Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 4.8 NEWTON’S METHOD ¤ 207
x 1 = −1.2
x 1 =1.1
x 1 =3
x 1 = −1.9
x 5 ≈−1.93822883 ≈ x 6
x 3 ≈−1.93828380
x 4 ≈−1.93822884
x 3 ≈−1.21997997 ≈ x 4
x 3 ≈ 1.13929741
x 4 ≈ 1.13929375 ≈ x 5
x 3 ≈ 2.98984106
x 4 ≈ 2.98984102 ≈ x 5
x 2 ≈−1.94278290 x 2 ≈−1.22006245 x 2 ≈ 1.14111662 x 2 ≈ 2.99
To eight decimal places, the roots of the equation are −1.93822883, −1.21997997, 1.13929375,and2.98984102.
25. From the graph, y = x 2√ 2 − x − x 2 and y =1intersect twice, at x ≈−2 and
at x ≈−1. f(x) =x 2 √ 2 − x − x 2 − 1
⇒
f 0 (x) =x 2 · 1
2 (2 − x − x2 ) −1/2 (−1 − 2x)+(2− x − x 2 ) 1/2 · 2x
= 1 2 x(2 − x − x2 ) −1/2 [x(−1 − 2x)+4(2− x − x 2 )]
= x(8 − 5x − 6x2 )
2 (2 + x)(1 − x) ,
√
so x n+1 = x n − x2 n 2 − xn − x 2 n − 1
x n(8 − 5x n − 6x 2 n)
try x 1 = −1.95.
2 (2 + x n)(1 − x n)
.Tryingx 1 = −2 won’t work because f 0 (−2) is undefined, so we’ll
x 1 = −1.95
x 1 = −0.8
x 2 ≈−1.98580357
x 2 ≈−0.82674444
x 3 ≈−1.97899778
x 3 ≈−0.82646236
x 4 ≈−1.97807848
x 4 ≈−0.82646233 ≈ x 5
x 5 ≈−1.97806682
x 6 ≈−1.97806681 ≈ x 7
To eight decimal places, the roots of the equation are −1.97806681 and −0.82646233.
27. Solving 4e −x2 sin x = x 2 − x +1is the same as solving
f(x) =4e −x2 sin x − x 2 + x − 1=0.
f 0 (x) =4e −x2 (cos x − 2x sin x) − 2x +1
⇒
x n+1 = x n −
4e −x2 n sin x n − x 2 n + x n − 1
4e −x2 n (cos x n − 2x n sin x n ) − 2x n +1 .
From the figure, we see that the graphs intersect at approximately x =0.2 and x =1.1.
x 1 =0.2
x 1 =1.1
x 2 ≈ 0.21883273
x 2 ≈ 1.08432830
x 3 ≈ 0.21916357
x 3 ≈ 1.08422462 ≈ x 4
x 4 ≈ 0.21916368 ≈ x 5
To eight decimal places, the roots of the equation are 0.21916368 and 1.08422462.