Chapter 4 | Applications of Derivatives
477
Figure 4.80 If the initial guess x 0 is too far from the root sought, it may lead to approximations that approach a different root.
Example 4.48 When Newton’s Method Fails
Consider the function f ( x ) = x 3 −2 x +2. Let x
0 =0. Show that the sequence x 1 , x 2 ,… fails to approach a
root of f .
Solution For f ( x ) = x 3 −2 x +2, the derivative is f ′( x ) =3 x 2 −2. Therefore, x 1 = x 0 − f ( x 0 ) f ′( x 0 ) =0− f (0) f ′(0) = − 2 −2 =1. In the next step, x 2 = x 1 − f ( x 1 ) f ′( x 1 ) =1− f (1) f ′(1) =1− 1 1 =0.
Consequently, the numbers x 0 , x 1 , x 2 ,… continue to bounce back and forth between 0 and 1 and never get closer to the root of f which is over the interval [−2, −1] (see Figure 4.81 ). Fortunately, if we choose an initial approximation x 0 closer to the actual root, we can avoid this situation.
Made with FlippingBook - professional solution for displaying marketing and sales documents online