Solução_Calculo_Stewart_6e

F.
208 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
29. (a) f(x) =x 2 − a ⇒ f 0 (x) =2x, so Newton’s method gives
x n+1 = x n − x2 n − a
2x n
= x n − 1 2 x n + a
2x n
= 1 2 x n +
a
2x n
= 1 2
x n + a
x n
.
(b) Using (a) with a = 1000 and x 1 = √ 900 = 30,wegetx 2 ≈ 31.666667, x 3 ≈ 31.622807,andx 4 ≈ 31.622777 ≈ x 5 .
So √ 1000 ≈ 31.622777.
31. f(x) =x 3 − 3x +6 ⇒ f 0 (x) =3x 2 − 3. Ifx 1 =1,thenf 0 (x 1)=0and the tangent line used for approximating x 2 is
horizontal. Attempting to find x 2 results in trying to divide by zero.
33. For f(x) =x 1/3 , f 0 (x) = 1 3 x−2/3 and
x n+1 = x n − f(x n)
f 0 (x n ) = x n − x1/3 n
1
3 x−2/3 n
= x n − 3x n = −2x n .
Therefore, each successive approximation becomes twice as large as the
previous one in absolute value, so the sequence of approximations fails to
converge to the root, which is 0. Inthefigure, we have x 1 =0.5,
x 2 = −2(0.5) = −1,andx 3 = −2(−1) = 2.
35. (a) f(x) =x 6 − x 4 +3x 3 − 2x ⇒ f 0 (x) =6x 5 − 4x 3 +9x 2 − 2 ⇒
f 00 (x) =30x 4 − 12x 2 +18x. Tofind the critical numbers of f,we’llfind the
zeros of f 0 .Fromthegraphoff 0 , it appears there are zeros at approximately
x = −1.3, −0.4,and0.5. Tryx 1 = −1.3 ⇒
x 2 = x 1 − f 0 (x 1)
≈−1.293344 ⇒
f 00 x3 ≈−1.293227 ≈ x4.
(x 1 )
Now try x 1 = −0.4 ⇒ x 2 ≈−0.443755 ⇒ x 3 ≈−0.441735 ⇒ x 4 ≈−0.441731 ≈ x 5 . Finally try
x 1 =0.5 ⇒ x 2 ≈ 0.507937 ⇒ x 3 ≈ 0.507854 ≈ x 4 . Therefore, x = −1.293227, −0.441731,and0.507854 are
all the critical numbers correct to six decimal places.
(b) There are two critical numbers where f 0 changes from negative to positive, so f changes from decreasing to increasing.
f(−1.293227) ≈−2.0212 and f(0.507854) ≈−0.6721,so−2.0212 is the absolute minimum value of f correct to four
decimal places.
37. From the figure, we see that y = f(x) =e cos x is periodic with period 2π. To
find the x-coordinates of the IP, we only need to approximate the zeros of y 00
on [0,π]. f 0 (x) =−e cos x sin x ⇒ f 00 (x) =e cos x sin 2 x − cos x .Since
e cos x 6=0, we will use Newton’s method with g(x) =sin 2 x − cos x,
g 0 (x) =2sinx cos x +sinx,andx 1 =1. x 2 ≈ 0.904173,
x 3 ≈ 0.904557 ≈ x 4.Thus,(0.904557, 1.855277) is the IP.