Multivariable Calculus 9th Edition by James Stewart solution manual

 ∈

−2sin

3
2sin
 
3
+

3

−2cos
 
3
+

6

and  = sin −1
 27 − 2
2

. The inverse for  =  6 +  4
( ≥ 0) is  =
√ 3
3
√  − 1 with  = 2sin  
3
+

3

, but because the domain of  is
 0
4
27
 , this expression is only
valid for  ∈
 0
4
27
 .
If we solve  =  6 +  4 for  using Maple, we get the two real solutions ±
√ 6
6
  13
( 23 − 2 13 + 4)
 13
,
where  = 108 + 12
√ 3  (27 − 4), and the inverse for  =  6
+  4 ( ≥ 0) is the positive solution, whose domain
is

4
27 ∞
 .
[continued]
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SECTION 1.5 INVERSEFUNCTIONSAND LOGARITHMS
¤
57
Mathematica also gives two real solutions, equivalent to those of Maple.
The positive one is
√ 6
6

3
√ 4 13
+ 2
3
√ 2 −13
− 2

, where
 = −2 + 27 + 3 √ 3 √  √ 27 − 4. Although this expression also has domain

4
27 ∞
 , Mathematica is mysteriously able to plot the solution for all  ≥ 0.
(b)
67. (a)  = () = 100 · 2 3 ⇒

100
= 2 3 ⇒ log 2


100

=

3
⇒  = 3log 2


100

. Using the Change of Base
Formula, we can write this as  =  −1 () = 3 ·
ln(100)
ln2
. This function tells us how long it will take to obtain
 bacteria (given the number ).
(b)  = 50,000 ⇒  =  −1 (50,000) = 3 ·
ln  50,000
100

ln2
= 3
 ln500
ln2

≈ 269 hours
68. (a) We write  =  0 (1 −  − ) and solve for :

 0
= 1 −  − ⇒  − = 1 −

 0
⇒
−


= ln

1 −

 0

⇒  = −ln

1 −

 0

. This formula gives the time (in seconds) needed after a discharge to
obtain a given charge .
(b) We set  = 09 0 and  = 50 to get  = −50ln

1 −
09 0
 0

= −50ln(01) ≈ 1151 seconds. It will take
approximately 115 seconds—just shy of two minutes—to recharge the capacitors to 90% of capacity.
69. (a) cos −1 (−1) =  because cos = −1 and  is in the interval [0] (the range of cos −1 ).
(b) sin −1 (05) =

6
because sin

6
= 05 and

6
is in the interval
 − 
2 

2

(the range of sin −1 ).
70. (a) tan −1 √ 3 =

3
because tan

3
=
√ 3 and

3
is in the interval
 − 
2 

2

(the range of tan −1 ).
(b) arctan(−1) = − 
4
because tan  − 
4

= −1 and − 
4
is in the interval
 − 
2 

2

(the range of arctan).
71. (a) csc
−1 √ 2 = 
4
because csc

4
=
√ 2 and

4
is in
 0

2

∪
 
3
2

(the range of csc −1 ).
(b) arcsin1 =

2
because sin

2
= 1 and

2
is in
 − 
2 

2

(the range of arcsin).
72. (a) sin
−1 (−1 √ 2) = − 
4
because sin  − 
4

= −1 √ 2 and − 
4
is in
 − 
2 

2
 .
(b) cos −1 √ 32  =

6
because cos

6
=
√ 32 and

6
is in [0].
73. (a) cot −1  − √ 3