Posted by **Missj** on Thursday, October 27, 2011 at 9:35pm.

Find the interval which contains a zero for the given function. The use Newton's method to approximate the zero of the function within 0.001.

a.) f(x)=x^3+x-1

b.)f(x)=2x^3+x^2-x-+1

- calculus -
**Steve**, Friday, October 28, 2011 at 11:13am
pick some value reasonably close to a root. Start too far away, and you may converge to a different root, or none at all.

a) since f(0) = -1 and f(1) = 1, start with x = 1.

The iterations converge quickly:

1: 1.0000 -> 0.7500 f(x) = 0.1719

2: 0.7500 -> 0.6860 f(x) = 0.0089

3: 0.6860 -> 0.6823 f(x) = 0.0000

4: 0.6823 -> 0.6823 f(x) = 0.0000

b) Is that -1 or +1 at the end? I'll assume -1.

f(1) = 1

f(0) = -1

1: 1.0000 -> 0.8571 f(x) = 0.1370

2: 0.8571 -> 0.8304 f(x) = 0.0044

3: 0.8304 -> 0.8295 f(x) = 0.0000

4: 0.8295 -> 0.8295 f(x) = 0.0000

If we use +1 instead of -1, the root has shifted away to the left:

1: 1.0000 -> 0.5714 f(x) = 1.1283

2: 0.5714 -> 0.0347 f(x) = 0.9666

3: 0.0347 -> 1.0814 f(x) = 3.6175

4: 1.0814 -> 0.6392 f(x) = 1.2916

5: 0.6392 -> 0.1660 f(x) = 0.8707

6: 0.1660 -> 1.8980 f(x) = 16.3803

7: 1.8980 -> 1.2270 f(x) = 4.9735

8: 1.2270 -> 0.7528 f(x) = 1.6672

9: 0.7528 -> 0.3260 f(x) = 0.8496

10: 0.3260 -> -2.6072 f(x) = -25.0408

11: -2.6072 -> -1.8829 f(x) = -6.9226

12: -1.8829 -> -1.4635 f(x) = -1.6637

13: -1.4635 -> -1.2771 f(x) = -0.2575

14: -1.2771 -> -1.2357 f(x) = -0.0112

15: -1.2357 -> -1.2338 f(x) = -0.0000

16: -1.2338 -> -1.2338 f(x) = -0.0000

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