Solução_Calculo_Stewart_6e

F.
206 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
17. From the graph, we see that there appear to be points of intersection near
x = −0.7 and x =1.2. Solving x 4 =1+x is the same as solving
f(x) =x 4 − x − 1=0. f(x) =x 4 − x − 1 ⇒ f 0 (x) =4x 3 − 1,
so x n+1 = x n − x4 n − x n − 1
.
4x 3 n − 1
x 1 = −0.7 x 1 =1.2
x 2 ≈−0.725253 x 2 ≈ 1.221380
x 3 ≈−0.724493 x 3 ≈ 1.220745
x 4 ≈−0.724492 ≈ x 5 x 4 ≈ 1.220744 ≈ x 5
To six decimal places, the roots of the equation are −0.724492 and 1.220744.
19. From the graph, we see that there appear to be points of intersection near
x =1.5 and x =3. Solving (x − 2) 2 =lnx is the same as solving
f(x) =(x − 2) 2 − ln x =0. f(x) =(x − 2) 2 − ln x
⇒
f 0 (x) =2(x − 2) − 1/x,sox n+1 = x n − (x n − 2) 2 − ln x n
2(x n − 2) − 1/x n
.
x 1 =1.5 x 1 =3
x 2 ≈ 1.406721 x 2 ≈ 3.059167
x 3 ≈ 1.412370 x 3 ≈ 3.057106
x 4 ≈ 1.412391 ≈ x 5 x 4 ≈ 3.057104 ≈ x 5
To six decimal places, the roots of the equation are 1.412391 and 3.057104.
21. From the graph, there appears to be a point of intersection near x =0.6.
Solving cos x = √ x is the same as solving f(x) =cosx − √ x =0.
f(x) =cosx − √
x ⇒ f 0 (x) =− sin x − 1/ 2 √
x ,so
cos x n − √ x n
x n+1 = x n −
− sin x n − 1/ 2 √ .Nowx 1 =0.6
x
⇒ x 2 ≈ 0.641928,
x 3 ≈ 0.641714 ≈ x 4 . To six decimal places, the root of the equation is 0.641714.
23. f(x) =x 6 − x 5 − 6x 4 − x 2 + x +10 ⇒
f 0 (x) =6x 5 − 5x 4 − 24x 3 − 2x +1
⇒
x n+1 = x n − x6 n − x 5 n − 6x 4 n − x 2 n + x n +10
6x 5 n − 5x 4 n − 24x 3 n − 2x n +1 .
From the graph of f, there appear to be roots near −1.9, −1.2, 1.1,and3.