# calculus

posted by Missj

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

1. Steve

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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