ()
=
lim
= =1
→0 − 1
→0
1
lim
Note that lim 0 () = 1 since the graph of has the same slope as the line = at = 0.
→0
8. This limit has the form 00 .
lim
−3
→3 2 − 9
−3
= lim
→3 ( + 3)( − 3)
= lim
1
→3 + 3
=
1
1
=
3+3
6
Note: Alternatively, we could apply l’Hospital’s Rule.
9. This limit has the form 00 .
2 − 2 − 8
( − 4)( + 2)
= lim
= lim ( + 2) = 4 + 2 = 6
→4
→4
→4
−4
−4
lim
Note: Alternatively, we could apply l’Hospital’s Rule.
10. This limit has the form 00 .
lim
→−2
3 + 8 H
32
= 3(−2)2 = 12
= lim
→−2 1
+2
Note: Alternatively, we could factor and simplify.
11. This limit has the form 00 .
lim
7
7 − 1 H
76
= lim
=
→1 32
3
→1 3 − 1
Note: Alternatively, we could factor and simplify.
12. This limit has the form 00 .
√
√
√
−2
−2
+2
−4
1
1
1
√
= lim √
= lim
·√
=
= lim
= √
→4 − 4
→4 − 4
→4
4
4+2
+ 2 →4 ( − 4) + 2
+2
lim
Note: Alternatively, we could apply l’Hospital’s Rule.
√
√
2
2
√
+
sin − cos H
cos + sin
2
2 = 2
=
lim
13. This limit has the form 00 .
lim
= lim
√ 2
→4
→4 tan − 1
→4
sec2
2
2
14. This limit has the form 00 .
3(1)2
3
tan 3 H
3 sec2 3
=
=
= lim
→0 sin 2
→0 2 cos 2
2(1)
2
15. This limit has the form 00 .
2(1)
2 − 1 H
22
=
=2
= lim
→0 sin
→0 cos
1
16. This limit has the form 00 .
17. This limit has the form 00 .
lim
lim
lim
2
→0 1 − cos
lim
H
= lim
2
→0 sin
= lim
2
→0 (sin )
=
2
=2
1
cos 0
1
sin( − 1) H
cos( − 1)
=
=
= lim
→1 32 + 1
3(1)2 + 1
4
→1 3 + − 2
18. The limit can be evaluated by substituting for .
lim
1 + cos
→ 1 − cos
=
1 + (−1)
0
= =0
1 − (−1)
2
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°
SECTION 6.8
INDETERMINATE FORMS AND L’HOSPITAL’S RULE
¤
667
1
√
√
1
H
√ =0
19. This limit has the form ∞
. lim
= lim 2 = lim
∞
→∞ 1 +
→∞
→∞ · 2
1
+ 2 H
1 + 2 H
2
=− .
= lim
= lim
→∞ 1 − 22
→∞ −4
→∞ −4
2
20. This limit has the form ∞
∞ . lim
1
+1
1
0+1
+ 2
A better method is to divide the numerator and the denominator by : lim
=− .
= lim
=
2
1
→∞ 1 − 2
→∞
0−2
2
−2
2
2
21. lim [(ln )] = −∞ since ln → −∞ as → 0+ and dividing by small values of just increases the magnitude of the
→0+
quotient (ln ). L’Hospital’s Rule does not apply.
22. This limit has the form ∞
∞.
1
√
1
ln
1
2 ln H
2
= lim
lim
= lim
= lim =
=0
→∞
→∞
→∞
→∞ 42
2
2
2
23. This limit has the form 00 .
1
1
ln(3) H
1(3) · (13)
−
=
lim
=−
lim
= lim
→3 3 −
→3
→3
−1
3
24. This limit has the form 00 .
8
8 − 5 H
8 ln 8 − 5 ln 5
= ln 8 − ln 5 = ln
= lim
→0
→0
1
5
lim
25. This limit has the form 00 .
√
√
1
(1 + 2)−12 · 2 − 12 (1 − 4)−12 (−4)
1 + 2 − 1 − 4 H
= lim 2
→0
→0
1
2
2
1
1
+√
= lim √
= √ +√ =3
→0
1 + 2
1 − 4
1
1
lim
26. This limit has the form ∞
∞.
1
1
1
10 · 10
10 · 10
10 · 10
10 H
H 1
H 1
1
= 6000
= lim
= 30
lim
= 600
lim
lim 10 = ∞
3
2
→∞
→∞
→∞
→∞
→∞
3
2
1
lim
27. This limit has the form 00 .
28. This limit has the form 00 .
1+1
+ − − 2 H
− − H
+ −
=2
=
lim
=
lim
=
→0 − − 1
→0 − 1
→0
1
lim
lim
→0
1
sinh − H
cosh − 1 H
sinh H
cosh
=
= lim
= lim
= lim
→0
→0
→0
3
32
6
6
6
sech2 0
1
tanh H
sech 2
=
= =1
= lim
2
→0 tan
→0 sec
sec2 0
1
29. This limit has the form 00 . lim
30. This limit has the form 00 .
cos
sin
1
− sin H
1 − cos H
−(− sin )
sin
= − lim
= lim
= lim
lim
→0 − tan
→0 1 − sec2
→0 −2 sec (sec tan )
2 →0
sec2
= − 12 lim cos3 = − 12 (1)3 = − 12
→0
sin
.
tan
→0
1−
Another method is to write the limit as lim
1−
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
668
¤
CHAPTER 6
INVERSE FUNCTIONS
31. This limit has the form 00 .
√
1
sin−1 H
1 1 − 2
1
= lim √
= lim
= =1
→0
→0
→0
1
1
1 − 2
lim
32. This limit has the form ∞
.
∞
33. This limit has the form 00 .
lim
→∞
(ln )2 H
2(ln )(1)
ln H
1
= 2 lim
= 2(0) = 0
= lim
= 2 lim
→∞
→∞
→∞ 1
1
3
lim
H
= lim
→0 3 − 1
→0
1
3 ln 3 + 3
3 ( ln 3 + 1)
ln 3 + 1
= lim
= lim
=
→0
→0
3 ln 3
3 ln 3
ln 3
ln 3
34. This limit has the form 00 .
1+1+2
4
+ − − 2 cos H
− − + 2 sin H
+ − + 2 cos
=
= =2
= lim
= lim
→0
→0
→0 − sin + cos + cos
sin
cos + sin
0+1+1
2
lim
35. This limit can be evaluated by substituting 0 for .
36. This limit has the form 00 .
lim
ln(1 + )
lim
→0 cos + − 1
=
ln 1
0
= =0
1+1−1
1
cos 0
1
sin( − 1) H
cos( − 1) + sin( − 1)
=
=
= lim
→1
4 − 1
4−1
3
→1 22 − − 1
0
37. This limit has the form ∞
, so l’Hospital’s Rule doesn’t apply. As → 0+ , arctan 2 → 0 and ln → −∞,
so lim
→0+
arctan 2
= 0.
ln
38. This limit has the form 00 .
2 sin H
2 cos + 2 sin H
−2 sin + 2 cos + 2 cos + 2 sin
= lim
= lim
→0 sin −
→0
→0
cos − 1
− sin
lim
(2 − 2 ) sin + 4 cos
→0
− sin
= lim
H
= lim
→0
39. This limit has the form 00 .
40. This limit has the form 00 .
2(1) − 0 − 0 + 4(1)
(2 − 2 ) cos − 2 sin − 4 sin + 4 cos
=
= −6
− cos
−1
− 1
→1 − 1
H
lim
(1)
−1
=
=
→1 −1
(1)
[for 6= 0] = lim
lim
−
H
→∞ (2) − tan−1
= lim
→∞
−−
1 + 2 H
2 H
2
= lim
= lim = lim = 0
1
→∞
→∞
→∞
−
1 + 2
41. This limit has the form 00 .
cos − 1 + 12 2 H
1
− sin + H
− cos + 1 H
sin H
cos
=
= lim
= lim
= lim
= lim
→0
→0
→0
→0 24
→0 24
4
43
122
24
lim
42. This limit has the form 00 .
lim
− sin H
= lim
→0 sin(2 )
1 − cos
→0 cos(2 ) · 2 + sin(2 )
H
= lim
1 − cos
= lim
→0 22 cos(2 ) + sin(2 )
sin
→0 22 [− sin(2 ) · 2] + cos(2 ) · 4 + cos(2 ) · 2
H
= lim
= lim
sin
→0 6 cos(2 ) − 43 sin(2 )
cos
→0 6 [− sin(2 ) · 2] + cos(2 ) · 6 − 43 cos(2 ) · 2 − sin(2 ) · 122
=
1
1
=
0+1·6−0−0
6
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°
SECTION 6.8
INDETERMINATE FORMS AND L’HOSPITAL’S RULE
¤
669
43. This limit has the form ∞ · 0. We’ll change it to the form 00 .
sin() H
cos()(−2 )
= lim
= lim cos() = (1) =
→∞
→∞
→∞
1
−12
lim sin() = lim
→∞
44. This limit has the form ∞ · 0. We’ll change it to the form ∞
∞.
lim
→∞
√
1 −12
H
1
= lim 21 2 = lim √ 2 = 0
→∞ 2
→∞
→∞
2
√ −2
= lim
45. This limit has the form 0 · ∞. We’ll change it to the form 00 .
lim sin 5 csc 3 = lim
5·1
5
sin 5 H
5 cos 5
=
=
= lim
→0 3 cos 3
3·1
3
→0 sin 3
→0
46. This limit has the form (−∞) · 0.
1
1
1
ln 1 −
·
H
−1
1
1 − 1 2
−1
= lim
= −1
=
= lim
= lim
lim ln 1 −
1
1
1
→−∞
→−∞
→−∞
→−∞
1
− 2
1−
3 H
32
3 H
3
2 = lim
2 = lim
2 = lim
2 = 0
→∞
→∞ 2
→∞ 2
→∞ 4
2
47. This limit has the form ∞ · 0. lim 3 − = lim
→∞
48. This limit has the form ∞ · 0.
lim 32 sin(1) = lim 12 ·
→∞
and
→∞
sin(1)
1 sin
= lim √
1
→0+
1
[where = 1] = ∞ since as → 0+ , √ → ∞
sin
→ 1.
49. This limit has the form 0 · (−∞).
ln
lim ln tan(2) = lim
50. This limit has the form 0 · ∞.
lim
→(2)−
1
H
→1+ cot(2)
→1+
= lim
→1+ (−2) csc2 (2)
cos sec 5 =
lim
1
2
=−
(−2)(1)2
=
cos
→(2)− cos 5
H
=
lim
− sin
→(2)− −5 sin 5
=
−1
1
=
−5
5
51. This limit has the form ∞ − ∞.
lim
→1
1
−
−1
ln
= lim
→1
H
= lim
ln − ( − 1) H
(1) + ln − 1
ln
= lim
= lim
→1 ( − 1)(1) + ln
→1 1 − (1) + ln
( − 1) ln
1
→1 12 + 1
·
2
1
1
=
=
= lim
→1 1 +
2
1+1
2
52. This limit has the form ∞ − ∞. lim (csc − cot ) = lim
→0
→0
cos
1
−
sin
sin
= lim
→0
1 − cos H
sin
=0
= lim
→0 cos
sin
53. This limit has the form ∞ − ∞.
lim
→0+
1
1
−
−1
= lim
→0+
1
1
− 1 − H
− 1
H
=
= lim
= lim
=
+
+
( − 1)
0+1+1
2
→0 + − 1
→0 + +
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°
670
¤
CHAPTER 6
INVERSE FUNCTIONS
54. This limit has the form ∞ − ∞.
lim
→0+
1
1
−
tan−1
= lim
tan−1 − H
1(1 + 2 ) − 1
1 − (1 + 2 )
= lim
= lim
−1
2
+
+
tan−1
(1
+
)
+
tan
+
(1 + 2 ) tan−1
→0
→0
= lim
−2
−2
H
= lim
+ (1 + 2 ) tan−1 →0+ 1 + (1 + 2 )(1(1 + 2 )) + (tan−1 )(2)
→0+
→0+
−2
= lim
−1
→0+ 2 + 2 tan
=
0
=0
2+0
55. This limit has the form ∞ − ∞.
lim
→0+
1
1
−
tan
tan − H
sec2 − 1
2 sec · sec tan
H
= lim
= lim
→0+ tan
→0+ sec2 + tan
→0+ · 2 sec · sec tan + sec2 + sec2
0
=
=0
0+1+1
= lim
56. The limit has the form ∞ − ∞ and we will change the form to a product by factoring out .
ln
ln H
1
lim ( − ln ) = lim 1 −
= ∞ since lim
= 0.
= lim
→∞
→∞
→∞
→∞ 1
√
57. =
⇒ ln =
√
ln = lim
lim ln = lim
→0+
√
ln , so
→0+
→0+
√
ln H
1
= lim
=0 ⇒
1 −32 = −2 lim+
−12
+
→0 −
→0
2
√
lim = lim ln = 0 = 1.
→0+
→0+
58. = (tan 2)
⇒ ln = · ln tan 2, so
lim ln = lim · ln tan 2 = lim
→0+
→0+
= lim
→0+
2
→0+ sin 2
· lim
−
→0+ cos 2
ln tan 2 H
(1 tan 2)(2 sec2 2)
−22 cos 2
= lim
= lim
2
2
+
+
1
−1
→0
→0 sin 2 cos 2
=1·0=0
⇒
lim (tan 2) = lim ln = 0 = 1.
→0+
→0+
59. = (1 − 2)1
⇒ ln =
1
ln(1 − 2) H
−2(1 − 2)
ln(1 − 2), so lim ln = lim
= −2 ⇒
= lim
→0
→0
→0
1
lim (1 − 2)1 = lim ln = −2 .
→0
→0
, so
⇒ ln = ln 1 +
1
− 2
1 +
ln(1 + ) H
lim ln = lim
= ⇒
= lim
= lim
→∞
→∞
→∞
→∞ 1 +
1
−12
60. =
1+
= lim ln = .
lim 1 +
→∞
→∞
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°
SECTION 6.8
61. = 1(1−)
⇒ ln =
1
1
ln H
1
ln , so lim ln = lim
ln = lim
= −1 ⇒
= lim
+
+
+
+
1−
1
−
1
−
−1
→1
→1
→1
→1
lim 1(1−) = lim ln = −1 =
→1+
→1+
62. = ( + 10)1
INDETERMINATE FORMS AND L’HOSPITAL’S RULE
⇒ ln =
1
.
1
ln( + 10), so
1
· ( + 10)
+ 10
ln( + 10) H
1
= lim
lim ln = lim ln( + 10) = lim
→∞
→∞
→∞
→∞
1
+ 10 H
H
=
lim
=
lim
= lim (1) = 1
→∞ + 10
→∞ + 10
→∞
→∞
= lim
⇒
lim ( + 10)1 = lim ln = 1 =
→∞
→∞
63. = 1
⇒ ln = (1) ln ⇒
lim ln = lim
→∞
→∞
ln H
1
=0 ⇒
= lim
→∞ 1
lim 1 = lim ln = 0 = 1
→∞
64. =
→∞
−
−
lim
→∞
⇒ ln = − ln ⇒
lim ln = lim
→∞
→∞
ln H
1
1
= lim = lim
=0 ⇒
→∞
→∞
= lim ln = 0 = 1
→∞
4
ln(4 + 1) H
4
+1 = 4 ⇒
⇒ ln = cot ln(4 + 1), so lim ln = lim
= lim
2
tan
→0+
→0+
→0+ sec
cot
65. = (4 + 1)
lim (4 + 1)cot = lim ln = 4 .
→0+
→0+
66. = (1 − cos )sin
⇒ ln = sin ln(1 − cos ), so
1
· sin
ln(1 − cos ) H
1
−
cos
= lim
lim ln = lim sin ln(1 − cos ) = lim
csc
− csc cot
→0+
→0+
→0+
→0+
2
sin
sin
sin
= − lim
·
= − lim
2
sin2
→0+ (1 − cos ) csc cot
→0+ csc cot − cot
sin3
sin3
=
−
lim
→0+ cos − cos2
→0+ (1 − cos ) cos
= − lim
H
= − lim
→0+
0
3 sin2 cos
3 sin cos
= − lim
=−
=0
(1 − cos )(− sin ) + cos (sin )
0+1
→0+ (cos − 1) + cos
⇒
lim (1 − cos )sin = lim ln = 0 = 1
→0+
→0+
67. = (1 + sin 3)1
lim ln = lim
→0+
→0+
⇒ ln =
1
ln(1 + sin 3) ⇒
3·1
ln(1 + sin 3) H
[1(1 + sin 3)] · 3 cos 3
3 cos 3
= lim
=
=3 ⇒
= lim
+
+
1
1
+
sin
3
1
+0
→0
→0
lim (1 + sin 3)1 = lim ln = 3
→0+
→0+
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°
¤
671
672
¤
CHAPTER 6
68. = (cos )1
2
INVERSE FUNCTIONS
⇒ ln =
1
ln cos , so
2
1
· (− sin )
1
1
ln cos H
− tan H
− sec2
cos
= lim
=−
= lim
= lim
lim ln = lim 2 ln cos = lim
2
→0
→0
→0
→0
→0
→0
2
2
2
2
⇒
2
1
lim (cos )1 = lim ln = −12 = √
→0
→0
− 1
. Note that =
→0+ ln + − 1
69. The given limit is lim
⇒ ln = ln , so
1
ln H
lim ln = lim ln = lim
= lim
= lim (−) = 0 ⇒
1
→0+
→0+
→0+ 1
→0+
→0+
− 2
lim = lim ln = 0 = 1.
→0+
→0+
Therefore, the numerator of the given limit has limit 1 − 1 = 0 as → 0+ . The denominator of the given limit → −∞ as
− 1
= 0.
→0+ ln + − 1
→ 0+ since ln → −∞ as → 0+ . Thus, lim
70. =
2 − 3
2 + 5
2+1
⇒
2 − 3
⇒
ln = (2 + 1) ln
2 + 5
ln(2 − 3) − ln(2 + 5) H
2(2 − 3) − 2(2 + 5)
−8(2 + 1)2
= lim
= lim
2
→∞
→∞
→∞ (2 − 3)(2 + 5)
1(2 + 1)
−2(2 + 1)
2+1
2
−8(2 + 1)
2 − 3
= −8 ⇒ lim
= lim ln = −8
= lim
→∞ (2 − 3)(2 + 5)
→∞
→∞
2 + 5
lim ln = lim
→∞
71.
From the graph, if = 500, ≈ 736. The limit has the form 1∞ .
2
2
Now = 1 +
⇒ ln = ln 1 +
⇒
2
1
− 2
1 + 2
ln(1 + 2) H
= lim
lim ln = lim
→∞
→∞
→∞
1
−12
= 2 lim
1
→∞ 1 + 2
= 2(1) = 2
⇒
2
1+
= lim ln = 2 [≈ 739]
→∞
→∞
lim
72.
From the graph, as → 0, ≈ 055. The limit has the form 00 .
ln 54
ln 5 − ln 4
5 − 4 H
5 ln 5 − 4 ln 4
=
=
= lim
[≈ 055]
→0 3 − 2
→0 3 ln 3 − 2 ln 2
ln 3 − ln 2
ln 32
lim
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°
SECTION 6.8
INDETERMINATE FORMS AND L’HOSPITAL’S RULE
From the graph, it appears that lim
73.
()
→0 ()
= lim
0 ()
→0 0 ()
¤
673
= 025
1
()
− 1 H
= lim 3
= .
= lim 2
→0 ()
→0 + 4
→0 3 + 4
4
We calculate lim
()
0 ()
= lim 0
= 4. We calculate
→0 ()
→0 ()
From the graph, it appears that lim
74.
lim
()
→0 ()
= lim
2 sin
H
→0 sec − 1
= lim
→0
2( cos + sin )
sec tan
2(− sin + cos + cos )
H
= lim
→0 sec (sec2 ) + tan (sec tan )
75. lim
→∞
H
= lim
H
→∞ −1
= lim
H
→∞ ( − 1)−2
76. This limit has the form ∞
. lim
∞
→∞
H
= · · · = lim
→∞ !
=
4
=4
1
=∞
ln H
1
1
= lim
= lim
= 0 since 0.
→∞ −1
→∞
1
H
= lim
77. lim √
= lim 1 2
−12 (2)
2
→∞
→∞
→∞
(
+
1)
+1
2
√
2 + 1
. Repeated applications of l’Hospital’s Rule result in the
original limit or the limit of the reciprocal of the function. Another method is to try dividing the numerator and denominator
1
1
= lim
= lim
= =1
by : lim √
2
2
2
2
2
→∞
→∞
→∞
1
+1
+ 1
1 + 1
78.
lim
sec H
=
→(2)− tan
lim
→(2)−
sec tan
tan
=
lim
. Repeated applications of l’Hospital’s Rule result in the
sec2
→(2)− sec
original limit or the limit of the reciprocal of the function. Another method is to simplify first:
lim
sec
→(2)− tan
=
79. = () = −
=
lim
1
→(2)− sin
=
1
=1
1
A. = B. Intercepts are 0 C. No symmetry
D. lim − = lim
→∞
→∞
1cos
lim
→(2)− sin cos
H
= lim
1
→∞
= 0, so = 0 is a HA.
lim − = −∞
→−∞
E. 0 () = − − − = − (1 − ) 0 ⇔ 1, so is increasing
H.
on (−∞ 1) and decreasing on (1 ∞) F. Absolute and local maximum
value (1) = 1 G. 00 () = − ( − 2) 0 ⇔ 2, so is CU
on (2 ∞) and CD on (−∞ 2). IP at 2 22
80. = () =
ln
2
A. = (0 ∞) B. ­intercept: none; ­intercept: () = 0 ⇔ ln = 0 ⇔ = 1
C. No symmetry D. lim () = −∞, so = 0 is a VA; lim
→0+
→∞
ln H
1
= 0, so = 0 is a HA.
= lim
→∞ 2
2
[continued]
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°
674
¤
CHAPTER 6
INVERSE FUNCTIONS
2 (1) − (ln )(2)
(1 − 2 ln )
1 − 2 ln 0
=
=
. () 0 ⇔ 1 − 2 ln 0 ⇔ ln 12
(2 )2
4
3
√
√
∞ .
0 12 and 0 () 0 ⇒ 12 , so is increasing on 0 and decreasing on
E. 0 () =
F. Local maximum value (12 ) =
G. 00 () =
=
1
12
=
2
⇒
H.
3 (−2) − (1 − 2 ln )(32 )
(3 )2
2 [−2 − 3(1 − 2 ln )]
−5 + 6 ln
=
6
4
00 () 0 ⇔ −5 + 6 ln 0 ⇔ ln 56
⇒ 56 [ is CU]
and 00 () 0 ⇔ 0 56 [ is CD]. IP at (56 5(653 ))
81. = () = −
2
A. = B. Intercepts are 0 C. (−) = − (), so the curve is symmetric
about the origin. D.
2
2
lim − = lim
2
2
→±∞
→±∞
1
H
= lim
2
→±∞ 2
2
E. 0 () = − − 22 − = − (1 − 22 ) 0 ⇔ 2 12
= 0, so = 0 is a HA.
⇔ || √12 , so is increasing on − √12 √12
√
and decreasing on −∞ − √12 and √12 ∞ . F. Local maximum value √12 = 1 2, local minimum
√
2
2
2
value − √12 = −1 2 G. 00 () = −2− (1 − 22 ) − 4− = 2− (22 − 3) 0 ⇔
3
or −
2
3
0, so is CU on
2
3
∞
2
H.
and
− 32 0 and CD on −∞ − 32 and 0 32 .
IP are (0, 0) and ± 32 ± 32 −32 .
82. = () =
A. = { | 6= 0} B. No intercept C. No symmetry
H
= ∞, = lim
= 0, so = 0 is a HA. lim
= ∞, lim
= −∞, so = 0 is a VA.
= lim
→∞
→∞ 1
→−∞
→0+
→0−
D. lim
E. 0 () =
−
0 ⇔ ( − 1) 0 ⇔ 1, so is increasing on (1 ∞), and decreasing
2
on (−∞ 0) and (0 1). F. (1) = is a local minimum value.
G. 00 () =
H.
2 ( ) − 2( − )
(2 − 2 + 2)
=
0 ⇔ 0
4
3
since 2 − 2 + 2 0 for all . So is CU on (0 ∞) and CD on (−∞ 0).
No IP
83. = () =
1
+ ln A. = (0 ∞) [same as ln ] B. No ­intercept; no ­intercept [1 |ln | on (0 1), and 1
and ln are both positive on (1 ∞)] C. No symmetry D. lim () = ∞, so = 0 is a VA.
→0+
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°
[continued]
SECTION 6.8
INDETERMINATE FORMS AND L’HOSPITAL’S RULE
−1
1
1
E. 0 () = − 2 + =
. 0 () 0 for 1, so is increasing on
2
¤
675
H.
(1 ∞) and is decreasing on (0 1).
F. Local minimum value (1) = 1 G. 00 () =
2
1
2−
− 2 =
.
3
3
00 () 0 for 0 2, so is CU on (0 2), and is CD on (2 ∞).
IP at 2 12 + ln 2
84. = () = (2 − 3)−
A. = B. ­intercept: (0) = −3; ­intercepts: () = 0 ⇔ 2 − 3 = 0 ⇔
√
2 − 3 H
2 H
2
= ± 3 C. No symmetry D. lim (2 − 3)− = lim
= lim = lim = 0, so = 0 is a HA.
→∞
→∞
→∞
→∞
E. 0 () = (2 − 3)(−− ) + − (2) = −− [(2 − 3) − 2] = −− ( − 3)( + 1). 0 () 0 ⇔ −1 3
and 0 () 0 ⇔ −1 or 3, so is increasing on (−1 3) and decreasing on (−∞ −1) and (3 ∞).
F. Local maximum value (3) = 6−3 ; local minimum value (−1) = −2
G. 00 () = (−− )(2 − 2) + (2 − 2 − 3)(− ) = − [−(2 − 2) + (2 − 2 − 3)] = − (2 − 4 − 1).
√
√
√
00 () = 0 ⇔ = 4 ±2 20 = 2 ± 5, so 00 () 0 ⇔ 2 − 5 or
√
√
√
2 + 5 and 00 () 0 ⇔ 2 − 5 2 + 5, so is CU on
√
√
√
√
−∞ 2 − 5 and 2 + 5 ∞ and is CD on 2 − 5 2 + 5 .
√
√
IP at 2 − 5 2 − 5 ≈ (−024 −373) and
√
√
2 + 5 2 + 5 ≈ (424 022)
H.
85. (a) () = −
(b) = () = − . We note that ln () = ln − = − ln = −
H
lim ln () = lim −
→0+
→0+
ln
, so
1
1
= lim = 0. Thus lim () = lim ln () = 0 = 1.
−−2
→0+
→0+
→0+
(c) From the graph, it appears that there is a local and absolute maximum of about (037) ≈ 144. To find the exact value, we
1
differentiate: () = − = − ln ⇒ 0 () = − ln −
+ ln (−1) = −− (1 + ln ). This is 0 only
when 1 + ln = 0 ⇔ = −1 . Also 0 () changes from positive to negative at −1 . So the maximum value is
(1) = (1)−1 = 1 .
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°
676
¤
CHAPTER 6
INVERSE FUNCTIONS
(d)
We differentiate again to get
00 () = −− (1) + (1 + ln )2 (− ) = − [(1 + ln )2 − 1]
From the graph of 00 (), it seems that 00 () changes from negative to
positive at = 1, so we estimate that has an IP at = 1.
86. (a) () = (sin )sin is continuous where sin 0, that is, on intervals
of the form (2 (2 + 1)), so we have graphed on (0 ).
(b) = (sin )sin
⇒ ln = sin ln sin , so
lim ln = lim sin ln sin = lim
→0+
→0+
→0+
= lim (− sin ) = 0
→0+
⇒
ln sin H
cot
= lim
csc
→0+ − csc cot
lim = 0 = 1
→0+
(c) It appears that we have a local maximum at (157 1) and local minima at (038 069) and (276 069).
= (sin )sin
⇒ ln = sin ln sin ⇒
cos
0
= (sin )
+ (ln sin ) cos = cos (1 + ln sin ) ⇒
sin
0 = (sin )sin (cos )(1 + ln sin ). 0 = 0 ⇒ cos = 0 or ln sin = −1 ⇒ 2 = 2 or sin = −1 .
On (0 ), sin = −1
⇒ 1 = sin−1 (−1 ) and 3 = − sin−1 (−1 ). Approximating these points gives us
(1 (1 )) ≈ (03767 06922), (2 (2 )) ≈ (15708 1), and (3 (3 )) ≈ (27649 06922). The approximations
confirm our estimates.
From the graph, we see that 00 () = 0 at ≈ 094 and ≈ 220.
(d)
Since 00 changes sign at these values, they are x­coordinates of inflection
points.
87. (a) () = 1
(b) Recall that = ln . lim 1 = lim (1) ln . As → 0+ ,
→0+
→0+
ln
→ −∞, so 1 = (1) ln → 0. This
indicates that there is a hole at (0 0). As → ∞, we have the indeterminate form ∞0 . lim 1 = lim (1) ln ,
→∞
but lim
→∞
→∞
ln H
1
= 0, so lim 1 = 0 = 1. This indicates that = 1 is a HA.
= lim
→∞ 1
→∞
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
SECTION 6.8
INDETERMINATE FORMS AND L’HOSPITAL’S RULE
¤
(c) Estimated maximum: (272 145). No estimated minimum. We use logarithmic differentiation to find any critical
0
1 1
1
1
1 − ln
= · + (ln ) − 2
numbers. = 1 ⇒ ln = lnx ⇒
⇒ 0 = 1
=0 ⇒
2
ln = 1 ⇒ = . For 0 , 0 0 and for , 0 0, so () = 1 is a local maximum value. This
point is approximately (27183 14447), which agrees with our estimate.
From the graph, we see that 00 () = 0 at ≈ 058 and ≈ 437. Since 00
(d)
changes sign at these values, they are ­coordinates of inflection points.
The first figure shows representative examples
88.
of () = − with odd. is even in the
second figure. All curves pass through the
origin and approach = 0 as → ∞.
0 () =
( − )
= 0 ⇔ = or = 0 (the latter for 1). At = 0, we have a local minimum for even.
At = , we have a local maximum for all . As increases, ( ()) gets farther away from the origin.
2 − 2 + 2 −
√
= 0 ⇔ = ± or = 0 (the latter for 2). As increases, the IP move
00 () =
2
farther away from the origin—they are symmetric about the line = .
⇒ 0 () = − = 0 ⇔ = ⇔ = ln , 0. 00 () = 0, so is CU on
H
−
= 1 . Now lim
= ∞, so 1 = ∞, regardless
= lim
(−∞ ∞). lim ( − ) = lim
→∞
→∞
→∞
→∞ 1
89. () = −
of the value of . For = lim ( − ), → 0, so is determined
→−∞
by −. If 0, − → ∞, and = ∞. If 0, − → −∞, and
= −∞. Thus, has an absolute minimum for 0. As increases, the
minimum points (ln − ln ) get farther away from the origin.
1 − − =
lim 1 − − =
(1 − 0) [because − → −∞ as → ∞]
→∞
→∞
=
, which is the speed the object approaches as time goes on, the so­called limiting velocity.
90. (a) lim = lim
→∞
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°
677
678
¤
CHAPTER 6
INVERSE FUNCTIONS
1 − −
(1 − − ) = lim
→0+
→0+
(b) lim = lim
→0+
H
[form is 00 ]
(−− ) · (−)
=
lim − = (1) =
1
→0+
→0+
= lim
The velocity of a falling object in a vacuum is directly proportional to the amount of time it falls.
1+
, so
, which is of the form 1∞ . = 1 +
⇒ ln = ln 1 +
→∞
−2
ln(1 + ) H
= lim
= lim
=
lim ln = lim ln 1 +
= lim
→∞
→∞
→∞
→∞ (1 + )(−12 )
→∞ 1 +
1
lim = . Thus, as → ∞, = 0 1 +
→ 0 .
→∞
91. First we will find lim
⇒
2
92. (a) = 3, = 005
⇒ =
(b) = 2, = 005 ⇒ =
1 − 10−045
1 − 10−
=
≈ 062, or about 62%.
2
ln 10
045 ln 10
1 − 10−02
≈ 080, or about 80%.
02 ln 10
Yes, it makes sense. Since measured brightness decreases with light entering farther from the center of the pupil, a smaller
pupil radius means that the average brightness measurements are higher than when including light entering at larger radii.
2
2
1 − 10− H
−10− (ln 10)(−2)
1
= lim
= lim
2 = 1, or 100%.
2
+
+
ln 10
2(ln 10)
→0
→0
→0+ 10
(c) lim = lim
→0+
We might expect that 100% of the brightness is sensed at the very center of the pupil, so a limit of 1 would make sense in
this context if the radius could approach 0. This result isn’t physically possible because there are limitations on how
small the pupil can shrink.
93. (a) lim () = lim
→∞ 1 + −
→∞
=
=
1+·0
It is to be expected that a population that is growing will eventually reach the maximum population size that can be
supported.
(b) lim () = lim
→∞
→∞
1
H
= lim
= lim
= 0
− 0 −
→∞
→∞ 1 −
1+
1+
− 1 −
0
0
0
0 is an exponential function.
2
1
1
2
2
= − lim
= −2 · 2 · ln 1 = − · 0 = 0
94. (a) lim = lim −
ln
ln
→+
→ +
→+
As the insulation of a metal cable becomes thinner, the velocity of an electrical impulse in the cable approaches zero.
2
[form is 0 · ∞]
(b) lim = lim −
ln
= − 2 lim 2 ln
→0+
→0+
→0+
1
2
·
ln
H
=0
= − 2 lim
[form is ∞∞] = − 2 lim = − 2 lim −
1
→0+
→0+ −2
→0+
2
2
3
As the radius of the metal cable approaches zero, the velocity of an electrical impulse in the cable approaches zero.
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°
SECTION 6.8
1
95. lim 2
→0
1
96. lim 2
→∞
0
2
√
= lim
3
→0
+1
0
ln(1 + ) = lim
→∞
0
INDETERMINATE FORMS AND L’HOSPITAL’S RULE
¤
679
2
2
√
√
3
1
H
+1
3 + 1
= lim √
=1
form 00 = lim
3
→0
→0
2
2
+1
0
ln(1 + )
ln(1
+
)
H
H
form ∞
= lim
= lim 1 +
∞
→∞
→∞
2
2
2
1
1
1
=
=
lim
=
= lim
→∞ 2(1 + )
→∞ 2(1 + )
→∞
2(0 + 1)
2
1
2 +1
= lim
97. Both numerator and denominator approach 0 as → 0, so we use l’Hospital’s Rule (and FTC1):
()
= lim
lim
→0 3
→0
0
sin(22) H
sin(22) H
cos(22)
= · cos 0 =
=
lim
=
lim
→0
→0
3
32
6
6
6
98. Both numerator and denominator approach 0 as → 0, so we use l’Hospital’s Rule. (Note that we are differentiating with
respect to , since that is the quantity which is changing.) We also use the Fundamental Theorem of Calculus, Part 1:
lim ( ) = lim
→0
→0
−( − )2(4)
2
2
H
− (4)
−( − ) (4)
0
√
√
= lim
= √
→0
4
4
4
99. We see that both numerator and denominator approach 0, so we can use l’Hospital’s Rule:
√
√
1
(23 − 4 )−12 (23 − 43 ) − 13 ()−23 2
23 − 4 − 3 H
2
√
lim
= lim
4
→
→
− 14 (3 )−34 (32 )
− 3
1
= 2
=
(23 − 4 )−12 (23 − 43 ) − 13 3 (2 )−23
− 14 (3 )−34 (32 )
(4 )−12 (−3 ) − 13 3 (3 )−23
− − 13
=
= 43 43 = 16
3 3 4 −34
3
9
− 4 ( )
−4
100. Let the radius of the circle be . We see that () is the area of the whole figure (a sector of the circle with radius 1), minus
the area of 4 . But the area of the sector of the circle is 12 2 (see Reference Page 1), and the area of the triangle is
1
| | = 12 ( sin ) = 12 2 sin . So we have () = 12 2 − 12 2 sin = 12 2 ( − sin ). Now by elementary
2
trigonometry, () = 12 || | | = 12 ( − ||) | | = 12 ( − cos )( sin ) = 12 2 (1 − cos ) sin .
So the limit we want is
1 2
( − sin )
()
1 − cos
H
= lim 1 22
= lim
+
+
+
()
(1
−
cos
)
cos + sin (sin )
→0
→0
→0
(1 − cos ) sin
2
lim
1 − cos
sin
H
= lim
2
→0+ cos − cos2 + sin
→0+ − sin − 2 cos (− sin ) + 2 sin (cos )
= lim
= lim
sin
→0+ − sin + 4 sin cos
= lim
1
→0+ −1 + 4 cos
=
1
1
=
−1 + 4 cos 0
3
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°
680
¤
CHAPTER 6
INVERSE FUNCTIONS
1+
1
− 2 ln
= lim − 2 ln
+ 1 . Let = 1, so as → ∞, → 0+ .
→∞
→∞
101. The limit, = lim
= lim
→0+
1
1−
1
1
1
− ln( + 1) H
1
+
1 = lim ( + 1) = lim
− 2 ln( + 1) = lim
=
= lim
2
2
2
2
→0+
→0+
→0+
→0+ 2 ( + 1)
Note: Starting the solution by factoring out or 2 leads to a more complicated solution.
102. = [ ()]()
⇒ ln = () ln (). Since is a positive function, ln () is defined. Now
lim ln = lim () ln () = −∞ since lim () = ∞ and lim () = 0 ⇒
→
→
→
→
lim ln () = −∞. Thus, if = ln ,
→
lim = lim = 0. Note that the limit, lim () ln (), is not of the form ∞ · 0.
→
→−∞
→
103. (a) We look for functions and whose individual limits are ∞ as → 0, but whose quotient has a limit of 7 as → 0.
One such pair of functions is () =
lim
()
→0 ()
= lim
72
→0 12
7
1
and () = 2 . We have lim () = lim () = ∞, and
→0
→0
2
= lim 7 = 7
→0
(b) We look for functions and whose individual limits are ∞ as → 0, but whose difference has a limit of 7 as → 0.
1
1
+ 7 and () = 2 . We have lim () = lim () = ∞, and
→0
→0
2
1
1
lim [ () − ()] = lim
+ 7 − 2 = lim 7 = 7
→0
→0
→0
2
One such pair of functions is () =
104. = lim
→0
sin 2
++ 2
3
sin 2 + 3 + H
2 cos 2 + 32 +
= lim
. As → 0, 32 → 0, and
3
→0
→0
32
= lim
(2 cos 2 + 32 + ) → + 2, so the last limit exists only if + 2 = 0, that is, = −2. Thus,
6 − 8
2 cos 2 + 32 − 2 H
−4 sin 2 + 6 H
−8 cos 2 + 6
=
, which is equal to 0 if and only
= lim
= lim
→0
→0
→0
32
6
6
6
lim
if = 43 . Hence, = 0 if and only if = −2 and = 43 .
105. (a) We show that lim
→0
1
()
= 0 for every integer ≥ 0. Let = 2 . Then
2
()
−1
H
−1 H
!
H
= lim
= lim
= · · · = lim = 0 ⇒
= lim
→0 2
→0 (2 )
→∞
→∞
→∞
lim
lim
→0
()
()
()
() − (0)
()
= lim
= 0.
= lim 2 = lim lim 2 = 0. Thus, 0 (0) = lim
→0
→0
→0
→0
→0
−0
(b) Using the Chain Rule and the Quotient Rule we see that () () exists for 6= 0. In fact, we prove by induction that for
each ≥ 0, there is a polynomial and a non­negative integer with () () = () () for 6= 0. This is
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°
CHAPTER 6 REVIEW
true for = 0; suppose it is true for the th derivative. Then 0 () = ()(23 ), so
(+1) () = [0 () () + () 0 ()] − −1 () () −2
= 0 () + ()(23 ) − −1 () ()−2
which has the desired form.
= +3 0 () + 2 () − +2 () ()−(2 +3)
Now we show by induction that () (0) = 0 for all . By part (a), 0 (0) = 0. Suppose that () (0) = 0. Then
() () − () (0)
() ()
() ()
() ()
= lim
= lim
= lim
→0
→0
→0
→0
−0
+1
(+1) (0) = lim
= lim () lim
→0
()
→0 +1
= (0) · 0 = 0
[by part (a)]
106. (a) For to be continuous, we need lim () = (0) = 1. We note that for 6= 0, ln () = ln || = ln ||.
→0
So lim ln () = lim ln || = lim
→0
→0
→0
ln || H
1
= lim
= 0. Therefore, lim () = lim ln () = 0 = 1.
→0 −12
→0
→0
1
So is continuous at 0.
(b) From the graphs, it appears that is differentiable at 0.
(c) To find 0 , we use logarithmic differentiation: ln () = ln || ⇒
1
0 ()
=
+ ln || ⇒
()
0 () = ()(1 + ln ||) = || (1 + ln ||), 6= 0. Now 0 () → −∞ as → 0 [since || → 1 and
(1 + ln ||) → −∞], so the curve has a vertical tangent at (0 1) and is therefore not differentiable there.
The fact cannot be seen in the graphs in part (b) because ln || → −∞ very slowly as → 0.
6 Review
1. True.
If is one­to­one, with domain , then −1 ( (6)) = 6 by the first cancellation equation [see (6.1.4)].
2. False.
By Theorem 6.1.7, ( −1 )0 (6) =
1
1
, not 0
unless −1 (6) = 6.
0 ( −1 (6))
(6)
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°
¤
681
682
¤
3. False.
4. False.
CHAPTER 6 INVERSE FUNCTIONS
For example, cos 2 = cos − 2 , so cos is not 1­1.
It is true that tan 3
= −1, but since the range of tan−1 is − 2 2 , we must have tan−1 (−1) = − 4 .
4
5. True.
The function = ln is increasing on (0 ∞), so if 0 , then ln ln .
6. True.
Since = ln , 5 = 5 ln .
7. True.
We can divide by since 6= 0 for every .
8. False.
For example, ln(1 + 1) = ln 2, but ln 1 + ln 1 = 0. In fact ln + ln = ln().
9. False.
Let = . Then (ln )6 = (ln )6 = 16 = 1, but 6 ln = 6 ln = 6 · 1 = 6 6= 1 = (ln )6 . What is true, however,
√
√
is that ln(6 ) = 6 ln for 0.
10. False.
(10 ) = 10 ln 10, which is not equal to 10−1 .
11. False.
ln 10 is a constant, so its derivative,
12. True.
= 3
13. False.
The “−1” is not an exponent; it is an indication of an inverse function.
14. False.
For example, tan−1 20 is defined; sin−1 20 and cos−1 20 are not.
15. True.
See Figure 2 in Section 6.7.
16. True.
−
17. True.
10
1
16
2
1
(ln 10), is 0, not 10
.
⇒ ln = 3 ⇒ = 13 ln
⇒ the inverse function is = 13 ln .
10
1
= − ln || 1 = − ln 10 + ln 1 = ln 10−1 + 0 = ln 10
16
16
= ln || 2 = ln 16 − ln 2 = ln
= ln 8 = ln 23 = 3 ln 2
2
tan
0
= 0.
2
18. False.
L’Hospital’s Rule does not apply since lim
19. False.
1
and () = . Then lim () = 1 and lim () = ∞, but
→∞
→∞
1
= , not 1.
lim [ ()]() = lim 1 +
→∞
→∞
→ − 1 − cos
=
Let () = 1 +
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°
CHAPTER 6 REVIEW
¤
1. No. is not 1­1 because the graph of fails the Horizontal Line Test.
2. (a) is one­to­one because it passes the Horizontal Line Test.
(b) When = 2, ≈ 02. So −1 (2) ≈ 02.
(c) The range of is [−1 35], which is the same as the domain of −1 .
(d) We reflect the graph of through the line = to obtain the graph of −1 .
3. (a) −1 (3) = 7 since (7) = 3.
4. We write =
(b) ( −1 )0 (3) =
1
1
1
= 0
=
0 ( −1 (3))
(7)
8
2 + 3
and solve for : (1 − 5) = 2 + 3 ⇒ − 5 = 2 + 3 ⇒ − 3 = 2 + 5
1 − 5
− 3 = (2 + 5) ⇒ =
⇒
−3
−3
−3
. Interchanging and gives =
, so −1 () =
.
2 + 5
2 + 5
2 + 5
5.
6.
= 5 − 1
= −
7. Reflect the graph of = ln about the ­axis to
obtain the graph of = − ln .
8. = ln( + 5). Start with the graph of = ln and shift
5 units to the left.
= ln
= ln
= −−
= ln( + 5)
= −ln
9.
= 2 arctan
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
683
684
¤
CHAPTER 6 INVERSE FUNCTIONS
10. We have seen that if 1, then for sufficiently large . (See Exercise 6.2.20.) In general, we could show that
lim = ∞ by using l’Hospital’s Rule repeatedly. Also, log increases much more slowly than either or .
→∞
[Compare the graph of log with those of and , or use l’Hospital’s Rule to show that lim (log ) = 0.]
→∞
So for large , log .
2
11. (a) 2 ln 5 = ln 5 = 52 = 25
(b) log6 4 + log6 54 = log6 (4 · 54) = log6 216 = log6 63 = 3
(c) Let = arcsin 45 , so sin = 45 . Draw a right triangle with angle as shown
in the figure. By the Pythagorean Theorem, the adjacent side has length 3,
4
opp
4
and tan arcsin
= tan =
= .
5
adj
3
12. (a) ln
1
= ln −3 = −3
3
(b) sin(tan−1 1) = sin
√
2
=
4
2
−3
(c) 10−3 log 4 = 10log 4
13. 2 = 3
1
1
=
43
64
⇒ ln(2 ) = ln 3 ⇒ 2 = ln 3 ⇒ = 12 ln 3 ≈ 0549
2
14. ln 2 = 5
= 4−3 =
15. = 10
⇒ ln = 5
⇒ 2 = 5
√
⇒ = ± 5 ≈ ±12182
⇒ ln = ln 10 ⇒ = ln 10 ⇒ ln = ln(ln 10) ⇒ = ln(ln 10) ≈ 0834
16. cos−1 = 2
⇒ cos(cos−1 ) = cos 2 ⇒ = cos 2 ≈ −0416
17. tan−1 (32 ) =
4
18. ln(1 + − ) = 3
⇒ tan(tan−1 (32 )) = tan
⇒ 1 + − = 3
4
⇒ 32 = 1 ⇒ 2 =
1
3
1
⇒ = ± √ ≈ ±0577
3
⇒ − = 3 − 1 ⇒ ln − = ln(3 − 1) ⇒ − = ln(3 − 1) ⇒
= − ln(3 − 1)
19. ln − 1 = ln(5 + ) − 4
= −3
5+
⇒ ln − ln(5 + ) = −4 + 1 ⇒ ln
⇒ = 5−3 + −3
or, multiplying by
⇒ − −3 = 5−3
= −3 ⇒ ln((5+)) = −3
5+
⇒ (1 − −3 ) = 5−3
⇒ =
5−3
1 − −3
3
5
≈ 0262.
, we have = 3
3
−1
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°
⇒
CHAPTER 6 REVIEW
20. log5 ( ) =
⇒ log5 = ⇒ =
Or: log5 ( ) = ⇒ 5 =
21. () = 2 ln
22. () =
⇒ 0 () = 2 ·
1 +
⇒ 0 () =
√
⇒ ln 5 = ln
⇒ ln 5 = ln ⇒ =
ln 5
.
ln
1
+ (ln )(2) = + 2 ln or (1 + 2 ln )
1 + −
2
(1 + )
√
⇒ 0 () = 10 · ln 10 ·
25. = ln |sec 5 + tan 5|
0 =
.
log5
=
(1 + )2
⇒ 0 () = tan 2 · sec2 2 · 2 = 2 sec2 (2) tan 2
23. () = tan 2
24. () = 10
¤
√
(ln 10)10
1
√ =
√
2
2
⇒
1
5 sec 5 (tan 5 + sec 5)
(sec 5 tan 5 · 5 + sec2 5 · 5) =
= 5 sec 5
sec 5 + tan 5
sec 5 + tan 5
26. = cos
⇒
0 = (cos )0 + cos ( )0 = (− sin · ) + cos ( · ) = ( cos − sin )
27. = ln(sec2 ) = 2 ln |sec |
28. =
⇒ 0 = (2 sec )(sec tan ) = 2 tan
ln(4 ) ⇒
0 =
ln(4 ) + 4
1
1
1
1
[ ln(4 )]−12
[ ln(4 )] =
· 1 · ln(4 ) + · 4 · 43 =
· [ln(4 ) + 4] =
4
4
2
2 ln( )
2 ln( )
2 ln(4 )
√
√
Or: Since y is only defined for 0, we can write = · 4 ln = 2 ln . Then
ln + 1
1
1
= √
· 1 · ln + ·
. This agrees with our first answer since
0 = 2 · √
2 ln
ln
ln(4 ) + 4
4 ln + 4
4(ln + 1)
ln + 1
√
= √
=
= √
.
2 ln(4 )
2 · 4 ln
2 · 2 ln
ln
29. =
1
2
⇒ 0 =
30. = (arcsin 2)2
1
31. = 5 arctan
32. = sec−1
0
2 (1 )0 − 1 2
2 (1 )(−12 ) − 1 (2)
−1 (1 + 2)
=
=
2
2
4
( )
4
1
4 arcsin 2
⇒ 0 = 2(arcsin 2) · (arcsin 2)0 = 2 arcsin 2 ·
·2= √
1 − 42
1 − (2)2
1
⇒ =5·
2 ·
1
1+
0
1
=
1
5
− 2 =− 2
1
+1
1+ 2
5
1
1
⇒ 0 = · √
+ sec−1 · 1 = √
+ sec−1
2
2
−1
−1
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°
685
686
¤
CHAPTER 6 INVERSE FUNCTIONS
⇒ 0 = 3 ln (ln 3)
33. = 3 ln
34. = cos + cos( )
⇒ 0 = − sin cos − sin( )
35. = tan−1 − 12 ln(1 + 2 )
0 = ·
1
( ln ) = 3 ln (ln 3) · + ln · 1 = 3 ln (ln 3)(1 + ln )
⇒
1
1
+ tan−1 · 1 −
1 + 2
2
36. () = log10 (1 + 2 )
⇒ 0 () =
2
1 + 2
=
+ tan−1 −
= tan−1
1 + 2
1 + 2
2
1
· 2 =
(ln 10)(1 + 2 )
(ln 10)(1 + 2 )
⇒ 0 = cosh(2 ) · 2 + sinh(2 ) · 1 = 22 cosh(2 ) + sinh(2 )
37. = sinh(2 )
38. = (cos )
⇒ ln = ln(cos ) = ln cos ⇒
1
0
=·
· (− sin ) + ln cos · 1 ⇒
cos
0 = (cos ) (ln cos − tan )
⇒ 0 =
39. = ln(arcsin 2 )
40. = arctan arcsin
1
1
1
2
√
(arcsin 2 ) =
·
·
· 2 =
2
2
arcsin 2
arcsin 2
(arcsin 2 ) 1 − 4
1 − ( )
√
⇒ 0 =
1
1
1
√ 2 · √1 − · √
2
1 + arcsin
1
1
= ln −1 + (ln )−1 = − ln + (ln )−1
+
ln
41. = ln
⇒ 0 = −1 ·
1
1
1
1
+ (−1)(ln )−2 · = − −
(ln )2
2
− 4
2
2( + 1)( + 4)
= ln 2 − 4 − ln |2 + 5| ⇒ 0 = 2 −
or
2 + 5
2 − 4
2 + 5
( + 2)( − 2)(2 + 5)
42. = ln
43. = ln(cosh 3)
44. =
⇒ 0 = (1 cosh 3)(sinh 3)(3) = 3 tanh 3
(2 + 1)4
(2 + 1)3 (3 − 1)5
ln = ln
⇒
(2 + 1)4
= ln(2 + 1)4 − ln[(2 + 1)3 (3 − 1)5 ] = 4 ln(2 + 1) − [ln(2 + 1)3 + ln(3 − 1)5 ]
(2 + 1)3 (3 − 1)5
= 4 ln(2 + 1) − 3 ln(2 + 1) − 5 ln(3 − 1) ⇒
1
1
1
(2 + 1)4
0
=4· 2
· 2 − 3 ·
·2−5·
· 3 ⇒ 0 =
+1
2 + 1
3 − 1
(2 + 1)3 (3 − 1)5
[The answer could be simplified to 0 = −
45. = cosh−1 (sinh )
6
15
8
−
−
.
2 + 1
2 + 1
3 − 1
(2 + 56 + 9)(2 + 1)3
, but this is unnecessary.]
(2 + 1)4 (3 − 1)6
⇒ 0 = (cosh ) sinh2 − 1
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
CHAPTER 6 REVIEW
√
√
√
1
1
−1 √
46. = tanh−1 ⇒ 0 = tanh−1 +
+
√ 2 √ = tanh
2(1 − )
1 − ( ) 2
√
47. = cos tan 3
⇒
√
0
√
√
√
0 = − sin tan 3 · tan 3 = − sin tan 3 tan 3 · 12 (tan 3)−12 · sec2 (3) · 3
√
√
−3 sin tan 3 tan 3 sec2 (3)
√
=
2 tan 3
48.
1 ( + 1)2
1
tan−1 + ln 2
2
4
+1
1
1
1
tan−1 + ln | + 1| − ln(2 + 1)
2
2
4
1 1
1 1
1 2
1
1
1
=
+
−
=
−
+
2 2 + 1
2+1
4 2 + 1
2 2 + 1
2 + 1
+1
2
2
1
1
+1
1 1−
1−
+
=
+ 2
=
2 2 + 1
+1
2 (2 + 1)(1 + )
( + 1)(1 + )
=
=
2
1
1
=
2 (2 + 1)(1 + )
(1 + )(2 + 1)
49. () = ()
⇒ 0 () = () 0 ()
50. () = ( )
⇒ 0 () = 0 ( )
1 0
0 ()
() =
()
()
51. () = ln |()|
⇒ 0 () =
52. () = (ln )
⇒ 0 () = 0 (ln ) ·
0 (ln )
1
=
⇒ 0 () = 2 ln 2 ⇒ 00 () = 2 (ln 2)2
53. () = 2
54. () = ln(2) = ln 2 + ln
⇒ ···
⇒ () () = 2 (ln 2)
⇒ 0 () = −1 , 00 () = −−2 , 000 () = 2−3 , (4) () = −2 · 3−4 , ,
() () = (−1)−1 ( − 1)! −
55. We first show it is true for = 1: 0 () = + = ( + 1) . We now assume it is true for = :
() () = ( + ) . With this assumption, we must show it is true for = + 1:
(+1) () =
()
() =
[( + ) ] = + ( + ) = [ + ( + 1)] .
Therefore, () () = ( + ) by mathematical induction.
56. Using implicit differentiation, = + arctan
0
2
1 + 2
= 1 ⇒ 0 =
⇒ 0 = 1 +
1
0
1 + 2
⇒ 0 1 −
1
1 + 2
=1 ⇒
1 + 2
1
= 2 + 1.
2
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CHAPTER 6 INVERSE FUNCTIONS
57. = (2 + )−
⇒ 0 = (2 + )(−− ) + − · 1 = − [−(2 + ) + 1] = − (− − 1). At (0 2), 0 = 1(−1) = −1,
so an equation of the tangent line is − 2 = −1( − 0), or = − + 2.
58. = () = ln
⇒ 0 () = ln + 1, so the slope of the tangent at ( ) is 0 () = 2 and an equation is
− = 2( − ) or = 2 − .
59. = [ln( + 4)]2
+ 4 = 0
⇒ 0 = 2[ln( + 4)]1 ·
⇒
60. () = sin
ln( + 4)
1
·1=2
and 0 = 0 ⇔ ln( + 4) = 0 ⇔
+4
+4
+ 4 = 1 ⇔ = −3, so the tangent is horizontal at the point (−3 0).
⇒ 0 () = [sin (cos )] + sin (1) = sin ( cos + 1). As a check on our work, we notice from the
graphs that 0 () 0 when is increasing. Also, we see in the larger viewing rectangle a certain similarity in the graphs of
and 0 : the sizes of the oscillations of and 0 are linked.
61. (a) The line − 4 = 1 has slope 14 . A tangent to = has slope 14 when 0 = = 14
⇒ = ln 14 = − ln 4.
Since = , the ­coordinate is 14 and the point of tangency is − ln 4 14 . Thus, an equation of the tangent line
is − 14 = 14 ( + ln 4) or = 14 + 14 (ln 4 + 1).
= . Thus, an equation of the tangent line is
(b) The slope of the tangent at the point ( ) is
=
− = ( − ). We substitute = 0, = 0 into this equation, since we want the line to pass through the origin:
0 − = (0 − ) ⇔ − = (−) ⇔ = 1. So an equation of the tangent line at the point ( ) = (1 )
is − = ( − 1) or = .
62. (a) lim () = lim [(− − − )] = lim (− − − ) = (0 − 0) = 0 because − → −∞ and − → −∞
→∞
→∞
→∞
as → ∞.
(b) () = (− − − ) ⇒ 0 () = (− (−) − − (−)) = (−− + − )
(c) 0 () = 0 ⇔ − = −
⇔
= (−+)
⇔ ln
ln()
= ( − ) ⇔ =
−
63. lim −3 = 0 since −3 → −∞ as → ∞ and lim = 0.
→∞
64.
→−∞
lim ln(100 − 2 ) = −∞ since as → 10− , (100 − 2 ) → 0+ .
→10−
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°
CHAPTER 6 REVIEW
65. Let = 2( − 3). As → 3− , → −∞.
lim 2(−3) = lim = 0
→−∞
→3−
66. If = 3 − = (2 − 1), then as → ∞, → ∞.
lim arctan(3 − ) = lim arctan = 2 .
→∞
→∞
67. As → 0, cosh → 1, so lim ln(cosh ) = 0.
→0
68. −1 ≤ sin ≤ 1
⇒ −− ≤ − sin ≤ − . Now lim ±− = 0, so by the Squeeze Theorem,
→∞
lim − sin = 0.
→∞
69. lim
1 + 2
→∞ 1 − 2
(1 + 2 )2
= lim
→∞ (1 − 2 )2
= lim
12 + 1
→∞ 12 − 1
=
0+1
= −1
0−1
4
4
4
1
1
70. Let = 4, so = 4. As → ∞, → ∞. lim 1 +
= lim 1 +
= lim 1 +
= 4
→∞
→∞
→∞
71. This limit has the form 00 .
72. This limit has the form 00 .
73. This limit has the form 00 .
74. This limit has the form ∞
∞.
1
− 1 H
= =1
= lim
→0 tan
→0 sec2
1
lim
lim
→0
0
1 − cos H
sin
= =0
= lim
→0 2 + 1
2 +
1
2+2
2 − −2 H
22 + 2−2
=
=4
= lim
→0 ln( + 1)
→0
1( + 1)
1
lim
lim
→∞
2 − −2 H
22 + 2−2
= lim 2( + 1)(2 + −2 ) = ∞
= lim
→∞
→∞
ln( + 1)
1( + 1)
since 2( + 1) → ∞ and (2 + −2 ) → ∞ as → ∞.
75. This limit has the form ∞ · 0.
H
2 − 3 ∞
2 − 32 ∞
form = lim
form
∞
∞
−2
−2
→−∞
−2
H
2 − 6 ∞
−6
H
= lim
= lim
=0
∞ form
→−∞ 4−2
→−∞ −8−2
lim (2 − 3 )2 = lim
→−∞
→−∞
76. This limit has the form 0 · (−∞).
77. This limit has the form ∞ − ∞.
lim
→1+
1
−
−1
ln
= lim
→1+
H
= lim
lim 2 ln = lim
ln − + 1
( − 1) ln
1
→1+ 12 + 1
ln
H
→0+ 12
→0+
=
H
= lim
= lim
1
→0+ −23
= lim
· (1) + ln − 1
→0+
→1+ ( − 1) · (1) + ln
1 2
−2 = 0
= lim
ln
→1+ 1 − 1 + ln
1
1
=
1+1
2
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CHAPTER 6 INVERSE FUNCTIONS
⇒ ln = cos ln tan , so
78. = (tan )cos
lim
→(2)−
so
lim
ln =
→(2)−
lim
→(2)−
(tan )cos =
0
ln tan H
(1 tan ) sec2
sec
cos
=
lim
=
lim
=
lim
= 2 = 0,
2
2
−
sec
sec tan
1
→(2)
→(2)− tan
→(2)− sin
lim
→(2)−
79. = () = sin , − ≤ ≤
ln = 0 = 1.
A. = [− ] B. ­intercept: (0) = 0; () = 0 ⇔ sin = 0 ⇒
= −, 0, . C. No symmetry D. No asymptote E. 0 () = cos + sin · = (cos + sin ).
0 () = 0 ⇔ − cos = sin ⇔ −1 = tan ⇒ = − 4 3
. 0 () 0 for − 4 3
and 0 () 0
4
4
for − − 4 and 3
and is decreasing on − − 4 and 3
, so is increasing on − 4 3
.
4
4
4
√
H.
F. Local minimum value − 4 = (− 22)−4 ≈ −032 and
34
3 √
22
≈ 746
local maximum value 4 =
G. 00 () = (− sin + cos ) + (cos + sin ) = (2 cos ) 0 ⇒
− 2 2 and 00 () 0 ⇒ − − 2 and 2 , so is
CU on − 2 2 , and is CD on − − 2 and 2 . There are inflection
points at − 2 −−2 and 2 2 .
80. = () = sin−1 (1)
A. = { | −1 ≤ 1 ≤ 1} = (−∞ −1] ∪ [1 ∞) B. No intercept
C. (−) = −(), symmetric about the origin D.
1
E. 0 () =
1 − (1)2
lim sin−1 (1) = sin−1 (0) = 0, so = 0 is a HA.
→±∞
1
−1
− 2 = √
0, so is decreasing on (−∞ −1) and (1 ∞).
4 − 2
F. No local extreme value, but (1) = 2 is the absolute maximum value
H.
and (−1) = − 2 is the absolute minimum value.
22 − 1
43 − 2
G. 00 () =
=
0 for 1 and 00 () 0
2(4 − 2 )32
(4 − 2 )32
for −1, so is CU on (1 ∞) and CD on (−∞ −1). No IP
81. = () = ln
A. = (0 ∞) B. No ­intercept; ­intercept 1. C. No symmetry D. No asymptote
H.
[Note that the graph approaches the point (0 0) as → 0+ .]
E. () = (1) + (ln )(1) = 1 + ln , so () → −∞ as → 0 and
0
0
+
0 () → ∞ as → ∞. 0 () = 0 ⇔ ln = −1 ⇔ = −1 = 1.
0 () 0 for 1, so is decreasing on (0 1) and increasing on
(1 ∞). F. Local minimum: (1) = −1. No local maximum.
G. 00 () = 1, so 00 () 0 for 0. The graph is CU on (0 ∞) and
there is no IP.
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°
CHAPTER 6 REVIEW
2
82. = () = 2−
A. = B. ­intercept 1; no ­intercept C. No symmetry D.
2
is a HA. E. = () = 2−
¤
2
lim 2− = 0, so = 0
→±∞
2
⇒ 0 () = 2(1 − )2− 0 ⇔ 1, so is increasing on (−∞ 1) and
decreasing on (1 ∞). F. (1) = is a local and absolute maximum value.
√
2
G. 00 () = 2 22 − 4 + 1 2− = 0 ⇔ = 1 ± 22 .
√
√
√
00 () 0 ⇔ 1 − 22 or 1 + 22 , so is CU on −∞ 1 − 22
√
√
√
√ √
and 1 + 22 ∞ , and CD on 1 − 22 1 + 22 . IP at 1 ± 22
83. = () = ( − 2)−
H.
A. = B. ­intercept: (0) = −2; ­intercept: () = 0 ⇔ = 2
C. No symmetry D. lim
→∞
−2 H
1
= lim = 0, so = 0 is a HA. No VA
→∞
E. 0 () = ( − 2)(−− ) + − (1) = − [−( − 2) + 1] = (3 − )− .
H.
() 0 for 3, so is increasing on (−∞ 3) and decreasing on (3 ∞).
0
F. Local maximum value (3) = −3 , no local minimum value
G. 00 () = (3 − )(−− ) + − (−1) = − [−(3 − ) + (−1)]
= ( − 4)− 0
for 4, so is CU on (4 ∞) and CD on (−∞ 4). IP at (4 2−4 )
84. = () = + ln(2 + 1)
A. = B. ­intercept: (0) = 0 + ln 1 = 0; ­intercept: () = 0 ⇔
ln(2 + 1) = − ⇔ 2 + 1 = −
⇒ = 0 since the graphs of = 2 + 1 and = − intersect only at = 0.
C. No symmetry D. No asymptote E. 0 () = 1 +
2
2 + 1
=
2 + 2 + 1
( + 1)2
= 2
. 0 () 0 if 6= −1 and
2
+1
+1
is increasing on . F. No local extreme values
G. 00 () =
(2 + 1)2 − 2(2)
2[(2 + 1) − 22 ]
2(1 − 2 )
=
= 2
.
2
2
2
2
( + 1)
( + 1)
( + 1)2
H.
00 () 0 ⇔ −1 1 and 00 () 0 ⇔ −1 or 1, so is
CU on (−1 1) and is CD on (−∞ −1) and (1 ∞). IP at (−1 −1 + ln 2)
and (1 1 + ln 2)
85. If 0, then lim () = lim − = lim
→−∞
→−∞
→−∞
H
If 0, then lim () = −∞, and lim () = lim
→−∞
→∞
H
= lim
1
→−∞
1
→∞
= 0, and lim () = ∞.
→∞
= 0.
If = 0, then () = , so lim () = ±∞, respectively.
→±∞
So we see that = 0 is a transitional value. We now exclude the case = 0, since we know how the function behaves
in that case. To find the maxima and minima of , we differentiate: () = −
691
⇒
0 () = (−− ) + − = (1 − )− . This is 0 when 1 − = 0 ⇔ = 1. If 0 then this
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°
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CHAPTER 6 INVERSE FUNCTIONS
represents a minimum value of (1) = 1(), since 0 () changes from negative to positive at = 1;
and if 0, it represents a maximum value. As || increases, the maximum or
minimum point gets closer to the origin. To find the inflection points, we
differentiate again: 0 () = − (1 − )
⇒
00 () = − (−) + (1 − )(−− ) = ( − 2)− . This changes sign
when − 2 = 0
⇔
= 2. So as || increases, the points of inflection get
closer to the origin.
86. We exclude the case = 0, since in that case () = 0 for all . To find the maxima and minima, we differentiate:
2
() = −
2
2
2
⇒ 0 () = − (−2) + − (1) = − (−22 + 1)
√
This is 0 where −22 + 1 = 0 ⇔ = ±1 2. So if 0, there are two maxima or minima, whose ­coordinates
approach 0 as increases. The negative solution gives a minimum and the positive solution gives a maximum, by the First
√
2
√
√
Derivative Test. By substituting back into the equation, we see that ±1 2 = ±1 2 −(±1 2 ) = ± 2.
So as increases, the extreme points become more pronounced. Note that if 0, then lim () = 0. If 0, then there
→±∞
are no extreme values, and lim () = ∓∞.
→±∞
2
To find the points of inflection, we differentiate again: 0 () = − −22 + 1 ⇒
2
2
2
00 () = − (−4) + (−22 + 1)(−2− ) = −22 − (3 − 22 ). This is 0 at = 0 and where
3 − 22 = 0 ⇔ = ± 3(2) ⇒ IP at ± 3(2) ± 32 −32 . If 0 there are three inflection points,
and as increases, the ­coordinates of the nonzero inflection points approach 0. If 0, there is only one inflection point,
the origin.
87. () = − cos( + )
⇒
() = 0 () = {− [− sin( + )] + cos( + )(−− )} = −− [ sin( + ) + cos( + )] ⇒
() = 0 () = −{− [2 cos( + ) − sin( + )] + [ sin( + ) + cos( + )](−− )}
= −− [2 cos( + ) − sin( + ) − sin( + ) − 2 cos( + )]
= −− [(2 − 2 ) cos( + ) − 2 sin( + )] = − [(2 − 2 ) cos( + ) + 2 sin( + )]
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°
CHAPTER 6 REVIEW
¤
693
88. (a) Let () = ln + − 3. Then 0 () = 1 + 1 [for 0)] and (2) ≈ −0307 and () ≈ 0718.
is differentiable on (2 ), continuous on [2 ] and (2) 0, () 0. Therefore, by the Intermediate Value Theorem
there exists a number in (2 ) such that () = 0. Thus, there is one solution. But 0 () 0 for ∈ (2 ), so is
increasing on (2 ), which means that there is exactly one solution.
(b) We use Newton’s method with () = ln + − 3, 0 () = 1 + 1, and 1 = 2.
2 = 1 −
ln 2 + 2 − 3
ln 1 + 1 − 3
=2−
≈ 220457. Similarly, 3 ≈ 220794, 4 = 220794. Thus, the solution
11 + 1
12 + 1
of the equation, correct to four decimal places, is 22079.
89. (a) () = (0) = 200
⇒ (05) = 20005 = 360 ⇒ 05 = 18 ⇒ 05 = ln 18 ⇒
= 2 ln 18 = ln(18)2 = ln 324
⇒
() = 200(ln 324) = 200(324)
(b) (4) = 200(324)4 ≈ 22,040 cells
(c) 0 () = 200(324) · ln 324, so 0 (4) = 200(324)4 · ln 324 ≈ 25,910 cells per hour
(d) 200(324) = 10,000 ⇒
(324) = 50 ⇒ ln 324 = ln 50 ⇒ = (ln 50)(ln 324) ≈ 333 hours
90. (a) If () is the mass remaining after years, then () = (0) = 100 . (524) = 100524 = 12 · 100
524 = 12
⇒
1
⇒ 524 = − ln 2 ⇒ = − 524
ln 2 ⇒ () = 100−(ln 2)524 = 100 · 2−524 . Thus,
(20) = 100 · 2−20524 ≈ 71 mg.
(b) 100 · 2−524 = 1 ⇒ 2−524 =
91. Let () =
1
100
⇒ −
1
ln 2 = ln
524
100
⇒ = 524
ln 100
≈ 348 years
ln 2
64
=
= (1 + )−1 , where = 64, = 31, and = −07944.
1 + 31−07944
1 +
0 () = −(1 + )−2 ( ) = − (1 + )−2
00 () = − [ − 2(1 + )−3 ( )] + (1 + )−2 ( − 2 )
= −2 (1 + )−3 [−2 + (1 + )] = −
2 (1 − )
(1 + )3
The population is increasing most rapidly when its graph changes from CU to CD; that is, when 00 () = 0 in this case.
00 () = 0 ⇒ = 1 ⇒ =
1
1
ln
=
1
⇒ = ln
1
⇒ =
ln(1)
ln(131)
=
≈ 432 days. Note that
−07944
=
= , one­half the limit of as → ∞.
=
=
1 + (1)
1+1
2
1 + (1) ln(1)
1 + ln(1)
92. Let = 4. Then = 4 and
4
0
1
=
16 + 2
1
0
1
1
· 4 =
16 + 162
4
1
0
1 −1 1
tan = 14 (tan−1 1 − tan−1 0) = 14 4 − 0 = 16
=
.
1 + 2
4
0
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°
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CHAPTER 6 INVERSE FUNCTIONS
93. Let = −2 2 . Then = −4 and
94.
5
2
1
0
2
−2 =
−2
−2 1
− 4 = − 14 0 = − 14 (−2 − 1) = 14 (1 − −2 ).
0
5
= 12 ln |1 + 2| 2 = 12 (ln 11 − ln 5) = 12 ln 11
5
1 + 2
95. Let = , so = When = 0 = 1; when = 1 = . Thus,
1
0
=
1 + 2
1
1
= arctan 1 = arctan − arctan 1 = arctan − 4 .
1 + 2
96. Let = sin . Then = cos , so
2
0
cos
=
1 + sin2
97. Let =
1
0
1
1
= tan−1 0 = tan−1 1 − tan−1 0 = 4 − 0 = 4 .
1 + 2
√
. Then = √
2
⇒
√ = 2
= 2 + = 2 + .
sin(ln )
=
sin = − cos + = − cos(ln ) + .
1
98. Let = ln . Then = , so
√
√
99. Let = 2 + 2. Then = (2 + 2) = 2( + 1) and
+1
=
2 + 2
1
2
= 12 ln || + = 12 ln 2 + 2 + .
100. Let = 1 + cot . Then = − csc2 , so
101. Let = ln(cos ). Then =
tan ln(cos ) = −
1
√
=
2
1 − 4
103. Let = tan . Then = sec2 and
105.
−1
sinh =
−2
107. cos ≤ 1
1
(−) = − ln || + = − ln |1 + cot | + .
√
= 12 sin−1 + = 12 sin−1 2 + .
1 − 2
2tan sec2 =
2 =
2tan
2
+ =
+ .
ln 2
ln 2
1
cosh +
2 + 1
=
106. 1 + 2 2
− sin
= − tan ⇒
cos
102. Let = . Then = 2 , so
csc2
=
1 + cot
= − 12 2 + = − 12 [ ln(cos )]2 + .
2
104.
−1
−1
3
1
1
1 2
+ ln ||
+ ln |−1| − 2 + ln |−2| = − − ln 2
=
+
=
2
2
2
−2
−2
⇒
√
√
1 + 2 2 =
⇒ cos ≤
⇒
1
0
⇒
1
1√
1
1 + 2 ≥ 0 = 0 = − 1
0
cos ≤
1
1
= 0 = − 1
0
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°
CHAPTER 6 REVIEW
108. For 0 ≤ ≤ 1, 0 ≤ sin−1 ≤ 2 , so
109. () =
√
1
110. () =
2
ln
0 () =
1
sin−1 ≤
0
⇒ 0 () =
√
1
1
1
2 = 4 2 0 = 4 .
0
√
√
¤
695
√
√
1
√ =
= √
= √
2
2
2
− ⇒
2
ln
2
− = −
ln
0
2
− +
2
0
2
5
1 ln 5
1
1
ln
=
() =
−
5−1 1
4 0
ln 5
= 14 12 2
= 18 (ln 5)2
111. (a) avg =
2
− = −−(ln )
2
2
2
−(ln )
1
+2−4
+−(2) (2) = −
= ln
= (1)
0
(b) () =
ln
(1) − (ln ) · 1
1 − ln
, [1 5]. 0 () =
=
= 0 ⇔ 1 − ln = 0 ⇔ ln = 1 ⇔ = .
2
2
(1) = 0, () = 1 ≈ 037, and (5) = 15 ln 5 ≈ 032. So () = 1 is the absolute maximum value and (1) = 0 is
the absolute minimum value.
112. =
0
0
1
1
(− − ) + 0 − − = −− − −2 + + − 0
1
2
by cylindrical shells. Let = 2
1 + 4
−2
= [(−1 − 1) − (−2 − −2 )] + [( + −1 ) − (1 + 1)] = 2 + + −1 + −2 − 4
113. =
0
=
1
0
2
1
=
.
= tan−1 0 = tan−1 1 − tan−1 0 =
2
1+
4
4
114. () = + 2 +
so 0 (1) =
116.
⇒ 0 () = 1 + 2 + and (0) = 1 ⇒ (1) = 0 [where = −1 ],
1
1
1
= 0
= .
0 ((1))
(0)
2
115. () = ln + tan−1
0 () =
⇒ = 2 . Then
⇒ (1) = ln 1 + tan−1 1 = 4
1
1
1
2
1
+
= 0
=
= .
, so 0
2
1+
4
(1)
32
3
⇒ 4 = 1 [where = −1 ].
The area of such a rectangle is just the product of its sides, that is, () = · − .
We want to find the maximum of this function, so we differentiate:
0 () = ( − − ) + − (1) = − (1 − ). This is 0 only at = 1, and changes
from positive to negative there, so by the First Derivative Test this gives a local
maximum. So the largest area is (1) = 1.
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°
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CHAPTER 6 INVERSE FUNCTIONS
We find the equation of a tangent to the curve = − , so that we can find the
117.
­ and ­intercepts of this tangent, and then we can find the area of the triangle.
−
= −− ,
The slope of the tangent at the point − is given by
=
and so the equation of the tangent is − − = −− ( − ) ⇔
= − ( − + 1).
The ­intercept of this line is = − ( − 0 + 1) = − ( + 1). To find the ­intercept we set = 0 ⇒
− ( − + 1) = 0 ⇒ = + 1. So the area of the triangle is () = 12 − ( + 1) ( + 1) = 12 − ( + 1)2 . We
differentiate this with respect to : 0 () = 12 − (2)( + 1) + ( + 1)2 − (−1) = 12 − 1 − 2 . This is 0
at = ±1, and the root = 1 gives a maximum, by the First Derivative Test. So the maximum area of the triangle is
(1) = 12 −1 (1 + 1)2 = 2−1 = 2.
118. Using Theorem 4.2.4 with = 0, = 1, ∆ =
1
1
1
, and = 0 + (1) = , we have 0 = lim
.
→∞ =1
This series is a geometric series with = = 1 , so
= 1
=1
1
= lim
1
→∞ =1
0
(1 ) − 1
−1
= 1 1
1 − 1
−1
1
. As → ∞, 1 → 0+ , so 1 → 0 = 1.
= lim ( − 1)1 1
→∞
−1
Let = 1. Then 1 − 1 = − 1 → 0+ as → ∞, so l’Hospital’s Rule gives lim
1
=
0
lim ( − 1)
→0+
⇒
→0+ − 1
= − 1.
→0+ − 1
= lim
1
→0+
= 1 and we have
lim
+ 1 − + 1 H
+ 1 ln − + 1 ln
= ln − ln = (−1), so is continuous at −1.
= lim
→−1
→−1
+1
1
119. lim () = lim
→−1
120. Let 1 = arccot , so cot 1 = = 1.
1
So sin(arccot ) = sin 1 = √
.
2 + 1
1
1
.
Let 2 = arctan √
, so tan 2 = √
2 + 1
2 + 1
2
√
2 + 1
+1
.
=
Hence, cos{arctan[sin(arccot )]} = cos 2 = √
2
2 + 2
+2
121. Using FTC1, we differentiate both sides of the given equation,
1
() = ( − 1)2 +
() = ( − 1) · 2 · 2 + 2 + − () ⇒ ()(1 − − ) = 2( − 1)2 + 2
() =
−
() , and get
1
⇒
2 [1 + 2( − 1)]
2 (2 − 1)
=
.
−
1−
1 − −
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
CHAPTER 6 REVIEW
122. The area () =
0
sin(2 ) , and the area () = 12 sin(2 ). Since lim () = 0 = lim (), we can use
→0+
→0+
l’Hospital’s Rule:
() H
sin(2 )
= lim 1
→0+ ()
→0+
sin(2 ) + 12 [2 cos(2 )]
2
lim
H
= lim
→0+
[by FTC1 and the Product Rule]
2 cos(2 )
cos(2 ) − 23 sin(2 ) + 2 cos(2 )
= lim
→0+
2
2
2 cos(2 )
=
=
3 cos(2 ) − 22 sin(2 )
3−0
3
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
¤
697
PROBLEMS PLUS
2
2
1. Let = () = − . The area of the rectangle under the curve from − to is () = 2− where ≥ 0. We maximize
2
2
2
(): 0 () = 2− − 42 − = 2− 1 − 22 = 0 ⇒ = √12 . This gives a maximum since 0 () 0
for 0 ≤ √12 and 0 () 0 for √12 . We next determine the points of inflection of (). Notice that
2
0 () = −2− = −(). So 00 () = −0 () and hence, 00 () 0 for − √12 √12 and 00 () 0 for − √12
and √12 . So () changes concavity at = ± √12 , and the two vertices of the rectangle of largest area are at the inflection
points.
2. We use proof by contradiction. Suppose that log2 5 is a rational number. Then log2 5 = where and are positive
integers ⇒ 2 = 5 ⇒ 2 = 5 . But this is impossible since 2 is even and 5 is odd. So log2 5 is irrational.
3. ln 2 − 2 − 2 ≤ 0
⇒ 2 − 2 − 2 ≤ 0 = 1 ⇒ 2 − 2 − 3 ≤ 0 ⇒ ( − 3)( + 1) ≤ 0 ⇒ ∈ [−1 3].
√
√
Since the argument must be positive, 2 − 2 − 2 0 ⇒ − 1 − 3 − 1 + 3 0 ⇒
√
√
√
√
∈ −∞ 1 − 3 ∪ 1 + 3 ∞ . The intersection of these intervals is −1 1 − 3 ∪ 1 + 3 3 .
4.
1
1
1
1
1
1
+
+
+
+
=
log
log
log
log2 log3 log5
log 2
log 3
log 5
[Change of Base formula]
=
log 2
log 3
log 5
+
+
log
log log
=
log(2 · 3 · 5)
log 2 + log 3 + log 5
=
log
log
[Law 1 of Logarithms]
1
log 30
1
=
=
log
log
log30
log 30
[Change of Base formula]
=
5. () has the form () , so it will have an absolute maximum (minimum) where has an absolute maximum (minimum).
2
() = 10| − 2| − =
0
() =
10( − 2) − 2
if − 2 0
10[−( − 2)] − 2 if − 2 0
=
−2 + 10 − 20 if 2
−2 − 10 + 20 if 2
⇒
−2 + 10 if 2
−2 − 10 if 2
0 () = 0 if = −5 or = 5, and 0 (2) does not exist, so the critical numbers of are −5, 2, and 5. Since 00 () = −2 for
all 6= 2, is concave downward on (−∞ 2) and (2 ∞), and will attain its absolute maximum at one of the critical
numbers. Since (−5) = 45, (2) = −4, and (5) = 5, we see that (−5) = 45 is the absolute maximum value of . Also,
lim () = −∞, so lim () = lim () = 0 But () 0 for all , so there is no absolute minimum value of .
→∞
→∞
→∞
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
699
700
¤
CHAPTER 6 PROBLEMS PLUS
6. For =
=
2
−2
4
0
4
(−2) , let = − 2 so that = + 2 and = . Then
4
( + 2) =
2
7. Consider the statement that
−2
2
4
4
4
4
+ −2 2 = 0 [by 4.5.6] + 2 0 (−2) = 2.
( sin ) = sin( + ). For = 1,
( sin ) = sin + cos , and
sin( + ) = [sin cos + cos sin ] =
since tan =
⇒ sin =
sin + cos = sin + cos
and cos = . So the statement is true for = 1.
Assume it is true for = . Then
+1
sin( + ) = sin( + ) + cos( + )
( sin ) =
+1
= [ sin( + ) + cos( + )]
But
sin[ + ( + 1)] = sin[( + ) + ] = sin( + ) cos + sin cos( + ) = sin( + ) + cos( + ).
Hence, sin( + ) + cos( + ) = sin[ + ( + 1)]. So
+1
( sin ) = [ sin(+ )+ cos(+)] = [ sin( +( + 1))] = +1 [sin(+ ( + 1))].
+1
Therefore, the statement is true for all by mathematical induction.
8. Let = tan−1 . Then tan = , so from the triangle we see that
sin(tan−1 ) = sin = √
Using this fact we have that
1 + 2
sinh
sinh
sin(tan−1 (sinh )) =
= tanh .
=
2
cosh
1 + sinh
Hence, sin−1 (tanh ) = sin−1 (sin(tan−1 (sinh ))) = tan−1 (sinh ).
9. We first show that
0 () =
tan−1 for 0. Let () = tan−1 −
. Then
1 + 2
1 + 2
1
1(1 + 2 ) − (2)
(1 + 2 ) − (1 − 2 )
22
−
=
=
0 for 0. So () is increasing
2
2
2
2
2
1+
(1 + )
(1 + )
(1 + 2 )2
on (0 ∞). Hence, 0 ⇒ 0 = (0) () = tan−1 −
. So
tan−1 for 0 . We next show that
1 + 2
1 + 2
tan−1 for 0. Let () = − tan−1 . Then 0 () = 1 −
1
2
=
0. Hence, () is increasing
1 + 2
1 + 2
on (0 ∞). So for 0 , 0 = (0) () = − tan−1 . Hence, tan−1 for 0, and we conclude that
tan−1 for 0.
1 + 2
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
CHAPTER 6 PROBLEMS PLUS
10. The shaded region has area
1
0
() = 13 . The integral
1
0
¤
701
−1 ()
gives the area of the unshaded region, which we know to be 1 − 13 = 23 .
So
1
0
−1 () = 23 .
11. By the Fundamental Theorem of Calculus, () =
√
√
1 + 3 ⇒ 0 () = 1 + 3 0 for −1.
1
So is increasing on (−1 ∞) and hence is one­to­one. Note that (1) = 0, so −1 (1) = 0 ⇒
( −1 )0(0) = 1 0 (1) = √12 .
12. = √
2 − 1
√
2
sin
√
−√
arctan
. Let = + 2 − 1. Then
2
2
−1
+ − 1 + cos
2
1
1
cos ( + cos ) + sin2
0 = √
−√
·
·
2
2
( + cos )2
2 − 1
2 − 1 1 + sin ( + cos )
2
cos + cos2 + sin2
2
cos + 1
1
1
−√
·
−√
·
= √
= √
( + cos )2 + sin2
2 − 1
2 − 1
2 − 1
2 − 1 2 + 2 cos + 1
=
2 − 1
2 + 2 cos + 1 − 2 cos − 2
√
= √
2 − 1 (2 + 2 cos + 1)
2 − 1 (2 + 2 cos + 1)
But 2 = 22 + 2
√
√
2 − 1 − 1 = 2 + 2 − 1 − 1 = 2 − 1, so 2 + 1 = 2, and 2 − 1 = 2( − 1).
√
√
2( − 1)
− 1
So 0 = √
= √
. But − 1 = 2 + 2 − 1 − 1 = 2 − 1,
2
2
− 1 (2 + 2 cos )
− 1 ( + cos )
so 0 = 1( + cos ).
13. If = lim
→∞
+
−
= lim
, then has the indeterminate form 1∞ , so
1
1
−
+
+
ln( + ) − ln( − ) H
+
−
= lim ln
= lim
ln = lim ln
= lim
→∞
→∞
→∞
→∞
−
−
1
−12
→∞
( − ) − ( + ) −2
·
( + )( − )
1
22
2
= lim
= 2
→∞ 2 − 2
→∞ 1 − 22
= lim
Hence, ln = 2, so = 2 . From the original equation, we want = 1
⇒ 2 = 1 ⇒ = 12 .
14. We first present some preliminary results that we will invoke when calculating the limit.
(1) If = (1 + ) , then ln = ln(1 + ), and lim ln = lim ln(1 + ) = 0. Thus, lim (1 + ) = 0 = 1.
→0+
→0+
(2) If = (1 + ) , then ln = ln(1 + ), and implicitly differentiating gives us
→0+
0
=·
+ ln(1 + ) ⇒
1 +
[continued]
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
702
¤
CHAPTER 6 PROBLEMS PLUS
0 =
(3) If =
+ ln(1 + ) . Thus, = (1 + )
1 +
⇒ 0 = (1 + )
+ ln(1 + ) .
1 +
(1 + ) − ()
+ 2 − 2
, then 0 =
=
=
.
2
1 +
(1 + )
(1 + )2
(1 + )2
( + 2)1 − 1
1 [(1 + 2)1 − 1]
= lim 1
1
1
→∞ ( + 3)
→∞
−
[(1 + 3)1 − 1]
lim
[factor out 1 ]
(1 + 2)1 − 1
→∞ (1 + 3)1 − 1
= lim
(1 + 2) − 1
→0+ (1 + 3) − 1
2
+ ln(1 + 2)
(1 + 2)
1 + 2
H
= lim
3
→0+
+ ln(1 + 3)
(1 + 3)
1 + 3
= lim
[let = 1, form 00 by (1)]
[by (2)]
2
+ ln(1 + 2)
(1 + 2)
1
+
2
· lim
= lim
3
→0+ (1 + 3)
→0+
+ ln(1 + 3)
1 + 3
2
+ ln(1 + 2)
1
1
+
2
= · lim
1 →0+ 3
+ ln(1 + 3)
1 + 3
[by (1), now form 00]
2
2
+
(1 + 2)2
1 + 2
= lim
3
3
→0+
+
(1 + 3)2
1 + 3
[by (3)]
H
=
2+2
4
2
= =
3+3
6
3
15. As in Exercise 4.3.70, assume that the integrand is defined at = 0 so that it is continuous there. By l’Hospital’s Rule and the
Fundamental Theorem, using the notation exp() = ,
lim
→0
0
(1 − tan 2)1 H
(1 − tan 2)1
ln(1 − tan 2)
= exp ln lim (1 − tan 2)1 = exp lim
= lim
→0
→0
→0
1
−2 sec2 2
−2 · 12
H
= exp lim
= exp
= −2
→0 1 − tan 2
1−0
16. Case (i) (first graph): For + ≥ 0, that is, ≥ −, | + | = + ≤
⇒ ≤ − .
Note that = − is always above the line = − and that = − is a slant asymptote.
Case (ii) (second graph): For + 0, that is, −, | + | = − − ≤
⇒ ≥ − − .
Note that − − is always below the line = − and = − is a slant asymptote.
Putting the two pieces together gives the third graph.
[continued]
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
CHAPTER 6 PROBLEMS PLUS
¤
703
17. Both sides of the inequality are positive, so cosh(sinh ) sinh(cosh )
⇔ cosh2 (sinh ) sinh2 (cosh ) ⇔ sinh2 (sinh ) + 1 sinh2 (cosh )
⇔ 1 [sinh(cosh ) − sinh(sinh )][sinh(cosh ) + sinh(sinh )]
− −
+ −
− −
+ −
− sinh
sinh
+ sinh
⇔ 1 sinh
2
2
2
2
[use the addition formulas and cancel]
⇔ 1 [2 cosh(2) sinh(2)][2 sinh(2) cosh(2)]
⇔ 1 [2 sinh(2) cosh(2)][2 sinh(2) cosh(2)] ⇔ 1 sinh sinh − ,
by the half­angle formula. Now both and − are positive, and sinh for 0, since sinh 0 = 0 and
(sinh − )0 = cosh − 1 0 for 0, so 1 = − sinh sinh − . So, following this chain of reasoning
backward, we arrive at the desired result.
18. First, we recognize some symmetry in the inequality:
that
+
≥ 2
⇔
·
≥ · . This suggests that we need to show
≥ for 0. If we can do this, then the inequality
≥ is true, and the given inequality follows. () =
0 () =
−
( − 1)
=
= 0 ⇒ = 1. By the First Derivative Test, we have a minimum of (1) = , so
2
2
≥ for all .
Let () = 2 and () =
19.
√
[ 0]. From the graphs of and ,
we see that will intersect exactly once when and share a tangent
line. Thus, we must have = and 0 = 0 at = .
() = () ⇒ 2 =
and
So we must have
= 212 = 2
√
=
√
4
√
≈ 3297.
⇒
√ 2
=
4
√
()
√
2
⇒ 2 =
⇒ = 14 . From (), 2(14) =
14 ⇒
0 () = 0 () ⇒ 22 =
√ .
4
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
⇒
704
¤
CHAPTER 6 PROBLEMS PLUS
20. We see that at = 0, () = = 1 + = 1, so if = is to lie above = 1 + ,
the two curves must just touch at (0 1), that is, we must have 0 (0) = 1. [To see this
analytically, note that ≥ 1 + ⇒ − 1 ≥ ⇒
− 1
≥ 1 for 0, so
− 1
≥ 1. Similarly, for 0, − 1 ≥ ⇒
→0+
0 (0) = lim
0 (0) = lim
→0−
− 1
≤ 1, so
− 1
≤ 1.
Since 1 ≤ 0 (0) ≤ 1, we must have 0 (0) = 1.] But 0 () = ln ⇒ 0 (0) = ln , so we have ln = 1 ⇔
= .
Another method: The inequality certainly holds for ≤ −1, so consider −1, 6= 0. Then ≥ 1 + ⇒
≥ (1 + )1 for 0 ⇒ ≥ lim (1 + )1 = , by Equation 6.4.8. Also, ≥ 1 + ⇒ ≤ (1 + )1
→0+
for 0 ⇒ ≤ lim (1 + )
1
→0−
= . So since ≤ ≤ , we must have = .
10
21. Suppose that the curve = intersects the line = . Then 0 = 0 for some 0 0, and hence = 0
. We find the
, 0, because if is larger than the maximum value of this function, then the curve =
1
1
1 1
(1 − ln ). This is 0 only where
= 1
does not intersect the line = . 0 () = (1) ln − 2 ln + ·
2
maximum value of () =
1
= , and for 0 , 0 () 0, while for , 0 () 0, so has an absolute maximum of () = 1 . So if
= intersects = , we must have 0 ≤ 1 . Conversely, suppose that 0 ≤ 1 . Then ≤ , so the graph of
= lies below or touches the graph of = at = . Also 0 = 1 0, so the graph of = lies above that of =
at = 0. Therefore, by the Intermediate Value Theorem, the graphs of = and = must intersect somewhere between
= 0 and = .
22.
= 3 +
⇒ 0 = 32 +
⇒ 00 = 6 + . The curve will have
inflection points when 00 changes sign. 00 = 0 ⇒ −6 = , so 00 will change
sign when the line = −6 intersects the curve = (but is not tangent to it).
Note that if = 0, the curve is just = , which has no inflection point.
The first figure shows that for 0, = −6 will intersect = once, so
= 3 + will have one inflection point.
The second figure shows that for 0, the line = −6 can intersect the curve
= in two points (two inflection points), be tangent to it (no inflection point), or not
intersect it (no inflection point). The tangent line at ( ) has slope , but from the
diagram we see that the slope is
. So
=
⇒ = 1. Thus, the slope is .
The line = −6 must have slope greater than , so −6 ⇒ −6.
Therefore, the curve = 3 + will have one inflection point if 0 and two inflection points if −6.
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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