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

348 Chapter 4 Applications of Differentiation
than 1, so let’s take x 1 − 1 as a convenient first approximation. Then, remembering to
put our calculator in radian mode, we get
x 2 < 0.75036387
x 3 < 0.73911289
x 4 < 0.73908513
x 5 < 0.73908513
1
Since x 4 and x 5 agree to six decimal places (eight, in fact), we conclude that the root of
the equation, correct to six decimal places, is 0.739085.
n
y=cos x
y=x
0 1
FIGURE 7
Instead of using the rough sketch in Figure 6 to get a starting approximation for
Newton’s method in Example 3, we could have used the more accurate graph that a
calculator or computer provides. Figure 7 suggests that we use x 1 − 0.75 as the initial
approximation. Then Newton’s method gives
x 2 < 0.73911114
x 3 < 0.73908513
x 4 < 0.73908513
and so we obtain the same answer as before, but with one fewer step.
1. The figure shows the graph of a function f. Suppose that
Newton’s method is used to approximate the root s of the
equation f sxd − 0 with initial approximation x 1 − 6.
(a) Draw the tangent lines that are used to find x 2 and x 3,
and esti mate the numerical values of x 2 and x 3.
(b) Would x 1 − 8 be a better first approximation? Explain.
(a) x 1 − 0 (b) x 1 − 1 (c) x 1 − 3
(d) x 1 − 4 (e) x 1 − 5
y
0 1 3 5 x
7et0408x04
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5. For which of the initial approximations x 1 − a, b, c, and d do
you think Newton’s method will work and lead to the root of
the equation f sxd − 0?
y
2. Follow the instructions for Exercise 1(a) but use x 1 − 1
as the starting approximation for finding the root r.
3. Suppose the tangent line to the curve y − f sxd at the point
s2, 5d has the equation y − 9 2 2x. If Newton’s method is
used to locate a root of the equation f sxd − 0 and the initial
approximation is x 1 − 2, find the second approximation x 2.
4. For each initial approximation, determine graphically what
happens if Newton’s method is used for the function whose
graph is shown.
7et0408x05
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MasterID: 03017
a 0 b c d x
6–8 Use Newton’s method with the specified initial approximation
x 1 to find x 3, the third approximation to the root of the given
equation. (Give your answer to four decimal places.)
6. 2x 3 2 3x 2 1 2 − 0 , x 1 − 21
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