Programacion

Section 4.7 Newton's Method
13. For x! œ c!Þ$, the procedure converges to the root c!Þ$##")&$&ÞÞÞÞ (a)

265

(b)

(c)

(d) Values for x will vary. One possible choice is x! œ !Þ1.

(e) Values for x will vary. œ xn c
$

and c0.34730 (b) The estimated solutions of x$ c 3x c 1 œ 0 are c1.53209, c0.34730, 1.87939.

(d) The estimated x-values where g(x) œ 0.25x% c 1.5x# c x b 5 hashorizontal tangents are the roots of gw (x) œ x$ c 3x c 1, and these are c1.53209, c0.34730, 1.87939.

Ê x# œ 1.155327774 Ê x16 œ x17 œ 1.165561185 œ xn c
# $ %

if x! œ 0.5, then x% œ 0.630115396; if x! œ 2.5, then x% œ 2.57327196

#

$

b

16. f(x) œ x% c 2x$ c x# c 2x b 2 Ê f w (x) œ 4x$ c 6x# c 2x c 2 Ê xn

#

b

15. f(x) œ tan x c 2x Ê f w (x) œ sec# x c 2 Ê xn

1

œ xn ctan axn b c 2xn sec axn b

#

b

14. (a) f(x) œ x$ c 3x c 1 Ê f w (x) œ 3x# c 3 Ê xn

1

xn c 3xn c 1 3xn c 3

Ê the two negative zeros are c1.53209

; x! œ 1 Ê x" œ 12920445

1

xn c 2xn c xn c 2xn b 2 4xn c 6xn c 2xn c 2

;

266

Chapter 4 Applications of Derivatives

17. (a) The graph of f(x) œ sin 3x c 0.99 b x# in the window c2 Ÿ x Ÿ 2, c2 Ÿ y Ÿ 3 suggests threeroots. However, when you zoom in on the x-axis near x œ 1.2, you can see that the graph lies above the axis there. There are only two roots, one near x œ c1, the other near x œ 0.4. (b) f(x) œ sin 3x c 0.99 b x# Ê f w (x) œ 3 cos 3x b 2x are approximately 0.35003501505249 and c1.0261731615301 18. (a) Yes, three times as indicted by the graphs (b) f(x) œ cos 3x c x Ê f w (x) œ c3 sin 3x c 1 Ê xn 1œ xn c
cos a3xn b c xn c3 sin a3xn b c 1

; at

approximately c0.979367, c0.887726, and 0.39004 we have cos 3x œ x œ xn c
# %

x! œ c0.5, then x$ œ c0.5411961; the roots are approximately „ 0.5411961 and „ 1.30656296 because f(x) is an even function. 20. f(x) œ tan x Ê f w (x) œ sec# x Ê xn approximate 1 to be 3.14159. œ xn c
tan axn b sec axn b
#

1

; x! œ 3 Ê x" œ 3.13971 Ê x# œ3.14159 and we

21. From the graph we let x! œ 0.5 and f(x) œ cos x c 2x Ê x# œ .45018 Ê at x ¸ 0.45 we have cos x œ 2x.
b

Ê xn

1

œ xn c

cos axn b c 2xn csin axn b c 2

Ê x" œ .45063

22. From the graph we let x! œ c0.7 and f(x) œ cos x b x Ê x# œ c.73908 Ê at x ¸ c0.74 we have cos x œ cx.
b

Ê xn

1

œ xn c

xn b cos axn b 1 c sin axn b

Ê x" œ c.73944

$

b

19.f(x) œ 2x% c 4x# b 1 Ê f w (x) œ 8x$ c 8x Ê xn

b

b

b

Ê xn

1

œ xn c

#

sin (3xn ) c 0.99bxn 3 cos (3xn ) b 2xn

and the solutions

1

2xn c 4xn b1 8xn c 8xn

; if x! œ c2, then x' œ c1.30656296; if

Section 4.7 Newton's Method
23. If f(x) œ x$ b 2x c 4, then f(1) œ c1  0 and f(2) œ 8 € 0 Ê by the Intermediate Value Theorem the equation Then x! œ 1 Ê x" œ 1.2 Êx# œ 1.17975 Ê x$ œ 1.179509 Ê x% œ 1.1795090 Ê the root is approximately 1.17951. 24. We wish to solve 8x% c 14x$ c 9x# b 11x c 1 œ 0. Let f(x) œ 8x% c 14x$ c 9x# b 11x c 1, then
# $ #

267

1

œ xn c

1

œ xn c

x! c1.0 0.1 0.6 2.0

approximation of corresponding root c0.976823589 0.100363332 0.642746671 1.983713587 œ xi c œ xi c
$

procedure in problem 13 in this section. (a)For x! œ # or x! œ c!Þ), xi Ä c" as i gets large. (b) For x! œ c!Þ& or x! œ !Þ#&, xi Ä ! as i gets large. (c) For x! œ !Þ) or x! œ #, xi Ä " as i gets large. (d) (If your calculator has a CAS, put it in exact mode, otherwise approximate the radicals with a decimal value.) For x! œ c x! œ
È c 721 È21 7

or x! œ c
È c 721

È21 7 ,

Newton's method does not converge. The values of xi alternatebetween

or x! œ

as i increases.

26. (a) The distance can be represented by
# D(x) œ É(x c 2)# b ˆx# b " ‰ , where x   0. The #

distance D(x) is minimized when # f(x) œ (x c 2)# b ˆx# b " ‰ is minimized. If # f w (x) œ 4 ax$ b x c 1b and f w w (x) œ 4 a3x# b 1b € 0. Now f w (x) œ 0 Ê x$ b x c 1 œ 0 Ê x ax# b 1b œ 1 Ê xœ x"1. b
#

# f(x) œ (x c 2)# b ˆx# b " ‰ , then #

gave...