Solução_Calculo_Stewart_6e

F.
TX.10
40 ¤ CHAPTER 1 PRINCIPLES OF PROBLEM SOLVING
9. |x| + |y| ≤ 1. The boundary of the region has equation |x| + |y| =1. In quadrants
I,II,III,andIV,thisbecomesthelinesx + y =1, −x + y =1, −x − y =1,and
x − y =1respectively.
11. (log 2 3)(log 3 4)(log 4 5) ···(log 31 32) =
ln 3
ln 2
ln 4
ln 3
ln 5
···
ln 4
ln 32 ln 32 ln 25
= =
ln 31 ln 2 ln 2 = 5ln2
ln 2 =5
13. ln x 2 − 2x − 2 ≤ 0 ⇒ x 2 − 2x − 2 ≤ e 0 =1 ⇒ x 2 − 2x − 3 ≤ 0 ⇒ (x − 3)(x +1)≤ 0 ⇒ x ∈ [−1, 3].
Since the argument must be positive, x 2 − 2x − 2 > 0 ⇒ x − 1 − √ 3 x − 1+ √ 3 > 0 ⇒
x ∈ −∞, 1 − √ 3 ∪ 1+ √ 3, ∞ . The intersection of these intervals is −1, 1 − √ 3 ∪ 1+ √ 3, 3 .
15. Let d be the distance traveled on each half of the trip. Let t 1 and t 2 be the times taken for the first and second halves of the trip.
For the first half of the trip we have t 1 = d/30 and for the second half we have t 2 = d/60. Thus, the average speed for the
entire trip is
is 40 mi/h.
total distance
total time
= 2d
t 1 + t 2
=
2d
d
30 + d 60
17. Let S n be the statement that 7 n − 1 is divisible by 6.
• S 1 is true because 7 1 − 1=6is divisible by 6.
· 60
60 = 120d
2d + d = 120d =40. The average speed for the entire trip
3d
• Assume S k is true, that is, 7 k − 1 is divisible by 6. Inotherwords,7 k − 1=6m for some positive integer m. Then
7 k+1 − 1=7 k · 7 − 1=(6m +1)· 7 − 1=42m +6=6(7m +1), which is divisible by 6,soS k+1 is true.
• Therefore, by mathematical induction, 7 n − 1 is divisible by 6 for every positive integer n.
19. f 0 (x) =x 2 and f n+1 (x) =f 0 (f n (x)) for n =0, 1, 2,....
f 1 (x) =f 0 (f 0 (x)) = f 0
x
2 = x 22 = x 4 , f 2 (x) =f 0 (f 1 (x)) = f 0 (x 4 )=(x 4 ) 2 = x 8 ,
f 3(x) =f 0(f 2(x)) = f 0(x 8 )=(x 8 ) 2 = x 16 , .... Thus, a general formula is f n(x) =x 2n+1 .