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+x1
b.)f(x)=2x^3+x^2x+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