5128_ch04ansTE_pp652-661
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Additional Answers 659<br />
65. x 2 2, x 3 4, x 4 8, and x 5 16;<br />
⏐x n ⏐ 2 n1 .<br />
[–10, 10] by [–3, 3]<br />
66. (a) b 0 f(a), b 1 f (a), and b 2 f (a)<br />
.<br />
2<br />
(b) 1 x x 2<br />
(c) As one zooms in, the two graphs quickly become indistinguishable.<br />
They appear to be identical.<br />
(d) The quadratic approximation is<br />
1 (x 1) (x 1) 2 .<br />
As one zooms in, the two graphs quickly become indistinguishable.<br />
They appear to be identical.<br />
(e) The quadratic approximation is<br />
x x2<br />
1 . 2 8<br />
As one zooms in, the two graphs quickly become indistinguishable.<br />
They appear to be identical.<br />
(f) For f :1 x ;<br />
for g:1 (x 1) 2 x;<br />
x<br />
for h:1 2<br />
67. Finding a zero of sin x by Newton’s method would use the recursive formula<br />
sin(<br />
xn)<br />
x n1 x n x c os(<br />
xn)<br />
n tan x n , and that is exactly what the calculator<br />
would be doing. Any zero of sin x would be a multiple of p.<br />
69. lim tan x<br />
lim sin x/ cos x<br />
<br />
x→0 x x→0 x<br />
lim<br />
x→0 1<br />
sin x<br />
cos x x<br />
lim 1<br />
<br />
x→0 cos x lim sin x<br />
<br />
x→0 x<br />
(1)(1) 1.<br />
70. g(a) c, so if E(a) 0, then g(a) f(a) and<br />
c f(a). Then<br />
E(x) f(x) g(x) f(x) f(a) m(x a).<br />
E(x)<br />
Thus, f (x) f (a)<br />
m.<br />
x a x a<br />
lim f(x ) f(a)<br />
f (a), so if the limit of<br />
x→a x a<br />
E(x)<br />
is zero, then m f (a) and g(x) L(x).<br />
x a<br />
Section 4.6<br />
Exercises 4.6<br />
45. (a) The point being plotted would correspond to a point on the edge of the<br />
wheel as the wheel turns.<br />
(b) One possible answer:<br />
16pt, where t is in seconds.<br />
(c) Assuming counterclockwise motion, the rates are as follows.<br />
p 4 ; d x<br />
71.086 ft/sec<br />
dt<br />
d y<br />
71.086 ft/sec<br />
dt<br />
2 : d x<br />
100.531 ft/sec<br />
dt<br />
d y<br />
0 ft/sec<br />
dt<br />
p: d x<br />
0 ft/sec<br />
dt<br />
d y<br />
100.531 ft/sec<br />
dt<br />
46. (a) One possible answer:<br />
x 30 cos , y 40 30 sin <br />
(b) Assuming counterclockwise motion, the rates are as follows.<br />
When t 5: d x<br />
0 ft/sec<br />
dt<br />
d y<br />
18.850 ft/sec<br />
dt<br />
when t 8: d x<br />
17.927 ft/sec<br />
dt<br />
d y<br />
5.825 ft/sec<br />
dt<br />
Quick Quiz (Sections 4.4–4.6)<br />
4. (a) Since d y<br />
1 1<br />
dx<br />
2 x1/2 , the slope at (25, 5) is . 1 0<br />
1<br />
The linearization is L(x) (x 25) 5, so L(26) 5.1.<br />
1 0<br />
f(<br />
5)<br />
(b) x 2 5 5 52 26<br />
5.1.<br />
f (<br />
5)<br />
2(5)<br />
(c) The appropriate linearization is for the curve y 3 x at the point (27, 3).<br />
Since d y<br />
1 1<br />
dx<br />
3 x2/3 , the slope at (27, 3) is . The linearization is<br />
2 7<br />
1<br />
L(x) (x 27) 3, so L(26) 3 – 1/27 2.963.<br />
2 7<br />
Review Exercises<br />
5. (a) Approximately (, 0.385]<br />
(b) Approximately [0.385, )<br />
(c) None (d) (, )<br />
(e) Local maximum at (0.385, 1.215)<br />
(f) None<br />
6. (a) [1, ) (b) (, 1]<br />
(c) (, )<br />
(d) None<br />
(e) Local minimum at (1, 0)<br />
(f ) None<br />
7. (a) [0, 1) (b) (1, 0]<br />
(c) (1, 1)<br />
(d) None<br />
(e) Local minimum at (0, 1)<br />
(f) None<br />
8. (a) (, 2 1/3 ] (, 0.794]<br />
(b) [2 1/3 ,1) [0.794, 1) and (1, )<br />
(c) (, 2 1/3 ) (, 1.260) and (1, )<br />
(d) (1.260, 1)<br />
(e) Local maximum at<br />
21/3 , 2 3 21/3 (0.794, 0.529)<br />
(f)<br />
21/3 , 1 3 21/3 (1.260, 0.420)<br />
<br />
9. (a) None (b) [1, 1]<br />
(c) (1, 0) (d) (0, 1)<br />
(e) Local maximum at (1, p);<br />
local minimum at (1, 0)<br />
(f)<br />
0, p 2