✔ 最佳答案
We can have another iteration at different range, say[-1,0].
By Secant Method,
x0 = -0.492886437
x1 = -0.66054111
x2 = -0.606739564
x3 = -0.609309055
x4 = -0.609363987
x5 = -0.609363927
...
By Newton Method,
x0 = -1
x1 = -0.648832461
x2 = -0.609987937
x3 = -0.609364094
x4 = -0.609363927
...
By Method of Bisection,
(-1,0)==>(-1,-0.5)==>(-0.75,-0.5)==>...==>(-0.609375,-0.609354741)
By Method of False Position,
x0 = 0
x1 = -0.492886437
x2 = -0.593233123
x3 = -0.607266076
x4 = -0.609093437
x5 = -0.609329099
x6 = -0.609359442
x7 = -0.609363349
Therefore another A is -0.6094
You can also find the third A = 3.6384