James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

346 Chapter 4 Applications of Differentiation
y
{x¡, f(x¡)}
{x, f(x)}
r
0 x£ x x¡ x
x¢
Next we repeat this procedure with x 1 replaced by the second approximation x 2 , using
the tangent line at sx 2 , f sx 2 dd. This gives a third approximation:
x 3 − x 2 2 f sx 2d
f 9sx 2 d
If we keep repeating this process, we obtain a sequence of approximations x 1 , x 2 , x 3 , x 4 , . . .
as shown in Figure 3. In general, if the nth approximation is x n and f 9sx n d ± 0, then
the next approximation is given by
FIGURE 3
2
x n11 − x n 2 f sx nd
f 9sx n d
Sequences were briefly introduced
in A Preview of Calculus on page 5.
A more thorough discussion starts
in Section 11.1.
If the numbers x n become closer and closer to r as n becomes large, then we say that
the sequence converges to r and we write
lim
n l ` xn − r
y
0
x£ x¡
r
x
x
Although the sequence of successive approximations converges to the desired root for
fun ctions of the type illustrated in Figure 3, in certain circumstances the sequence may
not converge. For example, consider the situation shown in Figure 4. You can see that
x 2 is a worse approximation than x 1 . This is likely to be the case when f 9sx 1 d is close to
0. It might even happen that an approximation (such as x 3 in Figure 4) falls outside the
domain of f. Then Newton’s method fails and a better initial approximation x 1 should
be chosen. See Exercises 31–34 for specific examples in which Newton’s method works
very slowly or does not work at all.
FIGURE 4
tec In Module 4.8 you can investigate
how Newton’s method works for
several functions and what happens
when you change x 1.
Figure 5 shows the geometry behind
the first step in Newton’s method in
Example 1. Since f 9s2d − 10, the
tangent line to y − x 3 2 2x 2 5 at
s2, 21d has equation y − 10x 2 21
so its x-intercept is x 2 − 2.1.
1
y=˛-2x-5
1.8 2.2
x
Example 1 Starting with x 1 − 2, find the third approximation x 3 to the root of the
equation x 3 2 2x 2 5 − 0.
SOLUtion We apply Newton’s method with
f sxd − x 3 2 2x 2 5 and f 9sxd − 3x 2 2 2
Newton himself used this equation to illustrate his method and he chose x 1 − 2 after
some experimentation because f s1d − 26, f s2d − 21, and f s3d − 16. Equation 2
becomes
x n11 − x n 2 f sx nd
f 9sx n d − x n 2 x n 3 2 2x n 2 5
3x 2 n 2 2
With n − 1 we have
x 2 − x 1 2 f sx 1d
f 9sx 1 d − x 1 2 x 1 3 2 2x 1 2 5
3x 2 1 2 2
− 2 2 23 2 2s2d 2 5
3s2d 2 2 2
− 2.1
_2
y=10x-21
Then with n − 2 we obtain
FIGURE 5
x 3 − x 2 2 x 2 3 2 2x 2 2 5
3x 2 2 2 2
− 2.1 2 s2.1d3 2 2s2.1d 2 5
3s2.1d 2 2 2
< 2.0946
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