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Chapter 4 | Applications of Derivatives
Example 4.46 Finding a Root of a Polynomial
Use Newton’s method to approximate a root of f ( x ) = x 3 −3 x +1 in the interval [1, 2]. Let x
0 =2 and find
x 1 , x 2 , x 3 , x 4 , and x 5 .
Solution From Figure 4.78 , we see that f has one root over the interval (1, 2). Therefore x 0 =2 seems like a reasonable first approximation. To find the next approximation, we use Equation 4.8 . Since f ( x ) = x 3 −3 x +1, the derivative is f ′( x ) =3 x 2 −3. Using Equation 4.8 with n =1 (and a calculator that displays 10 digits), we obtain x 1 = x 0 − f ( x 0 ) f ′( x 0 ) =2− f (2) f ′(2) =2− 3 9 ≈ 1.666666667. To find the next approximation, x 2 , we use Equation 4.8 with n =2 and the value of x 1 stored on the calculator. We find that x 2 = x 1 = f ( x 1 ) f ′( x 1 ) ≈ 1.548611111. Continuing in this way, we obtain the following results: x 1 ≈ 1.666666667 x 2 ≈ 1.548611111 x 3 ≈ 1.532390162 x 4 ≈ 1.532088989 x 5 ≈ 1.532088886 x 6 ≈ 1.532088886. We note that we obtained the same value for x 5 and x 6 . Therefore, any subsequent application of Newton’s method will most likely give the same value for x n .
Figure 4.78 The function f ( x ) = x 3 −3 x +1 has one root over the interval [1, 2].
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