5128_ch04ansTE_pp652-661

Additional Answers 659
65. x 2 2, x 3 4, x 4 8, and x 5 16;
⏐x n ⏐ 2 n1 .
[–10, 10] by [–3, 3]
66. (a) b 0 f(a), b 1 f (a), and b 2 f (a)
.
2
(b) 1 x x 2
(c) As one zooms in, the two graphs quickly become indistinguishable.
They appear to be identical.
(d) The quadratic approximation is
1 (x 1) (x 1) 2 .
As one zooms in, the two graphs quickly become indistinguishable.
They appear to be identical.
(e) The quadratic approximation is
x x2
1 . 2 8
As one zooms in, the two graphs quickly become indistinguishable.
They appear to be identical.
(f) For f :1 x ;
for g:1 (x 1) 2 x;
x
for h:1 2
67. Finding a zero of sin x by Newton’s method would use the recursive formula
sin(
xn)
x n1 x n x c os(
xn)
n tan x n , and that is exactly what the calculator
would be doing. Any zero of sin x would be a multiple of p.
69. lim tan x
lim sin x/ cos x
x→0 x x→0 x
lim
x→0 1
sin x
cos x x
lim 1
x→0 cos x lim sin x
x→0 x
(1)(1) 1.
70. g(a) c, so if E(a) 0, then g(a) f(a) and
c f(a). Then
E(x) f(x) g(x) f(x) f(a) m(x a).
E(x)
Thus, f (x) f (a)
m.
x a x a
lim f(x ) f(a)
f (a), so if the limit of
x→a x a
E(x)
is zero, then m f (a) and g(x) L(x).
x a
Section 4.6
Exercises 4.6
45. (a) The point being plotted would correspond to a point on the edge of the
wheel as the wheel turns.
(b) One possible answer:
16pt, where t is in seconds.
(c) Assuming counterclockwise motion, the rates are as follows.
p 4 ; d x
71.086 ft/sec
dt
d y
71.086 ft/sec
dt
2 : d x
100.531 ft/sec
dt
d y
0 ft/sec
dt
p: d x
0 ft/sec
dt
d y
100.531 ft/sec
dt
46. (a) One possible answer:
x 30 cos , y 40 30 sin
(b) Assuming counterclockwise motion, the rates are as follows.
When t 5: d x
0 ft/sec
dt
d y
18.850 ft/sec
dt
when t 8: d x
17.927 ft/sec
dt
d y
5.825 ft/sec
dt
Quick Quiz (Sections 4.4–4.6)
4. (a) Since d y
1 1
dx
2 x1/2 , the slope at (25, 5) is . 1 0
1
The linearization is L(x) (x 25) 5, so L(26) 5.1.
1 0
f(
5)
(b) x 2 5 5 52 26
5.1.
f (
5)
2(5)
(c) The appropriate linearization is for the curve y 3 x at the point (27, 3).
Since d y
1 1
dx
3 x2/3 , the slope at (27, 3) is . The linearization is
2 7
1
L(x) (x 27) 3, so L(26) 3 – 1/27 2.963.
2 7
Review Exercises
5. (a) Approximately (, 0.385]
(b) Approximately [0.385, )
(c) None (d) (, )
(e) Local maximum at (0.385, 1.215)
(f) None
6. (a) [1, ) (b) (, 1]
(c) (, )
(d) None
(e) Local minimum at (1, 0)
(f ) None
7. (a) [0, 1) (b) (1, 0]
(c) (1, 1)
(d) None
(e) Local minimum at (0, 1)
(f) None
8. (a) (, 2 1/3 ] (, 0.794]
(b) [2 1/3 ,1) [0.794, 1) and (1, )
(c) (, 2 1/3 ) (, 1.260) and (1, )
(d) (1.260, 1)
(e) Local maximum at
21/3 , 2 3 21/3 (0.794, 0.529)
(f)
21/3 , 1 3 21/3 (1.260, 0.420)
9. (a) None (b) [1, 1]
(c) (1, 0) (d) (0, 1)
(e) Local maximum at (1, p);
local minimum at (1, 0)
(f)
0, p 2