Given f(x)=tanx+tanhx
f(a)=f(1.6)=tan(1.6)+tanh(1.6)=−33.3109
f(b)=f(3)=tan(3)+tanh(3)=0.8525
Using iterative Regula-Falsi Formula, the first approximation,
x1=a−f(b)−f(a)b−af(a)==1.6−0.8525+33.31093−1.6(−33.3109)=2.9324
f(2.9324)=0.7821
a=1.6,b=2.9324
x2=1.6−0.7821+33.31092.9324−1.6(−33.3109)=2.9018 f(2.9018)=0.7495
a=1.6,b=2.9018
x3=1.6−0.7495+33.31092.9018−1.6(−33.3109)=2.8732 f(2.8732)=0.7186
a=1.6,b=2.8732
x4=1.6−0.7186+33.31092.8732−1.6(−33.3109)=2.8463 f(2.8463)=0.6891
a=1.6,b=2.8463
x5=1.6−0.6891+33.31092.8463−1.6(−33.3109)=2.8211 f(2.8211)=0.6609
...
x149=2.36550
x150=2.36548
x151=2.36546
...
x499=2.36502
Root of f(x) is 2.365
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